From triangle ACE, we have
Joint C:
Consider FBD of joint C as shown in fig(a)
Since three forces are acting, so apply lami,s theorem at joint C.
TBC/sin(90 – θ) = TCD/sin 270 = 2000/sin θ
TBC/cos θ = TCD/sin 270 = 2000/sin θ
TBC = 2000/tan θ = 2000/0.66 = 3000.3N ...(v)
TBC = 3000.3N (Tensile)
TCD = – 2000/sin θ = 2000/0.55 = 3604.9N ...(vi)
TCD = 3604.9N (Compressive)
Joint B:
Consider FBD of joint B as shown in fig (b)
Since, TBC = 3000.3N
Let, TAB = Force in the member AB
TDB = Force in the member DB
Since four forces are acting at joint B, So apply resolution of forces at joint B
RH = TAB – TBC = 0, TAB = TBC
= 3000.03 = TAB
TAB = 3000.03 ...(vii)
TAB = 3000.03N (Tensile)
RV = – TDB – 2000 = 0
TDB = -2000N ...(viii)
TDB = 2000N (compressive)
Joint D:
Consider FBD of joint D as shown in fig(c)
Since, TDB = – 2000N
TCD = 3604.9N
Let, TAD = Force in the member AD
TDE = Force in the member DE
Since four forces are acting at joint D, So apply resolution of forces at joint D.
RV = 2000 + TCD sin θ + TAD sin θ – TED sin θ = 0
RV = 2000 + TCD sin θ + TAD sin θ – TED sin θ = 0
TAD – TED = 7241.26N ...(ix)
RH = TCD cos θ – TAD cos θ – TEDcos θ = 0
= 3604.9 = TAD + TED
TAD + TED = 3604.9 ...(x)
Solving equation (ix) and (x), we get
TED = 55423.1N ... (xi)
TED = 5542.31N (Tensile)
TAD = -1818.18N ...(xii)
TAD = 1818.18N (compressive)
Member |
AB |
BC |
CD |
DE |
DB |
AD |
Force in N |
3000.03 |
3000.03 |
3604.9 |
5542.31 |
2000 |
1818.18 |
Nature C = Compression T = Tension |
T |
T |
C |
T |
C |
C |