---
title: "Consistent deformation: bent frame with support movement"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:consistent-deformation-bent-frame-settlement-20knm"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/consistent-deformation-bent-frame-settlement-20knm
site: BCT Engineering
---

# Consistent deformation: bent frame with support movement

Theory of Structures II · Chapter 2: Force method

- Asked in: 2065 Shrawan, 20 marks
- Video: [SA22: 3.9 Consistent Deformations - Frames: support movement](https://www.youtube.com/watch?v=BzBAmyRR4i0) — khaled Elaswany
- 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/consistent-deformation-bent-frame-settlement-20knm
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Use the consistent deformation method to analyze the bent frame shown in the figure below. Draw Axial Force Diagram, Shear Force Diagram, and Bending Moment Diagram for the shown system, if support A settles down by 10 mm, shifts towards left by 10 mm and rotates clockwise by 0.002 radians. [Figure: bent frame; column AC (3EI) of 4 m (2 m x 2 m) with 30 kN horizontal at mid-height, A fixed; beam CB (5EI) of 5 m carrying 20 kN/m UDL, B fixed.]

## Short answer

Release B; redundants: H_B = -19.29 + 0.00398 EI kN, V_B = 54.57 + 0.00416 EI kN, M_B = -49.29 - 0.01093 EI kNm. For EI = 20 000 kNm²: H_B = 60.3 kN, V_B = 137.7 kN, M_B = -267.9 kNm.

## Answer

**Assumptions:** A(0,0), C(0,4), B(5,4). Column AC (3EI) is 4 m high with the 30 kN horizontal load (to the right) at mid-height; beam CB (5EI) is 5 m with 20 kN/m downward; B is fixed. Support A moves: 10 mm down, 10 mm to the left and rotates 0.002 rad clockwise. The value of EI is not given in the question, so the answer is derived in terms of EI (kNm² units, E in kN/m², I in m⁴), and a numerical example with $EI=20\,000$ kNm² is added. Axial deformations are ignored.

#### Step 1: Degree of indeterminacy and primary structure
Reactions = 3 + 3 = 6, equilibrium = 3, so $D_s=3$. Release B. Redundants: $X_1=H_B$ (right +), $X_2=V_B$ (up +), $X_3=M_B$ (anticlockwise +). Primary structure: cantilever B-C-A fixed at A (which moves with the support).

#### Step 2: Displacement of B in the primary structure
**(a) Due to the loads** (using the moment diagrams of the primary structure): the 20 kN/m UDL on CB and the 30 kN force on AC give
$$
\{\Delta_L\}=\frac{1}{EI}\begin{Bmatrix}733.33\\-2079.17\\-436.67\end{Bmatrix}
$$
**(b) Due to the movement of A.** The primary structure moves as a rigid body with A: for A displaced $(u_A,v_A)=(-0.010,-0.010)$ m and rotated $\theta_A=-0.002$ rad (anticlockwise +), the point B at $(5,4)$ from A moves by
$$
u_B=u_A-\theta_A\,y=-0.010+0.002\times4=-0.002\ \text{m},\quad v_B=v_A+\theta_A\,x=-0.010-0.002\times5=-0.020\ \text{m},\quad \theta_B=-0.002\ \text{rad}
$$
$$
\{\Delta_S\}=\{-0.002,\;-0.020,\;-0.002\}^T\ (\text{m, m, rad})
$$

#### Step 3: Flexibility matrix of the primary structure (unit forces at B)
$$
[F]=\frac{1}{EI}\begin{bmatrix}7.111&-13.333&-2.667\\-13.333&41.667&9.167\\-2.667&9.167&2.333\end{bmatrix}
$$
(e.g. $f_{11}=\dfrac{1}{3EI}\displaystyle\int_0^4(4-y)^2dy=\dfrac{7.111}{EI}$; $f_{33}=\dfrac{4}{3EI}+\dfrac{5}{5EI}=\dfrac{2.333}{EI}$.)

