Engineering Mechanics of Materials |
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Page 214
... moments of inertia can be made mathematically identical with the plane stress equations 2.4 and 2.6 , respectively , by satisfying the following equalities : 6 x = 1x , 6 , = l ,, Txy = P xy σn = Ins Tnt = Pnt ( 5.3 ) Therefore , using ...
... moments of inertia can be made mathematically identical with the plane stress equations 2.4 and 2.6 , respectively , by satisfying the following equalities : 6 x = 1x , 6 , = l ,, Txy = P xy σn = Ins Tnt = Pnt ( 5.3 ) Therefore , using ...
Page 224
... moments of inertia I , and I , if they are known to be equal in magnitude . Find the polar moments of inertia with respect to axes through points C and O. Y 5.8-5.10 The sections shown in Figures H5.8 , H5.9 , and H5.10 have one axis of ...
... moments of inertia I , and I , if they are known to be equal in magnitude . Find the polar moments of inertia with respect to axes through points C and O. Y 5.8-5.10 The sections shown in Figures H5.8 , H5.9 , and H5.10 have one axis of ...
Page 269
... moments at the ends of the beam of finite length , as shown in Figure 6.1 ( c ) . This cylindrical surface , which is unstrained , is referred to as the neutral surface . Rigorously , Eq . 6.3 is only valid for a beam subjected to pure ...
... moments at the ends of the beam of finite length , as shown in Figure 6.1 ( c ) . This cylindrical surface , which is unstrained , is referred to as the neutral surface . Rigorously , Eq . 6.3 is only valid for a beam subjected to pure ...
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 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 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