---
title: "Frame horizontal deflection at free end"
subject: "Theory of Structures I"
chapter: 3
key: "semester_4/theory_of_structures_i:frame-virtual-work-free-end-2077"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/frame-virtual-work-free-end-2077
site: BCT Engineering
---

# Frame horizontal deflection at free end

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

- Asked in: 2077 Chaitra, 7 marks
- Video: [Strain Energy - Problem No 11 (Horizontal Deflection in a Frame)](https://www.youtube.com/watch?v=WdUnhdQDN3A) — Stan Academy
- 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/frame-virtual-work-free-end-2077
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3

## Question

Determine the horizontal deflection at free end using method of virtual work. [Figure: frame fixed at the bottom of the left column (2EI, 4 m high, 50 kN horizontal load at mid-height, 2 m above the base); top beam of 6 m (3 m + 3 m, 1.5EI) with 20 kN/m UDL; right column EI, 3 m high, hanging down to a free end where a 30 kN horizontal load acts.]

## Short answer

The horizontal deflection at the free end equals 3633.3/EI (kN·m³) towards the left, obtained by integrating M·m/EI along the three cantilever members for the given loading.

## Answer

#### Given data
Frame fixed at the base A (0,0). Left column AB: 4 m high, $2EI$, with 50 kN horizontal (to the right) at 2 m above the base. Beam BC: 6 m, $1.5EI$, UDL 20 kN/m (down). Right column CD: 3 m, $EI$, hanging down from C to the free end D, where a 30 kN horizontal load acts towards the left. Required: horizontal deflection at the free end D by virtual work.

```text
    B =========== C        UDL 20 kN/m on BC
    |             |
 50->| (2 m)      | 3 m
    |             D <- 30 kN (free end)
    A (fixed)
```

#### Virtual load
Apply a unit horizontal load at D, to the left (the same direction as the 30 kN). Distances $s$ are measured from D along DC, from C along CB, and from B along BA. Since the structure is a cantilever fixed at A, no reactions are needed to find $M$ and $m$ (moments are taken on the free-end side).

| Member | $s$ | EI | $M$ (kNm) | $m$ |
|---|---|---|---|---|
| DC | 0-3 | $EI$ | $30s$ | $s$ |
| CB | 0-6 | $1.5EI$ | $90 + 10s^2$ | 3 |
| BA (top 2 m) | 0-2 | $2EI$ | $450 - 30s$ | $3 - s$ |
| BA (lower 2 m) | 2-4 | $2EI$ | $350 + 20s$ | $3 - s$ |

The column is split at the 50 kN load, where the expression for $M$ changes.

#### Integrals
| Member | $\int Mm\,ds$ | divide by | contribution |
|---|---|---|---|
| DC | 270 | $EI$ | 270.00 |
| CB | 3780 | $1.5EI$ | 2520.00 |
| BA (0-2) | 1700 | $2EI$ | 850.00 |
| BA (2-4) | -13.33 | $2EI$ | -6.67 |
| **Total** | | | **3633.33** |

$$
\Delta_{Dh} = \frac{3633.33}{EI}
$$

**Answer: horizontal deflection of the free end $= 3633.3/EI$ (kNm$^3$/EI), towards the left** (the direction of the unit load and the 30 kN force). Assumed: the 50 kN load acts to the right and the 30 kN load to the left.

## Check yourself

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

1. For the horizontal deflection of free end D, the unit load is:
   A. Horizontal at D, in the direction of the 30 kN force
   B. Vertical at D
   C. Horizontal at A
   D. A couple at A
   Answer: A. The unit load is at the required point and direction, here horizontal at D to the left. (Key point: Virtual load)

2. Why are reactions not needed to find M and m?
   A. The frame is cantilevered from A, so moments are taken from the free-end side
   B. The frame is a truss
   C. Reactions are zero
   D. The loads cancel out
   Answer: A. Moments are computed from the free end toward the fixed support, so the fixed-end reactions never enter. (Key point: Free-end side moments)

3. Why is column BA split into two portions?
   A. The moment expression changes at the 50 kN load
   B. It has two materials
   C. The unit load changes
   D. EI changes
   Answer: A. The 50 kN horizontal load at mid-height changes M's expression at that point. (Key point: Splitting column)

4. How is the beam BC (1.5EI) integral treated?
   A. 3780 divided by 1.5EI = 2520/EI
   B. 3780 times 1.5EI
   C. 3780 divided by EI
   D. Ignored
   Answer: A. Each member's integral is divided by its own flexural stiffness: 3780/1.5 = 2520. (Key point: Stiffness division)

5. The horizontal deflection of D is:
   A. 3633.3/EI to the left
   B. 3633.3/EI to the right
   C. 810/EI
   D. 2520/EI
   Answer: A. Total of 270 + 2520 + 850 - 6.67 = 3633.33, positive, so along the unit load direction (left). (Key point: Result)

## Related in Theory of Structures I

- [Frame rotation and vertical deflection at free end](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-rotation-vertical-free-end-2073m) — Theory of Structures I, chapter 3
- [Overhang beam free end deflection by virtual work](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Abeam-vw-free-end-d-2065) — Theory of Structures I, chapter 3
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- [Frame horizontal deflection at roller support](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-virtual-work-roller-2076) — Theory of Structures I, chapter 3

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