---
title: "Fixed shaft: end couples, diameter, section of no twist"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:shaft-fixed-ends-couples-no-twist"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/shaft-fixed-ends-couples-no-twist
site: BCT Engineering
---

# Fixed shaft: end couples, diameter, section of no twist

Strength of Materials · Chapter 6: Torsion

- Asked in: 2078 Bhadra, 8 marks
- Video: [The steel shaft has a diameter of 40mm and is fixed at its ends A and B. If it is subjected to....](https://www.youtube.com/watch?v=Z49KwEwyMD4) — Engineering Made Possible
- 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-couples-no-twist
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

Determine the end fixing couples, diameter of the shaft if the maximum shearing stress is not to exceed $50\ \text{MN/m}^2$ and the position of the section where the shaft suffers no angular twist. [Figure: shaft fixed at both ends A and D; torque $T_B = 15\ \text{kNm}$ at B and torque $T_B = 25\ \text{kNm}$ (printed as $T_B$, read as the torque at C) at C in the opposite sense; A to B 1.5 m, B to C 1.2 m, C to D 1.5 m.]

## Short answer

End fixing couples: at A 0.714 kN·m opposite to TB, at D 10.714 kN·m in the same sense as TB. Maximum torque 14.286 kN·m gives shaft diameter 113.3 mm. Zero twist occurs 0.075 m from B towards C.

## Answer

**Reading of the figure:** shaft of uniform diameter fixed at A and D; $T_B = 15$ kN·m at B and $T_C = 25$ kN·m at C (opposite sense to $T_B$); $AB = 1.5$ m, $BC = 1.2$ m, $CD = 1.5$ m; $\tau_{max} = 50\ \text{MN/m}^2$.

#### End fixing couples
Let $T_A$ be the reaction at A (positive in the sense of $T_B$). The torques in the segments are

$$
T_{AB} = T_A, \qquad T_{BC} = T_A + 15, \qquad T_{CD} = T_A + 15 - 25 = T_A - 10 \quad(\text{kN·m})
$$

The ends are fixed, so the total twist is zero. $G$ and $J$ are constant:

$$
T_A(1.5) + (T_A+15)(1.2) + (T_A-10)(1.5) = 0 \Rightarrow 4.2\,T_A + 3 = 0 \Rightarrow T_A = -0.714\ \text{kN·m}
$$

The negative sign means the reaction at A acts opposite to $T_B$. Also $T_D = 10.714$ kN·m from equilibrium ($T_A + T_D + 15 - 25 = 0$ with the signs above).

| Segment | Torque (kN·m) |
|---|---|
| AB | -0.714 |
| BC | 14.286 |
| CD | -10.714 |

**Answer (end couples):** $|T_A| = 0.714$ kN·m, acting opposite to $T_B$; $|T_D| = 10.714$ kN·m, acting in the same sense as $T_B$.

#### Diameter
The maximum torque is $T_{BC} = 14.286$ kN·m.

$$
d^3 = \frac{16T_{max}}{\pi\tau} = \frac{16(1.4286\times10^{7})}{\pi(50)} = 1.4551\times10^{6}\ \text{mm}^3 \Rightarrow d = 113.32\ \text{mm}
$$

#### Section with no angular twist
The rotation of a section relative to A is $\varphi = \sum \dfrac{TL}{GJ}$. At B (taking $GJ = 1$ units, kN·m²): $\varphi_B = 0.714\times1.5 = 1.071$. In BC the twist changes at the rate $T_{BC} = 14.286$ in the opposite direction, so it reaches zero at a distance

$$
x = \frac{1.071}{14.286} = 0.075\ \text{m from B}
$$

**Answer:** the section $\approx 0.075$ m from B (i.e. 1.575 m from A, inside BC) suffers no angular twist; $d \approx 113.3$ mm.

## Check yourself

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

1. What condition determines T_A?
   A. T_A = T_D
   B. Total twist zero: sum of T L = 0 (G, J constant)
   C. Torques equal
   D. Stress equal
   Answer: B. Both ends are fixed. (Key point: Fixed ends)

2. What is T_A?
   A. 15 kN.m
   B. 10 kN.m
   C. -0.714 kN.m, opposite to T_B
   D. 0.714 same sense as T_B
   Answer: C. 4.2 T_A + 3 = 0. (Key point: Result)

3. Which segment governs the diameter?
   A. AB
   B. CD, 10.714
   C. Both equal
   D. BC, with 14.286 kN.m
   Answer: D. Largest torque magnitude is in BC. (Key point: Design torque)

4. Where is the no-twist section?
   A. 0.075 m from B, inside BC
   B. At B
   C. At C
   D. At midspan of AB
   Answer: A. Twist 1.071 at B reduces at 14.286 per metre. (Key point: No-twist section)

## Related in Strength of Materials

- [Fixed shaft, two torques: couples, diameter](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ashaft-fixed-ends-torques-30-375) — Strength of Materials, chapter 6
- [Stepped shaft, three torques: twist at free end](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Astepped-shaft-three-torques-twist-at) — Strength of Materials, chapter 6
- [Steel shaft 200 hp: torque, shear, twist](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asteel-shaft-200hp-150rpm) — Strength of Materials, chapter 6
- [Stepped shaft: angle of twist at free end](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ashaft-angle-of-twist-free-end) — Strength of Materials, chapter 6
- [Stepped shaft: maximum torque at C](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Astepped-shaft-fixed-max-torque-at) — Strength of Materials, chapter 6
- [Solid shaft 300 kW, 120 rpm with twist limit](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asolid-shaft-300kw-120rpm-twist-limit) — Strength of Materials, chapter 6

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