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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 

 

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