A semi-circular two hinged arch of radius R is subjected to a uniformly distributed load of w/unit length over the entire span. Assuming EI to be constant, determine the horizontal thrust.
Answers
Answer:
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Given: A semi-circular two hinged arch of radius R is subjected to a uniformly distributed load of w/unit length over the entire span
To find: the horizontal thrust
Explanation:
V(A) = V(B) = (1/2)×w×2R = wR
x = R( 1 - sinθ), y = R cosθ and ds = R dθ
M' = wRx - wx²/2
= wR ×R(1-sinθ) - [(w/2)R²(1 - sinθ)²]
= wR²/2(1-sinθ) + (1 + sinθ)
= wR²( 1 - sin²θ)
= (wR²/2) cos²θ
Horizontal thrust given by H = ∫M' y ds / ∫y² ds
Numerator ∫M' ds= 2∫(wR²/ 2)cos²θ×Rcosθ×Rdθ
= wR⁴∫ cos²θ dθ
Now,, consider I = ∫cos²θ dθ
= ∫(1 - sin²θ dθ
Let t = sinθ
then dt = cosθ dθ
∴ I = ∫( 1 - t²) dθ
= [ t - (t³/3)
= sinθ - (sin³θ/3)
∴ Numerator = wR⁴[ sinθ - (sin³θ/3)]
= WR⁴[1 - 1/3]= 2/3wR⁴
Denominator = 2∫R² cos²θR dθ = R⁴π/2
∴ H = 2/3 wR⁴×{2/πR³
= 4/3 (wR/π)
Answer = 2/3wR⁴ , 4/3 (wR/π)