Chp 4: Moment Mathematics Engineering 36 Chp 4: Moment Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

Moments are VECTORS As Described Last Lecture a Moment is a measure of “Twisting Power” A Moment has Both MAGNITUDE & Direction and can be Represented as a Vector, M, with Normal Vector properties

Moments are VECTORS Describe M in terms of a unit vector, û, directed along the LoA for M Find the θk by Direction CoSines

M = r X F MAGNITUDE of M measures the tendency of a force to cause rotation of a body about an Axis thru the pivot-Pt O

M = r X F The SENSE of the moment may be determined by the Right-Hand Rule If the fingers of the RIGHT hand are curled from the direction of r toward the direction of F, then the THUMB points in the direction of the Moment Moment Direction

M = r X F Combining (1) & (2) yields the Definition of the vector CROSS PRODUCT (c.f. MTH3) Engineering Mechanics uses the Cross Product to Define the Moment Vector Vectors MUST be placed TAIL-to-TAIL to find Q in rXF = |r|*|F|*sinQ û is a unit vector directed by the Rt-Hand Rule θ is the Angle Between the LoA’s for r & F

M = r X F → θ by Tail-toTail When Finding Moment Magnitudes using: The Angle θ MUST be determined by placing Vectors r & F in the TAIL-to-TAIL Orientation See Diagram at Right

Cross Product Math Properties Recall Vector ADDITION Behaved As Algebraic Addition BOTH Commutative and Associative. The Vector PRODUCT Math-Properties do NOT Match Algebra - Vector Products: Are NOT Commutative Are NOT Associative ARE Distributive

Vector Prod: Rectangular Comps Vector Products Of Cartesian Unit Vectors Vector Product In Terms Of Rectangular Coordinates

rxF in 3D  Deteriminant Notation Consider 3D versions of r & F Taking the Cross Product Yields M Determinant Notation provides a convenient Tool For the Calculation Don’t Forget the MINUS sign in the Middle ( 𝑗 ) Term See also TextBook pg123 Basic Minor for 𝑖

Varignon’s Theorem The Moment About a Point O Of The Resultant Of Several ConCurrent Forces Is Equal To The Sum Of The Moments Of The Various Forces About The Same Point O Stated Mathematically Varignon’s Theorem Makes It Possible To Replace The Direct Determination Of The Moment of a Force F By The Moments of Its Components (which are concurrent)

rxF in 3D  Vector Properties Cartesian CoOrds for a 3D M vector The Magnitude of a 3D M vector Direction CoSines Unit Vector

rxF in 2D  r & F in XY Plane If r & F Lie in the XY Plane, then rz = Fz = 0. Thus the rxF Determinant So in this case M is confined to the Z-Direction:

Direction for r in rF Consider the CrowBar Below We Want to find the Torque (Moment) About pt-B due to Pull, P, applied at pt-A using rP We have Two Choices for r: r points A→B r points B→A Which is Correct?

Direction for r in rF We can find the Direction for r by considering the SIGN of the Moment In this case it’s obvious (to me, anyway) that P will cause CLOCKwise Rotation about Pt-B Recall That In the x-y Plane ClockWise Rotation is defined as NEGATIVE Test rP and rP y x

Direction for r in rF Put r and r into Component form Equal but Opposite Then the two r’s y Now let x

  Direction for r in rF then the rxP calculations noting y  Thus rB→A is the CORRECT position vector x

FROM the PIVOT TO the FORCE Direction for r in rF To Calc the Moment about pt-B use: The position Vector points FROM the PIVOT-point TO the Force APPLICATION-point on the Force LoA Summarize this as FROM the PIVOT TO the FORCE

Unit Vector Notation: u ≡ λ Our Text uses u to denote the unit vector While u is quite popular as the unit vector notation, other symbols are often used (kind of like 𝜃 & 𝜙 for angles) On Occasion I will use λ to represent the unit vector This is usually apparent from the problem or situation context

Example: 3D Moment A Rectangular Plate Is Supported By The Brackets At A and B and By A Wire CD. Knowing That The Tension In The Wire is 200 N, Determine The Moment About A Of The Force Exerted By The Wire At connection-point C. Solution Plan The Moment MA Of The Force F Exerted By The Wire Is Obtained By Evaluating The Vector Product

Example 3D Moment - Solution Resolve Both F and rAC into Cartesian Components Take Cross-Product Using Determinant The 28.8k moment will cause the L brackets to “fold up”. The −7.68i moment will cause the screws to SHEAR. The 28.8j moment will cause the screws to PULL OUT. Which Moment will Most Likely Cause DEFORMATION?

