In this problem we look at determining the reactions of the two supports of an arch. The arch is shown above. Since the arch is symmetric, we can assume the distributed weight can be concentrated at the centre, and that weight is P = 80 kN. There is also a force on the arch Q = 40 kN which is located as shown and the angle of inclination α = 30 degrees. Finally there is a couple MD as shown whose value is 120 kN-m. The dimensions are as follows: a = 10 m (the distance between the two supports,) b = 2 m, h = 5 m.

We start by summing forces in the x-direction, thus

{\it X_b}+Q\cos(\alpha)=0 (1)

thus

X_b = -Q\cos(\alpha) (2)

We now sum the forces in the y-direction:

{\it N_a}+{\it Y_b}-P-Q\sin(\alpha)=0 (3)

and take moments around Point A:

a{\it Y_b}-Q\cos(\alpha)h-Q\sin(\alpha)b+{\it M_D}-1/2\,Pa=0 (4)

We solve Equation (4) for Yb:

Y_b = {\frac {Q\cos(\alpha)h}{a}}+{\frac {Q\sin(\alpha)b}{a}}-{\frac {{\it M_D}}{a}}+1/2\,P (5)

and with this result Equation (3) for Na:

N_a = -{\frac {Q\cos(\alpha)h}{a}}-{\frac {Q\sin(\alpha)b}{a}}+{\frac {{\it M_D}}{a}}+1/2\,P+Q\sin(\alpha) (6)

Substituting the given parameters yields Xb = -34.6 kN, Yb = 49.3 kN and Na = 50.7 kN.

The problem is taken from Theoretical Mechanics – A Short Course -Targ.