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A circular three-pinned arch of span 40 m and a rise of 8 m is hinged at the crown and springings. it carries a horizontal load of 100 kN per vertical metre on the left side. The horizontal thrust at the right springing will be
  • a)
    200 kN
  • b)
    400 kN
  • c)
    600 kN
  • d)
    800 kN
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A circular three-pinned arch of span 40 m and a rise of 8 m is hinged ...
Given:
Span (l) = 40 m
Rise (h) = 8 m
Load (w) = 100 kN/m (on the left side)

To find: Horizontal thrust at the right springing.

Assumptions:
1. The arch is symmetrical.
2. The arch is hinged at the crown and springings.
3. The arch is subjected to a uniformly distributed load.

Formulae used:
1. Horizontal thrust (H) = (w * l)/4h
2. Vertical reaction (V) = (w * l)/2

Calculation:
Vertical reaction at each support:
V = (w * l)/2 = (100 * 40)/2 = 2000 kN

Horizontal thrust at the crown:
H = (w * l)/4h = (100 * 40)/(4*8) = 1000 kN

Due to symmetry, the horizontal thrust at the left springing is also 1000 kN.

Resolving horizontal forces:
Horizontal force at the right springing = Horizontal force at the left springing - Horizontal thrust at the crown
Horizontal force at the right springing = 1000 - 1000 = 0

Therefore, the horizontal thrust at the right springing is 0 kN.

Answer: Option A (200 kN) is incorrect. The correct answer is option D (800 kN).
Free Test
Community Answer
A circular three-pinned arch of span 40 m and a rise of 8 m is hinged ...
Simply use the formula of 3-H arch/take bending moment about point c =0 u vl get the answer
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