Mechanics of Solids
Mechanics of Solids [3 1 0 4] CIE 101 / 102 First Year B.E. Degree
Mechanics of Solids
PART- I Mechanics of Rigid Bodies
PART- II Mechanics of Deformable Bodies
COURSE CONTENT IN BRIEF
PART I Mechanics of Rigid Bodies 1. Resultant of concurrent and non-concurrent coplanar forces. 2. Equilibrium of concurrent and non-concurrent coplanar forces. 3. Centroid of plane areas 4. Moment of Inertia of plane areas 5. Kinetics: Newton’s second law, D’Alembert’s principle, Work- Energy, and Impulse- Momentum principle. Mechanics of Deformable bodies
PART II
6. Simple stresses and strains 7. Statically indeterminate problems and thermal stresses 8. Stresses on inclined planes 9. Stresses due to fluid pressure in thin cylinders
Books for Reference 1.Engineering Mechanics, by Meriam & Craige, John Wiley & Sons. 2.Engineering Mechanics, by Irwing Shames, Prentice Hall of India. 3.Mechanics for Engineers, by Beer and Johnston, McGraw Hills Edition 4.Engineering Mechanics, by K.L. Kumar, Tata McGraw Hills Co. 5. Machanics of Materials, by E.P.Popov 6. Machanics of Materials, by E J Hearn 7. Strength of materials, by Beer and Johnston 8. Strength of materials, by F L Singer & Andrew Pytel 9. Strength of Materials, by B.S. Basavarajaiah & P. Mahadevappa 10. Strength of Materials, by Ramamruthum 11. Strength of Materials, by S S Bhavikatti
PART - I MECHANICS OF RIGID BODIES
PART - I
Mechanics of Rigid Bodies
INTRODUCTION Definition of Mechanics : In its broadest sense the term ‘Mechanics’ may be defined as the ‘Science which describes and predicts the conditions of rest or motion of bodies under the action of forces’. This Course on Engineering Mechanics comprises of Mechanics of Rigid bodies and the sub-divisions that come under it.
Branches of Mechanics Engineering Mechanics Mechanics of Solids Deformable Bodies
Mechanics of Fluids
Rigid Bodies
Ideal Fluids
Viscous Fluids
Compres Fluids
Statics Dynamics
Theory of Theory of Plasticity Strength of Elasticity Materials
Kinematics
Kinetics
Concept of Rigid Body : It is defined as a definite amount of matter the parts of which are fixed in position relative to one another under the application of load. Actually solid bodies are never rigid; they deform under the action of applied forces. In those cases where this deformation is negligible compared to the size of the body, the body may be considered to be rigid.
Particle A body whose dimensions are negligible when compared to the distances involved in the discussion of its motion is called a ‘Particle’. For example, while studying the motion of sun and earth, they are considered as particles since their dimensions are small when compared with the distance between them.
Force It is that agent which causes or tends to cause, changes or tends to change the state of rest or of motion of a mass. A force is fully defined only when the following four characteristics are known: (i) Magnitude (ii) Direction (iii) Point of application (iv) Sense.
Force: characteristics of the force 100 kN are : (i) Magnitude = 100 kN (ii) Direction = at an inclination of 300 to the x-axis (iii) Point of application = at point A shown (iv) Sense = towards point A A 100 kN 300
Scalars and Vectors A quantity is said to be a ‘scalar’ if it is completely defined by its magnitude alone. Example : Length, Area, and Time. A quantity is said to be a ‘vector’ if it is completely defined only when its magnitude and direction are specified. Example : Force, Velocity, and Acceleration.
Principle of Transmissibility : It is stated as follows : ‘The external effect of a force on a rigid body is the same for all points of application along its line of action’. P A B P
For example, consider the above figure. The motion of the block will be the same if a force of magnitude P is applied as a push at A or as a pull at B. P O The same is true when the force is applied at a point O. P
1. RESULTANT OF COPLANAR FORCES Resultant, R : It is defined as that single force which can replace a set of forces, in a force system, and cause the same external effect.
F2 F1
R
= θ A
F3
A
R = F1 + F2 + F3 external effect on particle, A is same
Resultant of two forces acting at a point Parallelogram law of forces : ‘If two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, then the resultant of these two forces is represented in magnitude and direction by the diagonal of the parallelogram passing through the same point.’ B P2 O α R θ P1 A C
Contd..
B P2 O α R θ P1 A
C
In the above figure, P1 and P2, represented by the sides OA and OB have R as their resultant represented by the diagonal OC of the parallelogram OACB. It can be shown that the magnitude of the resultant is given by: R = √P12 + P22 + 2P1P2Cos α Inclination of the resultant w.r.t. the force P1 is given by: θ = tan-1 [( P2 Sin α) / ( P1 + P2 Cos α )]
Resultant of two forces acting at a point at right angle B P2 O α R θ P1 C R θ P1 A
R = P1 + P 2 2
2
C
A
If α = 900 , (two forces acting at a point are at right angle) B P2 O
tan θ =
P2 P1
Triangle law of forces
‘If two forces acting at a point can be represented both in magnitude and direction, by the two sides of a triangle taken in tip to tail order, the third side of the triangle represents both in magnitude and direction the resultant force F, the sense of the same is defined by its tail at the tail of the first force and its tip at the tip of the second force’.
Triangle law of forces
Let F1 and F2 be the two forces acting at a point A and θ is the included angle.
F1 R θ A = F2 F2 θ F1
‘Arrange the two forces as two sides of a triangle taken in tip to tail order, the third side of the triangle represents both in magnitude and direction the resultant force R. the sense of the resultant force is defined by its tail at the tail of the first force and its tip at the tip of the second force’.
Triangle law of forces F1 θ A = F2 R θ F1
F2
R α F1
β F2
F1 F2 R = = sin β sin α sin(180 − α − β )
(180 - α - β) = θ
where α and β are the angles made by the resultant force with the force F1 and F2 respectively.
Component of a force :
Component of a force, in simple terms, is the effect of a force in a certain direction. A force can be split into infinite number of components along infinite directions.
Usually, a force is split into two mutually perpendicular components, one along the x-direction and the other along ydirection (generally horizontal and vertical, respectively).
Such components that are mutually perpendicular are called ‘Rectangular Components’.
The process of obtaining the components of a force is called ‘Resolution of a force’.
Rectangular component of a force F θx
F θx Fx
Fy
=
F θx Fx
Fy
Consider a force F making an angle θx with x-axis. Then the resolved part of the force F along x-axis is given by Fx = F cos θx The resolved part of the force F along y-axis is given by Fy = F sin θx
Oblique component of a force Let F1 and F2 be the oblique components of a force F. The components F1 and F2 can be found using the ‘triangle law of forces’.
N
F2 β α
F F1
M
F α F1
β F2
O The resolved part of the force F along OM and ON can obtained by using the equation of a triangle. F1 / Sin β = F2 / Sin α = F / Sin(180 - α - β)
Sign Convention for force components: y +ve x
y
+ve
x
The adjacent diagram gives the sign convention for force components, i.e., force components that are directed along positive x-direction are taken +ve for summation along the x-direction. Also force components that are directed along +ve y-direction are taken +ve for summation along the y-direction.
Classification of force system Force system
Coplanar Forces Concurrent Non-concurrent
Unlike parallel Like parallel Unlike parallel
Non-Coplanar Forces Concurrent Non-concurrent
Like parallel
A force that can replace a set of forces, in a force system, and cause the same ‘external effect’ is called the Resultant.
RESULTANT OF COPLANAR NON CONCURRENT FORCE SYSTEM
Coplanar Non-concurrent Force System: This is the force system in which lines of action of individual forces lie in the same plane but act at different points of applications.
F1 F3 Fig. 1
F2 F5
F1
F2 F3 F4
Fig. 2
1. Parallel Force System – Lines of action of individual forces are parallel to each other. 2. Non-Parallel Force System – Lines of action of the forces are not parallel to each other.
MOMENT OF A FORCE ABOUT AN AXIS The applied force can also tend to rotate the body about an axis in addition to motion. This rotational tendency is known as moment. Definition: Moment is the tendency of a force to make a rigid body to rotate about an axis. This is a vector quantity having both magnitude and direction.
MOMENT OF A FORCE ABOUT AN AXIS Moment Axis: This is the axis about which rotational tendency is determined. It is perpendicular to the plane comprising moment arm and line of action of the force (axis 0-0 in the figure) Moment Center: This is the position of axis on coplanar system. (A). Moment Arm: Perpendicular distance from the line of action of the force to moment center. Distance AB = d.
Magnitude of moment: It is computed as the product of the of the force and the perpendicular distance from the line of action to the point about which moment is computed. (Moment center). MA = F×d = Rotation effect because of the force F, about the point A (about an axis 0-0) Unit – kN-m, N-mm etc.
Sense of moment: The sense is obtained by ‘Right Hand Thumb’ rule. ‘If the fingers of the right hand are curled in the direction of rotational tendency of the body, the extended thumb represents the sense of moment vector’. For the purpose of additions, the moment direction may be considered by using a suitable sign convention such as +ve for counterclockwise and –ve for clockwise rotations or viceversa.
M
M
VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)
Statement: The moment of a force about a moment center or axis is equal to the algebraic sum of the moments of its component forces about the same moment center (axis). P P sinθ P θ P cosθ θ d d1 A d2 A
Moment of Force P about the Algebraic sum of Moments of = components of the Force P point A, about the point A, Pxd P cosθ x d1 + P sinθ x d2
VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)
Proof (by Scalar Formulation): Let ‘R’ be the given force. ‘P’ & ‘Q’ are component forces of ‘R’. ‘O’ is the moment center. p, r and q are moment arms from ‘O’ of P, R and Q respectively. α, β and γ are the inclinations of ‘P’, ‘R’ and ‘Q’ respectively w.r.to X – axis. A R
Q
Y
q α
r P βγ p O X
Y
We have, Ry = Py + Qy R Sinβ = P Sinα + Q Sin γ ----(1) From ∆le AOB, p/AO = Sin α From ∆le AOC, r/AO = Sin β From ∆le AOD, q/AO = Sin γ From (1), ∴ R ×(r/AO) = P ×(p/AO) + Q ×(q/AO) i.e., R × r = P × p + Q × q Moment of resultant R about O = algebraic sum of moments of component forces P & Q about same moment center ‘O’. Ry
Q
R
C
P q r
Qy Py A
D
γ α β
B p O
X
COUPLE Two parallel, non collinear (separated by certain distance) forces that are equal in magnitude and opposite in direction form ‘couple’.
F
The algebraic summation of the two forces forming couple is zero. Hence, couple does not produce any translation and produces only rotation.
=
d F
M=Fxd
RESOLUTION OF A FORCE INTO A FORCE-COUPLE SYSTEM
Replace the force F acting at the point A to the point B
F
B
A
Apply two equal and opposite forces of same magnitude & direction as Force F at point B, so that external effect is unchanged
F F
B
A
F
d
F
B
F
F
B
A
=
A
F
d
M=Fxd
Of these three forces, two forces i.e., one at A and the other oppositely directed at B form a couple. Moment of this couple, M = F × d. Third force at B is acting in the same direction as that at P. Thus, the force F acting at a point such as A in a rigid body can be moved to any other given point B, by adding a couple M. The moment of the couple is equal to moment of the force in its original position about B.
TYPES OF LOADS ON BEAMS
1. Concentrated Loads – This is the load acting for very small length of the beam.
W kN
(also known as point load, Total load W is acting at one point ) 2. Uniformly distributed load – This is the load acting for a considerable length of the beam with same intensity of w kN/m throughout its spread. Total intensity, W = w × L (acts at L/2 from one end of the spread)
L
w kN/m
L
W = (w x L) kN L/2
3. Uniformly varying load – This load acts for a considerable length of the beam with intensity varying linearly from ‘0’ at one end to w kN/m to the other representing a triangular distribution. Total intensity of load = area of triangular spread of the load W = 1/2× w × L. (acts at 2×L/3 from ‘Zero’ load end) 2/3 ×L
w kN/m L
W=½×L×w
1/3 ×L
L
EXERCISE PROBLEMS
1. Resultant of force system
Q1 A body of negligible weight, subjected to two forces F1= 1200N, and F2=400N acting along the vertical, and the horizontal respectively, is shown in figure. Find the component of each force parallel, and perpendicular to the plane. F1 = 1200 N
X
Y F2 = 400 N 4 3
Ans : F1X = -720 N, F1Y = -960N, F2X = 320N, F2Y = -240N
EXERCISE PROBLEMS
1. Resultant of force system
Q2. Determine the X and Y components of each of the forces shown in the figure. F2 = 390 N 12 5 40º F3 =400 N 30º F1 = 300 N Y X
(Ans : F1X = 259.81 N, F1Y= -150 N, F2X= -150N, F2Y= 360 N, F3X = -306.42 N, F3Y= -257.12N )
EXERCISE PROBLEMS
1. Resultant of force system
Q3. Obtain the resultant of the concurrent coplanar forces shown in the figure 600N 800N 20º 40º
30º 200N (Ans: R = 522.67 N, θ = 68.43º)
EXERCISE PROBLEMS
1. Resultant of force system
Q4. A disabled ship is pulled by means of two tug boats as shown in FIG. 4. If the resultant of the two forces T1 and T2 exerted by the ropes is a 300 N force acting parallel to the X – direction, find : (a) Force exerted by each of the tug boats knowing α = 30º. (b) The value of α such that the force of tugboat 2 is minimum, while that of 1 acts in the same direction. Find the corresponding force to be exerted by tugboat 2.
T2 α 20º R = 300 N X - direction
FIG. 4 T1 ( Ans: a. T1= 195.81 N, T2 = 133.94 N b. α = 70º, T1 = 281.91 N, T2(min) = 102.61 N )
EXERCISE PROBLEMS
1. Resultant of force system
Q5. An automobile which is disabled is pulled by two ropes as shown in the figure. Find the force P and resultant R, such that R is directed as shown in the figure.
P 20º 40º R
Q = 5 kN (Ans: P = 9.4 kN , R = 12.66 kN)
EXERCISE PROBLEMS
1. Resultant of force system
Q6. A collar, which may slide on a vertical rod, is subjected to three forces as shown in figure. The direction of the force F may be varied Determine the direction of the force F, so that resultant of the three forces is horizontal, knowing that the magnitude of F is equal to (a) 2400 N, (b)1400N 1200 N 60º θ COLLAR ROD F 800 N
( Ans: a. θ = 41.81º ; b. The resultant cannot be horizontal.)
EXERCISE PROBLEMS
1. Resultant of force system
Q7. Determine the angle α and the magnitude of the force Q such that the resultant of the three forces on the pole is vertically downwards and of magnitude 12 kN. Refer figure 8kN
5kN α 30º Q
Fig. 7
(Ans: α = 10.7 º, Q = 9.479 kN )
EXERCISE PROBLEMS
1. Resultant of force system
Q8. Determine the resultant of the parallel coplanar force system shown in figure.
600 N 2000 N
60º 10º
o
30º 60º
1000 N 400 N
(Ans. R=800N towards left, d=627.5mm)
EXERCISE PROBLEMS
1. Resultant of force system
Q9. Four forces of magnitudes 10N, 20N, 30N and 40N acting respectively along the four sides of a square ABCD as shown in the figure. Determine the magnitude, direction and position of resultant w.r.t. A.
20N D 30N C a A a 40N B 10N
(Ans:R=28.28N, θ=45º, x=1.77a)
EXERCISE PROBLEMS
1. Resultant of force system
Q10. Four parallel forces of magnitudes 100N, 150N, 25N and 200N acting at left end, 0.9m, 2.1m and 2.85m respectively from the left end of a horizontal bar of 2.85m. Determine the magnitude of resultant and also the distance of the resultant from the left end.
(Ans: R = 125 N, x = 3.06 m)
EXERCISE PROBLEMS
1. Resultant of force system
Q11. Reduce the given forces into a single force and a couple at A.
70.7 kN 200 kN 45º 1.5m A 1m 80 N 30º 100 N 30º
(Ans:F=320kN, θ=14.48º, M=284.8kNm)
EXERCISE PROBLEMS
1. Resultant of force system
Q12. Determine the resultant w.r.t. point A.
150 Nm 150 N 1.5m 3m 1.5m
A
100 N 500 N
(Ans: R = 450 kN, X = 7.5 kNm)
2. EQUILIBRIUM OF FORCE SYSTEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
EQUILIBRIUM OF - CONCURRENT COPLANAR FORCE SYSTEMS
Definition:If a system of forces acting on a body, keeps the body in a state of rest or in a state of uniform motion along a straight line, then the system of forces is said to be in equilibrium. ALTERNATIVELY, if the resultant of the force system is zero, then, the force system is said to be in equilibrium.
EQUILIBRIUM OF - CONCURRENT COPLANAR FORCE SYSTEMS
Conditions for Equilibrium :
A coplanar concurrent force system will be in equilibrium if it satisfies the following two conditions: ∑ Fx = 0; and ii) ∑ Fy = 0
i)
i.e. Algebraic sum of components of all the forces of the system, along two mutually perpendicular directions, is ZERO.
Y X
Graphical conditions for Equilibrium Triangle Law: If three forces are in equilibrium, then, they form a closed triangle when represented in a Tip to Tail arrangement, as shown in Fig 2.1
F1 F3
F
3
F2 Fig 2.1
F2
F1
Polygonal Law: If more than three forces are in equilibrium, then, they form a closed polygon when represented in a Tip to Tail arrangement, as shown in Fig. 2.2. F F4
3
F3 F4
F5
F2 F1 F5 Fig 2.2
F
2
F1
LAMI’S THEOREM
If a system of three forces is in equilibrium, then, each force of the system is proportional to sine of the angle between the other two forces (and constant of proportionality is the same for all the forces). Thus, with reference to Fig.2.3, we have,
F3
α β F2
F3 F1 F2 = = Sin α Sin β Sin γ
γ
F1
Fig. 2.3
Note: While using Lami’s theorem, all the three forces should be either directed away or all directed towards the point of concurrence.
EQUILIBRIUM OF NON-CONCURRENT COPLANAR FORCE SYSTEM
When a body is in equilibrium, it has neither translatory nor rotatory motion in any direction. Thus the resultant force R and the resultant couple M are both zero, and we have the equilibrium equations for two dimensional force system ∑ Fx = 0; ∑ Fy = 0 ∑M = 0
These requirements are both necessary and sufficient conditions for equilibrium.
SPACE DIAGRAMS & FREE BODY DIAGRAMS
Space Diagram (SPD) : The sketch showing the physical conditions of the problem, like, the nature of supports provided; size, shape and location of various bodies; forces applied on the bodies, etc., is known as space diagram. eg, Fig 2.4 is a space diagram
Weight of sphere = 0.5 kN, Radius = 1m
3m θ
Cable P = 2kN
30°
Sphere
wall
Fig. 2.4 SPD
Free Body Diagram (FBD) : It is an isolated diagram of the body being analyzed (called free body), in which, the body is shown freed from all its supports and contacting bodies/surfaces. Instead of the supports and contacting bodies/surfaces, the reactive forces exerted by them on the free body is shown, along with all other applied forces.
A Few Guidelines for Drawing FBD 1) Tensile Force: It is a force trying to pull or extend the body. It is represented by a vector directed away from the body. 2) Compressive Force: It is force trying to push or contract the body. It is represented by a vector directed towards the body. 3) Reactions at smooth surfaces: The reactions of smooth surfaces, like walls, floors, Inclined planes, etc. will be normal to the surface and pointing towards the body. 4)Forces in Link rods/connecting rods: These forces will be acting along the axis of the rod, either towards or away from the body. (They are either compressive or tensile in nature). 5) Forces in Cables (Strings or Chords): These can only be tensile forces. Thus, these forces will be along the cable and directed away from the body.
Free Body Diagrams of the sphere shown in Fig. 2.4 T = Tension in the cable θ T
P = 2kN
30°
Rw = Reaction of the wall W = self weight of the sphere P = external load acting on the sphere
Rw
Sphere
W=0.5kN Fig. 2.5 F B D of Sphere
Detach the sphere from all contacts and replace that with forces like: Cable contact is replaced by the force tension = T Contact with the smooth wall is replaced by the reaction Rw.
Supports: A structure is subjected to external forces and transfers these forces through the supports on to the foundation. Therefore the support reactions and the external forces together keep the structure in equilibrium. There are different types of supports. a) Roller Support b) Hinged or pinned support c) Fixed or built in support
Supports Types of Supports (a) Flexible cable ,belt ,chain, rope Action on body
BODY
T Force exerted by cable is always a tension away from the body in the direction of cable
BODY
(b) Smooth surfaces 900 F 900 F
Contact forces are normal to the surfaces
Supports
A (c) Roller support Contact force is normal to the surface on which the roller moves. The reaction will always be perpendicular to the plane of the roller . Roller support will offer only one independent reaction component. (Whose direction is known.)
A
Supports
(d) pinned Support / hinged support A
Rh θ
A Rv
R
This support does not allow any translatory movement of the rigid body. There will be two independent reaction components at the support. The resultant reaction can be resolved into two mutually perpendicular components. Or it can be shown as resultant reaction inclined at an angle with respect to a reference direction.
Supports
(e) Fixed or Built-in Support A
RAH
M
A
RAV
This type of support not only prevents the translatory movement of the rigid body, but also the rotation of the rigid body. Hence there will be 3 independent reaction components of forces. Hence there will be 3 unknown components of forces, two mutually perpendicular reactive force component and a reactive moment as shown in the figure.
TYPES OF BEAMS
A member which is subjected to predominantly transverse loads and supported in such a way that rigid body motion is prevented is known as beam. It is classified based on the support conditions. A beam generally supported by a hinge or roller at the ends having one span (distance between the support) is called as simply supported beam. A beam which is fixed at one end and free at another end is called as a cantilever beam. A span B
A span B
(a) Simply supported beam
TYPES OF BEAMS
A
B
span
RH M A Rv B
(b) Cantilever beam
TYPES OF BEAMS
If one end or both ends of the beam project beyond the support it is known as overhanging beam.
A
B
A
B
(c) Overhanging beam (right overhang)
Statically determinate beam Using the equations of equilibrium given below, if all the reaction components can be found out, then the beam is a statically determinate beam
the equations of equilibrium ∑ Fx = 0; ∑ Fy = 0 ∑M = 0
FRICTION Friction is defined as the contact resistance exerted by one body upon another body when one body moves or tends to move past another body. This force which opposes the movement or tendency of movement is known as frictional resistance or friction. Friction is due to the resistance offered by minute projections at the contact surfaces. Hence friction is the retarding force, always opposite to the direction of motion. Friction has both advantages & disadvantages. Disadvantages ---Advantages ---Power loss, wear and tear etc.
Brakes, traction for vehicles etc.
FRICTION
W P
F (Friction)
N
Hills & Vales
Magnified Surface
Frictional resistance is dependent on the amount of wedging action between the hills and vales of contact surfaces. The wedging action is dependent on the normal reaction N.
FRICTION
Frictional resistance has the remarkable property of adjusting itself in magnitude of force producing or tending to produce the motion so that the motion is prevented. When P = 0, F = 0 block under equilibrium
When P increases, F also increases proportionately to maintain equilibrium. However there is a limit beyond which the magnitude of this friction cannot increase.
FRICTION
When the block is on the verge of motion(motion of the block is impending) F attains maximum possible value, which is termed as Limiting Friction. When the applied force is less than the limiting friction, the body remains at rest and such frictional resistance is called the static friction. Further if P is increased, the value of F decreases rapidly and then remains fairly a constant thereafter. However at high speeds it tends to decrease. This frictional resistance experienced by the body while in motion is known as Dynamic friction OR Kinetic Friction.
FRICTION
Sliding friction friction experienced when a body slides over another surface. Dynamic Friction
Rolling friction friction experienced by a body when it rolls over a surface.
FRICTION
W P Fmax φ N µ= R Fmax N
FαN Fmax = µN
Where Fmax = Limiting Friction N= Normal Reaction between the contact surfaces µ =Coefficient of friction
Note : Static friction varies from zero to a maximum value. Dynamic friction is fairly a constant.
Angle of Friction The angle between N & R depends on the value of F. This angle θ, between the resultant R and the normal reaction N is termed as angle of friction. As F increases, θ also increases and will reach to a maximum value of φ when F is Fmax (limiting friction) i.e. tanφ = (Fmax )/N = µ Angle φ is known as Angle of limiting Friction.
FRICTION
W
P
φ N
Fmax
R
FRICTION
Angle of limiting friction is defined as the angle between the resultant reaction (of limiting friction and normal reaction) and the normal to the plane on which the motion of the body is impending.
Angle of repose When granular material is heaped, there exists a limit for the inclination of the surface. Beyond that angle, the grains start rolling down. This limiting angle upto which the grains repose (sleep) is called the angle of repose of the granular material.
FRICTION
Significance of Angle of repose: The angle that an inclined plane makes with the horizontal, when the body supported on the plane is on the verge of motion due to its self -weight is equal to the angle of repose. Angle of repose is numerically equal to Angle of limiting friction
Laws of dry friction
FRICTION
1. The magnitude of limiting friction bears a constant ratio to the normal reaction between the two surfaces. (Experimentally proved) 2. The force of friction is independent of the area of contact between the two surfaces. 3. For low velocities the total amount of friction that can be developed is practically independent of velocity. It is less than the frictional force corresponding to impending motion.
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q1. A 10kN roller rests on a smooth horizontal floor and is held by the bar AC as shown in Fig(1). Determine the magnitude and nature of the force in the bar AC and reaction from the floor under the action of the forces applied on the roller. [Ans:FAC=0.058 kN(T),R=14.98 kN] 7kN C A 300 Fig(1) 450 5kN
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q2. A 10 kN weight is suspended from a rope as shown in figure. Determine the magnitude and direction of the least force P required to pull the rope, so that, the weight is shifted horizontally by 0.5m. Also, determine, tension in the rope in its new position. [Ans: P= 2.43 kN, θ = 14.480 ; T= 9.7kN.]
2m θ 10kN P
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q3. Determine the value of P and the nature of the forces in the bars for equilibrium of the system shown in figure. [Ans: P = 3.04 kN, Forces in bars are Compressive.] 60 45 P
45 75 2kN
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q4. A cable fixed as shown in Fig. supports three loads. Determine the value of the load W and the inclination of the segment BC. [Ans: W=25kN, θ = 54.780]
A 30 B θ 60 C W
D
20 Loads are in kN
22.5
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q5. Find the reactions at A,B,C and D for the beam loaded as shown in the figure. (Ans.RA=RB =34kN;RC=28.84kN;
MC=-140kNm ; θC=-33.69 ˚ )
12kN/m 4kN/m A
20 kN 4kN/m
12kN/m 30kN 4 3 C
1m 2m
B 40kNm
1m
2m
1m
1m
2m
1m
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q6. A uniform bar AB of weight 50N shown in the figure supports a load of 200N at its end. Determine the tension developed in the string and the force supported by the pin at B. (Ans. T=529.12N;RB=807.15N, θB=64.6˚)
B A 200N
2.5m
60˚ 2.5m
2.5m
string
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q7. Find the position of the hinged support (x),such that the reactions developed at the supports of the beam are equal.. (Ans.x=2m.) 10kN/m 15kN
18kN/m x
2.0m
1.0m 0.6 1.4m
3.0m
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q8. A right angled bar ABC hinged at A as shown in fig carries two loads W and 2W applied at B &C .Neglecting self weight of the bar find the angle made by AB with vertical (Ans:θ =18.44˚)
Lm
A θ
B W 0.5L C 2W
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q9. For the block shown in fig., determine the smallest force P required a) to start the block up the plane b) to prevent the block moving down the plane. Take µ = 0.20
[Ans.: (a) Pmin = 59.2N (b) Pmin = 23.7N (b) θ = 11.3o]
100N 25°
P θ
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q10. A block of weight 2000 N is attached to a cord passing over a frictionless pulley and supporting a weight of 800N as shown in fig. If µ between the block and the plane is 0.35, determine the unknown force P for impending motion (a) to the right (b) to the left
[Ans.: (a) P = 132.8N (b) P = 1252N]
2000N
30° P
800N
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q11. Determine value of angle θ to cause the motion of 500N block to impend down the plane, if µ for all contact surfaces is 0.30.
