Assume a beam of length L = 10 ft, supporting a load P = 2 k. The beam bending moment and shear force may be computed, like the external reactions, by equations of equilibrium ΣH=0, ΣV= 0, and ΣM=0. Bending moment and shear force cause bending-and shear stress, similar to axial load yielding axial stress f= P/A.

1 Cantilever beam with concentrated load

V Shear diagram (shear force at any point along beam)

M Bending moment diagram (bending moment at any point along beam)

∆ Deflection diagram (exaggerated for clarity)

**Reactions,**found by equilibrium, ΣV=0 (up +) and ΣM=0 (clockwise +)

**Shear V,**found by vertical equilibrium, ΣV=0 (up +)

Left of a and right of b, shear is zero because there is no beam to resist it (reaction at b reduces shear to zero). Shear may be checked, considering it starts and stops with zero. Concentrated loads or reactions change shear from left to right of them. Without load between a and b (beam DL assumed negligible) shear is constant.

**Bending moment M,**found by moment equilibrium, ΣM=0 (clockwise +)

The mid-span moment being half the moment at b implies linear distribution. The support reaction moment is equal and opposite to the beam moment.

**Deflection ∆**is described later. Diagrams visualize positive and negative bending by concave and convex curvature, respectively. They are drawn, visualizing a highly flexible beam, and may be used to verify bending.

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