A machine element is subjected to the biaxial state of stress σ_x = 80 MPa, σ_y = 20 MPa and τ_xy = 40 MPa. If the shear strength of the material is 100 MPa then factor of safety as per maximum shear stress theory is.

Solution:

Tresca's theory: 

• Tresca's criterion states that a material point yields when the maximum shear stress at that point reaches the maximum shear stress in a uniaxial tension specimen at yield. 
• This theory is suitable for ductile materials.

​In a uniaxial tensile test, the principal stress at the yield point will be σ1=σy, σ2=σ0, σ3=σ0, where σy= yield stress

• So, maximum shear stress at yielding for the uniaxial test,

\(\tau = \frac{\sigma_1}{2} = \frac{\sigma_Y}{2}\)

• Then according to maximum shear stress theory, for no failure: \(\tau_{max\ bi axial} \leq \tau_{max\ uni\ axial}\)

\(\tau_{max} = \frac{\sigma _Y}{2}\)

• For design, the maximum shear stress formula is given as : \(\tau_{max} = \frac{\sigma _{YT} / F.O.S.}{2}=\frac{\sigma _{YS}}{ F.O.S}\)

where, F.O.S = Factor of safety

Calculation:

Given:

σx = 80MPa ,  σY= 20 MPa , Τxy = 40Mpa

Principal stresses:

\(σ_{1,2} = \frac{σ{x}+σ{y}}{2} \pm \sqrt {({\frac{σ{x}-σ{y}}{2}})^2 +τ_{xy}^2}\)

\(σ_{1,2} = \frac{80+20}{2} \pm \sqrt {({\frac{80-20}{2}})^2 +40^2} \)

So, 

σ1 = 100 MPa,  σ2 = 0 MPa

Shear stress, \(τ = \frac{\sigma_1 -\sigma_2}{2} = \frac{100-0}{2} =50 MPa\)

As per Tresca's theory, 

\(F.O.S. = \frac{Material \ Strength}{ Shear\ stress} \Rightarrow F.O.S. =\frac{100}{50}=2\)