---
title: "Stiffness method: final end moments, frame"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-frame-20knm-8knm"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-20knm-8knm
site: BCT Engineering
---

# Stiffness method: final end moments, frame

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2078 Kartik, 12 marks
- Video: [Frame Direct Stiffness 2](https://www.youtube.com/watch?v=BSKOsbR3ddU) — Kiran Koraddi
- 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/stiffness-frame-20knm-8knm
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Compute the final end moments for the following loaded frame using the stiffness method. [Figure: frame with A fixed at the left end of beam AB (3EI, 4 m); 20 kNm moment at B; beam BC (2EI, 5 m) with 8 kN/m UDL; column BE (2EI) with 20 kN horizontal load, E hinged at the base; column CD (EI) of height 7 m, D fixed; dimensions 4 m, 5 m as marked.]

## Short answer

Using stiffness method with unknowns θ_B, θ_C, θ_E (hinge at E). Slope-deflection equations, equilibrium at joints B, C and M_EB=0 give a 3×3 matrix. Solution: EIθ_B=4.79, EIθ_C=-9.44, EIθ_E=2.61. Final moments (kNm): M_AB=7.18, M_BA=14.37, M_BC=-16.55, M_CB=5.39, M_BE=22.18, M_EB=0, M_CD=-5.39, M_DC=-2.70.

## Answer

**Assumptions (figure partly unclear):** A(0,0), B(4,0), C(9,0) lie on one horizontal girder: AB = 4 m (3EI), BC = 5 m (2EI) with 8 kN/m downward. Column BE (2EI) is 4 m long and carries the 20 kN horizontal load (to the right) at its mid-height; E is a hinge. Column CD (EI) is 7 m long and fixed at D. A is fixed. The 20 kNm moment at B is clockwise. Because A is fixed, the girder cannot sway, so the unknowns are only the joint rotations $\theta_B,\theta_C$ and the hinge rotation $\theta_E$.

#### Fixed-end moments (clockwise +)
- BC: $\mp\dfrac{wL^2}{12}=\mp\dfrac{8\times5^2}{12}=\mp16.67$ kNm
- BE (point load 20 kN at mid-height): $\mp\dfrac{PL}{8}=\mp\dfrac{20\times4}{8}=\mp10$ kNm (for the load acting to the right on this downward member: $FEM_{BE}=10.00$, $FEM_{EB}=-10.00$ kNm)
- AB and CD: no loads, so zero.

#### Slope-deflection (stiffness) equations
- $M_{AB}=\frac{2(3EI)}{4}\left(\theta_{B}\right)$
- $M_{BA}=\frac{2(3EI)}{4}\left(2\theta_{B}\right)$
- $M_{BC}=\frac{2(2EI)}{5}\left(2\theta_{B} + \theta_{C}\right) - 16.67$
- $M_{CB}=\frac{2(2EI)}{5}\left(2\theta_{C} + \theta_{B}\right) + 16.67$
- $M_{BE}=\frac{2(2EI)}{4}\left(2\theta_{B} + \theta_{E}\right) + 10.00$
- $M_{EB}=\frac{2(2EI)}{4}\left(2\theta_{E} + \theta_{B}\right) - 10.00$
- $M_{CD}=\frac{2EI}{7}\left(2\theta_{C}\right)$
- $M_{DC}=\frac{2EI}{7}\left(\theta_{C}\right)$

#### Equilibrium
- Joint B: $M_{BA}+M_{BC}+M_{BE}=20$ (applied clockwise moment)
- Joint C: $M_{CB}+M_{CD}=0$
- Hinge E: $M_{EB}=0$

$$
EI\begin{bmatrix}6.6000 & 0.8000 & 1.0000 \\ 0.8000 & 2.1714 & 0.0000 \\ 1.0000 & 0.0000 & 2.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \theta_{E}\end{Bmatrix}=\begin{Bmatrix}26.667 \\ -16.667 \\ 10.000\end{Bmatrix}
$$
$$
EI\theta_{B}=4.7900,\quad EI\theta_{C}=-9.4402,\quad EI\theta_{E}=2.6050
$$

#### Final end moments (kNm, clockwise on the member end +)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | 7.18 | BA | 14.37 |
| BC | -16.55 | CB | 5.39 |
| BE | 22.18 | EB | 0.00 |
| CD | -5.39 | DC | -2.70 |

Checks:
- Joint B: (14.37) + (-16.55) + (22.18) = 20.00 kNm (applied clockwise moment 20.00) ✓
- Joint C: (5.39) + (-5.39) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint E: (0.00) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Hinge: $M_{EB}=0.00$ ✓

Bending moment ordinates (sagging/inside + according to member direction):
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | 7.18 | -14.37 | - |
| BC | -16.55 | -5.39 | 14.34 at 2.78 m |
| BE | 22.18 | 0.00 | -8.91 at 2.00 m |
| CD | -5.39 | 2.70 | - |

Reactions: A: $H=-14.39$, $V=-5.39$, $M=-7.18$; E: $H=-4.45$, $V=27.62$, $M=0.00$; D: $H=-1.16$, $V=17.77$, $M=2.70$.

**Answer: $M_{AB}=7.18$, $M_{BA}=14.37$, $M_{BC}=-16.55$, $M_{CB}=5.39$, $M_{BE}=22.18$, $M_{EB}=0$, $M_{CD}=-5.39$, $M_{DC}=-2.70$ kNm.**

## Check yourself

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

1. What are the unknowns in this analysis, and why is there no sway unknown?
   A. theta_B, theta_C, theta_E; the girder cannot sway because A is fixed
   B. Sway only; the joints do not rotate
   C. theta_B, theta_C and sway Delta; D is a roller
   D. theta_A, theta_B, theta_C; E is fixed
   Answer: A. Rotations at B, C and the hinge E are unknown. Fixed support A prevents horizontal movement of the girder, so no sway equation is needed. (Key point: Unknowns)

2. What are the magnitudes of the fixed-end moments for BC (8 kN/m, 5 m) and for BE (20 kN at mid-height, 4 m)?
   A. 10.00 for BC and 16.67 for BE
   B. 33.33 for BC and 20 for BE
   C. 16.67 for BC and 10.00 for BE
   D. 40 for BC and 10.00 for BE
   Answer: C. BC: wL^2/12 = 8 x 25/12 = 16.67 kNm. BE: PL/8 = 20 x 4/8 = 10 kNm. (Key point: Fixed-end moments)

3. What is the right-hand side of the joint B equation M_BA + M_BC + M_BE = ?
   A. 0
   B. 20 kNm, the applied clockwise moment
   C. 8 kNm
   D. -20 kNm
   Answer: B. The sum of the member end moments at B equals the applied external clockwise moment at B, 20 kNm. (Key point: Joint B equilibrium)

4. What extra equation does the hinge at E provide?
   A. theta_E = 0
   B. M_EB = 0, since a hinge cannot carry moment
   C. V_E = 0
   D. M_EB equals M_BE
   Answer: B. At a hinge the end moment is zero, so M_EB = 0. This gives the equation that determines the hinge rotation theta_E. (Key point: Hinge condition)

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