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Inclined shear reinforcement is practical because seismic activity can cause the loads to reverse, which could cause the cracks to form parallel to the inclined reinforcement.

Aggregate interlock is when aggregate particles on two faces of a crack are interlocked. This friction resists the sliding of the faces past each other.

A simply-supported beam has a 2 kip/ft uniformly-distributed load over the entire 8-ft span. What is the magnitude of the maximum shear force in the beam?

If φVc = 44 kip and Vu = 39 kip, are stirrups required?

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.