---
title: "Overhang beam free end deflection by virtual work"
subject: "Theory of Structures I"
chapter: 3
key: "semester_4/theory_of_structures_i:beam-vw-free-end-d-2065"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/beam-vw-free-end-d-2065
site: BCT Engineering
---

# Overhang beam free end deflection by virtual work

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

- Asked in: 2065 Chaitra, 8 marks
- Video: [SA22: Virtual Work Method (Beams)](https://www.youtube.com/watch?v=V_gQ4Y03Flg) — Dr. Structure
- 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/beam-vw-free-end-d-2065
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3

## Question

Calculate the vertical deflection of free end D of the beam loaded as shown in fig-2 by using virtual work method. Take EI as constant throughout. [Figure: beam ABCD, A hinge, B 2 m from A with 20 kN downward, C roller 2 m from B, free end D 2 m from C with 10 kN downward.]

## Short answer

Applying the virtual work method with a unit load at D: the vertical deflection at the free end D is 40/EI (kN·m³) downward.

## Answer

#### Given data
Beam ABCD: A hinge, B at 2 m (20 kN down), C roller at 4 m, D free end at 6 m (10 kN down). $EI$ constant.

```text
    A     B     C     D
    ^-----+-----^-----+
        20 kN         10 kN
      2 m   2 m   2 m
```

#### Real loading
$\sum M_C = 0$: $R_A\times4 + 10\times2 = 20\times2 \Rightarrow R_A = 5$ kN (up); $R_C = 25$ kN (up).

Bending moments (sagging +, $x$ from A): AB: $M = 5x$ (10 kNm at B); BC: $M = 5x - 20(x-2) = 40 - 15x$, so $M_C = -20$ kNm; CD: $M = -10s$ ($s$ from D), giving $M_C = -20$ kNm.

#### Virtual load: unit downward load at D
Reactions: $R_A = -0.5$, $R_C = 1.5$. Moments: AB: $m = -0.5x$ (hogging), BC: $m = -0.5x$ (to C: $-2$), CD: $m = -s$.

#### Integrals
| Portion | $M$ | $m$ | $\int Mm\,dx$ |
|---|---|---|---|
| AB (0-2) | $5x$ | $-0.5x$ | $-2.5\times\frac{8}{3} = -6.667$ |
| BC (2-4) | $5x - 20(x-2)$ | $-0.5x$ | $+20.000$ (see below) |
| CD (0-2, from D) | $-10s$ | $-s$ | $10\times\frac{8}{3} = 26.667$ |

Evaluating the BC integral: $\int_2^4 (-0.5x)(40-15x)\,dx = \int_2^4(-20x + 7.5x^2)dx = -10(16-4) + 2.5(64-8) = -120 + 140 = 20$.

$$
\Delta_D = \frac{-6.667 + 20 + 26.667}{EI} = \frac{40}{EI}
$$

**Answer: vertical deflection of the free end D $= 40/EI$ (kNm$^3$/EI), downward.**

## Check yourself

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

1. Taking moments about C, what is R_A?
   A. 5 kN upward
   B. 20 kN
   C. 10 kN
   D. 25 kN
   Answer: A. R_A x 4 + 10 x 2 = 20 x 2 gives R_A = 5 kN upward; R_C = 25 kN. (Key point: Real reactions)

2. For a unit downward load at D, what are the virtual reactions?
   A. R_A = 1, R_C = 0
   B. R_A = 0.5, R_C = 1.5
   C. R_A = 0.5, R_C = 0.5
   D. R_A = -0.5 (down), R_C = 1.5 (up)
   Answer: D. Moments about C: R_A x 4 = -1 x 2, so R_A = -0.5 and R_C = 1.5. (Key point: Virtual reactions)

3. Why is m negative in AB and BC?
   A. The unit load at D makes the beam hog between A and C, while M is sagging in AB
   B. EI is negative
   C. The beam is hogging due to M
   D. The unit load acts upward
   Answer: A. The overhang load causes hogging in the virtual system, which is opposite in sense to the sagging real moment in AB, giving a negative product. (Key point: Sign of m)

4. What is the deflection of D?
   A. 20/EI upward
   B. 60/EI downward
   C. 40/EI downward
   D. 6.667/EI
   Answer: C. Sum of -6.667 + 20 + 26.667 = 40, so Delta_D = 40/EI downward. (Key point: Result)

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