3rd lecture shear and moment diagram for determinate beam

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3rd lecture shear and moment diagram for determinate beam

  1. 1. Theory of Structures 3rd Lecture (9/10/2012): Analysis of Determinate Beams; Shear & Moment Diagrams Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 1
  2. 2. Beams • Members that are slender and support loads applied perpendicular to their longitudinal axis. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 2 Span, L Distributed Load, w(x) Concentrated Load, P Longitudinal Axis
  3. 3. Types of Beams • Depends on the support configuration Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 3 M Fv FH Fixed FV FV FH Pin Roller PinRoller FVFV FH
  4. 4. Statically Indeterminate Beams • Can you guess how we find the “extra” reactions? Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 4 Continuous Beam Propped Cantilever Beam
  5. 5. Internal Loadings developed in a structure member • in order to properly design structural components, we must know the distribution of the internal forces within the component. • In this lecture, we will determine the normal (axial) force, shear, and moment at a point in a structural component. • Internal loads at a point within a structural component. For a coplanar structure, the internal load at a specified point will consist of a normal (Axial) force, N, a shear force, V, and a bending moment, M. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 5
  6. 6. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 6 Positive normal (Axial) force, N, tends to elongate the components. Again, note that the normal (axial) forces act in opposite directions on either side of the cut. Positive shear, V, tends to rotate the component clockwise. Note that the shear is in opposite directions on either side of a cut through the component. Positive moment, M, tends to deform the component into a dish-shaped configuration such that it would hold water. Again, note that the moment acts in opposite directions on either side of the cut.
  7. 7. Internal Reactions in Beams • At any cut in a beam, there are 3 possible internal reactions required for equilibrium: – Normal (Axial) force, – Shear force, – Bending moment. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 7 L P a b
  8. 8. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 8
  9. 9. Internal Reactions in Beams • At any cut in a beam, there are 3 possible internal reactions required for equilibrium: – normal force, – shear force, – bending moment. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 9 Pb/L x Left Side of Cut V M N Positive Directions Shown!!!
  10. 10. Internal Reactions in Beams • At any cut in a beam, there are 3 possible internal reactions required for equilibrium: – normal force, – shear force, – bending moment. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 10 Pa/L L - x Right Side of CutVM N Positive Directions Shown!!!
  11. 11. Finding Internal Reactions • Pick left side of the cut: – Find the sum of all the vertical forces to the left of the cut, including V. Solve for shear, V. – Find the sum of all the horizontal forces to the left of the cut, including N. Solve for axial force, N. It’s usually, but not always, 0. – Sum the moments of all the forces to the left of the cut about the point of the cut. Include M. Solve for bending moment, M • Pick the right side of the cut: – Same as above, except to the right of the cut. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 11
  12. 12. Example: Find the internal reactions at points indicated. All axial force reactions are zero. Points are 2-ft apart. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 12 20 ft P = 20 kips 12 kips8 kips 12 ft 1 7 10 6 2 3 94 5 8 Point 6 is just left of P and Point 7 is just right of P.
  13. 13. 20 ft P = 20 kips 12 kips8 kips 12 ft 1 7 10 6 2 3 94 5 8 V (kips) M (ft-kips) 8 kips -12 kips 96 48 64 48 72 24 80 16 32 x xDr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 13
  14. 14. 20 ft P = 20 kips 12 kips8 kips 12 ft V (kips) M (ft-kips) 8 kips -12 kips 96 ft-kips x x V & M Diagrams What is the slope of this line? a b c 96 ft-kips/12’ = 8 kips What is the slope of this line? -12 kips Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 14
  15. 15. 20 ft P = 20 kips 12 kips8 kips 12 ft V (kips) M (ft-kips) 8 kips -12 kips 96 ft-kips x x V & M Diagrams a b c What is the area of the blue rectangle? 96 ft-kips What is the area of the green rectangle? -96 ft-kips Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 15
  16. 16. Draw Some Conclusions • The magnitude of the shear at a point equals the slope of the moment diagram at that point. • The area under the shear diagram between two points equals the change in moments between those two points. • At points where the shear is zero, the moment is a local maximum or minimum. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 16
  17. 17.      dxxVxM dxxwxV functionloadthexw )()( )()( )( Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 17 The Relationship Between Load, Shear and Bending Moment
  18. 18. Shear Force & Bending Moment Diagram Structure under loads w dx dV  V dx dM  Bending moment diagram Shear force diagram Slope of V = load Slope of M = V Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 18
  19. 19. Load 0 Constant Linear Shear Constant Linear Parabolic Moment Linear Parabolic Cubic Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 19 Common Relationships
  20. 20. Load 0 0 Constant Shear Constant Constant Linear Moment Linear Linear Parabolic Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 20 Common Relationships M
  21. 21. Example: Draw Shear & Moment diagrams for the following beam Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 21 3 m 1 m1 m 12 kN 8 kN A C B D RA = 7 kN  RC = 13 kN 
  22. 22. 3 m 1 m1 m 12 kN A C B D V (kN) M (kN-m) 7 -5 8 8 kN 7 -15 8 7 -82.4 m Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 22
  23. 23. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 23 Because the form of the loading does not change anywhere along the beam, single equations will suffice for moment and shear: ∑Fy=0; 75-10*x-20/(2*9)*x^2 ∑M=0; 75*x-5*x^2-(10/27)*x^3 The reactions are VL = 75kN, VR = 105kN, and HL = 0. 75 kN 105 kN
  24. 24. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 24 -105 -85 -65 -45 -25 -5 15 35 55 75 0 1 2 3 4 5 6 7 8 9 Shear Diagram 0 20 40 60 80 100 120 140 160 180 200 220 0 1 2 3 4 5 6 7 8 9 Shear and moment are plotted :
  25. 25. Dr Hussein M. Al.Khuzaie; husseinma@coe- muth.org 25 Examples
  26. 26. Dr Hussein M. Al.Khuzaie; husseinma@coe- muth.org 26
  27. 27. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 27 4- Develop equations for shear and moment as a function of position for the following structural components. Plot the functions. 9 m
  28. 28. Dr Hussein M. Al.Khuzaie; husseinma@coe-muth.org 28 Any questions?

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