Solution Manual 3rd Ed. Metal Forming Mechanics And Metallurgy
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Solution Manual 3rd Ed. Metal Forming: Mechanics and MetallurgyChapter 1 Determine the principal stresses for the stress state 1 34 0 ij = 3 5 2. 4 2 7 Solution: I1 = 10+5+7=32, I2 = -(50+35+70)+9 +4 +16 = -126, I3 = 350 -48 -40 -80 3 -63 = 119; 222 -126 -119 =0. A trial and error solution gives -= 13.04. Factoring out 13.04,2 -8.96 + 9.16 = 0. Solving; 1 = 13.04, 2 = 7.785, 3 = 1.175. 1-2 A5-cm. diameter solid shaft is simultaneously subjected to an axialload of 80 kN and a torque of 400 Nm. a. Determine the principalstresses at the surface assuming elastic behavior. b. Find thelargest shear stress. Solution: a. The shear stress,, at a radius,r, is = sr/R where sis the shear stress at the surface R is theradius of the rod. The torque, T, is given by T = 2tr2dr = (2s/R)r3dr = sR3/2. Solving for = s, s = 2T/(R3) = 2(400N)/(0.0253) =16 MPa The axial stress is.08MN/(0.
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Solution Manual 3rd Ed. Metal Forming: Mechanics and MetallurgyChapter 1 Determine the principal stresses for the stress state 1 34 0 ij = 3 5 2. 4 2 7 Solution: I1 = 10+5+7=32, I2 = -(50+35+70)+9 +4 +16 = -126, I3 = 350 -48 -40 -80 3 -63 = 119; 222 -126 -119 =0. A trial and error solution gives -= 13.04. Factoring out 13.04,2 -8.96 + 9.16 = 0. Solving; 1 = 13.04, 2 = 7.785, 3 = 1.175. 1-2 A5-cm. diameter solid shaft is simultaneously subjected to an axialload of 80 kN and a torque of 400 Nm. a. Determine the principalstresses at the surface assuming elastic behavior. b. Find thelargest shear stress. Solution: a. The shear stress,, at a radius,r, is = sr/R where sis the shear stress at the surface R is theradius of the rod. The torque, T, is given by T = 2tr2dr = (2s/R)r3dr = sR3/2. Solving for = s, s = 2T/(R3) = 2(400N)/(0.0253) =16 MPa The axial stress is.08MN/(0.0252) = 4.07 MPa 1,2 = 4.07/2[(4.07/2)2 + (16/2)2)]1/2 = 1.029, -0.622 MPa b. the largest shearstress is (1.229 + 0.622)/2 = 0.925 MPa A long thin-wall tube,capped on both ends is subjected to internal pressure. Duringelastic loading, does the tube length increase, decrease or remainconstant? Solution: Let y = hoop direction, x = axial direction,and z = radial direction. ex = e2 = (1/E)[ - ( 3 + 1)] = (1/E)[2 -(22)] = (2/E)(1-2) Since u < 1/2 for metals, ex = e2 is positiveand the tube lengthens. 4 A solid 2-cm. 827ec27edc