The stresses on the surface of a machine component are shown in Figure P15.82. The ultimate failure strengths for this material are 200 MPa in tension and 600 MPa in compression. Use the Mohr failure criterion to determine whether this component is safe for the state of stress shown. Support your

Question

The stresses on the surface of a machine component are shown in Figure P15.82. The ultimate failure strengths for this material are 200 MPa in tension and 600 MPa in compression. Use the Mohr failure criterion to determine whether this component is safe for the state of stress shown. Support your answer with appropriate documentation.

Accepted Answer

Principal stresses:

[latex]\begin{aligned}\sigma_{p 1, p 2} &=\frac{(150 MPa )+(0 MPa )}{2} \pm \sqrt{\left(\frac{(150 MPa )-(0 MPa )}{2} ight)^{2}+(100 MPa )^{2}} \&=75 MPa \pm 125 MPa\end{aligned}[/latex]

therefore, [latex]\sigma_{p 1}=200 MPa[/latex] and [latex]\sigma_{p 2}=-50 MPa[/latex]

Mohr failure criterion:
If [latex]\sigma_{p 1}[/latex] is positive and [latex]\sigma_{p 2}[/latex] is negative, then failure will occur if the following interaction equation is greater than or equal to 1:

[latex]\frac{\sigma_{p 1}}{\sigma_{U T}}-\frac{\sigma_{p 2}}{\sigma_{U C}} \geq 1[/latex].

For the principal stresses existing in the component:

[latex]\begin{aligned}\frac{\sigma_{p 1}}{\sigma_{U T}}-\frac{\sigma_{p 2}}{\sigma_{U C}} &=\frac{200 MPa }{200 MPa }-\frac{(-50 MPa )}{600 MPa } \&=1.0-(-0.0833) \&=1.0833>1 ext{N.G.} \end{aligned}[/latex]

Therefore, the component fails according to the Mohr failure criterion.

The factor of safety corresponding to the Mohr failure criterion is:

[latex]FS =\frac{1}{1.0833}=0.923[/latex]

Question 15.82: The stresses on the surface of a machine component are shown...

Related Answered Questions