Engineering 36
Chp 5:
Equivalent Loads
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-36: Engineering Mechanics - Statics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
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.
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
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
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
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
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Transmissibility Limitations
 Principle of transmissibility may not always
apply in determining
• Internal Forces
• Deformations
Rigid
Deformed

TENSION

COMPRESSION
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
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
   

M  rA  F  rB   F

 
 rA  rB  F
 
 r F
M  F r sin    Fd d  the  distance
 
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
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
• The Couples Lie In Parallel Planes
• The Couples Have The Tendency
To Cause Rotation In The
Same Direction
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
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
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Couple Addition
 Consider Two Intersecting
Planes P1 and P2 With
Each Containing a Couple

 
M 1  r  F1 in plane P1

 
M 2  r  F2 in plane P2
 Resultants Of The Force Vectors
Also Form a Couple
     
M  r  R  r  F1  F2 
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Couple Addition
 By Varignon’s Distributive Theorem
for Vectors
    
M  r  F1  r  F2


 M1  M 2
 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
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
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
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Resolution of a Force Into a
Force at O and a Couple
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
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Force-Couple System at O’
 Moving F from A To a Different Point O’
Requires Addition of a Different
Couple Vector


M O'  r   F
 The Moments of F about O and O’ are Related
By The Vector S That Joins O and O’



        
M O '  r 'F  s  r  F  r  F  s  F


 
 M O'  M O  s  F
 
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Force-Couple System at O’


 
M O'  M O  s  F
 Moving The Force-couple System From O
to O’ Requires The Addition Of The Moment
About O’ Generated by the Force At O
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Example: Couples
 Determine The
Components Of The
Single Couple
Equivalent To The
Couples Shown
Engineering-36: Engineering Mechanics - Statics
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 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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Example: Couples
 Attach Equal And Opposite 20 lb
Forces In the ±x Direction at A
• No Net Change to the Structure
Mx
Mz
My
My
 The Three Couples May Be
Represented By 3 Vector Pairs
Mz
M x  30 lb18 in.  540lb  in.
M y  20 lb12 in.  240lb  in.
Mx
M z  20 lb9 in.  180lb  in.
M  540lb  in.iˆ  240lb  in. ˆj  180lb  in.kˆ
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Example: Couples
rDC
rDE
 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
M  M D  18 in. ˆj   30 lbkˆ
 9 in. ˆj  12 in.kˆ   20 lbiˆ


M  540lb  in.iˆ  240lb  in. ˆj  180lb  in.kˆ
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Reduction to Force-Couple Sys
 A SYSTEM OF FORCES May Be
REPLACED By A Collection Of FORCECOUPLE SYSTEMS Acting at Given Point O
 The Force And Couple Vectors May then Be
Combined Into a single Resultant ForceVector and a Resultant Couple-Vector


R  F
Engineering-36: Engineering Mechanics - Statics
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
R
 
MO   r  F

Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
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:
R
R  
M O'  M O  s  R
 Two Systems Of Forces Are
EQUIVALENT If They Can Be
Reduced To The SAME
Force-Couple System
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
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.
(a)
(b)
Force Systems That Can be
Reduced to a Single Force
a) Concurrent Forces
– Generates NO Moment
b) Coplanar Forces (next slide)
R

zR

M
c) The Forces Are Parallel
y
x
– CoOrds for Vertical Forces
Engineering-36: Engineering Mechanics - Statics
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xRy  M zR
(c)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
CoPlanar Force Systems
 System Of CoPlanar Forces Is
Reduced To A Force-couple System
That
Perpendicular
 Is Mutually



R   F and M  
R
O


r F

 System Can Be Reduced To a Single
Force By Moving The Line Of Action
R To Point-A Such That d:
d  M OR R
y0
 In Cartesian Coordinates use
transmissibility to slide the Force
PoA to Points on the X & Axes
xRy  yRx  M
Engineering-36: Engineering Mechanics - Statics
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x0
R
O
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Example: 2D Equiv. Sys.

Solution Plan
a) Compute
–
–

For The Beam, Reduce
The System Of Forces
Shown To
a) An Equivalent ForceCouple System At A
b) An Equivalent ForceCouple System At B
c) A Single Force applied
at the Correct Location
.
Engineering-36: Engineering Mechanics - Statics
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The Resultant Force
The Resultant Couple
About A
b) Find An Equivalent
Force-couple System at
B Based On The Forcecouple System At A
c) Determine The Point
Of Application For The
Resultant Force Such
That Its Moment About
A Is Equal To The
Resultant Couple at A
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Example: 2D Equiv. Sys. - Soln
a) Find the resultant force and the
resultant couple at A.


R  F
 150 N  ˆj  600 N  ˆj  100 N  ˆj  250 N  ˆj


R  600 N j

Now Calculate the Total Moment
About A as Generated by the
Individual Forces.

R
 
MA   r F

  
   
  

 1.6 iˆ   600ˆj  2.8 iˆ  100ˆj
 4.8 iˆ   250ˆj
Engineering-36: Engineering Mechanics - Statics
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R
M A  1880N  mkˆ
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt

Example: 2D Equiv. Sys. - Soln
b) Find An Equivalent Forcecouple 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

R  600 N ˆj
rBA
•
The Couple At B Is Equal To The
Moment About B Of The Forcecouple System Found At A
R R 

M B  M A  rBA  R
 1880N  mkˆ   4.8 miˆ   600 N  ˆj
 1880N  mkˆ  2880N  mkˆ
Engineering-36: Engineering Mechanics - Statics
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
R
M B  1000 N  mk
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Example: 2D Equiv. Sys. - Soln
c) Determine a SINGLE
Resultant Force (NO Couple)
•
•

Chk 1000 Nm at B
R


M B   rBx  R
 3.13  4.8iˆ   600 N  ˆj
 1.67iˆ   600 N  ˆj
 1002N  m kˆ


Engineering-36: Engineering Mechanics - Statics
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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
R

System

M A   rAx  R
 1880N  m kˆ  xiˆ   600 N  ˆj
 1880N  m kˆ  600x N kˆ
Then the Single-Force Resultant

R  600 N ˆj
x  3.13 m
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Example: 3D Equiv. Sys.
 Solution Plan:
 3 Cables Are Attached
To The Bracket As
Shown. Replace The
Forces With An
Equivalent ForceCouple System at A
Engineering-36: Engineering Mechanics - Statics
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• 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 

R  F
• Calculate The Equivalent
Couple
R
 
M A   r  F 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Example Equiv. Sys. - Solution
 Resolve The Forces Into
Rectangular Components

FB  700 N uˆ




r
75i  150 j  50k
uˆ  BE 
rBE
175
 Determine The Relative
Position Vectors w.r.t.
A



rAB  0.075i  0.050k m 



rAC  0.075i  0.050k m 



rAD  0.100i  0.100 j m
Engineering-36: Engineering Mechanics - Statics
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 0.429iˆ  0.857 ˆj  0.289kˆ

FB  300iˆ  600 ˆj  200kˆ N 

FC  1000N  cos 45iˆ  cos 45kˆ
 707iˆ  707kˆ N 





FD  1200N  cos60 iˆ  cos30 ˆj
 600iˆ  1039ˆj N 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Example Equiv. Sys. - Solution
 Compute Equivalent Force


R  F
 300  707  600iˆ
  600  1039 ˆj
 200  707 kˆ

R  1607iˆ  439ˆj  507kˆ N
 Compute Equivalent Couple

R
 
MA   r F

iˆ


rAB  F B 0.075
300
kˆ
ˆj
0.050  30iˆ  45kˆ
0
 600
200
iˆ
ˆj
kˆ
707
0
 707


rAC  F c 0.075 0  0.050  17.68 ˆj
iˆ
ˆj
kˆ


rAD  F D 0.100  0.100 0  163.9kˆ
600
1039 0

R


M A  30i 17.68 j 118.9k
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Distributed Loads
 The Load on an Object may be Spread
out, or Distributed over the surface.
Load Profile, w(x)
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
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
W  w x dx
Point Load, W, is
span
Determined
by Area
100
W
N
Under the
3
Profile Curve



Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
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
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Distributed Loads
 Now Use Centroidal Methodology
x 
 LeverArm Intensity    x wx dx 
n
span
 And also:
 x  xW
span
x is theCentroidLocation
 Equating the
Ω Expressions
x
find
Engineering-36: Engineering Mechanics - Statics
32
n
 x wx dx
n
n
span
W
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Distributed Loads on Beams
L
W   wdx   dA  A
0
OP W   xdW
L
OP  A   xdA  x A
0
• 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.
Engineering-36: Engineering Mechanics - Statics
33
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Engineering-36: Engineering Mechanics - Statics
34
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Integration Not Always Needed
 The Areas & Centroids of Common
Shapes Can be found on Inside BackCover of the Text Book
 Std Areas can be
added & subtracted
directly
 Std Centroids can
be combined using
[LeverArm]∙[Intensity]
methods
Engineering-36: Engineering Mechanics - Statics
35
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Example:Trapezoidal Load Profile
 Solution Plan
 A beam supports a
distributed load as
shown. Determine the
equivalent concentrated
load and its Location on
the Beam
Engineering-36: Engineering Mechanics - Statics
36
• 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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Example:Trapezoidal Load Profile
SOLUTION:
• The magnitude of the concentrated load is
equal to the total load, or the area under the
curve.
1500  4500 N
F
 6m F  18.0 kN
2
m
• The line of action of the concentrated load
passes through the area centroid of the curve.
63 kN  m
X 
18 kN
Engineering-36: Engineering Mechanics - Statics
37
X  3.5 m
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
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
Engineering-36: Engineering Mechanics - Statics
38
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Engineering 36
Appendix
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-36: Engineering Mechanics - Statics
39
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt