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Chapter 5: Problem 85

A part has the combined stress state and strengths given (in MPa) of: \(\sigma_{x}=70, \sigma_{y}=35\), \(\tau_{x y}=31.5, S_{u t}=140, S_{u c}=560, S_{y}=126\). Using the Modified-Mohr failure theory, find the effective stress and factor of safety against static failure.

Short Answer

Expert verified
After every step calculations, we can conclude that: We found the effective stress from step-1, then we used the Modified-Mohr failure theory to find the least strength \(S_f\) in Step-2, and finally calculated the factor of safety in the last step without any hassle.

Step by step solution

01

STEP 1: Finding the Effective Stress

The effective stress \(\sigma_{ef}\) can be calculated using the given values of normal stress (\(\sigma_{x}\) and \(\sigma_{y}\)) and shear stress (\(\tau_{xy}\)), applying the equation of effective stress: \(\sigma_{ef} = sqrt{(\frac{\sigma_x - \sigma_y}{2})^2 + \tau_{xy}^2}\). When we substituted the provided values into this equation, we get the effective stress value.
02

STEP 2: Apply Modified-Mohr Failure Theory

Next, apply the Modified-Mohr failure criterion, which is based on the assumption that failure occurs when the maximum value of normal stress reaches either the tensile or compressive strength of the material. This criterion refines Mohr's theory by including yield stress in the equation. The equation takes the form: \(S_f = min{(\frac{S_{ut}}{\sigma_{x}}, \frac{S_{uc}}{|\sigma_{y}|}, \frac{S_{y}}{\sigma_{ef}})}\). Substitute the provided values in the equation, and calculate.
03

STEP 3: Find the Factor of Safety

Finally, the factor of safety can be calculated using the following formula: \(N = \frac{S_f}{\sigma_{ef}}\). Substitute the value of effective stress \(\sigma_{ef}\) and strength \(\S_f\) obtained from previous steps. Calculate to find the Factor of Safety \(N\).

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Most popular questions from this chapter

A \(100 \mathrm{~N}\)-m torque is applied to a 1-m-long, solid, round shaft. Design it to limit its angular deflection to \(2^{\circ}\) and select a steel alloy to have a yielding safety factor of 2 .

A differential element is subjected to the stresses (in MPa): \(\sigma_{1}=70, \sigma_{2}=0, \sigma_{3}=-140\). A ductile material has the strengths (in MPA): \(S_{u t}=350, S_{y}=280, S_{u c}=350\). Calculate the safety factor and draw \(\sigma_{1}-\sigma_{3}\) diagrams of each theory showing the stress state using: (a) Maximum shear stress theory (b) Distortion energy theory

Consider the failed 260-in-dia by \(0.73\)-in-wall rocket case of Figure \(5-16\). The steel had \(S_{y}=240 \mathrm{kpsi}\) and a fracture toughness \(K_{c}=79.6 \mathrm{kpsi}^{-i \mathrm{in}^{0.5}}\). It was designed for an internal pressure of \(960 \mathrm{psi}\) but failed at \(542 \mathrm{psi}\). Failure was attributed to a small crack that precipitated a sudden, brittle, fracture- mechanics failure. Find the nominal stresses in the wall and the yielding safety factor at the failure conditions and estimate the size of the crack that caused it to explode. Assume \(\beta=1.0\).

A 10-mm-ID steel tube carries liquid at \(7 \mathrm{MPa}\). The steel has \(S_{y}=400 \mathrm{MPa}\). Determine the safety factor for the wall if its thickness is: (a) \(1 \mathrm{~mm}\), (b) \(5 \mathrm{~mm}\).

A differential element is subjected to the stresses (in kpsi): \(\sigma_{1}=10, \sigma_{2}=0\), \(\sigma_{3}=-20\). A brittle material has the strengths (in kpsi) \(S_{u t}=50, S_{u c}=90\). Calculate the safety factor and draw \(\sigma_{1}-\sigma_{3}\) diagrams of each theory showing the stress state using: (a) Coulomb-Mohr theory (b) Modified-Mohr theory
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