---
title: "Fixed shaft: torsional reactions, rotations, max shear"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:shaft-torsional-reactions-copper-steel"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/shaft-torsional-reactions-copper-steel
site: BCT Engineering
---

# Fixed shaft: torsional reactions, rotations, max shear

Strength of Materials · Chapter 6: Torsion

- Asked in: 2081 Baisakh, 8 marks
- Video: [Mechanics of Materials: Lesson 26 - Statically Indeterminate Angle of Twist Due to Torque](https://www.youtube.com/watch?v=BV-Q_duj2Dk) — Jeff Hanson
- 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-torsional-reactions-copper-steel
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

For the following shaft system with the torque as shown in figure, determine the torsional reactions. Also determine the rotations of joint B and D and maximum shear stress developed in each member. Take $G_{steel} = 78\ \text{GPa}$ and $G_{copper} = 44\ \text{GPa}$. [Figure: shaft fixed at both ends A and D, three segments of 1 m each: A-B copper $\phi 8\ \text{cm}$, B-C steel $\phi 6\ \text{cm}$, C-D copper $\phi 8\ \text{cm}$; a torque of 20 kNm applied at B and a torque of 4 kNm applied at C (opposite sense as drawn).]

## Short answer

Solve compatibility ΣTL/(GJ)=0: reactions T_A = 13.66 kNm, T_D = 2.34 kNm. Rotations φ_B = 0.0772 rad, φ_C = 0.0133 rad. Maximum shear stresses: AB (copper) 135.8 MPa, BC (steel) 149.6 MPa, CD (copper) 23.3 MPa.

## Answer

**Reading of the figure:** shaft fixed at A and D; AB and CD are copper ($\phi 80$ mm), BC is steel ($\phi 60$ mm), each 1 m long. A torque $T_B = 20$ kN·m acts at B and $T_C = 4$ kN·m acts at C in the opposite sense. The problem is statically indeterminate (one equation of equilibrium, one of compatibility).

#### Polar moments
$$
J_{80} = \frac{\pi}{32}(80)^4 = 4.0212\times10^{6}\ \text{mm}^4, \qquad J_{60} = \frac{\pi}{32}(60)^4 = 1.2723\times10^{6}\ \text{mm}^4
$$

#### Compatibility
Let $T_A$ be the reaction at A (positive in the same sense as $T_B$). The torques in the segments are

$$
T_{AB} = T_A, \qquad T_{BC} = T_A + 20, \qquad T_{CD} = T_A + 16 \quad(\text{kN·m})
$$

The fixed end D does not rotate relative to A, so the total twist is zero ($\varphi = \sum \frac{TL}{GJ} = 0$):

$$
\frac{T_A L}{G_cJ_{80}} + \frac{(T_A+20)L}{G_sJ_{60}} + \frac{(T_A+16)L}{G_cJ_{80}} = 0
$$

With $\dfrac{L}{G_cJ_{80}} = 5.6518\times10^{-9}$ and $\dfrac{L}{G_sJ_{60}} = 1.0076\times10^{-8}$ (per N·mm):

$$
T_A = -\frac{20\,b + 16\,a}{2a + b} = -13.656\ \text{kN·m}
$$

#### Torsional reactions
- At A: $|T_A| = 13.656$ kN·m, acting opposite to $T_B$.
- At D: $|T_D| = 2.344$ kN·m, acting opposite to $T_B$.

Check: $13.656 + 2.344 = 16.000 = 20 - 4$ kN·m.

#### Rotations of the joints (D is fixed, so its rotation is zero)
$$
\varphi_B = \frac{|T_{AB}|L}{G_cJ_{80}} = 0.07718\ \text{rad} = 4.422^\circ
$$

$$
\varphi_C = \varphi_B - \frac{T_{BC}L}{G_sJ_{60}} = 0.01325\ \text{rad} = 0.759^\circ
$$

Both are in the direction of the 20 kN·m torque, and $\varphi_D = 0$ (check: $\varphi_C - \frac{T_{CD}L}{G_cJ_{80}} = 0$).

#### Maximum shear stress ($\tau = \dfrac{16T}{\pi d^3}$)
| Member | Torque (kN·m) | $\tau_{max}$ (MPa) |
|---|---|---|
| AB (copper) | 13.656 | 135.8 |
| BC (steel) | 6.344 | 149.6 |
| CD (copper) | 2.344 | 23.3 |

## Check yourself

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

1. Why can this not be solved by equilibrium alone?
   A. Shaft is too long
   B. Both ends are fixed; two unknown reactions but one equilibrium equation
   C. Different materials
   D. Torques are unequal
   Answer: B. Compatibility is needed. (Key point: Indeterminate)

2. What compatibility condition is used?
   A. Twist in each segment is equal
   B. Torque in each segment is equal
   C. Total twist from A to D is zero
   D. Stress in each segment is equal
   Answer: C. Both ends are fixed so sum of TL/GJ = 0. (Key point: Compatibility)

3. What is the equilibrium check on reactions?
   A. T_A = T_D
   B. T_A + T_D = 24 kN.m
   C. T_A = 20 kN.m
   D. T_A + T_D = 20 - 4 = 16 kN.m
   Answer: D. Applied torques oppose, so 13.656 + 2.344 = 16. (Key point: Equilibrium check)

4. Which member has greatest shear stress?
   A. Steel BC, 149.6 MPa, despite smaller torque than AB
   B. Copper AB because torque is largest
   C. Copper CD
   D. All equal
   Answer: A. Stress is 16T/(pi d^3); the small 60 mm diameter raises it. (Key point: Greatest stress)

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