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Unformatted text preview: 2 . The moment of inertia of the beam is I all the way along its length. You are welcome to use the table in Appendix C whenever it would be useful. A B C M E 1 E 2 L/2 L/2 Problem 6: Consider the composite beam cross section shown below. The top part of the beam (shaded) has a Young’s modulus of E 1 , and the bottom part of the beam (hatched) has a Young’s modulus of E 2 = 2E 1 . The directions of y and z are marked; it’s up to you to decide where to locate the origin. First locate the neutral axis in the beam. You will now examine the values of stress and strain on the dotted line indicated in the figure. Calculate the strain ε xx and the stress σ xx (as a function of the distance y off the neutral axis) that would result from an applied moment M about the z axis. Then graph both ε xx and σ xx vs. y. b/2 b b/4 b/4 Direction of y a a Direction of z a...
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- Spring '09
- mechanics, Second moment of area