---
title: "Stiffness method: frame with 16 t, 5 t"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-frame-16t-5t-inextensible"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-16t-5t-inextensible
site: BCT Engineering
---

# Stiffness method: frame with 16 t, 5 t

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2066 Bhadra, 10 marks
- Video: [Stiffness matrix method for non sway frames](https://www.youtube.com/watch?v=p60pRwtjaZQ) — Dr Venkat Lute
- 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-16t-5t-inextensible
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyze the frame given below with inextensible members using the stiffness method. [Figure: frame; column AB (I) of 3 m + 2 m + 2 m heights with 5 t horizontal load; beam BC (2I) of 4 m (2 m + 2 m) with 16 t vertical load at mid-span; C on a roller; A fixed.]

## Short answer

Kinematic unknowns θB, θC, sway Δ. Solving equilibrium gives EIθB=10.64, EIθC=-9.32, EIΔ=27.20; member end moments: M_AB=-11.05, M_BA=-3.95, M_BC=3.95, M_CB=0 t·m.

## Answer

#### Assumptions
Column AB is 3 m high, with the 5 t horizontal load at joint B. Beam BC is 4 m with the 16 t load at its mid-point. Units: t and m. Axial deformation is ignored (inextensible), so B and C move horizontally by the same sway $\Delta$ and neither joint moves vertically.

#### Degrees of freedom
Unknowns: $\theta_B$, $\theta_C$ and the sway $\Delta$ (chord rotation of the column $\psi = \Delta/3$). Kinematic indeterminacy = 3. Clockwise end moments are positive.

#### Fixed-end moments
BC: $PL/8 = 16\times 4/8 = 8$ t·m, so $FEM_{BC} = -8$, $FEM_{CB} = +8$.

#### Stiffness (slope-deflection) equations
$$
\begin{aligned}
M_{AB} &= \tfrac{2EI}{3}\,(\theta_B - 3\psi) = \tfrac{2EI}{3}\theta_B - \tfrac{2EI}{3}\Delta \\
M_{BA} &= \tfrac{2EI}{3}\,(2\theta_B - 3\psi) = \tfrac{4EI}{3}\theta_B - \tfrac{2EI}{3}\Delta \\
M_{BC} &= \tfrac{2(2EI)}{4}(2\theta_B + \theta_C) - 8 = EI(2\theta_B+\theta_C) - 8 \\
M_{CB} &= EI(\theta_B + 2\theta_C) + 8
\end{aligned}
$$

#### Equilibrium equations
1. Joint B: $M_{BA} + M_{BC} = 0$
2. Joint C (roller, no moment): $M_{CB} = 0$
3. Horizontal force: the only horizontal restraint is at A, so the column shear equals the 5 t load: $(M_{AB} + M_{BA})/3 = -5$

Substituting:

$$
\begin{aligned}
\tfrac{10}{3}EI\,\theta_B + EI\,\theta_C - \tfrac{2}{3}EI\,\Delta &= 8 \\
EI\,\theta_B + 2EI\,\theta_C &= -8 \\
2EI\,\theta_B - \tfrac{4}{3}EI\,\Delta &= -15
\end{aligned}
$$

Solving:

$$
EI\,\theta_B = \tfrac{117}{11} = 10.636,\quad EI\,\theta_C = -\tfrac{205}{22} = -9.318,\quad EI\,\Delta = \tfrac{1197}{44} = 27.205
$$

#### Member end moments
| End | Moment (t·m) |
|---|---|
| $M_{AB}$ | -11.045 (anticlockwise) |
| $M_{BA}$ | -3.955 |
| $M_{BC}$ | +3.955 |
| $M_{CB}$ | 0 |

Joint B: $-3.955 + 3.955 = 0$ (check).

#### Reactions and diagrams
- Reaction at C (moments about B for beam BC, with $M_{BC} = 3.955$ clockwise): $R_C = (16\times 2 + 3.955)/4 = 8.989$ t up.
- At A: $V_A = 16 - 8.989 = 7.011$ t up, $H_A = 5$ t to the left, $M_A = 11.045$ t·m anticlockwise.
- BMD: column AB is a straight line from 11.045 at A, passing through zero 2.21 m above A, to 3.955 at B on the opposite face. Beam BC: 3.955 (sagging) at B, 17.977 sagging under the 16 t load, 0 at C.
- SFD: column 5 t constant; beam +7.011 from B up to the load, then -8.989 to C.
- Axial: column AB carries 7.011 t compression; beam BC has no axial force.

**Answer:** $\theta_B = 10.64/EI$, $\theta_C = -9.32/EI$, $\Delta = 27.20/EI$; $M_{AB}=-11.05$, $M_{BA}=-3.95$, $M_{BC}=3.95$, $M_{CB}=0$ t·m.

## Check yourself

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

1. What are the unknown displacements?
   A. theta_B, theta_C and horizontal sway Delta (3 DOF)
   B. Only theta_B
   C. Vertical deflections at B and C
   D. theta_B and theta_C only
   Answer: A. Inextensible members give one common sway; with two rotations the kinematic indeterminacy is 3. (Key point: Degrees of freedom)

2. What equation comes from C being a roller?
   A. M_BC = 0
   B. Delta = 0
   C. M_CB = 0 (no moment at the end)
   D. theta_C = 0
   Answer: C. A roller cannot resist moment, so M_CB = 0. (Key point: Roller condition)

3. The third equilibrium equation is...
   A. M_BA + M_BC = 5
   B. Sum of vertical forces = 16
   C. M_AB = M_BA
   D. Column shear (M_AB + M_BA)/3 equals the 5 t horizontal load (taken as -5 by sign)
   Answer: D. Only A restrains horizontally, so the column shear must balance the horizontal load. (Key point: Sway equation)

4. FEM_BC for the 16 t central load on a 4 m beam is...
   A. PL/8 = 8 tm
   B. PL/4 = 16 tm
   C. wL^2/12
   D. PL/2 = 32 tm
   Answer: A. A central point load gives PL/8 = 16x4/8 = 8. (Key point: Fixed-end moment of beam)

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