Replace the loading by an equivalent resultant force and specify its location on the beam,measured from point $B$ .

Units Used:

$kip$ = ${ 10 }^{ 3 }$ $Ib$

Given:

${ w }_{ 1 }$ = $800$ $\frac { Ib }{ ft }$

${ w }_{ 2 }$ = $500$ $\frac { Ib }{ ft }$

$a$ = $12$ $ft$

$b$ = $9$ $ft$

Question

Replace the loading by an equivalent resultant force and specify its location on the beam,measured from point [latex]B[/latex].

Units Used:

[latex]kip[/latex] = [latex]{ 10 }^{ 3 }[/latex] [latex]Ib[/latex]

Given:

[latex]{ w }_{ 1 }[/latex] = [latex]800[/latex] [latex]\frac { Ib }{ ft }[/latex]

[latex]{ w }_{ 2 }[/latex] = [latex]500[/latex] [latex]\frac { Ib }{ ft }[/latex]

[latex]a[/latex] = [latex]12[/latex] [latex]ft[/latex]

[latex]b[/latex] = [latex]9[/latex] [latex]ft[/latex]

Accepted Answer

[latex]{ F }_{ R }[/latex] = [latex]\frac { 1 }{ 2 } a{ w }_{ 1 }+\frac { 1 }{ 2 } ({ w }_{ 1 }-{ w }_{ 2 })b+{ w }_{ 2 }b\quad \quad \quad\quad\quad\quad[/latex] [latex]{ F }_{ R }[/latex] = [latex]10.65[/latex] [latex]kip[/latex]

[latex]{ F }_{ R }x[/latex] = [latex]-\frac { 1 }{ 2 } a{ w }_{ 1 }\frac { a }{ 3 } +\frac { 1 }{ 2 } \left( { w }_{ 1 }-{ w }_{ 2 } \right) b\frac { b }{ 3 } +{ w }_{ 2 }b\frac { b }{ 2 }[/latex]

[latex]x[/latex] = [latex]\frac { -\frac { 1 }{ 2 } a{ w }_{ 1 }\frac { a }{ 3 } +\frac { 1 }{ 2 } \left( { w }_{ 1 }-{ w }_{ 2 } \right) b\frac { b }{ 3 } +{ w }_{ 2 }b\frac { b }{ 2 } }{ { F }_{ R } } \quad \quad \quad x[/latex] = [latex]0.479[/latex] [latex]ft[/latex]

[latex]\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x[/latex] = [latex]0.479[/latex] [latex]ft[/latex]

( [latex]to[/latex] [latex]the[/latex] [latex]right[/latex] [latex]of[/latex] [latex]B[/latex] )

Question : Replace the loading by an equivalent resultant force and spe...

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