C1 me2113 lab.docx

ME2113-1 DEFLECTION AND BENDING STRESSES IN BEAMS (EA-02-21)

INFORMAL REPORT

SEMESTER 3 2011/2012

Name: Sahil Anil Patel Matric no. : A0071443M October 28, 2011 Group 202

NATIONAL UNIVERSITY OF SINGAPORE DEPARTMENT OF MECHANICAL ENGINEERING

OBJECTIVES: 1. To analyse beam theory experimentally by loading a cantilever and studying the resulting stresses and deflection. 2. To deduce the Young’s modulus and Poisson’s ratio of the beam from the experimental results. 3. To investigate in terms of their relation to each other, the magnitude and signs of the strains and stresses at two locations along the cantilever beam with respect to beam theory.

SAMPLE CALCULATIONS: Part 1 Sample calculations for the loading of 250g Calculation of second moment of area, IZ 1 bh 3 12 (25.6)(6.06) 3  12  474.76 mm4

Iz 

 4.7476 1010 m4



Calculation of Young’s Modulus, E vL  

3EI PL3  P  3 z (vL ) 3EI z L

From graph 1, gradient = 5.8096 N/mm = 5809.6 N/m





3EIz  5809.6 L3  L3    0.253 E  5809.6  5809.6  63.73109 Pa = 63.73 GPa 10  3I 3(4.7476 10    z 

At x = 50mm Theoretical Calculation for bending moment, MXZ Mxz

= - P (L – x) = - 2.45 (0.25 – 0.05) = - 0.49 Nm

Calculation of theoretical longitudinal stress, σxx1 M h  xz    Iz 2 

 xx1   

 6.06 103  0.49 =    4.7476 1010  2   3.127 106 Pa = 3.127 MPa

Calculation of experimental longitudinal stress, σxx1 

 xx1  Exx1 = 63.73109  46.5 10-6 = 2.96106 Pa = 2.96 MPa



At x = 150mm Calculation for bending moment, MXZ Mxz

= - P (L – x) = - 2.45 (0.25 – 0.15) = - 0.245 Nm

Calculation of theoretical longitudinal stress, σxx2 M h  xz    Iz 2 

 xx 2   

 0.245 6.06 103  =    4.7476 1010  2   1.564 106 Pa =1.564 MPa



Calculation of experimental longitudinal stress, σxx2  xx2  Exx2 = 63.7310 9  2110-6 = 1.34 106 Pa = 1.34 MPa



Calculation of Poisson’s ratio, v

zz = - υ xx

υ1 = 0.304 υ2 = 0.312

Hence, from the gradients ofthe two lines in graph 2,  υ1 = 0.304 υ2 = 0.312 Thus, poisson’s ratio, υ =

0.304  0.312  0.308 2

Part 2 

Slope of graph 3 Equation of graph 3 is y = 0.0539x – 0.0492 Since εxx1 is in microns , Therefore, slope of graph = 0.0539 x 106 N Highest reading εxx1 = 800 x 10-6 Evaluation of handgrip force From Graph3,

P = 0.0539 106 x ɛxx1

Thus, handgrip force = 0.0539 x 106 x 800 x 10-6 = 43.12 N

DISCUSSION:

Comment on the signs of the strains (xx1, zz1, xx2 and zz2) with respect to the location and orientation of the strain gauges and how the beam is loaded.

1.

In this experiment, there are 4 strain gauges, where both, εxx1 and εzz1 are mounted on the top and both εxx2 and εzz2 are mounted on the bottom of the beam. Strain gauges εxx are in the axial direction and εzz are in the transverse direction. Also, the beam is experiencing a loading such that it bends downwards. So, by beam theory axial loading will cause contraction in the transverse direction as shown by the equation of Poisson’s ratio  zz 



 xx E



  xx . Henceforth as strain along axial direction increases, so

will the strain along the transverse direction decreases.



For ɛxx1, the value is positive because this strain gauge is being placed on top of the beam. Since the beam is experiencing tensile stress on the top surface, a positive strain value indicates an elongation in length along the axial direction. ɛxx2 is negative as it is place on the bottom side of the beam. Since the beam is experiencing compressive at the bottom, a negative strain value indicates a reduction in length along the axial direction. Since ɛxx1, is positive, ɛzz1 will be negative as shown by the equation of Poisson’s ratio (

 zz  2. 



 xx E



  xx ). Similarly, since ɛxx2, is negative, ɛzz1 will be positive.

With reference to Graph 4, comment on the slopes of the six theoretical lines and also on how stress varies with beam location. Graph 4 shows steeper slope as the load applied increases. This is because when a larger load is applied, the cantilever beam is bent more, so the stress at a specific point increases. The negative gradient means that the stresses decrease as strain gauge location (x) increases. The stress decreases to 0 MPa when x = 0.25m since at the end of the beam the deflection equation has to satisfy boundary conditions so there is no stress experience at the free end.

3.

Comment on the accuracy of your handgrip force. The handgrip force is generally inaccurate due to possible experimental errors. Firstly, if the tensile or compressive force in the gripper is too high, it will not translate the handgrip force accurately into strain as strain gauges has only a certain range of accuracy, Secondly, we are unable to maintain a constant gripping force, thus, the strain value is always fluctuating.

Conclusion: Firstly, the objective for our experiment is met. For our experiment an aluminum specimen is provided. From experimental results Young’s modulus and Poisson’s ratio are 63.73 GPa and 0.308 respectively. Comparing this to a typical aluminum material the Young’s modulus and Poisson’s ratio are 63.73GPa and 0.308 respectively. Henceforth our results are relatively close and accurate. Secondly, in accordance to equation of Poisson’s ratio strain gauge values in the axial direction is always of opposite sign to the strain gauge value in the transverse direction on the same side of the beam. And, by beam theory axial loading will cause contraction along transverse direction. Thirdly, in accordance to beam theory we observed the relationship between strain and stress. It is when a larger load is applied a specific point, the cantilever beam is bent more, so the stress at a specific point increases. In addition, the stresses decrease as strain gauge location increases along the length of the beam, which evidently shows that the largest bending moment and stress occurs at the cantilever end. Finally, in this experiment we learnt that stress cannot be measured directly, however it can be calculated by obtaining its bending deflection and strain using dial gauge and strain gauges respectively. Also, we managed to learn that in the designing of beams, we have to consider a few factor namely, its cross-sectional area, length of beam and its material (for which different material will have a different Young’s modulus, E and Poisson’s ratio, v).