Question

A three-hinged arch of span \(8\,\mathrm{m}\) carries a central point load of \(32\,\mathrm{kN}\). If the central rise is \(2\,\mathrm{m}\), the horizontal thrust is

• A. \(12\,\mathrm{kN}\)
• B. \(16\,\mathrm{kN}\)
• C. \(24\,\mathrm{kN}\)
• D. \(32\,\mathrm{kN}\)

Hint:

For a three-hinged arch carrying a central point load, \[ \boxed{H=\frac{WL}{4h}.} \]

Answer 1

The Correct Option is D

Step 1: Find the support reactions.
The arch is symmetric and the load sits right at midspan, so each support carries half the load: $V_A = V_B = \dfrac{W}{2} = \dfrac{32}{2} = 16\,\mathrm{kN}$.
Step 2: Take moments about the crown hinge.
Cut the arch at the central hinge and look at the left half only. A hinge cannot carry moment, so the bending moment there is zero. The point load itself acts right at the hinge, so it contributes no moment about that point, leaving only $V_A$ and the horizontal thrust $H$: \[ V_A\left(\frac{L}{2}\right) - Hh = 0. \]
Step 3: Solve for the horizontal thrust.
\[ H = \frac{V_A (L/2)}{h} = \frac{16 \times 4}{2} = 32\,\mathrm{kN}. \]
\[ \boxed{32\,\mathrm{kN}} \]