BC = √(22+ 12 ) + = 2.23m
sin θ = 1/(2.23) = 0.447
sin θ = 2/(2.23) = 0.894
Let
TCD = Force in the member CD
TCB = Force in the member CB
TDB = Force in the member DB
TAB = Force in the member AB
TAD = Force in the member AD
TBD = Force in the member BD
Consider ** C:*
Consider FBD of ***** C as shown in fig(a)
There are three forces are acting so apply lami’ theorem at **** C
TCD/sin(90 – θ) = TBC/sin 270 = 15/sin θ
TCD = 30KN (Tensile)
TBC = – 33.56 ...(i)
TBC = 33.56 (Compressive)
Consider ** B:*
Consider FBD of ***** B as shown in fig (b)
There are three forces are acting so apply lami’ theorem at **** B
TAB/sin(90 – θ) = TBC/sin90 = TDB/sin(180 + θ)
T4 = –30KN ...(ii)
TAB = – 30KN(Compressive)
TDB = 15 ...(iii)
TDB = 15(Tensile)
Consider ** D:*
Consider FBD of ***** D as shown in fig(c)
There are four forces are acting so apply resolution of forces at ***** D
RH = 0, TCD – TAD cos θ – TED cos θ = 0
30 – (TAD + TED) cos θ = 0
TAD + TED = 30/cos θ = 30/0/89 = 33.56 ...(iv)
RV = 0
TEDsin θ - TAD sin θ – TDB = 0
(TED – TAD) sin θ = 15
TED – TAD = 15/sin θ
TED – TAD = 33.56 ...(v)
Solve equation (iv) and (v), we get
TAD = 0 ...(vi) TAD = 0
TED = 33.56 ...(vii)
TED = 33.56 (Tensile)
| Member | CD | BC | BD | BA | AD | DE |
| Force in kN | 30 | 33.56 | 15 | 30 | 0 | 33.56 |
| Nature C = Compression T = Tension | T | C | T | C | — | T |
