Strength of MaterialsSimple stress, simple strai, torsion, shear and moment in beams, beam deflections, continuous beams, combined stresses. |
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Page 147
... Repeat Prob . 529 if the distributed load is 12 kN / m and the length of the beam is 8 m . 531. A 15 - ft beam simply supported at the ends carries a concentrated load of 9000 lb at midspan . Select the lightest S section that can be ...
... Repeat Prob . 529 if the distributed load is 12 kN / m and the length of the beam is 8 m . 531. A 15 - ft beam simply supported at the ends carries a concentrated load of 9000 lb at midspan . Select the lightest S section that can be ...
Page 409
... Repeat Prob . 1134 for a column with an effective length of 4.5 m . 1136. A steel column 2 in . by 3 in . in section has an effective length of 5 ft . Compute the maximum load that can be carried at an eccentricity of 5 in . from the ...
... Repeat Prob . 1134 for a column with an effective length of 4.5 m . 1136. A steel column 2 in . by 3 in . in section has an effective length of 5 ft . Compute the maximum load that can be carried at an eccentricity of 5 in . from the ...
Page 500
... Repeat Prob . 1405 for a W8 X 40 beam . Ans . 1.09 1407. Repeat Prob . 1405 for the T section shown in Fig . 14-4 . Ans . 1.77 1408. The centroidal axis of the section shown in Fig . P - 1408 is 5.7 in . above the bottom , and the ...
... Repeat Prob . 1405 for a W8 X 40 beam . Ans . 1.09 1407. Repeat Prob . 1405 for the T section shown in Fig . 14-4 . Ans . 1.77 1408. The centroidal axis of the section shown in Fig . P - 1408 is 5.7 in . above the bottom , and the ...
Common terms and phrases
allowable stresses aluminum angle assumed axes axial load beam in Fig beam loaded beam shown bending bending moment bolts cantilever beam centroid column compressive stress Compute the maximum concentrated load connector cross section deformations Determine the maximum diameter elastic curve element equal equivalent Euler's formula fibers flange flexural stress flexure formula free-body diagram Hence Hooke's law horizontal Illustrative Problem kips kN·m kN/m lb.ft lb/ft length loaded as shown M₁ M₂ maximum shearing stress maximum stress method midspan mm² Mohr's circle moment of inertia neutral axis obtain P₁ plane product of inertia proportional limit R₂ R2 Figure radius reaction resisting resultant rivet segment shaft shear center shear diagram shearing force shown in Fig slope Solution span steel strain tensile stress thickness torque torsional U.S. Customary Units uniformly distributed load vertical shear weld zero ΕΙ