Engineering Mechanics of Materials |
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Page 414
... column is subjected to a loading equal to 0.6 of the Euler critical loading with an eccentricity of 0.1 in . What is the deflection of the column ? You are to assume that the column be- haves elastically . 8.25 A column is initially ...
... column is subjected to a loading equal to 0.6 of the Euler critical loading with an eccentricity of 0.1 in . What is the deflection of the column ? You are to assume that the column be- haves elastically . 8.25 A column is initially ...
Page 416
... Column end conditions discussed in Section 8.4 determine column effective lengths L. , and these lengths differ from the geometric length of the column except in the case where both ends of the columns are pinned . Care must be taken to ...
... Column end conditions discussed in Section 8.4 determine column effective lengths L. , and these lengths differ from the geometric length of the column except in the case where both ends of the columns are pinned . Care must be taken to ...
Page 424
... column is 12 ft long and has dressed cross - sectional dimen- sions 9.5 x 11.5 in . Solution . The least cross ... column has the W ( wide - flanged ) shape shown in Figure H8.27 and is fabricated of steel with a yield stress of 50 ksi ...
... column is 12 ft long and has dressed cross - sectional dimen- sions 9.5 x 11.5 in . Solution . The least cross ... column has the W ( wide - flanged ) shape shown in Figure H8.27 and is fabricated of steel with a yield stress of 50 ksi ...
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