---
title: "Truss joint D: loading, lack of fit, temperature"
subject: "Theory of Structures I"
chapter: 3
key: "semester_4/theory_of_structures_i:truss-defl-d-2074"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-defl-d-2074
site: BCT Engineering
---

# Truss joint D: loading, lack of fit, temperature

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

- Asked in: 2074 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-d-2074
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3

## Question

Determine the vertical deflection of joint D of the truss due to (i) loading shown, (ii) members DE and DC being 5 mm too long and (iii) temperature of member CD alone is rise up by 20°C. Take $E = 200\times10^3\ \text{N/mm}^2$, coefficient of thermal expansion $= 12\times10^{-6}/^\circ$C. [Figure: truss with A (hinge) and B (roller) at the bottom, D at the bottom middle, AD = DB = 4 m, top joints E (above left) and C (above right); all angles 60°; 15 kN downward at E and 30 kN downward at C.]

## Short answer

Vertical deflection of joint D is 7.23 mm downward (assuming member area 1000 mm²; load part scales as 1/A). Load causes 0.90 mm, DE and DC each 5 mm too long add 5.77 mm, and 20°C rise in CD adds 0.55 mm.

## Answer

#### Given data and assumptions
A(0,0) hinge, D(4,0), B(8,0) roller; top joints E(2, 3.464) and C(6, 3.464); all members are 4 m long (60$^\circ$). Loads: 15 kN at E and 30 kN at C, downward. $E = 200\times10^3$ N/mm$^2$, $\alpha = 12\times10^{-6}$/$^\circ$C. The member area is not given; it is assumed to be the same for all members, $A = 1000$ mm$^2 = 10^{-3}$ m$^2$ (the loading part scales as $1/A$; the other parts do not depend on $A$). A unit downward load is applied at D.

Reactions (moments about A): $B_y\times8 = 15\times2 + 30\times6 = 210 \Rightarrow B_y = 26.25$ kN, $A_y = 18.75$ kN.

#### Member forces (tension +)
| Member | L (m) | F (kN) | k | kFL (kN m) | Temp. term (mm) | Fit term (mm) |
|---|---|---|---|---|---|---|
| AD | 4.000 | 10.83 | 0.289 | 12.500 | 0 | 0 |
| DB | 4.000 | 15.16 | 0.289 | 17.500 | 0 | 0 |
| AE | 4.000 | -21.65 | -0.577 | 50.000 | 0 | 0 |
| ED | 4.000 | 4.33 | 0.577 | 10.000 | 0 | 2.8868 |
| EC | 4.000 | -12.99 | -0.577 | 30.000 | 0 | 0 |
| DC | 4.000 | -4.33 | 0.577 | -10.000 | 0.5543 | 2.8868 |
| CB | 4.000 | -30.31 | -0.577 | 70.000 | 0 | 0 |

#### (i) Loading
$$
\Delta_1 = \frac{\sum kFL}{AE} = \frac{180.000}{10^{-3}\times2\times10^8} = 0.900\ \text{mm}\ (\downarrow)
$$

#### (ii) DE and DC each 5 mm too long
$$
\Delta_2 = \sum k\,e = (0.577)(5) + (0.577)(5) = 5.774\ \text{mm}\ (\downarrow)
$$

#### (iii) Temperature rise of 20$^\circ$C in CD
$$
\Delta L = 12\times10^{-6}\times20\times4000 = 0.96\ \text{mm},\qquad \Delta_3 = k\,\Delta L = 0.577\times0.96 = 0.554\ \text{mm}\ (\downarrow)
$$

#### Total
$$
\Delta_D = 0.900 + 5.774 + 0.554 = 7.228\ \text{mm}
$$

**Answer: vertical deflection of D = 7.23 mm downward** (for $A = 1000$ mm$^2$; the load part scales as $1/A$).

## Check yourself

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

1. Moments about A give the roller reaction B_y as:
   A. 26.25 kN
   B. 18.75 kN
   C. 45 kN
   D. 30 kN
   Answer: A. B_y x 8 = 15 x 2 + 30 x 6 = 210, so B_y = 26.25 kN and A_y = 18.75 kN. (Key point: Reactions)

2. Because member area is not given, the answer assumes A = 1000 mm^2. Which part depends on A?
   A. Only the loading part, scaling as 1/A
   B. All three parts
   C. Only temperature
   D. Only lack of fit
   Answer: A. Lack of fit and temperature effects are geometric and independent of A. (Key point: Loading deflection scaling)

3. DE and DC each 5 mm too long with k = 0.577 give:
   A. 5.774 mm downward
   B. 2.887 mm
   C. 10 mm
   D. 0.554 mm
   Answer: A. Sum k e = 0.577 x 5 + 0.577 x 5 = 5.774 mm downward. (Key point: Lack of fit)

4. A 20 C rise in CD (4 m) causes a deflection of D of:
   A. 0.554 mm downward
   B. 0.96 mm
   C. 5.77 mm
   D. 0
   Answer: A. Delta L = 12e-6 x 20 x 4000 = 0.96 mm; times k = 0.577 gives 0.554 mm. (Key point: Temperature)

## 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×
- [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×

## Related in Theory of Structures I

- [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 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×
- [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×
- [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: 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
- [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: 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: 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