A circular bar made of steel is to resist an occasional torque of 2.2 kNm acting in 
transverse plane. If the allowable stresses in compression, tension, and shear are 100 
MN/m2
, 35 MN/m2
, and 50 MN/m2
respectively, find:
a. The diameter of the bar
b. The angle of twist under the applied torque per meter length of bar.
Take G = 40 GN/m2


To solve the problem of determining the diameter of the circular steel bar and the angle of twist, we can follow these steps:

### Given Data:
- Torque, \( T = 2.2 \, \text{kNm} = 2.2 \times 10^3 \, \text{Nm} \)
- Allowable shear stress, \( \tau_{\text{allowable}} = 50 \, \text{MN/m}^2 = 50 \times 10^6 \, \text{N/m}^2 \)
- Shear modulus, \( G = 40 \, \text{GN/m}^2 = 40 \times 10^9 \, \text{N/m}^2 \)
- We need to find:
  - Diameter of the bar (\( d \))
  - Angle of twist per meter length (\( \theta \))

### Step 1: Calculating the Diameter of the Bar

For a circular bar under a torque, the maximum shear stress (\( \tau \)) is related to the applied torque by the formula:

\[
T = \frac{\pi}{16} \cdot \tau \cdot d^3
\]

Where:
- \( T \) is the applied torque (Nm),
- \( \tau \) is the allowable shear stress (N/m²),
- \( d \) is the diameter of the bar (m).

Rearranging to solve for \( d \):

\[
d = \left( \frac{16T}{\pi \tau} \right)^{\frac{1}{3}}
\]

Substituting the values:

\[
d = \left( \frac{16 \times 2.2 \times 10^3}{\pi \times 50 \times 10^6} \right)^{\frac{1}{3}}
\]

Let me calculate that.

### Step 2: Angle of Twist Calculation

For the angle of twist per unit length, the formula is:

\[
\theta = \frac{T L}{G J}
\]

Where:
- \( \theta \) is the angle of twist per meter (radians),
- \( T \) is the applied torque (Nm),
- \( L = 1 \, \text{m} \) (length),
- \( G \) is the shear modulus (N/m²),
- \( J = \frac{\pi d^4}{32} \) is the polar moment of inertia for a circular section.

Substituting \( J \) and solving for \( \theta \):

\[
\theta = \frac{T L}{G \cdot \frac{\pi d^4}{32}} = \frac{32 T L}{G \pi d^4}
\]

Once we find the diameter \( d \), we can substitute it to find the angle of twist. Let me compute these values.### Results:
a. The diameter of the bar is approximately \( 60.74 \, \text{mm} \).

b. The angle of twist under the applied torque per meter length of the bar is approximately \( 0.0412 \, \text{radians} \).

Would you like further details on any part of the solution?

Here are 5 related questions:
1. How would the result change if the allowable shear stress was increased?
2. What would be the effect of reducing the torque on the required diameter?
3. How does changing the length of the bar affect the angle of twist?
4. Can you calculate the twist angle for different bar materials with other values of \( G \)?
5. How would the solution differ if the bar had a hollow circular section instead of a solid one?

**Tip:** In torsion problems, the polar moment of inertia is crucial for predicting both strength and deformation (twist).

A circular bar made of steel is to resist an occasional torque of 2.2 kNm acting in transverse plane. If the allowable stresses in compression, tension, and shear are 100 MN/m2, 35 MN/m2, and 50 MN/m2 respectively, find: a. The diameter of the bar b. The angle of twist under the applied torque per meter length of bar. Take G = 40 GN/m2

Learn how to calculate the diameter and angle of twist for a circular steel bar subjected to 2.2 kNm torque. This problem involves key concepts in material mechanics and torsion analysis. Find step-by-step solutions using the torsion formula and the angle of twist equation, suitable for undergraduate engineering students.

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