## Strength of materials |

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Page 366

In other words, the

replaces; the

wide, as in Fig. 10-1c. n The flexure formula can now be applied directly to either

the

wood section, the actual steel stress is n times the stress in the

with the ...

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 Fig. 10-1c. n The flexure formula can now be applied directly to either

the

**equivalent**wood section or the**equivalent**steel section. With the**equivalent**wood section, the actual steel stress is n times the stress in the

**equivalent**wood;with the ...

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 was based on the difference in normal forces between two adjacent

sections. Since the forces on the original composite section and on the

section. Deflections in ...

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 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 (see page 365), Eq. (5-4) is valid for eithersection. Deflections in ...

Page 382

... bending of reinforced concrete beams, the steel is prevented from sliding by

the grip of the enveloping concrete. The stress developed by dividing this

gripping force by the surface area of the reinforcing bars per (a)

section R0. 10-10. (b) Shear stress distribution linear inch is the bond stress. The

bond stress is analogous to the shearing stress in a homogeneous beam; it may

be computed by applying Eq. (5-4) (page 163) to the

concrete shown in Fig.

... bending of reinforced concrete beams, the steel is prevented from sliding by

the grip of the enveloping concrete. 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 distribution linear inch is the bond stress. The

bond stress is analogous to the shearing stress in a homogeneous beam; it may

be computed by applying Eq. (5-4) (page 163) to the

**equivalent**section ofconcrete shown in Fig.

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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