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 1 of 2.
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.
Engineer said:
2 years ago
W = 1000 kg convert this load into UDL of half span of the arch then UDL load,
w = 200 kg/m.
The horizontal thrust formula for half UDL of the arch is = wl * l/16h.
So, H = w(l*l)/16h
= 200 * 10/16 * 1
= 1250 kg.
w = 200 kg/m.
The horizontal thrust formula for half UDL of the arch is = wl * l/16h.
So, H = w(l*l)/16h
= 200 * 10/16 * 1
= 1250 kg.
(2)
Subhash chandra said:
5 years ago
W=1000 kg convert this load into UDL of half span of the arch then UDL load,w=200 kg/m
The horizontal thrust formula for half UDL of the arch is =wl/16h.
So, H=wl/16h=200*10/16*1=1250 kg.
The horizontal thrust formula for half UDL of the arch is =wl/16h.
So, H=wl/16h=200*10/16*1=1250 kg.
(2)
VIPIN SAINATH said:
5 years ago
The formula is H= W*X / 2h, Where X= L / sq.rt3, but X value is directly given in question, X=2.5m.
h=1m and Span is 10m.
H = (1000 kgx 2.5m)/ (2 x 1m) = 1250 kg.
h=1m and Span is 10m.
H = (1000 kgx 2.5m)/ (2 x 1m) = 1250 kg.
(1)
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.
MANASWINI said:
9 years ago
Va = 750kg & Vb = 250kg.
Moment at crown due to external loading.
= Vb *l/2.
= 250 * 5.
= 1250kg m.
Horizontal thrust = 250/1=1250 kg.
Moment at crown due to external loading.
= Vb *l/2.
= 250 * 5.
= 1250kg m.
Horizontal thrust = 250/1=1250 kg.
Achut Paudel said:
10 years ago
@Tej Deuba,
This formula is for UDL. For a concentrated load, formula should be different.
This formula is for UDL. For a concentrated load, formula should be different.
Kisore said:
5 years ago
I got 2250kg of horizontal thrust by solving equilibrium method as we did for beams.
Tej deuba said:
1 decade ago
H = wl2/8h this formula is not apply when apply H = 781kg why answer is 1250kg.
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.
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