Question
a.
\(l=\{\frac{eh^3}{12(1-μ^2 ) k_s}\}^\frac{1}{4}\)
b.
\(l=\{\frac{eh^3}{12(1-μ^4 ) k_s}\}^\frac{1}{4}\)
c.
\(l=\{\frac{eh^3}{12(1-μ^2 ) k_s}\}^\frac{1}{2}\)
d.
\(l=\{\frac{eh^3}{12(1-μ^2 ) k_s}\}^\frac{1}{3}\)
Posted under Pavement Design and Types Foundation Engineering
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