Step 1

We are given the distributed load $w = \left( {{x^2} + 3x + 100} \right)\;{{{\rm{lb}}} \mathord{\left/ {\vphantom {{{\rm{lb}}} {{\rm{ft}}}}} \right. } {{\rm{ft}}}}$ and the span of the beam is $15\;{\rm{ft}}$ as shown in the figure.

We are asked to determine the equivalent resultant force and its location on the beam measured from point A.

 
Step 2

To represent the differential force $\left( {d{F_R}} \right)$ acting on the beam at a distance $dx$, we need to draw the diagram of the beam as:

To calculate the differential force acting on the beam, we have:

\[d{F_R} = wdx\]

Here, w is the distributed load on the beam.

Substitute the values in the above expression, we get:

\[d{F_R} = \left( {{x^2} + 3x + 100} \right)\;{{{\rm{lb}}} \mathord{\left/ {\vphantom {{{\rm{lb}}} {{\rm{ft}}}}} \right. } {{\rm{ft}}}} \cdot dx\]

To calculate the equivalent resultant force of the distributed load, we need to integrate the differential force from the limit $0$ to $15$ as:

\[\begin{array}{c} {F_R} = \int {d{F_R}} \\ {F_R} = \int\limits_0^{15} {\left( {{x^2} + 3x + 100} \right) \cdot dx} \\ {F_R} = \left[ {\frac{{{x^3}}}{3} + \frac{{3{x^2}}}{2} + 100x} \right]_0^{15} \end{array}\]

Substitute the limit in x, we get:

\[\begin{array}{c} {F_R} = \left[ {\frac{{{{\left( {15} \right)}^3}}}{3} + \frac{{3 \times {{\left( {15} \right)}^2}}}{2} + 100 \times \left( {15} \right) - 0} \right]\\ {F_R} = 2962.5\;{\rm{lb}} \end{array}\]
 
Step 3

To calculate the location x of the resultant force acting from point A, we have:

\[x = \frac{{\int {xdA} }}{{\int\limits_A {dA} }}\]

Substitute the values in the above expression, we get:

\[\begin{array}{c} x = \frac{{\int\limits_0^{15} {x\left( {{x^2} + 3x + 100} \right)} \cdot dx}}{{\int\limits_A {dA} }}\\ x = \frac{{\int\limits_0^{15} {\left( {{x^3} + 3{x^2} + 100x} \right)} \cdot dx}}{{\int\limits_A {dA} }}\\ x = \frac{{\left[ {\frac{{{x^4}}}{4} + {x^3} + 50{x^2}} \right]_0^{15}}}{{2962.5\;{\rm{lb}}}} \end{array}\]

Substitute the limit in x, we get:

\[\begin{array}{c} x = \frac{{\left[ {\frac{{{{\left( {15} \right)}^4}}}{4} + {{\left( {15} \right)}^3} + 50 \times {{\left( {15} \right)}^2} - 0} \right]}}{{2962.5\;{\rm{lb}}}}\\ x = \frac{{\left( {27281.25\;{\rm{lb}} \cdot {\rm{ft}}} \right)}}{{\left( {2962.5\;{\rm{lb}}} \right)}}\\ x = 9.21\;{\rm{ft}} \end{array}\]