problem: determine the force & moment reactions at a for the beam which is subjected to the distributed load…

Explanation:

Step1: Find the total force due to the distributed - load

The distributed - load is given by (w = kx^{1.5}), and when (x = 16) ft, (w=400) lb/ft. So, (400=k(16)^{1.5}). First, calculate ((16)^{1.5}=16^{\frac{3}{2}}=\sqrt{16^{3}} = 64). Then (k=\frac{400}{64}=\frac{25}{4}) lb/ft². The total force (F) due to the distributed - load is (F=\int_{0}^{16}w\mathrm{d}x=\int_{0}^{16}kx^{1.5}\mathrm{d}x). Substitute (k = \frac{25}{4}) into the integral: (F=\frac{25}{4}\int_{0}^{16}x^{1.5}\mathrm{d}x). Using the power - rule for integration (\int x^{n}\mathrm{d}x=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)), we have (F=\frac{25}{4}\times\frac{x^{2.5}}{2.5}\big|{0}^{16}=\frac{25}{4}\times\frac{2}{5}x^{2.5}\big|{0}^{16}= \frac{5}{2}x^{2.5}\big|_{0}^{16}). [ \begin{align*} F&=\frac{5}{2}(16)^{2.5}\ &=\frac{5}{2}\times1024\ & = 2560\text{ lb} \end{align*} ]

Step2: Find the location of the centroid of the distributed - load

The centroid (x_{c}) of the distributed - load is given by (x_{c}=\frac{\int_{0}^{16}x\cdot w\mathrm{d}x}{\int_{0}^{16}w\mathrm{d}x}). We know that (\int_{0}^{16}w\mathrm{d}x = 2560) lb. And (\int_{0}^{16}x\cdot w\mathrm{d}x=\int_{0}^{16}x\cdot kx^{1.5}\mathrm{d}x=k\int_{0}^{16}x^{2.5}\mathrm{d}x). Substitute (k=\frac{25}{4}) into the integral: (\int_{0}^{16}x\cdot kx^{1.5}\mathrm{d}x=\frac{25}{4}\int_{0}^{16}x^{2.5}\mathrm{d}x). Using the power - rule for integration (\int x^{n}\mathrm{d}x=\frac{x^{n + 1}}{n + 1}+C) ((n\neq-1)), we have (\frac{25}{4}\times\frac{x^{3.5}}{3.5}\big|{0}^{16}=\frac{25}{4}\times\frac{2}{7}x^{3.5}\big|{0}^{16}=\frac{25}{14}x^{3.5}\big|{0}^{16}). [ \begin{align*} \int{0}^{16}x\cdot kx^{1.5}\mathrm{d}x&=\frac{25}{14}(16)^{3.5}\ &=\frac{25}{14}\times16384\ &=\frac{25\times16384}{14}\ &=\frac{409600}{14} \end{align*} ] Then (x_{c}=\frac{\int_{0}^{16}x\cdot w\mathrm{d}x}{\int_{0}^{16}w\mathrm{d}x}=\frac{\frac{409600}{14}}{2560}=\frac{409600}{14\times2560}=\frac{409600}{35840}=11.43) ft.

Step3: Calculate the reactions at A

For the vertical reaction (A_y) at A, by the equilibrium of vertical forces (\sum F_y = 0), (A_y=F = 2560) lb (upward). For the moment reaction (M_A) at A, by the equilibrium of moments about A (\sum M_A=0), (M_A=F\times x_{c}=2560\times11.43 = 29260.8) lb - ft (counter - clockwise).

Answer:

The vertical force reaction at A is (A_y = 2560) lb (upward) and the moment reaction at A is (M_A=29260.8) lb - ft (counter - clockwise)