---
title: "Flexibility matrix method: portal frame"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:flexibility-matrix-portal-12000"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/flexibility-matrix-portal-12000
site: BCT Engineering
---

# Flexibility matrix method: portal frame

Theory of Structures II · Chapter 2: Force method

- Asked in: 2078 Bhadra, 10 marks
- Video: [Flexibility Matrix Method, Analysis of Frame - Problem No 10](https://www.youtube.com/watch?v=OY32k4ncsI8) — Stan Academy
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/flexibility-matrix-portal-12000
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Analyse the portal frame ABCD shown in the figure below using the flexibility matrix method. Take $EI = 12000\ \text{kN/m}^2$. [Figure: frame with column AB of height 4 m (2 m + 2 m) with 50 kN horizontal at mid-height, beam BC of 8 m carrying 25 kN/m UDL, column CD of height 3 m; A fixed, D supported.]

## Short answer

Use the flexibility method with redundants at the hinge D. Compatibility gives X1 = 42.23 kN (left) and X2 = 104.55 kN (upward), so HD = 42.23 kN left and VD = 104.55 kN upward.

## Answer

Reading of the figure: A (0, 0) fixed; column AB is 4 m high with a 50 kN horizontal load (towards the right) at mid-height E; beam BC is 8 m long with 25 kN/m; column CD is 3 m high, so D is 1 m above the level of A, and D is hinged. $EI = 12000$ kNm$^2$ is constant.

$r = 3 + 2 = 5$, so $D_s = 2$. Release the hinge at D and take the reactions at D as the redundants: $X_1$ = horizontal ($\rightarrow$), $X_2$ = vertical ($\uparrow$). The primary structure is a cantilever fixed at A. In the flexibility method the compatibility condition is $[F]\{X\} + \{\Delta_0\} = 0$.

#### Moments in the primary structure (sign: + tension on the inner face)
- Loads only ($M_0$): at A, $-(50\times2 + 25\times8\times4) = -900$; at E, $-800$; at B, $-800$; the beam falls parabolically to 0 at C.
- Unit $X_1$ ($m_1$): $m_1 = $ lever arm of the unit horizontal force; at C and along BC it is $3$, at B it is $3$, at A it is $-1$.
- Unit $X_2$ ($m_2$): at B it is $8$, falling linearly to 0 at C; along AB it is $8$.

#### Flexibility matrix and load vector
$$
\begin{aligned}
f_{11} &= \frac{90.333}{EI} \\
f_{12} &= \frac{128.000}{EI} \\
f_{22} &= \frac{426.667}{EI} \\
\Delta_{10} &= \frac{-9566.7}{EI} \\
\Delta_{20} &= \frac{-39200.0}{EI}
\end{aligned}
$$

In units of m/kN with $EI = 12000$:

$$
[F] = \begin{bmatrix} 0.00753 & 0.01067 \\ 0.01067 & 0.03556 \end{bmatrix},\qquad
\{\Delta_0\} = \begin{Bmatrix} -0.79722 \\ -3.26667 \end{Bmatrix}\ \text{m}
$$

#### Compatibility equations
$$
\begin{aligned}
90.333\,X_1 + 128.000\,X_2 &= 9566.7 \\
128.000\,X_1 + 426.667\,X_2 &= 39200.0
\end{aligned}
$$

$$
\{X\} = -[F]^{-1}\{\Delta_0\} \ \Rightarrow\ X_1 = -42.23\ \text{kN},\quad X_2 = 104.55\ \text{kN}
$$

#### Final bending moments (kNm)
| Section | $M_0$ | $m_1$ | $m_2$ | $M=M_0+\sum X m$ |
|---|---|---|---|---|
| A | -900.00 | -1.00 | 8.00 | -21.41 |
| E | -800.00 | 1.00 | 8.00 | -5.87 |
| B | -800.00 | 3.00 | 8.00 | -90.34 |
| mid BC | -200.00 | 3.00 | 4.00 | 91.48 |
| C | 0.00 | 3.00 | 0.00 | -126.70 |
| D | 0.00 | 0.00 | 0.00 | 0.00 |

Maximum sagging moment in BC: 91.89 kNm at 3.82 m from B.

#### Reactions
- At D (hinge): $H_D = 42.23$ kN ($\leftarrow$), $V_D = 104.55$ kN ($\uparrow$)
- At A: $H_A = 7.77$ kN ($\leftarrow$), $V_A = 95.45$ kN ($\uparrow$), $M_A = 21.41$ kNm (anticlockwise)

Check: $\Sigma V = 200.00 = 25 \times 8 = 200$ kN; $\Sigma H = 50 - 7.77 - 42.23 = 0$.

**Answer: $X_1 = H_D = 42.23$ kN towards the left and $X_2 = V_D = 104.55$ kN upward.**

## Check yourself

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

1. What is the compatibility statement of the flexibility method here and what is released?
   A. [F]{X} + {Delta0} = 0, after releasing the hinge D and taking its horizontal and vertical reactions as redundants
   B. [F]{X} = {Delta0} with A released instead
   C. [K]{X} = {P}, solving for joint displacements
   D. {X} = [F]{Delta0} with the fixed end A released
   Answer: A. With Ds = 2, the displacements at D due to loads plus redundants must be zero, giving [F]{X} + {Delta0} = 0. (Key point: Method set-up)

2. How is M0 at A found for the cantilever primary structure?
   A. -(50 x 4 + 25 x 8) = -400 kNm
   B. -(50 x 2 + 25 x 8 x 4) = -900 kNm
   C. -(25 x 8 x 8) = -1600 kNm
   D. -(50 x 2) = -100 kNm
   Answer: B. The 50 kN horizontal load at 2 m height gives 100 kNm and the 200 kN UDL load at 4 m lever arm gives 800 kNm, both hogging. (Key point: Primary-structure moment)

3. Why is f12 = f21 = 0.01067 in the matrix [F]?
   A. Because X1 and X2 are equal
   B. Because the frame is symmetric
   C. Maxwell's reciprocal theorem makes the flexibility matrix symmetric
   D. Because of the equilibrium equations at A
   Answer: C. The deflection at point 1 due to a unit load at 2 equals that at 2 due to a unit load at 1. (Key point: Symmetry of F)

4. X1 = -42.23 and X2 = 104.55. What are the reactions at D and the check?
   A. H_D = 42.23 kN to the right and V_D = 104.55 kN down
   B. H_D = 7.77 kN and V_D = 95.45 kN
   C. H_D = 104.55 kN and V_D = 42.23 kN
   D. H_D = 42.23 kN to the left and V_D = 104.55 kN up; the check is sum H = 50 - 7.77 - 42.23 = 0 and sum V = 200 kN
   Answer: D. X1 is negative, so H_D acts leftwards. The vertical reactions V_A + V_D equal the 200 kN UDL load. (Key point: Reactions and check)

## Related in Theory of Structures II

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