ENGR-36_Lec-09_Moments_Equiv
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Transcript ENGR-36_Lec-09_Moments_Equiv
Engineering 36
Chp 5:
Equivalent Loads
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
1
Bruce Mayer, PE
[email protected] • 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
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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
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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
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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
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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
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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
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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
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Bruce Mayer, PE
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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
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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
[email protected] • 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
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Bruce Mayer, PE
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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
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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
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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
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Example: Couples
Determine The
Components Of The
Single Couple
Equivalent To The
Couples Shown
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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
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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 lb18 in. 540lb in.
M y 20 lb12 in. 240lb in.
Mx
M z 20 lb9 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
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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 lbkˆ
9 in. ˆj 12 in.kˆ 20 lbiˆ
M 540lb in.iˆ 240lb in. ˆj 180lb in.kˆ
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • 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
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R
MO r F
Bruce Mayer, PE
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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
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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
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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
y0
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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x0
R
O
Bruce Mayer, PE
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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
.
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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
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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 mkˆ
Bruce Mayer, PE
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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 mkˆ 4.8 miˆ 600 N ˆj
1880N mkˆ 2880N mkˆ
Engineering-36: Engineering Mechanics - Statics
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R
M B 1000 N mk
Bruce Mayer, PE
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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 4.8iˆ 600 N ˆj
1.67iˆ 600 N ˆj
1002N m kˆ
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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
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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
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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
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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
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0.429iˆ 0.857 ˆj 0.289kˆ
FB 300iˆ 600 ˆj 200kˆ N
FC 1000N cos 45iˆ cos 45kˆ
707iˆ 707kˆ N
FD 1200N cos60 iˆ cos30 ˆj
600iˆ 1039ˆj N
Bruce Mayer, PE
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Example Equiv. Sys. - Solution
Compute Equivalent Force
R F
300 707 600iˆ
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
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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
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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
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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
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Distributed Loads
Now Use Centroidal Methodology
x
LeverArm Intensity x wx dx
n
span
And also:
x xW
span
x is theCentroidLocation
Equating the
Ω Expressions
x
find
Engineering-36: Engineering Mechanics - Statics
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n
x wx dx
n
n
span
W
Bruce Mayer, PE
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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
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Engineering-36: Engineering Mechanics - Statics
34
Bruce Mayer, PE
[email protected] • 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
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Bruce Mayer, PE
[email protected] • 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
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• 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
[email protected] • 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
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X 3.5 m
Bruce Mayer, PE
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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
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt
Engineering 36
Appendix
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
39
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt