---
title: "Truss joint E: temperature and lack of fit"
subject: "Theory of Structures I"
chapter: 3
key: "semester_4/theory_of_structures_i:truss-defl-e-2072m"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-defl-e-2072m
site: BCT Engineering
---

# Truss joint E: temperature and lack of fit

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

- Asked in: 2072 Magh, 10 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-2072m
- 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 increase in temperature of 20°C of member CD and member CE being 5 mm too long. Take $E = 200\ \text{kN/mm}^2$, $\alpha = 12\times10^{-6}/^\circ$C, area = 1000 mm² for all members. [Figure: truss with A (hinge), E (bottom middle) and B (roller), AE = EB = 4 m; top joints C (above left) and D (above right); angles 60°; 20 kN downward at C.]

## Short answer

Virtual unit load at E. Sum kFL/AE for load = 0.400 mm down. Temperature rise in CD gives -0.554 mm (up); lack of fit in CE gives 2.887 mm down. Total deflection 2.73 mm downward.

## Answer

#### Given data
A(0,0) hinge, E(4,0), B(8,0) roller; top joints C(2, 3.464) and D(6, 3.464); all members 4 m (60$^\circ$). 20 kN down at C. $E = 200$ kN/mm$^2 = 2\times10^8$ kN/m$^2$, $A = 1000$ mm$^2 = 10^{-3}$ m$^2$, $\alpha = 12\times10^{-6}$/$^\circ$C. Virtual load: a unit downward load at E.

Reactions (real): $\sum M_B = 0$: $A_y\times8 = 20\times6 \Rightarrow A_y = 15$ kN, $B_y = 5$ kN.

#### Member forces (tension +)
| Member | L (m) | F (kN) | k | kFL (kN m) | Temp. term (mm) | Fit term (mm) |
|---|---|---|---|---|---|---|
| AE | 4.000 | 8.66 | 0.289 | 10.000 | 0.0000 | 0.0000 |
| EB | 4.000 | 2.89 | 0.289 | 3.333 | 0.0000 | 0.0000 |
| AC | 4.000 | -17.32 | -0.577 | 40.000 | -0.0000 | -0.0000 |
| CE | 4.000 | -5.77 | 0.577 | -13.333 | 0.0000 | 2.8868 |
| CD | 4.000 | -5.77 | -0.577 | 13.333 | -0.5543 | -0.0000 |
| ED | 4.000 | 5.77 | 0.577 | 13.333 | 0.0000 | 0.0000 |
| DB | 4.000 | -5.77 | -0.577 | 13.333 | -0.0000 | -0.0000 |

#### (i) Effect of the 20 kN load shown in the figure

$$
\Delta_1 = \frac{\sum kFL}{AE} = \frac{80.000}{10^{-3}\times2\times10^8} = 0.400\ \text{mm}\ (\downarrow)
$$

#### (ii) Temperature rise of 20$^\circ$C in CD (top chord)
$$
\Delta L_{CD} = 12\times10^{-6}\times20\times4000 = 0.96\ \text{mm},\qquad \Delta_2 = k_{CD}\,\Delta L = (-0.577)(0.96) = -0.554\ \text{mm}\ (\uparrow)
$$

#### (iii) CE 5 mm too long
$$
\Delta_3 = k_{CE}\,e = (0.577)(5) = 2.887\ \text{mm}\ (\downarrow)
$$

#### Total
$$
\Delta_E = 0.400 - 0.554 + 2.887 = 2.732\ \text{mm}
$$

**Answer: vertical deflection of E = 2.73 mm downward** (including the 20 kN load; the temperature change and lack of fit alone give 2.33 mm downward).

## Check yourself

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

1. Which effects are added to get the vertical deflection of E?
   A. Load and temperature in all members
   B. Load, temperature rise in CD and lack of fit in CE
   C. Load only
   D. Temperature in CD and lack of fit in CE only, since the load is not asked
   Answer: B. The 20 kN load is shown in the figure, so its effect is included with the temperature rise in CD and the CE lack of fit. (Key point: Three effects)

2. Why does the temperature rise of CD move E upward (-0.554 mm)?
   A. CD is in compression under the unit load (k negative) but it lengthens, giving a negative product
   B. CD is in tension under the unit load, and it lengthens
   C. The temperature rise shortens CD
   D. CD has no stress
   Answer: A. Delta_2 = k x alpha DeltaT L = (-0.577)(0.96) = -0.554 mm; the negative sign shows upward movement. (Key point: Temperature term sign)

3. What is the contribution of CE being 5 mm too long?
   A. 5 mm downward
   B. 2.887 mm downward, from k_CE x e = 0.577 x 5
   C. 0 because CE is a diagonal
   D. -2.887 mm
   Answer: B. With k_CE = +0.577 (tension under the unit load), Delta_3 = 0.577 x 5 = 2.887 mm downward. (Key point: Lack of fit)

4. Which of the following is the total vertical deflection of E?
   A. 0.400 mm up
   B. 2.33 mm upward
   C. 2.73 mm downward
   D. 3.84 mm downward
   Answer: C. 0.400 - 0.554 + 2.887 = 2.73 mm downward; the lack of fit dominates. (Key point: Total)

## Close variants set in other exams

- [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 decrease and lack of fit](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-defl-e-decrease-temp-2075bh) — Theory of Structures I, chapter 3, asked 2×

## 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 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×
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- [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 L2 by virtual work and temperature](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-vw-l2-temperature-2069p) — Theory of Structures I, chapter 3

## Related in other subjects

- [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
- [Truss: loads, temperature rise, lack of fit](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-loads-temp-lack-of-fit) — 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