200N 500N θ=? [Ans.: θ = 28.4°]
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q12. A horizontal bar 10m long and of negligible weight rests on rough inclines as shown in fig. If angle of friction is 15o, how close to B may the 200N force be applied before the motion impends.
100N A 2m 30°
200N X=? B 60° [Ans.: x = 3.5m]
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q13. Determine the vertical force P required to drive the wedge B downwards in the arrangements shown in fig. Angle of friction for all contact surfaces is 12o.Weight of block A= 1600 N.
P B 20°
A
[Ans.: P = 328.42N]
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q14. Determine the force P which is necessary to start the wedge to raise the block A weighing 1000N. Self weight of the wedge may be ignored. Take angle of friction, φ = 15o for all contact surfaces.
A P
[Ans.: P = 1192N]
20 ° wedge
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q15. A ladder of weight 200N, 6m long is supported as shown in fig. If µ between the floor and the ladder is 0.5 & between the wall and the ladder is 0.25 and it supports a vertical load of 1000N, determine a) the least value of α at which the ladder may be placed without slipping b) the reactions at A & B [Ans.: (a) α = 56.3o (b) RA = 1193 N, RB = 550N]
1000N 5m α
B
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q16. An uniform ladder of weight 250N is placed against a smooth vertical wall with its lower end 5m from the wall. µ between the ladder and the floor is 0.3. Show that the ladder remains in equilibrium in this position. What is the frictional resistance on the ladder at the point of contact between the ladder and the floor?
Smooth wall B 12m [Ans.: FA = 52 N] A
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q17. A ladder of length 5m weighing 500N is placed at 45o against a vertical wall. µ between the ladder and the wall is 0.20 & between ladder and ground is 0.50. If a man weighing 600N ascends the ladder, how high will he be when the ladder just slips. If a boy now stands on the bottom rung of the ladder, what must be his least weight so that the man can go to the top of the ladder. [Ans.: (a) x = 2.92m (b) Wboy = 458N]
3.
CENTROID OF PLANE AREA
3.
CENTROID
Centre of gravity : of a body is the point at which the whole weight of the body may be assumed to be concentrated. A body is having only one center of gravity for all positions of the body. It is represented by CG. or simply G or C.
Contd.
CENTRE OF GRAVITY
Consider a block of uniform thickness and having a uniform mass m. It is possible to support (hold) the block in stable position by a rod as shown in the figure provided rod must be positioned exactly at the point of intersection of the diagonals. Or the rod must be supported exactly below where the total weight of the block act.
Contd.
W
R
W
CENTRE OF GRAVITY
The block can be supported from any position provided the support rod and the line of action of weight are in same line. This indicates that the whole weight of the block act through one point. This point is called as centre of gravity.
R
Centre of gravity is that point about which the summation of the first moments of the weights of the elements of the body is zero.
Contd.
W x
CENTRE OF GRAVITY
W3 W2 W1 W4
To determine mathematically the location of the “centre of gravity” of any body, we apply the “principle of moments” to the parallel system of gravitational forces
X1 X2
The moment of the resultant gravitational force W, about any axis
=
the algebraic sum of the moments about the same axis of the gravitational forces dW acting on all infinitesimal elements of the body. Contd.
W x
CENTRE OF GRAVITY dW3 dW2 dW1 dW4
X1 X2
x ⋅ W = ∫ x × dW
x ⋅ W = dW1 × x1 + dW2 × x2 + dW3 × x3 + ................dWn × xn
Where
W =
∫ dW
Contd.
x
CENTRE OF GRAVITY
where
x = x- coordinate of centre of gravity
∫ x ⋅ dW x=
W
x
Similarly, y and z coordinates of the centre of gravity are
∫ y ⋅ dW y=
W
and
∫ z ⋅ dW z=
W
----(1)
x
CENTRE OF MASS With the substitution of W= m g and dW = g dm
(if ‘g’ is assumed constant for all particles, then )
∫ x ⋅ dm x= m ,
∫ y ⋅ dm y= m ,
∫ z ⋅ dm z= m ----(2)
Equation 2 is independent of g and therefore define a unique point in the body which is a function solely of the distribution of mass. This point is called the centre of mass and clearly coincides with the centre of gravity as long as the gravity field is treated as uniform and parallel
Contd.
When speaking of an actual physical body, we use the term “centre of mass”. The term centroid is used when the calculation concerns a geometrical shape only. Calculation of centroid falls within three distinct categories, depending on whether we can model the shape of the body involved as a line, an area or a volume.
Contd.
The centroid “C” of the Volume segment,
∫ x ⋅ dV x=
V
,
∫ y ⋅ dV y=
V
,
∫ z ⋅ dV z=
V
The centroid “C” of the line segment,
∫ x ⋅ dL x=
L
,
∫ y ⋅ dL y=
L
,
∫ z ⋅ dL z=
L
The centroid “C” of the Area segment, AREA: when the density ρ, is constant and the body has a small constant thickness t, the body can be modeled as a surface area. The mass of an element becomes dm = ρ t dA. If ρ and t are constant over entire area, the coordinates of the ‘centre of mass’ also becomes the coordinates of the centroid, C of the surface area and which may be written as
∫ x ⋅ dA , x=
A
y =
∫ y ⋅ dA , A
∫ z ⋅ dA z=
A
Contd.
Centroid of Simple figures: using method of moment ( First moment of area) Centroid of an area may or may not lie on the area in question. It is a unique point for a given area regardless of the choice of the origin and the orientation of the axes about which we take the moment.
The coordinates of the centroid of the surface area about any axis can be calculated by using the equn. (A) x = (a1) x1 + (a2) x2 + (a3) x3 + ……….+(an) xn = First moment of area
Moment of Total area ‘A’ about yaxis
=
Algebraic Sum of moment of elemental ‘dA’ about the same axis
where (A = a1 + a2 + a3 + a4 + ……..+ an)
AXIS of SYMMETRY: It is an axis w.r.t. which for an elementary area on one side of the axis , there is a corresponding elementary area on the other side of the axis (the first moment of these elementary areas about the axis balance each other) If an area has an axis of symmetry, then the centroid must lie on that axis. If an area has two axes of symmetry, then the centroid must lie at the point of intersection of these axes.
Contd.
For example: The rectangular shown in the figure has two axis of symmetry, X-X and Y-Y. Therefore intersection of these two axes gives the centroid of the rectangle.
Y
B/2 B/2 da da
D/2
X da × x = da × x Moment of areas,da about y-axis cancel each other da × x + da × x = 0
x
x
D
X
D/2
B Y
Contd.
AXIS of SYMMETYRY
‘C’ must lie on the axis of symmetry ‘C’ must lie on the axis of symmetry
‘C’ must lie at the intersection of the axes of symmetry
To locate the centroid of simple rectangular area from first principles
D
B
To locate the centroid w.r.t. the base line x-x Let the distance of centroid from the base line x-x be y
D y X X
Then from the Principle of Moments
A ⋅ y = ∫ y ⋅ da
Moment of Total area A about x-axis = Sum of moment of elemental area dA about the same axis
Contd.
Consider a elemental area dA at a distance y from the base line (x-x) Let the thickness of the element be ‘dy’ Area of small element = dA = B .dy Moment of this elemental area about x-x axis x = (area) x (distance) = (B.dy) . (y)
B
dA
dy
y x y
Contd.
Sum of Moment of all such elemental areas comprising the total area =
= ∫ y ⋅ da
= ∫ B ⋅ dy ⋅ y
⎡ By ⎤ =⎢ ⎥ 2 ⎦0 ⎣
2 D
⎡ BD 2 ⎤ =⎢ ⎥ 2 ⎦ ⎣
Then from the Principle of Moments
BD 2 Ay = 2
D y= 2
BD 2 y= 2A
,
BD 2 y= 2 BD
,
Contd.
Similarly we can show
Y
B
x Y x
dx
A ⋅ x = ∫ x ⋅ da x = B/2
= ∫ x ⋅ (D ⋅ dx )
⎡ Dx ⎤ =⎢ ⎥ 2 ⎦0 ⎣
2
B
To locate the centroid of simple right angle triangular area from first principles
H
B
To locate the centroid w.r.t. the base line x-x. Let the distance of centroid from the base line x-x be y y H
x
x
Then from the Principle of Moments
A ⋅ y = ∫ y ⋅ da
Moment of Total area A about x-axis
=
Algebraic Sum of moment of elementary area ‘dA’ about the same axis.
Contd.
Consider a small elemental area ‘dA’ at a distance ‘y’ from the base line (x-x) Let the thickness of the element be ‘dy’ Area of small element = dA = b .dy dy dA Moment of this small elemental area about xx axis = (area) x (distance) = (b.dy) . (y)
Contd.
b
y x
x
y
Then from the Principle of Moments
A ⋅ y = ∫ y ⋅ da A ⋅ y = ∫ b ⋅ dy ⋅ y
A⋅ y = ∫ B (H − y ) ⋅ dy ⋅ y H da = b × dy = B (H − y ) × dy H
b=
B (H − y ) H
y = H/3
H/3 H y
X
B
X
Contd.
Similarly we can prove that x = B/3
A ⋅ x = ∫ x ⋅ da da = h ⋅ dx = H (B − x ) ⋅ dx B
A ⋅ x = ∫ b ⋅ dx ⋅ x
A⋅ x = ∫ H (B − x ) ⋅ dx ⋅ x B
Y dx b=
H (B − x ) B
x = B/3
H
x b x
B/3
Y
B
Contd.
The centroid of simple right angled triangle area from the base
B/3
H
Centroid
B H/3
To locate the centroid of Semi Circular Area w.r.t. the diameter AB from first principles Consider a semicircle of radius R, πD 2 A = area =
8
D = 2R
Let ‘G’ be the centroid of the Semicircle, and y is its distance from the diameter AB.
Contd.
dA = r × dθ × dr
Consider a small elemental area da, located at distance y from the diameter AB, Let ‘r’ =radial distance of area ‘da’ from centre of the semi circle.
da = r × dθ × dr y = r × sin θ
Moment of this elemental area about the diameter AB =
= r 2 sin θ ⋅ dr ⋅ dθ
Contd.
Moment of all such elemental area ‘da’ about the diameter AB
= ∫ ∫ r 2 ⋅ sin θ ⋅ dr.dθ
OO
π R
⎡r ⎤ = ∫ ⎢ ⎥ ⋅ sin θ ⋅ .dθ 3 ⎦O O⎣
3
π
R
R3 π = (−[cos θ ]O ) 3
2.R 3 = 3
Contd.
Then from the Principle of Moments
A ⋅ y = ∫ y ⋅ da
A⋅ y =
π R
y
Centroid
OO
r 2 ⋅ sin θ ⋅ dr.dθ ∫∫
A⋅ y =
4R y= 3π
2.R 3 3
x
R
x y
4R y= 3π
Because of symmetry
x=0 R
To locate the centroid of Quarter Circular Area w.r.t. the boundary radial line AB from first principle
A ⋅ y = ∫ y ⋅ da π dr r dθ
A⋅ y = y B
OO
r 2 ⋅ sin θ ⋅ dr.dθ ∫∫
2 R
A
θ R
4R y= 3π
Contd.
Centroid of Quarter Circular Area w.r.t. the boundary radial lines
4R x= 3π y= 4R 3π
Centroid
4R y= 3π
To locate the centroid of Circular Sector w.r.t. the y-axis shown from first principle y Xc=(2/3)Rcosθ
da
Consider a triangle of differential area = da =
= 1 × base × height 2
1 = × R × dθ × R 2
Distance of the differential area ‘da’, from y-axis = xc =
2 = × R × cos θ 3
Contd.
To locate the centroid of Circular Sector w.r.t. the y-axis shown from first principle Consider a triangle of differential area = da =
= 1 × base × height 2
1 = × R × dθ × R 2
Distance of the differential area ‘da’, from y-axis = xc =
2 = × R × cos θ 3
Contd.
Then from the Principle of Moments α A=∫
1 −α 2
R × dθ × R = R × α
2
A × x = ∫ da × xc
(
2 1 2 R × α x = ∫ ⋅ R cos θ × R dθ 3 2 −α
2
)
α
(
2 R × α x = × R 3 × sin α 3
2
)
2 R × sin α x= 3 α
2 R × sin α x= α 3
For a semicircular area 2α = π, if we use this value in the above formula we get
y =
4 R 3π
To locate the centroid of area under the curve x = k y3 from x = 0 to x = a from first principle
Consider a vertical element of area da = y dx at a distance x from the y-axis. da To find x- coordinate,
x a x
X
A × x = ∫ da × x
At x = a, y = b, i.e. a = k b3, k = a/b3
Contd.
A = ∫ y ⋅ dx
0
Substituting y = (x/k)1/3 and k = a/b3
x ∫ ydx = ∫ xydx
0 0
a
a
⎛ x⎞ ⎛ x⎞ x ∫ ⎜ ⎟ dx = ∫ x⎜ ⎟ dx k⎠ k⎠ 0⎝ 0 ⎝
a
1 3
a
1 3
3ab 3a b ⋅x= 4 7
2
4 x= a 7
Contd.
To find y, Coordinate of centroid of the rectangular element is yc = y/2
A × y = ∫ yc × dA
⎛ y⎞ y ∫ ydx = ∫ ⎜ ⎟( ydx ) 2⎠ 0 0⎝ a a
Substituting,
y = b( x/a)1/3
3ab 3ab 2 ⋅y= 4 10 2 y= b 5
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.1: Locate the centroid of the shaded area shown
10 50 10 40
Ans: x=12.5, y=17.5
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.2: Locate the centroid of the shaded area shown
500 1000 mm
D=600 r=600
300 300 500
1000 mm
Ans: x=474mm, y=474mm
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.3: Locate the centroid of the shaded area w.r.t. to the axes shown y-axis 90 20 120 r=40 20 60 x-axis
Ans: x=34.4, y=40.3
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.4: Locate the centroid of the shaded area w.r.t. to the axes shown y-axis 250 mm
20 10 380
10
200 mm
10 x-axis
Ans: x= -5mm, y=282mm
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.5 Locate the centroid of the shaded area w.r.t. to the axes shown y 30 50
30
40 40 20 r=20 20 x
Ans:x =38.94, y=31.46
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.6 Locate the centroid of the shaded area w.r.t. to the axes shown y 1.0
2.4 m r=0.6 1.0 1.0 x 1.5 1.5
Ans: x=0.817, y=0.24
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.7 Locate the centroid of the shaded area w.r.t. to the axes shown
Ans: x= -30.43, y= +9.58
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.8 Locate the centroid of the shaded area.
20
Ans: x= 0, y= 67.22(about base)
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.9 Locate the centroid of the shaded area w.r.t. to the base line.
2
Ans: x=5.9, y= 8.17
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.10 Locate the centroid of the shaded area w.r.t. to the axes shown
Ans: x=21.11, y= 21.11
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.11 Locate the centroid of the shaded area w.r.t. to the axes shown
Ans: x= y= 22.22
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.12 Locate the centroid of the shaded area w.r.t. to the axes shown
Y R=25 R=25
80
X
50 50 75
Ans: x= 24.33 y= 4.723
4. MOMENT OF INERTIA OF PLANE AREA
4. MOMENT OF INERTIA Moment of Inertia( Second moment area) The product of the elemental area and square of the perpendicular distance between the centroid of area and the axis of reference is the “Moment of Inertia” about the reference axis Iox= da1 y12 + da2 y22+ da3 y32+ -= ∑ da y2 Ioy = da1 x12 + da2 x22 + da3 x32+ ---= ∑ da x2
y x
dA
y o Radius of Gyration
A r1 r2 r3 B Elemental area A
k k k B
Radius of gyration is defined as a constant distance of all elemental areas which have been rearranged with out altering the total moment of inertia.
IAB= da k2 + da k2 + ----------IAB = ∑ da k2 IAB= A k2 k=√IAB/A
Polar moment of Inertia (Perpendicular Axes theorem) The moment of inertia of an area about an axis perpendicular to the plane of the area is called “Polar Moment of Inertia” and it is denoted by symbol Izz or J or Ip. The moment of inertia of an area in xy plane w.r.to z. axis is Izz = Ip = J = ∫r2dA = ∫(x2 + y2) dA = ∫x2dA + ∫y2dA = Ixx +Iyy
Y
x
r
O
y
x
z
Polar moment of Inertia (Perpendicular Axes theorem) Hence polar M.I. for an area w.r.t. an axis perpendicular to its plane of area is equal to the sum of the M.I. about any two mutually perpendicular axes in its plane, passing through the point of intersection of the polar axis and the area.
Parallel Axis Theorem
dA x
* G
y
x
d A B
Moment of inertia of any area about an axis AB is equal to the M.I. about parallel centroidal axis plus the product of the total area and square of the distance between the two axes.
IAB =∑dA (d +y)2 = ∑dA (d2 + y2 + 2 × d × y) =∑dA. d2 +∑dA y2 + ∑ 2×d×dA y = ∑dA. d2 +∑dA y2 + 2×d. ∑ y dA In the above term (2×d) is constant & ∑ y dA = 0 IAB = Ixx + A.d2
MOMENT OF INERTIA BY DIRECT INTEGRATION Moment of inertia of rectangular area about centroidal horizontal axis by direct integration M.I. about its horizontal centroidal axis :
I x x = ∫ dA × y 2
−D / 2
+D / 2
= ∫ ( B × dy ) × y x −D / 2 BD 3 = 12
2
+D / 2
D/2 D y
dy
G
x
B
Moment of Inertia of rectangular area about its base(about the line AB) using Parallel Axis Theorem IAB = IXX + A(d)2 Where d = D/2, the distance between axes xx and AB
D/2
=BD /12+(BD)(D/2) =BD3/12+BD3/4 =BD3/3
3
.
dy y
2
x
D
G
x
A B
B
Moment of inertia of Triangular area about the base by direct integration
dy
(h-y)
h x h/3 A
x
y
x
B b From similar triangles b/h = x/(h-y) ∴ x = b . (h-y)/h
dy
(h-y)
h x h/3 A
x
y
x
B b IAB = ∫ dA.y2 = ∫ (x.dy)y2
Ixx = ∫
h 0
(b . (h-y) y2.dy) /h
= b[ h (y3/3) – y4/4 ]/h = bh3/12
Moment of inertia of Triangular area about the centroidal horizontal axis Using Parallel axis theorem . MI about any line(AB) = MI about cenroidal parallel axis + Ad2 IAB = Ixx + Ad2 h x A b Centroidal horizontal axis x Ixx = MI about centroidal axis x x IAB= MI about the Base line AB Ixx = IAB – Ad2 = bh3/12 – bh/2 . (h/3)2 = bh3/36
dy
(h-y)
h/3
y
x
B
Moment of inertia of Circular area about the centroidal horizontal axis Ixx = ∫ dA . y2 = ∫ ∫ (x.dθ.dr) r2Sin2θ
0 0 R 2π
dθ
=∫ ∫
R
R 2π
r3.dr
Sin2θ
2π
r dθ
0 0
x A
θ
y=rSinθ
x
=∫ r3 dr ∫ {(1- Cos2θ)/2} dθ
0
R B
=[r4/4]
R 0
0
[θ/2 – Sin2θ/4]
2π 0
= R4/4[π - 0] = πR4/4 IXX = π R4/4 = πD4/64
Moment of inertia of Semi-circular area about the Base & centroidal horizontal axis IAB = ∫ dA . y2 = ∫ ∫ (r.dθ.dr) r2Sin2θ
0 0 R π 3.dr ∫ Sin2θ =∫ r 0 R π =∫ ∫ r3 0 0 0 R π
dθ
y0
dr (1- Cos2θ)/2) dθ [θ/2 – Sin2θ/4] π 0
=[R4/4]
= R4/4[π/2 - 0] = πR4/8 = (πD4/32)
x 4R/3π A
R
x B
y0
Moment of inertia of Semi-circular area about the centriodal horizontal axis using parallel axis theorem: IAB = Ixx + A(d)
2
Ixx = IAB – A(d)2 = π R4/8 Ixx = 0.11R4 πR2/2 . (4R/3π)2
Moment of inertia of Quarter-circular area about the base & centroidal horizontal axis IAB = ICD IAB = ∫ ∫ (r.dθ.dr). r2Sin2θ
0 0 R π/2
D
y
=∫ r3.dr ∫ Sin2θ dθ
0 R 0 π/2 0
R
π/2
x A C
4R/3π
x
4R/3π
=∫ r3 dr ∫ (1- Cos2θ)/2) dθ
0
y
B
=[R4/4] [θ/2 – (Sin2 θ)/4] = R4 (π/16 – 0) = πR4/16
π/2 0
Moment of inertia about Centroidal axis, Ixx = IAB - Ad2 = πR4/16 - πR2. (0. 424R)2 = 0.055R4
Sl.No
Y
1 x0 x h x0 x x0
Figure b d x0
Xo
Ix
-x 0 0
Iy
-y 0 0
I xx bd3/3
I yy -
d/2
Y
bd3/12
-
x x0 bh3/36 h/3 x x0 4R/3π x0 x x0 4R/3π 0.11R4 πR4/8 πR4/8 πR4/4 πR4/4 bh3/12 -
2
b
O
3
y0
R
4 x0 x y
y0
y0
5
y0 y0
0.055R4
0.055R4
πR4/16
πR4/16
4R/3π
EXERCISE PROBLEMS
Q1. Determine the moment of inertia about the centroidal axes. 30mm 30mm
20
30mm
100mm
[Ans: Y = 27.69mm Ixx = 1.801 x 106mm4 Iyy = 1.855 x 106mm4]
EXERCISE PROBLEMS
Q2. Determine second moment of area about the centroidal horizontal and vertical axes. 300mm 300mm 200
200mm 900mm [Ans: X = 99.7mm from A, Y = 265 mm Ixx = 10.29 x 109mm4, Iyy = 16.97 x 109mm4]
EXERCISE PROBLEMS
Q3. Determine M.I. Of the built up section about the horizontal and vertical centroidal axes and the radii of gyration.
200mm 20 140mm 20 100mm
[Ans: Ixx = 45.54 x 106mm4, Iyy = 24.15 x 106mm4 rxx = 62.66mm, ryy = 45.63mm]
60
EXERCISE PROBLEMS
Q4. Determine the horizontal and vertical centroidal M.I. Of the shaded portion of the figure.
X
60
20 20
X
60
60
[Ans: X = 83.1mm Ixx = 2228.94 x 104mm4, Iyy = 4789.61 x 104mm4]
EXERCISE PROBLEMS
Q5. Determine the spacing of the symmetrically placed vertical blocks such that Ixx = Iyy for the shaded area.
200mm
400mm 200mm d 200mm 200mm
[Ans: d/2 = 223.9mm d=447.8mm]
600mm
EXERCISE PROBLEMS
Q6. Find the horizontal and vertical centroidal moment of inertia of the section shown in Fig. built up with R.S.J. (I-Section) 250 x 250 and two plates 400 x 16 mm each attached one to each. Properties of I section are Ixx = 7983.9 x 104mm4 Iyy = 2011.7 x 104mm4
160mm
2500mm
Cross sectional area=6971mm2
160mm 4000mm
[Ans: Ixx = 30.653 x 107mm4, Iyy = 19.078 x 107mm4]
EXERCISE PROBLEMS
Q7. Find the horizontal and vertical centroidal moment of inertia of built up section shown in Figure. The section consists of 4 symmetrically placed ISA 60 x 60 with two plates 300 x 20 mm2.
Properties of ISA
Cross sectional area = 4400mm2 18.5mm 18.5mm
Ixx = Iyy ;Cxx = Cyy =18.5mm
200mm 20mm
300mm
[Ans: Ixx = 111.078 x 107mm4, Iyy = 39.574 x 107mm4]
EXERCISE PROBLEMS
Q8. The R.S. Channel section ISAIC 300 are placed back to back with required to keep them in place. Determine the clear distance d between them so that Ixx = Iyy for the composite section. Properties of ISMC300 C/S Area = 4564mm2 104mm4 X d [Ans: d = 183.1mm] Y
23.6mm
Y
Lacing
Ixx = 6362.6 x 104mm4 Iyy = 310.8 x Cyy = 23.6mm
X 380mm
EXERCISE PROBLEMS
Q9. Determine horizontal and vertical centroidal M.I. for the section shown in figure.
40mm 160mm 40mm 40mm 90mm
[Ans: Ixx = 2870.43 x 104mm4, Iyy = 521.64 x 104mm4]
5. Kinetics of rectilinear motion
5. Kinetics of rectilinear motion In this chapter we will be studying the relationship between forces on a body/particle and the accompanying motion
Newton’s Second law of motion:
Newton’s first and third law of motion were used extensively in the study of statics (the bodies at rest) whereas Newton’s second law of motion is used extensively in the study of the kinetics.
Work done by force: Work done by a force is the product of the force and the distance moved by the point of application in the direction of the force. It is a scalar quantity.
F sinθ
F
F cosθ
F sinθ
F B α F cosθ
A
Work done = (F cosα ) × s
α
s
X- component of force F moves through distance a S, S = displacement of force from A to B Unit: Nm ( Joule )
POWER:It is defined as the time rate of doing work. Power = work done /Time= (force × distance) /Time = force × velocity Unit: (Nm)/s = [watt] (kN m)/s = [kilo watt] 1 metric H.P=735.75 watts.
Energy:It is defined as the capacity to do work. It is a scalar quantity. Unit :- N m (Joule)
Momentum:Quantity of motion possessed by a body is called momentum. It is the product of mass and velocity. It is a vector quantity. Unit:- N s. Impulse of a Force:It is defined as the product of force and the time over which it acts. It is a vector quantity. Unit:- N s.
Newton’s second law of motion. “If the resultant force acting on a particle is not zero , the particle will have an acceleration proportional to the magnitude of the resultant force and its direction is along that of the resultant force.” Fαa F =Resultant of forces a = Acceleration of the particle. F = ma m= mass of the particle.
The constant value obtained for the ratio of the magnitude of the force and acceleration is characteristic of the particle and is denoted by ‘m’. Where ‘m’ is mass of the particle Since ‘m’ is a +ve scalar, the vectors of force ‘F’and acceleration ‘a’ have the same direction.
Units Force in Newtons (N) Acceleration in m/s2
N = 1 Kgm/s2
F2 F1 F3
he n t nt e i lta a ov s u m = l m f re il o R y w on od cti B re di
R = Resultant of forces F1,F2 and F3
Using the rectangular coordinate system we have components along axes as, ΣFx = max ΣFy = may ΣFz = maz where Fx ,Fy Fz and ax , ay ,az are rectangular components of resultant forces and accelerations respectively.
Newton’s second law may also be expressed by considering a force vector of magnitude ‘ma’ but of sense opposite to that of the acceleration. This vector is denoted by (ma)rev. The subscript indicates that the sense of acceleration has been reversed and is called the inertia force vector.
ov m ll i w t dy ltan B o su re
in e
t
ed h
i ct e ir
on
o
R
=
m
a
e rc Fo a tia m er = In R
F2 F1 F3
R = Resultant of forces F1,F2 and F3
F2 F1 F3
F2
ce m or = F R tia = F Iner
a
F1 F3
If the inertia force vector is added to the forces acting on the particle we obtain a system of forces whose resultant is zero.
Resultant of forces F1,F2, F3 and Inertia force = 0
F1 + F2 + F3 + ma = 0
The particle may thus be considered to be in equilibrium. (THIS IS DYNAMIC EQUILIBRIUM)
It was pointed out by D’Alembert (Alembert, Jean le Rond d’ (1717-1783), French mathematician and philosopher) that problems of kinetics can be solved by using the principles of statics only (the equations of equilibrium) by considering an inertia force in a direction directly opposite to the acceleration in addition to the real forces acting on the system D’Alembert’s principle states that When different forces act on a system such that it is in motion with an acceleration in a particular direction, the vectorial sum of all the forces acting on the system including the inertia force (‘ma’ taken in the opposite direction to the direction of the acceleration) is zero.
