Engineering Mechanics of Materials |
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Page 218
... principal centroidal axes of 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 ...
... principal centroidal axes of 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 ...
Page 226
... principal axes of inertia and the second the v principal axes of inertia for ... axis of sym- metry for a cross section is a principal axis of inertia , it ... centroidal axis of inertia ( denoted as the v principal axis ) , and a ...
... principal axes of inertia and the second the v principal axes of inertia for ... axis of sym- metry for a cross section is a principal axis of inertia , it ... centroidal axis of inertia ( denoted as the v principal axis ) , and a ...
Page 258
... principal centroidal axis of inertia , of the neutral axis for unsymmetric bending in terms of the principal centroi- dal moments of inertia I , and I , and the orientation , angle 4 , of the plane of the loads with respect to the v ...
... principal centroidal axis of inertia , of the neutral axis for unsymmetric bending in terms of the principal centroi- dal moments of inertia I , and I , and the orientation , angle 4 , of the plane of the loads with respect to the v ...
Contents
T Shear and Moment at Specified Sections | 24 |
Stress Strain and Their Relationships | 53 |
Stresses and Strains in Axially Loaded Members | 115 |
Copyright | |
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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