---
title: "Frame: roller displacement and hinge rotation"
subject: "Theory of Structures I"
chapter: 3
key: "semester_4/theory_of_structures_i:frame-roller-hinge-rotation-2071b"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/frame-roller-hinge-rotation-2071b
site: BCT Engineering
---

# Frame: roller displacement and hinge rotation

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

- Asked in: 2071 Bhadra, 12 marks
- 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-roller-hinge-rotation-2071b
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3

## Question

Calculate horizontal displacement of the roller support and angular displacement of the fixed hinge of the given portal frame by using unit load (virtual work) method. Express the result in terms of sectional stiffness EI. [Figure: portal frame, joints 0 (hinge, bottom left), 1 (top left), 2 (top right), 3 (roller, bottom right); width 6 m, height 9 m; members 0-1 and 1-2 have 2EI, member 2-3 has EI; 20 kN/m UDL acts horizontally along the right column 2-3.]

## Short answer

Unit load at roller for horizontal displacement; unit couple at hinge for rotation. Integrate Mm/stiffness yields -71077.5/EI and -5670/EI. Roller moves 71077.5/EI left; hinge rotates 5670/EI rad anticlockwise.

## Answer

#### Given data
Portal frame: node 0 (hinge) at (0,0), node 1 at (0,9), node 2 at (6,9), node 3 (roller) at (6,0). Members 0-1 and 1-2: $2EI$; member 2-3: $EI$. UDL 20 kN/m acting horizontally (to the left) on the right column 2-3 (9 m). Required: horizontal displacement of the roller 3 and rotation at the hinge 0.

#### Real loading: reactions
Total horizontal load $= 20\times9 = 180$ kN (left), so $H_0 = 180$ kN (to the right). Moments about node 0: the UDL resultant acts at 4.5 m height, giving $180\times4.5 = 810$ kNm; the roller reaction $V_3\times6$ balances it, so $V_3 = 135$ kN (downward) and $V_0 = 135$ kN (up).

#### Virtual systems
1. Unit horizontal load at node 3 (to the right): $H_0 = 1$ (to the left), vertical reactions zero.
2. Unit couple at node 0 (clockwise): $V_0 = 1/6$ (down), $V_3 = 1/6$ (up), $H_0 = 0$.

#### Bending moments ($s$ from the first node of each member; $M$, $m_1$ and $m_2$ use the same sign convention)
| Member | $s$ (m) | stiffness | $M$ (kNm) | $m_1$ | $m_2$ |
|---|---|---|---|---|---|
| 0-1 | 0-9 | $2EI$ | $-180s$ | $s$ | 1 |
| 1-2 | 0-6 | $2EI$ | $135s - 1620$ | 9 | $1 - s/6$ |
| 2-3 | 0-9 | $EI$ | $-10s^2 + 180s - 810$ | $9 - s$ | 0 |

For the unit couple, $m_2 = 1$ in column 0-1 (the vertical reaction at 0 has no lever arm there), $m_2$ falls linearly to zero at joint 2 along the beam, and $m_2 = 0$ in column 2-3, because the only force on that side, the vertical reaction at the roller 3, acts along the column's axis.

#### Integrals ($\int Mm\,ds$ / stiffness, units $1/EI$)
| Member | Horizontal disp. | Rotation at 0 |
|---|---|---|
| 0-1 | -21870.0 | -3645.0 |
| 1-2 | -32805.0 | -2025.0 |
| 2-3 | -16402.5 | 0 |
| **Sum** | **-71077.5** | **-5670.0** |

$$
\Delta_{3h} = -\frac{71\,077.5}{EI},\qquad \theta_0 = -\frac{5670}{EI}
$$

The negative signs mean the displacements are opposite to the unit loads: the roller moves to the left and the hinge rotates anticlockwise.

**Answer: horizontal displacement of the roller $= 71\,077.5/EI$ (kNm$^3$/EI) to the left (in the direction of the UDL); rotation of the hinge $= 5670/EI$ rad anticlockwise**, with $EI$ the stiffness of column 2-3 (the other two members have $2EI$).

## Check yourself

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

1. The UDL acts horizontally to the left on column 2-3 (9 m). What is H_0?
   A. 180 kN to the right
   B. 180 kN to the left
   C. 90 kN to the right
   D. 20 kN
   Answer: A. Total horizontal load is 20 x 9 = 180 kN, so H_0 = 180 kN opposes it. (Key point: Horizontal reaction)

2. How are V_3 = 135 kN and V_0 = 135 kN found?
   A. The UDL resultant of 180 kN at 4.5 m height gives 810 kNm about node 0, balanced by V_3 x 6
   B. V_3 = H_0
   C. By dividing 180 equally
   D. V_3 x 9 = 180 x 6
   Answer: A. 810/6 = 135 kN, and V_0 is equal and opposite for vertical equilibrium. (Key point: Vertical reactions)

3. What virtual loads are used?
   A. A unit couple at node 3 and a unit load at node 0
   B. Unit vertical loads at both
   C. A unit horizontal load at node 3 and a unit couple at node 0
   D. A unit UDL on the column
   Answer: C. Horizontal displacement of the roller needs a unit horizontal load at node 3; rotation at the hinge needs a unit couple at node 0. (Key point: Virtual systems)

4. The sums come out negative (-71,077.5 and -13,770). What does this mean?
   A. The results are zero
   B. The calculation is wrong
   C. Displacements are opposite to the unit loads: the roller moves left and the hinge rotates clockwise
   D. The stiffness is negative
   Answer: C. A negative result means the actual displacement is opposite to the virtual unit load direction. (Key point: Sign of results)

## 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 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
- [Truss horizontal displacement at C by virtual work](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-virtual-work-horizontal-c) — 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 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
- [Steps of virtual (unit) load method](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aunit-load-method-steps) — Theory of Structures I, chapter 3, asked 4×

## Related in other subjects

- [Force method: fixed portal frame, 30 kN/m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aportal-fixed-force-method-4x4-30knm) — Theory of Structures II, chapter 2
- [Collapse value of P: portal frame with horizontal P/2](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acollapse-p-portal-p-half-ml) — Theory of Structures II, chapter 5
- [Stiffness matrix: frame with coordinates 1, 2, 3](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-coords-123-15knm-10kn) — Theory of Structures II, chapter 3
- [Fixed beam: end moments from support rotation or settlement](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Afixed-beam-end-moments-rotation-settlement) — Theory of Structures II, chapter 3, asked 4×