Strength of MaterialsSimple stress, simple strai, torsion, shear and moment in beams, beam deflections, continuous beams, combined stresses. |
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Page 201
... Equivalent cantilever loadings Figure 6-14 Moment diagram by parts . load included between B and b - b . Note that ... equivalent to the cantilever loading at ( a ) , whereas the moment effect of the uniform loading on any section of the ...
... Equivalent cantilever loadings Figure 6-14 Moment diagram by parts . load included between B and b - b . Note that ... equivalent to the cantilever loading at ( a ) , whereas the moment effect of the uniform loading on any section of the ...
Page 367
... Equivalent wood section ( c ) Equivalent steel section Figure 10-1 Equivalent sections . Furthermore , in order to be equivalent , the loads carried by the steel fiber and the equivalent wood fiber must be equal , so or , in terms of ...
... Equivalent wood section ( c ) Equivalent steel section Figure 10-1 Equivalent sections . Furthermore , in order to be equivalent , the loads carried by the steel fiber and the equivalent wood fiber must be equal , so or , in terms of ...
Page 372
... equivalent section of a composite beam because its derivation was based on the difference in normal forces between two adjacent sections . Since the forces on the original composite section and on the equivalent section are the same ...
... equivalent section of a composite beam because its derivation was based on the difference in normal forces between two adjacent sections . Since the forces on the original composite section and on the equivalent section are the same ...
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