<ul><li>IV .) Shear and Bending Moment in Beams </li></ul><ul><li>A.) Reaction Forces (Statics Review) </li></ul><ul><li>1...
<ul><li>a.) Roller - produces a reaction force  perpendicular to the support plane. </li></ul>R Y
<ul><li>b.) Pin (or Hinge) - produces a  vertical and horizontal reaction. </li></ul>R y R x
<ul><li>c.) Fixed - produces a reaction force in any direction and Moment. </li></ul>R y R x M
<ul><li>2.) Apply laws of equilibrium to find  R AX,  R AY,  R BY </li></ul><ul><li>  F x  = 0   F y  = 0   M z  = 0 ...
<ul><li>B.) Internal Shear </li></ul><ul><li>1.) Shear - find by cutting a section at the point of interest and   F y  = ...
<ul><li>B.) Internal Bending Moment </li></ul><ul><li>2.) Moment - find by cutting a section at the point of interest and ...
<ul><li>If you were to find the internal shear and moment at several locations along the length of a beam, you could plot ...
4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B V(k)
4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B M (k-ft)
<ul><li>C.) Shear Diagram - Simpler Way to Draw </li></ul><ul><li>1.)  Sketch the beam with loads and  supports shown (thi...
<ul><li>3.)  Draw Shear Diagram baseline </li></ul><ul><li>(shear = zero) below the load diagram </li></ul><ul><li>a horiz...
<ul><li>5.)  Working from  left  to  right , calculate  </li></ul><ul><li>the shear on each side of each  </li></ul><ul><l...
<ul><li>b.) Point loads (and reactions) cause </li></ul><ul><li>  a vertical jump in the shear  </li></ul><ul><li>diagram....
<ul><li>c.)  For portions of a beam under  </li></ul><ul><li>  distributed  loading: </li></ul><ul><li>i.) the slope of th...
<ul><li> V = wL (uniformly distributed) </li></ul><ul><li>V = (wL)/2 (triangular distribution) </li></ul><ul><li>Note: ...
4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B
<ul><li>D.) Moment Diagram - Simpler Method </li></ul><ul><li>1.)  Moment = 0 at ends of simply    supported beams. </li><...
<ul><li>3.)  Extend the vertical lines below the </li></ul><ul><li>shear diagram and draw the Moment </li></ul><ul><li>Dia...
<ul><li>a.)  the change in moments between  </li></ul><ul><li>  two points is equal to the area under </li></ul><ul><li>  ...
<ul><li>i.)  if the shear is positive and constant, </li></ul><ul><li> the slope of the moment diagram is </li></ul><ul><l...
Positive, constant slope Negative, constant slope (+) 0 (-) M
<ul><li>ii.)  if the shear is positive and increasing, </li></ul><ul><li> the slope of the moment diagram is  </li></ul><u...
Positive, decreasing slope Negative,  decreasing slope (+) 0 (-) M
4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B
 
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Shear and Bending Moment in Beams

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Shear and Bending Moment in Beams

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Shear and Bending Moment in Beams

