---
title: "Frame deflection at C due to temperature"
subject: "Theory of Structures I"
chapter: 3
key: "semester_4/theory_of_structures_i:frame-temperature-c-2074"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/frame-temperature-c-2074
site: BCT Engineering
---

# Frame deflection at C due to temperature

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

- Asked in: 2074 Bhadra, 6 marks
- Video: [Problem 3 Deflection Caused by Temperature Variations in Steel Bent Frame | Rigid Jointed Frame](https://www.youtube.com/watch?v=b4Hub20XF0k) — Ganesh Shegar
- 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/frame-temperature-c-2074
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3

## Question

Determine horizontal and vertical deflection of point C of the frame shown in figure below due to the temperature variation. [Figure: frame with fixed support A at the bottom, column AB of 6 m, top member BC of 4 m; temperature -20°C on the outer side and +20°C on the inner side; member depth 30 cm. Coefficient of expansion not stated, assume suitable data.]

## Short answer

Horizontal deflection of C is 28.8 mm left, vertical deflection 51.2 mm upward. Temperature difference 40°C across 0.30 m depth (α=12e-6/°C) gives curvature 1.6e-3 per m; virtual work yields these displacements.

## Answer

#### Given data and assumptions
Frame fixed at A: column AB 6 m, member BC 4 m. Outer face $-20^\circ$C, inner face $+20^\circ$C, depth $d = 0.30$ m. Coefficient of expansion not given; take $\alpha = 12\times10^{-6}$/$^\circ$C (usual value for steel/concrete).

#### Temperature effects
- Mean temperature $= \dfrac{-20 + 20}{2} = 0$, so there is **no axial deformation**.
- Temperature difference across depth $= 40^\circ$C, so curvature

$$
\kappa = \frac{\alpha\,\Delta T}{d} = \frac{12\times10^{-6}\times40}{0.30} = 1.6\times10^{-3}\ \text{per m}
$$

The inner face is hotter (longer), so each member curves with its inner face convex.

#### Virtual work
$$
\Delta = \int m\,\kappa\,ds
$$

**Horizontal deflection of C** (unit horizontal load at C, $m$ is the vertical distance from C to the section: zero in BC, and $m = 6 - y$ at height $y$ above A in AB):

$$
\Delta_h = \kappa\int_0^6 (6-y)\,dy = 1.6\times10^{-3}\times18 = 28.8\ \text{mm}
$$

The column AB bends with its inner (right) face longer, so C moves **to the left** (towards the outer side).

**Vertical deflection of C** (unit vertical load at C, $m$ = horizontal distance from C: 4 m in all of AB, and $x$ in BC):

$$
\Delta_v = \kappa\left[\int_0^6 4\,dy + \int_0^4 x\,dx\right] = 1.6\times10^{-3}\times(24 + 8) = 51.2\ \text{mm}
$$

**Answer: horizontal deflection of C $= 28.8$ mm to the left; vertical deflection of C $= 51.2$ mm upward.**

## Check yourself

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

1. Outer face is -20°C and inner face +20°C. What does the mean temperature of 0°C imply?
   A. Zero deflection of C
   B. No axial deformation of the members
   C. No curvature
   D. Maximum axial elongation
   Answer: B. Axial change depends on the mean temperature; with a mean of zero there is no axial deformation, and only bending from the gradient remains. (Key point: Mean temperature)

2. How is the curvature caused by the temperature difference found here?
   A. kappa = alpha x mean temperature / d
   B. kappa = alpha x DeltaT x d
   C. kappa = alpha x DeltaT / d = 1.6 x 10^-3 per m
   D. kappa = DeltaT / (alpha x d)
   Answer: C. Curvature is alpha times the temperature difference across the depth, divided by the depth: 12e-6 x 40 / 0.30 = 1.6e-3 per m. (Key point: Curvature)

3. Why does the member BC not contribute to the horizontal deflection of C?
   A. The virtual moment m for a horizontal unit load at C is zero along BC
   B. BC is at the same temperature on both faces
   C. Its curvature is zero
   D. BC is shorter than AB
   Answer: A. A horizontal unit load at C has no moment arm along the horizontal member BC, so m = 0 there; only AB contributes, giving 1.6e-3 x 18 = 28.8 mm. (Key point: Horizontal deflection)

## Related in Theory of Structures I

- [Frame deflection due to temperature variation](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-temperature-free-end-2076) — Theory of Structures I, chapter 3
- [Frame horizontal deflection at roller support](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-virtual-work-roller-2076) — Theory of Structures I, chapter 3
- [Frame vertical deflection at C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-vertical-defl-c-2065) — Theory of Structures I, chapter 3
- [Frame vertical deflection at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-vw-d-vertical-2071o) — Theory of Structures I, chapter 3
- [Frame horizontal deflection at E](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-horizontal-defl-e-2074) — Theory of Structures I, chapter 3
- [Frame horizontal deflection at free end](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-virtual-work-free-end-2077) — Theory of Structures I, chapter 3

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