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 + τxysin 2θ
2
2
dσx’ = - (σx –σy) (2sin 2θ) + 2 τxy 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 inplane shear stress:
σ avg = (σ x + σ y ) / 2