---
title: "Stiffness matrix: final end moments, 2.5EI frame"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-frame-20knm-8knm-15kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-20knm-8knm-15kn
site: BCT Engineering
---

# Stiffness matrix: final end moments, 2.5EI frame

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2075 Chaitra, 10 marks
- 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-15kn
- 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 matrix method. [Figure: frame ABCD; beam BC (2EI) of 10 m with 8 kN/m UDL and 20 kNm moment at B; column AB (2.5EI) of height 6 m with 15 kN horizontal load 4 m above A, A fixed; column CD (2EI) of height 6 m, D fixed.]

## Short answer

Stiffness method with unknowns theta_B, theta_C and sway Delta gives final moments M_AB=-3.78, M_BA=57.50, M_BC=-37.50, M_CB=65.57, M_CD=-65.57, M_DC=-48.14 kNm.

## Answer

**Assumptions:** A(0,0), B(0,6), C(10,6), D(10,0); the 20 kNm moment at B is clockwise; the 15 kN load acts to the right on AB at 4 m above A; the UDL of 8 kN/m acts downward on BC. A and D are fixed. Axial deformation is neglected.

#### Unknowns
$\theta_B,\theta_C$ and the sway $\Delta$ of the girder (right +); $\psi_{AB}=\psi_{CD}=\Delta/6$.

#### Fixed-end moments
- BC: $\mp\dfrac{8\times10^2}{12}=\mp66.67$ kNm
- AB (point load 15 kN at a = 4 m from A, b = 2 m): the load acts perpendicular to the member: $FEM_{AB}=-6.67$, $FEM_{BA}=13.33$ kNm (from $Pab^2/L^2$ and $Pa^2b/L^2$ with the sign for a load to the right)

#### Stiffness equations of the members
- $M_{AB}=\frac{2(2.5EI)}{6}\left(\theta_{B} - (0.5000)\Delta_1\right) - 6.67$
- $M_{BA}=\frac{2(2.5EI)}{6}\left(2\theta_{B} - (0.5000)\Delta_1\right) + 13.33$
- $M_{BC}=\frac{2(2EI)}{10}\left(2\theta_{B} + \theta_{C}\right) - 66.67$
- $M_{CB}=\frac{2(2EI)}{10}\left(2\theta_{C} + \theta_{B}\right) + 66.67$
- $M_{CD}=\frac{2(2EI)}{6}\left(2\theta_{C} - (0.5000)\Delta_1\right)$
- $M_{DC}=\frac{2(2EI)}{6}\left(\theta_{C} - (0.5000)\Delta_1\right)$

#### Equilibrium
- Joint B: $M_{BA}+M_{BC}=20$ (applied clockwise moment)
- Joint C: $M_{CB}+M_{CD}=0$
- Storey shear: taking moments about B for column AB (the 15 kN load is 2 m below B) gives the base shear $H_A=\dfrac{M_{AB}+M_{BA}-15\times2}{6}$; for CD, $H_D=\dfrac{M_{CD}+M_{DC}}{6}$. Horizontal equilibrium $H_A+H_D+15=0$ then gives $\dfrac{M_{AB}+M_{BA}}{6}+\dfrac{M_{CD}+M_{DC}}{6}+10=0$. When the member equations (which include the FEMs) are substituted, the constant on the right-hand side becomes $10+\dfrac{-6.67+13.33}{6}=11.11$ kN. This equals the equivalent joint load at B, i.e. the fixed-end reaction of the loaded column at B, $\dfrac{Pa^2(L+2b)}{L^3}=\dfrac{15\times4^2(6+4)}{6^3}=11.11$ kN.

$$
EI\begin{bmatrix}2.4667 & 0.4000 & -0.4167 \\ 0.4000 & 2.1333 & -0.3333 \\ -0.4167 & -0.3333 & 0.2500\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}73.333 \\ -66.667 \\ 11.111\end{Bmatrix}
$$
(Rows 1 and 2: joints B and C. Row 3: storey shear, right-hand side = 11.11 kN.)

#### Solution
$$
EI\theta_{B}=49.5343,\quad EI\theta_{C}=-26.1394,\quad EI\Delta_{1}=92.1490
$$

#### Final end moments (kNm, clockwise on the member end +)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -3.78 | BA | 57.50 |
| BC | -37.50 | CB | 65.57 |
| CD | -65.57 | DC | -48.14 |

Checks:
- Joint B: (57.50) + (-37.50) = 20.00 kNm (applied clockwise moment 20.00) ✓
- Joint C: (65.57) + (-65.57) = 0.00 kNm (applied clockwise moment 0.00) ✓

| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -3.78 | -57.50 | - |
| BC | -37.50 | -65.57 | 48.96 at 4.65 m |
| CD | -65.57 | 48.14 | - |

Reactions: A: $H=3.95$, $V=37.19$, $M=3.78$; D: $H=-18.95$, $V=42.81$, $M=48.14$.

**Answer: $M_{AB}=-3.78$, $M_{BA}=57.50$, $M_{BC}=-37.50$, $M_{CB}=65.57$, $M_{CD}=-65.57$, $M_{DC}=-48.14$ kNm.**

## Check yourself

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

1. The 20 kNm moment at B appears in the joint B equation as:
   A. MBA + MBC = 0
   B. MBA + MBC = 20
   C. MBA - MBC = 20
   D. Part of the sway equation
   Answer: B. An external clockwise moment at a joint equals the sum of member end moments there. (Key point: Applied joint moment)

2. How is the 15 kN load on column AB included in the sway equation?
   A. It is ignored
   B. It is added to the UDL
   C. Its fixed-end shear at B (11.11 kN) is taken as an equivalent joint load on the girder
   D. It becomes a moment at C
   Answer: C. The load is replaced by its fixed-end reactions, whose horizontal force at B acts as a joint load. (Key point: Load on column)

3. Why is ψAB = ψCD = Δ/6?
   A. Both have equal EI
   B. They are fixed
   C. The loads are equal
   D. Both columns are 6 m high and the girder sways equally
   Answer: D. Chord rotation is sway divided by height; equal heights give equal ψ. (Key point: Chord rotations)

4. FEM for BC (8 kN/m, 10 m) magnitude:
   A. 66.67 kNm
   B. 100 kNm
   C. 80 kNm
   D. 133.3 kNm
   Answer: A. wL^2/12 = 8 x 100 / 12 = 66.67 kNm. (Key point: FEM BC)

## Related in Theory of Structures II

- [Stiffness method: final end moments, frame](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-20knm-8knm) — Theory of Structures II, chapter 3
- [Stiffness matrix: frame with 100 kN, 15 kNm](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-100kn-20knm-15knm) — Theory of Structures II, chapter 3
- [Stiffness matrix: frame, end moments](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-60kn-50kn-10knm) — Theory of Structures II, chapter 3
- [Stiffness matrix: frame with 20 kN/m, 10 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-20knm-10kn-2i) — Theory of Structures II, chapter 3
- [Stiffness matrix: frame with 50 kNm moment](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-40kn-10knm-50knm) — Theory of Structures II, chapter 3
- [Stiffness method: bent frame member end moments](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-bent-frame-end-moments-inextensible) — Theory of Structures II, chapter 3

## Related in other subjects

- [Frame AFD, SFD, BMD (20 kNm, 15 kN)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-20knm-15kn) — Strength of Materials, chapter 1
- [Frame horizontal deflection at free end](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-virtual-work-free-end-2077) — Theory of Structures I, chapter 3
- [Frame AFD, SFD, BMD (30 kN/m, 50 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-30knm-50knm) — Strength of Materials, chapter 1
- [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