Checking for Flexure-Shear Failures

As a final check that a given structural member is safe, it is necessary to ensure that the flexural reinforcement does not yield under the combined effect of applied loading including the interaction
of shear and moment. The member will have been designed against flexural forces, but the addition of the shear term in the numerator of Equation (6) means that it may still be dominated by yield of the longitudinal reinforcement. To ensure that this does not occur, the following equation is derived from Figure 4 by taking moments about point O and solving for the necessary force in the reinforcement:

Flt = (Тзд + °‘5Nf + (Vf ~ °’5Vs ~ Vp) cot0‘ (7)

The reinforcement that is provided at the location where this equation is being checked must be sufficiently developed to resist the force calculated by Equation (7). This force need not be taken as greater than the reinforcement force calculated under maximum moment alone at the maximum moment location of the same member. The reason for this limit is that near applied loads, where the moment is maximum, the shear is carried by a “compression fan”, similar to a strut-and-tie model rather than by the sectional behaviour assumed in Figure 4.

If Equation (7) is not satisfied, then the applied loads (Mf, Nf, Vf), can be proportionately reduced until it is satisfied, and these new values are the final estimate of the strength of the member. A member which is controlled by this equation is predicted to fail in a flexure-shear mode rather than simply in flexure or shear alone.