# Civil Engineering - Theory of Structures - Discussion

Discussion Forum : Theory of Structures - Section 1 (Q.No. 17)

17.

The force in

*BF*of the truss shown in given figure, isDiscussion:

10 comments Page 1 of 1.
Mahesh said:
2 years ago

By equilibrium conditions.

All vertical forces=0,

Va + Vb = 5t, ----> eqn(1).

Taking moments about A,

Vb*2=5*1+4*1,

Vb = 4.5t

Va = 4.5t.

All vertical forces=0,

Va + Vb = 5t, ----> eqn(1).

Taking moments about A,

Vb*2=5*1+4*1,

Vb = 4.5t

Va = 4.5t.

(3)

Sk said:
3 years ago

Thanks everyone for explaining.

Michael said:
3 years ago

Why zero can't be? Explain, please.

Gajanan todeti said:
4 years ago

Answer is A) 4T tension.

Nath said:
5 years ago

Here, Assume x as the length of each segment.

Abhi said:
6 years ago

How to consider the length of this? Please explain me in detail.

Manoj paridwal said:
6 years ago

Is sign is - then compression?

If sign positive then tension.

If sign positive then tension.

Krunal said:
7 years ago

How to determine it as compression or tension?

P.K. Nayak said:
8 years ago

Taking moment about A.

Vb * 2 = 5 * 1 + 4 * 1.

=> Vb = 4.5T Upward.

Considering joint B

Fbc = 0 & Fbf = 4.5T c.

Vb * 2 = 5 * 1 + 4 * 1.

=> Vb = 4.5T Upward.

Considering joint B

Fbc = 0 & Fbf = 4.5T c.

(2)

Patel said:
1 decade ago

Reactions are va+ab = 5t.

Moment at A= 4t*x+5t*x-vb*2x = 0.

So 9t = 2vb.

vb = 4.5t.

So force in compression member BF = 4.5t compression.

Moment at A= 4t*x+5t*x-vb*2x = 0.

So 9t = 2vb.

vb = 4.5t.

So force in compression member BF = 4.5t compression.

(1)

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