Principle Stresses and Maximum In-Plane Shear Stress

To determine the maximum and minimum normal stress we

must differentiate σx’ with respect to θ and equate to zero.

σx' = (σx + σy ) + (σx-σy) cos 2 θ + xysin 2θ

2 2

dσx' = - (σx –σy) (2sin 2θ) + 2 xy cos 2θ = 0

dθ 2

Solving this equation will give the orientation of the

planes of maximum and minimum normal stress (θp)

tan 2θp = τxy

(σx-σy)/2

N.B. The solution has two roots θp1 and θp2, the values of 2θp1 and

2θp2 are 180o apart, which means that θp1 and θp2 will be 90o apart.

Principle normal stresses can be obtained by substituting

the values of θp1 and θp2 into equation of σx’.

Sin 2θp1 = τxy / √ [(σx-σy)/2]2 + τxy2

Cos 2θp1 = [ (σx-σy)/2 ] / √ [(σx-σy)/2]2 + τxy2

Getting values of sin 2θp2 and cos 2θp2 and

substituting into σx’ we will get:

σ1,2 = ( σx + σy ) /2 ± √ [(σx-σy)/2]2 + τxy2

Depending on the sign, this equation gives either the maximum

or minimum principle normal stresses (σ1> σ2).

N.B. If the the values of θp1 and θp2 are substituted into the

shear stress equation (τx’y’ ) it will give τx’y’ = 0.

Y’

x

X’

Maximum In Plane Shear Stress

To determine the maximum and minimum shear stress we

must differentiate τx’y’ with respect to θ and equate to zero.

We will get the values of θs which determines the orientation of an element subjected to maximum shear stress:

dτx’y’ = - σx –σy sin 2 θ + τxy cos 2θ

2

dτx’y’ = - (σx –σy ) 2 cos 2θ - 2 τxy sin 2 θ = 0

θ 2

tan 2 θs = - [(σx –σy )/2 ] / τxy

tan 2 θs = - [(σx –σy )/2 ] / τxy

N.B. Each root of 2θs is 90o from 2θp , which means that θs

and θp are 45o apart.

The planes of maximum shear stress can be determined by orienting

an element 45o from the position of an element that defines the

planes of principle stress.

By substituting the values of θs1 and θs2 into the

tx’y’ equation will give:

τmax = ± √ [(σx-σy)/2]2 + τxy2

There is also a normal stress on the planes of maximum in-

plane shear stress:

o avg = (σ x + σ y ) / 2