Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

[email protected] • ENGR-36_Lec-11_Mech_Equilibrium_2D-n-3D.pptx1

Bruce Mayer, PE Engineering-36: Vector Mechanics - Statics1

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp 5: MechEquilibriu

m



ReCall Equilibrium Conditions Rigid Body in Static Equilibrium

Characterized by• Balanced external forces and moments

– Will impart no Tendency toward Translational or Rotational motion to the body

The NECESSARY and SUFFICIENT condition for the static equilibrium of a body are that the RESULTANT Force and Couple from all external forces form a system equivalent to zero



Equilibrium cont. Rigid Body Equilibrium Mathematically

0and0 FrMF O

Resolving into Rectangular Components the Resultant Forces & Moments Leads to an Equivalent Definition of Rigid Body Equilibrium

000000

zyx

zyx

MMMFFF



2D Planar System 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



2D Planar System:

kFrFrjFriFrFFrr

kjiFr xyyxxxyy

yx

yxˆˆ00ˆ00

00

ˆˆˆ

So in this case M due to rxF is confined to the Z-Direction:

kFrFrkMFr xyyxzzXYˆˆM

If r Lies in the XY Plane, then rz = 0. With Fz = 0 the rxF Determinant:

000 zyx MFF

0xM 0yM 0zM



Example Crane problem

A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G.

Determine the components of the reactions at A and B.

Solution Plan• Create a free-body diagram

for the crane• Determine Rcns at B by

solving the equation for the sum of the moments of all forces about Pin-A – Note there will be no

Moment contribution from the unknown reactions at A.

• Determine the reactions at A by solving the equations for the sum of all horizontal force components and all vertical force components.




Solution Plan cont.• Check the values obtained

for the reactions by verifying that the moments about B of all forces Sum to zero.

Reaction Analysis• PIN at A

– X & Y Reactions• ROCKER at B

– NORMAL Reaction Only +X Rcn in This Case




Determine the reactions at Pt-A by solving the eqns for the sum of all horizontal & vertical forces

Determine Pt-B Rcn by solving the equation for the sum of the moments of all forces about Pt-A.

0m6kN5.23

m2kN81.9m5.1:0

BM A

kN1.107B

0:0 BAF xx

kN1.107xA

0kN5.23kN81.9:0 yy AF

kN 3.33yA

Draw the Free BodyDiagram (FBD)



Example Coal Car

A loaded coal car is at rest on an inclined track. The car has a gross weight of 5500 lb, for the car and its load as applied at G. The car is held in position by the cable.

Determine the TENSION in the cable and the REACTION at each pair of wheels.


for the car with the coord system ALIGNED WITH the CABLE

• Determine the reactions at the wheels by solving equations for the sum of moments about points above each axle, in the LoA of the Pull Cable

• Determine the cable tension by solving the equation for the sum of force components parallel to the cable



Example Coal Car Determine the reactions at

the wheels Create Free Body Dia.

lb 232025sinlb 5500

lb 498025coslb 5500

y

x

W

W

00in.5

in.6lb 9804in.25lb 2320:0

2

R

M A

lb 17582 R

Determine cable tension

00in.5

in.6lb 9804in.25lb 2320:0

1

R

M B

lb 5621 R

0Tlb 4980:0 xF

lb 4980T



Example Cable Braced Beam

The frame supports part of the roof of a small building. The tension in the cable is 150 kN.

Determine the reaction at the fixed end E.


for the frame and cable.• Solve 3 equilibrium

equations for the reaction force components and Moment at E.

Reaction Analysis• The Support at E is a

CANTILEVER; can Resist– Planar Forces– Planar Moment

Also

mDF 5.765.4 22



Triangle Trig Review

adjopp

hypopp

hypadj

tansincos

SohCahTo

a



Example Cable Braced Beam Create a free-body

diagram for the frame and cable

0kN1505.75.4:0 xx EF

kN 0.90xE

0kN1505.7

6kN204:0 yy EF

kN 200yE

:0EM

0m5.4kN1505.7

6

m8.1kN20m6.3kN20

m4.5kN20m7.2kN20

EM

mkN0.180 EM

Solve 3 equilibrium eqns for the reaction force components and couple

sin

cos



Rigid Body Equilibrium in 3D SIX scalar equations are required to

express the conditions for the equilibrium of a rigid body in the general THREE dimensional case

000000

zyx

zyxMMMFFF

These equations can be solved for NO MORE than 6 unknowns • The Unknowns generally represent

REACTIONS at Supports or Connections.



Rigid Body Equilibrium in 3D The scalar equations are often

conveniently obtained by applying the vector forms of the conditions for equilibrium

Solve the Above Eqns with the Usual Techniques; e.g., Determinant Operations • A Clever Choice for the Pivot Point, O,

Can Eliminate from the Calculation as Many as 3 Unknowns (a PoC somewhere)

00 FrMF O



Example 3D Ball-n-Socket Solution Plan

• Create a free-body diagram of the Sign

• Apply the conditions for static equilibrium to develop equations for the unknown reactions.

Reaction Analysis• Ball-n-Socket at A can

Resist Translation in 3D– Can NOT Resist TWIST

(moment) in Any Direction• Cables Pull on Their

Geometric Axis

A sign of Uniform Density weighs 270 lb and is supported by a BALL-AND-SOCKET joint at A and by two Cables.

Determine the tension in each cable and the reaction at A.



3D Ball-n-Socket Prob cont. The x-Axis rotation Means

that the Sign is Only Partially Constrained. Mathematically

Create a free-body diagram of the Sign

Note: The Sign can SWING about the x-Axis (Wind Accommodation?)

AnythingAxM• But If the Sign is NOT Swinging

(i.e., it’s hanging STILL) The Following Condition Must Exist

0AxM With The Absence of the Mx

Constraint This Problem Generates only 5 Unknowns• MX = 0 by X-Axis P.O.A.

for W, TEC & TBC



3D Ball-n-Socket Prob cont.2

State in Component-Form the Two Cable-Tension Vectors

kjiT

kjiT

ECECT

rrrrTT

kjiT

kjiT

rrT

rrrrTT

EC

EC

ECAEAC

AEACECEC

BD

BD

BD

BDBD

ABAD

ABADBDBD

ˆˆˆ7

ˆ2ˆ3ˆ6

ˆˆˆ12

ˆ8ˆ4ˆ8

72

73

76

32

31

32



3D Ball-n-Socket Prob cont.3

Apply the conditions

for static equilibrium to develop equations for the unknown reactions.

0lbft1080571.2667.2:ˆ0714.1333.5:ˆ0ˆlb 270ˆft 4

0:ˆ0lb 270:ˆ

0:ˆ0ˆlb 270

72

32

73

31

76

32

ECBD

ECBD

ECAEBDABA

ECBDz

ECBDy

ECBDx

ECBD

TTk

TTj

ji

TrTrM

TTAk

TTAj

TTAi

jTTAF

Solving 5 Eqns in 5 Unkwns

kjiA

TT ECBD

ˆˆˆ.

lb 22.5lb 101.2lb 338

lb 315lb 3101



WhiteBoard Work

Solve 3DProblem byMechanics& MATLAB

Determine the tensions in the cables and the components of reaction acting on the smooth collar at G necessary to hold the 2000 N sign in equilibrium. The sign weight may concentrated at the center of gravity.

>> r = [5 -3 -7]r = 5 -3 -7 >> F = [-13 8 11]F = -13 8 11 >> rxF = cross(r,F)rxF = 23 36 1









Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix



0yF

kMTTkRkMkTRkTR AAˆˆˆˆˆ 2121 OR0

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

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