fbpx

Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member AB, measured from A.

Given:

{ w }_{ 1 } = 200 \frac { N }{ m }

{ w }_{ 2 } = 100 \frac { N }{ m }

{ w }_{ 3 } = 200 \frac { N }{ m }

a = 5 m

b = 6 m

Step-by-step

{ F }_{ Rx } = -{ w }_{ 3 }a\quad \quad \quad \quad \quad \quad \quad \quad\quad\quad\quad\quad { F }_{ Rx } = -1000 N

{ F }_{ Ry } = \frac { -1 }{ 2 } \left( { w }_{ 1 }+{ w }_{ 2 } \right) b\quad \quad \quad \quad \quad \quad \quad \quad { F }_{ Ry } = -900 N

-y{ F }_{ Rx } = { w }_{ 3 }a\frac { a }{ 2 } -{ w }_{ 2 }b\frac { b }{ 2 } -\frac { 1 }{ 2 } \left( { w }_{ 1 }-{ w }_{ 2 } \right) b\frac { b }{ 3 }

y = \frac { { w }_{ 3 }a\frac { a }{ 2 } -{ w }_{ 2 }b\frac { b }{ 2 } -\frac { 1 }{ 2 } \left( { w }_{ 1 }-{ w }_{ 2 } \right) b\frac { b }{ 3 } }{ -{ F }_{ Rx } } \quad \quad \quad \quad \quad \quad \quad   y = 0.1 m

Trending

Online
Courses

Private
Teachers

Hire Our
Professionals

Study
Materials

Writing & Translation

View All
Services