Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

1. Engineering 36 Chp7:Beams-2 Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. Beam – What is it? • Beam Structural member designed to support loads applied at various points along its length • Beams can be subjected to CONCENTRATED loads or DISTRIBUTED loads or a COMBINATION of both. • Beam Design is 2-Step Process • Determine Axial/Shearing Forcesand Bending MomentsProduced By Applied Loads • Select Structural Cross-section and Material Best Suited To Resist the Applied Forces and Bending-Moments

3. Beam Loading and Supports • Beams are classified according to the Support Method(s); e.g., Simply-Supported, Cantilever • Reactions at beam supports are Determinate if they involve exactly THREE unknowns. • Otherwise, they are Statically INdeterminate

4. Shear & Bending-Moment • Goal = determine bending moment and shearing force at any point in a beam subjected to concentrated and distributed loads • Determine reactions at supports by treating whole beam as free-body. • Cut beam at C and draw free-body diagrams for AC and CB exposing V-M System • From equilibrium considerations, determine M & V or M’ & V’.

5. Consider a Conventionally (Gravity) Loaded Simply-Supported Beam with the X-Axis Origin Conventionally Located at the LEFT V & M Sign Conventions • Next Consider a Virtual Section Located at C • DEFINE this Case as POSITIVE • Shear, V • The Virtual Member LEFT of the Cut is pushed DOWN by the Right Virtual Member • Moment, M • The Beam BOWSUPward

6. V & M Sign Conventions (2) • Positive Shear • Right Member Pushes DOWN on Left Member • Positive Bending • Beam Bows UPward • POSITIVE Internal Forces, V & M • Note that at a Virtual Section the V’s & M’s MUST Balance

7. With the Signs of V&M Defined we Can now Determine the MAGNITUDE and SENSE for V&M at ANY arbitrary Virtual-Cut Location PLOTTING V&M vs. x Yields the Stacked Load-Shear-Moment (LVM) Diagram V & M Diagrams LOAD Diagram “Kinks” at Load-Application Points SHEAR Diagram MOMENT Diagram

8. Relations Between Load and V • On Element of Length Δx from C to C’; ΣFy = 0 • Separating the Variables and Integrating from Arbitrary Points C & D

9. Relations Between Ld and M • Now on C-C’ takeΣMC’ = 0 • Separating the Variables and Integrating from Arbitrary Points C & D

10. Summary: Load, V, M Relations • The 1st Derivative of V is the Negative of the Load • The 1st Derivative of M is the Shear • The Shear is the Negative of the Area under the Ld-Curve • The Moment is the Area under the V-Curve

11. Recall: Derivative = SLOPE • The SLOPE of the V-Curve is the Negative-VALUE of the Load Curve • The SLOPE of the M-Curve is the Positive-VALUE of the Shear Curve • Note that w is a POSITIVE scalar; i.e.; it is a Magnitude

12. Calculus Summary • The SLOPE of the V-curve is the negative MAGNITUDE of the w-Curve • The SLOPE of M-Curve is the VALUE of the V-Curve • The CHANGE in V between Pts a&b is the DEFINATE INTEGRAL between Pts a&b of the w-Curve • The CHANGE in M between Pts a&b is the DEFINATE INTEGRAL between Pts a&b of the V-Curve

13. Calculus Summary • When w is down • The negative VALUE of the w-Curve is the SLOPE of the V-curve • The Negative AREA Under w-Curve is the CHANGE in the V-Value • The VALUE of the V-Curve is the SLOPE of M-Curve • The AREA under the V-Curve is the CHANGE inthe M-Value

14. Example  V&M by Calculus • Solution Plan • Taking entire beam as free-body, calculate reactions at Support A and D. • Between concentrated load application points, dV/dx= −w = 0, and so the SLOPE is ZERO, and Thus Shear is Constant • For the Given Load & Geometry, Draw the shear and bending moment diagrams for the beam AE

15. Solution Plan (cont.) With UNIFORM loading between D and E, the shear variation is LINEAR mV = −1.5 kip/ft Between concentrated load application points, dM/dx = mM = V = const. The CHANGE IN MOMENT between load application points is equal to AREA UNDER SHEAR CURVE between Load-App points With a LINEAR shear variation between D and E, the bending moment diagram is a PARABOLA (i.e., 2nd degree in x). Example  V&M by Calculus

16. Example  V&M by Calculus • Taking entire beam as a free-body, determine reactions at supports

17. Example  V&M by Calculus • The VERTICAL Reactions • Between concentrated load application points, dV/dx = 0, and thus shear is Constant • With uniform loading between D and E, the shear variation is LINEAR. • SLOPE is constant at −w (−1.5 kip/ft in this case)

18. Example  V&M by Calculus • Between concentrated load application points, dM/dx = V = Const. And the change in moment between load application points is equal to AREA under the SHEAR CURVE between points.

19. Example  V&M by Calculus • With a Linear Shear variation between D and E, the bending moment diagram is PARABOLIC. • Note that the FREE End of a Cantilever Beam Cannot Support ANY Shear or Bending-Moment

20. WhiteBoard Work Let’s WorkThis Problemw/ Calculus& MATLAB

21. Engineering 36 Appendix Bruce Mayer, PE Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu