Question

ques no. 12 only

Answer 1

The force diagram for the whole truss is

5 m 60° AE

From the diagram

AC AE = tan 60° tan 60°

Now, at the equilibrium, the moment of force at point A is zero. i.e

F2 x 5 + F1 x 2.5 + FEy X - tan 6000

= Fax --(tat )

= Foy = - tan 60 (73+) - - tan 60" (152 + 12) – –20 tan 60°N

Now, at equilibrium, the sum of the forces in y-direction must be zero. i.e

FA+FEy = 0

FA= -FEy = 229 tan 60 N

And at equilibrium, the sum of the forces in x-direction also must be zero i.e

Fer - F1 - F2 = 0

Fer = Fi + F2 = 152 + 153 = 305 N

Now, at junction E, the situation can be visualized as

NOTE: We will take tension as positive from here on.

FED

At equilibrium, the forces in x and y direction must sum up to zero. i.e

FEy + FED sin 60 = 0

FED = -7 FEY sin 600 229 tan 60 sin 60° 229 cos 60° = 458 N

And

FE – FA – Fp cos 60 = 0

FEA = 305 – 458 cos 60% = 76 N

Now, at junction A, we have

FABT FEA

NOTE: The angle we took here is 60^o. It can easily be proved by simple geometry. Triangle BCD and BAD are similar so the angle BCD and BAD are the same and equal to 30^o. Which makes our angle 60^o.

At equilibrium, the forces in y-direction must sum up to zero. i.e

FE4 + Fap cos 60" = 0

FEA FAD= - cos 60° 76 cos 600 = -152 N

So, the correct answer is -152 N. The negative sign shows compression in the member.