solve completely with complete elaborationAt a point in a machine component, the stresses are:
e Normal stress in x-direction: 90 MPa (tensile)
* Normal stress in y-direction: 30 MPa (compressive)

Question

solve completely with complete elaborationAt a point in a machine component, the stresses are:
e Normal stress in x-direction: 90 MPa (tensile)
* Normal stress in y-direction: 30 MPa (compressive)
® Shear stress: 45 MPa
Tasks:
1. Calculate the principal stresses.
2. Determine the maximum shear stress.
3. State whether the stress state is safe if allowable stress is 160 MPa.

VIPSolutions · Accepted Answer

# Solution: Analysis

Given Data:
- Normal stress in x-direction, $ \sigma_x = 90 $ MPa (tensile)
- Normal stress in y-direction, $ \sigma_y = -30 $ MPa (compressive; negative sign used for compressive)
- Shear stress, $ \tau_{xy} = 45 $ MPa
- Allowable stress, $ \sigma_{\text{allow}} = 160 $ MPa

---

## 1. **Calculate the Principal Stresses**

The principal stresses ($\sigma_1$ and $\sigma_2$) are given by:

$$
\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 }
$$

### Substitute the values:

- $ \sigma_x = 90 $ MPa
- $ \sigma_y = -30 $ MPa
- $ \tau_{xy} = 45 $ MPa

#### Step 1: Calculate the average normal stress

$$
\frac{\sigma_x + \sigma_y}{2} = \frac{90 + (-30)}{2} = \frac{60}{2} = 30 \text{ MPa}
$$

#### Step 2: Calculate the radius (R) of Mohr's Circle

$$
R = \sqrt{ \left( \frac{90 - (-30)}{2} \right)^2 + (45)^2 }
= \sqrt{ \left( \frac{120}{2} \right)^2 + 2025 }
= \sqrt{ (60)^2 + 2025 }
= \sqrt{ 360 + 2025 }
= \sqrt{ 5625 }
= 75 \text{ MPa}
$$

#### Step 3: Principal Stresses

$$
\sigma_1 = 30 + 75 = 105 \text{ MPa}
$$
$$
\sigma_2 = 30 - 75 = -45 \text{ MPa}
$$

---

## 2. **Determine the Maximum Shear Stress**

Maximum shear stress ($\tau_{\text{max}}$) is:

$$
\tau_{\text{max}} = R = 75 \text{ MPa}
$$

Or, directly:

$$
\tau_{\text{max}} = \frac{\sigma_1 - \sigma_2}{2} = \frac{105 - (-45)}{2} = \frac{150}{2} = 75 \text{ MPa}
$$

---

## 3. **State Whether the Stress State is Safe**

The maximum absolute principal stress is $ |\sigma_1| = 105 $ MPa, $ |\sigma_2| = 45 $ MPa.

- **Maximum principal stress:** $ 105 $ MPa
- **Allowable stress:** $ 160 $ MPa

$$
\boxed{
\text{Since } 105 \text{ MPa} < 160 \text{ MPa, the component is SAFE.}
}
$$

---

## **Summary Table**

| Quantity                     | Value       |
|------------------------------|-------------|
| Principal stress ($\sigma_1$) | 105 MPa     |
| Principal stress ($\sigma_2$) | -45 MPa     |
| Maximum shear stress ($\tau_{\text{max}}$) | 75 MPa |
| Maximum principal stress      | 105 MPa     |
| Allowable stress              | 160 MPa     |
| **Safety**                    | **SAFE**    |

---

### **Conclusion**
- The principal stresses are $ 105 $ MPa and $ -45 $ MPa.
- The maximum shear stress is $ 75 $ MPa.
- The component is **safe** under the given allowable stress of $ 160 $ MPa.

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