Engineering Mechanics of Materials |
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Page 160
B. B. Muvdi, J. W. McNabb. 40 kN - m 400 kN - m Do = 0.20 m A 300 kN - m B Do = 0.40 m D = 0.12 m const D = 0.24 m -0.75 m + 1.20 m- ( a ) Torque ( kN - m ) +40 40 kN - m AB 29.3 MN / m2 -360 ( b ) 360 kN - m Do = 0.20 m D1 = 0.12 m BC ...
B. B. Muvdi, J. W. McNabb. 40 kN - m 400 kN - m Do = 0.20 m A 300 kN - m B Do = 0.40 m D = 0.12 m const D = 0.24 m -0.75 m + 1.20 m- ( a ) Torque ( kN - m ) +40 40 kN - m AB 29.3 MN / m2 -360 ( b ) 360 kN - m Do = 0.20 m D1 = 0.12 m BC ...
Page 163
... KN - m / m J J J J J J J J J J J J Do = 0.10 m , D , 0.06 m 20 kN - m 6 KN - m A D = 0.06 m 4 m B 6 kN - m B 20 kN - m 20 kN - m 2 m 2 m 2 m FIGURE H4.8 4.9 For the shaft depicted in Figure H4.9 , plot the variation of torque versus a ...
... KN - m / m J J J J J J J J J J J J Do = 0.10 m , D , 0.06 m 20 kN - m 6 KN - m A D = 0.06 m 4 m B 6 kN - m B 20 kN - m 20 kN - m 2 m 2 m 2 m FIGURE H4.8 4.9 For the shaft depicted in Figure H4.9 , plot the variation of torque versus a ...
Page 296
... kN - m 10 kN - m Elu = 6.00 X 103 KN - m2 Elu = 4.00 x 106 k - in.2 0.5 k / ft upward P = 8 k 10 ft- 10 ft- 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 loadings ...
... kN - m 10 kN - m Elu = 6.00 X 103 KN - m2 Elu = 4.00 x 106 k - in.2 0.5 k / ft upward P = 8 k 10 ft- 10 ft- 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 loadings ...
Contents
Internal Forces in Members | 1 |
Stress Strain and Their Relationships | 53 |
Stresses and Strains in Axially Loaded Members | 115 |
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absolute maximum shear aluminum angle of twist applied Assume axes axial force axially loaded beam shown bending C₁ cantilever beam Castigliano's second theorem column components compressive Compute coordinate cross section cross-sectional area cylinder deflection deformation depicted in Figure Determine diameter elastic curve equal equation equilibrium Example factor of safety flexural stress free-body diagram Homework Problems k-ft k-in kN-m length M₁ material maximum shear stress MN/m² modulus of elasticity Mohr's circle moment of inertia neutral axis normal stress obtained perpendicular plane stress plot principal centroidal axis principal stresses r₁ ratio Refer to Figure respect rotation section a-a segment shaft shear strain shown in Figure slope Solution Solve static statically indeterminate steel stress at point stress condition stress element T₁ tensile tension Tmax torque torsional V₁ yield strength yield stress zero σ₁