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

1. Engineering 36 Chp 5: Equivalent Loads Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. Introduction: Equivalent Loads • Any System Of Forces & Moments Acting On A Rigid Body Can Be Replaced By An Equivalent System Consisting of these “Intensities” acting at Single Point: • One FORCE (a.k.a. a Resultant) • One MOMENT (a.k.a. a Couple) Equiv. Sys.

3. Two Classes of Forces Act On Rigid Bodies: External forces Internal forces External vs. Internal Forces • The Free-Body Diagram Shows External Forces • UnOpposed External Forces Can Impart Accelerations (Motion) • Translation • Rotation • Both

4. Principle of Transmissibility Conditions Of Equilibrium Or Motion Are Not Affected By TRANSMITTING A Force Along Its LINE OF ACTION Transmissibility: Equivalent Forces Note: F & F’ Are Equivalent Forces • Moving the point of application of F tothe rear bumper doesnot affect the motion or the other forcesacting on the truck

5. Principle of transmissibility may not always apply in determining Internal Forces Deformations Transmissibility Limitations Rigid Deformed  TENSION  COMPRESSION

6. COUPLE  Two Forces F and −F With Same Magnitude Parallel Lines Of Action Distance separation Opposite Direction Moment of The Couple about O Moment of a Couple

7. Thus The Moment Vector Of The Couple is INDEPENDENT Of The ORIGIN Of The Coord Axes Thus it is a FREE VECTOR i.e., It Can Be Applied At Any Point on a Body With The Same Effect M of a Couple → Free Vector • Two Couples Are Equal If • F1d1 = F2d2 • The Couples Lie In Parallel Planes • The Couples Have The Tendency To Cause Rotation In The Same Direction

8. These Couples Exert Equal Twist on the Blk Some Equivalent Couples • For the Lug Wrench Twist • Shorter Wrench with greater Force Would Have the Same Result • Moving Handles to Vertical, WithSame Push/Pull Has Same Result

9. Consider Two IntersectingPlanes P1 and P2 WithEach Containing a Couple Couple Addition • Resultants Of The Force Vectors Also Form a Couple

10. By Varignon’s Distributive Theoremfor Vectors Couple Addition • Thus The Sum of Two Couples Is Also A Couple That Is Equal To The Vector Sum Of The Two individual Couples • i.e., Couples Add The Same as Force Vectors

11. Couples Are Vectors • Properties of Couples • A Couple Can Be Represented By A Vector With Magnitude & Direction Equal To The Couple-Moment • Couple Vectors Obey The Law Of Vector Addition • Couple Vectors Are Free Vectors • i.e., The Point Of Application or LoAIs NOT Significant • Couple Vectors May Be Resolved Into Component Vectors

12. Couple r x F Resolution of a Force Into a Force at O and a Couple • Force Vector F Can NOT Be Simply Moved From A To O Without Modifying Its Action On The Body • Attaching Equal & Opposite Force Vectors At O Produces NO Net Effect On The Body • But it DOES Produce a Couple • The Three Forces In The Middle Diagram May Be Replaced By An Equivalent Force Vector And Couple Vector; i.e., a FORCE-COUPLE SYSTEM

13. Force-Couple System at O’ • Moving F from A To a Different Point O’ Requires Addition of a Different Couple Vector • The Moments of F about O and O’ are Related By The Vector S That Joins O and O’

14. Force-Couple System at O’ • Moving The Force-couple System From O to O’ Requires The Addition Of The Moment About O’ Generated by the Force At O

15. Example: Couples • Solution Plan • Attach Equal And Opposite 20 Lb Forces In The ±x Direction At A, Thereby Producing 3 Couples For Which The Moment Components Are Easily Calculated • Alternatively, Compute The Sum Of The Moments Of The Four Forces About An Arbitrary Single Point. • The Point D Is A Good Choice As Only Two Of The Forces Will Produce Non-zero Moment Contributions • Determine The Components Of The Single Couple Equivalent To The Couples Shown

16. Attach Equal And Opposite 20 lb Forces In the ±x Direction at A No Net Change to the Structure The Three Couples May Be Represented By 3 Vector Pairs Mx My Mz My Mz Mx Example: Couples

17. Alternatively, Compute The Sum Of The Moments Of The Four Forces About D Only The Forces At C and E Contribute To The Moment About D i.e., The Position vector, r, for the Forces at D = 0 rDC rDE Example: Couples

18. Reduction to Force-Couple Sys • A SYSTEM OF FORCES May Be REPLACED By A Collection Of FORCE-COUPLE SYSTEMS Acting at Given Point O • The Force And Couple Vectors May then Be Combined Into a singleResultant Force-Vector and a Resultant Couple-Vector

19. Reduction to a Force-Couple Sys • The Force-Couple System at O May Be Moved To O’ With The Addition Of The Moment Of R About O’ as before: • Two Systems Of Forces Are EQUIVALENT If They Can Be Reduced To The SAME Force-Couple System

20. If the Resultant Force & Couple At O Are Perpendicular, They Can Be Replaced By A Single Force Acting With A New Line Of Action. Force Systems That Can be Reduced to a Single Force Concurrent Forces Generates NO Moment Coplanar Forces (next slide) The Forces Are Parallel CoOrds for Vertical Forces More Reduction of Force Systems (a) (b) (c)

21. System Of CoPlanar Forces Is Reduced To A Force-couple System That Is Mutually Perpendicular CoPlanar Force Systems • System Can Be Reduced To a Single Force By Moving The Line Of Action R To Point-A Such That d: • In Cartesian Coordinates use transmissibility to slide the Force PoA to Points on the X & Axes

22. Example: 2D Equiv. Sys. • Solution Plan • Compute • The Resultant Force • The Resultant Couple About A • Find An Equivalent Force-couple System at B Based On The Force-couple System At A • Determine The Point Of Application For The Resultant Force Such That Its Moment About A Is Equal To The Resultant Couple at A • For The Beam, Reduce The System Of Forces Shown To • An Equivalent Force-Couple System At A • An Equivalent Force-Couple System At B • A Single Force applied at the Correct Location .

23. Example: 2D Equiv. Sys. - Soln Find the resultant force and the resultant couple at A. • Now Calculate the Total Moment About A as Generated by the Individual Forces.

24. Find An Equivalent Force-couple System At B Based On The Force-couple System at A The Force Is Unchanged By The Movement Of The Force-Couple System From A to B rBA Example: 2D Equiv. Sys. - Soln • The Couple At B Is Equal To The Moment About B Of The Force-couple System Found At A

25. Determine a SINGLE Resultant Force (NO Couple) The Force Resultant Remains UNCHANGED from parts a) & b) The Single Force Must Generate the Same Moment About A (or B) as Caused by the Original Force System Example: 2D Equiv. Sys. - Soln • Chk 1000 Nm at B • Then the Single-Force Resultant 

26. 3 Cables Are Attached To The Bracket As Shown. Replace The Forces With An Equivalent Force-Couple System at A Solution Plan: Determine The Relative Position Vectors For The Points Of Application Of The Cable Forces With Respect To A. Resolve The Forces Into Rectangular Components Compute The Equivalent Force Example: 3D Equiv. Sys. • Calculate The Equivalent Couple

27. Determine The Relative Position Vectors w.r.t. A Example Equiv. Sys. - Solution • Resolve The Forces Into Rectangular Components

28. Compute Equivalent Force Example Equiv. Sys. - Solution • Compute Equivalent Couple

29. Distributed Loads • The Load on an Object may be Spread out, or Distributed over the surface. Load Profile, w(x)

30. Distributed Loads • If the Load Profile, w(x), is known then the distributed load can be replaced with at POINT Load at a SPECIFIC Location • Magnitude of thePoint Load, W, is Determined by Area Under the Profile Curve

31. Distributed Loads • To Determine the Point Load Location employ Moments • Recall: Moment = [LeverArm]•[Intensity] • In This Case • LeverArm = The distance from the Baseline Origin, xn • Intensity = The Increment of Load, dWn, which is that load, w(xn) covering a distance dx located at xn • That is: dWn = w(xn)•dx

32. Distributed Loads • Now Use Centroidal Methodology • And also: • Equating the Ω Expressionsfind

33. A distributed load is represented by plotting the load per unit length, w (N/m). The total load is equal to the area under the load curve. • A distributed load can be REPLACED by a concentrated load with a magnitude equal to the area under the load curve and a line of action passing through the areal centroid. Distributed Loads on Beams

35. Integration Not Always Needed • The Areas & Centroids of Common Shapes Can be found on Inside Back-Cover of the Text Book • Std Areas can be added & subtracted directly • Std Centroids can be combined using [LeverArm]∙[Intensity] methods

36. A beam supports a distributed load as shown. Determine the equivalent concentrated load and its Location on the Beam Solution Plan The magnitude of the concentrated load is equal to the total load (the area under the curve) The line of action of the concentrated load passes through the centroid of the area under the Load curve. The Equivalent Causes the SAME Moment about the beam-ends as does the Concentrated Loads Example:Trapezoidal Load Profile

37. Example:Trapezoidal Load Profile • SOLUTION: • The magnitude of the concentrated load is equal to the total load, or the area under the curve. • The line of action of the concentrated load passes through the area centroid of the curve.

38. WhiteBoard Work • For the Loading & Geometry shown Find: • The Equivalent Loading • HINT: Consider the Importance of the Pivot Point • The Scalar component of the Equivalent Moment about line OA Let’s WorkThis NiceProblem

39. Engineering 36 Appendix Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu