Arch: reactions, BM, thrust and shear at D by ILD

Data and assumptions: parabolic three-hinged arch, span $L = 50$ m, rise $f = 10$ m (the parabola is symmetrical about mid-span, $y = \dfrac{4\times 10\,x(50-x)}{50^2} = 0.016\,x(50-x)$). As dimensioned, the crown hinge $C$ is 30 m from $A$ and 20 m from $B$, where $y_C = 9.6$ m. UDL 20 kN/m on $AC$ ($x = 0$ to 30 m); 50 kN vertical at 20 m from $B$, i.e. at $C$ ($x = 30$ m). Section $D$ at $x = 15$ m: $y_D = 8.4$ m, $\tan\theta = 0.016\,(50-30) = 0.32$, $\theta = 17.74^\circ$.

Reactions

Vertical reactions (moments about $A$): the loads are 600 kN (UDL, at 15 m) and 50 kN (at 30 m).

$$V_B\times 50 = 600\times 15 + 50\times 30 = 10500 \Rightarrow V_B = 210\ \text{kN}, \qquad V_A = 650 - 210 = 440\ \text{kN}$$

Horizontal thrust (hinge at $C$, left part, load at $C$ itself has no moment about $C$):

$$H\times 9.6 = V_A\times 30 - 600\times 15 = 4200 \Rightarrow H = 437.5\ \text{kN}$$

(Check from the right part: $H\times 9.6 = V_B\times 20 = 4200$.)

Forces at D

$$\begin{aligned} M &= V_A\times 15 - 20\times 15\times 7.5 - H\,y_D = 6600 - 2250 - 437.5\times 8.4 = 675\ \text{kN m}\\ V_b &= V_A - 20\times 15 = 140\ \text{kN}\\ V &= V_b\cos\theta - H\sin\theta = 140(0.9523) - 437.5(0.3047) = 0\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 140(0.3047) + 437.5(0.9523) = 459.35\ \text{kN} \end{aligned}$$

Influence lines for M and N at D (unit load at x')

• Thrust (hinge at $x_C = 30$, $y_C = 9.6$): for $x'\le 30$ the right part $CB$ is a two-force member, so $H = \dfrac{R_B(L-x_C)}{y_C} = \dfrac{x'}{50}\cdot\dfrac{20}{9.6} = \dfrac{x'}{24}$; for $x'\ge 30$, $H = \dfrac{R_A\,x_C}{y_C} = \dfrac{50-x'}{16}$ (peak $1.25$ at $C$).
• $M = M_b - H\,y_D$ and $N = V_b\sin\theta + H\cos\theta$, where $V_b$ is the simple-beam shear.

The N-ILD is discontinuous at $D$ by $\sin\theta$ (the two rows at 15 m are the load just left and just right of $D$).

M and N at D from the ILDs

UDL on $AC$ (area of ILD from 0 to 30 m) plus the 50 kN load at $C$ (ordinate at 30 m):

$$\begin{aligned} M &= 20\times(45) + 50\times(-4.5) = 675\ \text{kN m}\\ N &= 20\times(19.687) + 50\times(1.3124) = 459.35\ \text{kN} \end{aligned}$$

These agree with the values from direct equilibrium.

Answer: reactions $V_A = 440$ kN, $V_B = 210$ kN, $H = 437.5$ kN (at both supports). At $D$: $M = 675$ kN m, $N = 459.35$ kN, $V = 0$ kN.