Engineering Mechanics of Materials |
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Page 214
... inertia can be made mathematically identical with the plane stress equations 2.4 and 2.6 , respectively , by satisfying the following equalities : 6 x = lx , σ = ly , Txy Py xy On = Ins nt Tn = Pnt ( 5.3 ) Therefore , using the ...
... inertia can be made mathematically identical with the plane stress equations 2.4 and 2.6 , respectively , by satisfying the following equalities : 6 x = lx , σ = ly , Txy Py xy On = Ins nt Tn = Pnt ( 5.3 ) Therefore , using the ...
Page 215
... inertia will be illustrated in Example 5.2 . It is desirable at this point to review briefly the question of determining moments and products of inertia of composite areas . If the moment of inertia of a composite area with respect to a ...
... inertia will be illustrated in Example 5.2 . It is desirable at this point to review briefly the question of determining moments and products of inertia of composite areas . If the moment of inertia of a composite area with respect to a ...
Page 218
... inertia . The second principal centroidal axis of inertia is perpendicular to the first and it is , obviously , not an axis of symmetry . If it is assumed that b < h , then the moment of inertia with respect to the horizontal principal ...
... inertia . The second principal centroidal axis of inertia is perpendicular to the first and it is , obviously , not an axis of symmetry . If it is assumed that b < h , then the moment of inertia with respect to the horizontal principal ...
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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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 longitudinal 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 force shear strain shown in Figure slope Solution Solve static statically indeterminate steel stress condition stress element T₁ tensile tension Tmax torque torsional V₁ vertical yield strength yield stress zero