#### Step 4: Consistent deformation (B is fixed)
$$
[F]\{X\}+\{\Delta_L\}+\{\Delta_S\}=\{0\}\quad\Rightarrow\quad \{X\}=\{X_L\}+EI\{x_S\}
$$
$$
\{X_L\}=\begin{Bmatrix}-19.286\ \text{kN}\\54.571\ \text{kN}\\-49.286\ \text{kNm}\end{Bmatrix},\qquad
\{x_S\}=\begin{Bmatrix}0.003978\\0.004157\\-0.010929\end{Bmatrix}\ (\text{per unit }EI)
$$
So $H_B=-19.29+0.00398\,EI$ kN, $V_B=54.57+0.00416\,EI$ kN and $M_B=-49.29-0.01093\,EI$ kNm. The settlement effects grow in proportion to EI.

#### Step 5: Results for the numerical example ($EI=20\,000$ kNm²)
$H_B=60.27$ kN, $V_B=137.71$ kN, $M_B=-267.86$ kNm.

Bending moments (sagging/inside tension +, kNm):

| Section | Loads only | Support movement (per unit EI) | Total, EI = 20 000 |
|---|---|---|---|
| M at A | -9.29 | -0.006054 | -130.36 |
| M at mid-height of AC | 12.14 | 0.001902 | 50.18 |
| M at C (column side) | -26.43 | 0.009857 | 170.71 |
| M at C (beam side) | -26.43 | 0.009857 | 170.71 |
| M at mid-span of CB | 24.64 | -0.000536 | 13.93 |
| M at B | -49.29 | -0.010929 | -267.86 |

Shear and axial forces (EI = 20 000; axial: + tension):
| Member | Axial N (+ tension) | Shear at start | Shear at end |
|---|---|---|---|
| AC | 37.71 | 90.27 | 60.27 |
| CB | 60.27 | -37.71 | -137.71 |

The BMD is drawn from the table: AC has a straight-line moment with a kink at the 30 kN load, CB is parabolic (UDL) between the C and B values.

**Answer: redundants at B: $H_B=-19.29+0.00398EI$ kN, $V_B=54.57+0.00416EI$ kN, $M_B=-49.29-0.01093EI$ kNm; for EI = 20 000 kNm²: $H_B=60.3$ kN, $V_B=137.7$ kN, $M_B=-267.9$ kNm.**

## Check yourself

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

1. For the bent frame with both A and B fixed, what are the redundants and the primary structure?
   A. Release A; take the three reactions at A as redundants; cantilever fixed at B
   B. Release C; take the moment at C as the only redundant
   C. Release B; take H_B, V_B and M_B as redundants; cantilever fixed at A
   D. Release both supports; D_s = 6
   Answer: C. Six reactions minus three equilibrium equations gives D_s = 3. B is released, leaving a cantilever B-C-A fixed at A, which moves with the support. (Key point: Primary structure and redundants)

2. A moves (-0.010 m, -0.010 m) and rotates 0.002 rad clockwise. With B at (5, 4) from A, what is the horizontal displacement of B in the primary structure?
   A. -0.010 m
   B. +0.002 m
   C. -0.018 m
   D. -0.002 m
   Answer: D. u_B = u_A - theta_A y = -0.010 + 0.002 x 4 = -0.002 m. The primary structure moves rigidly with A. (Key point: Support movement as rigid body motion)

3. Which equation gives the redundants when the support A moves?
   A. [F]{X} = {Delta_L}
   B. [F]{X} + {Delta_L} + {Delta_S} = 0
   C. [F]{X} = {Delta_S} only
   D. {X} = {Delta_L} + {Delta_S}
   Answer: B. B is fixed, so the displacements from the redundants, from the loads and from the movement of A must sum to zero. (Key point: Compatibility with settlement)

4. Why do the settlement contributions to H_B, V_B and M_B grow in proportion to EI?
   A. Because the flexibility matrix is independent of EI
   B. Because the loads are proportional to EI
   C. Because forces needed to impose a given displacement are proportional to member stiffness
   D. Because the movement of A is proportional to EI
   Answer: C. The settlement displacements are fixed numbers while [F] varies as 1/EI, so the redundants {X_S} = EI x {x_S} scale with EI. (Key point: Dependence of settlement effects on EI)

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