Question

(II) Assume that the supports of the uniform cantilever shown in Fig. 9–76 \(m{\bf{ = 2900}}\;{\bf{kg}}\) are made of wood. Calculate the minimum cross-sectional area required of each, assuming a safety factor of 9.0.

Short answer

The cross-sectional area of support A is \(1.6 \times {10^{ - 3}}\;{{\rm{m}}^{\rm{2}}}\).

The cross-sectional area of support B is \(8.0 \times {10^{ - 3}}\;{{\rm{m}}^{\rm{2}}}\).

Step-by-step solution

Concepts

You know that stress is the ratio of force and cross-sectional area.

For this problem, first, you have to find the forces acting on the supports.

Then, using the formula\({\bf{tensile strength = safety factor \times stress}}\), you get the cross-sectional area.

Explanation

The mass of the cantilever is \(m = 2900\;{\rm{kg}}\).

The safety factor is 9.0.

Let \({A_{\rm{A}}}\) and \({A_{\rm{B}}}\) be the cross-sectional areas of the two supports.

Calculation of the force on the support

If the cantilever is in equilibrium, the torque about the left support is zero. Then,

\(\begin{array}{c}{F_{\rm{B}}} \times \left( {20.0\;m} \right) - mg \times \left( {25.0\;m} \right) = 0\\{F_{\rm{B}}} \times \left( {20.0\;m} \right) = mg \times \left( {25.0\;m} \right)\\{F_{\rm{B}}} \times \left( {20.0\;m} \right) = \left( {2900\;{\rm{kg}}} \right) \times \left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {25.0\;m} \right)\\{F_{\rm{B}}} = 35525\;{\rm{N}}\end{array}\).

In equilibrium, the net vertical force is also zero. Then,

\(\begin{array}{c}{F_{\rm{A}}} + {F_{\rm{B}}} = mg\\{F_{\rm{A}}} + 35525\;{\rm{N = }}\left( {2900\;{\rm{kg}}} \right) \times \left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\{F_{\rm{A}}} = - 7105\;{\rm{N}}\end{array}\).

The negative value of the force on support A means that the force on support A acts downward.

Calculation of cross-sectional area

Now, the stress on support A is

\(\begin{array}{c}{\rm{tensile strength}} = {\rm{safety factor}} \times {\rm{stress}}\\\frac{{{\rm{tensile strength}}}}{{{\rm{safety factor}}}} = \frac{{\left| {{F_{\rm{A}}}} \right|}}{{{A_{\rm{A}}}}}\\{A_{\rm{A}}} = \left| {{F_{\rm{A}}}} \right|\left( {\frac{{{\rm{safety factor}}}}{{{\rm{tensile strength}}}}} \right)\end{array}\).

Substituting the values in the above equation,

\(\begin{array}{c}{A_{\rm{A}}} = \left( {7105\;{\rm{N}}} \right) \times \left( {\frac{{{\rm{9}}{\rm{.0}}}}{{{\rm{40}} \times {\rm{1}}{{\rm{0}}^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}}} \right)\\ = 1.6 \times {10^{ - 3}}\;{{\rm{m}}^{\rm{2}}}\end{array}\).

Hence, the cross-sectional area of support A is \(1.6 \times {10^{ - 3}}\;{{\rm{m}}^{\rm{2}}}\).

Now, the stress on support B is

\(\begin{array}{c}{\rm{tensile strength}} = {\rm{safety factor}} \times {\rm{stress}}\\\frac{{{\rm{tensile strength}}}}{{{\rm{safety factor}}}} = \frac{{\left| {{F_{\rm{B}}}} \right|}}{{{A_{\rm{B}}}}}\\{A_{\rm{B}}} = \left| {{F_{\rm{B}}}} \right|\left( {\frac{{{\rm{safety factor}}}}{{{\rm{tensile strength}}}}} \right)\end{array}\).

Substituting the values in the above equation,

\(\begin{array}{c}{A_{\rm{A}}} = \left( {35525\;{\rm{N}}} \right) \times \left( {\frac{{{\rm{9}}{\rm{.0}}}}{{{\rm{40}} \times {\rm{1}}{{\rm{0}}^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}}} \right)\\ = 8.0 \times {10^{ - 3}}\;{{\rm{m}}^{\rm{2}}}\end{array}\).

Hence, the cross-sectional area of support B is \(8.0 \times {10^{ - 3}}\;{{\rm{m}}^{\rm{2}}}\).