---
title: "Truss: lack of fit, temperature, bars BC BF"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:truss-lack-of-fit-temp-bc"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-lack-of-fit-temp-bc
site: BCT Engineering
---

# Truss: lack of fit, temperature, bars BC BF

Theory of Structures II · Chapter 2: Force method

- Asked in: 2066 Jestha, 20 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-lack-of-fit-temp-bc
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Determine forces in bars BC and BF of the truss shown below, if all inclined members are found to be 2 mm too long and all vertical members are subjected to a decrease in temperature of 15°C. Area of cross-section of all members is 40 cm². Take $E = 2\times10^{5}\ \text{N/mm}^2$, $\alpha = 10.8\times10^{-6}/°C$. [Figure: truss with joints A, B, C on the bottom chord (A hinged, B roller, C roller; 5 m + 5 m), D, E, F on the top chord, height 3 m, with vertical and inclined members.]

## Short answer

Take the B roller reaction as redundant and impose zero displacement at B. Compatibility gives X=179.80 kN, so BC=0.00 kN and BF=-174.73 kN compression.

## Answer

**Assumptions (figure read as follows):** bottom joints A(0,0), B(5,0), C(10,0); top joints D(0,3), E(5,3), F(10,3). Members: bottom chord AB, BC; top chord DE, EF; verticals AD, BE, CF; inclined members DB and BF. A is a hinge; B and C are rollers (vertical reactions only). All areas 40 cm², $E=2\times10^5$ N/mm², so $EA=8\times10^5$ kN.
The inclined members DB and BF are 2 mm too long; the verticals AD, BE, CF are cooled by 15°C ($\alpha=10.8\times10^{-6}$/°C).

#### Degree of indeterminacy
$m=9$, $r=4$, $j=6$: $D_s=9+4-12=1$. The truss is internally determinate, so the lack of fit and the temperature change only produce forces because of the extra support. Take the vertical reaction at B as the redundant $X$.

#### Free (initial) elongations $e_0$
- Inclined bars too long: $e_0=+2$ mm (DB and BF).
- Verticals cooled: $e_0=-\alpha\,\Delta T\,L=-10.8\times10^{-6}\times15\times3000=-0.486$ mm.

#### Unit load on the primary truss
The primary truss (B roller removed) is loaded by a unit upward force at B, giving $S_1$. Since no external load acts, $S_0=0$ and only the initial strains cause displacement of B.

| Bar | L (m) | $S_1$ | $e_0$ (mm) | $S_1e_0$ (mm) | $S_1^2L$ (m) | Final $S$ (kN) |
|---|---|---|---|---|---|---|
| AB | 5.000 | 0.000 | 0.000 | 0.0000 | 0.000 | 0.00 |
| BC | 5.000 | 0.000 | 0.000 | 0.0000 | 0.000 | 0.00 |
| DE | 5.000 | 0.833 | 0.000 | 0.0000 | 3.472 | 149.83 |
| EF | 5.000 | 0.833 | 0.000 | 0.0000 | 3.472 | 149.83 |
| AD | 3.000 | 0.500 | -0.486 | -0.2430 | 0.750 | 89.90 |
| BE | 3.000 | 0.000 | -0.486 | 0.0000 | 0.000 | 0.00 |
| CF | 3.000 | 0.500 | -0.486 | -0.2430 | 0.750 | 89.90 |
| DB | 5.831 | -0.972 | 2.000 | -1.9437 | 5.507 | -174.73 |
| BF | 5.831 | -0.972 | 2.000 | -1.9437 | 5.507 | -174.73 |

$$
\Delta_{B}=\sum S_1e_0=-3.8873+(-0.4860)=-4.3733\ \text{mm}\quad(\text{negative: B moves 4.373 mm downward in the primary truss})
$$
$$
f_{11}=\frac{\sum S_1^2L}{EA}=\frac{19.458}{8\times10^5}=0.02432\ \text{mm/kN}
$$
Compatibility ($B$ cannot move): $\Delta_B+f_{11}X=0$
$$
X=-\frac{-4.3733}{0.02432}=179.80\ \text{kN}
$$
Bar forces $S=XS_1$ ($S_0=0$, because the initial strains alone produce no forces in the statically determinate primary truss):

**Answer: force in BC = 0.00 kN and force in BF = -174.73 kN (+ tension, - compression).** Other bars: AB = 0.00, DE = 149.83, EF = 149.83, AD = 89.90, BE = 0.00, CF = 89.90, DB = -174.73 kN.

## Check yourself

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

1. The truss is internally determinate. Why do lack of fit and temperature produce forces?
   A. Because of the extra support, the redundant reaction at B restores the displacement
   B. They produce no forces at all
   C. Because the bars are overstressed
   D. Because the hinge at A resists them
   Answer: A. Initial strains only deform the primary truss freely; forces arise because the roller at B prevents B from moving. (Key point: Why forces develop)

2. What are the free elongations e0 used?
   A. +2 mm for the inclined bars and -alpha dT L = -0.486 mm for the verticals
   B. +2 mm for all bars
   C. -2 mm for inclined and +0.486 mm for verticals
   D. 0 for the inclined bars
   Answer: A. Too-long inclined bars give +2 mm; a cooling of 15 degrees C shortens the 3 m verticals by 0.486 mm. (Key point: Initial elongations)

3. Which equation gives X (the vertical reaction at B)?
   A. sum(S1 e0) + X sum(S1^2 L)/EA = 0
   B. X = sum(S1 e0)
   C. X sum(S1 e0) = 0
   D. sum(S0 S1 L) = X
   Answer: A. B must not move, so displacement from initial strains plus that from X is zero; S0 = 0 as no external loads act. (Key point: Compatibility equation)

4. What are the forces in BF and BC?
   A. BF = -174.73 kN (compression), BC = 0
   B. BF = +174.73 kN, BC = 149.83 kN
   C. BF = -89.90 kN, BC = 89.90 kN
   D. Both zero
   Answer: A. S = X S1 with X = 179.80 kN; BF has S1 = -0.972 giving -174.73 kN, and BC has S1 = 0. (Key point: Force in BF and BC)

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