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FIGURE 2.4 Bending moment and shear force diagrams.

FIGURE 2.4 Bending moment and shear force diagrams.

In the portion AC, the bending moment at a cross-section a distance x from point A is M = 55x. Therefore, the bending moment diagram starts at 0 at A and increases along a straight line to an ordinate of +165 at point C. In the portion CD, the bending moment at any point a distance x from C is M = 55(x+3) — 30x. Hence, the bending moment diagram in this portion is a straight line increasing from +165 at C to +265 at D. In the portion DE, the bending moment at any point a distance x from D is M = 55(x+7) — 30(x + 4) — 4x2/22. Hence, the bending moment diagram in this portion is a curve with an ordinate of +265 at D and +343 at E. In an analogous manner, the remainder of the bending moment diagram can easily be constructed.

Bending moment and shear force diagrams for beams with simple boundary conditions and subjected to some selected load cases are given in Figure 2.6.

### 2.2.3 Fixed-Ended Beams

When the ends of a beam are held so firmly that they are not free to rotate under the action of applied loads, the beam is known as a built-in or fixed-ended beam and it is statically indeterminate. The bending moment diagram for such a beam can be considered to consist of two parts, namely, the free bending moment diagram obtained by treating the beam as if the ends are simply supported and the fixing moment diagram resulting from the restraints imposed at the ends of the beam. The solution of a fixed beam is greatly simplified by considering Mohr's principles, which state that

1. The area of the fixing bending moment diagram is equal to that of the free bending moment diagram.

2. The centers of gravity of the two diagrams lie in the same vertical line; that is, they are equidistant from a given end of the beam.

The construction of the bending moment diagram for a fixed beam is explained with an example shown in Figure 2.7. PQUT is the free bending moment diagram, Ms and PQRS is the fixing moment diagram, M;. The net bending moment diagram, M, is shown shaded. If As is the area of the free bending moment diagram and Ai the area of the fixing moment diagram then, from the first Mohr principle we have As = Ai and

From the second principle, equating the moment about A of As and Ai, we have

Solving Equations 2.7 and 2.8 for MA and MB, we get

Wab2

Wa2b L2

Shear force can be determined once the bending moment is known. The shear force at the ends of the beam, that is, at A and B, will be

Bending moment and shear force diagrams for fixed-ended beams subjected to some typical loading cases are shown in Figure 2.8.

Shear force

Bending moment

0 0