Simple Shear Stress and Shear Strain

Types of Stress

Generally the internal resisting force per unit area is called stress. 

If this stress is normal (perpendicular) to the resisting cross section is called Normal stress or Axial stress and corresponding Strain is called Normal Strain.

If the stress is tangential (parallel) to the resisting cross section is called Tangential Stress or Shear stress and corresponding strain is called Shear strain.

Shear Stress

The internal resistance of material per unit area in the tangential direction of the cross section is called Tangential Stress or Shear stress. Generally the internal tangential resistance is called Shear force. So the resistance of shear force per unit area is called Shear stress.

The tangential force or Shear force acting on surface  = S

The cross sectional area of face  = A

Therefore the Shear stress

   Ƭ= Shear force /Area of cross   section = S/A

The unit of Shear stress is expressed as

1 N/m2 ,kN/m2 ,  MN/m2 

Note: Unlike normal stress, the distribution of shearing stresses Ƭ across a section cannot be taken as uniform. Dividing the total shear force F by the cross-sectional area A over which it acts, we can determine the average shear stress in the section: Ƭave= F/A.

Example:

Shearing stresses are commonly found in rivets, pins and bolts. If the plates, which are connected by a rivit as shown in the following figure, are subjected to tension forces, shear stresses will develop in the rivet. 

The shear force P in the shear plane is equal to tension force F. 

The average shear stress in the plane is

 Ƭave= F/A. 

This joint is said to be in single shear.

If the plates, which are connected by a rivet as shown in the following figure, are subjected to tension forces, shear stresses will develop in the rivet. This joint is said to be in double shear. To determine the average shear stress in each shear plane, free-body diagrams of rivet and of the portion of rivet located between the two planes are drawn. 

Observing that the shear P in each of the planes is P = F/2, the average shearing stress is

τave = F/2A

This joint is said to be in Double shear.

Shear strain

Shearing stresses in a material give rise to shearing strains. Consider a rectangular block of material, Figure, subjected to shearing stresses τ in one plane. 

The shearing stresses distort the rectangular face ABCD of the block into a parallelogram of ABC'D'. 

If the right-angles at the corners of the face change by amounts  ø, then ø is the shearing strain. 

Tanø = DD'/ AD = DD'/ h = ø 

{for small values of  ø , tan ø =  ø }

The angle ø is measured in radians, and is non-dimensional.

Modulus of Rigidity

With in the elastic limit the ratio of shear stress to shear strain is always constant, witch is called Rigidity Modulus or Shear Modulus.

It is also defined as the Shear Stress required to produce one unit Shear strain within the elastic limit is also called Shear Modulus.

Rigidity Modulus G = Shear stress/ Shear strain

G =  Ƭ/ø 

Complementary shear stresses

Principle of complementary shear stresses states that a set of shear stresses across a plane is always accompanied by a set of balancing shear stresses (i.e., of the same intensity) across the plane and normal to it.

Let us consider an element ABCD of uniform thickness of one unit, and subjected to pure shear τ in face AB and CD

Shear force on face AB = F = τ.AB 

Shear force on face CD = F = τ.CD ...(i) 

These parallel and equal forces form a couple. The moment of couple =

                         M = τ.AB.AD or τ.CD.AD ...(ii) 

For equilibrium of element there must be a restoring couple on the element.

Let assume the shear stress acting on the face AD and BC is τ' 

Shear force on face AD or BC 

                            F'= τ'.AD or τ'.BC ...(iii) 

They also form a couple as 

                  M' = τ'.AD.AB or τ'.BC.AB ...(iv) 

The moment of two couple must be equal;

M = M'

                        τ.AB.AD = τ'.AD.AB or 

                             τ = τ'

 τ' is called complementary shear stress

Example:

1.A force of 31.68 kN is required to punch a circular hole of 14 mm diameter in metal plate of 2 mm thickness. Calculate the ultimate shear stress and compressive stress developed in the punching rod.

Solution:

The punching surface area of the hole As = perimeter of the hole x thickness of the plate

As= p dxt = p x14x2 = 87.9645 mm2

The Ultimate shear stress = F/As

                                = 31.68x1000 / 87.9645

                                = 360.14 N/mm2

The compressive stress in the rod = F/A

                     = Force in the rod/ area of the road

                     = 31680/ 153.938

                     = 205.79 N/mm2