Chapter 4.9, Problem 37FP

Determine the resultant force and specify where it acts on the beam measured from A.

To determine

The resultant force and the distance of resultant force that acts on the beam measured from A.

Answer to Problem 37FP

The resultant force $F_R$ is $\underset{\_}{40.5\text{kN}(\downarrow )}$ and the distance (d) of resultant force that acts on the beam measured from A is $\underset{\_}{1.25\text{m}}$.

Explanation of Solution

Given information:

• The intensity of uniformly varying load $w_1$ is $6.0\text{kN}/\text{m}$.
• The intensity of uniformly varying load $w_1$ is $9.0\text{kN}/\text{m}$.
• The intensity of uniformly varying load $w_1$ is $3.0\text{kN}/\text{m}$.

Explanation:

Assumption: Upward is positive.

Determine the resultant force using the equation of equilibrium.

$\begin{array}{l}-F_R={\displaystyle \sum F_y}\\ -F_R=-\left(w_1\times l_1\right)-\left(w_2\times l_2\right)-\left(w_3\times l_3\right)\end{array}$ (I)

Here, the length of $w_1$ is $l_1$, the length of $w_2$ is $l_2$, and the length of $w_3$ is $l_3$.

Assumption: Anticlockwise is positive.

Determine the distance of resultant force which acts on the beam measured from A by taking the moment about A.

$\begin{array}{l}{\left(M_R\right)}_A={\displaystyle \sum M_A}\\ -F_R\left(d\right)=\left(w_1\times l_1\times \frac{l_1}{2}\right)-\left(w_2\times l_2\times \frac{l_2}{2}\right)-\left(w_3\times l_3\times \left(\frac{l_3}{2}+3.0\right)\right)\end{array}$ (II)

Show the calculation of resultant force and the distance as follows:

Conclusion:

Substitute $6.0\text{kN}/\text{m}$ for $w_1$, 1.5 m for $l_1$, $9.0\text{kN}/\text{m}$ for $w_1$, 3.0 m for $l_2$, $3.0\text{kN}/\text{m}$ for $w_1$, and 1.5 m for $l_3$ in Equation (I).

$\begin{array}{l}-F_R=-\left(6.0\times 1.5\right)-\left(9.0\times 3.0\right)-\left(3.0\times 1.5\right)\\ F_R=40.5\text{kN}(\downarrow )\end{array}$

Thus, the resultant force $F_R$ is $\underset{\_}{40.5\text{kN}(\downarrow )}$.

Substitute 40.5 kN for $F_R$, $6.0\text{kN}/\text{m}$ for $w_1$, 1.5 m for $l_1$, $9.0\text{kN}/\text{m}$ for $w_1$, 3.0 m for $l_2$, $3.0\text{kN}/\text{m}$ for $w_1$, and 1.5 m for $l_3$ in Equation (II).

$\begin{array}{l}-40.5\left(d\right)=\left(6.0\times 1.5\times \frac{1.5}{2}\right)-\left(9.0\times 3.0\times \frac{3.0}{2}\right)-\left(3.0\times 1.5\times \left(\frac{1.5}{2}+3.0\right)\right)\\ d=\frac{50.625}{40.5}\\ d=1.25\text{m}\end{array}$

Thus, the distance of resultant force which acts on the beam measured from A is $\underset{\_}{1.25\text{m}}$.