Arch 40 m: ILDs for thrust, BM at 8 m, shear

Data: symmetrical three-hinged parabolic arch, $L = 40$ m, $f = 10$ m, $y = \dfrac{4\times 10\,x\,(40-x)}{40^2} = \dfrac{x(40-x)}{40}$. Section $D$ at $x = 8$ m: $y_D = 6.4$ m, $\tan\theta = \dfrac{40-16}{40} = 0.6$, $\theta = 30.96^\circ$, $\sin\theta = 0.5145$, $\cos\theta = 0.8575$.

(i) ILD of horizontal thrust

For a unit load at $x'$, the simple-beam moment at the crown is $\dfrac{x'}{2}$ (left half) or $\dfrac{40-x'}{2}$ (right half), so

$$H = \frac{M_C}{f} = \frac{x'}{20}\ (x'\le 20), \qquad \frac{40-x'}{20}\ (x'\ge 20)$$

This is a triangle with a peak of $1.0$ at the crown and zero at $A$ and $B$.

(ii) ILD of bending moment at D

$M_D = M_b - H\,y_D$, with $M_b = \dfrac{x'(40-8)}{40} = 0.8x'$ for $x'\le 8$ and $\dfrac{8(40-x')}{40} = 0.2\,(40-x')$ for $x'\ge 8$.

• Load left of $D$: $M_D = 0.8x' - \dfrac{x'}{20}(6.4) = 0.48\,x'$ (so $+3.84$ m at $D$)
• Load between $D$ and $C$: $M_D = 0.2\,(40-x') - 0.32\,x' = 8 - 0.52\,x'$ (so $-2.4$ m at the crown)
• Load right of $C$: $M_D = 0.2\,(40-x') - 0.05\,(40-x')(6.4) = -0.12\,(40-x')$

The neutral point is at $s_0 = \dfrac{L^2}{3L-2a} = \dfrac{1600}{120-16} = 15.38$ m.

(iii) ILDs of radial shear and normal thrust at D

$$V = V_b\cos\theta - H\sin\theta, \qquad N = V_b\sin\theta + H\cos\theta$$

with $V_b = -\dfrac{x'}{40}$ for a load left of $D$ and $\dfrac{40-x'}{40}$ for a load right of $D$.

Ordinates

(The two rows at 8 m show the jumps in $V$ ($=\cos\theta = 0.8575$) and in $N$ ($=\sin\theta = 0.5145$) as the unit load crosses the section.)

Shapes

• H: triangle, 0 at $A$, 1.0 at the crown, 0 at $B$.
• M at D: 0 at $A$, rises to $+3.84$ m at $D$, falls to zero at the neutral point ($15.38$ m) and to $-2.4$ m at the crown, and returns to 0 at $B$.
• V at D: negative from $A$ to just left of $D$ ($-0.3773$), jumps up by $\cos\theta = 0.8575$ to $+0.4802$ just right of $D$, then decreases linearly to zero at $x' = 18.18$ m; it is negative from there to $B$ ($-0.0857$ at the crown, then rising linearly to 0 at $B$).
• N at D: positive (compressive) throughout; it rises to $0.2401$ just left of $D$, jumps by $\sin\theta = 0.5145$ to $0.7546$, rises to $1.1147$ at the crown and falls linearly to 0 at $B$.