---
title: "Frame horizontal deflection at E"
subject: "Theory of Structures I"
chapter: 3
key: "semester_4/theory_of_structures_i:frame-horizontal-defl-e-2074"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/frame-horizontal-defl-e-2074
site: BCT Engineering
---

# Frame horizontal deflection at E

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

- Asked in: 2074 Bhadra, 10 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-horizontal-defl-e-2074
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3

## Question

Determine horizontal deflection at E of the frame shown in figure below. [Figure: frame; left column A-B-C with A (hinge) at the bottom, B 3 m above A, C 3 m above B; 10 kN horizontal load at B; top beam CD of 4 m (2EI) with 20 kN/m UDL; right column D-E of 4 m (EI) with E roller; left column EI; width 4 m.]

## Short answer

Horizontal deflection at E is 921.7/EI to the right. Virtual work with a unit horizontal load at E gives ∫M m/EI = 921.7/EI kNm³.

## Answer

#### Given data
Frame: A(0,0) hinge, B(0,3), C(0,6), D(4,6), E(4,2) roller. Left column AC: $EI$ (3 m + 3 m); beam CD: $2EI$, 4 m, UDL 20 kN/m; right column DE: $EI$, 4 m. 10 kN horizontal at B (to the right). Required: horizontal deflection of E.

#### Real loading: reactions
$$
\sum M_A = 0:\; E_y\times4 = 10\times3 + 80\times2 \Rightarrow E_y = 47.5\ \text{kN},\quad A_y = 80 - 47.5 = 32.5\ \text{kN},\quad A_x = 10\ \text{kN (left)}
$$

(The 10 kN load gives 30 kNm clockwise about A, the UDL resultant 80 kN at $x = 2$ m gives 160 kNm.)

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

#### Moments and integrals ($s$ from the start of each member, A-B-C-D-E)
| Member | $s$ (m) | stiffness | $M$ (kNm) | $m$ | $\int Mm\,ds$ / stiffness |
|---|---|---|---|---|---|
| AB | 0-3 | $EI$ | $10s$ | $s$ | 90.00 |
| BC | 0-3 | $EI$ | 30 | $s + 3$ | 405.00 |
| CD | 0-4 | $2EI$ | $30 + 32.5s - 10s^2$ | $6 - s/2$ | 426.67 |
| DE | 0-4 | $EI$ | 0 | $4 - s$ | 0 |

$$
\Delta_{Eh} = \frac{90 + 405 + 426.67}{EI} = \frac{921.67}{EI}
$$

**Answer: horizontal deflection of E $= 921.7/EI$ (kNm$^3$/EI), to the right.**

## Check yourself

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

1. To find the horizontal deflection of E by the unit load method, which virtual loading is applied?
   A. A unit horizontal load at B
   B. A unit couple at E
   C. A unit horizontal load at E
   D. A unit vertical load at E
   Answer: C. The virtual load must correspond to the required displacement, so a unit horizontal load at E is used. (Key point: Virtual load choice)

2. Taking moments about A for the real loading, what gives E_y = 47.5 kN?
   A. E_y x 4 = 10 x 6 + 80 x 4
   B. E_y x 4 = 10 x 3 + 20 x 4
   C. E_y x 4 = 80 x 2 only, since the 10 kN load is horizontal
   D. E_y x 4 = 10 x 3 + 80 x 2, with the 80 kN UDL resultant acting 2 m from A
   Answer: D. The 10 kN load gives 30 kNm and the 80 kN UDL resultant at mid-span (2 m) gives 160 kNm; their sum 190 divided by 4 gives 47.5 kN. (Key point: Real reactions)

3. Why does member DE contribute nothing to the integral of Mm?
   A. Its virtual moment m is zero
   B. Its stiffness is infinite
   C. It is shorter than the other members
   D. Its real moment M is zero because no load acts beyond the section on the E side
   Answer: D. Measured from the roller end E, no real load acts on DE, so M = 0 and the integral vanishes even though m is non-zero. (Key point: Zero contribution of DE)

4. Beam CD has stiffness 2EI. How is this treated in the sum for the horizontal deflection of E?
   A. The integral of Mm over CD is divided by 2EI
   B. All members are assumed to have EI
   C. CD is ignored because it carries a UDL
   D. The integral over CD is multiplied by 2
   Answer: A. Each member's integral of Mm is divided by its own stiffness; the tabulated 426.67 is already the value for CD after dividing by 2. (Key point: Stiffness in the integral)

## Related in Theory of Structures I

- [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 E](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-defl-e-horizontal-2080) — 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: 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 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 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

## Related in other subjects

- [Frame AFD, SFD, BMD (30 and 20 kN/m, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-30-20) — Strength of Materials, chapter 1
- [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×
- [Frame AFD, SFD, BMD (5 kN/m, 10 kN, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-5knm-10kn) — Strength of Materials, chapter 1, asked 2×
- [Frame SFD, BMD (100 kN, 20 kN/m, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-sfd-bmd-100kn-hinge-60deg) — Strength of Materials, chapter 1