Shear Stress Distribution in Beams

When a beam is loaded, bending moment and shear force are developed at all section of the beam. In Unit 2, we have already learnt the methods of determining shear force and bending moment in a beam section under the given loading  conditions. 

Previously, we have seen the bending stress distribution at any cross section of the beam. Now, we will study the shear stress distribution at any section of the beam.

Shear Stress Distribution in Beams

The vertical shear force at any section of a beam produces shear stress at that section that varies along the depth of the section. This vertical shear stress is accompanied by a horizontal shear stress of equal magnitude, known as complementary shear stress. So at any point in a cross section of the beam, there is a  vertical shear stress and a horizontal shear stress of equal magnitude of τ. 

These two shear stresses cause the diagonal tension and compression inclined at 45 degrees to the horizontal.

Let us consider a element of length dx of the beam and subjected to bending moments M and M+dM shown in fig. 

Let dA be an elementary area at a distance y from the neutral axis. 

a = ∑dA - is the area of the beam above upper shaded portion of the beam

sy - is the bending stresses at a distance y from the neutral axis is equal to [My/I] and [(M+dM)y/I] 

For the upper shaded portion of the beam, 

FL - the total normal forces due to the bending stresses on the left side of the beam. 

FL = ∑dA.sy = ∑dA.(My/I)  --- (1)            FR - the total normal forces due to the bending stresses on the right side of the beam. 

FR = ∑dA.sy = ∑dA.[(M+dM)y/I] ---(2)

fv -  the net unbalanced force of FR and FL forces will be resisted by the shearing force, acting at the boundary surface between the shaded and the un shaded portions.

fV = τ b.dx  ---(3)

For equilibrium of the upper shaded portion 

FL+fV–FR = 0 

fV  =  FR - FL 

Substitute the the values in above equation  

 fV = τ b.dx 

        =  ∑dA.[(M+dM)y/I] - ∑dA.(My/I)

τ.b.dx  = ∑dA.y(dM)/I 

τ = ∑dA.y/Ib . (dM/dx)

But dx/dM  = Shear Force (Vx) at the section.

(Rate of change of bending moment is represents the shear Force at the section ) 

and ∑dA.y = a.y' = Q , represents the first moment of an area of the shaded section about N.A. 

The Shear stress at a layer is 

τy = Vxay'/ Ib = VxQ/ Ib

Where 

a = Area between the extreme face of beam and the plane at which the shear stress is τy 

y' = Distance of the centroid of area ‘a’ from N.A

Q = a.y' = the first moment of an area of the shaded section about N.A.

Vx = Shear force at the cross-section. 

I = Moment of Inertia of the beam cross section about N.A. 

b = Width of the fiber at the shear plane 

Note: 

1. The maximum shear stress occurs at the neutral axis and is zero at both the top and bottom surface of the beam. 

Shear flow has the units of force per unit distance.

2. For a rectangular section, the maximum shear stress τmax = 1.5 τave 

The average shear stress 

τave = Shear force / Area of cross section

3. For a circular section, the maximum shear stress τmax = 4/3 τave

The shear stress distribution across a circular section is parabolic curve. 

4.In case of triangular section , the shear stress is not maximum at N.A. The maximum shear stress is located at a height of h/2 

The maximum shear stress in Triangular section

 τmax = 1.5 τave = 3Vx/bh 

The  shear stress at N.A in Triangular section

τNA = 4/3 τave = 8Vx/3bh 

The shear stress distribution for unsymmetrical sections is obtained after the calculating the N.A.