A three hinged semicircular arch of radius R carries a uniformly distr...
The horizontal thrust in a three-hinged arch is given by the formula:
H = (wL^2)/(8R)
where w is the uniformly distributed load, L is the span of the arch, and R is the radius of curvature.
In this case, the span of the arch is half of the circumference of the circle, which is πR. Therefore, L = πR/2.
Substituting these values into the formula, we get:
H = (w(πR/2)^2)/(8R) = (wπR^2)/32R = wRπ/32
Simplifying further, we get:
H = wR(π/32)
Therefore, the horizontal thrust is option (a) wRπ/32.
A three hinged semicircular arch of radius R carries a uniformly distr...
The horizontal thrust in a three-hinged semicircular arch is given by the expression:
H = (5/8)*wR
Therefore, option c) 4wR/3 is not correct.
The derivation of the above formula is as follows:
Consider a small element of length dx at a distance x from the left support. The load on this element is w*dx. Let the vertical reaction at the left support be R1, and the vertical reactions at the hinges be R2 and R3, respectively.
Taking moments about the left support, we get:
R2*R2x + R3*R3x = w*x*dx/2 + R1*R1x
where R2x and R3x are the horizontal distances of the hinges from the left support, and R1x is the horizontal distance of the left support from the left end of the arch.
Since the arch is symmetrical, we have R2x = R3x = R/2 and R1x = R.
Substituting these values, we get:
R2^2 + R3^2 = w*x*dx/2 + R1*R
Taking moments about the hinge at the crown, we get:
R1*R1x + R2*R2x = R3*R3x
Substituting the values of R2x, R3x, and R1x, we get:
R1*R + R2*R/2 = R3*R/2
Solving these two equations for R2 and R3, we get:
R2 = (5/8)*w*x*dx*R/R1
R3 = (3/8)*w*x*dx*R/R1
Integrating these expressions over the span of the arch, we get:
H = R2 + R3 = (5/8)*w*R
Therefore, option a) wR and option b) wR/2 are not correct, and the correct answer is option c) 5wR/8.