Step By Step Solution
Step 1
We have the given diameter $350\;{\rm{mm}}$, length $2\;{\rm{m}}$, and the density of the concrete $2.45\;{{{\rm{Mg}}} \mathord{\left/ {\vphantom {{{\rm{Mg}}} {{{\rm{m}}^3}}}} \right. } {{{\rm{m}}^3}}}$.
We are required to determine the weight of column in pounds.
Step 2
Convert the unit of diameter $\left( d \right)$ into SI units.\[\begin{array}{c} d = 350\;{\rm{mm}}\\ = 350\;{\rm{mm}} \times \left( {\frac{{{{10}^{ – 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)\\ = 0.350\;{\rm{m}} \end{array}\]
Step 3
The equation for the volume of the column is given by,\[{V_c} = \frac{\pi }{4}{d^2}l\]
Here, $l$ is the length of the column.
Step 4
Substitute all the numeric values in the above equation.\[\begin{array}{c} {V_c} = \frac{\pi }{4}{\left( {0.350\;{\rm{m}}} \right)^2} \times 2\;{\rm{m}}\\ = 0.192\;{{\rm{m}}^3} \end{array}\]
Step 5
Convert the unit of density $\left( {{\rho _c}} \right)$ into SI units.\[\begin{array}{c} {\rho _c} = 2.45\;{{{\rm{Mg}}} \mathord{\left/ {\vphantom {{{\rm{Mg}}} {{{\rm{m}}^3}}}} \right. } {{{\rm{m}}^3}}}\\ = 2.45\;\left( {\frac{{{\rm{Mg}}}}{{{{\rm{m}}^3}}}} \right) \times \left( {\frac{{{{10}^3}\;{\rm{kg}}}}{{1\;{\rm{Mg}}}}} \right)\\ = 2.45 \times {10^3}\;{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right. } {{{\rm{m}}^3}}} \end{array}\]
Step 6
The equation for the mass of the column is given by,\[\begin{array}{c} {\rho _c} = \frac{{{m_c}}}{{{V_c}}}\\ {m_c} = {\rho _c} \times {V_c} \end{array}\]
Here, ${\rho _c}$ is the density of the concrete.
Step 7
Substitute all the numeric values in the above equation.\[\begin{array}{c} {m_c} = 2.45 \times {10^3}\;\left( {\frac{{{\rm{kg}}}}{{{{\rm{m}}^3}}}} \right) \times 0.192\;{{\rm{m}}^3}\\ = 4.70 \times {10^2}\;{\rm{kg}} \end{array}\]
Step 8
The equation for the weight of the concrete column is given by,\[{w_c} = {m_c}g\]
Here, $g$ is the gravitational acceleration, having a standard value of $9.81\;{\rm{m/}}{{\rm{s}}^2}$.
Step 9
Substitute all the numeric values in the above equation.\[\begin{array}{c} {w_c} = 4.70 \times {10^2} \times 9.81\\ = 4.61 \times {10^3}\;{\rm{N}}\\ = 4.61 \times {10^3}\;{\rm{N}} \times \left( {\frac{{0.2248\;{\rm{lbf}}}}{{1\;{\rm{N}}}}} \right)\\ = 1.03 \times {10^3}\;{\rm{lbf}} \end{array}\]
Step 9
Now, convert the units in kilo pounds.\[\begin{array}{c} {w_c} = 1.03 \times {10^3}\;{\rm{lbf}}\\ = 1.03 \times {10^3}\;{\rm{lbf}} \times \left( {\frac{{1\;{\rm{klbf}}}}{{{{10}^3}\;{\rm{lbf}}}}} \right)\\ = 1.03\;{\rm{klbf}} \end{array}\]