Shearing Stress Distribution in different cross Sections

1. Shearing Stress Distribution in Rectangular cross Section 

Let us consider the rectangular section ABCD of a beam as shown in figure. 

We have assumed one layer EF at a distance y from the neutral axis of the beam section.

The shear force acting on section is = F

b= Width of the rectangular section

d= Depth of the rectangular section

N.A: Neutral axis of the beam section

EF: Layer of the beam at a distance y from the neutral axis of the beam section

A- Area of section CDEF, where shear stress is to be determined 

A = b (d/2 - y)

ȳ = Distance of C.G of the area CDEF from neutral axis of the beam section

ȳ = y + (d/2 - y)/2 = (d/2 +y)/2

I = Moment of inertia of the given section about the neutral axis (mm4) 

I = bd3/12

Shear stress at a section will be given by following formula 

substitute the values in the above shear stress equation, the shear stress in the layer EF is,

when y= d/2 , the shear stress is zero

y = 0 , the shear stress  at neutral axis is maximum and equal to τmax

τmax = 3/2 F/bd

The average shear stress 

τave = F/bd =Shear force / Area of cross section

For a rectangular section,

The maximum shear stress τmax = 1.5 τave 

2. Shearing Stress Distribution in Triangular cross Section 

Let us consider a triangular section of base b and height h. Its centroid is located at a distance h/3 from the base. 

The shear force acting on section is = F.

Assume the shear stress at the plane EF, at a distance y from the neutral axis be τy.

Assume b′ & h′ be the width & height of the triangle above the plane EF.
From same triangles,
b / b′= h/ h′
b′ = bh′ / h
h′ = ((2/3) h - y )
b′ = (b/ h) ( (2/3) h - y )
Centroid of the triangle above the plane EF, through the neutral axis y' = ( y +( h′ /3 ))
On putting the value of h′ 

y' = y + (1/3)((2/3)h - y ) =(2/3)(y+(h/3))

Moment of this triangular area around the neutral axis Q = Ay'

The shear stress T, at the plane EF is

For shear stress τ to be maximum, dT / dy = 0 ,

(h /3) - 2 y = 0

∴ y = h/6

The maximum shear stress is at a distance form the base of the triangle is = h/3 +y

                                = h/ 3+ h/6 = h/2

∴ Maximum shear stress is at a distance of h/2 form the base of the triangle, that is also at a distance of h/6 from the centroidal axis.

Putting y = h/ 6 in the equation for the shear stress, the maximum shear stress

The shear stress at the neutral axis, (when y = 0), 

τNA = (F /3I ) (2h.h/9)
      = (8/3) ×(F/bh)

The shear stress distribution diagram is as shown in Figure.

Note:

The average shear stress = τave = F/A = 2F/bh

The maximum shear stress in Triangular section

 τmax  = 3Vx/bh = 1.5 τave

The  shear stress at N.A in Triangular section

τNA  = 8Vx/3bh = 4/3 τave