Horizontal Reaction in Three Hinged Arch

A three-hinged arch is a statically determinate structure. It has hinges at the two supports and one at the crown (usually at the highest point). When a three-hinged arch is subjected to a load, the supports exert reactions. These reactions typically have both vertical and horizontal components. The horizontal component of the reaction at the support is known as the horizontal thrust or horizontal reaction.

For a three-hinged arch spanning a distance 'L' and having a rise 'h', subjected to a uniformly distributed load (u.d.l.) 'W' per unit length over the entire span, the horizontal reaction at each support can be determined by considering the equilibrium of the arch. Since there is a hinge at the crown, the bending moment at the crown hinge is zero.

Calculation of Horizontal Reaction

Consider a three-hinged arch with supports at A and B and a hinge at the crown C. The span is L and the rise is h. A u.d.l. of intensity W acts over the entire span.

Let \( R_A \) and \( R_B \) be the reactions at supports A and B respectively. Each reaction can be resolved into vertical components \( V_A \) and \( V_B \) and horizontal components \( H_A \) and \( H_B \). For a symmetrical arch and symmetrical loading (u.d.l. over the entire span), the vertical reactions are equal, and the horizontal reactions are equal and opposite.

Thus, \( V_A = V_B = \frac{\text{Total Load}}{2} = \frac{W \times L}{2} \).

And \( H_A = H_B = H \).

To find the horizontal reaction \( H \), we take moments about the hinge C from one side (say, from support A). The coordinates of C are usually taken as \( (L/2, h) \) if A is at \( (0, 0) \).

The forces to the left of C are:

• Vertical reaction \( V_A \) acting upwards at A.
• Horizontal reaction \( H \) acting inwards at A.
• Uniformly distributed load \( W \) over the span from A to C (a length of \( L/2 \)). The total load over this section is \( W \times (L/2) \) acting at the midpoint of AC's horizontal projection, i.e., at a horizontal distance of \( (L/2)/2 = L/4 \) from A.

The moment at the crown hinge C is zero. Taking moments about C:

\( \Sigma M_C = 0 \)

Moment due to \( V_A \) about C: \( V_A \times (L/2) \) (clockwise)

Moment due to \( H \) about C: \( H \times h \) (anticlockwise)

Moment due to u.d.l. (W) over \( L/2 \) about C: \( (W \times L/2) \times (L/4) \) (anticlockwise)

Setting the sum of moments to zero:

\( V_A \times (L/2) - H \times h - (W \times L/2) \times (L/4) = 0 \)

Substitute \( V_A = WL/2 \):

\( (WL/2) \times (L/2) - H \times h - (WL/2) \times (L/4) = 0 \)

\( WL^2/4 - Hh - WL^2/8 = 0 \)

Rearranging the terms to solve for H:

\( Hh = WL^2/4 - WL^2/8 \)

\( Hh = \frac{2WL^2 - WL^2}{8} \)

\( Hh = \frac{WL^2}{8} \)

\( H = \frac{WL^2}{8h} \)

Thus, the horizontal reaction at each support for a three-hinged arch under u.d.l. over the entire span is \( \frac{WL^2}{8h} \).

Comparing with Options

Let's compare our derived formula with the given options:

• Option 1: \( WL^2/8h \)
• Option 2: \( WL^2/10h \)
• Option 3: \( WL^2/6h \)
• Option 4: \( WL^2/4h \)

Our derived formula \( H = \frac{WL^2}{8h} \) matches Option 1.

Therefore, the horizontal reaction at each support is given by \( WL^2/8h \).