Question

part d for the left figure below, replace the distributed loads by an equivalent resultant force and a couple moment acting at point a. (see the right figure below.) let a = 3.15 m, w1 = 5.50 kn/m, and w2 = 4.70 kn/m. calculate the resultant force, fr, and the couple moment, mr,a. dont forget to include the appropriate signs (consistent with the right - figure) with your numerical answers. express your answers numerically in kilonewtons and kilonewton - meters to three significant figures separated by a comma. view available hint(s) fr, mr,a =

Explanation:

Step1: Calculate resultant force of triangular - load

The resultant force of a triangular load with maximum intensity $w_1$ over a length $a$ is $F_{1}=\frac{1}{2}w_1a$. Substituting $w_1 = 5.50\ kN/m$ and $a = 3.15\ m$, we get $F_{1}=\frac{1}{2}\times5.50\times3.15=8.6625\ kN$. The resultant force of a rectangular - load with intensity $w_2$ over a length $a$ is $F_{2}=w_2a$. Substituting $w_2 = 4.70\ kN/m$ and $a = 3.15\ m$, we get $F_{2}=4.70\times3.15 = 14.705\ kN$. The total resultant force $F_R=F_1 + F_2$. So, $F_R=8.6625+14.705=23.3675\ kN\approx23.4\ kN$.

Step2: Calculate the couple - moment about point A

The moment due to the triangular load about point A: The line of action of the resultant of the triangular load is at $\frac{1}{3}a$ from the end with zero load intensity. So, $M_1=F_1\times\frac{1}{3}a$. Substituting $F_1 = 8.6625\ kN$ and $a = 3.15\ m$, we get $M_1=8.6625\times\frac{1}{3}\times3.15 = 9.095625\ kN\cdot m$. The moment due to the rectangular load about point A: The line of action of the resultant of the rectangular load is at $\frac{a}{2}$ from point A. So, $M_2=F_2\times\frac{a}{2}$. Substituting $F_2 = 14.705\ kN$ and $a = 3.15\ m$, we get $M_2=14.705\times\frac{3.15}{2}=23.113875\ kN\cdot m$. The total couple - moment $M_{R,A}=M_1+M_2$. So, $M_{R,A}=9.095625 + 23.113875=32.2095\ kN\cdot m\approx32.2\ kN\cdot m$.

Answer:

$23.4\ kN,32.2\ kN\cdot m$