[email protected] ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

[email protected] • ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx1

Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp 5: 2D Equil

Special Cases



2D Equil → Special Cases

PARTICLE: Size & Shape of the Object can be neglected as long as all applied Forces have a Point of Concurrency• Covered in Detail in Chp03

TWO-FORCE MEMBER: A Structural Element of negligible Wt with only 2 Forces acting on it

https://ecourses.ou.edu/cgi-bin/view_anime.cgi?file=m5521.swf&course=st&chap_sec=05.5','yes','240','196




THREE-FORCE MEMBER: A structural Element of negligible Wt with only 3 Forces acting on it• The forces must be either

concurrent or parallel.– In the PARALLEL Case the

PoC is located at Infinity– The NONparallel Case can

be Very Useful in LoadAnalysis

https://ecourses.ou.edu/cgi-bin/view_anime.cgi?file=m5522.swf&course=st&chap_sec=05.5','yes','240','196




FRICTIONLESS PULLEY: For a frictionless pulley in static equilibrium, the tension in the cable is the same on both sides of the pulley• Discussed Briefly in Chp03

– Will Prove the T1 = T2 = T Behavior Today



2D Planar System Equilibrium

In 2D systems it is assumed that• The System Geometry resides

completely the XY Plane• There is NO Tendency to

– Translate in the Z-Direction– Rotate about the X or Y Axes

These Conditions Simplify The Equilibrium Equations

000 zyx MFF



2D Planar System: 000 zyx MFF

No Z-Translation → NO Z-Directed Force:

000 zyx FFF

No X or Y Rotation → NO X or Y Applied Moments

000 zyx MMM



Special Case: 2-Force Member

A 2-Force Member/Element is a Body with negligible Weight and Only two applied Forces.

Some Special Properties of 2-Frc Ele’s• the LoA’s of the Two Forces MUST Cross

and thus Produce a PoC – Treat as a PARTICLE

• The Crossed LoA’s Define a PLANE– Treat as PLANAR System



2-Force Element Equilibrium Consider a L-Bracket plate subjected

to two forces F1 and F2

For static equilibrium, the sum of moments about Pt-A must be zero. Thus the moment of F2 About Pt-A must be zero. It follows that the line of action of F2 must pass through Pt-A

Similarly, the line of action of F1 must pass through Pt-B for the sum of moments about Pt-B to be zero.

Requiring that the sum of forces in any direction be zero leads to the conclusion that F1 and F2 must have equal magnitude but opposite sense.



Special Case: 2-Force Element

Mathematically• Since the Two Forces Must be Concurrent

00 F'sall F'sall

FFrM FPoCPoC

• Since the System is in Equilibrium ΣF’s =0.

ABBA FFFFF 0 F'sall

– Thus the two force are Equal and Opposite; that is, the forces CANCEL



Special Case: 3-Force Member

A 3-Force Element is a PLANAR Body with negligible Weight with Exactly 3 applied Forces (No applied Moments).

Claim: If a Planar 3-Force Element is in Equilibrium, Then the LoA’s for the

3-Forces must be CONCURRENT• If the Claim is TRUE, then the 3-Force

Element can be treated as a PARTICLE



3-Force 2D Body Equilibrium

Consider a Planar rigid body subjected to forces acting at only 3 points.

The lines of action of intersect F1 & F2, at Pt-D. The moment of F1 and F2 about this point of intersection is zero.

Since the rigid body is in equilibrium, the sum of the moments of F1, F2, and F3 about ANY Pivot-Pt must be zero. It follows that the moment of F3 about D must be zero as well and that the line of action of F3 must pass through D. The lines of action of the three forces must be Concurrent OR Parallel.



3-Force 2D Body: Parallel Forces

F1

F2

F3

d1d2

d3

O

If 3 Parallel Forces Maintain a Rigid Body in Static Equilibrium, The following Conditions MUST be Satisfied• For Translation Equilibrium

2130 FFFFx • For Rotation Equilibrium

223311

00

FdFdFd

O

FrM

x




Mathematically for ||-Forces• Since a Body in Equil. Has NO Net Moment

0sd' s,F' allsF' all

mmFOOFdFrM

• Since the System is in Equilibrium ΣF’s =0.

CBA FFFF 0 F'sall

• In Summary: The dmFm products and, 3 Forces, Sum to Zero




A Graphical Summary

AB is 3F Member(BC is 2F Member)



Example Pole Raising Solution Plan

• Create a free-body diagram of the joist. – Note that the joist is a 3

force body acted upon by the ROPE, its WEIGHT, and the REACTION at A

• The three forces must be concurrent for static equilibrium. Therefore, the reaction R must pass through the intersection of the lines of action of the weight and rope forces. Determine the direction of the reaction force R.

A man Raises a 10 kg Joist, of Length 4 m, by pulling on a rope.

Find the TENSION in the rope and the REACTION at A.



