---
title: "Slope deflection: continuous beam, 75 kN/m, 190 kN"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-deflection-beam-75knm-190kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-75knm-190kn
site: BCT Engineering
---

# Slope deflection: continuous beam, 75 kN/m, 190 kN

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2076 Asoj, 10 marks
- Video: [Slope Deflection Method for Beam Analysis - Problem No 8](https://www.youtube.com/watch?v=tuiMskZlEXA) — Stan Academy
- 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/slope-deflection-beam-75knm-190kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Draw BMD using the slope deflection method. [Figure: continuous beam with supports A, B, C, D and a fixed end at the right; spans 3 m (cantilever with 20 kN/m), 5 m (75 kN/m UDL, stiffness 2.5EI), 0.5 m + 1.5 m with 190 kN point load (3EI), and 4 m (1.5EI) to the fixed end.]

## Short answer

Slope-deflection gives final moments M_BA=90.00, M_BC=-90.00, M_CB=158.71, M_CD=-158.71, M_DC=-8.71, M_DE=8.71, M_ED=4.35 kNm. Sagging maximum in BC is 111.28 kNm at 2.32 m from B; CD gives 45.60 kNm hogging under the 190 kN load.

## Answer

**Assumptions:** A is the free end of a 3 m cantilever carrying 20 kN/m; B, C and D are simple supports; E is fixed. Spans: BC = 5 m (2.5EI) with 75 kN/m, CD = 2 m (3EI) with 190 kN at 0.5 m from C, DE = 4 m (1.5EI). No support settlement.

#### Step 1: Cantilever AB
$M_B=\dfrac{20\times3^2}{2}=90$ kNm (hogging). It is a known moment applied to joint B.

#### Step 2: Fixed-end moments (clockwise +)
- BC: $\mp\dfrac{75\times5^2}{12}=\mp156.25$ kNm
- CD: $FEM_{CD}=-\dfrac{190\times0.5\times1.5^2}{2^2}=-53.44$, $FEM_{DC}=+\dfrac{190\times0.5^2\times1.5}{2^2}=17.81$ kNm
- DE: no load

#### Step 3: Slope-deflection equations
- $M_{BC}=\frac{2(2.5EI)}{5}\left(2\theta_{B} + \theta_{C}\right) - 156.25$
- $M_{CB}=\frac{2(2.5EI)}{5}\left(2\theta_{C} + \theta_{B}\right) + 156.25$
- $M_{CD}=\frac{2(3EI)}{2}\left(2\theta_{C} + \theta_{D}\right) - 53.44$
- $M_{DC}=\frac{2(3EI)}{2}\left(2\theta_{D} + \theta_{C}\right) + 17.81$
- $M_{DE}=\frac{2(1.5EI)}{4}\left(2\theta_{D}\right)$
- $M_{ED}=\frac{2(1.5EI)}{4}\left(\theta_{D}\right)$

#### Step 4: Joint equilibrium
- Joint B: $M_{BC}+M_{BA}=0$ with $M_{BA}=+90$ kNm from the cantilever, so $M_{BC}=-90$ kNm
- Joint C: $M_{CB}+M_{CD}=0$
- Joint D: $M_{DC}+M_{DE}=0$

$$
EI\begin{bmatrix}2.0000 & 1.0000 & 0.0000 \\ 1.0000 & 8.0000 & 3.0000 \\ 0.0000 & 3.0000 & 7.5000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \theta_{D}\end{Bmatrix}=\begin{Bmatrix}66.250 \\ -102.812 \\ -17.812\end{Bmatrix}
$$
$$
EI\theta_{B}=43.3482,\quad EI\theta_{C}=-20.4464,\quad EI\theta_{D}=5.8036
$$

#### Step 5: Final moments (kNm, clockwise on the member end +)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | 0.00 | BA | 90.00 |
| BC | -90.00 | CB | 158.71 |
| CD | -158.71 | DC | -8.71 |
| DE | 8.71 | ED | 4.35 |

Joint checks:
- Joint B: (90.00) + (-90.00) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (158.71) + (-158.71) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint D: (-8.71) + (8.71) = 0.00 kNm (applied clockwise moment 0.00) ✓

#### Bending moment diagram ordinates (sagging +, kNm)
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | 0.00 | -90.00 | - |
| BC | -90.00 | -158.71 | 111.28 at 2.32 m |
| CD | -158.71 | 8.71 | -45.60 at 0.50 m (under the 190 kN load) |
| DE | 8.71 | -4.35 | - |

Reactions: B: $H=0.00$, $V=233.76$, $M=0.00$; C: $H=0.00$, $V=427.45$, $M=0.00$; D: $H=0.00$, $V=-39.47$, $M=0.00$; E: $H=0.00$, $V=3.26$, $M=-4.35$.

**Answer: bending moments 90.00 (hogging) at B (cantilever), 158.71 (hogging) at C, 8.71 (sagging) at D and 4.35 (hogging) at E (kNm); span BC sagging maximum 111.28 kNm at 2.32 m from B; in CD the moment under the 190 kN load is 45.60 kNm (hogging).**

## Check yourself

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

1. How is the cantilever AB handled?
   A. As a fixed-end span
   B. Its moment 20 x 3^2/2 = 90 kNm is applied as a known hogging moment on joint B
   C. It is ignored
   D. It adds an unknown θA
   Answer: B. A cantilever is determinate, so its root moment is directly known at B. (Key point: Cantilever)

2. Joint B equilibrium gives MBC = -90 because:
   A. MBC = FEM
   B. θB = 0
   C. MBA = +90 from the cantilever and MBA + MBC = 0
   D. MBC equals MCB
   Answer: C. The moment from the cantilever balances the moment of BC at the joint. (Key point: Joint B equation)

3. The FEM for BC (75 kN/m, 5 m) is:
   A. wL^2/8 = 234.4 kNm
   B. wL^2/2 = 937.5 kNm
   C. wL/2 = 187.5 kNm
   D. wL^2/12 = 156.25 kNm
   Answer: D. For UDL on a fixed-fixed span FEM = wL^2/12 = 75 x 25 / 12 = 156.25 kNm. (Key point: FEM BC)

4. Why do the slope-deflection equations contain no chord rotation terms?
   A. The supports are not settled, so ψ = 0
   B. The beam is symmetric
   C. Axial strain is large
   D. Stiffness is constant
   Answer: A. With no relative support displacement the term -3ψ vanishes. (Key point: Settlement)

## Related in Theory of Structures II

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- [Slope deflection: continuous beam, SFD and BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-50kn-10knm-3ei) — Theory of Structures II, chapter 3
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## Related in other subjects

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- [Beam reactions, deflection, BMD (30 kN, 20 kN/m)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-reactions-midspan-bmd-a2) — Computational Techniques in Civil Engineering, chapter 4, asked 2×
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