The problem under consideration may be solved by using the method developed earlier in statics. The particle is said to be in dynamic equilibrium. If ΣFx = 0 ΣFy= 0 ΣFz = 0 This principle is known as D’Alembert’s principle including inertia force vector
Work-Energy relation for translation From Newton’s second law of motion ∑F=m×a = Also
-----------(1)
a = dv/dt =( dv/dt) ×( ds/ds) = v × dv/ds substituting in (1)
∑ F = m × v × dv/ds ΣF × ds = m × v × dv ------------------(2) Let the initial velocity be u and the final velocity after it moves through a distance ‘ s’ be v
Integrating both sides, we get
∑ F ∫ ds = m ∫ v dv
0 u
s
v
v2 v ∑ F × s = (m ) 2 u 1 F × s = m(v 2 − u 2 ) ∑ 2
Therefore work done by a system of forces acting on a body while causing a displacement is equal to the change in kinetic energy of the body during the displacement.
Impulse-momentum relationship F=m×a F = m × (v - u)/t = (mv – mu)/t Force = Rate of change of momentum F × t = mv – mu Impulse = final momentum – Initial momentum The component of the resultant linear impulse along any direction is equal to change in the component of momentum in that direction.
EXERCISE PROBLEMS 5. Kinetics
Q1. Blocks A and B of mass 10 kg and 30 kg respectively are connected by an inextensible cord passing over a smooth pulley as shown in Fig. Determine the velocity of the system 4 sec. after starting from rest. Assume coefficient of friction =0.3 for all surfaces in contact.
B
60o
A
30o
Ans: v=13.6m/s
EXERCISE PROBLEMS 5. Kinetics
Q2.
A tram car weighs 150kN. The tractive resistance being 1% of the weight of car. What power will be required to move the car at uniform speed of 20 kmph
(i) Up an incline 1 in 300 (ii) (ii) Down an incline 1 in 250. Take efficiency 75%.
Ans: Pull = 2 kN
a) Output power=11.12kW
EXERCISE PROBLEMS 5. Kinetics
Q3. Two masses of 5 kg and 3 kg rest on two smooth inclined plane, each of inclination 30º and are connected by a string passing over a common apex. Find the velocity of 3 kg mass after 2 sec when released from rest. Find the distance it will cover before changing direction of motion, if 5kg mass is cut off after two sec of its release from rest.
5kg 30º
V = 4.45 m/s s = 0.61 m
3kg 30º
EXERCISE PROBLEMS 5. Kinetics
Q4. A locomotive weighing 900 kN pulls a train of 10 coaches each weighing 300 kN at 72 Kmph on a level track against a resistance of 7 N/kN. If the rear 4 coaches get snapped from the train, find the speed of the engine and the remaining coaches after 120 secs. Assume no change in resistance and draw bar pull. Find also distance traveled by detached coaches before coming to rest.
4x300=1200kN
6x300=1800kN
900kN
P
V = 23.66 m/s s = 2.9 km
EXERCISE PROBLEMS 5. Kinetics
Q5. Find the tension in the cord supporting body C in Fig. below. The pulleys are frictionless and of negligible weight.
A C 150 kN B 450 kN 300 kN
Assume all blocks moving either downward or upward and accordingly draw FBD 0=aA +2aB +cC
Ans : T=211.72 kN
EXERCISE PROBLEMS 5. Kinetics
Q6. Two blocks A and B are released from rest on a 30o inclined plane with horizontal, when they are 20m apart. The coefficient of friction under the upper block is 0.2 and that under lower block is 0.4. compute the time elapsed until the block touch. After they touch and move as a unit what will be the constant forces between them.
(Ans : t = 4.85 s, contact force=8.65 N)
EXERCISE PROBLEMS 5. Kinetics
Q7. An elevator cage of a mine shaft weighing 8kN when empty is lifted or lowered by means of rope. Once a man weighing 600N entered it and lowered at uniform acceleratin such that when a distance of 187.5 m was covered, the velocity of the cage was 25m/s. Determine the tension in the cable and force exerted by man on the floor of the cage.
(Ans: T=7139 N and R=498 N)
EXERCISE PROBLEMS 5. Kinetics
Q8. A small block starts from rest at point a and slides down the inclined plane. At what distance along the horizontal will it travel before coming to rest . Take µk=0.3 [Ans :s=6m ]
5m
A 3 4 B s C
EXERCISE PROBLEMS 5. Kinetics
Q9. The system starts from rest in the position shown . How much further will block ‘A’ move up the incline after block B hits the ground . assume the pulley to be frictionless and massless and µ is 0.2 .WA=1000N, WB=2000N. [ Answer s =1.27m]
A 4 3 B 3m
EXERCISE PROBLEMS 5. Kinetics
Q10. A 1500Kg automobile travels at a uniform rate of 50kmph to 75kmph . During the entire motion, the automobile is traveling on a level horizontal road and rolling resistance is 2 % of weight of automobile . Find (i) maximum power developed (ii) power required to maintain a constant speed of 75kmph.
[ ANSWER: power developed = 6.131KN]
EXERCISE PROBLEMS 5. Kinetics
Q11. Two bodies A and B weighing 2000N and 5200N are connected as shown in the figure . find the further distance moved by block a after the block B hits Wall. µ=0.2 .[ Answer s=1.34m]
A 5 12 B
3m
EXERCISE PROBLEMS 5. Kinetics
Q12. A spring is used to stop 60kg package which is sliding on a horizontal surface . the spring has a constant k = 20kN/m and is held by cable such that it is initially compressed at 120mm. knowing that the package has a velocity of 2.5m/s in position shown and maximum additional displacement of spring is 40mm . Determine the coefficient of kinetic friction between package and surface. (Answer µk=0.2) 2.5 m/s
60kg
600m
EXERCISE PROBLEMS 5. Kinetics
Q13. The system shown in figure has a rightward velocity of 4m/s, just before force P is applied. Determine the value of P that will give a leftward velocity of 6m/s in a time interval of 20sec. Take µ = 0.2 & assume ideal pulley. [Answer P=645.41N] P 1000N
400N
EXERCISE PROBLEMS 5. Kinetics
Q14. A locomotive of weight 500kN pulls a train of weight of 2500kN. The tractive resistance, due to friction is 10N/kN. The train can go with a maximum speed of 27kmph on a grade of 1in100. Determine (a) Power of the locomotive. (b) Maximum speed it can attain on a straight level track with the tractive resistance remaining same. [Answer (a) Power= 450kN (b) v=15m/s] Q15. A wagon weighing 400kN starts from rest, runs 30m down a 1% grade & strikes a post. If the rolling resistance of the track is 5N/kN, find the velocity of the wagon when it strikes the post. If the impact is to be cushioned by means of one bumper string, which compresses 1mm per 20 kN weight, determine how much the bumper spring will be compressed. [Answer v=1.716m/s, x=77.5mm]
EXERCISE PROBLEMS 5. Kinetics
Q16. A train whose weight is 20kN moves at the rate of 60kmph. After brakes are applied, it is brought to rest in 500m. Find the force exerted, assuming it to be uniform a) Use work-energy relation b) Use D’Alemberts equation.
Ans: F = 5.663 kN
EXERCISE PROBLEMS 5. Kinetics
Q17. The blocks A and B having weights 100 N and 300 N start from rest. The horizontal plane and the pulleys are frictionless. Determine the acceleration and the tension in the string. string Frictionless pulley
A string
Frictionless pulley
Ans: aA=8.403m/s2 aB=4.201 m/s2 T= 85.71 N
B
EXERCISE PROBLEMS 5. Kinetics
Q18. The blocks A and B having weights 100 N and 300 N start from rest. The horizontal plane and the pulleys are frictionless. Determine the velocity of block B after 0.5 seconds and the tension in the string. Use impulsemomentum relation. string Frictionless pulley
A string
Frictionless pulley B
EXERCISE PROBLEMS 5. Kinetics
Q19. The blocks A and B having weights 100 N and 300 N start from rest when a load of 100 N is applied on the block A as shown in the figure. The horizontal plane and the pulleys are frictionless. Determine the acceleration and the tension in the string. string 100 N
A
Frictionless pulley
string
Frictionless pulley B
EXERCISE PROBLEMS 5. Kinetics
Q20. Two blocks A and B are connected as shown in the figure. At the instant of their release if the block A, which is on smooth horizontal plane has a left ward velocity of 2 m/s, what would be its velocity 5 seconds after their release. The blocks A and B weigh 100 N and 300 N respectively. string Frictionless pulley
A string
Frictionless pulley B
EXERCISE PROBLEMS 5. Kinetics
Q21. An engine weighing 500 kN drags carriages weighing 1500kN up an incline of 1 in 100 against a resistance of 5N/kN starting from rest. It attains a velocity of 36 kmph (10m/s) in 1 km distance with a constant draw bar pull supplied by the engine. What is the power required for the same ? What is the tension developed in the link connecting the engine and carriages? 1500N 500N P 100 1
Ans: Pull=40.19kN, power=401kW, T=30.15kN
EXERCISE PROBLEMS 5. Kinetics
Q22. what velocity the block A will attain after 2 seconds starting from rest? Take µ = 0.2. WA = 1500N, WB = 2000N. Use impulse-momentum relation.
A 3
4
3 4
B
PART - II Mechanics of Deformable Bodies
COURSE CONTENT IN BRIEF
6. Simple stresses and strains 7. Statically indeterminate problems and thermal stresses 8. Stresses on inclined planes 9. Stresses due to fluid pressure in thin cylinders
6. Simple stresses and strains
The subject strength of materials deals with the relations between externally applied loads and their internal effects on bodies. The bodies are no longer assumed to be rigid and the deformations, however small, are of major interest The subject, strength of materials or mechanics of materials involves analytical methods for determining the strength , stiffness (deformation characteristics), and stability of various load carrying members. Alternatively the subject may be called the mechanics of solids.
GENERAL CONCEPTS STRESS No engineering material is perfectly rigid and hence, when a material is subjected to external load, it undergoes deformation. While undergoing deformation, the particles of the material offer a resisting force (internal force). When this resisting force equals applied load the equilibrium condition exists and hence the deformation stops. These internal forces maintain the externally applied forces in equilibrium.
STRESS
The internal force resisting the deformation per unit area is called as stress or intensity of stress.
Stress = internal resisting force / resisting cross sectional area
R = A
STRESS SI unit for stress N/m2 also designated as a pascal (Pa) Pa = N/m2 kilopascal, 1kPa = 1000 N/m2 megapascal, 1 MPa = 1×106 N/m2 = 1×106 N/(106mm2) = 1N/mm2 1 MPa = 1 N/mm2 gigapascal, 1GPa = 1×109 N/m2 = 1×103 MPa = 1×103 N/mm2
AXIAL LOADING – NORMAL STRESS P P R
STRESS
Consider a uniform bar of cross sectional area A, subjected to a tensile force P. Consider a section AB normal to the direction of force P Let R is the total resisting force acting on the cross section AB. Then for equilibrium condition,
A
B
R
R=P Then from the definition of stress, normal stress = σ = R/A = P/A
P
P
Symbol:
σ = Normal Stress
AXIAL LOADING – NORMAL STRESS Direct or Normal Stress: Intensity of resisting force perpendicular to or normal to the section is called the normal stress. Normal stress may be tensile or compressive
STRESS
Tensile stress: stresses that cause pulling on the surface of the section, (particles of the materials tend to pull apart causing extension in the direction of force) Compressive stress: stresses that cause pushing on the surface of the section, (particles of the materials tend to push together causing shortening in the direction of force)
STRESS • The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis. • The force intensity on that section is defined as the normal stress.
∆F σ = lim ∆A→0 ∆A
P σ ave = A
STRAIN STRAIN : when a load acts on the material it will undergo deformation. Strain is a measure of deformation produced by the application of external forces. If a bar is subjected to a direct load, and hence a stress, the bar will changes in length. If the bar has an original length L and change in length by an amount δL, the linear strain produced is defined as, δL Change in length ε= = Linear strain, Original length L Strain is a dimensionless quantity.
Linear Strain
P σ = = stress A
ε=
δ
2P P = σ= 2A A
σ=
L
= normal strain
ε=
δ
P A 2δ δ ε= = 2L L
L
STRESS-STRAIN DIAGRAM In order to compare the strength of various materials it is necessary to carry out some standard form of test to establish their relative properties. One such test is the standard tensile test in which a circular bar of uniform cross section is subjected to a gradually increasing tensile load until failure occurs. Measurement of change in length over a selected gauge length of the bar are recorded throughout the loading operation by means of extensometers. A graph of load verses extension or stress against strain is drawn as shown in figure.
STRESS-STRAIN DIAGRAM
Proportionality limit
Typical tensile test curve for mild steel
STRESS-STRAIN DIAGRAM
Typical tensile test curve for mild steel showing upper yield point and lower yield point and also the elastic range and plastic range
Stress-strain Diagram Limit of Proportionality : From the origin O to a point called proportionality limit the stress strain diagram is a straight line. That is stress is proportional to strain. Hence proportional limit is the maximum stress up to which the stress – strain relationship is a straight line and material behaves elastically. From this we deduce the well known relation, first postulated by Robert Hooke, that stress is proportional to strain. Beyond this point, the stress is no longer proportional to strain
PP σP = = Load at proportionality limit A Original cross sectional area
Stress-strain Diagram Elastic limit: It is the stress beyond which the material will not return to its original shape when unloaded but will retain a permanent deformation called permanent set. For most practical purposes it can often be assumed that points corresponding proportional limit and elastic limit coincide. Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will be thus some permanent deformation when load is removed.
PE Load at proportional limit σE = = A Original cross sectional area
Stress-strain Diagram Yield point: It is the point at which there is an appreciable elongation or yielding of the material without any corresponding increase of load.
PY σY = = A
Load at yield point Original cross sectional area
Ultimate strength: It is the stress corresponding to maximum load recorded during the test. It is stress corresponding to maximum ordinate in the stress-strain graph.
P σU = U = A
Maximum load taken by the material Original cross sectional area
Stress-strain Diagram Rupture strength (Nominal Breaking stress): It is the stress at failure. For most ductile material including structural steel breaking stress is somewhat lower than ultimate strength because the rupture strength is computed by dividing the rupture load (Breaking load) by the original cross sectional area.
PB σB = = A
load at breaking (failure) Original cross sectional area load at breaking (failure) Actual cross sectional area
True breaking stress =
Stress-strain Diagram After yield point the graph becomes much more shallow and covers a much greater portion of the strain axis than the elastic range. The capacity of a material to allow these large plastic deformations is a measure of ductility of the material Ductile Materials: The capacity of a material to allow large extension i.e. the ability to be drawn out plastically is termed as its ductility. Material with high ductility are termed ductile material. Example: Low carbon steel, mild steel, gold, silver, aluminum
Stress-strain Diagram
A measure of ductility is obtained by measurements of the percentage elongation or percentage reduction in area, defined as, increase in gauge length (up to fracture) ×100 = original gauge length Percentage elongation Reduction in cross sectional area of necked portion (at fracture)
Percentage reduction in = area original area
×100
Cup and cone fracture for a Ductile Material
Stress-strain Diagram Brittle Materials : A brittle material is one which exhibits relatively small extensions before fracture so that plastic region of the tensile test graph is much reduced. Example: steel with higher carbon content, cast iron, concrete, brick
Stress-strain diagram for a typical brittle material
HOOKE’S LAW Hooke’s Law For all practical purposes, up to certain limit the relationship between normal stress and linear strain may be said to be linear for all materials stress (σ) α strain (ε) stress (σ) constant strain (ε) = Thomas Young introduced a constant of proportionality that came to be known as Young’s modulus. stress (σ) E strain (ε) = = Young’s Modulus or Modulus of Elasticity
HOOKE’S LAW Young’s Modulus is defined as the ratio of normal stress to linear strain within the proportionality limit. stress (σ) = E = strain (ε)
P δL PL ÷ = A L AδL
The value of the Young’s modulus is a definite property of a material From the experiments, it is known that strain is always a very small quantity, hence E must be large. For Mild steel, E = 200GPa = 2×105MPa = 2×105N/mm2
Deformations Under Axial Loading • From Hooke’s Law: σ = Eε ε= σ
E =
P AE
• From the definition of strain:
L • Equating and solving for the deformation, PL δ = AE
• With variations in loading, crosssection or material properties, Pi Li δ =∑ i Ai Ei
ε=
δ
Consider an element of length, δx at a distance x from A
W A W
d1
x
B
Diameter at x, = d1
L = d1 + k × x
(d − d ) + 2 1 ×x
dx
d2 c/s area at x, πd12 =
4 =
π
4
(d1 + kx )2
Change in length over a length dx is Change in length over a length L is
⎛ ⎞ ⎜ ⎟ Wdx ⎛ PL ⎞ ⎟ =⎜ ⎟ =⎜ ⎝ AE ⎠ dx ⎜ π (d + kx )2 × E ⎟ ⎜ ⎟ 1 ⎝4 ⎠ ⎛ ⎞ ⎜ ⎟ L Wdx ⎟ =∫ ⎜ 0 ⎜ π (d + kx )2 × E ⎟ ⎜ ⎟ 1 4 ⎝ ⎠
Consider an element of length, δx at a distance x from A Change in length over a length L is
⎛ ⎞ ⎜ ⎟ L Wdx ⎟ =∫ ⎜ 0 π ⎜ (d + kx )2 × E ⎟ ⎜ ⎟ 1 ⎝4 ⎠ dt ⎞ ⎛ ⎜ W ⎟ L k ⎟ =∫ ⎜ 0 ⎜ π (t )2 × E ⎟ ⎜ ⎟ ⎝4 ⎠
Put d1+kx = t, Then k dx = dt
L
⎤ 4W ⎡ t 4W = ⎢ ⎥ = πEk ⎣ − 1 ⎦ 0 πEk
− 2 +1
L
− 4W ⎡ − 1⎤ ⎢ t ⎥ = πEk ⎣ ⎦0
⎡ ⎤ 1 ⎢ ⎥ (d1 + kx) ⎦ 0 ⎣
L
=
4WL WL = πEd1d 2 πd1d 2 × E 4
Derive an expression for the total extension of the tapered bar AB of rectangular cross section and uniform thickness, as shown in the figure, when subjected to an axial tensile load ,W. W B L b
d1 b
W A
d2
d1 b
W A x dx B
W
d2 b
Consider an element of length, δx at a distance x from A depth at x,
= d1 +
L = d1 + k × x
(d 2 − d1 ) × x
c/s area at x, = (d1 + kx )b
Change in length over a length dx is
⎛ ⎞ Wdx ⎛ PL ⎞ =⎜ =⎜ ⎟ ⎜ (d + kx )b × E ⎟ ⎟ ⎝ AE ⎠ dx ⎝ 1 ⎠
Change in length over a length L is
=∫
L
0
⎛ ⎞ Wdx ⎜ ⎜ (d + kx )b × E ⎟ ⎟ ⎝ 1 ⎠
P (log e d 2 − log e d1 ) = b× E ×k
2.302 × P × L (log d 2 − log d1 ) = b × E × (d 2 − d1 )
Derive an expression for the total extension produced by self weight of a uniform bar, when the bar is suspended vertically.
L
Diameter d
dx x
P1 = weight of the bar below element the section, = volume × specific weight dx = (π d2/4)× x × γ P1 = A× x ×γ
Diameter d
Extension of the element due to weight of the bar below that,
P1 dx ( A × x × ρ ) dx ⎡ PL ⎤ =⎢ ⎥ = AE = AE ⎣ AE ⎦ dx
Hence the total extension entire bar
=∫
L
0
( A × x × γ )dx ⎡ γx ⎤ γL2 =⎢ ⎥ = AE ⎣ 2E ⎦ 0 2E
2
L
The above expression can also be written as
A ( ALγ ) × L 1 PL = × = = × 2E A 2 AE 2 AE
γL2
Where, P = (AL)×γ = total weight of the bar
SHEAR STRESS Consider a block or portion of a material shown in Fig.(a) subjected to a set of equal and opposite forces P. then there is a tendency for one layer of the material to slide over another to produce the form failure as shown in Fig.(b) P P P P R Fig. b Fig. c R
Fig. a
The resisting force developed by any plane ( or section) of the block will be parallel to the surface as shown in Fig.(c). The resisting forces acting parallel to the surface per unit area is called as shear stress.
Shear stress (τ) =
Shear resistance Area resisting shear
P = A
This shear stress will always be tangential to the area on which it acts
Shear strain
If block ABCD subjected to shearing stress as shown in Fig.(d), then it undergoes deformation. The shape will not remain rectangular, it changes into the form shown in Fig.(e), as AB'C'D. τ τ B' C' C C B B
A
τ Fig. d
D
A
τ Fig. e
D
B
B'
τ
C
φ
A
τ Fig. e
D
Shear strain is defined as C' the change in angle between two line element which are originally right angles to one another.
BB′ shear strain = = tan φ ≈ φ AB
The angle of deformation
φ
is then termed as shear strain
The angle of deformation is measured in radians and hence is non-dimensional.
SHEAR MODULUS For materials within the proportionality limit the shear strain is proportional to the shear stress. Hence the ratio of shear stress to shear strain is a constant within the proportionality limit. Shear Modulus Shear stress (τ) = constant = G = or Shear strain (φ) Modulus of Rigidity
The value of the modulus of rigidity is a definite property of a material For Mild steel, G= 80GPa = 80,000MPa = 80,000N/mm2
example: Shearing Stress • Forces P and P‘ are applied transversely to the member AB. • Corresponding internal forces act in the plane of section C and are called shearing forces. • The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load P. • The corresponding average shear stress is, τ ave =
P A
• The shear stress distribution cannot be assumed to be uniform.
State of simple shear
Consider an element ABCD in a strained material subjected to shear stress, τ as shown in the figure A τ B
D
C
τ Force on the face AB = P = τ × AB × t Where, t is the thickness of the element. Force on the face DC is also equal to P
State of simple shear Now consider the equilibrium of the element. (i.e., ΣFx = 0, ΣFy = 0, ΣM = 0.) For the force diagram shown, ΣFx = 0, & ΣFy = 0, But ΣM = 0 The element is subjected force to a clockwise moment P × AD = (τ × AB × t) × AD D P C A P B
But, as the element is actually in equilibrium, there must be another pair of forces say P' acting on faces AD and BC, such that they produce a anticlockwise moment equal to ( P × AD )
State of simple shear P ' × AB = P × AD = (τ × AB × t)× AD ----- (1) If τ1 is the intensity of the shear stress on the faces AD and BC, then P ' can be written as, P ' = τ ' × AD × t Equn.(1) can be written as τ'
D A
P
B
P'
D
P' P τ
C
A
B
τ' τ
C
(τ ' × AD× t ) × AB = (τ × AB × t) × AD ----- (1) τ' =τ
State of simple shear Thus in a strained material a shear stress is always accompanied by a balancing shear of same intensity at right angles to itself. This balancing shear is called “complementary shear”. τ
A
B
The shear and the complementary shear together constitute a state of simple shear
τ'= τ
D
τ'= τ τ
C
Direct stress due to pure shear Consider a square element of side ‘a’ subjected to shear stress as shown in the Fig.(a). Let the thickness of the square be unity. τ τ A
A B B
τ
D
a a τ Fig.(a). τ
C
τ
D
a a τ Fig.(b).
C
τ
Fig.(b) shows the deformed shape of the element. The length of diagonal DB increases, indicating that it is subjected to tensile stress. Similarly the length of diagonal AC decreases indicating that compressive stress.
Direct stress due to pure shear Now consider the section, ADC of the element, Fig.(c). X
A
σn a a
A
τ
D
a a
D
( 2 )a
C
τ
C
Fig.(c).
Resolving the forces in σn direction, i.e., in the X-direction shown
For equilibrium
∑ Fx = 0
=σn ×
(
2 × a ×1 − 2(τ × a × cos 45)
)
σn =τ
Direct stress due to pure shear Therefore the intensity of normal tensile stress developed on plane BD is numerically equal to the intensity of shear stress.
Similarly it can be proved that the intensity of compressive stress developed on plane AC is numerically equal to the intensity of shear stress.
POISSON’S RATIO Poisson’s Ratio: Consider the rectangular bar shown in Fig.(a) subjected to a tensile load. Under the action of this load the bar will increase in length by an amount δL giving a longitudinal strain in the bar of δl εl = l Fig.(a)
POISSON’S RATIO The bar will also exhibit, reduction in dimension laterally, i.e. its breadth and depth will both reduce. These change in lateral dimension is measured as strains in the lateral direction as given below. δb δd ε lat = − = − b d The associated lateral strains will be equal and are of opposite sense to the longitudinal strain. Provided the load on the material is retained within the elastic range the ratio of the lateral and longitudinal strains will always be constant. This ratio is termed Poisson’s ratio (µ)
(− δd ) d = POISSON’S RATIO = δl Longitudinal strain l
Lateral strain
(− δb ) b OR δl l
Poisson’s Ratio = µ
For most engineering metals the value of µ lies between 0.25 and 0.33 y Lz P Lx z Poisson’s Ratio Lateral strain Strain in the direction of load applied
− δl y = ly
− δl z
In general
Ly P x
=
δl x
lx
OR = δl x
lz
lx
Poisson’s Ratio = µ
In general Px
Lx
y
Lz Ly Px
x
z
Strain in Y-direction =
Strain in X-direction = εx
=
δl x lx εy
=
δl y ly =µ
δl x lx Strain in Z-direction = εz
=
δl z lz =µ
δl x lx Load applied in Y-direction y
Lz
Py
Ly
Lx
x Py
z
Poisson’s Ratio
=
Lateral strain Strain in the direction of load applied
− δl x =
δl y
lx
− δl z
ly
OR = δl y
lz
ly
Strain in X-direction = εx
=
δl x lx =µ
δl y ly Load applied in Z-direction y
Lz
Pz
Ly
x z Pz Lateral strain Strain in the direction of load applied
Lx
− δl x =
Poisson’s Ratio
=
δl z
lx
− δl y
OR
=
lz
δl z
ly
lz
Strain in X-direction = εx
=
δl x lx =µ
δl z lz Load applied in X & Y direction y
Lz
Py
Ly Px
Strain in X-direction = εx x
Px
Lx
=
σx
E
−µ
σy
E
z
Py
Strain in Y-direction = εy
=
σy
E
−µ
σx
E
Strain in Z-direction = εz
= −µ
σy
E
−µ
σx
E
Py
y
Px
Pz
General case: Strain in X-direction = εx
Px
x
z
Pz
Py
εx = σy σx σz
σx
E
−µ
σy
E
−µ
σz
E
Strain in Y-direction = εy
E Strain in Z-direction = εz σy σx σz εz = −µ −µ
E E E
εy =
σy
E
−µ
σx
E
−µ
σz
σz σx
σy
Bulk Modulus Bulk Modulus A body subjected to three mutually perpendicular equal direct stresses undergoes volumetric change without distortion of shape. If V is the original volume and dV is the change in volume, then dV/V is called volumetric strain. A body subjected to three mutually perpendicular equal direct stresses then the ratio of stress to volumetric strain is called Bulk Modulus. σ = Bulk modulus, K ⎛ dV ⎞ ⎜ ⎟ V ⎠ ⎝
Relationship between volumetric strain and linear strain Consider a cube of side 1unit, subjected to three mutually perpendicular direct stresses as shown in the figure. Relative to the unstressed state, the change in volume per unit volume is dV = 1 − (1 + ε x )(1 + ε y )(1 + ε z ) = 1 − 1 + ε x + ε y + ε z 1 = εx +εy +εz = change in volume per unit volume
[
]
[
]
Relationship between volumetric strain and linear strain
Volumetric strain dV = εx +εy +εz V σy ⎛σx σ ⎜ =⎜ −µ −µ z E E ⎝ E
⎞ ⎛σ y σ σ ⎟ +⎜ −µ x −µ z ⎟ ⎜ E E E ⎠ ⎝
⎞ ⎛σ ⎟ +⎜ z −µ σy −µ σx ⎞ ⎟ ⎟ ⎜ E E ⎟ ⎠ ⎝E ⎠
1 − 2µ (σ x + σ y + σ z ) = E
For element subjected to uniform hydrostatic pressure,
σx =σy =σz =σ dV 1 − 2 µ (σ x + σ y + σ z ) = V E dV 1 − 2 µ (3σ ) = V E
K=
σ
⎛ dV ⎞ ⎜ ⎟ ⎝ V ⎠
E K= = bulk modulus 3(1 − 2 µ ) or E = 3K (1 - 2 µ )
Relationship between young’s modulus of elasticity (E) and modulus of rigidity (G) :A1 A H
45˚
a
B1 B
φ
D
a φτ τ C
Consider a square element ABCD of side ‘a’ subjected to pure shear ‘τ’. DA'B'C is the deformed shape due to shear τ. Drop a perpendicular AH to diagonal A'C. Strain in the diagonal AC = τ /E – µ (- τ /E) [ σn= τ ] = τ /E [ 1 + µ ] -----------(1) Strain along the diagonal AC=(A'C–AC)/AC=(A'C–CH)/AC=A'H/AC
In ∆le AA'H Cos 45˚ = A'H/AA' A'H= AA' × 1/√2 AC = √2 × AD ( AC = √ AD2 +AD2) Strain along the diagonal AC = AA'/ (√2 × √2 × AD)=φ/2 ----(2) Modulus of rigidity = G = τ /φ φ = τ /G Substituting in (2) Strain along the diagonal AC = τ /2G -----------(3) Equating (1) & (3) τ /2G = τ /E[1+µ] E=2G(1+ µ)
Relationship between E, G, and K:We have E = 2G( 1+ µ) -----------(1) E = 3K( 1-2µ) -----------(2) Equating (1) & (2) 2G( 1+ µ) =3K( 1- 2µ) 2G + 2Gµ=3K- 6Kµ µ= (3K- 2G) /(2G +6K) Substituting in (1) E = 2G[ 1+(3K – 2G)/ (2G+6K)] E = 18GK/( 2G+6K) E = 9GK/(G+3K)
Working stress: It is obvious that one cannot take risk of loading a member to its ultimate strength, in practice. The maximum stress to which the material of a member is subjected to in practice is called working stress. This value should be well within the elastic limit in elastic design method. Factor of safety: Because of uncertainty of loading conditions, design procedure, production methods, etc., designers generally introduce a factor of safety into their design, defined as follows
Factor of safety = Maximum stress or Yield stress (or proof stress) Allowable working Allowable working stress stress
Malleability: A property closely related to ductility, which defines a material’s ability to be hammered out in to thin sheets Homogeneous: A material which has a uniform structure throughout without any flaws or discontinuities. Isotropic: If a material exhibits uniform properties throughout in all directions ,it is said to be isotropic. Anisotropic: If a material does not exhibit uniform properties throughout in all directions ,it is said to be anisotropic or nonisotropic.
