BMayer@ChabotCollege.edu ENGR-36_Lec-08_Moments_Equiv-Loads.ppt 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Licensed.
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BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt1
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering 36
Chp 5: Equivalent
Loads
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt2
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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.
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt3
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt4
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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 tothe rear bumper doesnot affect the motion or the other forcesacting on the truck
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt5
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Transmissibility Limitations
Principle of transmissibility may not always apply in determining• Internal Forces• Deformations
Rigid Deformed
TENSION
COMPRESSION
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt6
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
distancethesin
dFdrFM
Fr
Frr
FrFrM
BA
BA
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt7
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt8
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt9
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Couple Addition
Consider Two IntersectingPlanes P1 and P2 WithEach Containing a Couple
222
111
planein
planein
PFrM
PFrM
21 FFrRrM
Resultants Of The Force Vectors Also Form a Couple
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt10
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Couple Addition
By Varignon’s Distributive Theoremfor Vectors
21
21
MM
FrFrM
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
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt11
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt12
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt13
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Force-Couple System at O’
FrMO
'
FsMM
FsFrFrsFrM
OO
O
'
' '
The Moments of F about O and O’ are Related By The Vector S That Joins O and O’
Moving F from A To a Different Point O’ Requires Addition of a Different Couple Vector
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt14
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Force-Couple System at O’
FsMM OO
'
Moving The Force-couple System From O to O’ Requires The Addition Of The Moment About O’ Generated by the Force At O
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt15
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: Couples
Determine The Components Of The Single Couple Equivalent To The Couples Shown
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
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt16
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
in.lb 540in. 18lb 30 xM
kji ˆˆˆM in.lb 180in.lb240in.lb 540
Mx
Mx
My
My
Mz
Mz
in.lb240in. 12lb 20 yM
in.lb 180in. 9lb 20 zM
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt17
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
ikj
kjD
ˆˆˆ
ˆˆMM
lb 20in. 12in. 9
lb 30in. 18
rDErDC
kji ˆˆˆM in.lb 180in.lb240in.lb 540
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt18
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Reduction to Force-Couple Sys
FrMFR RO
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
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt19
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Reduction to a Force-Couple Sys
RsMM RO
RO
'
Two Systems Of Forces Are EQUIVALENT If They Can Be Reduced To The SAME Force-Couple System
The Force-Couple System at O May Be Moved To O’ With The Addition Of The Moment Of R About O’ as before:
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt20
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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 Forcea) Concurrent Forces
– Generates NO Moment
b) Coplanar Forces (next slide)
c) The Forces Are Parallel– CoOrds for Vertical Forces
(a)
(b)
(c)
Rzy
Rxy
MxR
MzR
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt21
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
CoPlanar Force Systems System Of CoPlanar Forces Is
Reduced To A Force-couple System That Is Mutually Perpendicular
ROxy MyRxR
RMd RO
FrMFR RO
and
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
0y 0x
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt22
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: 2D Equiv. Sys.
.
For The Beam, Reduce The System Of Forces Shown To
a) An Equivalent Force-Couple System At A
b) An Equivalent Force-Couple System At B
c) A Single Force applied at the Correct Location
Solution Plana) Compute
– The Resultant Force – The Resultant Couple
About A
b) Find An Equivalent Force-couple System at B Based On The Force-couple 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
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt23
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: 2D Equiv. Sys. - Soln
jjjj
FR
ˆˆˆˆ N 250N 100N 600N 150
jR
N600 Now Calculate the Total Moment
About A as Generated by the Individual Forces.
ji
jiji
FrM RA
ˆˆ.
ˆˆ.ˆˆ.
25084
1008260061
kM RA
ˆmN 1880
a) Find the resultant force and the resultant couple at A.
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt24
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: 2D Equiv. Sys. - Solnb) 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
jR ˆN 600
kk
jik
RrMM BARA
RB
ˆˆ
ˆˆ.ˆ
mN 2880mN 1880
N 600m 84mN 1880
kM RB
mN 1000
• The Couple At B Is Equal To The Moment About B Of The Force-couple System Found At A
rBA
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt25
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: 2D Equiv. Sys. - Solnc) 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
m 133N 600 .ˆ xjR
kxk
jixk
RrM AxRA
ˆN 600ˆmN 1880
ˆN 600ˆˆmN 1880
Then the Single-Force Resultant
kji
ji
RrM BxRB
ˆmN 1002
ˆN 600ˆ67.1
ˆN 600ˆ8.413.3
Chk 1000 Nm at B
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt26
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: 3D Equiv. Sys.
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
FR
FrM RA
• Calculate The Equivalent Couple
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt27
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example Equiv. Sys. - Solution
Determine The Relative Position Vectors w.r.t. A
m 10001000
m 05000750
m 05000750
jir
kir
kir
AD
AC
AB
..
..
..
N ˆ200ˆ600ˆ300
ˆ289.0ˆ857.0ˆ429.0
175
5015075ˆ
ˆN 700
kjiF
kji
kji
r
ru
uF
B
BE
BE
B
N ˆ707ˆ707
ˆ45cosˆ45cosN 1000
ki
kiFC
N ˆ1039ˆ600
ˆ30cosˆ60cosN 1200
ji
jiFD
Resolve The Forces Into Rectangular Components
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt28
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example Equiv. Sys. - Solution Compute Equivalent Force
k
j
i
FR
ˆ
ˆ
ˆ
707200
1039600
600707300
N 5074391607 kjiR ˆˆˆ
k
kji
Fr
j
kji
Fr
ki
kji
Fr
FrM
DAD
cAC
BAB
RA
ˆ...
ˆˆˆ
ˆ...
ˆˆˆ
ˆˆ..
ˆˆˆ
9163
01039600
010001000
6817
7070707
050000750
4530
200600300
050000750
kjiM RA
9.11868.1730
Compute Equivalent Couple
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt29
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Distributed Loads
The Load on an Object may be Spread out, or Distributed over the surface.
Load Profile, w(x)
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt30
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
span
dxxwW
N 3
100W
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt31
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt32
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Distributed Loads
Now Use Centroidal Methodology
span
nn
span
x dxxwxIntensityLeverArm
And also:
Location Centroid theis xWxx Equating the
Ω Expressionsfind
W
dxxwx
x span
nn
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt33
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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.
AdAdxwWL
0
AxdAxAOP
dWxWOP
L
0
• 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.
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt34
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt35
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt36
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example:Trapezoidal Load Profile
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
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt37
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example:Trapezoidal Load ProfileSOLUTION:
• The magnitude of the concentrated load is equal to the total load, or the area under the curve.
kN 0.18F
• The line of action of the concentrated load passes through the area centroid of the curve.
kN 18
mkN 63 X m5.3X
m6m
N
2
45001500
F
BMayer@ChabotCollege.edu • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt38
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
WhiteBoard Work
Let’s WorkThis NiceProblem 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
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