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, is
Discussion:
10 comments Page 1 of 1.
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.
(2)
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)
Krunal said:
8 years ago
How to determine it as compression or tension?
Manoj paridwal said:
7 years ago
Is sign is - then compression?
If sign positive then tension.
If sign positive then tension.
Abhi said:
6 years ago
How to consider the length of this? Please explain me in detail.
Nath said:
6 years ago
Here, Assume x as the length of each segment.
Gajanan todeti said:
5 years ago
Answer is A) 4T tension.
Michael said:
4 years ago
Why zero can't be? Explain, please.
Sk said:
3 years ago
Thanks everyone for explaining.
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.
(6)
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