Maximum Shear Stress:

We have the common expression for shear stress as

τ_nt =τ_xy cos 2θ -(σ_x-σ_y/2) sin 2θ

Differentiating w.r.t. θ, and equating the derivative to zero,

dτ_nt/dθ  = -2 τ_xy sin 2θ - (σ_x - σ_y) cos 2θ = 0

∴  tan 2θ = -(σ_x -σ_y)/2τ_xy                                                                                    

Because the planes on which maximum shear stresses occur are specific set of planes we may denote them distinctly by Ψ (instead of general aspect angle θ).

Comparing Eqs. (2) and (7), we conclude that 2Ψ = 2Φ ± 90º as tan 2Φ. tan 2Ψ = - 1.

∴          Ψ = Φ ± 45^o

Eq. (8) denotes that the planes of maximum shear stress bisect the right angles between the main and minor principal planes.

The normals to the main and minor principal planes may now be described as the major and minor principal axes. At one the principal stresses and principal planes are known, further analysis may be simplified by expressing the state of stress w.r.t. a new coordinate system with major and minor principal axes as coordinate axes themselves. These axes are usually called axes 1 and 2 respectively.

The general expressions for stress components on arbitrary planes whose aspect angle θ¯ may now be measured with axis-1 as reference axis.

Hence,

σ_n = (σ₁+σ₂/2) +(σ₁ +σ₂/2) cos 2 θ¯

τ_nt = (σ₁ -σ₂/2) sin 2 θ¯

Eq. (8) already defines that θ should be ± 45º for τnt to be maximum.

Thus, τ_{max, min}  =  σ₁ -σ₂/2  sin (±90^o )

∴  τ_max, =  σ₁ -σ₂/2

And  τ_min, =  σ₁ -σ₂/2 

Since the sign of maximum shear stress is not significant, expression for τ_min is not generally used.