  1. 1. <ul><li>IV .) Shear and Bending Moment in Beams </li></ul><ul><li>A.) Reaction Forces (Statics Review) </li></ul><ul><li>1.) Replace Supports with unknown </li></ul><ul><li> reaction forces (free body diagram) </li></ul>
  2. 2. <ul><li>a.) Roller - produces a reaction force perpendicular to the support plane. </li></ul>R Y
  3. 3. <ul><li>b.) Pin (or Hinge) - produces a vertical and horizontal reaction. </li></ul>R y R x
  4. 4. <ul><li>c.) Fixed - produces a reaction force in any direction and Moment. </li></ul>R y R x M
  5. 5. <ul><li>2.) Apply laws of equilibrium to find R AX, R AY, R BY </li></ul><ul><li>  F x = 0  F y = 0  M z = 0 </li></ul>R AX R AY R BY
  6. 6. <ul><li>B.) Internal Shear </li></ul><ul><li>1.) Shear - find by cutting a section at the point of interest and  F y = 0 on the FBD. </li></ul>R y R x F.B.D. V
  7. 7. <ul><li>B.) Internal Bending Moment </li></ul><ul><li>2.) Moment - find by cutting a section at the point of interest and  M = 0 on the FBD. </li></ul>R y R x F.B.D. V M
  8. 8. <ul><li>If you were to find the internal shear and moment at several locations along the length of a beam, you could plot a graph shear vs. length and a graph of moment vs. length and find where the maximum shear and moment occur. </li></ul>
  9. 9. 4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B V(k)
  10. 10. 4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B M (k-ft)
  11. 11. <ul><li>C.) Shear Diagram - Simpler Way to Draw </li></ul><ul><li>1.) Sketch the beam with loads and supports shown (this is the LOAD DIAGRAM). </li></ul><ul><li>2.) Compute the reactions at the supports and show them on the sketch. </li></ul>
  12. 12. <ul><li>3.) Draw Shear Diagram baseline </li></ul><ul><li>(shear = zero) below the load diagram </li></ul><ul><li>a horizontal line. </li></ul><ul><li>4.) Draw vertical lines down from the load </li></ul><ul><li>diagram to the shear diagram at: </li></ul><ul><li>a.) supports </li></ul><ul><li>b.) point loads </li></ul><ul><li>c.) each end of distributed loads </li></ul>
  13. 13. <ul><li>5.) Working from left to right , calculate </li></ul><ul><li>the shear on each side of each </li></ul><ul><li>support and point load and at each </li></ul><ul><li>end of distributed loads: </li></ul><ul><li>a.) For portions of a beam that have </li></ul><ul><li>no loading, the shear diagram is </li></ul><ul><li>a horizontal line. </li></ul>
  14. 14. <ul><li>b.) Point loads (and reactions) cause </li></ul><ul><li> a vertical jump in the shear </li></ul><ul><li>diagram. </li></ul><ul><li> - The magnitude of the </li></ul><ul><li>jump is equal to the magnitude of the load (or reaction). </li></ul><ul><li>- Downward loads cause a negative change in shear. </li></ul>
  15. 15. <ul><li>c.) For portions of a beam under </li></ul><ul><li> distributed loading: </li></ul><ul><li>i.) the slope of the shear diagram is equal to the intensity (magnitude) of the uniformly distributed load (w). </li></ul><ul><li>ii.) the change in shear between two </li></ul><ul><li> points is equal to the area under the load diagram between those two points. </li></ul>
  16. 16. <ul><li> V = wL (uniformly distributed) </li></ul><ul><li>V = (wL)/2 (triangular distribution) </li></ul><ul><li>Note: If the distributed load is acting </li></ul><ul><li> downward “w” is negative. </li></ul><ul><li>6.) Locate points of zero shear using a </li></ul><ul><li> known shear value at a known location </li></ul><ul><li>and the slope of the shear diagram(w) </li></ul>
  17. 17. 4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B
  18. 18. <ul><li>D.) Moment Diagram - Simpler Method </li></ul><ul><li>1.) Moment = 0 at ends of simply supported beams. </li></ul><ul><li>2.) Peak Moments occur where the shear </li></ul><ul><li>diagram crosses through zero. There </li></ul><ul><li>can be more than one peak moment </li></ul><ul><li>on the diagram. </li></ul>
  19. 19. <ul><li>3.) Extend the vertical lines below the </li></ul><ul><li>shear diagram and draw the Moment </li></ul><ul><li>Diagram baseline (moment = 0), a </li></ul><ul><li>horizontal line. Also, extend vertical </li></ul><ul><li>lines down from points of zero shear. </li></ul><ul><li>4.) Working left to right, calculate the </li></ul><ul><li>moment at each point the shear was </li></ul><ul><li>calculated and at points of zero shear: </li></ul>
  20. 20. <ul><li>a.) the change in moments between </li></ul><ul><li> two points is equal to the area under </li></ul><ul><li> the shear diagram between those </li></ul><ul><li> points. </li></ul><ul><li>b.) determine the slope of the moment </li></ul><ul><li> diagrams as follows: </li></ul>
  21. 21. <ul><li>i.) if the shear is positive and constant, </li></ul><ul><li> the slope of the moment diagram is </li></ul><ul><li> positive and constant. </li></ul><ul><li> </li></ul>Negative, constant shear (-) (+) Positive, constant shear (+) 0 (-) V
  22. 22. Positive, constant slope Negative, constant slope (+) 0 (-) M
  23. 23. <ul><li>ii.) if the shear is positive and increasing, </li></ul><ul><li> the slope of the moment diagram is </li></ul><ul><li> positive and increasing. </li></ul><ul><li>Positive, decreasing shear </li></ul>Negative, decreasing shear (+) 0 (-) V
  24. 24. Positive, decreasing slope Negative, decreasing slope (+) 0 (-) M
  25. 25. 4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B
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