Determine the largest intensity wo of the distributed load that the beam can support if...

Considering moments about \(A\), we get

$$ \begin{aligned} \sum M_{A} &=0 \\ \left(R_{B} \times 12\right) &=\left(w_{0} \times 6 \times 3\right)+\left(2 \times w_{0} \times 6\right)(6+3) \\ 12 R_{B} &=18 w_{0}+108 w_{0} \\ R_{B} &=10.5 w_{0} \end{aligned} $$

By applying the equation of equilibrium for beam in \(y\)-direction:

$$ \begin{aligned} \sum F_{y} &=0 \\ R_{A}+R_{B} &=6 w_{0}+\left(6 \times 2 w_{0}\right) \\ R_{A} &=18 w_{0}-10.5 w_{0} \\ R_{A} &=7.5 w_{0} \end{aligned} $$

For \(A C\)

By applying the equation of equilibrium for beam in \(y\)-direction:

$$ \begin{aligned} \sum F_{y} &=0 \\ R_{A} &=\left(w_{0} \times x\right)+v \\ 7.5 w_{0} &=w_{0} x+v \\ v &=7.5 w_{0}-w_{0} x \end{aligned} $$

Considering moments, we get

$$ \begin{aligned} \sum M_{A} &=0 \\ M &=\left(w_{0} \times x \times \frac{x}{2}\right)+(v \times x) \\ M &=\frac{w_{0} x^{2}}{2}+7.5 w_{0} x-w_{0} x^{2} \\ M &=7.5 w_{0} x-\frac{w_{0} x^{2}}{2} \end{aligned} $$

For \(C B\)

By applying the equation of equilibrium for beam in \(y\)-direction:

$$ \begin{aligned} \sum F_{y} &=0 \\ v &=2 w_{0}(12-x)-R_{B} \\ v &=24 w_{0}-2 w_{0} x-10.5 w_{0} \\ v &=13.5 w_{0}-2 w_{0} x \end{aligned} $$

Considering moments, we get

$$ \begin{aligned} &\sum M=0 \\ &M=R_{B}(12-x)-2 w_{0}(12-x) \frac{(12-x)}{2} \\ &M=10.5(12-x)-w_{0}(12-x)^{2} \end{aligned} $$

By comparing the shear force values, the maximum value is at \(B\), and its value is given as 1200

lb. Hence

$$ \begin{aligned} v_{\max } &=10.5 w_{0} \\ 1200 &=10.5 w_{0} \\ w_{0}=& 114.286 \mathrm{lb} \ldots \ldots \end{aligned} $$

The bending moment reaches maximum value at zero shear force. Now from the shear force function in \(C B\) region, the zero shear force is located at

$$ \begin{aligned} &v=13.5 w_{0}-2 w_{0} x \\ &0=13.5(114.286)-2(114.286) x \\ &x=6.75 \mathrm{ft} \end{aligned} $$

Now the maximum moment value is

$$ \begin{aligned} &M_{\max }=-w_{0}(12-6.75)^{2}+10.5(12-6.75) \\ &M_{\max }=-27.5625 w_{0}+55.125 w_{0} \\ &M_{\max }=27.5625 w_{0} \end{aligned} $$

But given that magnitude of maximum moment is 600 lb-ft. Hence

$$ \begin{gathered} M_{\max }=27.5625 w_{0} \\ 600=27.5625 w_{0} \\ w_{0}=21.7687 \mathrm{lb} \ldots \ldots (2) \end{gathered} $$

By comparing the values of distributed load from equations (1) and (2), the largest is \(w_{0}=21.7687 \mathrm{lb}\)