How is the cantilever AB handled?

Its moment 20 x 3^2/2 = 90 kNm is applied as a known hogging moment on joint B

Joint B equilibrium gives MBC = -90 because:

MBA = +90 from the cantilever and MBA + MBC = 0

Why do the slope-deflection equations contain no chord rotation terms?

The supports are not settled, so ψ = 0

2076 Asoj · 10 marks

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

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.

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.]

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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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