The document discusses the concepts of shear and bending moment in beams. It covers reaction forces at supports, calculating internal shear and bending moment by cutting sections of the beam, and how to construct shear and moment diagrams by sketching the beam, supports, and loads and calculating shear and moment values at points along the beam from left to right. The diagrams allow identifying locations of maximum shear and moment.

1. IV .) Shear and Bending Moment in Beams A.) Reaction Forces (Statics Review) 1.) Replace Supports with unknown reaction forces (free body diagram)

2. a.) Roller - produces a reaction force perpendicular to the support plane. R Y

3. b.) Pin (or Hinge) - produces a vertical and horizontal reaction. R y R x

4. c.) Fixed - produces a reaction force in any direction and Moment. R y R x M

5. 2.) Apply laws of equilibrium to find R AX, R AY, R BY   F x = 0  F y = 0  M z = 0 R AX R AY R BY

6. B.) Internal Shear 1.) Shear - find by cutting a section at the point of interest and  F y = 0 on the FBD. R y R x F.B.D. V

7. B.) Internal Bending Moment 2.) Moment - find by cutting a section at the point of interest and  M = 0 on the FBD. R y R x F.B.D. V M

8. 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.

9. 4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B V(k)

10. 4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B M (k-ft)

11. C.) Shear Diagram - Simpler Way to Draw 1.) Sketch the beam with loads and supports shown (this is the LOAD DIAGRAM). 2.) Compute the reactions at the supports and show them on the sketch.

12. 3.) Draw Shear Diagram baseline (shear = zero) below the load diagram a horizontal line. 4.) Draw vertical lines down from the load diagram to the shear diagram at: a.) supports b.) point loads c.) each end of distributed loads

13. 5.) Working from left to right , calculate the shear on each side of each support and point load and at each end of distributed loads: a.) For portions of a beam that have no loading, the shear diagram is a horizontal line.

14. b.) Point loads (and reactions) cause a vertical jump in the shear diagram. - The magnitude of the jump is equal to the magnitude of the load (or reaction). - Downward loads cause a negative change in shear.

15. c.) For portions of a beam under distributed loading: i.) the slope of the shear diagram is equal to the intensity (magnitude) of the uniformly distributed load (w). ii.) the change in shear between two points is equal to the area under the load diagram between those two points.

16.  V = wL (uniformly distributed) V = (wL)/2 (triangular distribution) Note: If the distributed load is acting downward “w” is negative. 6.) Locate points of zero shear using a known shear value at a known location and the slope of the shear diagram(w)

17. 4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B

18. D.) Moment Diagram - Simpler Method 1.) Moment = 0 at ends of simply supported beams. 2.) Peak Moments occur where the shear diagram crosses through zero. There can be more than one peak moment on the diagram.

19. 3.) Extend the vertical lines below the shear diagram and draw the Moment Diagram baseline (moment = 0), a horizontal line. Also, extend vertical lines down from points of zero shear. 4.) Working left to right, calculate the moment at each point the shear was calculated and at points of zero shear:

20. a.) the change in moments between two points is equal to the area under the shear diagram between those points. b.) determine the slope of the moment diagrams as follows:

21. i.) if the shear is positive and constant, the slope of the moment diagram is positive and constant. Negative, constant shear (-) (+) Positive, constant shear (+) 0 (-) V

22. Positive, constant slope Negative, constant slope (+) 0 (-) M

23. ii.) if the shear is positive and increasing, the slope of the moment diagram is positive and increasing. Positive, decreasing shear Negative, decreasing shear (+) 0 (-) V

24. Positive, decreasing slope Negative, decreasing slope (+) 0 (-) M

25. 4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B