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Replace the loading by an equivalent resultant force and couple moment acting at point O.

Units Used:

kN = { 10 }^{ 3 } N

Given:

{ w }_{ 1 } = 7.5 \frac { kN }{ m }

{ w }_{ 2 } = 20 \frac { kN }{ m }

a = 3 m

b = 3 m

c = 4.5 m

Step-by-step

{ F }_{ R } = \frac { 1 }{ 2 } \left( { w }_{ 2 }-{ w }_{ 1 } \right) c+{ w }_{ 1 }c+{ w }_{ 1 }b+\frac { 1 }{ 2 } { w }_{ 1 }a

{ F }_{ R } = 95.6 kN

{ M }_{ Ro } = -\frac { 1 }{ 2 } \left( { w }_{ 2 }-{ w }_{ 1 } \right) c\frac { c }{ 3 } -{ w }_{ 1 }c\frac { c }{ 2 } -{ w }_{ 1 }b\left( c+\frac { b }{ 2 } \right) -\frac { 1 }{ 2 } { w }_{ 1 }a\left( b+c+\frac { a }{ 3 } \right) \quad \quad \quad \quad { M }_{ Ro } = -349 kN.m

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