---
title: "Statically indeterminate shaft with torque"
subject: "Strength of Materials"
chapter: 4
key: "semester_4/strength_of_materials_mechanical:shaft-fixed-both-ends-torque"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/shaft-fixed-both-ends-torque
site: BCT Engineering
---

# Statically indeterminate shaft with torque

Strength of Materials · Chapter 4: Torsion

- Asked in: year not given, 8 marks
- Video: [04.3-1 Statically indeterminate shaft - EXAMPLE](https://www.youtube.com/watch?v=mqLwCYA7X-E) — Introductory Engineering Mechanics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/shaft-fixed-both-ends-torque
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-4

## Question

A circular shaft ACB is fixed at both ends A and B. Portion AC is 0.6 m long and 50 mm in diameter; portion CB is 0.4 m long and 40 mm in diameter. Both portions are of steel with G = 80 GPa. A torque of 2.5 kN m is applied at the junction C. Find (a) the reactive torques at A and B, (b) the maximum shear stress in each portion, (c) the angle of twist of section C.

## Short answer

Reactive torques: T_A = 1.549 kN m, T_B = 0.951 kN m. Max shear stress: τ_AC = 63.1 MPa, τ_CB = 75.7 MPa. Angle of twist at C: 0.0189 rad (1.08°).

## Answer

#### Method
The problem is statically indeterminate: only one equilibrium equation is available for two unknown reactions. Add a compatibility condition.

```text
 //|A-----50mm-----C---40mm---B|//
 //|<---- 0.6 m --->|<-0.4 m->|//
                    | T0 = 2.5 kN m
```

#### Equations
1. Equilibrium: $T_A + T_B = T_0 = 2.5\times10^6$ N mm
2. Compatibility: the angle of twist of AC and CB must be equal at C (both ends are fixed and the section C rotates by one angle $\phi_C$)

$$
\frac{T_A L_1}{GJ_1} = \frac{T_B L_2}{GJ_2}
$$

Polar moments:

$$
J_1 = \frac{\pi}{32}(50)^4 = 613.6\times10^3\ \text{mm}^4, \qquad J_2 = \frac{\pi}{32}(40)^4 = 251.3\times10^3\ \text{mm}^4
$$

Torsional flexibility of each part, $L/J$: $\dfrac{600}{613.6\times10^3} = 9.778\times10^{-4}$ and $\dfrac{400}{251.3\times10^3} = 15.915\times10^{-4}$.

#### (a) Reactions
$$
T_A = T_0\,\frac{L_2/J_2}{L_1/J_1 + L_2/J_2} = 2.5\times\frac{15.915}{9.778+15.915} = 1.549\ \text{kN m}
$$

$$
T_B = 2.5 - 1.549 = 0.951\ \text{kN m}
$$

(The stiffer portion AC takes the larger share.)

#### (b) Maximum shear stresses
$$
\tau_{AC} = \frac{16T_A}{\pi d_1^3} = \frac{16\times1.549\times10^6}{\pi\times50^3} = 63.1\ \text{MPa}
$$

$$
\tau_{CB} = \frac{16T_B}{\pi d_2^3} = \frac{16\times0.951\times10^6}{\pi\times40^3} = 75.7\ \text{MPa}
$$

#### (c) Angle of twist of C
$$
\phi_C = \frac{T_AL_1}{GJ_1} = \frac{1.549\times10^6\times600}{80000\times613.6\times10^3} = 0.0189\ \text{rad} = 1.08^\circ
$$

Check using CB: $\dfrac{0.951\times10^6\times400}{80000\times251.3\times10^3} = 0.0189$ rad.

**Answer: $T_A = 1.549$ kN m, $T_B = 0.951$ kN m; $\tau_{AC} = 63.1$ MPa, $\tau_{CB} = 75.7$ MPa; $\phi_C = 0.0189$ rad $= 1.08^\circ$.**

## Check yourself

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

1. Why is the fixed-fixed shaft indeterminate?
   A. Only one equilibrium equation for two unknown reactive torques
   B. It has three unknowns
   C. The material is not elastic
   D. The torque is too large
   Answer: A. T_A + T_B = T_0 is the only equilibrium equation, so a compatibility condition is needed. (Key point: Statically indeterminate)

2. What compatibility condition is used?
   A. T_A = T_B
   B. Angle of twist of AC equals that of CB (both equal phi_C)
   C. Shear stress in AC equals that in CB
   D. J1 = J2
   Answer: B. Both ends are fixed, so section C rotates through one angle: T_A L1/(G J1) = T_B L2/(G J2). (Key point: Compatibility condition)

3. How is the torque shared?
   A. The more flexible portion takes the larger share
   B. The stiffer portion AC takes the larger share (T_A = 1.549, T_B = 0.951 kN m)
   C. Equally
   D. All goes to B
   Answer: B. T_A is proportional to L2/J2, the flexibility of CB; the stiffer AC carries more. (Key point: Share of torque)

4. Which portion has the larger maximum shear stress?
   A. AC at 63.1 MPa
   B. CB at 75.7 MPa, despite its smaller torque, because of its smaller diameter
   C. Both equal
   D. AC at 75.7 MPa
   Answer: B. tau = 16T/(pi d^3): CB's smaller d outweighs its lower torque, giving 75.7 MPa versus 63.1 MPa. (Key point: Maximum shear stress)

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