Engineering Mechanics of Materials |
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Page 211
... respect to the x and y axes , respectively . Thus , by definition , the moment of inertia of an area with respect to a given axis requires that an element of area dA be multiplied by the square of its distance from this axis and the ...
... respect to the x and y axes , respectively . Thus , by definition , the moment of inertia of an area with respect to a given axis requires that an element of area dA be multiplied by the square of its distance from this axis and the ...
Page 217
... respect to the X axis , ry the rectangular radius of gyration with respect to the Y axis , and ro the polar radius of gyration with respect to an axis through point O as shown in Figure 5.2 . Note that while moments of inertia can only ...
... respect to the X axis , ry the rectangular radius of gyration with respect to the Y axis , and ro the polar radius of gyration with respect to an axis through point O as shown in Figure 5.2 . Note that while moments of inertia can only ...
Page 306
... respect to point B , the origin for x1 . Upon integration from zero to x1F , the right side of Eq . 6.24 represents the first area moment of the M / EI , diagram between A and B with respect to B. THEOREM II . The tangential deviation ...
... respect to point B , the origin for x1 . Upon integration from zero to x1F , the right side of Eq . 6.24 represents the first area moment of the M / EI , diagram between A and B with respect to B. THEOREM II . The tangential deviation ...
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