Question

A truss is shown in fig. Find forces in all the members of the truss and indicate whether it is tension or compression.

Answer 1

Let the reaction at joint A and E are RAV and REV. 

First we calculate the support reaction,

R_H = 0, RAH = 0 ...(i) 

R_V = 0, RAV + REV – 10 – 15 – 20 – 10 = 0 

RAV + REV = 55 ...(ii)

Taking moment about point A and equating to zero; we get

15 X 3 + 10 X 3 + 20 X 6 – REV X 6 = 0 ...(iii)

R_EV = 32.5 KN ...(iv)

Value of (iv) putting in equation (ii)

We get, RBV = 22.5 KN ...(v)

Consider FBD of Joint B, as shown in fig(a)

∑H = 0; T_BC = 0

∑V = 0; T_BA + 10 = 0; T_BA = –10KN(C)

Consider FBD of Joint F, as shown in fig (b)

∑H = 0; T_FC = 0

∑V = 0; T_FE + 20 = 0; T_FE = –20KN(C)

Consider FBD of Joint A, as shown in 

∑H = 0; T_AD + T_AC cos 45 = 0

T_AD = –T_AC cos 45 ...(i)

∑V = 0; T_CA sin 45 + 22.5 – 10 = 0

T_CA = –17.67KN(C)

Putting this value in equation (i), we get

T_AD = 12.5KN(T)

Consider FBD of Joint D, as shown in fig (c)

∑H = 0;

– T_AD + T_DE = 0

T_AD = T_DE = 12.5KN (T)

∑V = 0; T_DC – 10 = 0

T_DC = 10KN (T)

Consider FBD of Joint E, as shown in fig(d)

∑H = 0;

– TED – TE_Cn cos 45 = 0

– 12.5 – T_EC cos 45 = 0

T_EC = – 17.67KN(C)

∑V = 0; T_FE + T_EC sin 45 + 32.5 = 0

T_FE + (–17.67) sin 45 + 32.5 = 0

T_FE = – 20KN(C)

Forces in all the members can be shown as in fig below.