Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 4 (Q.No. 15)
15.
A three hinged arch is loaded with an isolated load 1000 kg at a horizontal distance of 2.5 m from the crown, 1 m above the level of hinges at the supports 10 metres apart. The horizontal thrust is
Discussion:
16 comments Page 2 of 2.
GAURAB said:
6 years ago
It's WL/4H.
W=1000KG.
L= CENTER TO CENTER=5.
H= RISE =1.
SOLVE IT =1250KG.
W=1000KG.
L= CENTER TO CENTER=5.
H= RISE =1.
SOLVE IT =1250KG.
Netha said:
6 years ago
Formula = wl2/4h.
Teja said:
5 years ago
@Subhash.
By using this equation we get 125 kg.
Option B.
By using this equation we get 125 kg.
Option B.
Kisore said:
5 years ago
I got 2250kg of horizontal thrust by solving equilibrium method as we did for beams.
Shah Sawar said:
4 years ago
1st find the vertical forces at both ends. For that take moment about A ( left end). 1000 x 2.5 - By x 10 = 0.
By = 250 kg.
Now sum of all vertical forces = 1000 kg, we get Ay = 750 kg.
Now take section of the arch at crown where moment = 0 and consider left section.
750 x 5 - Ax * 1 - 1000 x 2.5 = 0. ( Ax is the horizontal force at hinge support A).
Ax = 1250 kg.
By = 250 kg.
Now sum of all vertical forces = 1000 kg, we get Ay = 750 kg.
Now take section of the arch at crown where moment = 0 and consider left section.
750 x 5 - Ax * 1 - 1000 x 2.5 = 0. ( Ax is the horizontal force at hinge support A).
Ax = 1250 kg.
Aryan said:
3 years ago
@All.
Here, it's not mentioned which side is the load is; Like they only told 2.5m from the crown.
Then, how to calculate further? Anyone explain me.
Here, it's not mentioned which side is the load is; Like they only told 2.5m from the crown.
Then, how to calculate further? Anyone explain me.
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