SSC JE Civil Engineering 30-Oct-2020 Shift 1 Paper

Question : 111 of 200

Marks: +1, -0

A three-hinged parabolic arch of span 20 m and rise 4 m carries a concentrated load of 150 kN at 4 m from left support 'A'. Calculate the vertical reaction and the horizontal thrust, respectively, at support 'A'.

$\V_\A = 40 \kN$ and $\H_\A = 80 \kN$
$\V_\A = 75 \kN$ and $\H_\A = 120 \kN$
$\V_\A = 80 \kN$ and $\H_\A = 50 \kN$
$\V_\A = 120 \kN$ and $\H_\A = 75 \kN$

Validate

Solution:

\Σ\V = 0, \V_\A + \V_\B = 150 \kN

\Σ\H = 0, \H_\A = \H_\B = \H

At support

\B

\Σ\M_\B = 0

\V_\A × 20 - 150 × 16 = 0

\V_\A = 120 \kN

\V_\B = 150 - 120 = 30 \kN

At central hinge,

\Σ\M_\C = 0

\V_\A × 10 - \H × 4 - 150 × 6 = 0

\H × 4 = 120 × 10 - 150 × 6

\H = 75 \kN

So, the vertical reaction at A and the horizontal thrust are 120 kN and 75kN respectively.

Go to Question:

Prev Question

Next Question