Exercise Problems Q1. An aluminum tube is rigidly fastened between a brass rod and steel rod. Axial loads are applied as indicated in the figure. Determine the stresses in each material and total deformation. Take Ea=70GPa, Eb=100GPa, Es=200GPa
Aa=1000mm2
20kN
Ab=700mm2
brass 500mm
15kN
15kN
As=800mm2
aluminum 600mm
steel 700mm
10kN
Ans: σb=28.57MPa, σa=5MPa, σs=12.5MPa, δl = - 0.142mm
Q2.
A 2.4m long steel bar has uniform diameter of 40mm for a length of 1.2m and in the next 0.6m of its length its diameter gradually reduces to ‘D’ mm and for remaining 0.6m of its length diameter remains the same as shown in the figure. When a load of 200kN is applied to this bar extension observed is equal to 2.59mm. Determine the diameter ‘D’ of the bar. Take E =200GPa 200kN
1000mm 500mm 500mm
200kN
Ф = 40mm Ф = D mm
Q3. The diameter of a specimen is found to reduce by 0.004mm when it is subjected to a tensile force of 19kN. The initial diameter of the specimen was 20mm. Taking modulus of rigidity as 40GPa determine the value of E and µ Ans: E=110GPa, µ=0.36 Q.4 A circular bar of brass is to be loaded by a shear load of 30kN. Determine the necessary diameter of the bars (a) in single shear (b) in double shear, if the shear stress in material must not exceed 50MPa. Ans: 27.6, 19.5mm
Q.5 Determine the largest weight W that can be supported by the two wires shown. Stresses in wires AB and AC are not to exceed 100MPa and 150MPa respectively. The cross sectional areas of the two wires are 400mm2 for AB and 200mm2 for AC. Ans: 33.4kN B 300 A 450 C
W
Q.6 A homogeneous rigid bar of weight 1500N carries a 2000N load as shown. The bar is supported by a pin at B and a 10mm diameter cable CD. Determine the stress in the cable D Ans: 87.53MPa
A 3m 2000 N
C 3m
B
Q.7. A stepped bar with three different cross-sectional areas, is fixed at one end and loaded as shown in the figure. Determine the stress and deformation in each portions. Also find the net change in the length of the bar. Take E = 200GPa
300mm2 20kN 450mm2 250mm2 40kN 10kN
250mm
320mm
270mm
Ans: -33.33, -120, 22.2MPa, -0.042, -0.192, 0.03mm, -0.204mm
Q.8
a) b) c)
The coupling shown in figure is constructed from steel of rectangular cross-section and is designed to transmit a tensile force of 50kN. If the bolt is of 15mm diameter calculate: The shear stress in the bolt; The direct stress in the plate; The direct stress in the forked end of the coupling.
Ans: a)141.5MPa, b)166.7MPa, c)83.3MPa
Q.9 The maximum safe compressive stress in a hardened steel punch is limited to 1000MPa, and the punch is used to pierce circular holes in mild steel plate 20mm thick. If the ultimate shearing stress is 312.5MPa, calculate the smallest diameter of hole that can be pierced. Ans: 25mm Q.10 A rectangular bar of 250mm long is 75mm wide and 25mm thick. It is loaded with an axial tensile load of 200kN, together with a normal compressive force of 2000kN on face 75mm×250mm and a tensile force 400kN on face 25mm×250mm. Calculate the change in length, breadth, thickness and volume. Take E = 200GPa & µ=0.3 Ans: 0.15,0.024,0.0197mm, 60mm3
Q.11 A piece of 180mm long by 30mm square is in compression under a load of 90kN as shown in the figure. If the modulus of elasticity of the material is 120GPa and Poisson’s ratio is 0.25, find the change in the length if all lateral strain is prevented by the application of uniform lateral external pressure of suitable intensity. 90kN
30 180
30
Ans: 0.125mm
Q.12 Define the terms: stress, strain, elastic limit, proportionality limit, yield stress, ultimate stress, proof stress, true stress, factor of safety, Young’s modulus, modulus of rigidity, bulk modulus, Poisson's ratio, Q.13 Draw a typical stress-strain diagram for mild steel rod under tension and mark the salient points. Q.14 Diameter of a bar of length ‘L’ varies from D1 at one end to D2 at the other end. Find the extension of the bar under the axial load P Q.15 Derive the relationship between Young’s modulus and modulus of rigidity.
Q.16 Derive the relationship between Young’s modulus and Bulk modulus. Q.17 A flat plate of thickness ‘t’ tapers uniformly from a width b1at one end to b2 at the other end, in a length of L units. Determine the extension of the plate due to a pull P. Q.18 Find the extension of a conical rod due to its own weight when suspended vertically with its base at the top. Q.19 Prove that a material subjected to pure shear in two perpendicular planes has a diagonal tension and compression of same magnitude at 45o to the planes of shear.
Q.20
For a given material E=1.1×105N/mm2& G=0.43×105N/mm2 .Find bulk modulus & lateral contraction of round bar of 40mm diameter & 2.5m length when stretched by 2.5mm.
ANS: K=83.33Gpa, Lateral contraction=0.011mm Q.21 The modulus of rigidity of a material is 0.8×105N/mm2 , when 6mm×6mm bar of this material subjected to an axial pull of 3600N.It was found that the lateral dimension of the bar is changed to 5.9991mm×5.9991mm. Find µ & E. ANS: µ=0.31, E= 210Gpa.
7. STATICALLY INDETERMINATE MEMBERS & THERMAL STRESSES
STATICALLY INDETERMINATE MEMBERS Structure for which equilibrium equations are sufficient to obtain the solution are classified as statically determinate. But for some combination of members subjected to axial loads, the solution cannot be obtained by merely using equilibrium equations. The structural problems with number of unknowns greater than the number independent equilibrium equations are called statically indeterminate. The following equations are required to solve the problems on statically indeterminate structure. 1) Equilibrium equations based on free body diagram of the structure or part of the structure. 2) Equations based on geometric relations regarding elastic deformations, produced by the loads.
COMPOUND BAR
Material(2) Material(1)
L1
L2
W A compound bar is one which is made of two or more than two materials rigidly fixed, so that they sustain together an externally applied load. In such cases (i) Change in length in all the materials are same. (ii) Applied load is equal to sum of the loads carried by
(dL)1 = (dL)2 (σ1/ E1)L1 = (σ2 /E2)L2 σ1 = σ2 ×( E1/E2)(L1/L2) E1/E2 is called modular ratio Total load = load carried by material (1) + load carried by material(2) W = σ 1 A1 + σ 2 A2 (2) (1)
From Equation (1) & (2) σ1 and σ2 can be calculated
Temperature Stress
A L A L B B B
B´ P αTL
A
L Any material is capable of expanding or contracting freely due to rise or fall in temperature. If it is subjected to rise in temperature of T˚C, it expands freely by an amount ‘αTL’ as shown in figure. Where α is the coefficient of linear expansion, T˚C = rise in temperature and L = original length.
From the above figure it is seen that ‘B’ shifts to B' by an amount ‘αTL’. If this expansion is to be prevented a compressive force is required at B'. Temperature strain = αTL/(L + αTL) ≈ αTL/L= αT Temperature stress = αTE Hence the temperature strain is the ratio of expansion or contraction prevented to its original length. If a gap δ is provided for expansion then Temperature strain = (αTL – δ) / L Temperature stress = [(αTL – δ)/L] E
Temperature stress in compound bars:Material(1) Material(2)
α1TL
x P1
(dL)2 (dL)1 P2 α2TL
x ∆ When a compound bar is subjected to change in temperature, both the materials will experience stresses of opposite nature. Compressive force on material (1) = tensile force on material (2) load) σ1A1 = σ2A2 (there is no external σ1=( σ2A2)/A1 (1)
As the two bars are connected together, the actual position of the bars will be at XX. α1TL – (dL)1 = α2TL + (dL)2 α1TL – (σ1 / E1) L =α2TL + (σ2 / E2) L αT – (σ1 / E1) = α2T + σ2 / E2 --------------------------(2) From (1) and (2) magnitude of σ1 and σ2 can be found out.
Actual expansion in material (1) = actual expansion in material (2)
Exercise problems Q.1 A circular concrete pillar consists of six steel rods of 22mm diameter each reinforced into it. Determine the diameter of pillar required when it has to carry a load of 1000kN. Take allowable stresses for steel & concrete as 140Mpa & 8Mpa respectively. The modular ratio is 15 ANS: D=344.3mm
Q.2 Determine the stresses & deformation induced in Bronze & steel as shown in figure. Given As=1000mm2, Ab=600mm2, Es= 200Gpa, Eb= 83Gpa ANS: ( σb=55Mpa, σs=93.5Mpa, dLs=dLb=0.093mm)
Bronze
Bronze
Steel
160kN
Q..3 A cart wheel of 1.2m diameter is to be provided with steel tyre. Assume the wheel to be rigid. If the stress in steel does not exceed 140MPa, calculate minimum diameter of steel tyre & minimum temperature to which it should be heated before on to the wheel. ANS: d=1199.16mm T=58.330C Q.4 A brass rod 20mm diameter enclosed in a steel tube of 25mm internal diameter & 10mm thick. The bar & the tube are initially 2m long & rigidly fastened at both the ends. The temperature is raised from 200C to 800C. Find the stresses in both the materials.
If the composite bar is then subjected to an axial pull of 50kN, find the total stress. Es=200GPa, Eb=80GPa, αs=12×10-6/0C, αb=19×10-6/0C. ANS: σb=8.81N/mm2 ( C ) , σs=47.99N/mm2( T )
8. STRESSES ON INCLINED PLANES
INTRODUCTION
The state of stress on any plane in a strained body is said to be ‘Compound Stress’, if, both Normal and Shear stresses are acting on that plane. For, example, the state of stress on any vertical plane of a beam subjected to transverse loads will, in general, be a Compound Stress. In actual practice the state of Compound Stress is of more common occurrence than Simple state of stress. In a compound state of stress, the normal and shear stress may have a greater magnitude on some planes which are inclined (or, Oblique) to the given stress plane.
Hence in compound state of stresses it is necessary to find the following (i) The normal and shear stress on a plane which is inclined (Oblique) to the given stress plane; (ii) The inclination of max. and min. normal stress planes and values of the normal stress (max. / min.) on them; (iii) The inclination of max. shear stress planes and the values of the shear stress (max.) on them.
EQUATIONS, NOTATIONS & SIGN CONVENTIONS :
Consider an element ABCD subjected to a state of compound stress as shown in the Fig. Let: C Normal Stress in x- direction σx σx Normal Stress in y- direction σy D τ Shear Stresses in x & y – directions θ Angle made by inclined plane AE wrt vertical τ σθ Normal Stress on inclined plane τθ Shear Stress on inclined plane σy τθ E B σθ θ A
τ σx
σy
(i) Normal & Shear stress on plane inclined (Oblique) to given stress plane:
Normal Stress, σθ, and Shear stress, τθ, on inclined plane are given by:
⎛σ x +σ y σθ = ⎜ ⎜ 2 ⎝ ⎞ ⎛σ x −σ y ⎟+⎜ ⎟ ⎜ 2 ⎠ ⎝ y ⎞ ⎟ cos 2θ + τ sin 2θ − − − (1) ⎟ ⎠
τθ
⎛σ = ⎜ ⎜ ⎝
x
−σ 2
⎞ ⎟ sin 2 θ − τ cos 2 θ − − − ( 2 ) ⎟ ⎠
(ii) The inclination of max. and min. normal stress planes and the values of the normal stress (max. / min.) on them Let, θP be the inclination of the plane of max. or min. normal stress and σP be the value of the max. or min. normal stress on that plane, then, from Eqn. (1): ⎛σ x +σ y ⎞ ⎛σ x −σ y ⎞ ⎟+⎜ ⎟ cos 2θ P + τ sin 2θ P - - - (1) σP = ⎜ ⎜ ⎟ ⎜ ⎟
⎝ 2 ⎠ ⎝ 2 ⎠
Thus, the condition for max. or min. normal stress to occur on a dσ P =0 For σ P to be max. or min., plane is, shear stress on that dθ plane should be zero. ⎛ σ x −σ y ⎞ ⇒⎜ ⎟ ⎜ 2 ⎟(− 2 sin 2θ P ) + τ (2 cos 2θ P ) = 0 These planes on which shear ⎠ ⎝ stress is zero and the normal ⎛ σ x −σ y ⎞ stress on them being either the ⇒⎜ ⎜ 2 ⎟(sin 2θ P ) − τ (cos 2θ P ) = 0 ⎟ ⎝ ⎠ max. or the min. are called ⇒ τθ = 0 ‘PRINCIPAL PLANES’.
P
From,τ θ P = 0, we have, ⎛σ x −σ y ⎞ τ ⎜ ⎟(sin 2θ P ) − τ (cos 2θ P ) = 0 ⇒ tan 2θ P = - - - (3) ⎜ 2 ⎟ ⎛ σ x −σ y ⎞ ⎝ ⎠ ⎜ ⎜ 2 ⎟ ⎟ ⎝ ⎠
The above Eqn. (3), gives two values for θP, which differ by 900. Thus, there are two mutually perpendicular Principal planes, on which there are only normal stresses, shear stress being zero on them. On one of them, the value of the normal stress is the max.; it is called the ‘Major Principal plane’, the max. normal stress on it is called the ‘Major Principal Stress’. On the other principal plane, the value of the normal stress is the min.; it is called the ‘Minor Principal plane’, the min. normal stress on it is called the ‘Minor Principal Stress’.
From tan 2θ P =
τ
⎛σ x −σ y ⎜ ⎜ 2 ⎝ ⎞ ⎟ ⎟ ⎠
2
, we get,
] 2+
τ2
2θP
y /2
±√ [(σ
x
(σx-σy)/ 2
-σ
sin 2θ P = ±
τ
⎡ (σ X − σ Y ) ⎤ ⎢ ⎥ +τ 2 ⎣ ⎦ (σ X − σ Y ) / 2 ⎡ (σ X − σ Y ) ⎤ ⎢ ⎥ +τ 2 ⎣ ⎦
2
2
τ
cos 2θ P = ±
2
Substituting for sin 2θP and cos 2θP in Eqn. (1), and simplifying, we get the equation for principal stresses as: σP =
(σ
x
+σ 2
y
)±
⎛σ x −σ ⎜ ⎜ 2 ⎝
y
⎞ ⎟ +τ ⎟ ⎠
2
2
− − − (4)
The above equation (4) gives two values for principal stresses. The numerically max. of the two values (+ ve or − ve) is the Major Principal Stress, (σMajor or σMax); The numerically min. (+ ve or − ve) is the Minor Principal Stress (σMinor or σMin).
(iii) Inclination of max. shear stress planes, Max. shear stress Equation. Let, θS be the inclination of the plane of max. or min. shear stress and τS be the value of the max. or min. shear stress on that plane, then, from Eqn. (2):
τS
⎛σ = ⎜ ⎜ ⎝
x
−σ 2
y
⎞ ⎟ sin 2 θ S − τ cos 2 θ S ⎟ ⎠
− − − (2)
dτ For τ S to be max. or min., S = 0 dθS ⎛ σ x −σ y ⎞ ⇒⎜ ⎜ 2 ⎟(2 cos 2θ S ) − τ (− 2 sin 2θ S ) = 0 ⎟ ⎝ ⎠ ⎛ σ x −σ y ⎞ ⎜ ⎜ 2 ⎟ ⎟ ⎝ ⎠ - - - (5) ⇒ tan 2θ S = − NOTE : We have tan2θ P × tan 2θ S = −1
τ
Eqn. (5) gives two values for θS, which differ by 900. Thus, there are two mutually perpendicular planes, on which shear stress are max.; numerically equal but opposite in sense. The planes of Max. Shear stresses are inclined at 450 to the Principal planes.
2] 2 +
⎛σ x −σ y ⎜ ⎜ 2 From tan 2θ S = − ⎝
y )/
−σ
±√ [(σ
x
(σx-σy)/ 2
cos 2θ s = ±
τ
τ
⎞ ⎟ ⎟ ⎠ , we get,
τ2
2θS
τ
sin 2θ S = ±
⎡ (σ x − σ y ) ⎤ ⎥ +τ ⎢ 2 ⎦ ⎣ (σ x − σ y )/ 2 ⎡ (σ x − σ y ) ⎤ ⎢ ⎥ +τ 2 ⎣ ⎦
2
2
2
2
Substituting for sin 2θS and cos 2θS in Eqn. (2), we get the equation for Max. shear stresses as: τ max . = ± ⎜ ⎜
⎝ ⇒ τ max . ⎛σ x −σ y ⎞ 2
2 2
⎟ +τ ⎟ ⎠
⎛ σ Major − σ Minor = ±⎜ ⎜ 2 ⎝
⎞ ⎟ ⎟ ⎠
⎤ ⎥ ⎥ ⎥ − − − (6 ) ⎥ ⎥ ⎦
The above equation (6) gives two values for Max. shear stresses, which are numerically equal but opposite in sense.
NOTATIONS & SIGN CONVENTIONS σx σy Normal Stress in x- direction Normal Stress in y- direction Shear Stresses in x & y – directions Angle made by inclined plane wrt vertical Normal Stress on inclined plane AE σx τ C σθ D
σy τ τθ E B θ A
τ θ σθ
σx
τθ θP σP θS
Shear Stress on inclined plane AE σy Inclination of Principal planes Principal stresses Inclination of Max. shear stress planes [θS = θP + 450]. All the parameters are shown in their +ve sense in the Fig.
Normal stresses, σ Tensile stresses +ve. Shear Stresses, τ, in x – direction & Inclined Plane Clockwise +ve. Anti-Clockwise +ve. Shear Stresses, τ, in y – direction Angle, θ measured w r t vertical, Anti-Clockwise +ve.
Exercise Problems:
Q.1 The principal stresses at a point in a strained material are 80 MPa(C) and 40 MPa(T). Find the normal, tangential and resultant stress on a plane inclined at 50o to the major principal plane. [Ans: - 9.58MPa, -59MPa].
Q.2 The stresses in a strained material is as shown in Fig. Find the normal and shear stresses on plane inclined at 30o to the horizontal. Also determine the intensity and position of the plane upon which there is only shear stress. Sketch the plane.
60MPa 90 MPa
30o
Ans: σ = − 22.5MPa, τ = − 64.95MPa. Pure shear plane σ = 0, Ө = 39.23o w.r.t. horizontal. τpure = − 73.48MPa.
Q.3 A plane element is subjected to the system of stresses as shown in Fig. Determine (i) the principal stresses and inclination of their planes (ii) maximum shearing stresses and inclination of their planes. Represent your answers in neat sketches.
160 MPa 80 MPa 200 MPa
Ans: σ1 = 262.46MPa, σ2 = 97.54MPa. Өp = 37.98 or 127.98 with horizontal τtmax = 82.46, Өs = 82.98 or 172.98 with horizontal
Q.4 At a point in a structural member subjected to stresses as shown in fig. determine the principal stresses and the maximum shear stress. Also determine and sketch planes on which these stresses act. 80 [Ans:131.23, 48.77, − 37.98o, − 127.98o 41.23, 7o, 970 , Angles w.r.t. horizontal] 100
4 0 Q.5 At a point in a material under stress, the intensity of resultant stress on a certain plane is 60 MPa, directed outwards and inclined at 30o to the normal of that plane. The stress on the plane at right angles to this has a normal stress component of 40 Mpa (T). Find (i) the principal stresses and inclination of their planes, (ii) the maximum shear stresses and inclination of their planes . [Ans: 76.57MPa, 15.39MPa, 39.36o , 129.36o and 30.59 MPa, 84.36o , 174.36o, Angles w.r.t. horizontal]
9. Stresses due to fluid pressure in thin cylinders
9 -THIN CYLINDERS
INTRODUCTION: In many engineering applications, cylinders are frequently used for transporting or storing of liquids, gases or fluids. Eg: Pipes, Boilers, storage tanks etc. These cylinders are subjected to fluid pressures. When a cylinder is subjected to internal pressure, at any point on the cylinder wall, three types of stresses are induced on three are, They mutually perpendicular planes. 1. Hoop or Circumferential Stress (σC) – This is directed along the tangent to the circumference and tensile in nature. Thus, there will be increase in diameter.
2. Longitudinal Stress (σL) – This stress is directed along the length of the cylinder. This is also tensile in nature and tends to increase the length.
3. Radial pressure (σ r) – It is compressive in nature. Its magnitude is equal to fluid pressure on the inside wall and zero on the outer wall if it is open to atmosphere.
σC p σC
σC
σL p
σL p σL σr
σC
σL
1. Hoop /Circumferential Stress (σC)
2. Longitudinal Stress (σL) σr 3. Radial Stress (σr) σC Element on the cylinder wall subjected to these three stresses
σL
σL
σC
σr
A cylinder or spherical shell is considered to be thin when the metal thickness is small compared to internal diameter. i. e., when the wall thickness, ‘t’ is equal to or less than ‘d/20’, where ‘d’ is the internal diameter of the cylinder or shell, we consider the cylinder or shell to be thin, otherwise thick. Magnitude of radial pressure is very small compared to other two stresses in case of thin cylinders and hence neglected.
p×d Circumferential stress, σ c = ........................(1) 2× t
p×d Longitudin stress, σL = al ...................( 2) 4× t
Maximum Shear stress, τmax = (σc- σL) / 2 τmax = ( p x d) / (8 x t) Where p = internal fluid pressure d= internal diameter,
t = thickness of the wall
EVALUATION OF STRAINS σ L=(pd)/(4t)
σ C=(pd)/(2t)
σ C=(pd)/(2t)
σ L=(pd)/(4t) A point on the surface of thin cylinder is subjected to biaxial stress system, (Hoop stress and Longitudinal stress) mutually perpendicular to each other, as shown in the figure. The strains due to these stresses i.e., circumferential and longitudinal are obtained by applying Hooke’s law and Poisson’s theory for elastic materials.
Circumferential strain, ε C : σC σL σL σL −µ× = 2× −µ× εC = E E E E σL = × (2 − µ) E
C=(pd)/(2t)
L=(pd)/(4t)
C=(pd)/(2t)
L=(pd)/(4t)
i.e.,
δd p×d εC = = × (2 − µ)................................(3) d 4× t × E
Longitudinal strain, ε L : σC σL (2 × σ L ) σ L σL −µ× = −µ× = × (1 − 2 × µ) εL = E E E E E
i.e., δl p×d εL = = × (1 − 2 × µ)................................(4) L 4× t × E
dv VOLUMETRIC STRAIN, V
σL=(pd)/(4t)
π 2 σC=(pd)/(2t σC=(pd)/(2t) We have volume, V = × d × L. 4 π 2 π dV = × d × dL + L × × 2 × d × dd σL=(pd)/(4t) 4 4 π 2 π × d × dL + L × × 2 × d × dd dv 4 dL dd 4 = = + 2× π 2 V L d ×d ×L 4 p×d p×d dV = (1 − 2 × µ) + 2 × (2 − µ) = εL+2× εC V 4× t × E 4× t × E dv p×d i.e., = (5 − 4 × µ).................(5) V 4× t × E
JOINT EFFICIENCY The cylindrical shells like boilers are having two types of joints namely Longitudinal and Circumferential joints. Due to the holes for rivets, the net area of cross section decreases and hence the stresses increase. If the efficiencies of these joints are known, the stresses can be calculated as follows. Let ηL=Efficiency of Longitudinal joint and ηC=Efficiency of Circumferential joint. Circumferential stress is given by,
p×d σC = 2 × t × ηL
.............(1)
Longitudinal stress is given by,
p×d σL = .......... ...(2) 4×t×ηC
Note: In longitudinal joint, the circumferential stress is developed and in circumferential joint, longitudinal stress is developed.
Exercise Problems
Q.1 Calculate the circumferential and longitudinal strains for a boiler of 1000mm diameter when it is subjected to an internal pressure of 1MPa. The wall thickness is such that the safe maximum tensile stress in the boiler material is 35 MPa. Take E=200GPa and µ= 0.25. (Ans: ε C=0.0001531, ε L=0.00004375) Q.2 A water main 1m in diameter contains water at a pressure head of 120m. Find the thickness of the metal if the working stress in the pipe metal is 30 MPa. Take unit weight of water = 10 kN/m3. (Ans: t=20mm)
Q.3 A gravity main 2m in diameter and 15mm in thickness. It is subjected to an internal fluid pressure of 1.5 MPa. Calculate the hoop and longitudinal stresses induced in the pipe material. If a factor of safety 4 was used in the design, what is the ultimate tensile stress in the pipe material? (Ans: σC=100 MPa, σL=50 MPa, σU=400 MPa) Q.4 At a point in a thin cylinder subjected to internal fluid pressure, the value of hoop strain is 600×10-4 (tensile). Compute hoop and longitudinal stresses. How much is the percentage change in the volume of the cylinder? Take E=200GPa and µ= 0.2857. (Ans: σC=140 MPa, σL=70 MPa, %age change=0.135%.)
Q.5 A cylindrical tank of 750mm internal diameter and 1.5m long is to be filled with an oil of specific weight 7.85 kN/m3 under a pressure head of 365 m. If the longitudinal joint efficiency is 75% and circumferential joint efficiency is 40%, find the thickness of the tank required. Also calculate the error of calculation in the quantity of oil in the tank if the volumetric strain of the tank is neglected. Take permissible tensile stress as 120 MPa, E=200GPa and µ= 0.3 for the tank material. (Ans: t=12 mm, error=0.085%.)
