When support B sinks 19 mm, what is the chord rotation of span AB (L = 4 m)?

+0.00475 (clockwise), as B goes down relative to A

Why is psi_BC negative?

C is higher than B relative to the chord, so the chord rotates anticlockwise

Because BC carries a point load

Which joint equations determine the unknown rotations theta_B and theta_C?

M_BA + M_BC = 0 and M_CB + M_CD = 0

What does the settlement do to the moments at B compared with no settlement?

It changes M_BA by about -104.5 kNm (the -6EI psi/L term adds to the constants)

It only affects the fixed support A

2074 Asoj · 12 marks

Slope deflection: continuous beam, B sinks 19 mm

Short answer

Slope-deflection incorporates chord rotations from the 19 mm settlement. Final end moments: M_AB = -114.63, M_BA = -82.39, M_BC = 82.39, M_CB = 74.97, M_CD = -74.97, M_DC = -37.49 kNm.

Analyse the continuous beam loaded as shown in figure below and draw the bending moment diagrams using the slope deflection method. Support B sinks by 19 mm. Take $EI = 10{,}000\ \text{kN-m}^2$ . [Figure: beam ABCD, A fixed; AB = 4 m (1.5EI) with UDL (intensity printed as ':0 kN/m', likely 10 kN/m [?]); BC = 4 m (2EI) with a 50 kN point load at 2 m from C; CD = 2 m + 2 m (1.5EI), D fixed.]

Assumptions (figure partly illegible): the UDL printed as ':0 kN/m' is taken as 10 kN/m on AB; BC carries 50 kN at mid-span (2 m from C); CD (2 m + 2 m) carries no load; spans are 4 m each. A and D are fixed. Support B settles by 19 mm. $EI=10\,000$ kNm²; AB = 1.5EI, BC = 2EI, CD = 1.5EI.

Chord rotations (clockwise +, a right-hand end that goes down rotates the chord clockwise)

$\psi_{AB}=\dfrac{0.019}{4}=0.00475$ (B moves down relative to A)
$\psi_{BC}=-\dfrac{0.019}{4}=-0.00475$ (C is higher than B)
$\psi_{CD}=0$

Fixed-end moments (clockwise +, kNm)

AB: $\mp\dfrac{10\times4^2}{12}=\mp13.33$
BC: $\mp\dfrac{50\times4}{8}=\mp25.00$
CD: 0

Slope-deflection equations

(the constants contain the FEM and the term $-6EI\psi/L$ from the settlement)

$M_{AB}=7500.00\left(\theta_{B}\right) - 120.21$
$M_{BA}=7500.00\left(2\theta_{B}\right) - 93.54$
$M_{BC}=10000.00\left(2\theta_{B} + \theta_{C}\right) + 117.50$
$M_{CB}=10000.00\left(2\theta_{C} + \theta_{B}\right) + 167.50$
$M_{CD}=7500.00\left(2\theta_{C}\right)$
$M_{DC}=7500.00\left(\theta_{C}\right)$

Equilibrium and solution

Joint B: $M_{BA}+M_{BC}=0$ . Joint C: $M_{CB}+M_{CD}=0$ .

\begin{bmatrix}35000.0000 & 10000.0000 \\ 10000.0000 & 35000.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-23.958 \\ -167.500\end{Bmatrix}

\theta_{B}=0.000743519,\quad \theta_{C}=-0.00499815

Final end moments (kNm)

End	$M$ (kNm)	End	$M$ (kNm)
AB	-114.63	BA	-82.39
BC	82.39	CB	74.97
CD	-74.97	DC	-37.49

Checks:

Joint B: (-82.39) + (82.39) = 0.00 kNm (applied clockwise moment 0.00) ✓
Joint C: (74.97) + (-74.97) = 0.00 kNm (applied clockwise moment 0.00) ✓

Bending moment diagram (sagging +, kNm)

Member	BM at first end	BM at second end	Max within member
AB	-114.63	82.39	-
BC	82.39	-74.97	53.71 at 2.00 m (under the 50 kN load)
CD	-74.97	37.49	-

Reactions (kN, kNm): A: $H=0.00$ , $V=69.26$ , $M=114.63$ ; B: $H=0.00$ , $V=-43.60$ , $M=0.00$ ; C: $H=0.00$ , $V=92.45$ , $M=0.00$ ; D: $H=0.00$ , $V=-28.11$ , $M=37.49$ .

For comparison, if B did not settle the end moments at B would be $M_{BA}=22.11$ and $M_{BC}=-22.11$ kNm; the settlement therefore changes $M_{BA}$ by -104.50 kNm.

Answer: $M_{AB}=-114.63$ , $M_{BA}=-82.39$ , $M_{BC}=82.39$ , $M_{CB}=74.97$ , $M_{CD}=-74.97$ , $M_{DC}=-37.49$ kNm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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