Engineering Mechanics of Materials |
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Page 116
... cross section is perpendicular to that section and represents the resultant of an infinite number of minute forces perpendicular to the cross section and distributed in some manner throughout the cross - sectional area of the member ...
... cross section is perpendicular to that section and represents the resultant of an infinite number of minute forces perpendicular to the cross section and distributed in some manner throughout the cross - sectional area of the member ...
Page 241
... cross - sectional area whose inside diameter is 0.15 m and outside diameter is 0.30 m . The beam is subjected to a concentrated force of 50 kN at midspan . ( a ) For a location 3 m from the left support , show the variation of the ...
... cross - sectional area whose inside diameter is 0.15 m and outside diameter is 0.30 m . The beam is subjected to a concentrated force of 50 kN at midspan . ( a ) For a location 3 m from the left support , show the variation of the ...
Page 265
... cross- sectional area illustrated in Figure H5.20 . Assume the loads to act vertically downward and their plane to be normal to the 0.12 - m side of the section , which is the top of the beam . Compute the maximum tensile and maximum ...
... cross- sectional area illustrated in Figure H5.20 . Assume the loads to act vertically downward and their plane to be normal to the 0.12 - m side of the section , which is the top of the beam . Compute the maximum tensile and maximum ...
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