paraphrasing

The objective of this experiment is to determine the stress, deflection, and the strain of a simply supported beam under load. Also to experimentally verify the beam stress and flexure formulas. The experiment was done was using the machine to apply a load to a simply supported beam and measure the deflection and the strain of it. We notice that whenever the load increases, the deflection and the stress increase too. We can use the strain to find the theoretical stress and we can use moment, moment of inertia, and the neutral axis to find the experimental stress. The moment of inertia calculated was 0.05122 . The neutral axis was 0.515 in, and the maximum deflection was 0.000013 in. We also calculated the maximum stress using the maximum load applied on the beam, which turned out to equal 40049.5 (psi).

Table of Contents

Abstract……………………………………………………………..………..2

Table of Contents……………………………………….……………………3

Introduction and Theory…………………………………………………….4-8

Procedures………………………………………………….……………….9-10

Summary of Important Results…………………...………………………..11-12

Sample Calculations and Error Analysis……………….………………….13-14

Discussion and Conclusion………………………………………………..15-16

References……………………………………..…………………………….17

Appendix……………………………………………………….……………18

Introduction and Theory

Structural members are usually designed to carry tensile, compressive, or transverse loads. A member which carries load transversely to its length is called a beam. In this experiment, a beam will be symmetrically loaded as shown in Fig. III-1(a), where P is the applied load. Note that at any cross section of the beam there will be a shear force V (Fig. III-1(b)) and moment M (Fig. III-1c). Also, in the central part of the beam (between the loads P/2) V is zero and M has its maximum constant value. Notice the sign convention of a positive moment, M, causing a negative (downward) deflection, y.

If in this part a small slice EFGH of the beam is imagined to be cut out, as shown, then it is clear that the external applied moment, M, must be balanced by internal forces (stresses) at the sections (faces) EF and GH. For M applied as shown in Fig. III-2(a), these forces would be compressive near the top, EG, and tensile near the bottom, FH. Since the beam material is considered elastic, these forces would deform the beam such that the length EG would tend to become shorter, and FH would tend to become longer. The first fundamental assumption of the beam theory can be stated as follows:

“Sections, or cuts, which are plane (flat) before deformation, remain plane after deformation.”

Thus, under this assumption, the parallel and plane section EF and GH will deform into plane sections E’F’ and G’H’ which will intersect at point O, as shown in Fig. III-2(b). Since E’F’ and G’H’ are no longer parallel, they can be thought of as being sections of a circle at some radial distance from O. Convince yourself of this by drawing a square on an eraser and observe its shape when you bend the eraser. Since the forces near E’G’ are compressive, and those near F’H’ are tensile, there must be some radial distance r where the forces are neither compressive nor tensile, but zero. This axis, N-N, is called the neutral axis. Notice that N-N is not assumed to lie in the center of the beam.

Consider an arc of distance +η, from the neutral axis, or distance r + η from O (Fig. III-2(b)). At this radius, the length of arc is l’=(r + η) Δθ. As shown in Fig. III-2(a), the length of the arc was l before the deformation. This length is also equal to rΔθ (because at N-N there are no forces to change the length). Thus, the strain at distance +η from the neutral axis can be found by:

(III-1)

In other words, the axial strain is proportional to the distance from the neutral axis. It is remarked that this strain is positive, because positive η was taken on the tensile side of N-N in Figure. III-2(b). Had η been taken in the opposite direction, then the strain would have been negative, as appropriate for the compressive side.

The second fundamental assumption is that Hooke’s Law applies both in tension and compression with the same modulus of Elasticity. Therefore, from equation (I-3) and (III-1),

(III-2)

If c is the maximum distance from the neutral axis (largest positive or negative value of η), then

the maximum stress (compressive or tensile) is given by σm = Ec/r, and Eq. (III-2) can also be written as

(III-3)

That is, the stress at a section EF or GH, due to applied moment M, varies linearly from zero at

the neutral axis to some maximum value σm (positive or negative) when η = c. To obtain the

beam stress formula, it remains to define where the neutral axis is located, and to relate σm to M.

To locate the neutral axis, it is observed that the tensile and compressive forces on a section are equal to the stress times a differential element of area, as shown in Fig. III-2(c). For static equilibrium, the sum (or integral) of all these internal forces must be zero. That is,

where, the integrals are over the whole cross-sectional area. Thus, it is seen that the neutral axis is located such that the first moment of area about it is zero; that is, the neutral axis passes through the centroid of the cross-sectional area. In Figure. III-2(c), a rectangular area was used for illustration; however, any shape of vertically symmetric cross-sectional area is valid for the area integral.

In a similar fashion, the moment due to all the forces is the sum (or integral) of the forces times their moment arms about the neutral axis, and this must be equal to the external applied moment.

Thus,

(III-4)

If I is defined as the second moment of area about the neutral axis, commonly called the moment of inertia,

(III-5)

Then Eq. (III-4) can be written as:

(III-6)

Where Z = I/c is the section modulus, which depends only on the cross-sectional geometry of the beam. Equation (III-6) is the beam stress equation which relates the maximum (compressive or tensile) stress to the applied moment. Notice its similarity to Equation (I-1), the stress equation for uniaxial tension. It is understood, of course, that σm is the maximum bending stress at a particular location, x, along the beam. In general, both σm and M are functions of x, and are related by Equation (III-6).

