We accept a coordinate system at any point, for example at point A. We mark characteristic points.
Forces do not act in directions parallel to the coordinate system axes, so we need to project them onto the horizontal and vertical directions.

Force vectors

\begin{aligned} &\overline{F_i}=|F_i|\cdot\frac{\overline{r_i}}{|r_i|}\\ &\overline{r_i} - \text{vector specifying the direction of the force } \overline{P_i}\\ \end{aligned}

\begin{aligned} &\overline{F_1}=2\sqrt{2}\cdot\frac{(2,2,0)}{\sqrt{2^2+2^2}}=\frac{2\sqrt{2}(4,5,0)}{2\sqrt{2}}=(2,2,0) \ kN\\ &\overline{F_2}=|F_2|\cdot\frac{\overline{BD}}{|BD|}\\ &\overline{F_2}=3\sqrt{2}\cdot\frac{(-2,2,0)}{\sqrt{4+4}}=(-3,3,0) \ kN\\ &\overline{F_3}=4\sqrt{13}\cdot\frac{(-3,-2,0)}{\sqrt{9+4}}=(-12,-8,0) \ kN\\ &\overline{F_6}=17\sqrt{34}\cdot\frac{(3,5,0)}{\sqrt{9+25}}=(51,85,0) \ kN\\ &\overline{F_7}=10\sqrt{29}\cdot\frac{(-2,5,0)}{\sqrt{4+25}}=(-20,50,0) \ kN\\ \\ &\text{Resultant force vector: } \overline{S}=(18,132,0)\ kN\\ \end{aligned}

a) Reduction at point A

\begin{aligned} &\overline{M_1}=\overline{F_1}\times \overline{DA}=&(2,2,0)\\ &&\underline{(0,-7,0)}\\ &&(0,0,-14)\\ \\ &\overline{M_2}=\overline{F_2}\times \overline{BA}=&(-3,3,0)\\ &&\underline{(-2,-5,0)}\\ &&(0,0,21) \\ \\ &\overline{M_3}=\overline{F_3}\times \overline{EA}=&(-12,-8,0)\\ &&\underline{(-5,-7,0)}\\ &&(0,0,44) \\ \\ &\overline{M_4}=&(0,0,-200)\\ \\ &\overline{M_5}=&(0,0,9)\\ \\ &\overline{M_6}=\overline{F_6}\times \overline{CA}=&(51,85,0)\\ &&\underline{(-2,0,0)}\\ &&(0,0,170) \\ \\ &\overline{M_7}=\overline{F_7}\times \overline{CA}=&(-20,50,0)\\ &&\underline{(-2,0,0)}\\ &&(0,0,100) \\ \\ &\overline{M_A}=\sum_{i=1}^n{M_i}\\ &&\overline{M_A}=(0,0,130)\\ \end{aligned}

b) Reduction at point B

\begin{aligned} &\overline{M_1}=\overline{F_1}\times \overline{DB}=&(2,2,0)\\ &&\underline{(2,-2,0)}\\ &&(0,0,-8)\\ \\ &\overline{M_2}=\overline{F_2}\times \overline{BB}=&(0,0,0)\\ \\ &\overline{M_3}=\overline{F_3}\times \overline{EB}=&(0,0,0)\\ \\ &\overline{M_4}=&(0,0,-200)\\ \\ &\overline{M_5}=&(0,0,9)\\ \\ &\overline{M_6}=\overline{F_6}\times \overline{CB}=&(51,85,0)\\ &&\underline{(0,5,0)}\\ &&(0,0,255) \\ \\ &\overline{M_7}=\overline{F_7}\times \overline{CB}=&(-20,50,0)\\ &&\underline{(0,5,0)}\\ &&(0,0,-100) \\ \\ &\overline{M_B}=\sum_{i=1}^n{M_i}\\ &&\overline{M_B}=(0,0,-44)\\ \end{aligned}

Verification - from the theorem about changing the pole of the moment:

\begin{aligned} &\overline{M_B}=\overline{M_A}+\overline{S}\times\overline{AB}=&(0,0,130)+&(18,132,0)\\ &&&\underline{(2,5,0)}\\ &&&(0,0,-174)\\ &\overline{M_B}=(0,0,130)+(0,0,-174)\\ &\overline{M_B}=(0,0,-44)\\ &\text{the check gives a positive result}\\ \end{aligned}

c) reduction to the simplest form

\begin{aligned} &\text{since } S\ne 0, M_A\ne 0 \ \text{the system is reduced to the resultant force}\\ &\text{We need to transfer the resultant force to a line, which we call the central axis}\\ &\text{We will now find the equation of this line:}\\ &\text{from the theorem about changing the pole of the moment } \end{aligned}

\begin{aligned} &\overline{M_O}=\overline{M_A}+\overline{S}\times\overline{AO}\\ &\text{where: } &M_O - \text{zero moment}\\ &&O=(x,y,0)- \text{point with unknown coordinates}\\ &&A=(0,0,0) - \text{point at the beginning of the coordinate system} \end{aligned}

\begin{aligned} &\overline{AO}=(x-0,y-0,0-0)=(x,y,0)\\ &\overline{M_O}=\overline{M_A}+\overline{S}\times\overline{AO}=&(0,0,130)+&(18,132,0)\\ &&&\underline{(x,y,0)}\\ &(0,0,0)=(0,0,130)+(0,0,18y-132x)\\ &0=130+18y-132x\\ &y=7.333x-7.222\\ \end{aligned}

\begin{aligned} &\text{To draw the line, we determined any two points that belong to the line}\\ &\text{for } x=2, y=7.444\\ &\text{for } x=0, y=\frac{7.222}{7.333}=0.985\\ \end{aligned}

Statics - Planar force system

Engineering mechanics - Statics - Planar force system - Example 5

Example 0