---
title: "Truss joint E: temperature decrease and lack of fit"
subject: "Theory of Structures I"
chapter: 3
key: "semester_4/theory_of_structures_i:truss-defl-e-decrease-temp-2075bh"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-defl-e-decrease-temp-2075bh
site: BCT Engineering
---

# Truss joint E: temperature decrease and lack of fit

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

- Asked in: 2075 Bhadra, 12 marks
- Video: [Unit load method for truss with temperature and lack of fit effect](https://www.youtube.com/watch?v=HNprIHQrq7k) — Vasudeo Chaudhari
- 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/truss-defl-e-decrease-temp-2075bh
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3

## Question

Determine the vertical deflection of joint E due to the decrease in temperature by 20°C in member CD and member CE being 8 mm too long. Take $E = 200\times10^3\ \text{N/mm}^2$, $\alpha = 12\times10^{-6}/^\circ$C. Horizontal member area = 2000 mm², vertical/inclined member area = 1000 mm². [Figure: truss with A (hinge), E (bottom middle) and B (roller), AE = EB = 5 m; top joints D and C; all inclined members at 60°; 80 kN downward load at C.]

## Short answer

Vertical deflection of joint E is 6.98 mm downward. Load contributes 1.67 mm down, 20°C temperature drop in member CD adds 0.69 mm down, and member CE fabricated 8 mm too long adds 4.62 mm down.

## Answer

#### Given data
A(0,0) hinge, E(5,0), B(10,0) roller; top joints D(2.5, 4.330) and C(7.5, 4.330); all members are 5 m long (equilateral triangles). 80 kN acts downward at C. $E = 200\times10^3$ N/mm$^2 = 2\times10^8$ kN/m$^2$, $\alpha = 12\times10^{-6}$/$^\circ$C. Areas: horizontal members AE, EB, DC = 2000 mm$^2$; inclined members = 1000 mm$^2$. Apply a unit load (down) at E.

Reactions (moments about A): $B_y\times10 = 80\times7.5 \Rightarrow B_y = 60$ kN, $A_y = 20$ kN. For the unit load at E: $A_y = B_y = 0.5$.

#### Member forces (tension +)
| Member | L (m) | F (kN) | k | kFL (kN m) | Temp. term (mm) | Fit term (mm) |
|---|---|---|---|---|---|---|
| AE | 5.000 | 11.55 | 0.289 | 16.667 | 0 | 0 |
| EB | 5.000 | 34.64 | 0.289 | 50.000 | 0 | 0 |
| AD | 5.000 | -23.09 | -0.577 | 66.667 | 0 | 0 |
| DE | 5.000 | 23.09 | 0.577 | 66.667 | 0 | 0 |
| DC | 5.000 | -23.09 | -0.577 | 66.667 | 0.6928 | 0 |
| EC | 5.000 | -23.09 | 0.577 | -66.667 | 0 | 4.6188 |
| CB | 5.000 | -69.28 | -0.577 | 200.000 | 0 | 0 |

#### (i) Loading
Because the areas differ, use $\sum kFL/A$ with the area of each member:

$$
\Delta_1 = \frac{1}{E}\sum\frac{kFL}{A} = \frac{1}{2\times10^8}\left[\frac{133.33}{0.002} + \frac{266.67}{0.001}\right] = 1.667\ \text{mm}\ (\downarrow)
$$

#### (ii) Temperature fall of 20$^\circ$C in CD
$$
\Delta L_{CD} = 12\times10^{-6}\times(-20)\times5000 = -1.2\ \text{mm},\qquad \Delta_2 = k\,\Delta L = (-0.577)(-1.2) = 0.693\ \text{mm}\ (\downarrow)
$$

#### (iii) CE 8 mm too long
$$
\Delta_3 = k_{CE}\,e = (0.577)(8) = 4.619\ \text{mm}\ (\downarrow)
$$

#### Total
$$
\Delta_E = 1.667 + 0.693 + 4.619 = 6.978\ \text{mm}
$$

**Answer: vertical deflection of joint E = 6.98 mm downward.**

## Check yourself

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

1. Because member areas differ, the loading deflection is computed as:
   A. (1/E) sum of kFL/A with A for each member
   B. sum kFL / (A E) with one A
   C. Using the average area
   D. Ignoring horizontal members
   Answer: A. Each member's own area enters the term kFL/A. (Key point: Different areas)

2. For a 20 C fall in CD of 5 m, the length change and its effect are:
   A. -1.2 mm giving 0.693 mm downward at E, with k = -0.577
   B. +1.2 mm
   C. -12 mm
   D. No effect
   Answer: A. Delta L = 12e-6 x (-20) x 5000 = -1.2 mm; k x Delta L = (-0.577)(-1.2) = +0.693 mm downward. (Key point: Temperature fall in CD)

3. For CE 8 mm too long with k = 0.577:
   A. 4.619 mm downward
   B. 8 mm
   C. 4.619 mm up
   D. 0.577 mm
   Answer: A. k e = 0.577 x 8 = 4.619 mm, downward. (Key point: Lack of fit)

4. Total deflection of E is:
   A. 6.98 mm downward
   B. 6.98 mm upward
   C. 1.67 mm
   D. 4.6 mm
   Answer: A. 1.667 + 0.693 + 4.619 = 6.978 mm downward. (Key point: Total)

## Close variants set in other exams

- [Truss joint E: temperature and lack of fit](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-defl-e-2072m) — Theory of Structures I, chapter 3, asked 3×

## Related in Theory of Structures I

- [Truss joint E: loading, lack of fit, temperature](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-defl-e-2068) — Theory of Structures I, chapter 3, asked 2×
- [Truss deflection at B: load, temperature, lack of fit](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-defl-b-loading-temp-lackfit-2081) — Theory of Structures I, chapter 3
- [Truss joint D: loading, lack of fit, temperature](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-defl-d-2074) — Theory of Structures I, chapter 3, asked 3×
- [Truss joint E: temperature and lack of fit](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-defl-e-temp-lackfit-2075b) — Theory of Structures I, chapter 3
- [Truss joint B: temperature and vertical members long](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-defl-b-temp-lackfit-2073) — Theory of Structures I, chapter 3, asked 2×
- [Truss joint H: temperature and long vertical members](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-defl-h-temp-lackfit-2069b) — Theory of Structures I, chapter 3, asked 2×

## Related in other subjects

- [Truss: lack of fit, temperature, bars BC BF](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-lack-of-fit-temp-bc) — Theory of Structures II, chapter 2
- [Truss forces: member AC 10 mm too long](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-lack-of-fit-ac-10mm) — Theory of Structures II, chapter 2
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- [Truss: temperature decrease in vertical members](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-temp-decrease-vertical-ab-ef) — Theory of Structures II, chapter 2