Mechanics of Solids
PART- I Mechanics of Rigid Bodies
PART- II Mechanics of Deformable Bodies
COURSE CONTENT IN BRIEF
PART I Mechanics of Rigid Bodies 1. Resultant of concurrent and non-concurrent coplanar forces. 2. Equilibrium of concurrent and non-concurrent coplanar forces. 3. Centroid of plane areas 4. Moment of Inertia of plane areas 5. Kinetics: Newton’s second law, D’Alembert’s principle, Work- Energy, and Impulse- Momentum principle. Mechanics of Deformable bodies
PART II
6. Simple stresses and strains 7. Statically indeterminate problems and thermal stresses 8. Stresses on inclined planes 9. Stresses due to fluid pressure in thin cylinders
Books for Reference 1.Engineering Mechanics, by Meriam & Craige, John Wiley & Sons. 2.Engineering Mechanics, by Irwing Shames, Prentice Hall of India. 3.Mechanics for Engineers, by Beer and Johnston, McGraw Hills Edition 4.Engineering Mechanics, by K.L. Kumar, Tata McGraw Hills Co. 5. Machanics of Materials, by E.P.Popov 6. Machanics of Materials, by E J Hearn 7. Strength of materials, by Beer and Johnston 8. Strength of materials, by F L Singer & Andrew Pytel 9. Strength of Materials, by B.S. Basavarajaiah & P. Mahadevappa 10. Strength of Materials, by Ramamruthum 11. Strength of Materials, by S S Bhavikatti
PART - I MECHANICS OF RIGID BODIES
PART - I
Mechanics of Rigid Bodies
INTRODUCTION Definition of Mechanics : In its broadest sense the term ‘Mechanics’ may be defined as the ‘Science which describes and predicts the conditions of rest or motion of bodies under the action of forces’. This Course on Engineering Mechanics comprises of Mechanics of Rigid bodies and the sub-divisions that come under it.
Branches of Mechanics Engineering Mechanics Mechanics of Solids Deformable Bodies
Mechanics of Fluids
Rigid Bodies
Ideal Fluids
Viscous Fluids
Compres Fluids
Statics Dynamics
Theory of Theory of Plasticity Strength of Elasticity Materials
Kinematics
Kinetics
Concept of Rigid Body : It is defined as a definite amount of matter the parts of which are fixed in position relative to one another under the application of load. Actually solid bodies are never rigid; they deform under the action of applied forces. In those cases where this deformation is negligible compared to the size of the body, the body may be considered to be rigid.
Particle A body whose dimensions are negligible when compared to the distances involved in the discussion of its motion is called a ‘Particle’. For example, while studying the motion of sun and earth, they are considered as particles since their dimensions are small when compared with the distance between them.
Force It is that agent which causes or tends to cause, changes or tends to change the state of rest or of motion of a mass. A force is fully defined only when the following four characteristics are known: (i) Magnitude (ii) Direction (iii) Point of application (iv) Sense.
Force: characteristics of the force 100 kN are : (i) Magnitude = 100 kN (ii) Direction = at an inclination of 300 to the x-axis (iii) Point of application = at point A shown (iv) Sense = towards point A A 100 kN 300
Scalars and Vectors A quantity is said to be a ‘scalar’ if it is completely defined by its magnitude alone. Example : Length, Area, and Time. A quantity is said to be a ‘vector’ if it is completely defined only when its magnitude and direction are specified. Example : Force, Velocity, and Acceleration.
Principle of Transmissibility : It is stated as follows : ‘The external effect of a force on a rigid body is the same for all points of application along its line of action’. P A B P
For example, consider the above figure. The motion of the block will be the same if a force of magnitude P is applied as a push at A or as a pull at B. P O The same is true when the force is applied at a point O. P
1. RESULTANT OF COPLANAR FORCES Resultant, R : It is defined as that single force which can replace a set of forces, in a force system, and cause the same external effect.
F2 F1
R
= θ A
F3
A
R = F1 + F2 + F3 external effect on particle, A is same
Resultant of two forces acting at a point Parallelogram law of forces : ‘If two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, then the resultant of these two forces is represented in magnitude and direction by the diagonal of the parallelogram passing through the same point.’ B P2 O α R θ P1 A C
Contd..
B P2 O α R θ P1 A
C
In the above figure, P1 and P2, represented by the sides OA and OB have R as their resultant represented by the diagonal OC of the parallelogram OACB. It can be shown that the magnitude of the resultant is given by: R = √P12 + P22 + 2P1P2Cos α Inclination of the resultant w.r.t. the force P1 is given by: θ = tan-1 [( P2 Sin α) / ( P1 + P2 Cos α )]
Resultant of two forces acting at a point at right angle B P2 O α R θ P1 C R θ P1 A
R = P1 + P 2 2
2
C
A
If α = 900 , (two forces acting at a point are at right angle) B P2 O
tan θ =
P2 P1
Triangle law of forces
‘If two forces acting at a point can be represented both in magnitude and direction, by the two sides of a triangle taken in tip to tail order, the third side of the triangle represents both in magnitude and direction the resultant force F, the sense of the same is defined by its tail at the tail of the first force and its tip at the tip of the second force’.
Triangle law of forces
Let F1 and F2 be the two forces acting at a point A and θ is the included angle.
F1 R θ A = F2 F2 θ F1
‘Arrange the two forces as two sides of a triangle taken in tip to tail order, the third side of the triangle represents both in magnitude and direction the resultant force R. the sense of the resultant force is defined by its tail at the tail of the first force and its tip at the tip of the second force’.
Triangle law of forces F1 θ A = F2 R θ F1
F2
R α F1
β F2
F1 F2 R = = sin β sin α sin(180 − α − β )
(180 - α - β) = θ
where α and β are the angles made by the resultant force with the force F1 and F2 respectively.
Component of a force :
Component of a force, in simple terms, is the effect of a force in a certain direction. A force can be split into infinite number of components along infinite directions.
Usually, a force is split into two mutually perpendicular components, one along the x-direction and the other along ydirection (generally horizontal and vertical, respectively).
Such components that are mutually perpendicular are called ‘Rectangular Components’.
The process of obtaining the components of a force is called ‘Resolution of a force’.
Rectangular component of a force F θx
F θx Fx
Fy
=
F θx Fx
Fy
Consider a force F making an angle θx with x-axis. Then the resolved part of the force F along x-axis is given by Fx = F cos θx The resolved part of the force F along y-axis is given by Fy = F sin θx
Oblique component of a force Let F1 and F2 be the oblique components of a force F. The components F1 and F2 can be found using the ‘triangle law of forces’.
N
F2 β α
F F1
M
F α F1
β F2
O The resolved part of the force F along OM and ON can obtained by using the equation of a triangle. F1 / Sin β = F2 / Sin α = F / Sin(180 - α - β)
Sign Convention for force components: y +ve x
y
+ve
x
The adjacent diagram gives the sign convention for force components, i.e., force components that are directed along positive x-direction are taken +ve for summation along the x-direction. Also force components that are directed along +ve y-direction are taken +ve for summation along the y-direction.
Classification of force system Force system
Coplanar Forces Concurrent Non-concurrent
Unlike parallel Like parallel Unlike parallel
Non-Coplanar Forces Concurrent Non-concurrent
Like parallel
A force that can replace a set of forces, in a force system, and cause the same ‘external effect’ is called the Resultant.
RESULTANT OF COPLANAR NON CONCURRENT FORCE SYSTEM
Coplanar Non-concurrent Force System: This is the force system in which lines of action of individual forces lie in the same plane but act at different points of applications.
F1 F3 Fig. 1
F2 F5
F1
F2 F3 F4
Fig. 2
1. Parallel Force System – Lines of action of individual forces are parallel to each other. 2. Non-Parallel Force System – Lines of action of the forces are not parallel to each other.
MOMENT OF A FORCE ABOUT AN AXIS The applied force can also tend to rotate the body about an axis in addition to motion. This rotational tendency is known as moment. Definition: Moment is the tendency of a force to make a rigid body to rotate about an axis. This is a vector quantity having both magnitude and direction.
MOMENT OF A FORCE ABOUT AN AXIS Moment Axis: This is the axis about which rotational tendency is determined. It is perpendicular to the plane comprising moment arm and line of action of the force (axis 0-0 in the figure) Moment Center: This is the position of axis on coplanar system. (A). Moment Arm: Perpendicular distance from the line of action of the force to moment center. Distance AB = d.
Magnitude of moment: It is computed as the product of the of the force and the perpendicular distance from the line of action to the point about which moment is computed. (Moment center). MA = F×d = Rotation effect because of the force F, about the point A (about an axis 0-0) Unit – kN-m, N-mm etc.
Sense of moment: The sense is obtained by ‘Right Hand Thumb’ rule. ‘If the fingers of the right hand are curled in the direction of rotational tendency of the body, the extended thumb represents the sense of moment vector’. For the purpose of additions, the moment direction may be considered by using a suitable sign convention such as +ve for counterclockwise and –ve for clockwise rotations or viceversa.
M
M
VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)
Statement: The moment of a force about a moment center or axis is equal to the algebraic sum of the moments of its component forces about the same moment center (axis). P P sinθ P θ P cosθ θ d d1 A d2 A
Moment of Force P about the Algebraic sum of Moments of = components of the Force P point A, about the point A, Pxd P cosθ x d1 + P sinθ x d2
VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)
Proof (by Scalar Formulation): Let ‘R’ be the given force. ‘P’ & ‘Q’ are component forces of ‘R’. ‘O’ is the moment center. p, r and q are moment arms from ‘O’ of P, R and Q respectively. α, β and γ are the inclinations of ‘P’, ‘R’ and ‘Q’ respectively w.r.to X – axis. A R
Q
Y
q α
r P βγ p O X
Y
We have, Ry = Py + Qy R Sinβ = P Sinα + Q Sin γ ----(1) From ∆le AOB, p/AO = Sin α From ∆le AOC, r/AO = Sin β From ∆le AOD, q/AO = Sin γ From (1), ∴ R ×(r/AO) = P ×(p/AO) + Q ×(q/AO) i.e., R × r = P × p + Q × q Moment of resultant R about O = algebraic sum of moments of component forces P & Q about same moment center ‘O’. Ry
Q
R
C
P q r
Qy Py A
D
γ α β
B p O
X
COUPLE Two parallel, non collinear (separated by certain distance) forces that are equal in magnitude and opposite in direction form ‘couple’.
F
The algebraic summation of the two forces forming couple is zero. Hence, couple does not produce any translation and produces only rotation.
=
d F
M=Fxd
RESOLUTION OF A FORCE INTO A FORCE-COUPLE SYSTEM
Replace the force F acting at the point A to the point B
F
B
A
Apply two equal and opposite forces of same magnitude & direction as Force F at point B, so that external effect is unchanged
F F
B
A
F
d
F
B
F
F
B
A
=
A
F
d
M=Fxd
Of these three forces, two forces i.e., one at A and the other oppositely directed at B form a couple. Moment of this couple, M = F × d. Third force at B is acting in the same direction as that at P. Thus, the force F acting at a point such as A in a rigid body can be moved to any other given point B, by adding a couple M. The moment of the couple is equal to moment of the force in its original position about B.
TYPES OF LOADS ON BEAMS
1. Concentrated Loads – This is the load acting for very small length of the beam.
W kN
(also known as point load, Total load W is acting at one point ) 2. Uniformly distributed load – This is the load acting for a considerable length of the beam with same intensity of w kN/m throughout its spread. Total intensity, W = w × L (acts at L/2 from one end of the spread)
L
w kN/m
L
W = (w x L) kN L/2
3. Uniformly varying load – This load acts for a considerable length of the beam with intensity varying linearly from ‘0’ at one end to w kN/m to the other representing a triangular distribution. Total intensity of load = area of triangular spread of the load W = 1/2× w × L. (acts at 2×L/3 from ‘Zero’ load end) 2/3 ×L
w kN/m L
W=½×L×w
1/3 ×L
L
EXERCISE PROBLEMS
1. Resultant of force system
Q1 A body of negligible weight, subjected to two forces F1= 1200N, and F2=400N acting along the vertical, and the horizontal respectively, is shown in figure. Find the component of each force parallel, and perpendicular to the plane. F1 = 1200 N
X
Y F2 = 400 N 4 3
Ans : F1X = -720 N, F1Y = -960N, F2X = 320N, F2Y = -240N
EXERCISE PROBLEMS
1. Resultant of force system
Q2. Determine the X and Y components of each of the forces shown in the figure. F2 = 390 N 12 5 40º F3 =400 N 30º F1 = 300 N Y X
(Ans : F1X = 259.81 N, F1Y= -150 N, F2X= -150N, F2Y= 360 N, F3X = -306.42 N, F3Y= -257.12N )
EXERCISE PROBLEMS
1. Resultant of force system
Q3. Obtain the resultant of the concurrent coplanar forces shown in the figure 600N 800N 20º 40º
30º 200N (Ans: R = 522.67 N, θ = 68.43º)
EXERCISE PROBLEMS
1. Resultant of force system
Q4. A disabled ship is pulled by means of two tug boats as shown in FIG. 4. If the resultant of the two forces T1 and T2 exerted by the ropes is a 300 N force acting parallel to the X – direction, find : (a) Force exerted by each of the tug boats knowing α = 30º. (b) The value of α such that the force of tugboat 2 is minimum, while that of 1 acts in the same direction. Find the corresponding force to be exerted by tugboat 2.
T2 α 20º R = 300 N X - direction
FIG. 4 T1 ( Ans: a. T1= 195.81 N, T2 = 133.94 N b. α = 70º, T1 = 281.91 N, T2(min) = 102.61 N )
EXERCISE PROBLEMS
1. Resultant of force system
Q5. An automobile which is disabled is pulled by two ropes as shown in the figure. Find the force P and resultant R, such that R is directed as shown in the figure.
P 20º 40º R
Q = 5 kN (Ans: P = 9.4 kN , R = 12.66 kN)
EXERCISE PROBLEMS
1. Resultant of force system
Q6. A collar, which may slide on a vertical rod, is subjected to three forces as shown in figure. The direction of the force F may be varied Determine the direction of the force F, so that resultant of the three forces is horizontal, knowing that the magnitude of F is equal to (a) 2400 N, (b)1400N 1200 N 60º θ COLLAR ROD F 800 N
( Ans: a. θ = 41.81º ; b. The resultant cannot be horizontal.)
EXERCISE PROBLEMS
1. Resultant of force system
Q7. Determine the angle α and the magnitude of the force Q such that the resultant of the three forces on the pole is vertically downwards and of magnitude 12 kN. Refer figure 8kN
5kN α 30º Q
Fig. 7
(Ans: α = 10.7 º, Q = 9.479 kN )
EXERCISE PROBLEMS
1. Resultant of force system
Q8. Determine the resultant of the parallel coplanar force system shown in figure.
600 N 2000 N
60º 10º
o
30º 60º
1000 N 400 N
(Ans. R=800N towards left, d=627.5mm)
EXERCISE PROBLEMS
1. Resultant of force system
Q9. Four forces of magnitudes 10N, 20N, 30N and 40N acting respectively along the four sides of a square ABCD as shown in the figure. Determine the magnitude, direction and position of resultant w.r.t. A.
20N D 30N C a A a 40N B 10N
(Ans:R=28.28N, θ=45º, x=1.77a)
EXERCISE PROBLEMS
1. Resultant of force system
Q10. Four parallel forces of magnitudes 100N, 150N, 25N and 200N acting at left end, 0.9m, 2.1m and 2.85m respectively from the left end of a horizontal bar of 2.85m. Determine the magnitude of resultant and also the distance of the resultant from the left end.
(Ans: R = 125 N, x = 3.06 m)
EXERCISE PROBLEMS
1. Resultant of force system
Q11. Reduce the given forces into a single force and a couple at A.
70.7 kN 200 kN 45º 1.5m A 1m 80 N 30º 100 N 30º
(Ans:F=320kN, θ=14.48º, M=284.8kNm)
EXERCISE PROBLEMS
1. Resultant of force system
Q12. Determine the resultant w.r.t. point A.
150 Nm 150 N 1.5m 3m 1.5m
A
100 N 500 N
(Ans: R = 450 kN, X = 7.5 kNm)
2. EQUILIBRIUM OF FORCE SYSTEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
EQUILIBRIUM OF - CONCURRENT COPLANAR FORCE SYSTEMS
Definition:If a system of forces acting on a body, keeps the body in a state of rest or in a state of uniform motion along a straight line, then the system of forces is said to be in equilibrium. ALTERNATIVELY, if the resultant of the force system is zero, then, the force system is said to be in equilibrium.
EQUILIBRIUM OF - CONCURRENT COPLANAR FORCE SYSTEMS
Conditions for Equilibrium :
A coplanar concurrent force system will be in equilibrium if it satisfies the following two conditions: ∑ Fx = 0; and ii) ∑ Fy = 0
i)
i.e. Algebraic sum of components of all the forces of the system, along two mutually perpendicular directions, is ZERO.
Y X
Graphical conditions for Equilibrium Triangle Law: If three forces are in equilibrium, then, they form a closed triangle when represented in a Tip to Tail arrangement, as shown in Fig 2.1
F1 F3
F
3
F2 Fig 2.1
F2
F1
Polygonal Law: If more than three forces are in equilibrium, then, they form a closed polygon when represented in a Tip to Tail arrangement, as shown in Fig. 2.2. F F4
3
F3 F4
F5
F2 F1 F5 Fig 2.2
F
2
F1
LAMI’S THEOREM
If a system of three forces is in equilibrium, then, each force of the system is proportional to sine of the angle between the other two forces (and constant of proportionality is the same for all the forces). Thus, with reference to Fig.2.3, we have,
F3
α β F2
F3 F1 F2 = = Sin α Sin β Sin γ
γ
F1
Fig. 2.3
Note: While using Lami’s theorem, all the three forces should be either directed away or all directed towards the point of concurrence.
EQUILIBRIUM OF NON-CONCURRENT COPLANAR FORCE SYSTEM
When a body is in equilibrium, it has neither translatory nor rotatory motion in any direction. Thus the resultant force R and the resultant couple M are both zero, and we have the equilibrium equations for two dimensional force system ∑ Fx = 0; ∑ Fy = 0 ∑M = 0
These requirements are both necessary and sufficient conditions for equilibrium.
SPACE DIAGRAMS & FREE BODY DIAGRAMS
Space Diagram (SPD) : The sketch showing the physical conditions of the problem, like, the nature of supports provided; size, shape and location of various bodies; forces applied on the bodies, etc., is known as space diagram. eg, Fig 2.4 is a space diagram
Weight of sphere = 0.5 kN, Radius = 1m
3m θ
Cable P = 2kN
30°
Sphere
wall
Fig. 2.4 SPD
Free Body Diagram (FBD) : It is an isolated diagram of the body being analyzed (called free body), in which, the body is shown freed from all its supports and contacting bodies/surfaces. Instead of the supports and contacting bodies/surfaces, the reactive forces exerted by them on the free body is shown, along with all other applied forces.
A Few Guidelines for Drawing FBD 1) Tensile Force: It is a force trying to pull or extend the body. It is represented by a vector directed away from the body. 2) Compressive Force: It is force trying to push or contract the body. It is represented by a vector directed towards the body. 3) Reactions at smooth surfaces: The reactions of smooth surfaces, like walls, floors, Inclined planes, etc. will be normal to the surface and pointing towards the body. 4)Forces in Link rods/connecting rods: These forces will be acting along the axis of the rod, either towards or away from the body. (They are either compressive or tensile in nature). 5) Forces in Cables (Strings or Chords): These can only be tensile forces. Thus, these forces will be along the cable and directed away from the body.
Free Body Diagrams of the sphere shown in Fig. 2.4 T = Tension in the cable θ T
P = 2kN
30°
Rw = Reaction of the wall W = self weight of the sphere P = external load acting on the sphere
Rw
Sphere
W=0.5kN Fig. 2.5 F B D of Sphere
Detach the sphere from all contacts and replace that with forces like: Cable contact is replaced by the force tension = T Contact with the smooth wall is replaced by the reaction Rw.
Supports: A structure is subjected to external forces and transfers these forces through the supports on to the foundation. Therefore the support reactions and the external forces together keep the structure in equilibrium. There are different types of supports. a) Roller Support b) Hinged or pinned support c) Fixed or built in support
Supports Types of Supports (a) Flexible cable ,belt ,chain, rope Action on body
BODY
T Force exerted by cable is always a tension away from the body in the direction of cable
BODY
(b) Smooth surfaces 900 F 900 F
Contact forces are normal to the surfaces
Supports
A (c) Roller support Contact force is normal to the surface on which the roller moves. The reaction will always be perpendicular to the plane of the roller . Roller support will offer only one independent reaction component. (Whose direction is known.)
A
Supports
(d) pinned Support / hinged support A
Rh θ
A Rv
R
This support does not allow any translatory movement of the rigid body. There will be two independent reaction components at the support. The resultant reaction can be resolved into two mutually perpendicular components. Or it can be shown as resultant reaction inclined at an angle with respect to a reference direction.
Supports
(e) Fixed or Built-in Support A
RAH
M
A
RAV
This type of support not only prevents the translatory movement of the rigid body, but also the rotation of the rigid body. Hence there will be 3 independent reaction components of forces. Hence there will be 3 unknown components of forces, two mutually perpendicular reactive force component and a reactive moment as shown in the figure.
TYPES OF BEAMS
A member which is subjected to predominantly transverse loads and supported in such a way that rigid body motion is prevented is known as beam. It is classified based on the support conditions. A beam generally supported by a hinge or roller at the ends having one span (distance between the support) is called as simply supported beam. A beam which is fixed at one end and free at another end is called as a cantilever beam. A span B
A span B
(a) Simply supported beam
TYPES OF BEAMS
A
B
span
RH M A Rv B
(b) Cantilever beam
TYPES OF BEAMS
If one end or both ends of the beam project beyond the support it is known as overhanging beam.
A
B
A
B
(c) Overhanging beam (right overhang)
Statically determinate beam Using the equations of equilibrium given below, if all the reaction components can be found out, then the beam is a statically determinate beam
the equations of equilibrium ∑ Fx = 0; ∑ Fy = 0 ∑M = 0
FRICTION Friction is defined as the contact resistance exerted by one body upon another body when one body moves or tends to move past another body. This force which opposes the movement or tendency of movement is known as frictional resistance or friction. Friction is due to the resistance offered by minute projections at the contact surfaces. Hence friction is the retarding force, always opposite to the direction of motion. Friction has both advantages & disadvantages. Disadvantages ---Advantages ---Power loss, wear and tear etc.
Brakes, traction for vehicles etc.
FRICTION
W P
F (Friction)
N
Hills & Vales
Magnified Surface
Frictional resistance is dependent on the amount of wedging action between the hills and vales of contact surfaces. The wedging action is dependent on the normal reaction N.
FRICTION
Frictional resistance has the remarkable property of adjusting itself in magnitude of force producing or tending to produce the motion so that the motion is prevented. When P = 0, F = 0 block under equilibrium
When P increases, F also increases proportionately to maintain equilibrium. However there is a limit beyond which the magnitude of this friction cannot increase.
FRICTION
When the block is on the verge of motion(motion of the block is impending) F attains maximum possible value, which is termed as Limiting Friction. When the applied force is less than the limiting friction, the body remains at rest and such frictional resistance is called the static friction. Further if P is increased, the value of F decreases rapidly and then remains fairly a constant thereafter. However at high speeds it tends to decrease. This frictional resistance experienced by the body while in motion is known as Dynamic friction OR Kinetic Friction.
FRICTION
Sliding friction friction experienced when a body slides over another surface. Dynamic Friction
Rolling friction friction experienced by a body when it rolls over a surface.
FRICTION
W P Fmax φ N µ= R Fmax N
FαN Fmax = µN
Where Fmax = Limiting Friction N= Normal Reaction between the contact surfaces µ =Coefficient of friction
Note : Static friction varies from zero to a maximum value. Dynamic friction is fairly a constant.
Angle of Friction The angle between N & R depends on the value of F. This angle θ, between the resultant R and the normal reaction N is termed as angle of friction. As F increases, θ also increases and will reach to a maximum value of φ when F is Fmax (limiting friction) i.e. tanφ = (Fmax )/N = µ Angle φ is known as Angle of limiting Friction.
FRICTION
W
P
φ N
Fmax
R
FRICTION
Angle of limiting friction is defined as the angle between the resultant reaction (of limiting friction and normal reaction) and the normal to the plane on which the motion of the body is impending.
Angle of repose When granular material is heaped, there exists a limit for the inclination of the surface. Beyond that angle, the grains start rolling down. This limiting angle upto which the grains repose (sleep) is called the angle of repose of the granular material.
FRICTION
Significance of Angle of repose: The angle that an inclined plane makes with the horizontal, when the body supported on the plane is on the verge of motion due to its self -weight is equal to the angle of repose. Angle of repose is numerically equal to Angle of limiting friction
Laws of dry friction
FRICTION
1. The magnitude of limiting friction bears a constant ratio to the normal reaction between the two surfaces. (Experimentally proved) 2. The force of friction is independent of the area of contact between the two surfaces. 3. For low velocities the total amount of friction that can be developed is practically independent of velocity. It is less than the frictional force corresponding to impending motion.
EXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
Q1. A 10kN roller rests on a smooth horizontal floor and is held by the bar AC as shown in Fig(1). Determine the magnitude and nature of the force in the bar AC and reaction from the floor under the action of the forces applied on the roller. [Ans:FAC=0.058 kN(T),R=14.98 kN] 7kN C A 300 Fig(1) 450 5kN
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q2. A 10 kN weight is suspended from a rope as shown in figure. Determine the magnitude and direction of the least force P required to pull the rope, so that, the weight is shifted horizontally by 0.5m. Also, determine, tension in the rope in its new position. [Ans: P= 2.43 kN, θ = 14.480 ; T= 9.7kN.]
2m θ 10kN P
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q3. Determine the value of P and the nature of the forces in the bars for equilibrium of the system shown in figure. [Ans: P = 3.04 kN, Forces in bars are Compressive.] 60 45 P
45 75 2kN
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q4. A cable fixed as shown in Fig. supports three loads. Determine the value of the load W and the inclination of the segment BC. [Ans: W=25kN, θ = 54.780]
A 30 B θ 60 C W
D
20 Loads are in kN
22.5
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q5. Find the reactions at A,B,C and D for the beam loaded as shown in the figure. (Ans.RA=RB =34kN;RC=28.84kN;
MC=-140kNm ; θC=-33.69 ˚ )
12kN/m 4kN/m A
20 kN 4kN/m
12kN/m 30kN 4 3 C
1m 2m
B 40kNm
1m
2m
1m
1m
2m
1m
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q6. A uniform bar AB of weight 50N shown in the figure supports a load of 200N at its end. Determine the tension developed in the string and the force supported by the pin at B. (Ans. T=529.12N;RB=807.15N, θB=64.6˚)
B A 200N
2.5m
60˚ 2.5m
2.5m
string
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q7. Find the position of the hinged support (x),such that the reactions developed at the supports of the beam are equal.. (Ans.x=2m.) 10kN/m 15kN
18kN/m x
2.0m
1.0m 0.6 1.4m
3.0m
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q8. A right angled bar ABC hinged at A as shown in fig carries two loads W and 2W applied at B &C .Neglecting self weight of the bar find the angle made by AB with vertical (Ans:θ =18.44˚)
Lm
A θ
B W 0.5L C 2W
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q9. For the block shown in fig., determine the smallest force P required a) to start the block up the plane b) to prevent the block moving down the plane. Take µ = 0.20
[Ans.: (a) Pmin = 59.2N (b) Pmin = 23.7N (b) θ = 11.3o]
100N 25°
P θ
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q10. A block of weight 2000 N is attached to a cord passing over a frictionless pulley and supporting a weight of 800N as shown in fig. If µ between the block and the plane is 0.35, determine the unknown force P for impending motion (a) to the right (b) to the left
[Ans.: (a) P = 132.8N (b) P = 1252N]
2000N
30° P
800N
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q11. Determine value of angle θ to cause the motion of 500N block to impend down the plane, if µ for all contact surfaces is 0.30.
