Lecture #19

Reading: 7:1-3

 

Shear stress from transverse shear

 

Consider a piece of a beam with constant shear force.  Again, let’s assume there is a vertical axis of symmetry.  From equilibrium,

 

             S M = 0

 

             (M+dM) - M - Vdx = 0

 

             dM = V dx

 

             V = dM/dx

 

 

So the moments on the two sides are not the same.  If we consider the stress caused by the bending moments (note this assumes that the material is linearly elastic), the stresses must be larger on the right hand side.

 

 

 

 

Now, let’s measure an arbitrary distance y from the neutral axis and slice the piece horizontally.  If we look at the piece we have cut off, we see that because the net horizontal force on each side is different, a shear force develops on the cut horizontal face.

 

           F₁ =  ∫  s  dA  =  ∫ ( M y / I ) dA

 

           F₁ = ( M / I ) ∫  y dA

 

In the same way,

 

           F₂ = ( (M+dM) / I )  ∫  y dA

 

 

We recognize that   ∫  y dA = A y_c

 

Let’s define this quantity as  Q =   ∫  y dA = A y_c = S A y_c

 

 

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