---
title: "Cantilever tip deflection, temperature gradient"
subject: "Theory of Structures I"
chapter: 3
key: "semester_4/theory_of_structures_i:beam-temperature-gradient-cantilever"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/beam-temperature-gradient-cantilever
site: BCT Engineering
---

# Cantilever tip deflection, temperature gradient

Theory of Structures I · Chapter 3: Analysis by the Virtual Work Method

- Asked in: 2068 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/beam-temperature-gradient-cantilever
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3

## Question

The bottom of the beam shown below is subjected to a temperature of 200°C, while the temperature of its top is 50°C. If the coefficient of linear expansion $\alpha = 12\times10^{-6}/^\circ$C, determine the vertical displacement of its free end B due to temperature gradient. The beam has a rectangular cross-section with a depth of 30 cm. [Figure: cantilever AB of length 3 m, fixed at A; cross-section 5 cm wide and 30 cm deep.]

## Short answer

Temperature gradient 150°C over 0.3 m depth yields curvature 6e-3 /m. Unit load integral ∫κ x dx = κL^2/2 = 0.027 m. Free end displaces 27 mm upward.

## Answer

#### Given data
Cantilever AB, fixed at A, length $L = 3$ m, rectangular section 5 cm $\times$ 30 cm (depth $d = 0.30$ m). Bottom temperature $200^\circ$C, top $50^\circ$C, $\alpha = 12\times10^{-6}$/$^\circ$C. The temperature varies linearly over the depth.

#### Temperature effect
Mean temperature $T_m = \dfrac{200 + 50}{2} = 125^\circ$C: this causes only axial elongation, no vertical displacement.

The temperature difference across the depth, $\Delta T = 200 - 50 = 150^\circ$C, gives a uniform curvature:

$$
\kappa = \frac{\alpha\,\Delta T}{d} = \frac{12\times10^{-6}\times150}{0.30} = 6\times10^{-3}\ \text{per m}
$$

The bottom fibres are hotter, so they lengthen more and the beam curves **upward** (concave up, i.e. a sagging-type curvature).

#### Virtual work
Unit vertical load at B: $m = x$ ($x$ from B). The unit downward load causes hogging, while the thermal curvature is sagging, so B moves opposite to the unit load (upward). In magnitude:

$$
\Delta_B = \int_0^L m\,\kappa\,dx = \kappa\int_0^3 x\,dx = \kappa\,\frac{L^2}{2}
$$

$$
\Delta_B = 6\times10^{-3}\times\frac{3^2}{2} = 0.027\ \text{m}
$$

**Answer: vertical displacement of the free end $= 27$ mm upward.** (The mean temperature only lengthens the beam horizontally, by $\alpha T_m L = 12\times10^{-6}\times125\times3 = 4.5$ mm if measured from 0$^\circ$C; it does not move B vertically.)

## Check yourself

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

1. What does the mean temperature of 125°C cause?
   A. Rotation at A
   B. Only axial elongation
   C. Curvature of 6 x 10^-3
   D. Vertical displacement of B
   Answer: B. The mean temperature elongates the beam axially (4.5 mm) but does not bend it. (Key point: Mean temperature)

2. What is the curvature?
   A. 12e-6 x 125 / 0.30
   B. 12e-6 x 200 / 0.30
   C. 12e-6 x 150 / 0.30 = 6 x 10^-3 per m
   D. 12e-6 x 150 x 0.30
   Answer: C. kappa = alpha DeltaT / d with DeltaT = 200 - 50 = 150°C. (Key point: Curvature)

3. The bottom is hotter than the top. How does the free end move?
   A. Horizontally only
   B. Does not move
   C. Downward
   D. Upward, because the bottom fibres lengthen more
   Answer: D. The hotter bottom lengthens more, so the beam curves concave upward and the free end moves up. (Key point: Direction)

4. How is the free end deflection obtained by virtual work?
   A. Delta = kappa L^2 / 2 = 27 mm
   B. Delta = kappa L = 18 mm
   C. Delta = kappa L^2 / 8
   D. Delta = kappa L^3 / 3
   Answer: A. With m = x from the free end, the integral of m kappa dx is kappa L^2/2 = 6e-3 x 9/2 = 0.027 m. (Key point: Deflection formula)

## Related in Theory of Structures I

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