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Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING CAMBRIDGE, MASSACHUSETTS 02139 2.002 MECHANICS AND MATERIALS II SOLUTIONS FOR HOMEWORK NO. Problem 1 (20 points) (a) The equilibrium equations are 11 x 1 + 12 x 2 + 13 x 3 + b 1 = (1) 21 x 1 + 22 x 2 + 23 x 3 + b 2 = (2) 31 x 1 + 32 x 2 + 33 x 3 + b 3 = (3) All shear stresses are zero. Furthermore, the gravitational body force loading b has one nonzero component only, in the direction of e 3 . Therefore, from the third equilibrium equation we get: 33 33 x 3 + b 3 = x 3 = b 3 (4) We also know that 33 ( x ) = p ( x ) and b 3 = g . We can substitute these into Equation 4 and integrate both sides with respect to x 3 x 3 p ( x ) x 3 dp ( x ) dx 3 b 3 dx 3 dp ( x ) = gx 3 p ( x 3 ) + p = gx 3 (5) = dx 3 p p ( x 3 ) = p gx 3 (6) Note that p is only a function of x 3 . (b) 1. The traction vector on a surface can be found by multiplying the stress with the unit outward normal vector on that surface. In this case, the normal vector is n l = cos sin (7) The traction vector is then 1 3 = p ( x ) cos = p ( x ) cos t l p ( x ) p ( x ) sin t l (8) p ( x ) sin 2. Action and reaction, forces need to balance at the interface. p ( x ) cos p ( x ) sin (9) t l = 3. 11 12 13 21 22 23 cos sin (10) t d = 31 32 33 which gives 3 linear equations involving ij p cos = 11 cos + 13 sin (11) = 21 cos + 23 sin (12) p sin = 31 cos + 33 sin (13) Problem 2 (20 points) 1 + ij = ij ij E 1 + kk + T ij (14) 3 k =1 The procedure is similar to what was discussed in class for elastic constitutive relations without thermal effects. The idea...
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 Spring '04
 DavidParks
 Mechanical Engineering

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