Example Pole Raising Use LoA’s & Trigonometry to

Determine the direction of the reaction force R

Create a free-body diagram of the joist

636.1414.1

313.2tan

m 2.313m 515.0828.2

m 515.020tanm 414.1)2545cot(

m 414.1

m828.245cosm445cos

21

AE

CE

BDBFCE

CDBD

AFAECD

ABBFAF

6.58A LARGE, SCALED Diagram is REALLY

Useful in this Problem

70



70º Angle Analysis



Example Pole Raising Use the Law of the Sines to

Find the Reaction Force R Draw the Force

Triangle to Scale

38.6sin

N 1.98

110sin4.31sin RT

N 8.147

N9.81

R

T = 58.6° = 58.6°

Solving find



Special Case: Frictionless Pulley

A FrictionLess Pulley is Typically used to change the Direction of a Cable or Rope in Tension

Pulley with PERFECT Axel (FrictionLess)



Special Case: FrictionLess Pulley

A Perfect Axel Generates NO moment to Resist Turning.

Consider the FBD for a Perfect Pulley• Since the LoA’s for FAx & FAy

Pass Thru the Axel-Axis Pt-A they Generate No moment about this point .

• T1 and T2 have Exactly the SAME Lever arm, i.e., the Radius, R, of the Pulley



Special Case: FrictionLess Pulley

Since the Pulley is in Equilibrium ΣMA = 0

Writing the Moment Eqn

0or

0

0

21

21

T's all

TTkR

kTRkTR

A

ˆ

ˆˆ

TrM

Thus for the NO-Friction Perfect Pulley 21 TT



FritionFilled Pulley

Consider the case where we have a pulley that is NOT Free Wheeling; i.e., the pulley resists rotation

Example: Automobile alternator changes thermal-mechanical energy into electrical energy

1T

2T

http://images.google.com/imgres?imgurl=http://www.airpowersa.co.za/Plane%20power%20alternator.jpg&imgrefurl=http://www.airpowersa.co.za/indexmain.htm&h=1061&w=1326&sz=68&hl=en&start=21&tbnid=wftICpAOPbgnRM:&tbnh=120&tbnw=150&prev=/images?q=alternator&start=20&gbv=2&ndsp=20&svnum=10&hl=en&sa=N



FrictionFilled Pulley

In Alternator Operation the generation of electricity produces a resisting moment that counters the direction of spin; The FBD in this case →

The ΣMA = 0

MAz

021 kMkTRkTR Aˆˆˆ



FrictionFilled Pulley

Thus a RESISTING Moment causes a DIFFERENCE between the two Tensions

More on This when we Learn Chp08

MAz

RMTT

OR

MTTR

A

A

21

21 0



FrictionLess Pulley; 3F Mem In the System at

Right Member ABC, which is a FOUR-Force System, can be reduced to a 3-Force System using and Equivalent Resultant-Couple System at the Pulley



FrictionLess Pulley; 3F Mem

Recall that Forces Can be MOVED to a new point on a Body as long as the Rotation Tendency caused by the move is accounted for by the Addition of a COUPLE-Moment at the new Point



FrictionLess Pulley; 3F Mem Apply the Equivalent

Loading Method to a FrictionLess Pulley

From the Previous Discussion the MOMENT about the Axle (Pin) of a Frictionless pulley produced by the Tensions is ZERO

Thus Can Move the T’s to the Pin with a Couple of ZERO

1T

2T

21 TT




The Equivalent Systems by MA = 0

T

T

T

T

RT




Moving the FrictionLess Pulley Force-Resultant to the Pin at Pt-A produces the FBD Shown At Right• Now can Draw the

Force Triangle

RT

C

B



FrictionLess Pulley - Important

For a FrictionLess Pulley the Tension Forces and be to the Pulley Axel (Pin) WithOUT the Addition of a Couple

T

T

T

T=



Special Cases Summarized

Particle:

AB FF

3-Force Planar Element:

0 CBA FFF

2-Force Element:

0003D

002D

zyx

yx

FFF

FF

21 TT FrictionLess Pulley:



WhiteBoard Work

Lets WorkThese NiceProblems



Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix



Jib Problem The upper portion of the

crane boom consists of the jib AB, which is supported by the pin at A, the guy line BC, and the backstay CD, each cable being separately attached to the mast at C. If the 5-kN load is supported by the hoist line, which passes over the pulley at B, determine the magnitude of the resultant force the pin exerts on the jib at A for equilibrium, the tension in the guy line BC, and the tension T in the hoist line. Neglect the weight of the jib. The pulley at B has a radius of 0.1 m.



Disk Problem

The smooth disksD and E have a weight of 200 lb and 100 lb, respectively. Determine the largest horizontal force P that can be applied to the center of disk E without causing the disk D to move up the incline.

Documents

[email protected] ENGR-36_Lec-12_2D-Equilibrium_SpecialCases_2n3_F_members.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics