A three-hinged arch is a statically determinate structure. This means that its reactions and internal forces can be determined using only the equations of static equilibrium. A three-hinged arch has three hinges: typically one at each support and one at the crown (the highest point of the arch).

For a three-hinged arch with supports A and B at the same level and a hinge at the crown C, the three equilibrium equations are:

• Sum of horizontal forces is zero: $\sum F_x = 0$
• Sum of vertical forces is zero: $\sum F_y = 0$
• Sum of moments about any point is zero: $\sum M = 0$

Additionally, because of the hinge at the crown C, the bending moment at C is zero. This provides an extra equation, making the structure determinate.

Consider the given three-hinged arch with span $2l$ and rise $h$. A point load $W$ is acting at a distance $a$ from the left support. Let the left support be A and the right support be B. Let the crown hinge be C.

The horizontal reaction at the supports ($H_A$ and $H_B$) are equal in magnitude and opposite in direction, assuming no horizontal applied loads other than reactions. Let's denote the horizontal reaction as $H$. By $\sum F_x = 0$, $H_A = H_B = H$ (assuming outward positive).

We can determine the vertical reactions ($V_A$ and $V_B$) by considering the overall equilibrium of the arch, treating it like a simply supported beam with span $2l$ carrying the load $W$ at distance $a$ from A.

• $\sum M_A = 0$: $V_B \times (2l) - W \times a = 0$
• This gives $V_B = \frac{Wa}{2l}$
• $\sum F_y = 0$: $V_A + V_B - W = 0$
• Substituting $V_B$: $V_A + \frac{Wa}{2l} - W = 0$
• This gives $V_A = W - \frac{Wa}{2l} = \frac{W(2l-a)}{2l}$

To find the horizontal reaction $H$, we use the condition that the bending moment at the crown hinge C is zero. The crown C is typically at the mid-span, so its horizontal distance from A is $l$, and its height (rise) from the support level is $h$.

Consider the left section of the arch (from A to C). The external forces acting on this section are the reactions $V_A$ and $H_A$ at A, and the load $W$ if $a \le l$. The bending moment at C due to these forces must be zero.

The moment of the vertical reaction $V_A$ about C is $V_A \times l$ (clockwise). The moment of the horizontal reaction $H_A$ about C is $H_A \times h$ (anti-clockwise). The moment of the load $W$ (if $a \le l$) about C is $W \times (l-a)$ (clockwise).

Setting the sum of moments about C for the left section to zero (taking anti-clockwise moments as positive):

$-V_A \times l + H_A \times h - W \times (l-a) = 0$ (This equation is valid if the load $W$ is between A and C, i.e., $a \le l$)

Substitute the value of $V_A$ into this equation:

$-\left(\frac{W(2l-a)}{2l}\right) \times l + H \times h - W(l-a) = 0$

$-\frac{W(2l-a)}{2} + Hh - W(l-a) = 0$

$-Wl + \frac{Wa}{2} + Hh - Wl + Wa = 0$

$Hh = Wl - \frac{Wa}{2} + Wl - Wa$

$Hh = 2Wl - \frac{3Wa}{2}$

Wait, this result does not match the options. Let's re-check the moment calculation about C, or consider an alternative method using the bending moment in an equivalent simply supported beam.

The bending moment at any point in a three-hinged arch is equal to the bending moment at the corresponding point in a simply supported beam of the same span, minus the moment caused by the horizontal thrust acting on the arch's rise at that point. Let $M_{beam}(x)$ be the bending moment at a distance $x$ from the left support in an equivalent simply supported beam of span $2l$ with load $W$ at distance $a$. Let $y(x)$ be the rise of the arch at distance $x$. The bending moment in the arch $M_{arch}(x)$ is $M_{beam}(x) - H y(x)$. Since the moment at the crown C (at $x=l$) is zero, $M_{arch}(l) = 0$.

$M_{beam}(l) - H y(l) = 0$

At the crown, $x=l$, and the rise $y(l) = h$. So, $M_{beam}(l) - Hh = 0$, which means $H = \frac{M_{beam}(l)}{h}$.

Now, let's calculate the bending moment at $x=l$ for the simply supported beam of span $2l$ with load $W$ at distance $a$. The vertical reactions are $V_A^{beam} = \frac{W(2l-a)}{2l}$ and $V_B^{beam} = \frac{Wa}{2l}$.

If $a \le l$ (load is on the left half): The bending moment at $x=l$ is the moment of $V_A^{beam}$ about $x=l$ minus the moment of $W$ about $x=l$.

$M_{beam}(l) = V_A^{beam} \times l - W \times (l-a)$

$M_{beam}(l) = \frac{W(2l-a)}{2l} \times l - W(l-a)$

$M_{beam}(l) = \frac{W(2l-a)}{2} - W(l-a)$

$M_{beam}(l) = Wl - \frac{Wa}{2} - Wl + Wa$

$M_{beam}(l) = \frac{Wa}{2}$

Now substitute this into the equation for $H$:

$H = \frac{M_{beam}(l)}{h} = \frac{Wa/2}{h} = \frac{Wa}{2h}$

This matches one of the options. This result is valid when the load is located at a distance $a \le l$ from the left support. If the load were on the right half ($a > l$), the bending moment at $x=l$ in the equivalent beam would be calculated differently, leading to a different expression for $H$. However, since $\frac{Wa}{2h}$ is provided as an option, it is implied that the intended scenario or the expected formula corresponds to the load being on the left half ($a \le l$).

Thus, the horizontal reaction at the support is $\frac{Wa}{2h}$ when the point load $W$ is acting at a distance $a$ from the left support, assuming $a \le l$.

Revision Table: Three Hinged Arch Analysis

Additional Information: Three Hinged Arches

Three-hinged arches are simple yet important structural elements. They are often used for long spans like bridges and large roofs. Their static determinacy simplifies the analysis compared to two-hinged or fixed arches.

• The horizontal thrust $H$ is a key force in arch structures. It provides resistance to the outward spreading tendency caused by vertical loads.
• The shape of a three-hinged arch can vary (e.g., parabolic, circular), but the principle of zero moment at the crown hinge still holds, allowing for determinate analysis.
• For a parabolic arch with supports at the same level and crown at midspan, the equation is typically $y = \frac{4h}{L^2} x(L-x)$, where $L$ is the total span (here $L=2l$). Substituting $x=l$, $y=h$, we get $h = \frac{4h}{(2l)^2} l(2l-l) = \frac{4h}{4l^2} l^2 = h$, which is consistent.
• The formula $H = \frac{M_{beam}(l)}{h}$ is a general and powerful way to find the horizontal thrust in a three-hinged arch where the crown is at x=l, provided $M_{beam}(l)$ is the moment at the crown location in the corresponding simply supported beam.