Question

A concrete building is built of slabs 12 m long\({\rm{(15^\circ C)}}\). How wide should the expansion cracks between the slabs the at \({\rm{(15^\circ C)}}\) to prevent buckling if the range of temperature is \( - 30{\rm{^\circ C}}\)to \(50^\circ {\rm{C}}\)

Short answer

The cracks between the slabs should be \(1.15\,\,{\rm{cm}}\) to prevent buckling.

Step-by-step solution

Meaning of change in length

The change in length of material depends on the change in temperature and the original length of the material. The coefficient of thermal expansion of a material is different for different materials.

\(\Delta L = {L_{\rm{0}}}\alpha \left( {{T_{\rm{f}}} - {T_{\rm{i}}}} \right)\)

Here, \(\Delta L\)is the expansion of the slab; \({L_0}\)is the initial length of the slab; \({T_{\rm{f}}}\)is the final temperature of the slab; \({T_{\rm{i}}}\) is the initial temperature of the slab; \(\alpha \) is coefficient of thermal expansion.

Given information

The original length, \({L_0} = 12\,\,{\rm{m}}\).

The initial temperature of material, \({T_{\rm{i}}} = - 30{\rm{^\circ C}}\).

The final temperature of material, \({T_{\rm{f}}} = 50{\rm{^\circ C}}\).

Calculation of expansion of slab

The expansion of the slab to prevent buckling is calculated as follows:

\(\begin{aligned}{c}\Delta L &= \left( {12\,\,{\rm{m}}} \right)\left( {{{12 \times {{10}^{ - 6}}} \mathord{\left/{\vphantom {{12 \times {{10}^{ - 6}}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)\left( {50{\rm{^\circ C}} - \left( { - 30{\rm{^\circ C}}} \right)} \right)\\ &= {{144 \times {{10}^{ - 6}}\,{\rm{m}}} \mathord{\left/{\vphantom {{144 \times {{10}^{ - 6}}\,{\rm{m}}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}\left( {80{\rm{^\circ C}}} \right)\\ &= 0.0115\,\,{\rm{m}}\left( {\frac{{{\rm{100}}\,\,{\rm{cm}}}}{{{\rm{1}}\,\,\,{\rm{m}}}}} \right)\\ &= 1.15\,\,{\rm{cm}}\end{aligned}\)

Hence, the cracks between the slabs should be \(1.15\,\,{\rm{cm}}\)to prevent buckling.