---
title: "Stepped shaft, three torques: twist at free end"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:stepped-shaft-three-torques-twist-at"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/stepped-shaft-three-torques-twist-at
site: BCT Engineering
---

# Stepped shaft, three torques: twist at free end

Strength of Materials · Chapter 6: Torsion

- Asked in: 2068 Baisakh (old course), 8 marks
- Video: [Angle of Twist of Shaft with Torsion](https://www.youtube.com/watch?v=ogaJRMjItjA) — The Ryder Project
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/stepped-shaft-three-torques-twist-at
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

A stepped solid circular shaft of the dimensions shown in the figure is subjected to three torques. If the material has a shear modulus of elasticity G = 80 GPa, find the angle of twist in degrees at the free end. Also calculate the maximum shearing stress in the shaft. [Figure: shaft fixed at the left end: segment 1 diameter 8 cm, length 50 cm, with a 10 kNm torque at its right end; segment 2 diameter 6 cm, length 50 cm, with a 4 kNm torque at its right end; segment 3 diameter 4 cm, length 50 cm, with a 2 kNm torque at the free end.]

## Short answer

Sum twists using φ=TL/GJ for each segment. Torques: 2, 6, 16 kN·m. Twist = 0.104 rad = 5.96°. Max shear stress from τ=16T/πd³ is 159.2 MPa in segments 1 and 3.

## Answer

**Reading of the figure:** fixed at the left end; three segments each 500 mm long with diameters 80, 60 and 40 mm. Torques of 10 kN·m (at the end of segment 1), 4 kN·m (end of segment 2) and 2 kN·m (free end) act in the same sense. $G = 80$ GPa $= 8\times10^4\ \text{N/mm}^2$.

#### Torque in each segment (section method from the free end)
| Segment | Torque (kN·m) |
|---|---|
| 3 (free end, $d=40$) | 2 |
| 2 ($d=60$) | 2 + 4 = 6 |
| 1 ($d=80$, at the wall) | 2 + 4 + 10 = 16 |

#### Twist of each segment
$$
\varphi_i = \frac{T_iL_i}{GJ_i}, \qquad J_i = \frac{\pi}{32}d_i^4
$$

| Segment | $J$ (mm⁴) | $T$ (N·mm) | $\varphi$ (rad) |
|---|---|---|---|
| 1 | $4.0212\times10^{6}$ | $1.60\times10^{7}$ | 0.02487 |
| 2 | $1.2723\times10^{6}$ | $6.00\times10^{6}$ | 0.02947 |
| 3 | $2.5133\times10^{5}$ | $2.00\times10^{6}$ | 0.04974 |

$$
\varphi_{free} = 0.02487 + 0.02947 + 0.04974 = 0.10408\ \text{rad} = 5.963^\circ
$$

#### Maximum shear stress
$$
\tau = \frac{16T}{\pi d^3}: \qquad \tau_1 = 159.2,\quad \tau_2 = 141.5,\quad \tau_3 = 159.2\ \text{N/mm}^2
$$

The maximum, 159.2 N/mm², occurs in segments 1 and 3 (the 80 mm and 40 mm parts); segment 2 has the lowest stress.

**Answer:** angle of twist at the free end $= 5.96^\circ$ (0.1041 rad); $\tau_{max} = 159.2$ MPa, in segments 1 and 3.

## Check yourself

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

1. How is the internal torque in each segment found for the shaft fixed at the left?
   A. Average of all three torques
   B. Only the torque at the end of that segment is used
   C. Sum the torques lying between that section and the free end
   D. Sum of all torques for every segment
   Answer: C. Cutting a section and balancing from the free end gives 2, 6 and 16 kN·m in segments 3, 2 and 1. (Key point: Internal torque by sections)

2. What are the internal torques in the free-end, middle and wall-side segments?
   A. 16, 6, 2 kN·m
   B. 10, 4, 2 kN·m
   C. 2, 4, 10 kN·m
   D. 2, 6, 16 kN·m
   Answer: D. Torques act in the same sense and accumulate toward the wall: 2, 2+4 = 6, 2+4+10 = 16 kN·m. (Key point: Torques in segments)

3. How is the angle of twist at the free end obtained for a stepped shaft?
   A. Algebraic sum of T*L/(G*J) of each segment
   B. Using the average diameter
   C. Product of the segment twists
   D. Using only the thinnest segment
   Answer: A. Each segment twists by TL/GJ and the twists add up from the fixed end to the free end. (Key point: Total twist)

4. For each segment of the stepped shaft, which formula gives the maximum shear stress?
   A. tau = T/d
   B. tau = 16T/(pi d^3)
   C. tau = 32T/(pi d^4)
   D. tau = T L/(G J)
   Answer: B. For a solid circular shaft tau = T r/J = 16T/(pi d^3); it is evaluated for every segment and the largest value (159.2 MPa) is the maximum. (Key point: Stress depends on T and d)

## Related in Strength of Materials

- [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
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- [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
- [Fixed shaft: end couples, diameter, section of no twist](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ashaft-fixed-ends-couples-no-twist) — Strength of Materials, chapter 6
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