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Act like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer:1. Loads acting on a structural member produce stresses at critical point in the member as follows: 0,=125 MPa, 0,=-16 MPa and t= 55 MPa . Determine the factor of safety (SF) if the material is an isotropic material with a yield stress of 300 MPa using a) Tresca yield criterion, and b) Von Mises criterion (10 Marks).

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Act like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer:Uploaded Image1. Loads acting on a structural member produce stresses at critical point in the member as follows: 0,=125 MPa, 0,=-16 MPa and t= 55 MPa . Determine the factor of safety (SF) if the material is an isotropic material with a yield stress of 300 MPa using a) Tresca yield criterion, and b) Von Mises criterion (10 Marks).

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Asked by: manish kumar
Created at: 2025-10-04 22:43:20

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Let's solve the problem step-by-step: --- ## **Given Data** - \(\sigma_x = 125\) MPa - \(\sigma_y = -16\) MPa - \(\tau_{xy} = 55\) MPa - Yield stress, \(\sigma_{y,mat} = 300\) MPa We are to find the **factor of safety (SF)** using: a) **Tresca yield criterion** b) **Von Mises criterion** --- ## **Step 1: Find Principal Stresses (\(\sigma_1\) and \(\sigma_2\))** The principal stresses for a 2D stress state: \[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \] Substitute values: \[ \frac{\sigma_x + \sigma_y}{2} = \frac{125 + (-16)}{2} = \frac{109}{2} = 54.5~\text{MPa} \] \[ \frac{\sigma_x - \sigma_y}{2} = \frac{125 - (-16)}{2} = \frac{141}{2} = 70.5~\text{MPa} \] Calculate the square root term: \[ \sqrt{(70.5)^2 + (55)^2} = \sqrt{497.25 + 3025} = \sqrt{7995.25} \approx 89.49~\text{MPa} \] So, \[ \sigma_1 = 54.5 + 89.49 \approx 143.99~\text{MPa} \] \[ \sigma_2 = 54.5 - 89.49 \approx -34.99~\text{MPa} \] --- ## **Step 2: Tresca Yield Criterion** Tresca says yield occurs when the maximum shear stress reaches half the yield strength: \[ \tau_{max} = \frac{|\sigma_1 - \sigma_2|}{2} \] \[ \tau_{max} = \frac{|143.99 - (-34.99)|}{2} = \frac{178.98}{2} = 89.49~\text{MPa} \] Tresca criterion: \[ \tau_{max} = \frac{\sigma_{y,mat}}{2} \] **Factor of Safety (SF) using Tresca:** \[ SF_{Tresca} = \frac{\sigma_{y,mat}/2}{\tau_{max}} = \frac{300/2}{89.49} = \frac{150}{89.49} \approx 1.68 \] --- ## **Step 3: Von Mises Yield Criterion** Von Mises equivalent (octahedral) stress: \[ \sigma_{eq} = \sqrt{ \sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2 } \] \[ \sigma_{eq} = \sqrt{ (143.99)^2 - (143.99)(-34.99) + (-34.99)^2 } \] Calculate step by step: - \((143.99)^2 = 20741.6\) - \((143.99)(-34.99) = -5037.6\) - \((-34.99)^2 = 1224.3\) So, \[ \sigma_{eq} = \sqrt{20741.6 - (-5037.6) + 1224.3} = \sqrt{20741.6 + 5037.6 + 1224.3} \] \[ = \sqrt{27003.5} \approx 164.4~\text{MPa} \] **Factor of Safety (SF) using Von Mises:** \[ SF_{VM} = \frac{\sigma_{y,mat}}{\sigma_{eq}} = \frac{300}{164.4} \approx 1.83 \] --- ## **Step 4: Final Answers** ### **a) Tresca Yield Criterion** \[ \boxed{SF_{Tresca} \approx 1.68} \] ### **b) Von Mises Criterion** \[ \boxed{SF_{VM} \approx 1.83} \] --- ### **Summary Table** | Method | Factor of Safety (SF) | |-----------------|:--------------------:| | Tresca | **1.68** | | Von Mises | **1.83** |

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