Question

 Find out the axial forces in all the members of a truss with loading as shown in fig,

Answer 1

For Equilibrium

∑H = 0; R_AH = 15KN

∑V = 0; R_AV + R_BV = 0

∑MB = 0; R_AV X 4 + 10 X 4 + 5 X 8 = 0

R_AV = – 20 KN

And R_BV = 20KN

Consider Joint A as shown in fig(a)

Consider Joint B as shown in fig (b)

∑H = 0;

– T_AB – T_BF cos 45 = 0 

T_BF = – 15/cos 45 = – 21.21KN

T_BF = – 21.21KN(C)

∑V = 0

T_BC + T_BF sin 45 + 20 = 0

T_BC – 21.21 sin 45 + 20 = 0

T_BC = – 5KN(C) 

Answer 2

Consider Joint F as shown in fig (c)

∑H = 0;

T_FC + T_BF cos 45 + 10 = 0

T_FC – 21.21 cos 45 + 10 = 0 

T_FC = 5KN(T)

∑V = 0;

T_FE – T_FA – T_BF sin 45 = 0

T_FE – 20 + 21.21 sin 45 = 0 

T_FE = 5KN(T) 

Consider Joint C as shown in fig(d)

∑H = 0; 

– T_FC – T_CE cos 45 = 0

– 5 – T_CE cos 45 = 0

T_CE = – 7.071KN(C)

∑V = 0; 

T_CD – T_CB + T_CE sin 45 = 0

T_CD + 5 – 7.071 sin 45 + 20 = 0

T_CD = 0

Consider Joint D as shown in fig(e)

∑H = 0;

T_ED = 0

S.No.	Member	Force (KN)	Nature
1.	AB	15	T
2.	AD	20	T
3.	BD	21.21	C
4.	BC	5	C
5.	DC	5	T
6.	DE	5	T
7.	CE	7.071	C
8.	CF	0	—
9.	FE	0	—