What is Maximum shear stress theory?
Maximum shear stress theory is one of the theories of failure used for the safe design of mechanical components and it is suitable for a ductile material. The maximum shear stress theory is also called as Tresca theory of failure. The method is not suitable in hydrostatic stress conditions.
Statement of Maximum shear stress theory:
The maximum shear stress theory says that failure will occur when the maximum shear stress exceeds the shear stress at uniaxial loading.
It means that,
Maximum shear stress (Biaxial or Triaxial) ≤ `\tau_{\text{uniaxial}}`
`\tau_{max}` ≤ `\tau_{\text{uniaxial}}`
`\tau_{max}` ≤ `\frac{\sigma _{y}}{2}`
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Maximum shear stress theory in three-dimensional loading:
The maximum shear stress in triaxial loading is given by
`\tau_{max}`=`| \frac{\sigma _{1}-\sigma _{2}}{2},\frac{\sigma _{2}-\sigma _{3}}{2},\frac{\sigma _{1}-\sigma _{3}}{2}|`
`\frac{\sigma_{y}}{2}` = `| \frac{\sigma _{1}-\sigma _{2}}{2},\frac{\sigma _{2}-\sigma _{3}}{2},\frac{\sigma _{1}-\sigma _{3}}{2}|` —— [∵ `\tau_{max}` = `\frac{\sigma_{y}}{2}`]
Hence the conditions for the failure in three dimensional force are,
`\frac{\sigma _{1}-\sigma _{2}}{2}\leq \frac{\sigma _{y}}{2}`
`\frac{\sigma _{2}-\sigma _{3}}{2}\leq \frac{\sigma _{y}}{2}`
`\frac{\sigma _{1}-\sigma _{3}}{2}\leq \frac{\sigma _{y}}{2}`
Therefore the equations can written as,
`| \sigma _{1}-\sigma _{2} |\leq \sigma _{y}`
`| \sigma _{2}-\sigma _{3} |\leq \sigma _{y}`
`| \sigma _{1}-\sigma _{3} |\leq \sigma _{y}`
Where, `\sigma_{1}, \sigma_{2}, \sigma_{3}` are the principal stresses
Maximum shear stress theory for biaxial loading:
For biaxial loading one of the principal stress will become zero.
Hence by putting `\sigma_{3}` = 0 in above three equations we get,
`| \sigma _{1}-\sigma _{2} |\leq \sigma _{y}` —–(1)
`| \sigma _{2} |\leq \sigma _{y}` —–(2)
`| \sigma _{1} |\leq \sigma _{y}` —-(3)
Note:- Condition 1 is considered when values of `σ_{1}` and `σ_{2}` are different, while condition 2 or 3 are considered when `σ_{1}` and `σ_{2}` are same.
Yield surface for Maximum shear stress theory in biaxial loading:
For the tensile and compressive yield strength, the equations are written as,
Boundaries of region for tensile condition,
`| \sigma _{1}-\sigma _{2} |= \sigma _{y}` ——(a)
`| \sigma _{2} |= \sigma _{y}` ——(b)
`| \sigma _{1} |= \sigma _{y}` ——(c)
For the compressive condition,
`| \sigma _{1}-\sigma _{2} |= -\sigma _{y}` ——(d)
`| \sigma _{2} |= -\sigma _{y}` ——(e)
`| \sigma _{1} |= -\sigma _{y}` ——(f)
From these equations, yield surface can be plotted as,
Maximum shear stress theory formula in form of axial stresses `(\sigma _{x}` and `\sigma _{y})`:
The condition for maximum shear stress failure in biaxial loading is,
`\sigma _{1}-\sigma _{2}=\sigma _{y}`
But `\sigma _{1}` and `\sigma _{2}` are principal stresses which are given by,
`\sigma _{1}=\frac{\sigma _{x}+\sigma _{y}}{2}+\sqrt{(\frac{\sigma _{x}-\sigma _{y}}{2})^{2}+\tau ^{2}_{xy}}`
`\sigma _{2}=\frac{\sigma _{x}+\sigma _{y}}{2}-\sqrt{(\frac{\sigma _{x}-\sigma _{y}}{2})^{2}+\tau ^{2}_{xy}}`
Now, by putting the values of `\sigma_{1}` and `\sigma_{2}` in equation of maximum shear stress theory we get,
`\sigma _{1}-\sigma _{2}=\sigma _{y}`
`(\frac{\sigma _{x}+\sigma _{y}}{2}+\sqrt{(\frac{\sigma _{x}-\sigma _{y}}{2})^{2}+\tau_{xy}^{2}})-(\frac{\sigma _{x}+\sigma _{y}}{2}-\sqrt{(\frac{\sigma _{x}-\sigma _{y}}{2})^{2}+\tau_{xy}^{2}})=\sigma _{y}`
`2\times \sqrt{(\frac{\sigma _{x}-\sigma _{y}}{2})^{2}+\tau_{xy}^{2}}=\sigma _{y}`
`\sqrt{(\sigma _{x}-\sigma _{y})^{2}+4\times \tau_{xy}^{2}}=\sigma _{y}`
This is the equation of maximum shear stress theory for biaxial stresses.
Numerical on maximum shear stress theory:
The loading on the object has values `σ_{x}` = 120 Mpa, `σ_{y}` = 60 Mpa, `\tau_{\text{xy}}` = 30 Mpa while the yield strength of the material of the object is 300 Mpa. Find the F.O.S. by using maximum shear stress theory?
Given:-
`σ_{x}` = 120 Mpa
`σ_{y}` = 60 Mpa
`\tau_{\text{xy}}` = 30 Mpa
`S_{yt}` = 300 Mpa
Solution:-
By using the formula for maximum shear stress theory for biaxial loading.
`\sigma _{y}=\sqrt{(\sigma _{x}-\sigma _{y})^{2}+4\tau_{xy}^{2}}`
`\sigma _{y}=\sqrt{(120-60)^{2}+4\times (30)^{2}}`
`\sigma _{y}` = 84.85 Mpa
Now, the factor of safety is given by, (F.O.S.)=
`=\frac{S_{yt}}{\sigma _{y}}`
`=\frac{300}{84.85}`
= 3.53
F.O.S. = 3.53
Therefore the FOS by using the maximum shear stress theory is 3.53.
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