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.
B GURRAPPA said:
9 years ago
Formula is wl/8. So answer 1000 * 10/8 = 1250 kg.
Ranjan patra said:
9 years ago
Formula is WL/8H. So answer is 1250.
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.
Manaswin30781 said:
9 years ago
What is the formula for concentrated load?
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.
Tej deuba said:
1 decade ago
H = wl2/8h this formula is not apply when apply H = 781kg why answer is 1250kg.
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