---
title: "Force method: frame with 60 kN/m, BMD"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:force-method-frame-60knm-80kn-2i"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/force-method-frame-60knm-80kn-2i
site: BCT Engineering
---

# Force method: frame with 60 kN/m, BMD

Theory of Structures II · Chapter 2: Force method

- Asked in: 2072 Kartik, 10 marks
- Video: [Problem 7 Analyse Frame By Force Method / Flexibility Method [ FRAME ]](https://www.youtube.com/watch?v=Mlmp9s0tlgc) — Ganesh Shegar
- 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-60knm-80kn-2i
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Analyse the frame shown in figure below by using the force method and draw bending moment diagram. [Figure: frame; column (2I) of height 7 m, fixed at the base; 80 kN horizontal at the 3 m level; beam (I) of 10 m carrying 60 kN/m UDL, right end on a roller.]

## Short answer

Redundant roller reaction R_C found by compatibility: R_C = 266.05 kN upward. Fixed-end moment at A is M_A = 579.51 kNm (anticlockwise); maximum sagging moment in beam is 589.85 kNm.

## Answer

Reading of the figure: column AB is 7 m high ($2EI$) and fixed at A; the beam BC is 10 m long ($EI$) with 60 kN/m; C is a roller (vertical support); a horizontal load of 80 kN (towards the right) acts on the column at 3 m above A.

$r = 3 + 1 = 4$, so $D_s = 1$. Take the vertical reaction $R_C$ at the roller as the redundant.

```text
 B =================== C  (roller)
 |    60 kN/m
 |
 | 80 kN ->
 |
 A (fixed)
```

#### Primary structure (cantilever fixed at A)
| Section | $M_0$ (kNm) | $m$ (unit $R_C$) |
|---|---|---|
| A | -3240 | 10 |
| 80 kN level (3 m up) | -3000 | 10 |
| B | -3000 | 10 |
| mid BC | -750 | 5 |
| C | 0 | 0 |

($M_0$ at B $= -60\times10\times5 = -3000$; at A it is $-3000 - 80\times3 = -3240$.)

#### Flexibility coefficient and load term
$$
\begin{aligned}
f_{11} &= \frac{1}{2EI}\left[10^2(7)\right] + \frac{1}{EI}\left[\frac{10^2\times10}{3}\right] = \frac{350 + 333.33}{EI} = \frac{683.33}{EI} \\
\Delta_{10} &= \frac{1}{2EI}\left[-10\,\frac{3(3240+3000)}{2} - 10(3000)(4)\right] + \frac{1}{EI}\left[-30\int_0^{10}(10-x)^3dx\right] \\ &= \frac{-46800 - 60000 - 75000}{EI} = \frac{-181800}{EI}
\end{aligned}
$$

#### Compatibility ($\delta_C = 0$)
$$
f_{11}R_C + \Delta_{10} = 0 \ \Rightarrow\ R_C = \frac{181800}{683.33} = 266.05\ \text{kN}\ (\uparrow)
$$

#### Final bending moments $M = M_0 + R_C\,m$ (kNm)
| Section | BM | Tension side |
|---|---|---|
| A | -579.51 | outer face of the column |
| 80 kN level | -339.51 | outer face |
| B | -339.51 | outer (top of the beam) |
| Maximum in BC at 5.57 m from B | 589.85 | bottom of the beam |
| C | 0 | |

#### Reactions
- C: $R_C = 266.05$ kN ($\uparrow$)
- A: $V_A = 333.95$ kN ($\uparrow$), $H_A = 80$ kN ($\leftarrow$), $M_A = 579.51$ kNm (anticlockwise)

Check: $V_A + R_C = 600.00$ kN $= 60\times10 = 600$ kN.

**Answer: $R_C = 266.05$ kN; $M_A = 579.51$ kNm; $M_B = -339.51$ kNm (hogging); maximum sagging moment $589.85$ kNm in the beam.**

## Check yourself

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

1. The frame has r = 4 reaction components. What is D_s and the redundant?
   A. D_s = 1; vertical reaction R_C at the roller
   B. D_s = 2; both reactions at C
   C. D_s = 4; all reactions
   D. D_s = 0; no redundant
   Answer: A. D_s = 4 - 3 = 1, and removing the roller leaves a cantilever fixed at A. (Key point: Redundant)

2. In f11 = 10^2(7)/(2EI) + 10^2(10)/(3EI), why is the column term divided by 2EI?
   A. The column is shorter
   B. The column has twice the flexural rigidity (2EI)
   C. m is doubled in the column
   D. The load is doubled
   Answer: B. Each member's integral of m^2 is divided by its own EI, and the column is 2EI against EI for the beam. (Key point: Flexibility coefficient)

3. What does f11 R_C + D10 = 0 give for R_C?
   A. R_C = 600 kN
   B. R_C = 80 kN
   C. R_C = 266.05 kN upward
   D. R_C = 333.95 kN
   Answer: C. R_C = 181800/683.33 = 266.05 kN; then V_A = 600 - 266.05 = 333.95 kN. (Key point: Compatibility)

## Related in Theory of Structures II

- [Moment distribution: frame with 80 kN, 60 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-80kn-10knm-60kn) — Theory of Structures II, chapter 3
- [Moment distribution: frame with SFD, AFD, BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-sfd-afd-bmd) — Theory of Structures II, chapter 3
- [Moment distribution: frame, 150 kN, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-150kn-50kn-60knm) — Theory of Structures II, chapter 3
- [Plastic moment capacity of frame, 60 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aplastic-moment-capacity-frame-60kn-40kn) — Theory of Structures II, chapter 5
- [Moment distribution: frame with cantilever 15 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-50knm-15kn-cantilever) — Theory of Structures II, chapter 3
- [Moment distribution: frame, BMD and SFD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-20knm-50kn-150kn) — Theory of Structures II, chapter 3

## Related in other subjects

- [AFD, SFD, BMD of frame with 60 kN/m UDL](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-60knm-20knm-12kn) — Applied Mechanics, chapter 6
- [Frame AFD, SFD, BMD (10 kN/m, 60 kN, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-10knm-60kn) — Strength of Materials, chapter 1
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