Question

set w1 = 2.5 kip/ft and w2 = 2 kip/ft (figure 1). part b specify the location of the equivalent resultant force on the beam measured from point a. express your answer in feet to three significant figures.

Explanation:

Step1: Calculate the resultant of the triangular load

The triangular load has a base \(b = 4.5\) ft and height \(w_1=2.5\) kip/ft. The resultant of a triangular load \(F_1\) is given by \(F_1=\frac{1}{2}\times b\times w_1\). So \(F_1=\frac{1}{2}\times4.5\times2.5 = 5.625\) kip, and its line - of - action is at \(\frac{1}{3}\) of the base from the zero - load end, i.e., \(d_1=\frac{1}{3}\times4.5 = 1.5\) ft from the right - hand end of the triangular load part.

Step2: Calculate the resultant of the rectangular load

The rectangular load has a length \(L = 6\) ft and intensity \(w_2 = 2\) kip/ft. The resultant of the rectangular load \(F_2\) is \(F_2=w_2\times L=2\times6 = 12\) kip, and its line - of - action is at the mid - point of the rectangle, i.e., \(d_2 = 4.5+\frac{6}{2}=7.5\) ft from point \(A\).

Step3: Calculate the total resultant force \(F_R\)

\(F_R=F_1 + F_2=5.625+12=17.625\) kip.

Step4: Use the moment - equilibrium about point \(A\) to find the location of the resultant

Let the distance of the resultant force from point \(A\) be \(d\). The moment of the triangular load about \(A\) is \(M_1=F_1\times(4.5 - 1.5)=5.625\times3 = 16.875\) kip - ft. The moment of the rectangular load about \(A\) is \(M_2=F_2\times7.5=12\times7.5 = 90\) kip - ft.
We know that \(F_R\times d=M_1 + M_2\). So \(d=\frac{M_1 + M_2}{F_R}=\frac{16.875+90}{17.625}=\frac{106.875}{17.625}\approx6.06\) ft.

Answer:

\(6.06\) ft