---
title: "Frame: horizontal displacement and rotation at roller"
subject: "Theory of Structures I"
chapter: 3
key: "semester_4/theory_of_structures_i:frame-horizontal-rotation-roller-2079"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/frame-horizontal-rotation-roller-2079
site: BCT Engineering
---

# Frame: horizontal displacement and rotation at roller

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

- Asked in: 2079 Chaitra, 12 marks
- Video: [Unit load Method Frame | Virtual Work Method | Deflection And Slope at a Point | TOS I | SA - I](https://www.youtube.com/watch?v=wD61LB5bV08) — Parash Joshi - Civil Construction and Tutor
- 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-rotation-roller-2079
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3

## Question

Determine the horizontal displacement and rotation at roller support in the frame shown in the figure. [Figure: portal frame; A (hinge) at bottom of left column, B on the left column 1 m below the top joint C (left column 4 m above A as drawn), top beam CD of 5 m with 25 kN/m UDL and flexural rigidity 2EI, 15 kN horizontal load at B pointing right; right column DE of 3 m with EI, E roller at the bottom; left column EI.]

## Short answer

Horizontal displacement at the roller support equals 1435.8/EI (kN·m³) to the right, and the rotation is 90.10/EI rad anticlockwise, computed via unit load and unit couple methods.

## Answer

#### Given data
Frame with $A(0,0)$ hinge, $B(0,4)$ (15 kN to the right), $C(0,5)$, $D(5,5)$ and $E(5,2)$ roller. Left column $AC = 5$ m ($EI$), beam $CD = 5$ m ($2EI$) with a UDL of 25 kN/m, right column $DE = 3$ m ($EI$). Required: horizontal displacement and rotation at the roller E.

#### Real loading reactions
$$
\sum M_A = 0:\; E_y\times5 = 15\times4 + 125\times2.5 \Rightarrow E_y = 74.5\ \text{kN},\quad A_y = 125 - 74.5 = 50.5\ \text{kN},\quad A_x = 15\ \text{kN (left)}
$$

#### Virtual cases
- **Case 1:** unit horizontal load at E (to the right): $A_x = 1$, $E_y = 0.4$ (up), $A_y = 0.4$ (down).
- **Case 2:** unit couple (anticlockwise) at E: $E_y = 0.2$ (down), $A_y = 0.2$ (up), $A_x = 0$.

#### Bending moments ($s$ along each member, in the order A to B to C to D to E)
| Member | $s$ (m) | EI | $M$ | $m_1$ (unit $H$) | $m_2$ (unit couple) |
|---|---|---|---|---|---|
| AB | 0-4 | $EI$ | $15s$ | $s$ | 0 |
| BC | 0-1 | $EI$ | 60 | $s+4$ | 0 |
| CD | 0-5 | $2EI$ | $60 + 50.5s - 12.5s^2$ | $5 - 0.4s$ | $0.2s$ |
| DE | 0-3 | $EI$ | 0 | $3-s$ | 1 |

#### Integrals ($\int Mm\,ds/EI_{member}$, in units of $1/EI$)
| Member | Horizontal | Rotation |
|---|---|---|
| AB | 320.00 | 0 |
| BC | 270.00 | 0 |
| CD | 845.83 | 90.10 |
| DE | 0 | 0 |
| **Sum** | **1435.83** | **90.10** |

#### Results
$$
\Delta_{Eh} = \frac{1435.83}{EI}\ \text{(to the right)},\qquad \theta_E = \frac{90.10}{EI}\ \text{(anticlockwise)}
$$

**Answer: horizontal displacement of E $= 1435.8/EI$ (m, with $EI$ in kNm$^2$) to the right; rotation at E $= 90.10/EI$ rad anticlockwise.** Assumed geometry (as dimensioned in the figure): AB = 4 m and BC = 1 m, a 3 m right column, so the roller E is 2 m above A.

## Check yourself

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

1. To get horizontal displacement and rotation at E, you apply:
   A. Only a unit horizontal load
   B. A unit horizontal load at E in one case and a unit couple at E in another
   C. Two unit horizontal loads
   D. A unit vertical load
   Answer: B. Displacement needs a unit force; rotation needs a unit couple, each analysed separately. (Key point: Two virtual cases)

2. How is the member CD with stiffness 2EI treated?
   A. Its integral is divided by 2EI
   B. Its integral is multiplied by 2EI
   C. It is ignored
   D. EI is used for all members
   Answer: A. Integral M m ds is divided by the stiffness of each member, so CD uses 2EI. (Key point: Different EI)

3. The vertical reaction at the roller E from moments about A is:
   A. 50.5 kN
   B. 74.5 kN
   C. 125 kN
   D. 15 kN
   Answer: B. E_y x 5 = 15 x 4 + 125 x 2.5 gives E_y = 74.5 kN. (Key point: Real reaction)

4. The horizontal displacement of E is:
   A. 1435.8/EI to the right, with rotation 90.10/EI anticlockwise
   B. 90.10/EI right
   C. 1435.8/EI to the left
   D. 720/EI right
   Answer: A. The sums of integrals give 1435.83/EI horizontally and 90.10/EI for rotation. (Key point: Results)

## Related in Theory of Structures I

- [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 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 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 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
- [Parabolic arch 100 m: UDL left half and 80 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-100m-80kn-udl-d-20m) — Theory of Structures I, chapter 6, asked 4×
- [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 (12 kN, 25 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-12kn-25knm) — Strength of Materials, chapter 1
- [Frame AFD, SFD, BMD (100 kN, 25 kN/m, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-100kn-25knm) — Strength of Materials, chapter 1
- [Frame AFD, SFD, BMD (30 kN/m, 50 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-30knm-50knm) — 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×