Statically Determinate Arches

Neutral point

For a section $D$ of a three-hinged arch, the neutral point is the position of a unit load on the span for which the bending moment at $D$ is zero. In the ILD of $M_D$ it is the point where the influence line crosses the base line (ordinate zero). A load on one side of it produces a positive moment and on the other side a negative moment at $D$; this is why a rolling load must be put on different sides of the neutral point for the maximum positive and negative moments.

Graphical determination (section D in the left half)

When a load $P$ acts on the left half of the arch (on the segment $DC$), the right portion $CB$ carries no load, so it is a two-force member: the reaction at $B$ acts along the line $BC$. For $M_D = 0$ the reaction at $A$ must act along the line $AD$ (its moment about $D$ is then zero). The load $P$, $R_A$ and $R_B$ are in equilibrium, so the three lines must be concurrent.

1. Join $B$ to the crown hinge $C$ and extend it.
2. Join $A$ to the section $D$ and extend it.
3. Let the two lines meet at $E$.
4. The vertical through $E$ meets the span at the neutral point $N$.

            C
         .-' '-.        E  (intersection of AD and BC)
      D-'       '-.   .'
     /             '-B.
    A    . . . N . . . . 
         neutral point (vertical below E)

Derivation of the position

Let the span be $L$, the rise $f$, and let the section be at $x_D = a$ with ordinate $y_D$. Take a unit load at a distance $s$ from $A$ with $a < s < L/2$.

$$R_A = \frac{L-s}{L}, \qquad H = \frac{M_C}{f} = \frac{R_B\,(L/2)}{f} = \frac{s}{2f}$$ $$M_D = R_A\,a - H\,y_D = \frac{(L-s)\,a}{L} - \frac{s\,y_D}{2f}$$

Putting $M_D = 0$:

$$a - s\left(\frac{a}{L} + \frac{y_D}{2f}\right) = 0 \Rightarrow s_0 = \frac{a}{\dfrac{a}{L}+\dfrac{y_D}{2f}} = \frac{2f}{\dfrac{2f}{L}+\dfrac{y_D}{a}}$$

This is the same as the intersection of the lines $y = \dfrac{y_D}{a}x$ (line $AD$) and $y = \dfrac{2f}{L}(L-x)$ (line $BC$).

For a parabolic arch, $y_D = \dfrac{4fa(L-a)}{L^2}$, so

$$s_0 = \frac{L^2}{3L-2a}$$

Example: $L = 100$ m, $f = 10$ m, $a = 30$ m: $s_0 = \dfrac{100^2}{300-60} = 41.67$ m from $A$. A load to the left of $s_0$ gives positive $M_D$, a load to the right gives negative $M_D$ (this is confirmed by the ILD ordinates $+8.4$ m at $D$ and $-6$ m at the crown).

For the radial shear at $D$ the neutral point is found in the same way: it is the load position for which $V_D = R_A\cos\theta - H\sin\theta = 0$, i.e. where the reaction at $A$ is parallel to the tangent at $D$ (line through $A$ parallel to the tangent at $D$ meets line $BC$ at the point vertically above the neutral point).

Data: parabolic three-hinged arch, $L = 100$ m, rise $f = 10$ m at the crown $C$. UDL 50 kN/m on the left half ($x = 0$ to 50 m); 80 kN vertical load 25 m from $B$ ($x = 75$ m). Section $D$ at $x = 20$ m from $A$.

Step 1: Equation of the arch

$$y = \frac{4f\,x\,(L-x)}{L^2} = \frac{4\times 10\,x\,(100-x)}{100^2}$$ $$\tan\theta = \frac{dy}{dx} = \frac{4f\,(L-2x)}{L^2}$$

At the section $x = 20$ m: $y = 6.4$ m, $\tan\theta = 0.24$, so $\theta = 13.5^\circ$, $\sin\theta = 0.2334$, $\cos\theta = 0.9724$.

Step 2: Vertical reactions

Taking moments about $A$ for the whole arch:

$$R_B \times 100 = (50\times 50)\times 25 + 80\times 75 = 68500 \Rightarrow R_B = 685\ \text{kN}$$ $$R_A = 2580 - 685 = 1895\ \text{kN}$$

Step 3: Horizontal thrust

The bending moment at the crown hinge $C$ is zero. Taking moments of the forces on the left of $C$ about $C$ (rise $f = 10$ m):

$$H \times 10 = 1895\times 50 - (50\times 50)\times 25 = 32250 \Rightarrow H = 3225\ \text{kN}$$

Step 4: Forces at the section (x = 20 m)

Simple-beam values at the section: $M_b = 27900$ kN m, $V_b = 895$ kN.

$$\begin{aligned} M &= M_b - H\,y = 27900 - 3225\times 6.4 = 7260\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = 895\times 0.9724 - 3225\times 0.2334 = 117.66\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 895\times 0.2334 + 3225\times 0.9724 = 3344.82\ \text{kN} \end{aligned}$$

Answer: at $D$: bending moment $= 7260$ kN m, normal thrust $= 3344.82$ kN, radial shear $= 117.66$ kN.

Statement: a three-hinged parabolic arch carrying a UDL $w$ per unit horizontal length over the whole span has zero bending moment at every section.

Let the span be $L$ and the rise $f$, supports at the same level, origin at the left support.

Arch equation

$$y = \frac{4f\,x\,(L-x)}{L^2}$$

Reactions and thrust

By symmetry $R_A = R_B = \dfrac{wL}{2}$. The bending moment at the crown hinge is zero, so, taking moments of the left half about $C$:

$$H f = \frac{wL}{2}\times\frac{L}{2} - w\,\frac{L}{2}\times\frac{L}{4} = \frac{wL^2}{8} \Rightarrow H = \frac{wL^2}{8f}$$

Bending moment at a section at distance x

$$\begin{aligned} M_x &= R_A\,x - \frac{w\,x^2}{2} - H\,y \\ &= \frac{wLx}{2} - \frac{wx^2}{2} - \frac{wL^2}{8f}\cdot\frac{4f\,x\,(L-x)}{L^2} \\ &= \frac{w\,x\,(L-x)}{2} - \frac{w\,x\,(L-x)}{2} = 0 \end{aligned}$$

So $M_x = 0$ for every $x$. The beam bending moment $\dfrac{wx(L-x)}{2}$ is exactly cancelled by $H\,y$, because both are parabolic with the same shape.

Conclusion: the parabola is the funicular (equilibrium) shape for a UDL on the horizontal span. The arch is then in pure compression (thrust) with no bending and no radial shear.

The graphical method finds the reactions, then the resultant force on any section, from a force polygon and a linkage (funicular) polygon drawn to scale.

Steps

1. Draw the arch to a scale (span, rise, crown hinge $C$, supports $A$ and $B$) and mark the loads $P_1, P_2, \dots$ with their lines of action.
2. Force polygon: draw the load line $1\text{-}2\text{-}3\ldots$ to a force scale. Choose a pole $O$ and draw the rays $O1, O2, \dots$.
3. Reactions (using the hinge condition):
   • Combine the loads on the left of $C$ into their resultant $R_1$, and those on the right into $R_2$.
   • For $R_1$ alone, the right part $CB$ is unloaded, so the reaction at $B$ acts along $BC$. The line of $R_1$ meets line $BC$ at $E_1$; join $AE_1$, which is the direction of the reaction at $A$ for $R_1$. Close the triangle of forces.
   • Do the same for $R_2$ (reaction at $A$ along $AC$; its meeting point with $R_2$ joined to $B$).
   • Add the two sets of reactions vectorially. This gives $R_A$ and $R_B$ in magnitude and direction.
4. Linkage (thrust line) polygon: start from $A$ with a link parallel to the ray found from $R_A$ and continue the links parallel to the rays across each load. The polygon passes through $A$, $C$ and $B$ (this fixes the pole).
5. Bending moment at a section $D$: measure the vertical intercept $\delta$ between the arch axis and the linkage polygon at $D$. Then

$$M_D = H\times\delta$$

where $H$ is the pole distance (horizontal thrust) scaled from the force polygon. The sign is positive if the linkage polygon lies below the arch axis (arch in hogging) as per the usual convention, and negative otherwise. 6. Normal thrust and radial shear: read from the force polygon the resultant force $R_D$ on the section (the ray, or the sum of the reaction and the loads left of $D$). Draw the tangent to the arch at $D$. Resolve $R_D$:

$$N = R_D\cos\phi \ (\text{along tangent}), \qquad V = R_D\sin\phi \ (\text{perpendicular to tangent})$$

where $\phi$ is the angle between $R_D$ and the tangent.

   Space diagram                Force polygon
      P1   P2                     1 *----.
       |    |        C               |     '. O (pole)
    .--+----+--.   (hinge)         2 *------* 
   A     link polygon  B             |
   (through A, C, B)               3 *
   M = H x (vertical intercept)    H = pole distance

7. Check that the linkage polygon passes through all three points $A$, $C$, $B$ (the hinge condition) and that the force polygon closes.

Data: three-hinged symmetrical circular arch, $L = 12$ m, $f = 4$ m. Radius $R = \dfrac{L^2/4+f^2}{2f} = \dfrac{36+16}{8} = 6.5$ m (centre 2.5 m below the springing level). Section $D$ at $x = 3$ m from $A$:

$$y = \sqrt{6.5^2-(3-6)^2} - 2.5 = 3.266\ \text{m}, \qquad \tan\theta = \frac{6-3}{5.766} = 0.5203, \ \theta = 27.49^\circ$$

Influence lines (unit load at $x'$)

• Thrust: $H = \dfrac{x'}{2f} = \dfrac{x'}{8}$ for $x'\le 6$; $\dfrac{12-x'}{8}$ for $x'\ge 6$ (peak $0.75$ at the crown).
• Bending moment: $M = M_b - H\,y$, with $M_b = \dfrac{x'(12-3)}{12}$ for $x'\le 3$ and $\dfrac{3(12-x')}{12}$ for $x'\ge 3$.
• Radial shear: $V = V_b\cos\theta - H\sin\theta$ with $V_b = -\dfrac{x'}{12}$ (load left of $D$) or $\dfrac{12-x'}{12}$ (load right of $D$).
• Normal thrust: $N = V_b\sin\theta + H\cos\theta$.

The M-ILD is positive near $D$ and negative towards the crown; the V-ILD has a jump of $\cos\theta = 0.8871$ at $D$; the N-ILD has a jump of $\sin\theta = 0.4615$ at $D$ (shown by the two rows at 3 m).

Loading

UDL 20 kN/m on the left half ($x' = 0$ to 6 m) and a 40 kN point load at 3 m from the right support ($x' = 9$ m).

Each quantity $= 20\times(\text{area of ILD from 0 to 6}) + 40\times(\text{ordinate at 9 m})$. The areas are obtained by the trapezoidal rule on each straight segment (using the values immediately left and right of $D$):

$$\begin{aligned} M &= 14.02\ \text{kN m}\\ V &= 7.79\ \text{kN}\\ N &= 71.69\ \text{kN}\\ H &= 60\ \text{kN} \end{aligned}$$

Check (direct statics): $R_A = 100$ kN, $R_B = 60$ kN, $H = 60$ kN; $M = 100\times 3 - 20\times 3\times 1.5 - 60\times 3.266 = 14.02$ kN m.

Answer: bending moment $14.02$ kN m (sagging), radial shear $7.79$ kN, normal thrust $71.69$ kN at the section 3 m from the left support.

Data: three-hinged circular arch, span $L = 60$ m, rise $f = 5$ m (crown hinge $C$ at mid-span). UDL 25 kN/m on the left half ($x = 0$ to 30 m); 500 kN vertical load on the right half, 15 m from $B$ ($x = 45$ m). Section $D$ is 15 m horizontally from $A$.

Step 1: Equation of the arch

$$y = \sqrt{R^2-(x-30)^2} - (R-f),\quad R = \frac{L^2/4+f^2}{2f} = \frac{60^2/4+5^2}{2\times 5} = 92.5\ \text{m}$$ $$\tan\theta = \frac{dy}{dx} = \frac{30-x}{\sqrt{R^2-(x-30)^2}}$$

At the section $x = 15$ m: $y = 3.776$ m, $\tan\theta = 0.1643$, so $\theta = 9.33^\circ$, $\sin\theta = 0.1622$, $\cos\theta = 0.9868$.

Step 2: Vertical reactions

Taking moments about $A$ for the whole arch:

$$R_B \times 60 = (25\times 30)\times 15 + 500\times 45 = 33750 \Rightarrow R_B = 562.5\ \text{kN}$$ $$R_A = 1250 - 562.5 = 687.5\ \text{kN}$$

Step 3: Horizontal thrust

The bending moment at the crown hinge $C$ is zero. Taking moments of the forces on the left of $C$ about $C$ (rise $f = 5$ m):

$$H \times 5 = 687.5\times 30 - (25\times 30)\times 15 = 9375 \Rightarrow H = 1875\ \text{kN}$$

Step 4: Forces at the section (x = 15 m)

Simple-beam values at the section: $M_b = 7500$ kN m, $V_b = 312.5$ kN.

$$\begin{aligned} M &= M_b - H\,y = 7500 - 1875\times 3.776 = 420.6\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = 312.5\times 0.9868 - 1875\times 0.1622 = 4.31\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 312.5\times 0.1622 + 1875\times 0.9868 = 1900.86\ \text{kN} \end{aligned}$$

Answer: at $D$: bending moment $= 420.6$ kN m (sagging), radial shear $= 4.31$ kN, normal thrust $= 1900.86$ kN.

Data: three-hinged symmetrical circular arch, $L = 100$ m, $f = 10$ m. Radius $R = \dfrac{50^2+10^2}{2\times 10} = 130$ m. Rolling UDL 50 kN/m, length 25 m, moving left to right. Section at $x = 15$ m: $y_D = \sqrt{130^2-35^2} - 120 = 5.2$ m.

Maximum horizontal thrust

The ILD of $H$ is a triangle with peak $\dfrac{L}{4f} = 2.5$ at the crown. $H$ is maximum when the load is placed symmetrically about the crown (from 37.5 m to 62.5 m), where the ordinates are $1.875$ at the ends and $2.5$ at the middle:

$$\text{Area} = 2\times\frac{(1.875+2.5)}{2}\times 12.5 = 54.6875\ \text{m}, \qquad H_{max} = 50\times 54.6875 = 2734.4\ \text{kN}$$

ILDs for the section at x = 15 m

The ILD of the horizontal thrust is $H = \dfrac{x'}{2f}$ for a unit load at $x'$ left of the crown and $\dfrac{L-x'}{2f}$ right of it (peak $\dfrac{L}{4f} = 2.5$ at the crown). The ILD of the bending moment at the section is $M = M_b - H\,y$.

The ILD of $M$ crosses zero at the neutral point 36.59 m from $A$ ($\;$the vertical through the intersection of the line $A$–section (extended) and the line $B$–$C$). The ordinates are positive on one side of it and negative on the other.

Maximum positive bending moment

The UDL covers the region from x = 4.75 m to 29.75 m of the span (the whole load is on the span), which is the region of the ILD of the required sign.

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 50\times(145.66) = 7283\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times(21.5625) = 1078.13\ \text{kN} \end{aligned}$$

At the section: $y = 5.2$ m, $\theta = 15.62^\circ$, simple-beam shear $V_b = 521.88$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 521.88\times0.9631 - 1078.13\times0.2692 = 212.34\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 521.88\times0.2692 + 1078.13\times0.9631 = 1178.82\ \text{kN} \end{aligned}$$

Maximum negative bending moment

The UDL covers the region from x = 44.71 m to 69.71 m of the span (the whole load is on the span), which is the region of the ILD of the required sign.

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 50\times(-110.388) = -5519.4\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times(52.0883) = 2604.41\ \text{kN} \end{aligned}$$

At the section: $y = 5.2$ m, $\theta = 15.62^\circ$, simple-beam shear $V_b = 534.87$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 534.87\times0.9631 - 2604.41\times0.2692 = -186.06\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 534.87\times0.2692 + 2604.41\times0.9631 = 2652.25\ \text{kN} \end{aligned}$$

Absolute maximum bending moment

The absolute maximum is found by repeating the above for every section and every position of the UDL (the 25 m load fully on the span). The largest value occurs at a section about 22 m from the support, with the load covering about 7.9 m to 32.9 m of the span:

$$M_{abs} \approx 8061\ \text{kN m (sagging, at } x \approx 22.2\ \text{m})$$

Answer: $H_{max} = 2734.4$ kN; $M_{max}$ at 15 m $=+7283$ kN m (and $-5519$ kN m); absolute maximum bending moment $\approx 8061$ kN m.

