Now, from equilibrium,

 

             S F_x = 0 :          dF + F₁ = F₂

 

             dF = F₂ - F₁ = ((M + dM) Q / I)  -  (M Q / I) = dM Q / I = V dx Q / I

 

             dF/dx = VQ/I

 

This is a force per unit length of the beam.  If we divide by the thickness of the bottom cut surface, we get stress.

 

             t = (dF/dx)/t = VQ/(I t)

 

Note that by looking at an infinitesimal element at the corners of the bottom surface, we find the stress on the cross section is equal to that found on the bottom surface.  Also note that the shear stress on the cross section is in the same direction as the shear force on the cross section.

 

 

 

Let’s consider the shear stress variation on a rectangular cross section.  To calculate the stress at an arbitrary location y above the neutral axis, first draw a line perpendicular to V through the cross section at the point of interest.  Q is found from the area above this line.  The thickness t is the length of the line that cuts through material.

 

             Q = A y = b (h/2—y) (h/2 + y)/2 = b/2 ((h/2)² - y²)

 

             t = b

 

 

             t = VQ/(I t) = [ V (b/2)((h/2)² - y²)]/[(bh³/12) b]

 

We see immediately that the stress varies parabolically or quadradically with y.  If y = m h/2, we see that t = 0.  The maximum value of shear stress occurs when y = 0.

 

             t_max = [V b/2 (h/2)²]/[(bh³/12) b] = [12 V b h²]/[8 b²h³] = 3V/[2 bh] = 3V/(2A)

 

We have just shown that the maximum shear stress for a rectangular cross section is 50% larger than the average shear stress.

 

 

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