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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. 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

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  1. Class #24 Beams Shear and Bending Moment Diagrams Calculus Development Statics Spring 2006 Dr. Pickett

  2. 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

  3. If a whole beam is in equilibrium

  4. Then part of the beam is also in equilibrium A

  5. For left part: FB FB

  6. Sign Convention

  7. 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

  8. 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

  9. 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

  10. BEAM END CONDITIONS Pin-Roller Pin Fixed-Free Fixed-?

  11. BEAM END CONDITIONS Pin

  12. BEAM END CONDITIONS Roller Pin

  13. Problems • Draw the shear and bending moment diagrams of the following problems: (B & J 5th) 7.20,7.22

  14. 7.20 B&J 5th

  15. 7.20 contd

  16. 7.20 contdM x=a = M x=0 + ΔM x=0 to a

  17. 7.20 contd • AREA UNDER V DIAGRAM FROM TO MOMENT

  18. 7.22 B&J 5th

  19. 7.22 contd V(x) = -wx M(x) = -wx2/2

  20. ProblemB & J 7th #7.65

  21. 7.65 contd.

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