The remaining question about the beam concerns its degree of deformation, or flexure. That is, how is the radius of curvature, r, related to the moment M (or load P)? From calculus, it can be shown that the curvature of a function y(x) is given by

Thus, if x is the distance along the beam, y will be the deflection as indicated in Fig. III-1(a). For most beams of practical interest, this deflection will be small, so that the slope dy/dx will be very small compared to 1. Hence, a very good approximation is

But, since σm = Ec/r = Mc/I, there results the differential equation of the elastic curve:

(III-7)

To obtain the elastic curve of the beam, y(x), and the maximum deflection, ym, it is necessary to integrate Eq. (III-7) using the moment function M(x) in Fig. III-1(c). Thus, using M(x) = Px/2 for

0 ≤ x ≤ a and M(x) = Pa/2 for a ≤ x ≤ a + b, it is found that

And that the maximum deflection at x = a + b/2 is

(III-8)

In particular, for a = b = L/3,

(III-9)

Procedures

First, this test will be conducted on a 1018 steel beam (E = 30x10*6 psi) using the MTS testing machine. The position of the beam in the testing machine should be the same as the picture shown in the lab manual. Then we noticed that the cross-section of the I-beam is not symmetric: one flange of the beam is thicker than the other, therefore we measured the cross-sectional dimension carefully and the location of the loading points. After that we placed the beam in the testing machine and aligned the 12-inch black marks on the beam with roller supports of the lower fixture. We made sure the beam is squarely resting and centered on the lower support with strain gauge facing down. Then we went to the computer and entered the TestWork 4 software by double clicking on the icon. When prompted, we made sure the name field under login says “306A_Lab” then clicked OK to login then under the Open Method dialog, we selected “exp-3 4 Point Flex Mod X” after that we selected the Motor Reset button right corner by clicking on it. We zeroed the “load” readout by right clicking on the “load cell” icon and selecting “zero channel”. Next we used the handset to position the upper bending fixture over the beam. We did that by enabling the handset by pressing the unlock button, then slowly lowering the crosshead using the down arrow until the fixture is nearly touching the beam. While doing that, we made sure not to pinch the strain gauge lead wires. While observing the digital load readout on the screen, we used the thumb wheel of the handset to lower the fixture onto the beam. We watched for the load reading increase when the upper fixture makes contact. After that, we slowly raised the fixture until only a very slight preload of approximately 0.3 lb. is applied. Then we locked the handset. Then we connected the strain gauge wire to the #1 strain channel of the grey DAQ box, data acquisition is conducted by the LabVIEW software. We took the magnetic base holding the dial indicator, and positioned it with the dial indicator in the center of the beam on the bottom side. We made sure the dial indicator is not touching the strain gauge, after that we locked the MTS frame by activating the magnetic base then zeroed the dial indicator. Next, we started the LabVIEW software by double clocking the icon on the desktop, selecting “Open” and double clicking “exp2&3-Strain Mod-15 LV7.1” then we pressed the white arrow to start the strain gauge acquisition then we zeroed the strain. After that we pressed the green arrow on the TestWorks 4 GUI. We loaded the beam up to 1000 lb. in increments of 100 lb. and recorded the data needed. After reaching 1000 lb., we repeated the experiment after making sure it zeroed out.

Summary of Important Results

Figure 1: Experimental Deflection vs Load

Figure 2: Theoretical Deflection vs Load

Figure 3: Stress vs Load

All three graphs show that the relationships between stress vs. load, theoretical deflection vs. load, and experimental deflection vs. load are all directly proportional. To plot the second graph, we need to find the theoretical deflection by using the calculated length of the beam, which is 1.044 in, and the area moment of inertia, which is 0.05122 in4. The third graph is stress versus load. We get the stress by dividing the moment over the section modulus. The moment can be calculated which is the force applied multiplied by the distance 4 in. The section modulus is 0.0995, which is the moment of inertia over the neutral axis.

Sample Calculations and Error Analysis

A1 = 0.3178 in2 y1 = 0.1555 in

A2 = 0.0934 in2 y2 = 0.535 in

A3 = 0.292 in2 y3 = 0.901 in

c = = 0.515 in

Z (section modulus) = I/c = 0.0995

Error Analysis

Errors are avoidable in every experiment. One can only try to reduce them as much as possible while conducting an experiment. Therefore, the errors in this experiment could be due to wrong calculations, mistake in taking measurements, or a wrong setup for the sample in the machine. These are the errors that could happen by the person who is conducting the experiment. However, there are errors that not related to the one doing the experiment. The machine itself could be old or hasn’t had the proper maintenance and that could cause errors. Not only the machine could cause errors, but also the samples could be not manufactured well.

Discussion and Conclusion

After finishing the experiment and calculating the data and plotting them, we noticed a lot of things about the beam. Whenever the load increases, the deflection and the stress increase too. We can use the strain to find the theoretical stress and we can use moment, moment of inertia, and the neutral axis to find the experimental stress. And about the deflection, we used the dial indicator to find it directly from the experiment for the experimental, and we used load, length, modulus of elasticity, and the moment of inertia to find the theoretical data. The graph clearly shows that the load and deflection and load and stress have a directly proportional relationship.

Figure 5: shear force and bending moment diagrams

2000 lb

500 lb

-500 lb

REFRENCES

[1] CSUF EGME 306A Lab Manual.

Appendix

Stress vs Load

Load (lb)

Stress (psi)

Experimental Deflection vs Load

Load (lb)

Deflection (in)

Theoretical Deflection vs Load

Load (lb)

Deflection (in)

14