---
title: "Fixed shaft, two torques: couples, diameter"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:shaft-fixed-ends-torques-30-375"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/shaft-fixed-ends-torques-30-375
site: BCT Engineering
---

# Fixed shaft, two torques: couples, diameter

Strength of Materials · Chapter 6: Torsion

- Asked in: 2074 Asoj, 10 marks
- Video: [3.7 - Indeterminate Torsion II](https://www.youtube.com/watch?v=-UO1AicP9C0) — PatJHeffernan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/shaft-fixed-ends-torques-30-375
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

A horizontal shaft securely fixed at each ends has a free length of 11.25 m. Viewed from end "A" of the shaft, axial couples of 30 KN.m clockwise and 37.5 KN.m counter clockwise act on the shaft at a distance 4.5 m and 7.5 m from left respectively. Determine the end fixing couples in magnitude and direction and find the diameter of shaft (solid) for a maximum shearing stress of $60\ \text{N/mm}^2$. [Figure: shaft fixed at A and B; $T_C = +30$ kNm at 4.5 m from A, $T_D = -37.5$ kNm at 3.0 m beyond C (7.5 m from A); D to B 3.75 m; total 11.25 m.]

## Short answer

End couples: 5.50 kN·m anticlockwise at A, 13.00 kN·m clockwise at B (viewed from A). Maximum torque 24.50 kN·m in CD gives solid shaft diameter 127.6 mm.

## Answer

**Data:** length $= 11.25$ m, fixed at both ends. Viewed from A: $T_C = 30$ kN·m clockwise at 4.5 m from A, $T_D = 37.5$ kN·m anticlockwise at 7.5 m from A. Segments: $L_1 = 4.5$ m (A to C), $L_2 = 3.0$ m (C to D), $L_3 = 3.75$ m (D to B). $\tau_{max} = 60\ \text{N/mm}^2$.

#### End fixing couples
Take clockwise (viewed from A) as positive. Let $T_A$ be the end couple at A. The torques in the segments are

$$
T_1 = T_A, \qquad T_2 = T_A + 30, \qquad T_3 = T_A + 30 - 37.5 = T_A - 7.5 \quad(\text{kN·m})
$$

For a shaft fixed at both ends the total twist is zero ($G$, $J$ constant):

$$
T_A(4.5) + (T_A+30)(3.0) + (T_A-7.5)(3.75) = 0
$$

$$
11.25\,T_A + 90 - 28.125 = 0 \Rightarrow T_A = -5.50\ \text{kN·m}
$$

The negative sign means the couple at A is anticlockwise. From overall equilibrium, $T_A + 30 - 37.5 + T_B = 0$:

$$
T_B = 13.00\ \text{kN·m}\ \text{(clockwise)}
$$

| Segment | Torque in the shaft (kN·m) |
|---|---|
| A to C | -5.50 (anticlockwise) |
| C to D | 24.50 (clockwise) |
| D to B | -13.00 (anticlockwise) |

**End couples:** at A, $5.50$ kN·m **anticlockwise**; at B, $13.00$ kN·m **clockwise** (viewed from A). Check: $13.00 - 5.50 = 7.50 = 37.5 - 30$.

#### Diameter
The maximum torque is in segment CD: $T_{max} = 24.50$ kN·m.

$$
d^3 = \frac{16T_{max}}{\pi\tau} = \frac{16(2.4500\times10^{7})}{\pi(60)} = 2.0796\times10^{6}\ \text{mm}^3
$$

$$
d = 127.64\ \text{mm}
$$

**Answer:** end couples $5.5$ kN·m (A, anticlockwise) and $13$ kN·m (B, clockwise); diameter $\approx 127.6$ mm.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What condition finds T_A?
   A. T_A = T_B
   B. Total twist zero: sum T L = 0
   C. Sum of torques zero only
   D. Stress equal
   Answer: B. Fixed ends. (Key point: Compatibility)

2. Directions?
   A. A: 5.5 clockwise
   B. Both 7.5
   C. A: 5.5 kN.m anticlockwise; B: 13 kN.m clockwise
   D. A: 13 anticlockwise
   Answer: C. T_A = -5.5; T_B = 13. (Key point: End couples)

3. Where is torque greatest?
   A. A to C
   B. D to B
   C. Equal
   D. C to D, 24.5 kN.m
   Answer: D. T_2 = T_A + 30 = 24.5. (Key point: Max segment)

4. Result?
   A. 127.6 mm
   B. 200 mm
   C. 60 mm
   D. 100 mm
   Answer: A. d^3 = 16T/(pi tau). (Key point: Diameter)

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