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

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.

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

Transmissibility: Equivalent Forces Principle of Transmissibility Conditions Of Equilibrium Or Motion Are Not Affected By TRANSMITTING A Force Along Its LINE OF ACTION Note: F & F’ Are Equivalent Forces Moving the point of application of F to the rear bumper does not affect the motion or the other forces acting on the truck

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

Moment of a Couple COUPLE  Two Forces F and −F With Same Magnitude Parallel Lines Of Action Distance Separation Opposite Direction Moment of The Couple about O The origin is arbitrary

M of a Couple → Free Vector 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 Two Couples Are Equal If F1d1 = F2d2 (d1 & d2 are ┴ distances) The Couples Lie In Parallel Planes The Couples Have The Tendency To Cause Rotation In The Same Direction

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

Couple Addition Consider Two Intersecting Planes P1 and P2 With Each Containing a Couple Resultants Of The Force Vectors Also Form a Couple R is F1 + F2 by tip to tail

Couple Addition By Varignon’s Distributive Theorem for Vectors 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 The resultant Couple is Also a FREE Vector

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 LoA Is NOT Significant Couple Vectors May Be Resolved Into Component Vectors

Resolution of a Force Into a Force at O and a Couple Can be FREELY Moved Couple r x F 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

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’ r’ is the tip-to-tail sum of r & s

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 Remember that MO is a FREE Vector, and must be ADDED at O’ to obtain MO’

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

Example: Couples 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

Example: Couples 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

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 single Resultant Force-Vector and a Resultant Couple-Vector

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

More Reduction of Force Systems 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 (a) (b) Solve − 𝑧 ∗ ∙ 𝑅 𝑦 = 𝑀 𝑥 𝑅 for 𝑧 ∗ • Solve 𝑥 ∗ ∙ 𝑅 𝑦 = 𝑀 𝑧 𝑅 for 𝑥 ∗ • Then locate 𝑅 𝑦 = 𝐹 1 + 𝐹 2 + 𝐹 3 at a location in the 𝑥𝑧 of 𝑥 ∗ , 𝑧 ∗ (c)

CoPlanar Force Systems System Of CoPlanar Forces Is Reduced To A Force-couple System That Is Mutually Perpendicular 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 & Y Axes

Parallel Force Systems Proper Placement of the Resultant (at 𝑥 ∗ , 𝑧 ∗ in this case Produces the SAME MOMENT about the 𝑥 & 𝑧 Axes as did the 3 the ∥ Vectors Example of this methodology on next Slide 𝑅 𝑦 = 𝐹 1 + 𝐹 2 + 𝐹 3 𝑧 ∗ 𝑥 ∗

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 .

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.

Example: 2D Equiv. Sys. - Soln 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 The Couple At B Is Equal To The Moment About B Of The Force-couple System Found At A rBA Equivalent does NOT means the same numbers. It means the same MECHANICAL effect. • from slide-19 s is from the NEW pivot to the previous Force LoA Point

Example: 2D Equiv. Sys. - Soln 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 Chk 1000 Nm at B Chk at B: 600*(4.8 – 3.13) = 600*1.67 = 1000 N-m Then the Single-Force Resultant 

Example: 3D Equiv. Sys. 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 3 Cables Are Attached To The Bracket As Shown. Replace The Forces With An Equivalent Force-Couple System at A Calculate The Equivalent Couple

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

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

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

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 the Point Load, W, is Determined by Area Under the Profile Curve

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

Distributed Loads Now Use Centroidal Methodology And also: Equating the Ω Expressions find

Distributed Loads on Beams 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.

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

Scalene TriAngle Centroid 1/3 above the Flat Base * http://www.efunda.com/math/areas/triangle.cfm

Example:Trapezoidal Load Profile 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 A beam supports a distributed load as shown. Determine the equivalent concentrated load and its Location on the Beam

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. Trapezoidal Load-Profile

Let’s Work This Nice Problem WhiteBoard Work Let’s Work This Nice Problem 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 S&T 5.4(5, 29) P29 shown. ref. ENGR-36_Lab-08_Fa16_Lec-Notes.ppt

WhtBd Soln to Channel Bracke Engineering 36 Appendix WhtBd Soln to Channel Bracke Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu