91Ó°ÊÓ

The compressive strength may be described as the greatest compressive pressure that an object can sustain earlier to failure, divided by the object's cross-sectional area.

Given data:

The mass of the roof, \(m = 13600\;{\rm{kg}}\).

The area of the column per support, \(a = 4.0\;{\rm{cm}} \times 9.0\;{\rm{cm}}\).

The safety factor, \(N = 12\).

The number of supports required on each side can be calculated by using the following expression.

\(n = \frac{{mgN}}{{a{\sigma _{\rm{c}}}}}\)

Here, \({\sigma _{\rm{c}}}\) is the compressive stress on each column, and its value is \(35 \times {10^6}\;{\rm{Pa}}\).

Substitute the values in the above equation.

\(\begin{array}{c}n = \frac{{\left( {13600\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {12} \right)}}{{\left[ {\left( {4.0\;{\rm{cm}} \times 9.0\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}\;{{\rm{m}}^{\rm{2}}}}}{{{\rm{1}}\;{\rm{c}}{{\rm{m}}^{\rm{2}}}}}} \right)} \right]\left( {35 \times {{10}^6}\;{\rm{Pa}}} \right)}}\\n = 12.69 \approx 12\end{array}\)

Thus, the number of supports required is \(12\).

Since there are to be more than 12 supports and have the same number of supports on each side, there will be 14 or seven supports on each side. That means there will be six support-to-support spans, each of which would be given by spacing:

\(\begin{array}{l}D = \frac{{10\;{\rm{m}}}}{{6\;{\rm{gaps}}}}\\D = 1.66\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{\rm{gap}}}}} \right.} {{\rm{gap}}}}\end{array}\)

Thus, the gap between each support is \(1.66\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{\rm{gap}}}}} \right.} {{\rm{gap}}}}\).