Question

A symmetrical two-hinged parabolic arch rib has a span of 48 m between the abutment pins at the same level and a central rise of 10 m. Moment of inertia at any section follows the law $I=I_0\sec \theta .$ If a load of 80 kN crosses the span then maximum horizontal thrust at the hinges will be ______ kN.

• A. 131.25

Answer 1

The correct option is A 131.25


ILD for horizontal reaction at supports
$\therefore$ $H=\frac{25}{128}\times \frac{84}{10}\times 80=131.25kN$