Member JK
Equation to use: ΣMC = 0
Location of Unit Load = Point C
UJK=−abLd=−8(8)16(8sin60∘)
UJK=−√3/3
Influence Diagram for Member JK
FJK=10[12(16)(−√3/3)]+30(−√3/3)
FJK=−63.51 kN ← [ B ] answer for part 1
Member BC
Equation to use: ΣMJ = 0
Location of Unit Load = Point B
UBC=+abLd=4(12)16(8sin60∘)
UBC=√3/4
Influence Diagram for Member BC
FBC=10[12(16)(√3/4)]+30(√3/4)
FBC=47.63 kN ← [ A ] answer for part 2
Member CG
Equation to use: ΣFV = 0
Location of Unit Load = Point B (Within the Section)
UCGv=−aL
UCGsin60∘=−416
UCG=−√3/6
Location of Unit Load = Point C (Outside the Section)
UCGv=+bL
UCGsin60∘=816
UCG=√3/3
Influence Diagram for Member CG
y=√3/3−(−√3/6)=√3/2
x−8√3/3=4√3/2 → x=32/3
Maximum Compression Force:
FCG=10[12(16−323)(−√3/6)]+30(−√3/6)
FCG=−16.36 kN
Maximum Tension Force:
FCG=10[12(323)(√3/3)]+30(√3/3)
FBC=47.63 kN
Answer for part 3 = [ C ]