---
title: "Truss: flexibility matrix method"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:truss-flexibility-matrix-100kn-50kn-4x4"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/truss-flexibility-matrix-100kn-50kn-4x4
site: BCT Engineering
---

# Truss: flexibility matrix method

Theory of Structures II · Chapter 2: Force method

- Asked in: 2068 Chaitra, 8 marks
- Video: [Truss Analysis by Flexibility Matrix Method - Lack of Fit, Temperature Change](https://www.youtube.com/watch?v=A2zICmQDjSQ) — Stan Academy
- 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/truss-flexibility-matrix-100kn-50kn-4x4
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Use the force method (flexibility matrix) to solve the truss as shown in figure below. [Figure: square truss 4 m x 4 m with both diagonals; 100 kN downward and 50 kN horizontal at the top-left joint; bottom-left hinged, bottom-right roller.]

## Short answer

Using flexibility matrix with diagonal BD as redundant, the member forces (kN) are: AB 35.36, BC -14.64, CD -14.64, DA -64.64, AC 20.71, BD -50.00 (compression).

## Answer

**Assumptions:** square truss ABCD, 4 m x 4 m: A bottom-left (hinge), B bottom-right (roller, vertical reaction only), C top-right, D top-left. 100 kN acts downward and 50 kN horizontally (to the right) at D. Both diagonals AC and BD are present; EA is the same for all members.

#### Degree of indeterminacy
$m=6$, $r=3$, $j=4$: $D_s=6+3-8=1$. Choose the diagonal BD as redundant $X_1$ (coordinate 1). The flexibility matrix is $1\times1$: $[F]=[f_{11}]$.

#### Reactions (statically determinate)
Taking moments about A (the 50 kN force acts 4 m above A, the 100 kN force passes through A's vertical line):
$$
R_B\times4=50\times4\ \Rightarrow\ R_B=50\ \text{kN}\uparrow
$$
$\sum F_x=0$: $H_A=50$ kN to the left. $\sum F_y=0$: $V_A=100-50=50$ kN upward.

#### Member forces by joint method
- $S_0$: forces in the released truss (BD removed) due to the loads.
- $S_1$: forces due to a unit tension pair applied along BD.

| Member | L (m) | $S_0$ (kN) | $S_1$ | $S_0S_1L$ | $S_1^2L$ | Final $S$ (kN) |
|---|---|---|---|---|---|---|
| AB | 4.000 | 0.00 | -0.707 | 0.00 | 2.000 | 35.36 |
| BC | 4.000 | -50.00 | -0.707 | 141.42 | 2.000 | -14.64 |
| CD | 4.000 | -50.00 | -0.707 | 141.42 | 2.000 | -14.64 |
| DA | 4.000 | -100.00 | -0.707 | 282.84 | 2.000 | -64.64 |
| AC | 5.657 | 70.71 | 1.000 | 400.00 | 5.657 | 20.71 |
| BD | 5.657 | 0.00 | 1.000 | 0.00 | 5.657 | -50.00 |
| **Sum** | | | | **965.69** | **19.314** | |

#### Flexibility and compatibility
$$
f_{11}=\frac{\sum S_1^2L}{EA}=\frac{19.314}{EA},\qquad \Delta_{10}=\frac{\sum S_0S_1L}{EA}=\frac{965.69}{EA}
$$
$$
f_{11}X_1+\Delta_{10}=0\ \Rightarrow\ X_1=-\frac{965.69}{19.314}=-50.00\ \text{kN}
$$
The diagonal BD carries 50.00 kN (compression).

#### Final forces ($S=S_0+X_1S_1$)
AB = 35.36, BC = -14.64, CD = -14.64, DA = -64.64, AC = 20.71, BD = -50.00 kN (+ tension, - compression).

**Answer: BD = -50.00 kN, AC = 20.71 kN, AB = 35.36 kN, BC = -14.64 kN, CD = -14.64 kN, DA = -64.64 kN.**

## Check yourself

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

1. Why is the 1x1 flexibility matrix enough for this truss?
   A. m = 6, r = 3, j = 4 gives D_s = 1, so one redundant (diagonal BD)
   B. The truss has no redundants
   C. The truss has 2 redundants
   D. Only the reactions are indeterminate by 3
   Answer: A. D_s = 6 + 3 - 8 = 1, so one redundant, here the diagonal BD, giving [F] = [f11]. (Key point: Degree of indeterminacy)

2. Which expressions give f11 and Delta_10 for equal EA?
   A. f11 = sum S1^2 L/EA, Delta_10 = sum S0 S1 L/EA
   B. f11 = sum S0 L/EA, Delta_10 = sum S1 L/EA
   C. f11 = sum S1 L, Delta_10 = sum S0 L
   D. f11 = sum S0^2 L/EA, Delta_10 = sum S1^2 L/EA
   Answer: A. f11 is the displacement due to a unit redundant, and Delta_10 is the displacement due to the loads, both by unit load method. (Key point: Flexibility and compatibility)

3. What value of X1 (force in BD) is found?
   A. +50 kN tension
   B. -50 kN, compression
   C. -965.69 kN
   D. 0 kN
   Answer: B. X1 = -965.69/19.314 = -50.00 kN, so BD is in compression. (Key point: Redundant force)

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