## Question:

The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location $(x, y)$ on the slab.

Units Used:

$kN$ = ${ 10 }^{ 3 }$ $N$

Given:

${ F }_{ 1 }$ = $30$ $kN\quad a$ = $3$ $m$

${ F }_{ 2 }$ = $40$ $kN\quad b$ = $8$ $m$

${ F }_{ 3 }$ = $20$ $kN\quad c$ = $2$ $m$

${ F }_{ 4 }$ = $50$ $kN\quad d$ = $6$ $m$

$\quad \quad \quad\quad\quad\quad e$ = $4$ $m$

## Step-by-step

$+\uparrow { F }_{ R }$ = $\sum { { F }_{ x } }$ ;$\quad { F }_{ R }={ F }_{ 1 }{ +F }_{ 2 }{ +F }_{ 3 }{ +F }_{ 4 }$

$\quad \quad \quad \quad \quad \quad\quad { F }_{ R }$ = $140$ $kN$

${ M }_{ Rx }$ = $\sum { { M }_{ X } }$; $-{ F }_{ R }(y)$ = $-\left( { F }_{ 4 } \right) (a)$ $-\left[ ({ F }_{ 1 })(a+b) \right] –$ $\left[ ({ F }_{ 2 })(a+b+c) \right]$

$\quad \quad \quad \quad \quad \quad y$ = $\frac { { F }_{ 4 }a+{ F }_{ 1 }a+{ F }_{ 1 }b+{ F }_{ 2 }a+{ F }_{ 2 }b+{ F }_{ 2 }c }{ { F }_{ R } }$

$\quad \quad \quad \quad \quad \quad y$ = $7.14$ $m$

${ M }_{ Ry }$ = $\sum { { M }_{ y } }$; $\quad \left( { F }_{ R } \right) x$ = $\left( { F }_{ 4 } \right) \left( e \right) +\left( { F }_{ 3 } \right) \left( d+e \right) +\left( { F }_{ 2 } \right) \left( b+c \right)$

$\quad \quad \quad \quad \quad \quad x$ = $\frac { { F }_{ 4 }e+{ F }_{ 3 }d+{ F }_{ 3 }e+{ F }_{ 2 }b{ F }_{ 2 }c }{ { F }_{ R } }$

$\quad \quad \quad \quad \quad \quad x$ = $5.71$ $m$