Civil Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 2 (Q.No. 4)
4.
A three hinged parabolic arch hinged at the crown and springings, has a horizontal span of 4.8 m and a central rise of 1 m. It carries a uniformly distributed load of 0.75 tonne per metre over half left hand span. The horizontal thrust at the support will be
10.8 tonnes
1.08 tonnes
1.8 tonnes
0.8 tonnes
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
10 comments Page 1 of 1.

Emily said:   1 decade ago
Support reactions Va and Vb will be 1.35 and 0.45. Then apply equilibrium equation at crown.

Mounika Reddy said:   1 decade ago
convert distributed load to point load for easy calculations = 0.75*2.4=1.8 acts in the middle of left hand side.

Take moment at c = 0. Vay and Vby are 1.35 and 0.45.

Then take moment at crown is zero. Vax = 1.08 tonnes.

Nikhil said:   1 decade ago
Due to the u.d.l the reactions at the supports are Ra=1.35&Rb=0.45.

Take the left half or right half of the arch. Take the moments about hinge C.

There fore £Mc=Vb*2.4 - H*Y = 0.

0.45*2.4-H*1 = 0.

H= 1.08 tonnes.
(1)

RAMKUMAR said:   1 decade ago
Give the detailed work out for solve the problem.

Akscivilian said:   9 years ago
Guys take it simple,

H = wl2/16h.
H = 1.08 tonnes.
(1)

Bhavin said:   8 years ago
Hers, H = PL/8h.
(1)

Ziya said:   8 years ago
H = 2Pl^2/8h.

H = 1.08 tonnes.

Tanmoy karmakar said:   7 years ago
Please give the detailed answer.

Antaryami said:   7 years ago
H = (wl^2/16h).

W-UDL over the left or right portion of the arch.
h-Central rise.
L-horizontal length.
Ans - 1.08t.
(2)

Asita said:   5 years ago
Thanks everyone.
(1)

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