Question

For the beam and loading shown, write the equations for the internal shear force V and bending moment Mas function of x and then draw the shear and moment diagrams. Also state the values of the internal shear force and bending moment at x = 1.4 m and x = 4.9 m. (Answer: At x = 1.4 m, V = 8.51 kN and M = -23.5 kN m and at x = 4.9 m, V = 2.93 kN and M = -2.77 kN m] Note: The best way to handle this is the integration method. Remember there is a reaction moment at the wall at A. That moment acts like an applied moment and that means there is an initial bending moment which is opposite in sign than the applied moment. In other words, the bending moment shifts immediately upward from 0 at x = 0 if the reaction moment at the wall is CW and downward if is CCW. w 1 w = k1x – k2x2 1.8 kN/m 1 4.1 m -2.9 m

Answer 1

The placement of the bars in a beam will depend on the shear force and the bending moment diagram.

The diagram of bending and shearing forces shows the moments and shearing forces acting on any part of the beam. The layout of reinforcement is dependent on the size and sign (or direction) of diagrams in order to withstand these active forces. For instance, shear reinforcement or stirrups are tightly spaced on the region of the beam where shear force is greater. Additionally, maximal longitudinal reinforcement is supplied in the area with the highest bending moment. Similar to this, some longitudinal reinforcement is reduced when the influence of the bending moment decreases, when the direction of the bending stresses changes, or when the stresses in the web reinforcement increase.

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