Arch 20 m: ILDs and check by equilibrium

Data: symmetrical parabolic three-hinged arch, $L = 20$ m, $f = 4$ m: $y = \dfrac{4\times 4\,x(20-x)}{400} = 0.04\,x\,(20-x)$. A 4 kN point load acts at $x = 4$ m from the left hinge (support) $A$, and the internal forces are required at the section where the load acts. The section is taken just to the left of the load, so the load itself is not part of the free body. At $x = 4$: $y = 2.56$ m, $\tan\theta = 0.04\,(20-8) = 0.48$, $\theta = 25.64^\circ$, $\sin\theta = 0.4327$, $\cos\theta = 0.9015$.

Influence lines (unit load at x')

• $R_A = \dfrac{20-x'}{20}$, $H = \dfrac{x'}{2f} = \dfrac{x'}{8}$ for $x'\le 10$ and $\dfrac{20-x'}{8}$ for $x'\ge 10$ (peak $\dfrac{L}{4f} = 1.25$ at the crown).
• $M = M_b - H\,y$, with $M_b = \dfrac{x'(20-4)}{20} = 0.8x'$ ($x'\le 4$) or $\dfrac{4(20-x')}{20}$ ($x'\ge 4$).
• $V = V_b\cos\theta - H\sin\theta$ and $N = V_b\sin\theta + H\cos\theta$, with $V_b = -\dfrac{x'}{20}$ (load left of the section) or $\dfrac{20-x'}{20}$ (load right of the section).

(The row at 4 m is for the load just to the right of the section.)

Values from the ILDs for P = 4 kN at the section

$$\begin{aligned} M &= 4\times 1.92 = 7.68\ \text{kN m}\\ V &= 4\times 0.5049 = 2.019\ \text{kN}\\ N &= 4\times 0.7969 = 3.188\ \text{kN} \end{aligned}$$

Check by first principles (equilibrium)

$$R_B = \frac{4\times 4}{20} = 0.8\ \text{kN}, \qquad R_A = 3.2\ \text{kN}$$

Moment at the crown hinge (right part): $H f = R_B\times 10 = 8 \Rightarrow H = \dfrac{8}{4} = 2\ \text{kN}$.

$$\begin{aligned} M &= R_A\times 4 - H\times y = 12.8 - 2\times 2.56 = 7.68\ \text{kN m}\\ V &= R_A\cos\theta - H\sin\theta = 3.2(0.9015) - 2(0.4327) = 2.019\ \text{kN}\\ N &= R_A\sin\theta + H\cos\theta = 3.2(0.4327) + 2(0.9015) = 3.188\ \text{kN} \end{aligned}$$

The ILD values and the equilibrium values agree.

Answer: $M = 7.68$ kN m, radial shear $V = 2.019$ kN, normal thrust $N = 3.188$ kN at the section.