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Class #24. Beams Shear and Bending Moment Diagrams Calculus Development Statics Spring 2006 Dr. Pickett. Shear and Bending Moment Diagrams. Long and Slender beam It lies in a plane Is loaded only in that plane Have only 3 equilibrium equations. If a whole beam is in equilibrium.

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Class 24

Class #24

Beams

Shear and Bending Moment Diagrams

Calculus Development

Statics

Spring 2006

Dr. Pickett


Shear and bending moment diagrams
Shear and Bending Moment Diagrams

  • Long and Slender beam

    • It lies in a plane

    • Is loaded only in that plane

    • Have only 3 equilibrium equations






Steps for v and bm diagrams
Steps for V and BM diagrams

1.Draw FBD

2.Obtain reactions:

SM (@left support) to obtain reaction at right;

SM (@Right support) to obtain reaction at left;

Check SFy = 0

3. Cut a section ;

Obtain internal P,V,M at cut section ;

SM, SFy, SFx

4. Record, draw internal P, V, M on both sides of cut sections ;

- magnitude

- units

- direction on both sides of cut


If a whole beam is in equilibrium then part of the beam is also in equilibrium
If a whole beam is in equilibrium then part of the beam is also in equilibrium

  • Draw a free body diagram

  • Slope of shear ( V ) diagram @ X equals

    value of load diagram @ X

  • Integrating across the length of the beam

  • Not valid if concentrated load between x1 and x2.

  • The change in shear ( ΔV ) from

    equals the area under the load

    diagram from


  • As very very small

  • Slope of the moment ( M ) diagram @ X equals

    the value of the shear ( V ) diagram @ X

  • Integrating across the length of the beam

    Yes valid with concentrated load between x1 and x2

    Not valid if a couple is applied between x1 and x2.

  • The change in moment from equals the area under the shear diagram from


Beam end conditions
BEAM END CONDITIONS

Pin-Roller Pin

Fixed-Free

Fixed-?




Problems
Problems

  • Draw the shear and bending moment diagrams of the following problems: (B & J 5th)

    7.20,7.22




7 20 contd m x a m x 0 m x 0 to a
7.20 contdM x=a = M x=0 + ΔM x=0 to a


7 20 contd1
7.20 contd

  • AREA UNDER V DIAGRAM

    FROM

    TO

MOMENT



7 22 contd
7.22 contd

V(x) = -wx

M(x) = -wx2/2


Problem b j 7 th 7 65
ProblemB & J 7th #7.65



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