---
title: "Force method: frame with support settlement"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:force-method-frame-settlement-200-ei"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/force-method-frame-settlement-200-ei
site: BCT Engineering
---

# Force method: frame with support settlement

Theory of Structures II · Chapter 2: Force method

- Asked in: 2076 Chaitra, 10 marks
- Video: [force method -support settlement](https://www.youtube.com/watch?v=B8bH_CXgyCo) — battle structures
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/force-method-frame-settlement-200-ei
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Determine reactions at the hinged support using the force method when support D settles vertically downward by $200/EI$. Take EI to be constant. [Figure: frame ABCD; beam BC (2EI) of 4 m with 10 kN/m UDL; column AB (EI) of height 4 m, A fixed; column CD (EI) of 4 m with D hinged and a further 2 m dimension marked below.]

## Short answer

Use the force method with horizontal and vertical redundants at D and include the settlement in compatibility. Reactions at D are HD = 1.28 kN to the right and VD = 15.78 kN upward.

## Answer

Reading of the figure: A (0, 0) fixed, column AB 4 m ($EI$), beam BC 4 m ($2EI$) with 10 kN/m, column CD 4 m ($EI$) with D hinged at the level of A (the extra 2 m dimension is not needed). Support D settles by $200/EI$ (downward), where $EI$ is the stiffness of the columns.

$D_s = 2$. Release the hinge at D; redundants: $X_1$ = horizontal reaction ($\rightarrow$ positive) and $X_2$ = vertical reaction ($\uparrow$ positive).

#### Primary structure moments (sign: + tension on the inner face)
- $M_0$ (UDL on the cantilever A-B-C): $-80$ kNm at B and constant down the column; in the beam it falls parabolically from $-80$ at B to 0 at C.
- $m_1$ (unit horizontal force at D): CD from 0 at D to 4 at C, beam constant 4, column falling from 4 at B to 0 at A.
- $m_2$ (unit vertical force at D): zero in CD, falling linearly from 4 at B to 0 at C in the beam, and constant at 4 in the column.

#### Flexibility coefficients (in $1/EI$ units)
$$
\begin{aligned}
f_{11} &= \frac{74.667}{EI} \\
f_{12} &= \frac{48.000}{EI} \\
f_{22} &= \frac{74.667}{EI} \\
\Delta_{10} &= \frac{-853.3}{EI} \\
\Delta_{20} &= \frac{-1440.0}{EI}
\end{aligned}
$$

#### Compatibility with settlement
The displacement of D along $X_1$ is zero. The displacement along $X_2$ (upward) is $-200/EI$ because D moves down. The right side of each equation is $-\Delta_{i0} + \delta_i$ (second equation: $1440 - 200 = 1240$):

$$
\begin{aligned}
74.667\,X_1 + 48.000\,X_2 &= 853.3 \\
48.000\,X_1 + 74.667\,X_2 &= 1240.0
\end{aligned}
$$

$$
X_1 = 1.28\ \text{kN},\qquad X_2 = 15.78\ \text{kN}
$$

#### Final bending moments (kNm)
| Section | $M_0$ | $m_1$ | $m_2$ | $M=M_0+\sum X m$ |
|---|---|---|---|---|
| A | -80.00 | 0.00 | 4.00 | -16.87 |
| B (column) | -80.00 | 4.00 | 4.00 | -11.74 |
| mid BC | -20.00 | 4.00 | 2.00 | 16.70 |
| C (beam) | 0.00 | 4.00 | 0.00 | 5.13 |
| D | 0.00 | 0.00 | 0.00 | 0.00 |

#### Reactions
- At D: $H_D = 1.28$ kN ($\rightarrow$), $V_D = 15.78$ kN ($\uparrow$)
- At A: $H_A = 1.28$ kN ($\leftarrow$), $V_A = 24.22$ kN ($\uparrow$), $M_A = 16.87$ kNm (anticlockwise)

Check: $V_A + V_D = 40.00$ kN $= 10\times4 = 40$ kN.

**Answer: reactions at the hinged support D: horizontal $1.28$ kN (to the right) and vertical $15.78$ kN (upward).**

## Check yourself

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

1. How many redundants are there, and what are they?
   A. Two (Ds = 2): the horizontal and vertical reactions at the hinge D
   B. One: the vertical reaction at D
   C. Three: all the reactions at D and A
   D. Two: the moments at B and C
   Answer: A. Releasing the hinge D leaves a cantilever; X1 and X2 are the horizontal and vertical reactions. (Key point: Redundants)

2. What is M0 in the column and at B?
   A. -40 kNm, from the total load
   B. -80 kNm, from 10 x 4 x 2 (UDL on the 4 m beam)
   C. -160 kNm, from 10 x 4 x 4
   D. 0, since the load acts only on the beam
   Answer: B. The 40 kN load acts at 2 m from B, giving 80 kNm hogging, constant down the column. (Key point: M0 in primary structure)

3. How does the 200/EI settlement at D enter the equations?
   A. Both right sides increase by 200
   B. The right side becomes 1440 + 200 = 1640
   C. The right side of the second (vertical) equation becomes 1440 - 200 = 1240, while the first stays 853.3
   D. It is added to the left side as a new unknown
   Answer: C. D moves down by 200/EI, so the required compatible vertical displacement of D is not zero; only the vertical equation changes. (Key point: Effect of settlement)

4. What check confirms the reactions V_A = 24.22 kN and V_D = 15.78 kN?
   A. Their difference is 10 kN
   B. Their sum equals 20 kN, the load on one column
   C. V_A equals V_D
   D. Their sum is 40 kN, equal to the load 10 x 4
   Answer: D. Vertical equilibrium requires V_A + V_D to equal the total UDL of 40 kN. (Key point: Check on reactions)

## Related in Theory of Structures II

- [Force method: frame reactions at hinged support](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aforce-method-frame-50knm-hinged-d) — Theory of Structures II, chapter 2
- [Force method: reactions at D, frame with 110 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aforce-method-frame-110kn-30knm-d) — Theory of Structures II, chapter 2
- [Force method: frame with hinged support](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aforce-method-frame-30knm-50kn-hinged) — Theory of Structures II, chapter 2
- [Force method: L-frame reactions at A](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Alframe-force-reaction-a) — Theory of Structures II, chapter 2
- [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×
- [Flexibility matrix: reactions at support D](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aflexibility-matrix-frame-2knm-5kn-coords) — Theory of Structures II, chapter 2

## Related in other subjects

- [Beam reactions and mid-span deflection (2EI)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-support-reactions-midspan-k8) — Computational Techniques in Civil Engineering, chapter 4, asked 2×
- [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×
- [AFD, SFD, BMD of frame with 60 kN/m UDL](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-60knm-20knm-12kn) — Applied Mechanics, chapter 6
- [Frame AFD, SFD, BMD (10 kN/m, 60 kN, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-10knm-60kn) — Strength of Materials, chapter 1