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The building slab is subjected to four parallel column loadings. Determine { F }_{ 1 } and { F }_{ 2 } if the resultant force acts through point (12 m, 10 m).

Step-by-step

Equivalent Resultant Force. Sum the forces along z axis by referring to Fig. a,

+\uparrow  { ({ F }_{ R }) }_{ z }  =  \sum { F}_{ z } ;              –{ F }_{ R } = –{ F }_{ 1 }{ F }_{ 2 } – 12 – 6                  { F }_{ R } = { F }_{ 1 } + { F }_{ 2 } + 18

Location of the Resultant Force. Sum the moments about the x and y axes by
referring to Fig. a,

{ ({ M }_{ R }) }_{ x }\sum { M}_{ x } ;              -({ F }_{ 1 } + { F }_{ 2 } + 18)(10) = -12(8) – 6(20) – { F }_{ 2 }(20)

10{ F }_{ 1 } – 10{ F }_{ 2 } = 36

{ ({ M }_{ R }) }_{ y }\sum { M}_{ y } ;              ({ F }_{ 1 } + { F }_{ 2 } + 18)(12) =  12(6) + 6(26) + { F }_{ 1 }(22)

12{ F }_{ 2 } – 10{ F }_{ 1 } = 12

Solving Eqs (1) and (2),

{ F }_{ 1 } = 27.6 kN { F }_{ 2 } = 24.0 kN

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