The horizontal component of reaction developed at support A in th

SOLUTION

Concept:

A three-hinged arch is a statically determinate structure with two supports hinged and an internal hinge provided in the body of the arch generally at the crown. Such arches are analyzed using equations of static equilibrium only.

Calculation:

Let the reactions developed be as shown below:

For horizontal equilibrium, ∑ F_x = 0

⇒ H_A = H_B     ...1)

For vertical equilibrium, ∑F_y = 0

⇒ R_A + R_B = 2 × 6 + 20 = 32 kN – m     ...2)

∵ There is a hinge at C, bending moment at C must be zero.

Bending moment at C due to forces on left of C = 0

\(\Rightarrow {H_A} \times 6 - {R_A} \times 6 + 2 \times 6 \times \frac{6}{2} = 0\)

⇒ H_A + R_A = 6     ...3)

Also, bending moment at C due to forces on the right of C = 0

⇒ 20 × 6 + H_B × 8 - R_B × 10 = 0

⇒ 8 H_B + 10 R_B = 120     ...4)

Using equation (1), (2), (3) and (4)

R_A = 13.78, R_B = 18.22 kN

∴ H_A = 6 – 13.78 = -7.78 kN

& H_B = H_A = -7.78 kN

The negative value indicates opposite direction from that shown in the diagram.