Lecture #14

Reading: 6:2

 

Shear and Bending Moment Diagrams

 

If we consider a small piece of beam with a distributed load and draw a free body diagram with positive internal forces, we get

 

 

The shear and moment on the right side will be different from the left because of the distributed load, but not much different because dx is small.  Let V₂ = V₁ + dV and M₂ = M₁ + dM, then apply equilibrium equations.

 

 

 

S F_y = 0 :          V₁ + w dx - ( V₁ + dV ) = 0

 

                          w dx - dV = 0

 

                          dV = w dx                     or                       dV/dx = w

 

S M = 0 :          ( M₁ + dM ) - M₁ - V₁ dx - w dx ( dx / 2 ) = 0

 

                          dM - V₁ dx - w dx² / 2 = 0

 

Since dM and dx are small, dx² is very small and can be neglected.  We can also drop the subscript on V at this time.

 

                          dM = V dx                     or                       dM/dx = V

 

These equations are the basis of what I call the summation method.

 

dV/dx = w  says that the slope of the shear diagram is equal to the distributed load.

 

Integrating dV = w dx , we get

 

        ∫  dV =   ∫  w dx

 

             V₂ - V₁ =  ∫  w dx                       or                       V₂ = V₁ +  ∫  w dx

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