Principles

Let P1 and P2, in Fig. 50 (a), represent the resultants of all the loads on the left and right halves of the arch, respectively, and let P1 be equal to P2, and located equally distant from the crown. Let R1 and R2 represent the vertical reactions, which, since the loads are symmetrically placed, are equal. Let b be the horizontal distance between R1 and P1; also between R1 and P2, or, in other words, the leverage of P1 and P2 with respect to R1 and R2.

Arches 289

Fig. 50.

Let Q be the horizontal thrust, which is equal at all points; also, let c be the leverage of Q with respect to the crown. Now assume the right-hand half of the arch to be taken away, as in (6). To preserve equilibrium in the left side, the force Q1 must be supplied at the crown. The algebraic sum of the vertical forces, and, likewise, the sum of the horizontal forces.

must equal zero, or there will not be equilibrium. Then R1 must equal P1 and Q1 must equal Q. Also, the sum of the moments about any point must equal zero. Hence, taking moments about the abutments in (a), Qc = P1 b, and

Q =P1b / c ; also, since the arch is symmetrically loaded,

Q2c = P2b / c and P1b / c= P2b / cIn this case P1 or P2 may represent any number of loads, provided they are equal and symmetrically placed.