---
title: "Frame horizontal deflection at E"
subject: "Theory of Structures I"
chapter: 3
key: "semester_4/theory_of_structures_i:frame-defl-e-horizontal-2080"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/frame-defl-e-horizontal-2080
site: BCT Engineering
---

# Frame horizontal deflection at E

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

- Asked in: 2080 Chaitra, 8 marks
- Video: [Deflection of Frames using Principle of Virtual Work - Intro to Structural Analysis](https://www.youtube.com/watch?v=JAD-8ATyu6k) — Structures with Prof. H
- 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-defl-e-horizontal-2080
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3

## Question

Calculate the horizontal deflection at point E for the following frame. Assume flexural rigidity EI as constant throughout. [Figure: portal frame; left column A-B-C with A (hinge) at the bottom, B 2 m above A, C 2 m above B; top beam CD of length 4 m carrying 10 kN/m UDL; right column D-E-F with E 2 m below D and F (roller) 2 m below E; a 20 kN horizontal load acts at B pointing towards the frame.]

## Short answer

Using the virtual work method, the horizontal deflection at E is 720/EI (kN·m³) to the right, obtained by integrating the product of real and virtual moments along all frame members.

## Answer

#### Given data
Portal frame (EI constant): A(0,0) hinge, B(0,2), C(0,4), D(4,4), E(4,2), F(4,0) roller. 20 kN horizontal at B (to the right); UDL 10 kN/m on CD (down). Required: horizontal deflection at E using the unit load method.

```text
   C ============== D
   |   10 kN/m      |
   |                |
 20->B              E  <- unit load (virtual)
   |                |
   A (hinge)        F (roller)
```

#### Real loading: reactions
$\sum F_x = 0$: $A_x = 20$ kN (left). $\sum M_A = 0$: $F_y\times4 = 20\times2 + 40\times2 \Rightarrow F_y = 30$ kN; $A_y = 40 - 30 = 10$ kN.

#### Virtual load: unit horizontal load at E (to the right)
$A_x = 1$ (left). Moments about A: $F_y\times4 = 1\times2$, so $F_y = 0.5$ (up) and $A_y = 0.5$ (down).

#### Bending moments ($s$ from the first end of each member, in the order A-B-C-D-E-F)
| Member | $s$ range (m) | $M$ (kNm) | $m$ | $\int M m\,ds$ |
|---|---|---|---|---|
| AB | 0-2 | $20s$ | $s$ | 53.33 |
| BC | 0-2 | 40 | $s+2$ | 240.0 |
| CD | 0-4 | $40 + 10s - 5s^2$ | $4 - s/2$ | 426.67 |
| DE | 0-2 | 0 | $2-s$ | 0 |
| EF | 0-2 | 0 | 0 | 0 |

(The right column has no moment because the roller gives no horizontal force.)

#### Deflection
$$
\Delta_{Eh} = \frac{1}{EI}\sum\int Mm\,ds = \frac{53.33 + 240 + 426.67}{EI} = \frac{720}{EI}
$$

**Answer: horizontal deflection of E $= \dfrac{720}{EI}$ kNm$^3$/EI, 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. To find the horizontal deflection at E, the virtual load is:
   A. A unit vertical load at E
   B. A unit horizontal load at E
   C. A unit couple at A
   D. The 20 kN load repeated at E
   Answer: B. A unit load is placed at the point and in the direction of the required displacement. (Key point: Virtual load)

2. Why is there no bending moment in the right column DEF in the real loading?
   A. The roller gives no horizontal force, and the column carries only axial load
   B. It is fixed
   C. It has zero length
   D. The UDL is not on it
   Answer: A. With a roller at F no horizontal reaction exists on that side, so no moment arises in the column. (Key point: Roller effect on right column)

3. The vertical reaction at the roller F for the real loads is:
   A. 10 kN
   B. 30 kN
   C. 40 kN
   D. 50 kN
   Answer: B. Moments about A: F_y x 4 = 20 x 2 + 40 x 2 = 120, so F_y = 30 kN. (Key point: Real reactions)

4. The horizontal deflection at E equals:
   A. 360/EI to the left
   B. 720/EI to the right
   C. 720/EI to the left
   D. 53.33/EI
   Answer: B. Sum of integral M m ds = 53.33 + 240 + 426.67 = 720, positive, so in the direction of the unit load (right). (Key point: Final result)

## Related in Theory of Structures I

- [Frame horizontal deflection at E](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-horizontal-defl-e-2074) — Theory of Structures I, chapter 3
- [Frame: horizontal displacement and rotation at roller](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-horizontal-rotation-roller-2079) — Theory of Structures I, chapter 3
- [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
- [Frame horizontal deflection at free end](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-virtual-work-free-end-2077) — Theory of Structures I, chapter 3
- [Frame: roller displacement and hinge rotation](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-roller-hinge-rotation-2071b) — Theory of Structures I, chapter 3
- [Frame vertical deflection at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-vw-d-vertical-2071o) — Theory of Structures I, chapter 3

## Related in other subjects

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- [Frame AFD, SFD, BMD (5 and 10 kN/m, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-5knm-10knm) — Strength of Materials, chapter 1, asked 2×
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