---
title: "Force method: frame with hinged support"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:force-method-frame-30knm-50kn-hinged"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/force-method-frame-30knm-50kn-hinged
site: BCT Engineering
---

# Force method: frame with hinged support

Theory of Structures II · Chapter 2: Force method

- Asked in: 2075 Chaitra, 10 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/force-method-frame-30knm-50kn-hinged
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Determine reactions at the hinged support in the frame shown in figure below using the force method. [Figure: frame with beam of 10 m (1.5I) carrying 30 kN/m UDL, right end on a hinged support with redundants 1 (vertical) and 2 (horizontal); column of 2I and height 6 m, fixed at the base; 50 kN horizontal load at 2 m below the beam.]

## Short answer

Releasing the hinge and solving the force method equations yields the reactions at C: vertical reaction = 120.84 kN upward and horizontal reaction = 98.82 kN toward the left.

## Answer

Reading of the figure: column AB is 6 m high (stiffness $2I$) and fixed at A; the beam BC is 10 m long ($1.5I$) with 30 kN/m; the beam end C is on a hinged support at the level of the beam; a 50 kN horizontal load (towards the right) acts on the column 2 m below the beam, at E (4 m above A). Here $I$ stands for $EI$ units: the member stiffnesses are $2EI$ and $1.5EI$.

$r = 3 + 2 = 5$, so $D_s = 2$. Release the hinge at C: $X_1$ = vertical reaction at C ($\uparrow$), $X_2$ = horizontal reaction at C ($\rightarrow$). The primary structure is a cantilever fixed at A.

```text
 B ================= C (hinge)
 |    30 kN/m          X1 up, X2 horizontal
 |
 E ->  50 kN
 |
 A (fixed)
```

#### Moments in the primary structure (+ tension on the inner face)
- $M_0$: at A, $-(50\times4 + 30\times10\times5) = -1700$; at E and B, $-1500$ (the column between E and B carries only the beam load); falling parabolically to 0 at C.
- $m_1$ (unit vertical at C): $10$ in the column (from B down to A) and falling linearly from 10 at B to 0 at C.
- $m_2$ (unit horizontal at C): zero in the beam; in the column it grows linearly from 0 at B to $-6$ at A ($-2$ at E).

#### Flexibility coefficients (in $1/EI$ units)
$$
\begin{aligned}
f_{11} &= \frac{522.222}{EI} \\
f_{12} &= \frac{-90.000}{EI} \\
f_{22} &= \frac{36.000}{EI} \\
\Delta_{10} &= \frac{-72000.0}{EI} \\
\Delta_{20} &= \frac{14433.3}{EI}
\end{aligned}
$$

#### Compatibility
$$
\begin{aligned}
522.222\,X_1 - 90.000\,X_2 &= 72000.0 \\
-90.000\,X_1 + 36.000\,X_2 &= -14433.3
\end{aligned}
$$

$$
X_1 = V_C = 120.84\ \text{kN},\qquad X_2 = H_C = -98.82\ \text{kN}
$$

#### Final bending moments (kNm)
| Section | $M_0$ | $m_1$ | $m_2$ | $M=M_0+\sum X m$ |
|---|---|---|---|---|
| A | -1700.00 | 10.00 | -6.00 | 101.35 |
| E (50 kN level) | -1500.00 | 10.00 | -2.00 | -93.94 |
| B (column) | -1500.00 | 10.00 | 0.00 | -291.59 |
| mid BC | -375.00 | 5.00 | 0.00 | 229.21 |
| C | 0.00 | 0.00 | 0.00 | 0.00 |

Maximum sagging moment in the beam: 243.38 kNm at 5.97 m from B.

#### Reactions
- C (hinge): $V_C = 120.84$ kN ($\uparrow$), $H_C = 98.82$ kN (towards the left)
- A (fixed): $V_A = 179.16$ kN ($\uparrow$), $H_A = 48.82$ kN (towards the right), $M_A = 101.35$ kNm (clockwise; the moment at A puts the inner face of the column in tension)

Check: $V_A + V_C = 300.00$ kN $= 30\times10 = 300$ kN; $\Sigma H$: $48.82 + 50 - 98.82 = 0$. Moments about A: the loads give $50(4) + 300(5) = 1700$ kNm clockwise, while $V_C$ ($120.84\times10 = 1208.4$) and $H_C$ ($98.82\times6 = 592.9$) give $1801.3$ kNm anticlockwise, so the support couple is $101.35$ kNm clockwise.

**Answer: at the hinged support C, $V_C = 120.84$ kN upward and $H_C = 98.82$ kN towards the left.**

## Check yourself

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

1. Why are two redundants (vertical and horizontal at C) chosen for this frame?
   A. r = 3 + 2 = 5, so D_s = 2
   B. The frame has two members
   C. The hinge at C gives 3 reactions so D_s = 3
   D. The 50 kN load is horizontal
   Answer: A. Fixed A gives 3 reactions and hinged C gives 2, so r = 5 and D_s = 5 - 3 = 2. (Key point: Degree of indeterminacy)

2. What is the primary structure used?
   A. Simply supported beam
   B. A cantilever fixed at A with C released
   C. Three-hinged frame
   D. Portal frame with both bases fixed
   Answer: B. Releasing the hinge at C leaves the frame as a cantilever fixed at A, for which M0 is easily found. (Key point: Primary structure)

3. What do the two compatibility equations express?
   A. Moments at A and C are equal
   B. Total load equals total reaction
   C. Shear at B is zero
   D. Displacements at C in vertical and horizontal directions are zero
   Answer: D. C is a hinged support, so its vertical and horizontal displacements must vanish: f11X1 + f12X2 + Δ10 = 0 and f21X1 + f22X2 + Δ20 = 0. (Key point: Compatibility equations)

4. What are the reactions at the hinge C?
   A. VC = 179.16 kN up, HC = 48.82 kN left
   B. VC = 120.84 kN up, HC = 98.82 kN left
   C. VC = 120.84 kN up, HC = 98.82 kN right
   D. VC = 98.82 kN up, HC = 120.84 kN left
   Answer: B. Solving the two equations gives X1 = 120.84 kN up and X2 = -98.82 (i.e. 98.82 kN to the left); check: VA + VC = 300 kN. (Key point: Final 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
- [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
- [Force method: frame with 60 kN/m, BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aforce-method-frame-60knm-80kn-2i) — Theory of Structures II, chapter 2
- [Force method: frame with support settlement](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aforce-method-frame-settlement-200-ei) — 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

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