311 Ch13
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311 Ch13 311 Ch13 Presentation Transcript

    • IV .) Shear and Bending Moment in Beams
    • A.) Reaction Forces (Statics Review)
    • 1.) Replace Supports with unknown
    • reaction forces (free body diagram)
    • a.) Roller - produces a reaction force perpendicular to the support plane.
    R Y
    • b.) Pin (or Hinge) - produces a vertical and horizontal reaction.
    R y R x
    • c.) Fixed - produces a reaction force in any direction and Moment.
    R y R x M
    • 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
    • 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
    • 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
    • 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.
  • 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)
    • 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.
    • 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
    • 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.
    • 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.
    • 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.
    •  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)
  • 4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B
    • 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.
    • 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:
    • 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:
    • 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
  • Positive, constant slope Negative, constant slope (+) 0 (-) M
    • 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
  • Positive, decreasing slope Negative, decreasing slope (+) 0 (-) M
  • 4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B
  •