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
1250 kg
125 kg
750 kg
2500 kg
2325 kg.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
16 comments Page 1 of 2.

Tej deuba said:   1 decade ago
H = wl2/8h this formula is not apply when apply H = 781kg why answer is 1250kg.

Achut Paudel said:   10 years ago
@Tej Deuba,

This formula is for UDL. For a concentrated load, formula should be different.

Manaswin30781 said:   9 years ago
What is the formula for concentrated load?

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.

Ranjan patra said:   9 years ago
Formula is WL/8H. So answer is 1250.

B GURRAPPA said:   9 years ago
Formula is wl/8. So answer 1000 * 10/8 = 1250 kg.

Argade pooja said:   7 years ago
Wl/2 = 1000*2.5/2 = 1250kg.

GAURAB said:   6 years ago
It's WL/4H.
W=1000KG.
L= CENTER TO CENTER=5.
H= RISE =1.
SOLVE IT =1250KG.

Netha said:   6 years ago
Formula = wl2/4h.

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.
(1)


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