---
title: "Frame vertical deflection at D"
subject: "Theory of Structures I"
chapter: 3
key: "semester_4/theory_of_structures_i:frame-vw-d-vertical-2071o"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/frame-vw-d-vertical-2071o
site: BCT Engineering
---

# Frame vertical deflection at D

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

- Asked in: 2071 Magh (old course), 12 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-vw-d-vertical-2071o
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3

## Question

Determine the vertical deflection at point D of the frame loaded as shown in figure below. [Figure: portal frame; A (hinge) at the bottom left, B and C top joints, D at the foot of the right column; column height 6 m, beam BC = 6 m (3 m + 3 m) with 4 kN/m UDL, 15 kN horizontal load at B; cross-section 25 cm × 45 cm; $E = 150\ \text{kN/mm}^2$.]

## Short answer

Unit vertical load at D. Actuating moments: beam 2s^3 gives 648, column 6(72+15s) gives 4212; total 4860. With EI = 284766 kNm², deflection = 4860/EI = 17.07 mm downward.

## Answer

#### Given data and assumptions
Frame: A(0,0) at the base of the left column, B(0,6), C(6,6), and D at the foot of the right column, hanging from C (free end at D). Column height 6 m, beam BC = 6 m with UDL 4 kN/m (down), 15 kN horizontal at B (taken to the right). Section 25 cm $\times$ 45 cm, $E = 150$ kN/mm$^2 = 1.5\times10^8$ kN/m$^2$. Since a vertical deflection is asked for at D, D is taken as a free end; the frame is then stable only if A is **fixed**, so it is treated as a cantilever frame from A (a hinge at A with D free would be a mechanism).

$$
I = \frac{0.25\times0.45^3}{12} = 1.898\times10^{-3}\ \text{m}^4,\qquad EI = 284\,766\ \text{kNm}^2
$$

#### Virtual load
Unit vertical load (down) at D. $s$ is measured from D up the right column, from C along CB, and from B down the left column ($EI$ constant).

| Member | $s$ (m) | $M$ (kNm) | $m$ | $\int Mm\,ds$ |
|---|---|---|---|---|
| DC | 0-6 | 0 | 0 | 0 |
| CB | 0-6 | $2s^2$ | $s$ | 648 |
| BA | 0-6 | $72 + 15s$ | 6 | 4212 |

Details: in CB, $M = 4s^2/2 = 2s^2$ and $m = s$, so $\int_0^6 2s^3\,ds = 648$. In BA, $M = 4\times6\times3 + 15s = 72 + 15s$ and $m = 6$, so $\int_0^6 6(72+15s)\,ds = 6(432 + 270) = 4212$. (The UDL and the 15 kN load both bend the left column in the same sense as the unit load.)

$$
\Delta_{Dv} = \frac{648 + 4212}{EI} = \frac{4860}{EI} = \frac{4860}{284\,766} = 17.07\ \text{mm}
$$

**Answer: vertical deflection at D $= 4860/EI = 17.07$ mm downward.**

## Check yourself

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

1. What virtual load gives the vertical deflection at D?
   A. A unit vertical load at D
   B. A unit couple at D
   C. A unit horizontal load at D
   D. A unit vertical load at B
   Answer: A. The virtual load matches the displacement required. (Key point: Virtual load)

2. Why are both M and m zero in the right column DC?
   A. The loads and the unit load at D act along the line of the column, giving no bending moment
   B. It is not part of the frame
   C. It carries the UDL
   D. It is rigid
   Answer: A. Vertical forces at the free end D act along the axis of the vertical column, so there is no moment there. (Key point: Zero moment in DC)

3. Why is M = 72 + 15s in the left column BA?
   A. The UDL gives 4 x 6 = 24 only
   B. The UDL resultant 24 kN acts 3 m from the column (72 kNm) and the 15 kN horizontal load adds 15s
   C. The 15 kN horizontal load gives 72
   D. Both loads give 15s
   Answer: B. The beam UDL: 4 x 6 x 3 = 72 kNm, constant along the column; the horizontal load at B gives 15s at distance s below B. (Key point: Real moment in BA)

4. What is the vertical deflection at D?
   A. 648/EI
   B. 4860/EI = 17.07 mm downward
   C. 4860/EI = 1.707 mm
   D. 4212/EI upward
   Answer: B. The sum of the CB (648) and BA (4212) integrals gives 4860, which divided by EI = 284,766 kNm^2 gives 17.07 mm. (Key point: 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-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
- [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: 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 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

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