A two-hinged semi-circular arch of radius 12 m is subjected to ri

SOLUTION

Concept:

Horizontal thrust due to rise of temperature is given by:

\({\rm{H}} = \frac{{4{\rm{EI\alpha T}}}}{{{\rm{\pi \;}}{{\rm{R}}^2}}}\)

E = Modulus of Elasticity of the material

I = Moment of Inertia of the arch section

α = Coefficient of thermal expansion

T = Rise in temperature

R = Radius of the arch

Calculation:

As, E, I, α, π remains same

\({\rm{H}} \propto \frac{{\rm{T}}}{{{{\rm{R}}^2}}}\)

\({{\rm{H}}_1} = \frac{{{{\rm{T}}_1}}}{{{\rm{R}}_1^2}}\)                                …(i)

\({{\rm{H}}_2} = \frac{{{{\rm{T}}_2}}}{{{\rm{R}}_2^2}}\)                                …(ii)

R₁ = 12 m

H₂ = 0.50 H₁

R₂ = 9 m

From (i) and (ii)

\(\frac{{{{\rm{H}}_1}}}{{{{\rm{H}}_2}}} = \frac{{{{\rm{T}}_1}{\rm{R}}_2^2}}{{{{\rm{T}}_2}{\rm{R}}_1^2}}\)

\(\frac{{{{\rm{H}}_1}}}{{0.50{\rm{\;}}{{\rm{H}}_1}}} = \frac{{50 \times {9^2}}}{{{{\rm{T}}_2} \times {{12}^2}}}\)