---
title: "Force method: L-frame reactions at A"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:lframe-force-reaction-a"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/lframe-force-reaction-a
site: BCT Engineering
---

# Force method: L-frame reactions at A

Theory of Structures II · Chapter 2: Force method

- Asked in: 2079 Baishakh, 6 marks
- Video: [Indeterminate Frame by Force Method](https://www.youtube.com/watch?v=Z7BR7BvpoLU) — STATICS THE EASY WAY
- 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/lframe-force-reaction-a
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Determine reactions at A using the force method. [Figure: L-shaped frame; horizontal member AB of length l carrying 10 kN/m UDL, vertical member BC of height l, EI constant; A hinged, C fixed.]

## Short answer

Releasing the hinge at A gives a cantilever. Using redundants X₁ (horizontal) and X₂ (vertical) at A, computing flexibility coefficients and load displacements, zero-displacement compatibility gives X₂ = 3wl/7, X₁ = 3wl/28. With w=10 kN/m, V_A = 4.29l kN upward, H_A = 1.07l kN to the right.

## Answer

L-frame: A hinged (end of horizontal member AB, length $l$, UDL $w = 10$ kN/m), B rigid corner, vertical member BC of height $l$ down to C fixed. EI is constant.

```text
  w = 10 kN/m
 vvvvvvvvvvvvv
 A o----------B
               |
               | l
               |
              /// C (fixed)
 |<---- l ---->|
```

The frame is statically indeterminate to the 2nd degree ($r = 2 + 3 = 5$). Release the hinge at A to get the cantilever C-B-A, and take $X_1$ = horizontal reaction at A and $X_2$ = vertical reaction at A as redundants.

#### Moment expressions
In AB, $x$ measured from A; in BC, $y$ measured down from B.

| Member | $M_0$ (load) | $m_1$ (unit $X_1$) | $m_2$ (unit $X_2$) |
|---|---|---|---|
| AB | $-\dfrac{wx^2}{2}$ | $0$ | $x$ |
| BC | $-\dfrac{wl^2}{2}$ | $y$ | $l$ |

#### Flexibility coefficients
$$
\begin{aligned}
f_{11} &= \frac{1}{EI}\int_0^l y^2\,dy = \frac{l^3}{3EI} \\
f_{12} &= \frac{1}{EI}\int_0^l y\,l\,dy = \frac{l^3}{2EI} \\
f_{22} &= \frac{1}{EI}\left[\int_0^l x^2\,dx + \int_0^l l^2\,dy\right] = \frac{4l^3}{3EI}
\end{aligned}
$$

#### Load displacements
$$
\begin{aligned}
\Delta_{10} &= \frac{1}{EI}\int_0^l y\left(-\frac{wl^2}{2}\right)dy = -\frac{wl^4}{4EI} \\
\Delta_{20} &= \frac{1}{EI}\left[\int_0^l x\left(-\frac{wx^2}{2}\right)dx + \int_0^l l\left(-\frac{wl^2}{2}\right)dy\right] = -\frac{5wl^4}{8EI}
\end{aligned}
$$

#### Compatibility (zero displacement at the hinge A)
$$
\begin{aligned}
\tfrac13 X_1 + \tfrac12 X_2 &= \frac{wl}{4} \\
\tfrac12 X_1 + \tfrac43 X_2 &= \frac{5wl}{8}
\end{aligned}
$$

Solving: $X_2 = \dfrac{3wl}{7}$ and $X_1 = \dfrac{3wl}{28}$.

With $w = 10$ kN/m:

$$
V_A = \frac{30\,l}{7} = 4.29\,l\ \text{kN}\ (\uparrow), \qquad H_A = \frac{15\,l}{14} = 1.07\,l\ \text{kN}\ (\rightarrow)
$$

(with $l$ in metres). Check from the support at C: $V_C = 10l - 4.29l = 5.71\,l$ kN up, $H_C = 1.07\,l$ kN to the left, $M_C = 5l^2/14 = 0.357\,l^2$ kNm.

**Answer: at A, $H_A = 3wl/28 = 1.07\,l$ kN (to the right) and $V_A = 3wl/7 = 4.29\,l$ kN upward.**

## Check yourself

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

1. The L-frame is indeterminate to degree:
   A. 2
   B. 1
   C. 3
   D. 5
   Answer: A. r = 5, so 5 - 3 = 2. (Key point: Degree)

2. Redundants chosen are:
   A. X1 and X2, horizontal and vertical reactions at A
   B. Moment at C only
   C. Shear in BC
   D. The UDL
   Answer: A. Releasing the hinge at A leaves a cantilever from C. (Key point: Redundants)

3. The compatibility condition is:
   A. Zero displacement at hinge A in both directions
   B. Zero rotation at B
   C. Zero moment at C
   D. Equal reactions
   Answer: A. A is a hinge support so it cannot move. (Key point: Compatibility)

4. The solution is:
   A. X2 = 3wl/7, X1 = 3wl/28
   B. X2 = wl/2, X1 = 0
   C. X2 = 3wl/28, X1 = 3wl/7
   D. X2 = wl, X1 = wl/2
   Answer: A. Solving the two equations gives V_A = 3wl/7 and H_A = 3wl/28. (Key point: Result)

## 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: frame reactions and BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aforce-method-frame-100kn-30knm-reactions) — Theory of Structures II, chapter 2
- [Stiffness matrix: frame with 100 kN, end reactions](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-matrix-frame-100kn-end-reactions) — Theory of Structures II, chapter 3
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