# X f f x 0 f m a y 121010 208206 3020 a y 10 kn y f 10

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0 x F F x =0 0 F M -A y ×12+10×10-20×8+20×6 +30×2=0 A y =10 KN 0 y F 10-10+20-20-30+F y =0 F y =30 KN Segment AB 2 0 x 0 y F 10-V=0 V=10 KN 0 M M-10×x=0 M=10x Segment BC 4 2 x 0 y F 10-10-V=0 V=0 0 M M-10x+10(x-2)=0 M=20 KN.m Segment CD 6 4 x 0 y F A KN 10 C m 4 m 8 KN 20 KN 20 KN 30 B D E F m 4 m 2 m 2 A KN 10 C m 4 m 8 KN 20 KN 20 KN 30 B D E F m 4 m 2 m 2 y A y F x F KN 10 x V M KN 10

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7 10-10+20-V=0 V=20 KN 0 M M-10x+10(x-2)-20(x-4)=0 M=20(x-3) Segment DE 10 6 x 0 y F 10-10+20-20-V=0 V=0 0 M M-10x+10(x-2)-20(x-4)+20(x-6)=0 M=60 KN.m Segment EF 12 10 x 0 y F 10-10+20-20-30-V=0 V=-30 KN 0 M M-10x+10(x-2)-20(x-4)+20(x-6) +30(x-10)=0 M=30(12-x) KN 10 x V M KN 10 KN 20 KN 10 x V M KN 10 KN 20 KN 20 KN 10 x V M KN 10 KN 20 KN 20 KN 30
8 S.F Diagram B.M. Diagram A KN 10 C m 4 m 8 KN 20 KN 20 KN 30 B D E F m 4 m 2 m 2 x F KN 10 KN 30 KN 10 KN 20 KN 30 m KN . 20 m KN . 60

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9 Example 35: Draw the shear and moment diagrams for the beam shown below. 0 x F A x =0 0 B M wL 2 L -A y L=0 A y = 2 wL 0 y F 2 wL +B y -wL=0 B y = 2 wL 0 y F 2 wL -wx-V=0 V=-w(x- 2 L ) 0 M M- 2 wL x+wx( 2 x )=0 M= 2 w (xL-x 2 ) A B L w A B L wL y A y B x A 2 wL x V M wx 2 / x w Maximum moment occur when 0 dx dM 2 0 2 0 ) 2 ( 2 L x x L x L w dx dM Location of maximum moment
0 S.F Diagram B.M Diagram A B L w 2 wL 2 wL 2 wL 2 / L 2 / L 8 2 max wL M

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1 Example 36: Draw the shear and moment diagrams for the beam shown below. 0 x F A x =0 0 y F A y = 2 L w o 0 A M M A - 2 L w o L 3 2 =0 M A = 3 2 L w o 0 y F 2 L w o - L x w o 2 2 -V=0 V= 2 o w (L- L x 2 ) V max = 2 L w o 0 M M+ 3 2 L w o - 2 L w o x+ L x w o 2 2 x 3 1 =0 A B L o w L 3 2 L 3 1 2 L w o x A y A A M L x w o 2 2 3 2 L w o 2 L w o x L x w w o 3 / x V M Maximum shear force occur at 0 dx dV 0 L x w dx dV o x=0
2 M= L w o 6 (3L 2 x-x 3 -2L 3 ) M max = S.F Diagram L o w 3 2 L w o 2 L w o 2 L w o 3 2 L w o 3 2 L w o

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3 Example 37: The horizontal beam AD is loaded by a uniform distributed load of 5 KN per meter of length and is also subjected to the concentrated force of 10 KN applied as shown below. Determine the shearing force and bending moment diagrams. 0 x F A x =0 0 A M C y ×3-30×2=0 C y =20 KN 0 y F A y +20-30=0 A y =10 KN Segment AB 2 0 x 0 y F 10-5x-V=0 V=10-5x 0 M M-10x+5x 2 x =0 M=5x(2- 2 x ) Segment BC 3 2 x 0 y F 10-5x-10-V=0 V=-5x 0 M M-10x+5x 2 x +10(x-2)=0 M=20- 2 5 x 2 Segment CD 4 3 x m 2 m 1 m 1 m KN / 5 KN 10 A B C D KN 10 x A y A y C KN 20 KN 10 x V M x 5 2 / x m KN / 5 x V M x 5 m KN / 5 KN 10 2 / x KN 10
4 0 y F 10-5x-10+20-V=0 V=20-5x 0 M M-10x+5x 2 x +10(x-2)-20(x-3)=0 M=-40+20x- 2 5 x 2 S.F Diagram B.M Diagram ` x V M x 5 m KN / 5 KN 10 2 / x KN 10 KN 20 m 2 m 1 m 1 m KN / 5 KN 10 A B C D KN 10 KN 20 KN 10 KN 10 KN 15 KN 5 m KN . 10 m KN . 5 . 2

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5 Example 38: A beam ABC is simply supported at A and B and has an overhang BC . The beam is loaded by two forces P and a clockwise couple of moment Pa that act through the arrangement shown. Draw the shear force and bending moment diagrams for beam
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• Winter '15
• MAhmoudali

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