Arch 20 m: reactions, BM, thrust, shear, BMD

Data: three-hinged parabolic arch, span $L = 20$ m, rise $f = 4$ m ($y = 0.04\,x(20-x)$), supports at the same level; UDL 2 kN/m on the left 8 m ($x = 0$ to 8 m), total 16 kN acting at 4 m from $A$.

(a) Reactions

$$V_B\times 20 = 16\times 4 \Rightarrow V_B = 3.2\ \text{kN}, \qquad V_A = 16 - 3.2 = 12.8\ \text{kN}$$

Crown hinge ($M_C = 0$), left part:

$$H\times 4 = 12.8\times 10 - 16\times 6 = 32 \Rightarrow H = 8\ \text{kN}$$

The reaction at each support is the resultant of its vertical component and the thrust:

• At $A$: $R_A = \sqrt{12.8^2 + 8^2} = 15.09$ kN, inclined at $\tan^{-1}\dfrac{12.8}{8} = 57.99^\circ$ above the horizontal, directed up and to the right (towards the span).
• At $B$: $R_B = \sqrt{3.2^2 + 8^2} = 8.62$ kN, inclined at $\tan^{-1}\dfrac{3.2}{8} = 21.8^\circ$ above the horizontal, directed up and to the left.

(b) Forces at 4 m from the left end

At $x = 4$ m: $y = 0.04\times 4\times 16 = 2.56$ m, $\tan\theta = 0.04\,(20-8) = 0.48$, $\theta = 25.64^\circ$, $\sin\theta = 0.4327$, $\cos\theta = 0.9015$. Simple-beam shear $V_b = V_A - 2\times 4 = 4.8$ kN.

$$\begin{aligned} M &= V_A\times 4 - 2\times 4\times 2 - H\,y = 12.8\times 4 - 16 - 8\times 2.56 = 14.72\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = 4.8(0.9015) - 8(0.4327) = 0.87\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 4.8(0.4327) + 8(0.9015) = 9.29\ \text{kN} \end{aligned}$$

(c) Bending moment diagram

$M = M_b - H\,y$ along the span:

The bending moment is zero at $A$, at the crown hinge $C$ ($x = 10$ m) and at $B$. The maximum positive (sagging) value is about +15.06 kN m at $x \approx 4.7$ m (under the UDL) and the maximum negative (hogging) value is about -8 kN m at $x \approx 15$ m (in the right half).

 M (kN m)
 +15.06      ___
        _.-'   '-._
 0 ---A'---------C'-._----------B---
                       '-._ 
                           '--- -8

Answer: (a) $R_A = 15.09$ kN at 57.99° to the horizontal, $R_B = 8.62$ kN at 21.8° (thrust $H = 8$ kN); (b) $M = 14.72$ kN m, $N = 9.29$ kN, $V = 0.87$ kN at 4 m from the left; (c) maximum $M = +15.06$ kN m and minimum $M = -8$ kN m.