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III-1 Experiment III The Beam OBJECTIVES The objectives of this experiment are (a) to determine the stress, deflection and strain of a simply supported beam under load, and (b) to experimentally verify the beam stress and flexure formulas.
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 sections EF and GH will deform into plane sections EF′′ and GH′′ which will intersect at point O, as shown in Fig. III-2(b). Since EF′ ′ and GH′′ 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 EG′ ′are compressive, and those near FH′′ 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: ε η θ θ θ η = l - l l = (r + ) - r r = r ′ Δ Δ Δ (III-1) III-2 (a) (b) (c) ym b - Figure III-1. Symmetrically Loaded Beam (a), with Shear Force Diagram (b) and Bending Moment Diagram (c) L b c a 2 P 2 P V x M x 2 Pa 2 P + III-3 Figure III-2. Stresses and Strains of a Beam III-4 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 Fig. 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. Thus, from Eqs. (I-3) and (III-1), σ η = E r (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 σ m = Ec /r, and Eq. (III-2) can also be written as σ σ η = c m (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 σ m (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 σ m 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, m AA dF = dA = dA = 0 c σ ση ∫∫ 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 Fig. 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, M = dF = dA = cdA m 2 ∫∫ ∫ηησ σ η (III-4) If I is defined as the second moment of area about the neutral axis, commonly called the moment of inertia, I = dA 2 ∫η (III-5) then Eq. (III-4) can be written as:
m = Mc I = M Z σ (III-6) III-5 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 Eq. (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 1 r = dy dx (1 + dy dx) 2 2 2 23 2 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 1 r = dy dx 2 2 But, since σm = Ec /r = Mc /I, there results the differential equation of the elastic curve: EIdy dx = M(x) 2 2 (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 for 3 Pax(a + b) x y(x) = - , 0 x a 2EI 6 2 ⎛⎞ ≤ ≤ ⎜⎟ ⎝⎠ () for 32 P ax(2a + b) aax y x = - + , a x a + b 2EI 6 2 2 ⎛⎞ ≤≤ ⎜⎟ ⎝⎠ and that the maximum deflection at x = a + b/2 is ) ( 8 b + 2 ab + 3 a 2EIPa = y - 2 2 m (III-8) In particular, for a = b = L/ 3, -y = 23 Pa 48EI = 23 PL 1296EI m33 (III-9) III-6 Although the above stress and flexure formulas are quite simple, it took some of the best minds of the 17 th and 18 th centuries to derive them correctly. Part of the difficulty in obtaining the correct results at that time was that there were no methods of verifying the results experimentally. Today, with the advent of sensitive displacement dial gauges, the verification is more convenient.
Figure III-3. MTS Insight Tensile Testing Machine PROCEDURE 1. The 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 is shown in Fig. III-3. 2. Observe that the cross-section of the I-beam is not a symmetric; one flange of the beam is thicker than the other. Measure carefully the cross-sectional dimensions and the location of the loading and support points.
3. Carefully place the beam into the MTS machine if it is not already installed. Align the 12- inch black marks on the beam with the roller supports of the lower bending fixture. Make sure the beam is squarely resting and centered on the lower support with the strain gauge facing down. Beam Deflection Dial indicator Beam to Be Tested III-7 4.
Enter the TestWorks 4 software by double clicking on the icon on the desktop.
a. When prompted, make sure the Name field under the User Login says “306A_lab”.
b.
Click OK to login.
c.
Under the Open Method dialog, select “exp-3 4 Point Flex Mod X”.
d.
Now, select the Motor Reset button in the bottom right corner by clicking on it.
5. Zero the “load” readout by right clicking on the “Load cell” icon and selecting “zero channel”.
6.
Next we will use the handset to position the upper bending fixture over the beam.
a.
Enable the handset by pressing the “unlock” button at the top right of the handset b.
Slowly lower the crosshead using the down arrow until the fixture is NEARLY touching the beam.
c.
Make sure not to pinch the strain gauge lead wires between the fixture and the beam.
d.
While observing the digital load readout on the screen, use the thumb wheel of the handset to lower the fixture onto the beam. Watch for the load reading to increase when the upper loading component of the fixture makes contact with the beam.
e.
Now, slowly raise the fixture with the thumb wheel until only a very slight pre-load of approximately 0.2 lb is applied.
f. When finished, return control to the computer software by locking the handset using the same button as before.
Note that the beam has a strain gauge bonded to its surface. See Figure III-4. This strain gauge has a strain gauge factor of 2.14 ± 0.5% and 350 ohm resistance. Connect the strain gauge lead wires to the “#1 Strain” channel of the grey DAQ box. Data acquisition is conducted by the LabVIEW software. 7. Take the magnetic base holding the dial indicator, and position it with the dial indicator in the center of the beam on the bottom side. Make sure the dial indicator is not touching the strain gauge. Also, be VERY CAREFUL not to damage the dial indicator while doing this!
When the dial indicator is in position, lock it to the MTS frame by activating the magnetic base. Finally, zero the dial indicator by turning the dial of the indicator. Note the divisions on the dial indicator and the accuracy with which it can be read.
8.
Start the LabVIEW software.
a. Double click the LabVIEW icon on the desktop, or follow the procedure suggested for Experiment II.
b. Select “Open” and double click “exp2&3-Strain Mod 9-15 LV7.1”.
c.
Press the white arrow near the upper left corner of the screen to start the strain gauge acquisition.
d.
Press the “Zero Strain” button to zero-out the strain indicator.
9. You are now ready to run the experiment.
a.
Press the Green Arrow on the TestWorks 4 GUI.
b. Give the sample ID for your group. III-8 c. Under the Required Inputs dialog, enter 0.125 for the thickness and 0.500 for the width. DO NOT press OK yet!
10. You will load the beam up to 1000 lb in increments of 100 lb. To do this:
a.
“OK” the Required Inputs dialog from above and the test will begin.
b.
When the digital load readout reaches approximately 100 lb, press the “Pause” button to temporarily stop the test.
c.
Record the actual load on your data sheet from the digital readout.
d.
Toggle over to the LabVIEW screen and record the strain reading from the software.
e. Record the deflection of the beam from the dial indicator.
f. Return to the TestWorks 4 screen and press the “Pause” button again to resume the test.
g. Continue pausing and taking readings approximately every 100 lb from 0 to 1000 lb.
11. At 1000 lb the test will stop. Collect your last data point before pressing “OK” at the last dialog. 12. Before repeating the experiment, make sure your load readout, strain indicator, and dial indicator are all re-zeroed. 13. Repeat steps 7 through 9. Figure III-4. Beam with Bonded Strain Gauge REPORT REQUIREMENTS 1. Determine the neutral axis, moment of inertia, and section modulus of the beam cross- section. 2. Draw shear force and bending moment diagrams for the beam for the maximum load. 3. Compute the maximum bending stress and deflection of the beam for the maximum load. 4. Plot deflection versus load, and also stress versus load. Each graph should contain a curve based completely on theoretical calculations, and another using the experimental data points. III-9 5. By error calculation, determine if the beam theory is vindicated within the precision of the instruments. REFERENCES [1] Beckwith, T. G., Buck, N. L. and Marangoni, R. D., Mechanical Measurements , Addison- Wesley.
[2] Popov, E.P. Mechanics of Materials , Prentice-Hall Inc.
[3] Thomas, G.B., and Finney R.L., Calculus and Analytical Geometry, Addison-Wesley,