Data: symmetrical circular three-hinged arch, $L = 100$ m, $f = 10$ m, $R = 130$ m. Two loads, 20 kN (leading) and 30 kN, 5 m apart, roll from left to right (the 30 kN load is 5 m behind the 20 kN load). Section at $x = 25$ m: $y = \sqrt{130^2-25^2} - 120 = 7.574$ m.

ILDs for the section at x = 25 m

Thrust ILD: $H = \dfrac{x'}{2f}$ (load left of the crown) or $\dfrac{L-x'}{2f}$ (right of the crown), peak $\dfrac{L}{4f} = 2.5$ at the crown. Moment ILD: $M = M_b - H\,y$.

The ILD of $M$ is zero at the neutral point, 39.77 m from $A$; it is positive on one side and negative on the other.

Maximum positive bending moment

The critical position is: 20 kN at 25 m, 30 kN at 20 m (measured from $A$).

$$M = \sum P\,y = 408.46\ \text{kN m}, \qquad H = \sum P\,y_H = 55\ \text{kN}$$

With $y = 7.574$ m, $\theta = 11.09^\circ$ and the simple-beam shear $V_b = 9$ kN at the section:

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = -1.74\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 55.7\ \text{kN} \end{aligned}$$

Maximum negative bending moment

The critical position is: 20 kN at 55 m, 30 kN at 50 m (measured from $A$).

$$M = \sum P\,y = -308.82\ \text{kN m}, \qquad H = \sum P\,y_H = 120\ \text{kN}$$

With $y = 7.574$ m, $\theta = 11.09^\circ$ and the simple-beam shear $V_b = 24$ kN at the section:

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 0.48\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 122.38\ \text{kN} \end{aligned}$$

Answer: maximum positive moment at 25 m $= +408.5$ kN m with $N = 55.7$ kN and $V = -1.74$ kN; maximum negative moment $= -308.8$ kN m with $N = 122.4$ kN and $V = 0.48$ kN.

Data: symmetrical circular three-hinged arch, $L = 50$ m, $f = 10$ m, $R = \dfrac{25^2+10^2}{2\times 10} = 36.25$ m. Rolling UDL 50 kN/m, 10 m long, from left to right. Section at $x = 15$ m: $y = \sqrt{36.25^2-10^2} - 26.25 = 8.593$ m.

ILDs for the section at x = 15 m

The ILD of the horizontal thrust is $H = \dfrac{x'}{2f}$ for a unit load at $x'$ left of the crown and $\dfrac{L-x'}{2f}$ right of it (peak $\dfrac{L}{4f} = 1.25$ at the crown). The ILD of the bending moment at the section is $M = M_b - H\,y$.

The ILD of $M$ crosses zero at the neutral point 20.56 m from $A$ ($\;$the vertical through the intersection of the line $A$–section (extended) and the line $B$–$C$). The ordinates are positive on one side of it and negative on the other.

Maximum positive bending moment

The UDL covers the region from x = 7.7 m to 17.7 m of the span (the whole load is on the span), which is the region of the ILD of the required sign.

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 50\times(30.687) = 1534.3\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times(6.35) = 317.5\ \text{kN} \end{aligned}$$

At the section: $y = 8.593$ m, $\theta = 16.01^\circ$, simple-beam shear $V_b = 8$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 8\times0.9612 - 317.5\times0.2759 = -79.9\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 8\times0.2759 + 317.5\times0.9612 = 307.39\ \text{kN} \end{aligned}$$

Maximum negative bending moment

The UDL covers the region from x = 23.49 m to 33.49 m of the span (the whole load is on the span), which is the region of the ILD of the required sign.

$$\begin{aligned} M &= w\times(\text{area of ILD}) = 50\times(-26.912) = -1345.6\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times(10.641) = 532.05\ \text{kN} \end{aligned}$$

At the section: $y = 8.593$ m, $\theta = 16.01^\circ$, simple-beam shear $V_b = 215.1$ kN.

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 215.1\times0.9612 - 532.05\times0.2759 = 59.98\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 215.1\times0.2759 + 532.05\times0.9612 = 570.74\ \text{kN} \end{aligned}$$

Answer: maximum positive bending moment $= +1534.3$ kN m (radial shear $-79.9$ kN, normal thrust $307.4$ kN); maximum negative bending moment $= -1345.6$ kN m (radial shear $59.98$ kN, normal thrust $570.7$ kN).

Data: parabolic three-hinged arch, $L = 90$ m, $f = 6$ m. 180 kN vertical load at 25 m from $A$; UDL 6 kN/m on the right half ($x = 45$ to 90 m). Section $D$ at $x = 30$ m from $A$.

Step 1: Equation of the arch

$$y = \frac{4f\,x\,(L-x)}{L^2} = \frac{4\times 6\,x\,(90-x)}{90^2}$$ $$\tan\theta = \frac{dy}{dx} = \frac{4f\,(L-2x)}{L^2}$$

At the section $x = 30$ m: $y = 5.333$ m, $\tan\theta = 0.0889$, so $\theta = 5.08^\circ$, $\sin\theta = 0.0885$, $\cos\theta = 0.9961$.

Step 2: Vertical reactions

Taking moments about $A$ for the whole arch:

$$R_B \times 90 = 180\times 25 + (6\times 45)\times 67.5 = 22725 \Rightarrow R_B = 252.5\ \text{kN}$$ $$R_A = 450 - 252.5 = 197.5\ \text{kN}$$

Step 3: Horizontal thrust

The bending moment at the crown hinge $C$ is zero. Taking moments of the forces on the left of $C$ about $C$ (rise $f = 6$ m):

$$H \times 6 = 197.5\times 45 - 180\times 20 = 5287.5 \Rightarrow H = 881.25\ \text{kN}$$

Step 4: Forces at the section (x = 30 m)

Simple-beam values at the section: $M_b = 5025$ kN m, $V_b = 17.5$ kN.

$$\begin{aligned} M &= M_b - H\,y = 5025 - 881.25\times 5.333 = 325\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = 17.5\times 0.9961 - 881.25\times 0.0885 = -60.59\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 17.5\times 0.0885 + 881.25\times 0.9961 = 879.34\ \text{kN} \end{aligned}$$

Answer: at $D$: bending moment $= 325$ kN m, radial shear $= -60.59$ kN, normal thrust $= 879.34$ kN.

Data and assumptions: parabolic three-hinged arch, span $L = 80$ m, rise $f = 8$ m, $y = \dfrac{4\times 8\,x\,(80-x)}{80^2} = 0.005\,x\,(80-x)$. The 250 kN inclined load acts at $x = 30$ m (10 m left of the crown), pointing downward and to the right at 30° to the horizontal (assumed). Its components: horizontal $250\cos30^\circ = 216.51$ kN (to the right), vertical $250\sin30^\circ = 125$ kN (downward). 100 kN vertical load at 15 m from $B$ ($x = 65$ m). Section $D$ at $x = 15$ m.

Heights of the loads: $y(30) = 7.5$ m, $y(65) = 4.875$ m; at $D$: $y = 4.875$ m, $\tan\theta = 0.005\,(80-30) = 0.25$, $\theta = 14.04^\circ$.

Let the thrusts be $H_A$ (to the right at $A$) and $H_B$ (to the left at $B$), and the vertical reactions $V_A$, $V_B$.

Reactions

Horizontal equilibrium:

$$H_A - H_B + 216.51 = 0 \Rightarrow H_B = H_A + 216.51$$

Moments about $A$:

$$V_B\times 80 = 100\times 65 + 125\times 30 + 216.51\times 7.5 \Rightarrow V_B = \frac{11873.8}{80} = 148.42\ \text{kN}$$ $$V_A = 225 - 148.42 = 76.58\ \text{kN}$$

Hinge condition (moment about $C$ of the forces on the left part, $C$ at $(40, 8)$):

$$V_A\times 40 - H_A\times 8 - 125\times 10 - 216.51\times(8-7.5) = 0$$ $$H_A = \frac{76.58\times 40 - 1250 - 216.51\times 0.5}{8} = 213.11\ \text{kN}, \qquad H_B = 429.61\ \text{kN}$$

Forces at D (left part, 15 m from A, no load on it)

$$\begin{aligned} M &= V_A\times 15 - H_A\times y_D = 76.58\times 15 - 213.11\times 4.875 = 109.77\ \text{kN m}\\ V &= V_A\cos\theta - H_A\sin\theta = 76.58(0.9701) - 213.11(0.2425) = 22.61\ \text{kN}\\ N &= V_A\sin\theta + H_A\cos\theta = 76.58(0.2425) + 213.11(0.9701) = 225.32\ \text{kN} \end{aligned}$$

Answer: at $D$: bending moment $= 109.77$ kN m (sagging), normal thrust $= 225.32$ kN, radial shear $= 22.61$ kN.

Data: three-hinged circular arch, $L = 120$ m, $f = 15$ m. Radius $R = \dfrac{60^2+15^2}{2\times 15} = 127.5$ m. Two loads, 8 kN (leading) and 12 kN, 10 m apart, roll from left to right (the 12 kN load is 10 m behind the 8 kN load). Section at $x = 30$ m: $y = \sqrt{127.5^2-30^2} - 112.5 = 11.42$ m.

ILDs for the section at x = 30 m

Thrust ILD: $H = \dfrac{x'}{2f}$ (load left of the crown) or $\dfrac{L-x'}{2f}$ (right of the crown), peak $\dfrac{L}{4f} = 2$ at the crown. Moment ILD: $M = M_b - H\,y$.

The ILD of $M$ is zero at the neutral point, 47.57 m from $A$; it is positive on one side and negative on the other.

Maximum positive bending moment

The critical position is: 8 kN at 30 m, 12 kN at 20 m (measured from $A$).

$$M = \sum P\,y = 177.27\ \text{kN m}, \qquad H = \sum P\,y_H = 16\ \text{kN}$$

With $y = 11.42$ m, $\theta = 13.61^\circ$ and the simple-beam shear $V_b = 4$ kN at the section:

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 0.12\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 16.49\ \text{kN} \end{aligned}$$

Maximum negative bending moment

The critical position is: 8 kN at 70 m, 12 kN at 60 m (measured from $A$).

$$M = \sum P\,y = -146.36\ \text{kN m}, \qquad H = \sum P\,y_H = 37.33\ \text{kN}$$

With $y = 11.42$ m, $\theta = 13.61^\circ$ and the simple-beam shear $V_b = 9.33$ kN at the section:

$$\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 0.29\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 38.48\ \text{kN} \end{aligned}$$

Answer: at 30 m from the left support: maximum positive bending moment $=+177.27$ kN m with normal thrust $16.49$ kN and radial shear $0.12$ kN; maximum negative bending moment $=-146.36$ kN m with normal thrust $38.48$ kN and radial shear $0.29$ kN.

Principle: a three-hinged arch has the crown hinge $C$ where the bending moment is zero. If a single load $P$ acts on one half, the other half carries no load and is therefore a two-force member: the reaction at its support acts along the line joining the support to the hinge $C$.

Steps (load P on the left half)

1. Draw the arch to scale and mark the line of action of $P$.
2. The right half $CB$ is unloaded, so the reaction $R_B$ acts along the line $BC$. Extend the line $BC$ until it meets the line of action of $P$ at the point $E$.
3. The three forces $P$, $R_A$ and $R_B$ are in equilibrium, so they are concurrent. Hence $R_A$ passes through $A$ and $E$: join $AE$.
4. Draw the force triangle: draw $P$ to scale (a vertical line $ab$), then from $a$ draw a line parallel to $R_A$ ($AE$) and from $b$ a line parallel to $R_B$ ($BC$). They meet at $c$.
5. Read the reactions: $R_A = ca$ in magnitude and direction, $R_B = bc$. The horizontal components are equal and opposite, which gives the thrust $H$; the vertical components give $V_A$ and $V_B$.

   Space diagram (P on left half)        Force triangle
                  | P
        E *-------+---------           a
         /|  \                         |\
     C  / |    \                      P | \ R_B (parallel BC)
      .'  |     '.                    |  \
   A '    |       ' B                 b---c
   R_A along AE;  R_B along BC           R_A (parallel AE)

6. If $P$ lies on the right half, interchange the roles: $R_A$ acts along $AC$, extend it to meet $P$ at $E'$, and $R_B$ passes through $B$ and $E'$.
7. If several loads act, apply the method for the loads of each half separately and add the reactions vectorially.

Check: the vertical components satisfy $V_A + V_B = P$ and the moment of the reactions about any point equals that of $P$; with the numerical values, $H = \dfrac{M_C}{f}$ as in the analytical method.

Arches are curved structures that carry loads mainly by compression (thrust), which makes them suitable for masonry, concrete and steel. They are classified in several ways.

1. According to the number of hinges

2. According to the shape of the axis

• Parabolic arch: ideal for uniformly distributed load (bridges, dams).
• Circular (segmental) arch: easy to construct (masonry arches, culverts).
• Elliptical arch: low rise over a long span (viaducts).
• Catenary arch: funicular for self-weight (gateways, arch dams).
• Pointed (Gothic) arch: used in buildings and masonry structures.

3. According to use in Civil Engineering structures

• Arch bridges: deck arch (roadway above the arch), through arch (roadway below, hung by hangers) and half-through arch.
• Culverts and tunnels: semicircular or horseshoe masonry/concrete arches carrying earth load.
• Arch dams: curved in plan to transfer water pressure to the valley sides.
• Building openings: arches over doors and windows (flat, segmental, semicircular).
• Roofs and halls: steel or timber arch trusses for sheds, hangars and stadiums.
• Aqueducts and viaducts: series of masonry arches.

4. According to support level

• Arches with supports at the same level and arches with supports at different levels (skew-back arches on hill sides).

Data: three-hinged symmetrical parabolic arch, span $L$, rise $f$, supports at the same level, origin at the left support. UDL $w$ per unit horizontal length over the whole span (if the load is given per m², multiply by the width of the arch to get $w$ per unit length). Self-weight neglected.

Arch equation and slope

$$y = \frac{4f\,x\,(L-x)}{L^2}, \qquad \tan\theta = \frac{dy}{dx} = \frac{4f\,(L-2x)}{L^2}$$

Reactions and thrust

$$R_A = R_B = \frac{wL}{2}, \qquad H = \frac{M_C}{f} = \frac{1}{f}\left[\frac{wL}{2}\cdot\frac{L}{2} - \frac{wL}{2}\cdot\frac{L}{4}\right] = \frac{wL^2}{8f}$$

Bending moment

$$M = \frac{wLx}{2} - \frac{wx^2}{2} - H\,y = \frac{w\,x\,(L-x)}{2} - \frac{wL^2}{8f}\cdot\frac{4f\,x\,(L-x)}{L^2} = 0$$

Radial shear

The simple-beam shear at $x$ is $V_b = w\left(\dfrac{L}{2} - x\right)$.

$$V = V_b\cos\theta - H\sin\theta = \cos\theta\left[w\left(\frac{L}{2}-x\right) - \frac{wL^2}{8f}\cdot\frac{4f\,(L-2x)}{L^2}\right] = \cos\theta\left[\frac{w\,(L-2x)}{2} - \frac{w\,(L-2x)}{2}\right] = 0$$

Normal thrust

$$N = V_b\sin\theta + H\cos\theta = H\tan\theta\sin\theta + H\cos\theta = \frac{H}{\cos\theta}$$ $$N = \frac{wL^2}{8f}\sqrt{1 + \frac{16f^2\,(L-2x)^2}{L^4}}$$

Results: $M = 0$, $V = 0$ and $N = \dfrac{H}{\cos\theta}$ at every section. $N$ is smallest at the crown ($N = H = \dfrac{wL^2}{8f}$) and largest at the springings, where $N = H\sqrt{1+\left(\dfrac{4f}{L}\right)^2}$.