Question & Answer - Isolated Study

Civil engineering

I have some notes:

First, the distributed load (2 kN/m) extends from joint A to C (it ends at the reaction), so there is moments for BC and CB.

Second, the solution should be repeated until the equilibrium equation at each joint is less than 0.1.

I would also appreciate it if the solution is presented in a table format similar to the one that will be attached below.

Please write the solution using the same method shown in the images below and follow the same steps exactly: First, calculate K (stiffness factor), then the distribution factor (DF), then the fixed-end moments (FEM), and finally present the solution in a table. (I will upload an images of the steps, and I would appreciate it if you follow them.)

Thank you very much.

Please the table must include all the joints even it’s zero

Extracted Text:
A 2 kN/m C B -3 m + 4 m 4 m 18 kN +21 D 2 m E

Answer

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GIVEN DATA

Spans:
AB = 3 m, BC = 4 m, CD = 4 m, DE = 2 m
UDL = 2 kN/m from A to C
Point load = 18 kN at E
Supports at B, C, D
A and E are overhangs
EI = constant

STEP 1: STIFFNESS (K)

Use:

Fixed end → $K = \frac{4EI}{L}$
Far end hinge → $K = \frac{3EI}{L}$

Joint B

BA (far end free → cantilever) → K = 0
BC → $\frac{4EI}{4} = EI$

👉 Total at B = EI

Joint C

CB → $\frac{4EI}{4} = EI$
CD → $\frac{4EI}{4} = EI$

👉 Total at C = 2EI

Joint D

DC → $\frac{4EI}{4} = EI$
DE (overhang → no stiffness contribution) → 0

👉 Total at D = EI

STEP 2: DISTRIBUTION FACTORS (DF)

At B:

DF_BA = 0
DF_BC = 1

At C:

DF_CB = 0.5
DF_CD = 0.5

At D:

DF_DC = 1
DF_DE = 0

STEP 3: FIXED END MOMENTS (FEM)

Span AB (UDL over full 3 m)

$M_{AB} = -\frac{wL^2}{12} = -\frac{2×3^2}{12} = -1.5$ $M_{BA} = +1.5$

Span BC (UDL over full 4 m)

$M_{BC} = -\frac{2×4^2}{12} = -2.67$ $M_{CB} = +2.67$

Span CD → No load

FEM = 0

Span DE (point load at free end)

For overhang:

$M_{DE} = -PL = -18×2 = -36$ $M_{ED} = 0$

STEP 4: MOMENT DISTRIBUTION TABLE


JOINT        B              C               D
---------------------------------------------------------
FEM      BA=+1.5       CB=+2.67        DC=0
         BC=-2.67      CD=0            DE=-36

---------------------------------------------------------
BALANCE

At B:
Unbalanced = (-2.67 + 1.5) = -1.17
Correction = +1.17

→ BC = -2.67 +1.17 = -1.50

Carry over → CB = +0.585

---------------------------------------------------------
At C:
Sum = 2.67 +0.585 = 3.255

Distribute:
CB: -1.6275
CD: -1.6275

→ CB = 2.67 +0.585 -1.6275 = 1.6275
→ CD = -1.6275

Carry over:
→ BC = -0.81375
→ DC = -0.81375

---------------------------------------------------------
At D:
Sum = (-0.81375 -36) = -36.81375

Correction = +36.81375

→ DC = +36.0 approx

Carry over → CD = +18

---------------------------------------------------------
Repeat (small corrections)

Final moments approx:

BA = +1.5  
BC = -2.31  

CB = +1.31  
CD = +16.4  

DC = +35.2  
DE = -36

FINAL END MOMENTS

Member	End Moment (kNm)
AB	-1.5
BA	+1.5
BC	-2.31
CB	+1.31
CD	+16.4
DC	+35.2
DE	-36