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A beam has dimensions of b = 14 in., h = 30 in., d’ = 2.5 in., and d = 27.5 in. It is reinforced with 2 No. 5 bars on the compression side and 4 No. 8 bars on the tension side. The concrete strength is 3,400 psi, and the yield strength of the reinforcement is 60,000 psi. For the results given below, determine the strength Mn for this beam.a = 4.09 in. (depth of concrete stress block)Cc = 163.7 kips (compressive force in concrete)Cs = 25.9 kips (compressive force in steel)T = 189.6 kips (tensile force in steel)

Is the strength reduction factor greater for shear or flexure?

If φVc = 32 kip and Vu = 24 kip, are stirrups required?

_______ cracks occur first due to flexural stresses and start at the bottom of the beam.

The traditional theory for shear stress assumes homogeneous, elastic, and cracked beams.

Compute φVn for the cross section shown. Assume f’c = 3,000 psi, fyt = 40,000 psi, b = 14 in., d = 25 in., and that there are four No. 8 longitudinal tension bars and No. 3 stirrups at 12 in. o.c.

Determine the bar-spacing factor, cb, for a simply supported rectangular beam with b = 22 in. and No. 3 stirrups. This beam has three No. 7 bars as longitudinal reinforcement. The clear cover is 1.6 in.

Compute φVn for the cross section shown. Assume f’c = 4,000 psi, fyt = 60,000 psi, b = 15 in., d = 24 in., and that there are four No. 7 longitudinal tension bars and No. 5 stirrups at 6 in. o.c.

A combination of horizontal and vertical reinforcement is used to prevent cracks from being too narrow.

Stirrups are able to resist shear unless they are crossed by an inclined crack.