SAN2602
PROJECT 2025
UNIQUE NO.
DUE DATE: 2025
,Structural Analysis
QUESTION 1: MOMENT DISTRIBUTION METHOD
Fixed support at A
Pinned support at F
25 kN point load at B
40 kN point load at D
UDL of 10 kN/m on EF
Step 1: ASSUMED MEMBER LENGTHS
Member Length (m)
AB 4
BC 3
CD 3
DE 4
EF 3
Step 2: FIXED-END MOMENTS (FEM)
For AB: 25kN point load at midspan
For BC: No external load = FEM = 0
, For CD: 40kN point load at midspan
For DE: No load = FEM = 0
For EF: UDL of 10kN/m over 3m
Step 3: STIFFNESS FACTORS (Assuming EI = constant)
Member Support A Support B K values
4EI/4=EI4EI/4 =
AB Fixed Intermediate
EI4EI/4=EI
Both 4EI/3=1.33EI4EI/3 =
BC
Intermediate 1.33EI4EI/3=1.33EI
Both 4EI/3=1.33EI4EI/3 =
CD
Intermediate 1.33EI4EI/3=1.33EI
PROJECT 2025
UNIQUE NO.
DUE DATE: 2025
,Structural Analysis
QUESTION 1: MOMENT DISTRIBUTION METHOD
Fixed support at A
Pinned support at F
25 kN point load at B
40 kN point load at D
UDL of 10 kN/m on EF
Step 1: ASSUMED MEMBER LENGTHS
Member Length (m)
AB 4
BC 3
CD 3
DE 4
EF 3
Step 2: FIXED-END MOMENTS (FEM)
For AB: 25kN point load at midspan
For BC: No external load = FEM = 0
, For CD: 40kN point load at midspan
For DE: No load = FEM = 0
For EF: UDL of 10kN/m over 3m
Step 3: STIFFNESS FACTORS (Assuming EI = constant)
Member Support A Support B K values
4EI/4=EI4EI/4 =
AB Fixed Intermediate
EI4EI/4=EI
Both 4EI/3=1.33EI4EI/3 =
BC
Intermediate 1.33EI4EI/3=1.33EI
Both 4EI/3=1.33EI4EI/3 =
CD
Intermediate 1.33EI4EI/3=1.33EI