The shear here referred to is the shear of the beam as a whole on any vertical section. It does not refer to the shearing stresses between the slab and the rib.

The theoretical computation of the shear of a T-beam is a very complicated problem. Fortunately it is unnecessary to attempt to solve it exactly. The shearing resistance is certainly far greater in the case of a T-beam than in the case of a plain beam of the same width and total depth and loaded with the same total load. Therefore, if the shearing strength is sufficient, according to the rule, for a plain beam, it is certainly sufficient for the T-beam. In the first example of Article 291, the total load on the beam is 30,000 pounds. Therefore the maximum shear V at the end of the beam, is 15,000 pounds. In this particular case, d - x = 12.25. For this beam, d= 13.75 inches, and 6 = 11 inches. Substituting these values in Equation 31, we have: v = V = 15,000 = 111 pounds per square inch.

b (d-x) 11 x 12.25

Although this is probably a very safe stress for direct shearing, it is more than double the allowable direct tension due to the diagonal stresses; and therefore ample reinforcement must be provided. If only two of the 7/8-inch bars are turned at an angle of 45° at the end, these two bars will have an area of 1.54 square inches, and will have a working tensile strength (at the unit-stress of 16,000 pounds) of 24,640 pounds. This is more than the total vertical shear at the ends of the beam; and we may therefore consider that the beam is protected against this form of failure.