The correct answer is $\boxed{\frac{{100}}{{{\text{E}}{I_{\text{c}}}}}}$.
The horizontal deflection of a parabolic curved beam of span 10 m and rise 3 m when loaded with a uniformly distributed load $l$ t per horizontal length is given by the following formula:
$$\delta = \frac{{50}}{{{\text{E}}{I_{\text{c}}}}}$$
where $E$ is the Young’s modulus of the beam material and $I_c$ is the moment of inertia of the beam cross-section at the crown.
The moment of inertia of a parabolic beam cross-section at the crown is given by the following formula:
$$I_c = \frac{{1}{4}}{{b^3}}$$
where $b$ is the width of the beam.
The Young’s modulus of steel is approximately $200 \times {10^9}$ Pa.
The width of a typical steel beam is approximately 0.3 m.
Therefore, the horizontal deflection of
42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z"/> Subscribe on YouTube