A solid machine shaft, 60 mm in diameter, transmits a torsional moment of 1200 Nm. A

square key is used which one side is equal to one quarter of the shaft diameter and its length

is equal to 1.5 times the shaft diameter. Find the dimensions of key and check the key for

its induced shearing and crushing (compressive) stresses. Check the factor of the safety (fs)

of the key in shear and crushing when yield stresses in shearing & crushing are 350 N/mm2

and 425 N/mm2 respectively.

### Given Data:
- Shaft diameter, \( d = 60 \) mm
- Torsional moment, \( T = 1200 \) Nm = \( 1200 \times 10^3 \) Nmm
- Square key dimensions:
  - Width, \( w = \frac{d}{4} = \frac{60}{4} = 15 \) mm
  - Height, \( h = 15 \) mm (since it is a square key)
  - Length, \( l = 1.5 \times d = 1.5 \times 60 = 90 \) mm
- Yield stress in:
  - Shear, \( \tau_y = 350 \) N/mm²
  - Crushing, \( \sigma_c = 425 \) N/mm²

### Step 1: Calculate Torque Transmitted by the Shaft
The torque transmitted is given by:

\[
T = F \times r
\]

Where:
- \( r \) is the shaft radius, \( r = \frac{d}{2} = \frac{60}{2} = 30 \) mm
- \( F \) is the force exerted by the key

\[
F = \frac{T}{r} = \frac{1200 \times 10^3}{30} = 40000 \text{ N}
\]

### Step 2: Check Shear Stress in the Key
Shear stress in the key is given by:

\[
\tau = \frac{F}{A} = \frac{F}{w \times l}
\]

\[
\tau = \frac{40000}{15 \times 90} = \frac{40000}{1350} = 29.63 \text{ N/mm}^2
\]

Factor of safety in shear:

\[
FS_{\text{shear}} = \frac{\tau_y}{\tau} = \frac{350}{29.63} = 11.81
\]

### Step 3: Check Crushing Stress in the Key
Crushing stress in the key is given by:

\[
\sigma_c = \frac{F}{A} = \frac{F}{h \times l / 2}
\]

Since the key is in double shear, the effective area is \( \frac{h \times l}{2} \):

\[
\sigma_c = \frac{40000}{\frac{15 \times 90}{2}}
\]

\[
\sigma_c = \frac{40000}{675} = 59.26 \text{ N/mm}^2
\]

Factor of safety in crushing:

\[
FS_{\text{crushing}} = \frac{\sigma_c}{\sigma_{c,\text{actual}}} = \frac{425}{59.26} = 7.17
\]

### Final Answers:
- **Key dimensions:** \( 15 \times 15 \times 90 \) mm
- **Shear stress:** \( 29.63 \) N/mm²
- **Crushing stress:** \( 59.26 \) N/mm²
- **Factor of safety in shear:** 11.81
- **Factor of safety in crushing:** 7.17

The key is safe under the given conditions.

---
### Do you need further clarification or detailed steps?  
Here are 5 related questions to expand on this topic:
1. How does the key length affect the stress distribution in the key?
2. What happens if a key is improperly designed for shear and crushing stresses?
3. How does increasing the torque affect the key design requirements?
4. Why is a square key commonly used in shaft coupling applications?
5. What are the advantages of using a tapered key instead of a square key?

**Tip:** Always ensure the factor of safety is well above 1 to prevent failure under unexpected load conditions! 🚀

A solid machine shaft, 60 mm in diameter, transmits a torsional moment of 1200 Nm. A square key is used which one side is equal to one quarter of the shaft diameter and its length is equal to 1.5 times the shaft diameter. Find the dimensions of key and check the key for its induced shearing and crushing (compressive) stresses. Check the factor of the safety (fs) of the key in shear and crushing when yield stresses in shearing & crushing are 350 N/mm2 and 425 N/mm2 respectively.

Learn how to calculate the dimensions of a square key and check its safety factors under shear and crushing stresses in a shaft transmission system. This detailed solution involves torque analysis, shear stress, and crushing stress calculations, using yield stresses in shear (350 N/mm²) and crushing (425 N/mm²). Suitable for engineering students and professionals.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation