From the shown beam using the moment distribution method to: (1) Determine all the reactions at supports, (2) Draw its and bending moment diagrams, and shear force diagrams

Distribution factor:

KAB = 3(2EI)/8= 6 EI/ 8

KBC = 4(3EI)/8=12EI / 8

KCD= 3(3EI)/8=9 EI/ 8

Stiffens of member:

At joint B: d BA = KAB/(KAB+ KBC) =6/8 = 0.333 ; d CB = KAB/(KAB+ KBC) = 2/3 = 0.667

At joint C: d CB = KBC/(KBC+ KCD) = 12/21 = 0.571 ; d CD = KCD/(KBC+ KCD) = 9/21 =0.429

B C BA BC CB CD

DF 0.333 0.667 0.571 0.429

Joint couple 16.65 33.35

COFEM( due to load )

D.M15

3.885-26.6677.782

16.67526.667-1.905

-40-1.437

COD.M 0.317

-0.9530.636

3.891-2.218 -1.673

COD.M 0.369

-1.1090.740

0.318-0.181 -0.137

Σ 36.221 13.78 43.247 -43.247

9.53+12.87=22.4 KN

27.13+25.41=52.54 KN

Mohammad Saeed Anwar Khawam201401379 Thank you