# ENGR-36_Lec-19_Beams-2.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Licensed Electrical.

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BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 36 Chp 7: Beams-2

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 2 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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 1.Determine Axial/Shearing Forces and Bending Moments Produced By Applied Loads 2.Select Structural Cross-section and Material Best Suited To Resist the Applied Forces and Bending-Moments

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 3 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 4 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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’.

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 5 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics V & M Sign Conventions  Consider a Conventionally (Gravity) Loaded Simply-Supported Beam with the X- Axis Origin Conventionally Located at the LEFT  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 BOWS UPward

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 6 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 7 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics V & M Diagrams  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 LOAD Diagram SHEAR Diagram MOMENT Diagram “Kinks” at Load- Application Points

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 8 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Relations Between Load and V  On Element of Length Δx from C to C’; ΣF y = 0  Separating the Variables and Integrating from Arbitrary Points C & D

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 9 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Relations Between Ld and M  Now on C-C’ take ΣM C’ = 0  Separating the Variables and Integrating from Arbitrary Points C & D

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 10 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Summary: Load, V, M Relations  The 1 st Derivative of V is the Negative of the Load  The Shear is the Negative of the Area under the Ld-Curve  The 1 st Derivative of M is the Shear  The Moment is the Area under the V-Curve

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 11 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 12 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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 

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 13 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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 in the M-Value

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 14 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example V&M by Calculus  For the Given Load & Geometry, Draw the shear and bending moment diagrams for the beam AE  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

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 15 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example V&M by Calculus  Solution Plan (cont.) With UNIFORM loading between D and E, the shear variation is LINEAR –m V = −1.5 kip/ft Between concentrated load application points, dM/dx = m M = 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., 2 nd degree in x).

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 16 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Example  V&M by Calculus  Taking entire beam as a free-body, determine reactions at supports

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 17 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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)

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 18 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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.

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 19 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 20 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics WhiteBoard Work Let’s Work This Problem w/ Calculus & MATLAB

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 21 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 36 Appendix

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 22 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 23 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 24 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

BMayer@ChabotCollege.edu ENGR-36_Lec-19_Beams-2.pptx 25 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

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