Moment About an Axis (§4.5) Moment MO Of A Force F , Applied at The Point A, About a Point O, Recall Scalar Moment MOL About An AXIS OL Is The Projection Of The Moment Vector MO Onto The OL Axis using the Dot Product MOL it the tendency of the applied force to cause a rotation about the AXIS OL

Moment About an Axis – cont. Moments of F About The CoOrd Origin Moment Of A Force About An Arbitrary Axis BL Similar Analysis for CL, Starting With MC, Shows That MCL = MBL; i.e., the Result is Independent of the Location of the Point ON the Line

Moment About an Axis – cont. Since the moment, ML, about an arbitrary axis is INDEPENDENT of position vector, r, that runs from ANY Point on the axis to ANY point on the LoA of the force we can choose the MOST CONVENIENT Points on the Axis and the Force LoA to determine ML

MOL Physical Significance MOL Measures the Tendency of an Applied Force to Impart to a Rigid Body Rotation about a fixed Axis OL i.e., How Much will the Applied Force Cause The body to Rotate about an AXLE MOL can be Considered as the Component of M directed along “axis” OL

Example: MOL A Cube With Side Length a is Acted On By a Force P as Shown Determine The Moment Of P: About Pt A About The Edge (Axis) AB About The Diagonal (Axis) AG of The Cube For Lines AG and FC Determine The Perpendicular Distance Between them

(From the Pivot to the Force) Example MOL - Solution Moment of P about Pivot-Pt A (From the Pivot to the Force) Moment of P about Axis AB For b) ALREADY Know the moment about a pt (pt-A) on axis/line AB

Example MOL - Solution Alternative Moment of P about A

Example MOL - Solution Moment of P About Diagonal AG (From Part a)

Example MOL - Solution Perpendicular distance between AG and FC Notice That Plane OFC Appears To Be  to Line AG, And FC Resides In this Plane Since P Has Line-of-Action FC We Can Test Perpendicularity with Dot Product  Then the Moment (or twist) Caused by P About AG = Pd; Thus

Mixed Triple Product Do Find MOL we used the Qty û•(r x F). Formalize this Operation as the Mixed Triple Product for vectors S, P, & Q Associativity and Communtivity for the Mixed Triple Product Of Three Vectors

Evaluate the Mixed Triple Prod Let V = PxQ, Then And Thus Determinant Notation Yet Again

Mixed Triple Product vs MOL The Mixed Triple Product can be used to find the Magnitude of the Moment about an Axis.

Let’s Work This Nice Problem WhiteBoard Work Let’s Work This Nice Problem S&T 5.2.37 (ref. ENGR-36_Lab-07_Fa07_Lec-Notes.ppt) UseFul Moment is the Moment about the ScrewDriving Axis, Z. Any other Moment tends to BEND the Screw Determine MA as caused by application of the 120 N force

Alternative

Alternative

Alternative The Moment that BENDS the wrench handle is that which exists in the XY plane. Thus Adding the x & y moment components 𝑀 𝑏𝑒𝑛𝑑 = −4800 2 + −8314 2 =9600 N∙mm

Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engr/Math/Physics 25 Appendix Time For Live Demo Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

Vector/Cross Product TWISTING Power of a Force  MOMENT of the Force Quantify Using VECTOR PRODUCT or CROSS PRODUCT Vector Product Of Two Vectors P And Q Is Defined As The Vector V Which Satisfies: Line of Action of V Is Perpendicular To Plane Containing P and Q. Rt Hand Rule Determines Direction for V |V| =|P|•|Q|•sin

Moment About an Axis – cont. Moments of F About The CoOrd Origin Moment Of A Force About An Arbitrary Axis BL Similar Analysis for CL, Starting With MC, Shows That MCL = MBL; i.e., the Result is Independent of the Location of the Point ON the Line