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

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. 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

4. 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

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

6. 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

7. 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

8. 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

9. 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

10. 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

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

12. 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

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’
r’ is the tip-to-tail sum of r & s

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

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. 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

17. 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

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 single Resultant 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. 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)

21. 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

22. 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 ∗

24. 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 .

25. 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.

26. 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

27. 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 

28. 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

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

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

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

32. 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

33. 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

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

35. 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.

37. 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

38. Scalene TriAngle Centroid
1/3 above the Flat Base *

39. 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

40. 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

41. 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

42. WhtBd Soln to Channel Bracke
Engineering 36
Appendix
WhtBd Soln to Channel Bracke
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer