Wednesday, April 23, 2014


Cantilever beam with point load

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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