Experiment Title: Bukling of StrutsObjectives:To determine the bucking load for a pinned ended strut.Theory:Consider a simply-supported strut with length L and flexural rigidity EI loaded axially with a force P (Figure 1). At buckling it assumes a profile (mode y(x), where at x we have a bending moment M introduced by the mode shape eccentricity,M=-Py (1)

Bending theory states that M=EI curvature, which from linear bending theory can be written: (2)

Combining equation (1) and (2) we obtain the differential equation: (3) Rewriting this in the form: (4)gives the form of the equation for a free undamped oscillator, or the equation for simple harmonic motion. It is useful to differentiate this equation twice with respect to x so that two boundary conditions can be applied at each support. (5)

The solution of this equation has the form:

where A, B, C and D are constants dependent on the boundary conditions of the strut and w is defined:(6)

In the case shown in Figure 1. (7) (8)where primes denote differentiation with respect to x, a nontrivial solution (A0), is given by: (9)Unless the strut is restrained somewhere the lowest value of P (n=1) is the practical value for buckling-the Euler load (PE). (10)In reality the equilibrium paths are not flat, they exhibit a stable-symmetric profile. However, this stable nature is not significant until deflections get very large, so they are assumed to be flat and the critical load is a fair estimate of the failure load when strut buckle. We could consider any boundary condition combination by obtaining a solution of y(x). However, there is a more intuitive way of analyzing different cases using the concept of Effective length (Le). Common cases follow, the adjusted Euler loads take account of the effective lengths thus: (11)where Le is the effective length of column and varies depend on end conditions. The effective lengths corresponding to the simple support ends and fixed both ends are shown in Figure 2

Procedures:1. The digital indicator is switched to warm up for 10 minutes.2. A specimen is chosen and measured its length. The width and thickness of the beam is 3.1 mm and 19.8 mm respectively.3. The theoretical buckling load for a strut with pinned end condition is calculated. This is to ensure that the load applied to the strut does not exceed the buckling load.4. The grooved support is placed into the slot of the attachment for the end conditions and the side screws were tightened. 5. The top platen is moved to bring the distance between the two supports closer to the length of the struts.6. The tare button on the digital indicator is pressed to set the reading to zero.7. The specimen is placed in the groove of the top support.8. The jack is adjusted while holding the specimen so that the lower end of the specimen just rest in the groove of the bottom support.9. The position of the dial gauge is check to ensure that it is at the mid-length of the specimen. The dial gauge is set to zero.10. The tare button is pressed to set to zero.11. The specimen is loaded in small increments by turning the screw jack handle slowly in the clockwise direction.12. The load and the corresponding mid-span deflection is recorded for each load increment.

Results and Calculation:Length of member: 400 mm = 0.4 mWidth of member, b: 19.6 mmThickness of member, d: 3.1 mmMoment of inertia of member, I: 48.6586 Dial gauge reading: 1 div = 0.01 mmPcr : 600.3014 NLoad, PMid-Span Deflection, dd/P

Ndivmmmm/N

5080.080.001600000

100160.160.001600000

150260.260.001733333

200460.460.002300000

250670.670.002680000

Calculation:I = = = 48.6586 = 48.6586 x m

Pcl = EL = = 600.3014 N

Percentage of error, % = = x 100%= 19.2739%

Length of member: 450 mm = 0.45 mWidth of member, b: 19.6 mmThickness of member, d: 3.1 mmMoment of inertia of member, I: 48.6586 Dial gauge reading: 1 div = 0.01 mmPcr : 474.3122 NLoad, PMid-Span Deflection, dd/P

Ndivmmmm/N

255.00.0500.00200

5010.00.1000.00200

7518.00.1800.00240

10025.00.2500.00250

12538.00.3800.00304

Calculation:I = = = 48.6586 = 48.6586 x m

Pcl = EL = = 474.3122 N

Percentage of error, % = = x 100%= 37.3198%

Length of member: 500 mm = 0.5 mWidth of member, b: 19.6 mmThickness of member, d: 3.1 mmMoment of inertia of member, I: 48.6586 Dial gauge reading: 1 div = 0.01 mmPcr : 384.1929 N

Load, PMid-Span Deflection, dd/P

Ndivmmmm/N

255.00.0500.002000000

5016.00.1600.003200000

7534.00.3400.004533333

10059.00.5900.005900000

125101.01.0100.008080000

Calculation:I = = = 48.6586 = 48.6586 x m

Pcl = EL = = 384.1929N

Percentage of error, % = = x 100%= 58.2501%

Discussion:Euler theory states that provided the critical buckling load is not exceeded a strut will not undergo any excessive deflection and will remain in equilibrium. For pin-ended strut, it is assumed to undergo only axial loading and will remain in its elastic range prior to the critical load being reached. Deflection will only occur when the critical load is reached. Any deflections prior to this load are maintained inside the structureIn this experiment, 3 different length of struts (400mm, 450mm and 500mm) with the same width (19.8mm) and thickness (3.1mm) are tested. Each showing different results which have shown linear graph of deflection versus (deflection/load). For 400 mm strut, the theoretical critical buckling load calculated form formula Pcr = ( * El)/() is 600.3014N. The experimental buckling load obtained from the graph is 484.62N which has a percentage error of 19.2739%. For 450 mm strut, the theoretical critical buckling load is 474.3122N and the experimental buckling load is 297.37N. The percentage of error for this strut is 37.3198%. As for strut of length 500 mm, the theoretical critical buckling load is 382.1929N and the experimental buckling load is 160.48N. It has a percentage error of 58.2501%. According to the results obtained, the experimental buckling load is less than the theoretical critical buckling load. The shorter struts has the higher buckling resistance which the longer strut has smaller buckling resistance. From this point, it can be assumed that larger ratio between the second moment of area and length of the strut will result in better buckling resistance. As for the errors, shorter strut has lesser percentage of error while longer strut has higher percentage of error. This may be mostly caused by human errors. This is because the struts specimens and equipment has been used for many years. For example, the strut may have already bent slightly due to the same strut specimen has been used by other groups to do the experiment for many years. This will affect the accuracy of the experimental results. Besides, it may also been caused by systematic errors. This is because the gauge is so highly sensitive, a slight touch on the table or winds will also cause the readings to be inaccurate. To improve the accuracy of the results, the reading should be taken more than one times. In addition, the specimens used in the experiment should avoid reusing in the future.

Conclusion:In conclusion, the buckling of load for pin-ended conditions has been determined through this experiment. The results show that shorter strut has higher buckling load resistance while longer strut has lower buckling load resistance. It has predicted that struts with higher ratio of second moment of area and length can perform better when both ends are pinned under compression.

References: Srivastava, A. and Gope, P. (2007) Strength Of Materials. New Delhi, India: Prentice-Hall. Lecture 37. [online] Available at: http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-ROORKEE/strength%20of%20materials/lects%20&%20picts/image/lect37/lecture37.htm [Accessed: 7 Mar 2013]. Jones, R. M. (2006). Buckling of bars, plates, and shells. Blacksburg, Va, Bull Ridge Publishing. Ptumech.loremate.com (n.d.) Chapter 7 : Columns and Struts | Strength of Materials - Part 1. [online] Available at: http://ptumech.loremate.com/som1/node/9 [Accessed: 7 Mar 2013].