200N 500N θ=? [Ans.: θ = 28.4°]
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q12. A horizontal bar 10m long and of negligible weight rests on rough inclines as shown in fig. If angle of friction is 15o, how close to B may the 200N force be applied before the motion impends.
100N A 2m 30°
200N X=? B 60° [Ans.: x = 3.5m]
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q13. Determine the vertical force P required to drive the wedge B downwards in the arrangements shown in fig. Angle of friction for all contact surfaces is 12o.Weight of block A= 1600 N.
P B 20°
A
[Ans.: P = 328.42N]
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q14. Determine the force P which is necessary to start the wedge to raise the block A weighing 1000N. Self weight of the wedge may be ignored. Take angle of friction, φ = 15o for all contact surfaces.
A P
[Ans.: P = 1192N]
20 ° wedge
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q15. A ladder of weight 200N, 6m long is supported as shown in fig. If µ between the floor and the ladder is 0.5 & between the wall and the ladder is 0.25 and it supports a vertical load of 1000N, determine a) the least value of α at which the ladder may be placed without slipping b) the reactions at A & B [Ans.: (a) α = 56.3o (b) RA = 1193 N, RB = 550N]
1000N 5m α
B
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q16. An uniform ladder of weight 250N is placed against a smooth vertical wall with its lower end 5m from the wall. µ between the ladder and the floor is 0.3. Show that the ladder remains in equilibrium in this position. What is the frictional resistance on the ladder at the point of contact between the ladder and the floor?
Smooth wall B 12m [Ans.: FA = 52 N] A
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q17. A ladder of length 5m weighing 500N is placed at 45o against a vertical wall. µ between the ladder and the wall is 0.20 & between ladder and ground is 0.50. If a man weighing 600N ascends the ladder, how high will he be when the ladder just slips. If a boy now stands on the bottom rung of the ladder, what must be his least weight so that the man can go to the top of the ladder. [Ans.: (a) x = 2.92m (b) Wboy = 458N]
3.
CENTROID OF PLANE AREA
3.
CENTROID
Centre of gravity : of a body is the point at which the whole weight of the body may be assumed to be concentrated. A body is having only one center of gravity for all positions of the body. It is represented by CG. or simply G or C.
Contd.
CENTRE OF GRAVITY
Consider a block of uniform thickness and having a uniform mass m. It is possible to support (hold) the block in stable position by a rod as shown in the figure provided rod must be positioned exactly at the point of intersection of the diagonals. Or the rod must be supported exactly below where the total weight of the block act.
Contd.
W
R
W
CENTRE OF GRAVITY
The block can be supported from any position provided the support rod and the line of action of weight are in same line. This indicates that the whole weight of the block act through one point. This point is called as centre of gravity.
R
Centre of gravity is that point about which the summation of the first moments of the weights of the elements of the body is zero.
Contd.
W x
CENTRE OF GRAVITY
W3 W2 W1 W4
To determine mathematically the location of the “centre of gravity” of any body, we apply the “principle of moments” to the parallel system of gravitational forces
X1 X2
The moment of the resultant gravitational force W, about any axis
=
the algebraic sum of the moments about the same axis of the gravitational forces dW acting on all infinitesimal elements of the body. Contd.
W x
CENTRE OF GRAVITY dW3 dW2 dW1 dW4
X1 X2
x ⋅ W = ∫ x × dW
x ⋅ W = dW1 × x1 + dW2 × x2 + dW3 × x3 + ................dWn × xn
Where
W =
∫ dW
Contd.
x
CENTRE OF GRAVITY
where
x = x- coordinate of centre of gravity
∫ x ⋅ dW x=
W
x
Similarly, y and z coordinates of the centre of gravity are
∫ y ⋅ dW y=
W
and
∫ z ⋅ dW z=
W
----(1)
x
CENTRE OF MASS With the substitution of W= m g and dW = g dm
(if ‘g’ is assumed constant for all particles, then )
∫ x ⋅ dm x= m ,
∫ y ⋅ dm y= m ,
∫ z ⋅ dm z= m ----(2)
Equation 2 is independent of g and therefore define a unique point in the body which is a function solely of the distribution of mass. This point is called the centre of mass and clearly coincides with the centre of gravity as long as the gravity field is treated as uniform and parallel
Contd.
When speaking of an actual physical body, we use the term “centre of mass”. The term centroid is used when the calculation concerns a geometrical shape only. Calculation of centroid falls within three distinct categories, depending on whether we can model the shape of the body involved as a line, an area or a volume.
Contd.
The centroid “C” of the Volume segment,
∫ x ⋅ dV x=
V
,
∫ y ⋅ dV y=
V
,
∫ z ⋅ dV z=
V
The centroid “C” of the line segment,
∫ x ⋅ dL x=
L
,
∫ y ⋅ dL y=
L
,
∫ z ⋅ dL z=
L
The centroid “C” of the Area segment, AREA: when the density ρ, is constant and the body has a small constant thickness t, the body can be modeled as a surface area. The mass of an element becomes dm = ρ t dA. If ρ and t are constant over entire area, the coordinates of the ‘centre of mass’ also becomes the coordinates of the centroid, C of the surface area and which may be written as
∫ x ⋅ dA , x=
A
y =
∫ y ⋅ dA , A
∫ z ⋅ dA z=
A
Contd.
Centroid of Simple figures: using method of moment ( First moment of area) Centroid of an area may or may not lie on the area in question. It is a unique point for a given area regardless of the choice of the origin and the orientation of the axes about which we take the moment.
The coordinates of the centroid of the surface area about any axis can be calculated by using the equn. (A) x = (a1) x1 + (a2) x2 + (a3) x3 + ……….+(an) xn = First moment of area
Moment of Total area ‘A’ about yaxis
=
Algebraic Sum of moment of elemental ‘dA’ about the same axis
where (A = a1 + a2 + a3 + a4 + ……..+ an)
AXIS of SYMMETRY: It is an axis w.r.t. which for an elementary area on one side of the axis , there is a corresponding elementary area on the other side of the axis (the first moment of these elementary areas about the axis balance each other) If an area has an axis of symmetry, then the centroid must lie on that axis. If an area has two axes of symmetry, then the centroid must lie at the point of intersection of these axes.
Contd.
For example: The rectangular shown in the figure has two axis of symmetry, X-X and Y-Y. Therefore intersection of these two axes gives the centroid of the rectangle.
Y
B/2 B/2 da da
D/2
X da × x = da × x Moment of areas,da about y-axis cancel each other da × x + da × x = 0
x
x
D
X
D/2
B Y
Contd.
AXIS of SYMMETYRY
‘C’ must lie on the axis of symmetry ‘C’ must lie on the axis of symmetry
‘C’ must lie at the intersection of the axes of symmetry
To locate the centroid of simple rectangular area from first principles
D
B
To locate the centroid w.r.t. the base line x-x Let the distance of centroid from the base line x-x be y
D y X X
Then from the Principle of Moments
A ⋅ y = ∫ y ⋅ da
Moment of Total area A about x-axis = Sum of moment of elemental area dA about the same axis
Contd.
Consider a elemental area dA at a distance y from the base line (x-x) Let the thickness of the element be ‘dy’ Area of small element = dA = B .dy Moment of this elemental area about x-x axis x = (area) x (distance) = (B.dy) . (y)
B
dA
dy
y x y
Contd.
Sum of Moment of all such elemental areas comprising the total area =
= ∫ y ⋅ da
= ∫ B ⋅ dy ⋅ y
⎡ By ⎤ =⎢ ⎥ 2 ⎦0 ⎣
2 D
⎡ BD 2 ⎤ =⎢ ⎥ 2 ⎦ ⎣
Then from the Principle of Moments
BD 2 Ay = 2
D y= 2
BD 2 y= 2A
,
BD 2 y= 2 BD
,
Contd.
Similarly we can show
Y
B
x Y x
dx
A ⋅ x = ∫ x ⋅ da x = B/2
= ∫ x ⋅ (D ⋅ dx )
⎡ Dx ⎤ =⎢ ⎥ 2 ⎦0 ⎣
2
B
To locate the centroid of simple right angle triangular area from first principles
H
B
To locate the centroid w.r.t. the base line x-x. Let the distance of centroid from the base line x-x be y y H
x
x
Then from the Principle of Moments
A ⋅ y = ∫ y ⋅ da
Moment of Total area A about x-axis
=
Algebraic Sum of moment of elementary area ‘dA’ about the same axis.
Contd.
Consider a small elemental area ‘dA’ at a distance ‘y’ from the base line (x-x) Let the thickness of the element be ‘dy’ Area of small element = dA = b .dy dy dA Moment of this small elemental area about xx axis = (area) x (distance) = (b.dy) . (y)
Contd.
b
y x
x
y
Then from the Principle of Moments
A ⋅ y = ∫ y ⋅ da A ⋅ y = ∫ b ⋅ dy ⋅ y
A⋅ y = ∫ B (H − y ) ⋅ dy ⋅ y H da = b × dy = B (H − y ) × dy H
b=
B (H − y ) H
y = H/3
H/3 H y
X
B
X
Contd.
Similarly we can prove that x = B/3
A ⋅ x = ∫ x ⋅ da da = h ⋅ dx = H (B − x ) ⋅ dx B
A ⋅ x = ∫ b ⋅ dx ⋅ x
A⋅ x = ∫ H (B − x ) ⋅ dx ⋅ x B
Y dx b=
H (B − x ) B
x = B/3
H
x b x
B/3
Y
B
Contd.
The centroid of simple right angled triangle area from the base
B/3
H
Centroid
B H/3
To locate the centroid of Semi Circular Area w.r.t. the diameter AB from first principles Consider a semicircle of radius R, πD 2 A = area =
8
D = 2R
Let ‘G’ be the centroid of the Semicircle, and y is its distance from the diameter AB.
Contd.
dA = r × dθ × dr
Consider a small elemental area da, located at distance y from the diameter AB, Let ‘r’ =radial distance of area ‘da’ from centre of the semi circle.
da = r × dθ × dr y = r × sin θ
Moment of this elemental area about the diameter AB =
= r 2 sin θ ⋅ dr ⋅ dθ
Contd.
Moment of all such elemental area ‘da’ about the diameter AB
= ∫ ∫ r 2 ⋅ sin θ ⋅ dr.dθ
OO
π R
⎡r ⎤ = ∫ ⎢ ⎥ ⋅ sin θ ⋅ .dθ 3 ⎦O O⎣
3
π
R
R3 π = (−[cos θ ]O ) 3
2.R 3 = 3
Contd.
Then from the Principle of Moments
A ⋅ y = ∫ y ⋅ da
A⋅ y =
π R
y
Centroid
OO
r 2 ⋅ sin θ ⋅ dr.dθ ∫∫
A⋅ y =
4R y= 3π
2.R 3 3
x
R
x y
4R y= 3π
Because of symmetry
x=0 R
To locate the centroid of Quarter Circular Area w.r.t. the boundary radial line AB from first principle
A ⋅ y = ∫ y ⋅ da π dr r dθ
A⋅ y = y B
OO
r 2 ⋅ sin θ ⋅ dr.dθ ∫∫
2 R
A
θ R
4R y= 3π
Contd.
Centroid of Quarter Circular Area w.r.t. the boundary radial lines
4R x= 3π y= 4R 3π
Centroid
4R y= 3π
To locate the centroid of Circular Sector w.r.t. the y-axis shown from first principle y Xc=(2/3)Rcosθ
da
Consider a triangle of differential area = da =
= 1 × base × height 2
1 = × R × dθ × R 2
Distance of the differential area ‘da’, from y-axis = xc =
2 = × R × cos θ 3
Contd.
To locate the centroid of Circular Sector w.r.t. the y-axis shown from first principle Consider a triangle of differential area = da =
= 1 × base × height 2
1 = × R × dθ × R 2
Distance of the differential area ‘da’, from y-axis = xc =
2 = × R × cos θ 3
Contd.
Then from the Principle of Moments α A=∫
1 −α 2
R × dθ × R = R × α
2
A × x = ∫ da × xc
(
2 1 2 R × α x = ∫ ⋅ R cos θ × R dθ 3 2 −α
2
)
α
(
2 R × α x = × R 3 × sin α 3
2
)
2 R × sin α x= 3 α
2 R × sin α x= α 3
For a semicircular area 2α = π, if we use this value in the above formula we get
y =
4 R 3π
To locate the centroid of area under the curve x = k y3 from x = 0 to x = a from first principle
Consider a vertical element of area da = y dx at a distance x from the y-axis. da To find x- coordinate,
x a x
X
A × x = ∫ da × x
At x = a, y = b, i.e. a = k b3, k = a/b3
Contd.
A = ∫ y ⋅ dx
0
Substituting y = (x/k)1/3 and k = a/b3
x ∫ ydx = ∫ xydx
0 0
a
a
⎛ x⎞ ⎛ x⎞ x ∫ ⎜ ⎟ dx = ∫ x⎜ ⎟ dx k⎠ k⎠ 0⎝ 0 ⎝
a
1 3
a
1 3
3ab 3a b ⋅x= 4 7
2
4 x= a 7
Contd.
To find y, Coordinate of centroid of the rectangular element is yc = y/2
A × y = ∫ yc × dA
⎛ y⎞ y ∫ ydx = ∫ ⎜ ⎟( ydx ) 2⎠ 0 0⎝ a a
Substituting,
y = b( x/a)1/3
3ab 3ab 2 ⋅y= 4 10 2 y= b 5
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.1: Locate the centroid of the shaded area shown
10 50 10 40
Ans: x=12.5, y=17.5
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.2: Locate the centroid of the shaded area shown
500 1000 mm
D=600 r=600
300 300 500
1000 mm
Ans: x=474mm, y=474mm
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.3: Locate the centroid of the shaded area w.r.t. to the axes shown y-axis 90 20 120 r=40 20 60 x-axis
Ans: x=34.4, y=40.3
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.4: Locate the centroid of the shaded area w.r.t. to the axes shown y-axis 250 mm
20 10 380
10
200 mm
10 x-axis
Ans: x= -5mm, y=282mm
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.5 Locate the centroid of the shaded area w.r.t. to the axes shown y 30 50
30
40 40 20 r=20 20 x
Ans:x =38.94, y=31.46
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.6 Locate the centroid of the shaded area w.r.t. to the axes shown y 1.0
2.4 m r=0.6 1.0 1.0 x 1.5 1.5
Ans: x=0.817, y=0.24
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.7 Locate the centroid of the shaded area w.r.t. to the axes shown
Ans: x= -30.43, y= +9.58
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.8 Locate the centroid of the shaded area.
20
Ans: x= 0, y= 67.22(about base)
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.9 Locate the centroid of the shaded area w.r.t. to the base line.
2
Ans: x=5.9, y= 8.17
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.10 Locate the centroid of the shaded area w.r.t. to the axes shown
Ans: x=21.11, y= 21.11
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.11 Locate the centroid of the shaded area w.r.t. to the axes shown
Ans: x= y= 22.22
EXERCISE PROBLEMS 3. Centroid of plane area
Problem No.12 Locate the centroid of the shaded area w.r.t. to the axes shown
Y R=25 R=25
80
X
50 50 75
Ans: x= 24.33 y= 4.723
4. MOMENT OF INERTIA OF PLANE AREA
4. MOMENT OF INERTIA Moment of Inertia( Second moment area) The product of the elemental area and square of the perpendicular distance between the centroid of area and the axis of reference is the “Moment of Inertia” about the reference axis Iox= da1 y12 + da2 y22+ da3 y32+ -= ∑ da y2 Ioy = da1 x12 + da2 x22 + da3 x32+ ---= ∑ da x2
y x
dA
y o Radius of Gyration
A r1 r2 r3 B Elemental area A
k k k B
Radius of gyration is defined as a constant distance of all elemental areas which have been rearranged with out altering the total moment of inertia.
IAB= da k2 + da k2 + ----------IAB = ∑ da k2 IAB= A k2 k=√IAB/A
Polar moment of Inertia (Perpendicular Axes theorem) The moment of inertia of an area about an axis perpendicular to the plane of the area is called “Polar Moment of Inertia” and it is denoted by symbol Izz or J or Ip. The moment of inertia of an area in xy plane w.r.to z. axis is Izz = Ip = J = ∫r2dA = ∫(x2 + y2) dA = ∫x2dA + ∫y2dA = Ixx +Iyy
Y
x
r
O
y
x
z
Polar moment of Inertia (Perpendicular Axes theorem) Hence polar M.I. for an area w.r.t. an axis perpendicular to its plane of area is equal to the sum of the M.I. about any two mutually perpendicular axes in its plane, passing through the point of intersection of the polar axis and the area.
Parallel Axis Theorem
dA x
* G
y
x
d A B
Moment of inertia of any area about an axis AB is equal to the M.I. about parallel centroidal axis plus the product of the total area and square of the distance between the two axes.
IAB =∑dA (d +y)2 = ∑dA (d2 + y2 + 2 × d × y) =∑dA. d2 +∑dA y2 + ∑ 2×d×dA y = ∑dA. d2 +∑dA y2 + 2×d. ∑ y dA In the above term (2×d) is constant & ∑ y dA = 0 IAB = Ixx + A.d2
MOMENT OF INERTIA BY DIRECT INTEGRATION Moment of inertia of rectangular area about centroidal horizontal axis by direct integration M.I. about its horizontal centroidal axis :
I x x = ∫ dA × y 2
−D / 2
+D / 2
= ∫ ( B × dy ) × y x −D / 2 BD 3 = 12
2
+D / 2
D/2 D y
dy
G
x
B
Moment of Inertia of rectangular area about its base(about the line AB) using Parallel Axis Theorem IAB = IXX + A(d)2 Where d = D/2, the distance between axes xx and AB
D/2
=BD /12+(BD)(D/2) =BD3/12+BD3/4 =BD3/3
3
.
dy y
2
x
D
G
x
A B
B
Moment of inertia of Triangular area about the base by direct integration
dy
(h-y)
h x h/3 A
x
y
x
B b From similar triangles b/h = x/(h-y) ∴ x = b . (h-y)/h
dy
(h-y)
h x h/3 A
x
y
x
B b IAB = ∫ dA.y2 = ∫ (x.dy)y2
Ixx = ∫
h 0
(b . (h-y) y2.dy) /h
= b[ h (y3/3) – y4/4 ]/h = bh3/12
Moment of inertia of Triangular area about the centroidal horizontal axis Using Parallel axis theorem . MI about any line(AB) = MI about cenroidal parallel axis + Ad2 IAB = Ixx + Ad2 h x A b Centroidal horizontal axis x Ixx = MI about centroidal axis x x IAB= MI about the Base line AB Ixx = IAB – Ad2 = bh3/12 – bh/2 . (h/3)2 = bh3/36
dy
(h-y)
h/3
y
x
B
Moment of inertia of Circular area about the centroidal horizontal axis Ixx = ∫ dA . y2 = ∫ ∫ (x.dθ.dr) r2Sin2θ
0 0 R 2π
dθ
=∫ ∫
R
R 2π
r3.dr
Sin2θ
2π
r dθ
0 0
x A
θ
y=rSinθ
x
=∫ r3 dr ∫ {(1- Cos2θ)/2} dθ
0
R B
=[r4/4]
R 0
0
[θ/2 – Sin2θ/4]
2π 0
= R4/4[π - 0] = πR4/4 IXX = π R4/4 = πD4/64
Moment of inertia of Semi-circular area about the Base & centroidal horizontal axis IAB = ∫ dA . y2 = ∫ ∫ (r.dθ.dr) r2Sin2θ
0 0 R π 3.dr ∫ Sin2θ =∫ r 0 R π =∫ ∫ r3 0 0 0 R π
dθ
y0
dr (1- Cos2θ)/2) dθ [θ/2 – Sin2θ/4] π 0
=[R4/4]
= R4/4[π/2 - 0] = πR4/8 = (πD4/32)
x 4R/3π A
R
x B
y0
Moment of inertia of Semi-circular area about the centriodal horizontal axis using parallel axis theorem: IAB = Ixx + A(d)
2
Ixx = IAB – A(d)2 = π R4/8 Ixx = 0.11R4 πR2/2 . (4R/3π)2
Moment of inertia of Quarter-circular area about the base & centroidal horizontal axis IAB = ICD IAB = ∫ ∫ (r.dθ.dr). r2Sin2θ
0 0 R π/2
D
y
=∫ r3.dr ∫ Sin2θ dθ
0 R 0 π/2 0
R
π/2
x A C
4R/3π
x
4R/3π
=∫ r3 dr ∫ (1- Cos2θ)/2) dθ
0
y
B
=[R4/4] [θ/2 – (Sin2 θ)/4] = R4 (π/16 – 0) = πR4/16
π/2 0
Moment of inertia about Centroidal axis, Ixx = IAB - Ad2 = πR4/16 - πR2. (0. 424R)2 = 0.055R4
Sl.No
Y
1 x0 x h x0 x x0
Figure b d x0
Xo
Ix
-x 0 0
Iy
-y 0 0
I xx bd3/3
I yy -
d/2
Y
bd3/12
-
x x0 bh3/36 h/3 x x0 4R/3π x0 x x0 4R/3π 0.11R4 πR4/8 πR4/8 πR4/4 πR4/4 bh3/12 -
2
b
O
3
y0
R
4 x0 x y
y0
y0
5
y0 y0
0.055R4
0.055R4
πR4/16
πR4/16
4R/3π
EXERCISE PROBLEMS
Q1. Determine the moment of inertia about the centroidal axes. 30mm 30mm
20
30mm
100mm
[Ans: Y = 27.69mm Ixx = 1.801 x 106mm4 Iyy = 1.855 x 106mm4]
EXERCISE PROBLEMS
Q2. Determine second moment of area about the centroidal horizontal and vertical axes. 300mm 300mm 200
200mm 900mm [Ans: X = 99.7mm from A, Y = 265 mm Ixx = 10.29 x 109mm4, Iyy = 16.97 x 109mm4]
EXERCISE PROBLEMS
Q3. Determine M.I. Of the built up section about the horizontal and vertical centroidal axes and the radii of gyration.
200mm 20 140mm 20 100mm
[Ans: Ixx = 45.54 x 106mm4, Iyy = 24.15 x 106mm4 rxx = 62.66mm, ryy = 45.63mm]
60
EXERCISE PROBLEMS
Q4. Determine the horizontal and vertical centroidal M.I. Of the shaded portion of the figure.
X
60
20 20
X
60
60
[Ans: X = 83.1mm Ixx = 2228.94 x 104mm4, Iyy = 4789.61 x 104mm4]
EXERCISE PROBLEMS
Q5. Determine the spacing of the symmetrically placed vertical blocks such that Ixx = Iyy for the shaded area.
200mm
400mm 200mm d 200mm 200mm
[Ans: d/2 = 223.9mm d=447.8mm]
600mm
EXERCISE PROBLEMS
Q6. Find the horizontal and vertical centroidal moment of inertia of the section shown in Fig. built up with R.S.J. (I-Section) 250 x 250 and two plates 400 x 16 mm each attached one to each. Properties of I section are Ixx = 7983.9 x 104mm4 Iyy = 2011.7 x 104mm4
160mm
2500mm
Cross sectional area=6971mm2
160mm 4000mm
[Ans: Ixx = 30.653 x 107mm4, Iyy = 19.078 x 107mm4]
EXERCISE PROBLEMS
Q7. Find the horizontal and vertical centroidal moment of inertia of built up section shown in Figure. The section consists of 4 symmetrically placed ISA 60 x 60 with two plates 300 x 20 mm2.
Properties of ISA
Cross sectional area = 4400mm2 18.5mm 18.5mm
Ixx = Iyy ;Cxx = Cyy =18.5mm
200mm 20mm
300mm
[Ans: Ixx = 111.078 x 107mm4, Iyy = 39.574 x 107mm4]
EXERCISE PROBLEMS
Q8. The R.S. Channel section ISAIC 300 are placed back to back with required to keep them in place. Determine the clear distance d between them so that Ixx = Iyy for the composite section. Properties of ISMC300 C/S Area = 4564mm2 104mm4 X d [Ans: d = 183.1mm] Y
23.6mm
Y
Lacing
Ixx = 6362.6 x 104mm4 Iyy = 310.8 x Cyy = 23.6mm
X 380mm
EXERCISE PROBLEMS
Q9. Determine horizontal and vertical centroidal M.I. for the section shown in figure.
40mm 160mm 40mm 40mm 90mm
[Ans: Ixx = 2870.43 x 104mm4, Iyy = 521.64 x 104mm4]
5. Kinetics of rectilinear motion
5. Kinetics of rectilinear motion In this chapter we will be studying the relationship between forces on a body/particle and the accompanying motion
Newton’s Second law of motion:
Newton’s first and third law of motion were used extensively in the study of statics (the bodies at rest) whereas Newton’s second law of motion is used extensively in the study of the kinetics.
Work done by force: Work done by a force is the product of the force and the distance moved by the point of application in the direction of the force. It is a scalar quantity.
F sinθ
F
F cosθ
F sinθ
F B α F cosθ
A
Work done = (F cosα ) × s
α
s
X- component of force F moves through distance a S, S = displacement of force from A to B Unit: Nm ( Joule )
POWER:It is defined as the time rate of doing work. Power = work done /Time= (force × distance) /Time = force × velocity Unit: (Nm)/s = [watt] (kN m)/s = [kilo watt] 1 metric H.P=735.75 watts.
Energy:It is defined as the capacity to do work. It is a scalar quantity. Unit :- N m (Joule)
Momentum:Quantity of motion possessed by a body is called momentum. It is the product of mass and velocity. It is a vector quantity. Unit:- N s. Impulse of a Force:It is defined as the product of force and the time over which it acts. It is a vector quantity. Unit:- N s.
Newton’s second law of motion. “If the resultant force acting on a particle is not zero , the particle will have an acceleration proportional to the magnitude of the resultant force and its direction is along that of the resultant force.” Fαa F =Resultant of forces a = Acceleration of the particle. F = ma m= mass of the particle.
The constant value obtained for the ratio of the magnitude of the force and acceleration is characteristic of the particle and is denoted by ‘m’. Where ‘m’ is mass of the particle Since ‘m’ is a +ve scalar, the vectors of force ‘F’and acceleration ‘a’ have the same direction.
Units Force in Newtons (N) Acceleration in m/s2
N = 1 Kgm/s2
F2 F1 F3
he n t nt e i lta a ov s u m = l m f re il o R y w on od cti B re di
R = Resultant of forces F1,F2 and F3
Using the rectangular coordinate system we have components along axes as, ΣFx = max ΣFy = may ΣFz = maz where Fx ,Fy Fz and ax , ay ,az are rectangular components of resultant forces and accelerations respectively.
Newton’s second law may also be expressed by considering a force vector of magnitude ‘ma’ but of sense opposite to that of the acceleration. This vector is denoted by (ma)rev. The subscript indicates that the sense of acceleration has been reversed and is called the inertia force vector.
ov m ll i w t dy ltan B o su re
in e
t
ed h
i ct e ir
on
o
R
=
m
a
e rc Fo a tia m er = In R
F2 F1 F3
R = Resultant of forces F1,F2 and F3
F2 F1 F3
F2
ce m or = F R tia = F Iner
a
F1 F3
If the inertia force vector is added to the forces acting on the particle we obtain a system of forces whose resultant is zero.
Resultant of forces F1,F2, F3 and Inertia force = 0
F1 + F2 + F3 + ma = 0
The particle may thus be considered to be in equilibrium. (THIS IS DYNAMIC EQUILIBRIUM)
It was pointed out by D’Alembert (Alembert, Jean le Rond d’ (1717-1783), French mathematician and philosopher) that problems of kinetics can be solved by using the principles of statics only (the equations of equilibrium) by considering an inertia force in a direction directly opposite to the acceleration in addition to the real forces acting on the system D’Alembert’s principle states that When different forces act on a system such that it is in motion with an acceleration in a particular direction, the vectorial sum of all the forces acting on the system including the inertia force (‘ma’ taken in the opposite direction to the direction of the acceleration) is zero.
