BUCKLING OF STRUTS
EXPERIMENT 2 – BUCKLING OF STRUTS
Aim
We can find two failure points in members exposed to axial compressive forces, also called struts. The first one is squashing of the member, which will only happened if a member is “Stocky”. The second, and most common one, is buckling of the member. This failure happens without much warning, and therefore it is important for engineers to be able to calculate what load on structural element will cause buckling. In this experiment, we will observe the failure mode of buckling, by loading five slender aluminium elements in compression.
Procedure
In this experiment, we used a Digital Force Display (DFD), to observe the failure mode of buckling, of five slender aluminium elements. We used de DFD to put load on the elements in compression, starting with the longest one, and kept putting on load until there were no further increase in the load. We repeated this for each of the five slender aluminium elements four times.
Result
STRUT NUMBER
LENGTH (mm)
BUCKLING LOAD
(N)
1 2 3 4
AVERAGE
BUCKLING LOAD (N)
1
520
-35
-40
-36
-40
-37.75
2
470
-40
-56
-40
-52
-47
3
420
-53
-64
-52
-64
-58.25
4
370
-75
-76
-75
-77
-75.75
5
320
-99
-90
-98
-90
-94.25
Table 1 - The result of the experiment
Table 2 – Graph of buckling load vs. 1/L2 By finding the gradient of the graph, we can calculate π2*E*I : Where I = b3d/12 = (23*20)/12 = 13.33 mm4
We chose two points on the line and get:
1. x = 0.000004 and y = 42.0598
2. x = 0.000008 and y = 79.51968 then
G = π2*E*I y2-y1/x2-x1 = π2*E*(b3d/12) (42.1-79.52) / (4.0E-6 – 8.0E-6) = π2*E*13.33 E = 65482.29 N/mm2 E = 65.5 GPa
Discussion
We know that the actual Young’s modulus of Aluminium is 69GPa, and we can see that we got a slightly different answer, 65.5GPa, when calculating it from the experiment results. We get an error of 5.1%, which is small in this case. This difference can come from the fact that the struts have been used many times,
Aim
We can find two failure points in members exposed to axial compressive forces, also called struts. The first one is squashing of the member, which will only happened if a member is “Stocky”. The second, and most common one, is buckling of the member. This failure happens without much warning, and therefore it is important for engineers to be able to calculate what load on structural element will cause buckling. In this experiment, we will observe the failure mode of buckling, by loading five slender aluminium elements in compression.
Procedure
In this experiment, we used a Digital Force Display (DFD), to observe the failure mode of buckling, of five slender aluminium elements. We used de DFD to put load on the elements in compression, starting with the longest one, and kept putting on load until there were no further increase in the load. We repeated this for each of the five slender aluminium elements four times.
Result
STRUT NUMBER
LENGTH (mm)
BUCKLING LOAD
(N)
1 2 3 4
AVERAGE
BUCKLING LOAD (N)
1
520
-35
-40
-36
-40
-37.75
2
470
-40
-56
-40
-52
-47
3
420
-53
-64
-52
-64
-58.25
4
370
-75
-76
-75
-77
-75.75
5
320
-99
-90
-98
-90
-94.25
Table 1 - The result of the experiment
Table 2 – Graph of buckling load vs. 1/L2 By finding the gradient of the graph, we can calculate π2*E*I : Where I = b3d/12 = (23*20)/12 = 13.33 mm4
We chose two points on the line and get:
1. x = 0.000004 and y = 42.0598
2. x = 0.000008 and y = 79.51968 then
G = π2*E*I y2-y1/x2-x1 = π2*E*(b3d/12) (42.1-79.52) / (4.0E-6 – 8.0E-6) = π2*E*13.33 E = 65482.29 N/mm2 E = 65.5 GPa
Discussion
We know that the actual Young’s modulus of Aluminium is 69GPa, and we can see that we got a slightly different answer, 65.5GPa, when calculating it from the experiment results. We get an error of 5.1%, which is small in this case. This difference can come from the fact that the struts have been used many times,