HW_5 Solutions - ft kip w kip w V MAX 2 8 4 = = = Check...

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First Find support reactions. ) 15 ( 20 ) 45 ( 20 ) 30 ( ft kip ft kip ft A M Y B - + - = 0 ) 15 )( 30 ( / 4 = + ft ft ft kip kip A Y 80 = kip kip B kip F Y Y 20 20 80 - - + = 0 ) 30 ( / 4 = - ft ft kip kip B Y 80 = Shear Diagram shown to right Need to find magnitudes of moments Use symmetry. At support kipft M ft kip M M 300 0 ) 15 ( 20 - = = + = Also find center of beam due to symmetry kipft M ft ft ft kip ft kip ft kip M M 150 0 ) 5 . 7 )( 15 ( / 4 ) 15 ( 80 ) 30 ( 20 = = + - + =
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First Find support reactions. ) 3 )( 6 ( 6 6 3 1 ) 6 ( 2 1 ft ft w ft ft ft w M B + + = 0 ) 6 ( = - ft A Y w A Y 7 = w B ft w ft w F Y Y 7 ) 6 ( ) 6 ( 2 1 + + - - = w B Y 2 = Shear Diagram shown to right Find distance when shear is zero, will be local max moment ft x w xft w ft w F Y 4 0 7 ) ( ) 6 ( 2 1 = = + - - = Max Shear at A magnitude 4w
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Unformatted text preview: ft kip w kip w V MAX / 2 8 4 = = = Check Moments: The two magnitudes are w M M ft ft w ft w ft ft ft w M w M M ft ft w M A A 2 ) 2 )( 4 ( ) 4 ( 7 4 6 3 1 ) 6 ( 2 1 6 6 3 1 ) 6 ( 2 1 = = + +- + =-= = + = ∑ ∑ Max Moment failure will need load of. ft kip w ft fip w M / 5 / 30 6 = = = Since less load is needed for shear max load is based off shear. ft kip w / 2 =...
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HW_5 Solutions - ft kip w kip w V MAX 2 8 4 = = = Check...

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