The problem under consideration may be solved by using the method developed earlier in statics. The particle is said to be in dynamic equilibrium. If ΣFx = 0 ΣFy= 0 ΣFz = 0 This principle is known as D’Alembert’s principle including inertia force vector
Work-Energy relation for translation From Newton’s second law of motion ∑F=m×a = Also
-----------(1)
a = dv/dt =( dv/dt) ×( ds/ds) = v × dv/ds substituting in (1)
∑ F = m × v × dv/ds ΣF × ds = m × v × dv ------------------(2) Let the initial velocity be u and the final velocity after it moves through a distance ‘ s’ be v
Integrating both sides, we get
∑ F ∫ ds = m ∫ v dv
0 u
s
v
v2 v ∑ F × s = (m ) 2 u 1 F × s = m(v 2 − u 2 ) ∑ 2
Therefore work done by a system of forces acting on a body while causing a displacement is equal to the change in kinetic energy of the body during the displacement.
Impulse-momentum relationship F=m×a F = m × (v - u)/t = (mv – mu)/t Force = Rate of change of momentum F × t = mv – mu Impulse = final momentum – Initial momentum The component of the resultant linear impulse along any direction is equal to change in the component of momentum in that direction.
EXERCISE PROBLEMS 5. Kinetics
Q1. Blocks A and B of mass 10 kg and 30 kg respectively are connected by an inextensible cord passing over a smooth pulley as shown in Fig. Determine the velocity of the system 4 sec. after starting from rest. Assume coefficient of friction =0.3 for all surfaces in contact.
B
60o
A
30o
Ans: v=13.6m/s
EXERCISE PROBLEMS 5. Kinetics
Q2.
A tram car weighs 150kN. The tractive resistance being 1% of the weight of car. What power will be required to move the car at uniform speed of 20 kmph
(i) Up an incline 1 in 300 (ii) (ii) Down an incline 1 in 250. Take efficiency 75%.
Ans: Pull = 2 kN
a) Output power=11.12kW
EXERCISE PROBLEMS 5. Kinetics
Q3. Two masses of 5 kg and 3 kg rest on two smooth inclined plane, each of inclination 30º and are connected by a string passing over a common apex. Find the velocity of 3 kg mass after 2 sec when released from rest. Find the distance it will cover before changing direction of motion, if 5kg mass is cut off after two sec of its release from rest.
5kg 30º
V = 4.45 m/s s = 0.61 m
3kg 30º
EXERCISE PROBLEMS 5. Kinetics
Q4. A locomotive weighing 900 kN pulls a train of 10 coaches each weighing 300 kN at 72 Kmph on a level track against a resistance of 7 N/kN. If the rear 4 coaches get snapped from the train, find the speed of the engine and the remaining coaches after 120 secs. Assume no change in resistance and draw bar pull. Find also distance traveled by detached coaches before coming to rest.
4x300=1200kN
6x300=1800kN
900kN
P
V = 23.66 m/s s = 2.9 km
EXERCISE PROBLEMS 5. Kinetics
Q5. Find the tension in the cord supporting body C in Fig. below. The pulleys are frictionless and of negligible weight.
A C 150 kN B 450 kN 300 kN
Assume all blocks moving either downward or upward and accordingly draw FBD 0=aA +2aB +cC
Ans : T=211.72 kN
EXERCISE PROBLEMS 5. Kinetics
Q6. Two blocks A and B are released from rest on a 30o inclined plane with horizontal, when they are 20m apart. The coefficient of friction under the upper block is 0.2 and that under lower block is 0.4. compute the time elapsed until the block touch. After they touch and move as a unit what will be the constant forces between them.
(Ans : t = 4.85 s, contact force=8.65 N)
EXERCISE PROBLEMS 5. Kinetics
Q7. An elevator cage of a mine shaft weighing 8kN when empty is lifted or lowered by means of rope. Once a man weighing 600N entered it and lowered at uniform acceleratin such that when a distance of 187.5 m was covered, the velocity of the cage was 25m/s. Determine the tension in the cable and force exerted by man on the floor of the cage.
(Ans: T=7139 N and R=498 N)
EXERCISE PROBLEMS 5. Kinetics
Q8. A small block starts from rest at point a and slides down the inclined plane. At what distance along the horizontal will it travel before coming to rest . Take µk=0.3 [Ans :s=6m ]
5m
A 3 4 B s C
EXERCISE PROBLEMS 5. Kinetics
Q9. The system starts from rest in the position shown . How much further will block ‘A’ move up the incline after block B hits the ground . assume the pulley to be frictionless and massless and µ is 0.2 .WA=1000N, WB=2000N. [ Answer s =1.27m]
A 4 3 B 3m
EXERCISE PROBLEMS 5. Kinetics
Q10. A 1500Kg automobile travels at a uniform rate of 50kmph to 75kmph . During the entire motion, the automobile is traveling on a level horizontal road and rolling resistance is 2 % of weight of automobile . Find (i) maximum power developed (ii) power required to maintain a constant speed of 75kmph.
[ ANSWER: power developed = 6.131KN]
EXERCISE PROBLEMS 5. Kinetics
Q11. Two bodies A and B weighing 2000N and 5200N are connected as shown in the figure . find the further distance moved by block a after the block B hits Wall. µ=0.2 .[ Answer s=1.34m]
A 5 12 B
3m
EXERCISE PROBLEMS 5. Kinetics
Q12. A spring is used to stop 60kg package which is sliding on a horizontal surface . the spring has a constant k = 20kN/m and is held by cable such that it is initially compressed at 120mm. knowing that the package has a velocity of 2.5m/s in position shown and maximum additional displacement of spring is 40mm . Determine the coefficient of kinetic friction between package and surface. (Answer µk=0.2) 2.5 m/s
60kg
600m
EXERCISE PROBLEMS 5. Kinetics
Q13. The system shown in figure has a rightward velocity of 4m/s, just before force P is applied. Determine the value of P that will give a leftward velocity of 6m/s in a time interval of 20sec. Take µ = 0.2 & assume ideal pulley. [Answer P=645.41N] P 1000N
400N
EXERCISE PROBLEMS 5. Kinetics
Q14. A locomotive of weight 500kN pulls a train of weight of 2500kN. The tractive resistance, due to friction is 10N/kN. The train can go with a maximum speed of 27kmph on a grade of 1in100. Determine (a) Power of the locomotive. (b) Maximum speed it can attain on a straight level track with the tractive resistance remaining same. [Answer (a) Power= 450kN (b) v=15m/s] Q15. A wagon weighing 400kN starts from rest, runs 30m down a 1% grade & strikes a post. If the rolling resistance of the track is 5N/kN, find the velocity of the wagon when it strikes the post. If the impact is to be cushioned by means of one bumper string, which compresses 1mm per 20 kN weight, determine how much the bumper spring will be compressed. [Answer v=1.716m/s, x=77.5mm]
EXERCISE PROBLEMS 5. Kinetics
Q16. A train whose weight is 20kN moves at the rate of 60kmph. After brakes are applied, it is brought to rest in 500m. Find the force exerted, assuming it to be uniform a) Use work-energy relation b) Use D’Alemberts equation.
Ans: F = 5.663 kN
EXERCISE PROBLEMS 5. Kinetics
Q17. The blocks A and B having weights 100 N and 300 N start from rest. The horizontal plane and the pulleys are frictionless. Determine the acceleration and the tension in the string. string Frictionless pulley
A string
Frictionless pulley
Ans: aA=8.403m/s2 aB=4.201 m/s2 T= 85.71 N
B
EXERCISE PROBLEMS 5. Kinetics
Q18. The blocks A and B having weights 100 N and 300 N start from rest. The horizontal plane and the pulleys are frictionless. Determine the velocity of block B after 0.5 seconds and the tension in the string. Use impulsemomentum relation. string Frictionless pulley
A string
Frictionless pulley B
EXERCISE PROBLEMS 5. Kinetics
Q19. The blocks A and B having weights 100 N and 300 N start from rest when a load of 100 N is applied on the block A as shown in the figure. The horizontal plane and the pulleys are frictionless. Determine the acceleration and the tension in the string. string 100 N
A
Frictionless pulley
string
Frictionless pulley B
EXERCISE PROBLEMS 5. Kinetics
Q20. Two blocks A and B are connected as shown in the figure. At the instant of their release if the block A, which is on smooth horizontal plane has a left ward velocity of 2 m/s, what would be its velocity 5 seconds after their release. The blocks A and B weigh 100 N and 300 N respectively. string Frictionless pulley
A string
Frictionless pulley B
EXERCISE PROBLEMS 5. Kinetics
Q21. An engine weighing 500 kN drags carriages weighing 1500kN up an incline of 1 in 100 against a resistance of 5N/kN starting from rest. It attains a velocity of 36 kmph (10m/s) in 1 km distance with a constant draw bar pull supplied by the engine. What is the power required for the same ? What is the tension developed in the link connecting the engine and carriages? 1500N 500N P 100 1
Ans: Pull=40.19kN, power=401kW, T=30.15kN
EXERCISE PROBLEMS 5. Kinetics
Q22. what velocity the block A will attain after 2 seconds starting from rest? Take µ = 0.2. WA = 1500N, WB = 2000N. Use impulse-momentum relation.
A 3
4
3 4
B
PART - II Mechanics of Deformable Bodies
COURSE CONTENT IN BRIEF
6. Simple stresses and strains 7. Statically indeterminate problems and thermal stresses 8. Stresses on inclined planes 9. Stresses due to fluid pressure in thin cylinders
6. Simple stresses and strains
The subject strength of materials deals with the relations between externally applied loads and their internal effects on bodies. The bodies are no longer assumed to be rigid and the deformations, however small, are of major interest The subject, strength of materials or mechanics of materials involves analytical methods for determining the strength , stiffness (deformation characteristics), and stability of various load carrying members. Alternatively the subject may be called the mechanics of solids.
GENERAL CONCEPTS STRESS No engineering material is perfectly rigid and hence, when a material is subjected to external load, it undergoes deformation. While undergoing deformation, the particles of the material offer a resisting force (internal force). When this resisting force equals applied load the equilibrium condition exists and hence the deformation stops. These internal forces maintain the externally applied forces in equilibrium.
STRESS
The internal force resisting the deformation per unit area is called as stress or intensity of stress.
Stress = internal resisting force / resisting cross sectional area
R = A
STRESS SI unit for stress N/m2 also designated as a pascal (Pa) Pa = N/m2 kilopascal, 1kPa = 1000 N/m2 megapascal, 1 MPa = 1×106 N/m2 = 1×106 N/(106mm2) = 1N/mm2 1 MPa = 1 N/mm2 gigapascal, 1GPa = 1×109 N/m2 = 1×103 MPa = 1×103 N/mm2
AXIAL LOADING – NORMAL STRESS P P R
STRESS
Consider a uniform bar of cross sectional area A, subjected to a tensile force P. Consider a section AB normal to the direction of force P Let R is the total resisting force acting on the cross section AB. Then for equilibrium condition,
A
B
R
R=P Then from the definition of stress, normal stress = σ = R/A = P/A
P
P
Symbol:
σ = Normal Stress
AXIAL LOADING – NORMAL STRESS Direct or Normal Stress: Intensity of resisting force perpendicular to or normal to the section is called the normal stress. Normal stress may be tensile or compressive
STRESS
Tensile stress: stresses that cause pulling on the surface of the section, (particles of the materials tend to pull apart causing extension in the direction of force) Compressive stress: stresses that cause pushing on the surface of the section, (particles of the materials tend to push together causing shortening in the direction of force)
STRESS • The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis. • The force intensity on that section is defined as the normal stress.
∆F σ = lim ∆A→0 ∆A
P σ ave = A
STRAIN STRAIN : when a load acts on the material it will undergo deformation. Strain is a measure of deformation produced by the application of external forces. If a bar is subjected to a direct load, and hence a stress, the bar will changes in length. If the bar has an original length L and change in length by an amount δL, the linear strain produced is defined as, δL Change in length ε= = Linear strain, Original length L Strain is a dimensionless quantity.
Linear Strain
P σ = = stress A
ε=
δ
2P P = σ= 2A A
σ=
L
= normal strain
ε=
δ
P A 2δ δ ε= = 2L L
L
STRESS-STRAIN DIAGRAM In order to compare the strength of various materials it is necessary to carry out some standard form of test to establish their relative properties. One such test is the standard tensile test in which a circular bar of uniform cross section is subjected to a gradually increasing tensile load until failure occurs. Measurement of change in length over a selected gauge length of the bar are recorded throughout the loading operation by means of extensometers. A graph of load verses extension or stress against strain is drawn as shown in figure.
STRESS-STRAIN DIAGRAM
Proportionality limit
Typical tensile test curve for mild steel
STRESS-STRAIN DIAGRAM
Typical tensile test curve for mild steel showing upper yield point and lower yield point and also the elastic range and plastic range
Stress-strain Diagram Limit of Proportionality : From the origin O to a point called proportionality limit the stress strain diagram is a straight line. That is stress is proportional to strain. Hence proportional limit is the maximum stress up to which the stress – strain relationship is a straight line and material behaves elastically. From this we deduce the well known relation, first postulated by Robert Hooke, that stress is proportional to strain. Beyond this point, the stress is no longer proportional to strain
PP σP = = Load at proportionality limit A Original cross sectional area
Stress-strain Diagram Elastic limit: It is the stress beyond which the material will not return to its original shape when unloaded but will retain a permanent deformation called permanent set. For most practical purposes it can often be assumed that points corresponding proportional limit and elastic limit coincide. Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will be thus some permanent deformation when load is removed.
PE Load at proportional limit σE = = A Original cross sectional area
Stress-strain Diagram Yield point: It is the point at which there is an appreciable elongation or yielding of the material without any corresponding increase of load.
PY σY = = A
Load at yield point Original cross sectional area
Ultimate strength: It is the stress corresponding to maximum load recorded during the test. It is stress corresponding to maximum ordinate in the stress-strain graph.
P σU = U = A
Maximum load taken by the material Original cross sectional area
Stress-strain Diagram Rupture strength (Nominal Breaking stress): It is the stress at failure. For most ductile material including structural steel breaking stress is somewhat lower than ultimate strength because the rupture strength is computed by dividing the rupture load (Breaking load) by the original cross sectional area.
PB σB = = A
load at breaking (failure) Original cross sectional area load at breaking (failure) Actual cross sectional area
True breaking stress =
Stress-strain Diagram After yield point the graph becomes much more shallow and covers a much greater portion of the strain axis than the elastic range. The capacity of a material to allow these large plastic deformations is a measure of ductility of the material Ductile Materials: The capacity of a material to allow large extension i.e. the ability to be drawn out plastically is termed as its ductility. Material with high ductility are termed ductile material. Example: Low carbon steel, mild steel, gold, silver, aluminum
Stress-strain Diagram
A measure of ductility is obtained by measurements of the percentage elongation or percentage reduction in area, defined as, increase in gauge length (up to fracture) ×100 = original gauge length Percentage elongation Reduction in cross sectional area of necked portion (at fracture)
Percentage reduction in = area original area
×100
Cup and cone fracture for a Ductile Material
Stress-strain Diagram Brittle Materials : A brittle material is one which exhibits relatively small extensions before fracture so that plastic region of the tensile test graph is much reduced. Example: steel with higher carbon content, cast iron, concrete, brick
Stress-strain diagram for a typical brittle material
HOOKE’S LAW Hooke’s Law For all practical purposes, up to certain limit the relationship between normal stress and linear strain may be said to be linear for all materials stress (σ) α strain (ε) stress (σ) constant strain (ε) = Thomas Young introduced a constant of proportionality that came to be known as Young’s modulus. stress (σ) E strain (ε) = = Young’s Modulus or Modulus of Elasticity
HOOKE’S LAW Young’s Modulus is defined as the ratio of normal stress to linear strain within the proportionality limit. stress (σ) = E = strain (ε)
P δL PL ÷ = A L AδL
The value of the Young’s modulus is a definite property of a material From the experiments, it is known that strain is always a very small quantity, hence E must be large. For Mild steel, E = 200GPa = 2×105MPa = 2×105N/mm2
Deformations Under Axial Loading • From Hooke’s Law: σ = Eε ε= σ
E =
P AE
• From the definition of strain:
L • Equating and solving for the deformation, PL δ = AE
• With variations in loading, crosssection or material properties, Pi Li δ =∑ i Ai Ei
ε=
δ
Consider an element of length, δx at a distance x from A
W A W
d1
x
B
Diameter at x, = d1
L = d1 + k × x
(d − d ) + 2 1 ×x
dx
d2 c/s area at x, πd12 =
4 =
π
4
(d1 + kx )2
Change in length over a length dx is Change in length over a length L is
⎛ ⎞ ⎜ ⎟ Wdx ⎛ PL ⎞ ⎟ =⎜ ⎟ =⎜ ⎝ AE ⎠ dx ⎜ π (d + kx )2 × E ⎟ ⎜ ⎟ 1 ⎝4 ⎠ ⎛ ⎞ ⎜ ⎟ L Wdx ⎟ =∫ ⎜ 0 ⎜ π (d + kx )2 × E ⎟ ⎜ ⎟ 1 4 ⎝ ⎠
Consider an element of length, δx at a distance x from A Change in length over a length L is
⎛ ⎞ ⎜ ⎟ L Wdx ⎟ =∫ ⎜ 0 π ⎜ (d + kx )2 × E ⎟ ⎜ ⎟ 1 ⎝4 ⎠ dt ⎞ ⎛ ⎜ W ⎟ L k ⎟ =∫ ⎜ 0 ⎜ π (t )2 × E ⎟ ⎜ ⎟ ⎝4 ⎠
Put d1+kx = t, Then k dx = dt
L
⎤ 4W ⎡ t 4W = ⎢ ⎥ = πEk ⎣ − 1 ⎦ 0 πEk
− 2 +1
L
− 4W ⎡ − 1⎤ ⎢ t ⎥ = πEk ⎣ ⎦0
⎡ ⎤ 1 ⎢ ⎥ (d1 + kx) ⎦ 0 ⎣
L
=
4WL WL = πEd1d 2 πd1d 2 × E 4
Derive an expression for the total extension of the tapered bar AB of rectangular cross section and uniform thickness, as shown in the figure, when subjected to an axial tensile load ,W. W B L b
d1 b
W A
d2
d1 b
W A x dx B
W
d2 b
Consider an element of length, δx at a distance x from A depth at x,
= d1 +
L = d1 + k × x
(d 2 − d1 ) × x
c/s area at x, = (d1 + kx )b
Change in length over a length dx is
⎛ ⎞ Wdx ⎛ PL ⎞ =⎜ =⎜ ⎟ ⎜ (d + kx )b × E ⎟ ⎟ ⎝ AE ⎠ dx ⎝ 1 ⎠
Change in length over a length L is
=∫
L
0
⎛ ⎞ Wdx ⎜ ⎜ (d + kx )b × E ⎟ ⎟ ⎝ 1 ⎠
P (log e d 2 − log e d1 ) = b× E ×k
2.302 × P × L (log d 2 − log d1 ) = b × E × (d 2 − d1 )
Derive an expression for the total extension produced by self weight of a uniform bar, when the bar is suspended vertically.
L
Diameter d
dx x
P1 = weight of the bar below element the section, = volume × specific weight dx = (π d2/4)× x × γ P1 = A× x ×γ
Diameter d
Extension of the element due to weight of the bar below that,
P1 dx ( A × x × ρ ) dx ⎡ PL ⎤ =⎢ ⎥ = AE = AE ⎣ AE ⎦ dx
Hence the total extension entire bar
=∫
L
0
( A × x × γ )dx ⎡ γx ⎤ γL2 =⎢ ⎥ = AE ⎣ 2E ⎦ 0 2E
2
L
The above expression can also be written as
A ( ALγ ) × L 1 PL = × = = × 2E A 2 AE 2 AE
γL2
Where, P = (AL)×γ = total weight of the bar
SHEAR STRESS Consider a block or portion of a material shown in Fig.(a) subjected to a set of equal and opposite forces P. then there is a tendency for one layer of the material to slide over another to produce the form failure as shown in Fig.(b) P P P P R Fig. b Fig. c R
Fig. a
The resisting force developed by any plane ( or section) of the block will be parallel to the surface as shown in Fig.(c). The resisting forces acting parallel to the surface per unit area is called as shear stress.
Shear stress (τ) =
Shear resistance Area resisting shear
P = A
This shear stress will always be tangential to the area on which it acts
Shear strain
If block ABCD subjected to shearing stress as shown in Fig.(d), then it undergoes deformation. The shape will not remain rectangular, it changes into the form shown in Fig.(e), as AB'C'D. τ τ B' C' C C B B
A
τ Fig. d
D
A
τ Fig. e
D
B
B'
τ
C
φ
A
τ Fig. e
D
Shear strain is defined as C' the change in angle between two line element which are originally right angles to one another.
BB′ shear strain = = tan φ ≈ φ AB
The angle of deformation
φ
is then termed as shear strain
The angle of deformation is measured in radians and hence is non-dimensional.
SHEAR MODULUS For materials within the proportionality limit the shear strain is proportional to the shear stress. Hence the ratio of shear stress to shear strain is a constant within the proportionality limit. Shear Modulus Shear stress (τ) = constant = G = or Shear strain (φ) Modulus of Rigidity
The value of the modulus of rigidity is a definite property of a material For Mild steel, G= 80GPa = 80,000MPa = 80,000N/mm2
example: Shearing Stress • Forces P and P‘ are applied transversely to the member AB. • Corresponding internal forces act in the plane of section C and are called shearing forces. • The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load P. • The corresponding average shear stress is, τ ave =
P A
• The shear stress distribution cannot be assumed to be uniform.
State of simple shear
Consider an element ABCD in a strained material subjected to shear stress, τ as shown in the figure A τ B
D
C
τ Force on the face AB = P = τ × AB × t Where, t is the thickness of the element. Force on the face DC is also equal to P
State of simple shear Now consider the equilibrium of the element. (i.e., ΣFx = 0, ΣFy = 0, ΣM = 0.) For the force diagram shown, ΣFx = 0, & ΣFy = 0, But ΣM = 0 The element is subjected force to a clockwise moment P × AD = (τ × AB × t) × AD D P C A P B
But, as the element is actually in equilibrium, there must be another pair of forces say P' acting on faces AD and BC, such that they produce a anticlockwise moment equal to ( P × AD )
State of simple shear P ' × AB = P × AD = (τ × AB × t)× AD ----- (1) If τ1 is the intensity of the shear stress on the faces AD and BC, then P ' can be written as, P ' = τ ' × AD × t Equn.(1) can be written as τ'
D A
P
B
P'
D
P' P τ
C
A
B
τ' τ
C
(τ ' × AD× t ) × AB = (τ × AB × t) × AD ----- (1) τ' =τ
State of simple shear Thus in a strained material a shear stress is always accompanied by a balancing shear of same intensity at right angles to itself. This balancing shear is called “complementary shear”. τ
A
B
The shear and the complementary shear together constitute a state of simple shear
τ'= τ
D
τ'= τ τ
C
Direct stress due to pure shear Consider a square element of side ‘a’ subjected to shear stress as shown in the Fig.(a). Let the thickness of the square be unity. τ τ A
A B B
τ
D
a a τ Fig.(a). τ
C
τ
D
a a τ Fig.(b).
C
τ
Fig.(b) shows the deformed shape of the element. The length of diagonal DB increases, indicating that it is subjected to tensile stress. Similarly the length of diagonal AC decreases indicating that compressive stress.
Direct stress due to pure shear Now consider the section, ADC of the element, Fig.(c). X
A
σn a a
A
τ
D
a a
D
( 2 )a
C
τ
C
Fig.(c).
Resolving the forces in σn direction, i.e., in the X-direction shown
For equilibrium
∑ Fx = 0
=σn ×
(
2 × a ×1 − 2(τ × a × cos 45)
)
σn =τ
Direct stress due to pure shear Therefore the intensity of normal tensile stress developed on plane BD is numerically equal to the intensity of shear stress.
Similarly it can be proved that the intensity of compressive stress developed on plane AC is numerically equal to the intensity of shear stress.
POISSON’S RATIO Poisson’s Ratio: Consider the rectangular bar shown in Fig.(a) subjected to a tensile load. Under the action of this load the bar will increase in length by an amount δL giving a longitudinal strain in the bar of δl εl = l Fig.(a)
POISSON’S RATIO The bar will also exhibit, reduction in dimension laterally, i.e. its breadth and depth will both reduce. These change in lateral dimension is measured as strains in the lateral direction as given below. δb δd ε lat = − = − b d The associated lateral strains will be equal and are of opposite sense to the longitudinal strain. Provided the load on the material is retained within the elastic range the ratio of the lateral and longitudinal strains will always be constant. This ratio is termed Poisson’s ratio (µ)
(− δd ) d = POISSON’S RATIO = δl Longitudinal strain l
Lateral strain
(− δb ) b OR δl l
Poisson’s Ratio = µ
For most engineering metals the value of µ lies between 0.25 and 0.33 y Lz P Lx z Poisson’s Ratio Lateral strain Strain in the direction of load applied
− δl y = ly
− δl z
In general
Ly P x
=
δl x
lx
OR = δl x
lz
lx
Poisson’s Ratio = µ
In general Px
Lx
y
Lz Ly Px
x
z
Strain in Y-direction =
Strain in X-direction = εx
=
δl x lx εy
=
δl y ly =µ
δl x lx Strain in Z-direction = εz
=
δl z lz =µ
δl x lx Load applied in Y-direction y
Lz
Py
Ly
Lx
x Py
z
Poisson’s Ratio
=
Lateral strain Strain in the direction of load applied
− δl x =
δl y
lx
− δl z
ly
OR = δl y
lz
ly
Strain in X-direction = εx
=
δl x lx =µ
δl y ly Load applied in Z-direction y
Lz
Pz
Ly
x z Pz Lateral strain Strain in the direction of load applied
Lx
− δl x =
Poisson’s Ratio
=
δl z
lx
− δl y
OR
=
lz
δl z
ly
lz
Strain in X-direction = εx
=
δl x lx =µ
δl z lz Load applied in X & Y direction y
Lz
Py
Ly Px
Strain in X-direction = εx x
Px
Lx
=
σx
E
−µ
σy
E
z
Py
Strain in Y-direction = εy
=
σy
E
−µ
σx
E
Strain in Z-direction = εz
= −µ
σy
E
−µ
σx
E
Py
y
Px
Pz
General case: Strain in X-direction = εx
Px
x
z
Pz
Py
εx = σy σx σz
σx
E
−µ
σy
E
−µ
σz
E
Strain in Y-direction = εy
E Strain in Z-direction = εz σy σx σz εz = −µ −µ
E E E
εy =
σy
E
−µ
σx
E
−µ
σz
σz σx
σy
Bulk Modulus Bulk Modulus A body subjected to three mutually perpendicular equal direct stresses undergoes volumetric change without distortion of shape. If V is the original volume and dV is the change in volume, then dV/V is called volumetric strain. A body subjected to three mutually perpendicular equal direct stresses then the ratio of stress to volumetric strain is called Bulk Modulus. σ = Bulk modulus, K ⎛ dV ⎞ ⎜ ⎟ V ⎠ ⎝
Relationship between volumetric strain and linear strain Consider a cube of side 1unit, subjected to three mutually perpendicular direct stresses as shown in the figure. Relative to the unstressed state, the change in volume per unit volume is dV = 1 − (1 + ε x )(1 + ε y )(1 + ε z ) = 1 − 1 + ε x + ε y + ε z 1 = εx +εy +εz = change in volume per unit volume
[
]
[
]
Relationship between volumetric strain and linear strain
Volumetric strain dV = εx +εy +εz V σy ⎛σx σ ⎜ =⎜ −µ −µ z E E ⎝ E
⎞ ⎛σ y σ σ ⎟ +⎜ −µ x −µ z ⎟ ⎜ E E E ⎠ ⎝
⎞ ⎛σ ⎟ +⎜ z −µ σy −µ σx ⎞ ⎟ ⎟ ⎜ E E ⎟ ⎠ ⎝E ⎠
1 − 2µ (σ x + σ y + σ z ) = E
For element subjected to uniform hydrostatic pressure,
σx =σy =σz =σ dV 1 − 2 µ (σ x + σ y + σ z ) = V E dV 1 − 2 µ (3σ ) = V E
K=
σ
⎛ dV ⎞ ⎜ ⎟ ⎝ V ⎠
E K= = bulk modulus 3(1 − 2 µ ) or E = 3K (1 - 2 µ )
Relationship between young’s modulus of elasticity (E) and modulus of rigidity (G) :A1 A H
45˚
a
B1 B
φ
D
a φτ τ C
Consider a square element ABCD of side ‘a’ subjected to pure shear ‘τ’. DA'B'C is the deformed shape due to shear τ. Drop a perpendicular AH to diagonal A'C. Strain in the diagonal AC = τ /E – µ (- τ /E) [ σn= τ ] = τ /E [ 1 + µ ] -----------(1) Strain along the diagonal AC=(A'C–AC)/AC=(A'C–CH)/AC=A'H/AC
In ∆le AA'H Cos 45˚ = A'H/AA' A'H= AA' × 1/√2 AC = √2 × AD ( AC = √ AD2 +AD2) Strain along the diagonal AC = AA'/ (√2 × √2 × AD)=φ/2 ----(2) Modulus of rigidity = G = τ /φ φ = τ /G Substituting in (2) Strain along the diagonal AC = τ /2G -----------(3) Equating (1) & (3) τ /2G = τ /E[1+µ] E=2G(1+ µ)
Relationship between E, G, and K:We have E = 2G( 1+ µ) -----------(1) E = 3K( 1-2µ) -----------(2) Equating (1) & (2) 2G( 1+ µ) =3K( 1- 2µ) 2G + 2Gµ=3K- 6Kµ µ= (3K- 2G) /(2G +6K) Substituting in (1) E = 2G[ 1+(3K – 2G)/ (2G+6K)] E = 18GK/( 2G+6K) E = 9GK/(G+3K)
Working stress: It is obvious that one cannot take risk of loading a member to its ultimate strength, in practice. The maximum stress to which the material of a member is subjected to in practice is called working stress. This value should be well within the elastic limit in elastic design method. Factor of safety: Because of uncertainty of loading conditions, design procedure, production methods, etc., designers generally introduce a factor of safety into their design, defined as follows
Factor of safety = Maximum stress or Yield stress (or proof stress) Allowable working Allowable working stress stress
Malleability: A property closely related to ductility, which defines a material’s ability to be hammered out in to thin sheets Homogeneous: A material which has a uniform structure throughout without any flaws or discontinuities. Isotropic: If a material exhibits uniform properties throughout in all directions ,it is said to be isotropic. Anisotropic: If a material does not exhibit uniform properties throughout in all directions ,it is said to be anisotropic or nonisotropic.
