---
title: "Truss horizontal displacement at C by virtual work"
subject: "Theory of Structures I"
chapter: 3
key: "semester_4/theory_of_structures_i:truss-virtual-work-horizontal-c"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-virtual-work-horizontal-c
site: BCT Engineering
---

# Truss horizontal displacement at C by virtual work

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

- Asked in: 2078 Chaitra, 6 marks
- Video: [Deflection of Trusses Virtual Work Method](https://www.youtube.com/watch?v=7AJsvPf_o_w) — Nadim Shbeeb
- 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-virtual-work-horizontal-c
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3

## Question

Using virtual work method, calculate the horizontal displacement at joint C of the given truss. Take AE to be constant. [Figure: square truss ABCD of side 5 m with diagonal BD; A and D are supports at the bottom, B above A, C above D; 50 kN vertical load at C and 100 kN horizontal load at C pointing away from the frame.]

## Short answer

The horizontal displacement at joint C is 2914.2/AE (kN·m) to the right, determined by summing kFL/AE for all members under real and unit virtual loading.

## Answer

#### Given data and assumptions
Square panel truss: A(0,0), D(5,0), B(0,5), C(5,5); members AB, BC, CD, AD and diagonal BD. A is a hinge and D a roller. Loads at C: 50 kN downward and 100 kN horizontal to the right (away from the frame). $AE$ constant. Required: horizontal displacement of C.

#### Step 1: Real forces (tension +)
Reactions: $\sum M_A = 0$: $D_y\times5 = 50\times5 + 100\times5 \Rightarrow D_y = 150$ kN (up); $A_y = -100$ kN (down); $A_x = -100$ kN.

Joint C: CD carries the 50 kN vertical load: $F_{CD} = -50$ kN. The horizontal 100 kN is taken by BC: $F_{BC} = +100$ kN.
Joint B: $F_{BD}\cos45^\circ = -F_{BC}\Rightarrow F_{BD} = -141.42$ kN; $F_{AB} = -F_{BD}\sin45^\circ = +100$ kN.
Joint A / D: $F_{AD} = +100$ kN.

#### Step 2: Virtual forces (unit horizontal load at C to the right)
$k_{BC} = 1$, $k_{BD} = -\sqrt2 = -1.414$, $k_{AB} = 1$, $k_{AD} = 1$, $k_{CD} = 0$.

#### Step 3: Table
| Member | $L$ (m) | $F$ (kN) | $k$ | $kFL$ (kNm) |
|---|---|---|---|---|
| AB | 5 | +100 | 1 | 500 |
| BC | 5 | +100 | 1 | 500 |
| CD | 5 | -50 | 0 | 0 |
| AD | 5 | +100 | 1 | 500 |
| BD | 7.071 | -141.42 | -1.414 | 1414.2 |
| **Sum** | | | | **2914.2** |

$$
\Delta_{Ch} = \sum\frac{kFL}{AE} = \frac{2914.2}{AE}
$$

**Answer: horizontal displacement of C $= \dfrac{2914.2\ \text{kNm}}{AE}$, towards the right (direction of the unit load).**

## Check yourself

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

1. The horizontal displacement of C for a truss is found by:
   A. Sum of F k L / AE
   B. Integral of M m / EI
   C. Sum of F L / AE only
   D. Sum of k^2 L / AE
   Answer: A. F is the real member force, k the force from a unit load at C in the horizontal direction. (Key point: Truss formula)

2. The real force in diagonal BD is:
   A. +141.42 kN
   B. -141.42 kN (compression)
   C. -100 kN
   D. 0
   Answer: B. Joint B equilibrium: F_BD cos45 = -F_BC, so F_BD = -141.42 kN. (Key point: Diagonal force)

3. Why does member CD not contribute to the sum?
   A. Its real force is zero
   B. Its k value is zero for a horizontal unit load at C
   C. Its length is zero
   D. It is in tension
   Answer: B. The horizontal unit load at C goes through BC, BD, AB and AD; CD has k = 0 so kFL = 0. (Key point: Zero k member)

4. Horizontal displacement at C is:
   A. 1500/AE
   B. 2914.2/AE to the right
   C. 2914.2/AE to the left
   D. 5000/AE
   Answer: B. Sum of kFL = 500 + 500 + 0 + 500 + 1414.2 = 2914.2, positive, so in the unit load direction (right). (Key point: Result)

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