## Strength of materials |

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Results 1-3 of 82

Page 366

In other words, the

replaces; the

wide, as in ...

In other words, the

**equivalent**wood area is n times as wide as the steel itreplaces; the

**equivalent**wood section is shown in Fig. 10-lb. If desired, an**equivalent**steel section can be set up by replacing the original wood by steel - aswide, as in ...

Page 370

Shearing Stress and Deflection in Composite Beams The formula for horizontal

shearing stress (Eq. 5-4, page 163) developed for homogeneous beams applies

equally well to the

derivation ...

Shearing Stress and Deflection in Composite Beams The formula for horizontal

shearing stress (Eq. 5-4, page 163) developed for homogeneous beams applies

equally well to the

**equivalent**section of a composite beam, because itsderivation ...

Page 382

The stress developed by dividing this gripping force by the surface area of the

reinforcing bars per (a)

linear inch is the bond stress. The bond stress is analogous to the shearing stress

...

The stress developed by dividing this gripping force by the surface area of the

reinforcing bars per (a)

**Equivalent**section R0. 10-10. (b) Shear stress distributionlinear inch is the bond stress. The bond stress is analogous to the shearing stress

...

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allowable stresses aluminum angle assumed axes axial load beam in Fig beam loaded beam shown bending bolt cantilever beam caused centroid CN CN column compressive stress Compute the maximum concentrated load concrete cover plate cross section deformation Determine the maximum diameter elastic curve end moments equal equivalent Euler's formula factor of safety fibers flange flexure formula free-body diagram ft long ft-lb Hence hinged Hooke's law horizontal ILLUSTRATIVE PROBLEMS lb/ft length loaded as shown main plate maximum shearing stress maximum stress midspan midspan deflection modulus Mohr's circle moments of inertia neutral axis obtain plane plastic positive product of inertia proportional limit radius ratio reaction Repeat Prob resisting restrained beam resultant segment shaft shear center shear diagram shearing force shown in Fig Solution Solve Prob span static steel strain tensile stress thickness torque torsional uniformly distributed load vertical shear weld zero