Exercise Problems Q1. An aluminum tube is rigidly fastened between a brass rod and steel rod. Axial loads are applied as indicated in the figure. Determine the stresses in each material and total deformation. Take Ea=70GPa, Eb=100GPa, Es=200GPa
Aa=1000mm2
20kN
Ab=700mm2
brass 500mm
15kN
15kN
As=800mm2
aluminum 600mm
steel 700mm
10kN
Ans: σb=28.57MPa, σa=5MPa, σs=12.5MPa, δl = - 0.142mm
Q2.
A 2.4m long steel bar has uniform diameter of 40mm for a length of 1.2m and in the next 0.6m of its length its diameter gradually reduces to ‘D’ mm and for remaining 0.6m of its length diameter remains the same as shown in the figure. When a load of 200kN is applied to this bar extension observed is equal to 2.59mm. Determine the diameter ‘D’ of the bar. Take E =200GPa 200kN
1000mm 500mm 500mm
200kN
Ф = 40mm Ф = D mm
Q3. The diameter of a specimen is found to reduce by 0.004mm when it is subjected to a tensile force of 19kN. The initial diameter of the specimen was 20mm. Taking modulus of rigidity as 40GPa determine the value of E and µ Ans: E=110GPa, µ=0.36 Q.4 A circular bar of brass is to be loaded by a shear load of 30kN. Determine the necessary diameter of the bars (a) in single shear (b) in double shear, if the shear stress in material must not exceed 50MPa. Ans: 27.6, 19.5mm
Q.5 Determine the largest weight W that can be supported by the two wires shown. Stresses in wires AB and AC are not to exceed 100MPa and 150MPa respectively. The cross sectional areas of the two wires are 400mm2 for AB and 200mm2 for AC. Ans: 33.4kN B 300 A 450 C
W
Q.6 A homogeneous rigid bar of weight 1500N carries a 2000N load as shown. The bar is supported by a pin at B and a 10mm diameter cable CD. Determine the stress in the cable D Ans: 87.53MPa
A 3m 2000 N
C 3m
B
Q.7. A stepped bar with three different cross-sectional areas, is fixed at one end and loaded as shown in the figure. Determine the stress and deformation in each portions. Also find the net change in the length of the bar. Take E = 200GPa
300mm2 20kN 450mm2 250mm2 40kN 10kN
250mm
320mm
270mm
Ans: -33.33, -120, 22.2MPa, -0.042, -0.192, 0.03mm, -0.204mm
Q.8
a) b) c)
The coupling shown in figure is constructed from steel of rectangular cross-section and is designed to transmit a tensile force of 50kN. If the bolt is of 15mm diameter calculate: The shear stress in the bolt; The direct stress in the plate; The direct stress in the forked end of the coupling.
Ans: a)141.5MPa, b)166.7MPa, c)83.3MPa
Q.9 The maximum safe compressive stress in a hardened steel punch is limited to 1000MPa, and the punch is used to pierce circular holes in mild steel plate 20mm thick. If the ultimate shearing stress is 312.5MPa, calculate the smallest diameter of hole that can be pierced. Ans: 25mm Q.10 A rectangular bar of 250mm long is 75mm wide and 25mm thick. It is loaded with an axial tensile load of 200kN, together with a normal compressive force of 2000kN on face 75mm×250mm and a tensile force 400kN on face 25mm×250mm. Calculate the change in length, breadth, thickness and volume. Take E = 200GPa & µ=0.3 Ans: 0.15,0.024,0.0197mm, 60mm3
Q.11 A piece of 180mm long by 30mm square is in compression under a load of 90kN as shown in the figure. If the modulus of elasticity of the material is 120GPa and Poisson’s ratio is 0.25, find the change in the length if all lateral strain is prevented by the application of uniform lateral external pressure of suitable intensity. 90kN
30 180
30
Ans: 0.125mm
Q.12 Define the terms: stress, strain, elastic limit, proportionality limit, yield stress, ultimate stress, proof stress, true stress, factor of safety, Young’s modulus, modulus of rigidity, bulk modulus, Poisson's ratio, Q.13 Draw a typical stress-strain diagram for mild steel rod under tension and mark the salient points. Q.14 Diameter of a bar of length ‘L’ varies from D1 at one end to D2 at the other end. Find the extension of the bar under the axial load P Q.15 Derive the relationship between Young’s modulus and modulus of rigidity.
Q.16 Derive the relationship between Young’s modulus and Bulk modulus. Q.17 A flat plate of thickness ‘t’ tapers uniformly from a width b1at one end to b2 at the other end, in a length of L units. Determine the extension of the plate due to a pull P. Q.18 Find the extension of a conical rod due to its own weight when suspended vertically with its base at the top. Q.19 Prove that a material subjected to pure shear in two perpendicular planes has a diagonal tension and compression of same magnitude at 45o to the planes of shear.
Q.20
For a given material E=1.1×105N/mm2& G=0.43×105N/mm2 .Find bulk modulus & lateral contraction of round bar of 40mm diameter & 2.5m length when stretched by 2.5mm.
ANS: K=83.33Gpa, Lateral contraction=0.011mm Q.21 The modulus of rigidity of a material is 0.8×105N/mm2 , when 6mm×6mm bar of this material subjected to an axial pull of 3600N.It was found that the lateral dimension of the bar is changed to 5.9991mm×5.9991mm. Find µ & E. ANS: µ=0.31, E= 210Gpa.
7. STATICALLY INDETERMINATE MEMBERS & THERMAL STRESSES
STATICALLY INDETERMINATE MEMBERS Structure for which equilibrium equations are sufficient to obtain the solution are classified as statically determinate. But for some combination of members subjected to axial loads, the solution cannot be obtained by merely using equilibrium equations. The structural problems with number of unknowns greater than the number independent equilibrium equations are called statically indeterminate. The following equations are required to solve the problems on statically indeterminate structure. 1) Equilibrium equations based on free body diagram of the structure or part of the structure. 2) Equations based on geometric relations regarding elastic deformations, produced by the loads.
COMPOUND BAR
Material(2) Material(1)
L1
L2
W A compound bar is one which is made of two or more than two materials rigidly fixed, so that they sustain together an externally applied load. In such cases (i) Change in length in all the materials are same. (ii) Applied load is equal to sum of the loads carried by
(dL)1 = (dL)2 (σ1/ E1)L1 = (σ2 /E2)L2 σ1 = σ2 ×( E1/E2)(L1/L2) E1/E2 is called modular ratio Total load = load carried by material (1) + load carried by material(2) W = σ 1 A1 + σ 2 A2 (2) (1)
From Equation (1) & (2) σ1 and σ2 can be calculated
Temperature Stress
A L A L B B B
B´ P αTL
A
L Any material is capable of expanding or contracting freely due to rise or fall in temperature. If it is subjected to rise in temperature of T˚C, it expands freely by an amount ‘αTL’ as shown in figure. Where α is the coefficient of linear expansion, T˚C = rise in temperature and L = original length.
From the above figure it is seen that ‘B’ shifts to B' by an amount ‘αTL’. If this expansion is to be prevented a compressive force is required at B'. Temperature strain = αTL/(L + αTL) ≈ αTL/L= αT Temperature stress = αTE Hence the temperature strain is the ratio of expansion or contraction prevented to its original length. If a gap δ is provided for expansion then Temperature strain = (αTL – δ) / L Temperature stress = [(αTL – δ)/L] E
Temperature stress in compound bars:Material(1) Material(2)
α1TL
x P1
(dL)2 (dL)1 P2 α2TL
x ∆ When a compound bar is subjected to change in temperature, both the materials will experience stresses of opposite nature. Compressive force on material (1) = tensile force on material (2) load) σ1A1 = σ2A2 (there is no external σ1=( σ2A2)/A1 (1)
As the two bars are connected together, the actual position of the bars will be at XX. α1TL – (dL)1 = α2TL + (dL)2 α1TL – (σ1 / E1) L =α2TL + (σ2 / E2) L αT – (σ1 / E1) = α2T + σ2 / E2 --------------------------(2) From (1) and (2) magnitude of σ1 and σ2 can be found out.
Actual expansion in material (1) = actual expansion in material (2)
Exercise problems Q.1 A circular concrete pillar consists of six steel rods of 22mm diameter each reinforced into it. Determine the diameter of pillar required when it has to carry a load of 1000kN. Take allowable stresses for steel & concrete as 140Mpa & 8Mpa respectively. The modular ratio is 15 ANS: D=344.3mm
Q.2 Determine the stresses & deformation induced in Bronze & steel as shown in figure. Given As=1000mm2, Ab=600mm2, Es= 200Gpa, Eb= 83Gpa ANS: ( σb=55Mpa, σs=93.5Mpa, dLs=dLb=0.093mm)
Bronze
Bronze
Steel
160kN
Q..3 A cart wheel of 1.2m diameter is to be provided with steel tyre. Assume the wheel to be rigid. If the stress in steel does not exceed 140MPa, calculate minimum diameter of steel tyre & minimum temperature to which it should be heated before on to the wheel. ANS: d=1199.16mm T=58.330C Q.4 A brass rod 20mm diameter enclosed in a steel tube of 25mm internal diameter & 10mm thick. The bar & the tube are initially 2m long & rigidly fastened at both the ends. The temperature is raised from 200C to 800C. Find the stresses in both the materials.
If the composite bar is then subjected to an axial pull of 50kN, find the total stress. Es=200GPa, Eb=80GPa, αs=12×10-6/0C, αb=19×10-6/0C. ANS: σb=8.81N/mm2 ( C ) , σs=47.99N/mm2( T )
8. STRESSES ON INCLINED PLANES
INTRODUCTION
The state of stress on any plane in a strained body is said to be ‘Compound Stress’, if, both Normal and Shear stresses are acting on that plane. For, example, the state of stress on any vertical plane of a beam subjected to transverse loads will, in general, be a Compound Stress. In actual practice the state of Compound Stress is of more common occurrence than Simple state of stress. In a compound state of stress, the normal and shear stress may have a greater magnitude on some planes which are inclined (or, Oblique) to the given stress plane.
Hence in compound state of stresses it is necessary to find the following (i) The normal and shear stress on a plane which is inclined (Oblique) to the given stress plane; (ii) The inclination of max. and min. normal stress planes and values of the normal stress (max. / min.) on them; (iii) The inclination of max. shear stress planes and the values of the shear stress (max.) on them.
EQUATIONS, NOTATIONS & SIGN CONVENTIONS :
Consider an element ABCD subjected to a state of compound stress as shown in the Fig. Let: C Normal Stress in x- direction σx σx Normal Stress in y- direction σy D τ Shear Stresses in x & y – directions θ Angle made by inclined plane AE wrt vertical τ σθ Normal Stress on inclined plane τθ Shear Stress on inclined plane σy τθ E B σθ θ A
τ σx
σy
(i) Normal & Shear stress on plane inclined (Oblique) to given stress plane:
Normal Stress, σθ, and Shear stress, τθ, on inclined plane are given by:
⎛σ x +σ y σθ = ⎜ ⎜ 2 ⎝ ⎞ ⎛σ x −σ y ⎟+⎜ ⎟ ⎜ 2 ⎠ ⎝ y ⎞ ⎟ cos 2θ + τ sin 2θ − − − (1) ⎟ ⎠
τθ
⎛σ = ⎜ ⎜ ⎝
x
−σ 2
⎞ ⎟ sin 2 θ − τ cos 2 θ − − − ( 2 ) ⎟ ⎠
(ii) The inclination of max. and min. normal stress planes and the values of the normal stress (max. / min.) on them Let, θP be the inclination of the plane of max. or min. normal stress and σP be the value of the max. or min. normal stress on that plane, then, from Eqn. (1): ⎛σ x +σ y ⎞ ⎛σ x −σ y ⎞ ⎟+⎜ ⎟ cos 2θ P + τ sin 2θ P - - - (1) σP = ⎜ ⎜ ⎟ ⎜ ⎟
⎝ 2 ⎠ ⎝ 2 ⎠
Thus, the condition for max. or min. normal stress to occur on a dσ P =0 For σ P to be max. or min., plane is, shear stress on that dθ plane should be zero. ⎛ σ x −σ y ⎞ ⇒⎜ ⎟ ⎜ 2 ⎟(− 2 sin 2θ P ) + τ (2 cos 2θ P ) = 0 These planes on which shear ⎠ ⎝ stress is zero and the normal ⎛ σ x −σ y ⎞ stress on them being either the ⇒⎜ ⎜ 2 ⎟(sin 2θ P ) − τ (cos 2θ P ) = 0 ⎟ ⎝ ⎠ max. or the min. are called ⇒ τθ = 0 ‘PRINCIPAL PLANES’.
P
From,τ θ P = 0, we have, ⎛σ x −σ y ⎞ τ ⎜ ⎟(sin 2θ P ) − τ (cos 2θ P ) = 0 ⇒ tan 2θ P = - - - (3) ⎜ 2 ⎟ ⎛ σ x −σ y ⎞ ⎝ ⎠ ⎜ ⎜ 2 ⎟ ⎟ ⎝ ⎠
The above Eqn. (3), gives two values for θP, which differ by 900. Thus, there are two mutually perpendicular Principal planes, on which there are only normal stresses, shear stress being zero on them. On one of them, the value of the normal stress is the max.; it is called the ‘Major Principal plane’, the max. normal stress on it is called the ‘Major Principal Stress’. On the other principal plane, the value of the normal stress is the min.; it is called the ‘Minor Principal plane’, the min. normal stress on it is called the ‘Minor Principal Stress’.
From tan 2θ P =
τ
⎛σ x −σ y ⎜ ⎜ 2 ⎝ ⎞ ⎟ ⎟ ⎠
2
, we get,
] 2+
τ2
2θP
y /2
±√ [(σ
x
(σx-σy)/ 2
-σ
sin 2θ P = ±
τ
⎡ (σ X − σ Y ) ⎤ ⎢ ⎥ +τ 2 ⎣ ⎦ (σ X − σ Y ) / 2 ⎡ (σ X − σ Y ) ⎤ ⎢ ⎥ +τ 2 ⎣ ⎦
2
2
τ
cos 2θ P = ±
2
Substituting for sin 2θP and cos 2θP in Eqn. (1), and simplifying, we get the equation for principal stresses as: σP =
(σ
x
+σ 2
y
)±
⎛σ x −σ ⎜ ⎜ 2 ⎝
y
⎞ ⎟ +τ ⎟ ⎠
2
2
− − − (4)
The above equation (4) gives two values for principal stresses. The numerically max. of the two values (+ ve or − ve) is the Major Principal Stress, (σMajor or σMax); The numerically min. (+ ve or − ve) is the Minor Principal Stress (σMinor or σMin).
(iii) Inclination of max. shear stress planes, Max. shear stress Equation. Let, θS be the inclination of the plane of max. or min. shear stress and τS be the value of the max. or min. shear stress on that plane, then, from Eqn. (2):
τS
⎛σ = ⎜ ⎜ ⎝
x
−σ 2
y
⎞ ⎟ sin 2 θ S − τ cos 2 θ S ⎟ ⎠
− − − (2)
dτ For τ S to be max. or min., S = 0 dθS ⎛ σ x −σ y ⎞ ⇒⎜ ⎜ 2 ⎟(2 cos 2θ S ) − τ (− 2 sin 2θ S ) = 0 ⎟ ⎝ ⎠ ⎛ σ x −σ y ⎞ ⎜ ⎜ 2 ⎟ ⎟ ⎝ ⎠ - - - (5) ⇒ tan 2θ S = − NOTE : We have tan2θ P × tan 2θ S = −1
τ
Eqn. (5) gives two values for θS, which differ by 900. Thus, there are two mutually perpendicular planes, on which shear stress are max.; numerically equal but opposite in sense. The planes of Max. Shear stresses are inclined at 450 to the Principal planes.
2] 2 +
⎛σ x −σ y ⎜ ⎜ 2 From tan 2θ S = − ⎝
y )/
−σ
±√ [(σ
x
(σx-σy)/ 2
cos 2θ s = ±
τ
τ
⎞ ⎟ ⎟ ⎠ , we get,
τ2
2θS
τ
sin 2θ S = ±
⎡ (σ x − σ y ) ⎤ ⎥ +τ ⎢ 2 ⎦ ⎣ (σ x − σ y )/ 2 ⎡ (σ x − σ y ) ⎤ ⎢ ⎥ +τ 2 ⎣ ⎦
2
2
2
2
Substituting for sin 2θS and cos 2θS in Eqn. (2), we get the equation for Max. shear stresses as: τ max . = ± ⎜ ⎜
⎝ ⇒ τ max . ⎛σ x −σ y ⎞ 2
2 2
⎟ +τ ⎟ ⎠
⎛ σ Major − σ Minor = ±⎜ ⎜ 2 ⎝
⎞ ⎟ ⎟ ⎠
⎤ ⎥ ⎥ ⎥ − − − (6 ) ⎥ ⎥ ⎦
The above equation (6) gives two values for Max. shear stresses, which are numerically equal but opposite in sense.
NOTATIONS & SIGN CONVENTIONS σx σy Normal Stress in x- direction Normal Stress in y- direction Shear Stresses in x & y – directions Angle made by inclined plane wrt vertical Normal Stress on inclined plane AE σx τ C σθ D
σy τ τθ E B θ A
τ θ σθ
σx
τθ θP σP θS
Shear Stress on inclined plane AE σy Inclination of Principal planes Principal stresses Inclination of Max. shear stress planes [θS = θP + 450]. All the parameters are shown in their +ve sense in the Fig.
Normal stresses, σ Tensile stresses +ve. Shear Stresses, τ, in x – direction & Inclined Plane Clockwise +ve. Anti-Clockwise +ve. Shear Stresses, τ, in y – direction Angle, θ measured w r t vertical, Anti-Clockwise +ve.
Exercise Problems:
Q.1 The principal stresses at a point in a strained material are 80 MPa(C) and 40 MPa(T). Find the normal, tangential and resultant stress on a plane inclined at 50o to the major principal plane. [Ans: - 9.58MPa, -59MPa].
Q.2 The stresses in a strained material is as shown in Fig. Find the normal and shear stresses on plane inclined at 30o to the horizontal. Also determine the intensity and position of the plane upon which there is only shear stress. Sketch the plane.
60MPa 90 MPa
30o
Ans: σ = − 22.5MPa, τ = − 64.95MPa. Pure shear plane σ = 0, Ө = 39.23o w.r.t. horizontal. τpure = − 73.48MPa.
Q.3 A plane element is subjected to the system of stresses as shown in Fig. Determine (i) the principal stresses and inclination of their planes (ii) maximum shearing stresses and inclination of their planes. Represent your answers in neat sketches.
160 MPa 80 MPa 200 MPa
Ans: σ1 = 262.46MPa, σ2 = 97.54MPa. Өp = 37.98 or 127.98 with horizontal τtmax = 82.46, Өs = 82.98 or 172.98 with horizontal
Q.4 At a point in a structural member subjected to stresses as shown in fig. determine the principal stresses and the maximum shear stress. Also determine and sketch planes on which these stresses act. 80 [Ans:131.23, 48.77, − 37.98o, − 127.98o 41.23, 7o, 970 , Angles w.r.t. horizontal] 100
4 0 Q.5 At a point in a material under stress, the intensity of resultant stress on a certain plane is 60 MPa, directed outwards and inclined at 30o to the normal of that plane. The stress on the plane at right angles to this has a normal stress component of 40 Mpa (T). Find (i) the principal stresses and inclination of their planes, (ii) the maximum shear stresses and inclination of their planes . [Ans: 76.57MPa, 15.39MPa, 39.36o , 129.36o and 30.59 MPa, 84.36o , 174.36o, Angles w.r.t. horizontal]
9. Stresses due to fluid pressure in thin cylinders
9 -THIN CYLINDERS
INTRODUCTION: In many engineering applications, cylinders are frequently used for transporting or storing of liquids, gases or fluids. Eg: Pipes, Boilers, storage tanks etc. These cylinders are subjected to fluid pressures. When a cylinder is subjected to internal pressure, at any point on the cylinder wall, three types of stresses are induced on three are, They mutually perpendicular planes. 1. Hoop or Circumferential Stress (σC) – This is directed along the tangent to the circumference and tensile in nature. Thus, there will be increase in diameter.
2. Longitudinal Stress (σL) – This stress is directed along the length of the cylinder. This is also tensile in nature and tends to increase the length.
3. Radial pressure (σ r) – It is compressive in nature. Its magnitude is equal to fluid pressure on the inside wall and zero on the outer wall if it is open to atmosphere.
σC p σC
σC
σL p
σL p σL σr
σC
σL
1. Hoop /Circumferential Stress (σC)
2. Longitudinal Stress (σL) σr 3. Radial Stress (σr) σC Element on the cylinder wall subjected to these three stresses
σL
σL
σC
σr
A cylinder or spherical shell is considered to be thin when the metal thickness is small compared to internal diameter. i. e., when the wall thickness, ‘t’ is equal to or less than ‘d/20’, where ‘d’ is the internal diameter of the cylinder or shell, we consider the cylinder or shell to be thin, otherwise thick. Magnitude of radial pressure is very small compared to other two stresses in case of thin cylinders and hence neglected.
p×d Circumferential stress, σ c = ........................(1) 2× t
p×d Longitudin stress, σL = al ...................( 2) 4× t
Maximum Shear stress, τmax = (σc- σL) / 2 τmax = ( p x d) / (8 x t) Where p = internal fluid pressure d= internal diameter,
t = thickness of the wall
EVALUATION OF STRAINS σ L=(pd)/(4t)
σ C=(pd)/(2t)
σ C=(pd)/(2t)
σ L=(pd)/(4t) A point on the surface of thin cylinder is subjected to biaxial stress system, (Hoop stress and Longitudinal stress) mutually perpendicular to each other, as shown in the figure. The strains due to these stresses i.e., circumferential and longitudinal are obtained by applying Hooke’s law and Poisson’s theory for elastic materials.
Circumferential strain, ε C : σC σL σL σL −µ× = 2× −µ× εC = E E E E σL = × (2 − µ) E
C=(pd)/(2t)
L=(pd)/(4t)
C=(pd)/(2t)
L=(pd)/(4t)
i.e.,
δd p×d εC = = × (2 − µ)................................(3) d 4× t × E
Longitudinal strain, ε L : σC σL (2 × σ L ) σ L σL −µ× = −µ× = × (1 − 2 × µ) εL = E E E E E
i.e., δl p×d εL = = × (1 − 2 × µ)................................(4) L 4× t × E
dv VOLUMETRIC STRAIN, V
σL=(pd)/(4t)
π 2 σC=(pd)/(2t σC=(pd)/(2t) We have volume, V = × d × L. 4 π 2 π dV = × d × dL + L × × 2 × d × dd σL=(pd)/(4t) 4 4 π 2 π × d × dL + L × × 2 × d × dd dv 4 dL dd 4 = = + 2× π 2 V L d ×d ×L 4 p×d p×d dV = (1 − 2 × µ) + 2 × (2 − µ) = εL+2× εC V 4× t × E 4× t × E dv p×d i.e., = (5 − 4 × µ).................(5) V 4× t × E
JOINT EFFICIENCY The cylindrical shells like boilers are having two types of joints namely Longitudinal and Circumferential joints. Due to the holes for rivets, the net area of cross section decreases and hence the stresses increase. If the efficiencies of these joints are known, the stresses can be calculated as follows. Let ηL=Efficiency of Longitudinal joint and ηC=Efficiency of Circumferential joint. Circumferential stress is given by,
p×d σC = 2 × t × ηL
.............(1)
Longitudinal stress is given by,
p×d σL = .......... ...(2) 4×t×ηC
Note: In longitudinal joint, the circumferential stress is developed and in circumferential joint, longitudinal stress is developed.
Exercise Problems
Q.1 Calculate the circumferential and longitudinal strains for a boiler of 1000mm diameter when it is subjected to an internal pressure of 1MPa. The wall thickness is such that the safe maximum tensile stress in the boiler material is 35 MPa. Take E=200GPa and µ= 0.25. (Ans: ε C=0.0001531, ε L=0.00004375) Q.2 A water main 1m in diameter contains water at a pressure head of 120m. Find the thickness of the metal if the working stress in the pipe metal is 30 MPa. Take unit weight of water = 10 kN/m3. (Ans: t=20mm)
Q.3 A gravity main 2m in diameter and 15mm in thickness. It is subjected to an internal fluid pressure of 1.5 MPa. Calculate the hoop and longitudinal stresses induced in the pipe material. If a factor of safety 4 was used in the design, what is the ultimate tensile stress in the pipe material? (Ans: σC=100 MPa, σL=50 MPa, σU=400 MPa) Q.4 At a point in a thin cylinder subjected to internal fluid pressure, the value of hoop strain is 600×10-4 (tensile). Compute hoop and longitudinal stresses. How much is the percentage change in the volume of the cylinder? Take E=200GPa and µ= 0.2857. (Ans: σC=140 MPa, σL=70 MPa, %age change=0.135%.)
Q.5 A cylindrical tank of 750mm internal diameter and 1.5m long is to be filled with an oil of specific weight 7.85 kN/m3 under a pressure head of 365 m. If the longitudinal joint efficiency is 75% and circumferential joint efficiency is 40%, find the thickness of the tank required. Also calculate the error of calculation in the quantity of oil in the tank if the volumetric strain of the tank is neglected. Take permissible tensile stress as 120 MPa, E=200GPa and µ= 0.3 for the tank material. (Ans: t=12 mm, error=0.085%.)
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