Engineering Mechanics of Materials |
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Page 160
B. B. Muvdi, J. W. McNabb. 40 kN - m Do = 0.20 m 400 kN - m 300 kN - m Do = 0.40 m D1 = 0.12 m = const Do = 0.24 m -0.75 m + 1.20 m ( a ) Torque ( kN - m ) +40 40 kN - m AB -360 ( b ) 29.3 MN / m2 360 kN - m 1.60 m 60 kN - m -60 ( m ) 60 ...
B. B. Muvdi, J. W. McNabb. 40 kN - m Do = 0.20 m 400 kN - m 300 kN - m Do = 0.40 m D1 = 0.12 m = const Do = 0.24 m -0.75 m + 1.20 m ( a ) Torque ( kN - m ) +40 40 kN - m AB -360 ( b ) 29.3 MN / m2 360 kN - m 1.60 m 60 kN - m -60 ( m ) 60 ...
Page 163
... kN - m 4.11 A uniformly distributed torque is applied to the shaft depicted in Figure H4.11 . Determine the maxi- mum shearing stresses in this shaft . Uniformly distributed torque 3 kN - m / m J Q J J J J J J J J } A D 0.06 m 4 m B 6 ...
... kN - m 4.11 A uniformly distributed torque is applied to the shaft depicted in Figure H4.11 . Determine the maxi- mum shearing stresses in this shaft . Uniformly distributed torque 3 kN - m / m J Q J J J J J J J J } A D 0.06 m 4 m B 6 ...
Page 296
... kN - m 10 kN - m El = 6.00 × 103 KN - m2 A - El = 4.00 x 106 k - in.2 0.5 k / ft upward P = 8 k 10 ft 10 ft- FIGURE H6.29 B L = 10 m FIGURE H6.26 6.27 Find the deflection of the simply supported beam at the center of the span due to the ...
... kN - m 10 kN - m El = 6.00 × 103 KN - m2 A - El = 4.00 x 106 k - in.2 0.5 k / ft upward P = 8 k 10 ft 10 ft- FIGURE H6.29 B L = 10 m FIGURE H6.26 6.27 Find the deflection of the simply supported beam at the center of the span due to the ...
Contents
T Shear and Moment at Specified Sections | 24 |
Stress Strain and Their Relationships | 53 |
Stresses and Strains in Axially Loaded Members | 115 |
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acting allowable angle of twist applied Assume axes axis beam bending cantilever centroidal circle column components compressive Compute Consider constant construct coordinate cross section curve deflection deformation depicted in Figure Determine developed diameter direction discussed elastic element energy equal equation equilibrium Example expressed factor failure flexural force free-body diagram function given inertia joint length limit load material maximum shear stress method modulus moment moments neutral axis normal stress Note obtained plane plot positive principal stresses Problem properties quantity ratio reactions Refer to Figure relation represents resist respect rotation segment shaft shown in Figure slope Solution Solve static steel strain strength structural subjected Substitution supported surface tensile tension theory tion torque unit yield zero