---
title: "Slope deflection: continuous beam, I = 4x10^7 mm^4"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-deflection-beam-i4e7-60kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-i4e7-60kn
site: BCT Engineering
---

# Slope deflection: continuous beam, I = 4x10^7 mm^4

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2074 Chaitra, 8 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/slope-deflection-beam-i4e7-60kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyze the continuous beam shown in figure below by the slope deflection method. Given $I = 4\times10^{7}\ \text{mm}^4$ and $E = 200\ \text{kN/mm}^2$. Draw Bending Moment diagram. [Figure: beam ABCD, A and D fixed; AB = 4 m (EI); BC = 4 m (2EI) with 10 kN/m UDL; CD = 6 m (EI) with a 60 kN point load at the 2 m mark.]

## Short answer

Using slope-deflection, fixed-end moments and equilibrium yield the end moments: M_AB = -0.32 kNm, M_BA = -0.63, M_BC = 0.63, M_CB = 43.17, M_CD = -43.17, M_DC = 31.75 kNm. The bending moment diagram shows these values with max sagging 40.63 kNm at 2 m in CD.

## Answer

**Data:** $EI=200\ \text{kN/mm}^2\times4\times10^7\ \text{mm}^4=8\times10^{3}$ kNm². Relative stiffness: AB = EI, BC = 2EI, CD = EI. A and D fixed. AB = 4 m (no load), BC = 4 m with 10 kN/m, CD = 6 m with 60 kN at 2 m from C. No settlement, so the moments do not depend on the numerical value of EI.

#### Fixed-end moments (clockwise +)
- BC: $\mp\dfrac{10\times4^2}{12}=\mp13.33$ kNm
- CD ($a=2$, $b=4$): $FEM_{CD}=-\dfrac{60\times2\times4^2}{6^2}=-53.33$, $FEM_{DC}=+\dfrac{60\times2^2\times4}{6^2}=26.67$ kNm

#### Slope-deflection equations ($\theta_A=\theta_D=0$)
- $M_{AB}=\frac{2EI}{4}\left(\theta_{B}\right)$
- $M_{BA}=\frac{2EI}{4}\left(2\theta_{B}\right)$
- $M_{BC}=\frac{2(2EI)}{4}\left(2\theta_{B} + \theta_{C}\right) - 13.33$
- $M_{CB}=\frac{2(2EI)}{4}\left(2\theta_{C} + \theta_{B}\right) + 13.33$
- $M_{CD}=\frac{2EI}{6}\left(2\theta_{C}\right) - 53.33$
- $M_{DC}=\frac{2EI}{6}\left(\theta_{C}\right) + 26.67$

#### Joint equilibrium
- Joint B: $M_{BA}+M_{BC}=0$
- Joint C: $M_{CB}+M_{CD}=0$

$$
EI\begin{bmatrix}3.0000 & 1.0000 \\ 1.0000 & 2.6667\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}13.333 \\ 40.000\end{Bmatrix}
$$
$$
EI\theta_{B}=-0.6349,\quad EI\theta_{C}=15.2381
$$
With $EI=8000$ kNm²: $\theta_B=-0.000079$ rad and $\theta_C=0.001905$ rad (clockwise positive).

#### Final end moments (kNm)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -0.32 | BA | -0.63 |
| BC | 0.63 | CB | 43.17 |
| CD | -43.17 | DC | 31.75 |

Checks:
- Joint B: (-0.63) + (0.63) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (43.17) + (-43.17) = 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 | -0.32 | 0.63 | - |
| BC | 0.63 | -43.17 | 4.73 at 0.91 m |
| CD | -43.17 | -31.75 | 40.63 at 2.00 m |

Reactions: A: $H=0.00$, $V=0.24$, $M=0.32$; B: $H=0.00$, $V=8.81$, $M=0.00$; C: $H=0.00$, $V=72.86$, $M=0.00$; D: $H=0.00$, $V=18.10$, $M=-31.75$.

**Answer: $M_{AB}=-0.32$, $M_{BA}=-0.63$, $M_{BC}=0.63$, $M_{CB}=43.17$, $M_{CD}=-43.17$, $M_{DC}=31.75$ kNm (clockwise end moments +).**

## Check yourself

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

1. Fixed A and D mean:
   A. Sway at B
   B. θA = θD = 0
   C. Zero moment at A
   D. Zero moment at D
   Answer: B. Fixed supports prevent rotation. (Key point: Boundary)

2. FEM for CD (60 kN, a=2, b=4, L=6) is:
   A. PL/8 = 45
   B. Pa^2b/L^2 = 26.67 at C
   C. Pab^2/L^2 = 53.33 kNm
   D. PL/4 = 90
   Answer: C. 60 x 2 x 16 / 36 = 53.33. (Key point: FEM CD)

3. Does the numeric EI affect the moments here?
   A. Yes, directly
   B. Only deflection and rotations
   C. Only FEM
   D. No, with no settlement moments depend on relative stiffness only
   Answer: D. EI cancels; it scales rotations only. (Key point: EI value)

4. The equations to solve are:
   A. MBA+MBC=0 and MCB+MCD=0
   B. MAB=0
   C. MBC=MCB
   D. MCD=0
   Answer: A. Unknowns θB and θC from joint moment equilibrium. (Key point: Joint equations)

## Related in Theory of Structures II

- [Slope deflection: continuous beam, B sinks 19 mm](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-sinks-19mm) — Theory of Structures II, chapter 3
- [Slope deflection: continuous beam, 90 kN and 30 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-15knm-90kn-30kn) — Theory of Structures II, chapter 3
- [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
- [Slope deflection: beam, 360 kN load, fixed end](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-10kn-20knm-360kn) — Theory of Structures II, chapter 3
- [Slope deflection: beam with 50 kN and 30 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-50kn-30kn-bmd) — Theory of Structures II, chapter 3
- [Slope deflection: beam with settlement of A and C](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-60kn-2tm-sinking) — Theory of Structures II, chapter 3

## Related in other subjects

- [Conjugate beam: deflection at B, slope at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-50kn-20knm-3-spans) — Theory of Structures I, chapter 4
- [Moment area: slope and deflection under 50 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-12-ss-defl-4m-d) — Theory of Structures I, chapter 4
- [Conjugate beam: slope and deflection at free end C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-udl-15-3kn-2076b) — Theory of Structures I, chapter 4, asked 2×
- [Conjugate beam: slope and deflection at free end C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-udl-2knm-10kn-5kn-2068) — Theory of Structures I, chapter 4, asked 2×