Give step-by-step solution with explanation and final answer:
Answer this question in 3 steps with explanation at the end of each step and final answer at the end.1. A machine element is made of AIS

Question

Give step-by-step solution with explanation and final answer: 
Answer this question in 3 steps with explanation at the end of each step and final answer at the end.1. A machine element is made of AISI 1080 carbon steel with a yield strength of Sy = 420 MPa. The
state of plane stress at five potential critical points was determined as listed. For cach of the points,
(i) sketch a stress clement and Mohr’s circle, and calculate the static factor of safety based upon
(if) the maximum shear stress theory of failure (nsusstr) and (if) the distortion energy theory of
failure (nopere). Use the Mohrs circles for the state of stress calculations.

(a) ox= 180 MPa, oy = 180 MPa

(b) ox = 140 MPa, Tay = 80 MPa cw

(¢) oy=-80 MPa, ty = 120 MPa cw

(d) Tey =-200 MPa

(¢) ©x= 100 MPa, y= 200 MPa, xy = -60 MPa

VIPSolutions · Accepted Answer

Let's solve **part (a)** step by step as requested:

---

### **Given (for part a):**
- $\sigma_x = 180 $ MPa
- $\sigma_y = 180 $ MPa
- $\tau_{xy} = 0 $ MPa
- Yield strength, $S_y = 420$ MPa

---

## **Step 1: Stress Element & Mohr's Circle**

**Stress Element:**
- $\sigma_x = 180$ MPa (acts on x-face)
- $\sigma_y = 180$ MPa (acts on y-face)
- $\tau_{xy} = 0$ (no shear stress)

**Mohr's Circle:**
- Center: $\left(\frac{\sigma_x + \sigma_y}{2}, 0\right) = (180, 0)$
- Radius: $\sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{0^2 + 0^2} = 0$

**Principal Stresses:**
$$
\sigma_1 = \sigma_2 = 180 \text{ MPa}
$$
$$
\tau_{max} = 0 \text{ MPa}
$$

**Explanation:** Both normal stresses are equal, and there's no shear stress, so Mohr's circle reduces to a point at (180, 0).

---

## **Step 2: Factor of Safety - Maximum Shear Stress Theory (MSST/F)**

**MSST (Tresca):**
$$
n_{s,\text{MSSTF}} = \frac{S_y}{\sigma_1 - \sigma_2}
$$

Here,
$$
\sigma_1 = 180\text{ MPa},\quad \sigma_2 = 180\text{ MPa}
$$
So,
$$
|\sigma_1 - \sigma_2| = 0
$$
$$
n_{s,\text{MSSTF}} = \frac{420}{0} = \infty
$$

**Explanation:** Since both principal stresses are equal, the maximum shear stress is zero, and theoretically, the factor of safety is infinite (no danger of yielding according to Tresca).

---

## **Step 3: Factor of Safety - Distortion Energy Theory (DETF/Von Mises)**

**Von Mises:**
$$
\sigma_{eq} = \sqrt{\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2}
$$
But since $\sigma_1 = \sigma_2$, this simplifies:
$$
\sigma_{eq} = \sqrt{(180)^2 - (180)(180) + (180)^2} = \sqrt{180^2} = 180 \text{ MPa}
$$

**Factor of Safety:**
$$
n_{s,\text{DETF}} = \frac{S_y}{\sigma_{eq}} = \frac{420}{180} = 2.33
$$

**Explanation:** The equivalent stress is just the applied normal stress, so the factor of safety is 2.33 by the Von Mises criterion.

---

## **Final Answers (for part a):**

- **(i) Mohr's Circle:** Reduces to a point at (180, 0).
- **(ii) Factor of Safety (MSST):** $n_{s,\text{MSSTF}} = \infty$
- **(iii) Factor of Safety (Von Mises/DETF):** $n_{s,\text{DETF}} = 2.33$

---

Would you like to continue with part (b)?

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