ME 201Engineering Mechanics: Statics

Chapter 5 – Part D

Equilibrium in Three Dimensions5.5 Free-Body Diagrams

5.6 Equations of Equilibrium

5.7 Constraints and Statical Determinancy

Constraints for a Rigid Body

Statically Determinate – reactive forces can be determined from equations of equilibrium

Statically Indeterminate – more unknown forces than equations of equilibrium. Requires use of deformation conditions to solve (Strength of Materials).

Redundant Constraints – more supports than necessary to hold body in equilibrium

Improper Constraints - # of unknowns may be okay, but improperly constrained – may cause instability

Equilibrium in 3-Dimensions

Support Reactions A force is developed by a support that restricts

the translation of the attached member A couple is developed when rotation of the

attached member is prevented

∑F = 0∑Fx = 0∑Fy = 0∑Fz = 0

∑M = 0∑Mx = 0∑My = 0∑Mz = 0

Common Support Reactions(See also Text pages 238-239)

2D 3DSimple or Roller Support

1 unknown: 1F, 0M

Smooth Surface Support

Roller Support

1 unknown: 1F, 0M

Pin or Hinge Support

2 unknowns: 2F, 0M

Ball & Socket Support

3 unknowns: 3F, 0M

Fixed Support

3 unknowns: 2F, 1M

Fixed Support

6 unknowns: 3F, 3M

Example Problem

Given:A-ball jointBC, BD-cablesF=1 kN

Find:

A, TBC, TBD

z

x

y6m6m A

C

B

D6m3m

F (0,0,6)

(-3,6,0)

(6,0,0)

How many unknown? 5 unknowns: Ax, Ay, Az, TBC, TBD

Example Problem SolutionGiven:A-ball jointBC, BD-cablesF=1 kN

Find:A, TBC, TBD

Solution:1-FBD2-Force (vectors)

z

x

y6m6mAx

C

B

D6m3m

1 kN(0,0,6)

(-3,6,0)

(6,0,0)

Ay

Az

TBC

TBD

F

z

x

y6m6m A

C

B

D6m3m

F(0,0,6)

(-3,6,0)

(6,0,0)

A

BCT

BDT

zyx AAA

010000

))6(6

606(

22

BCT BCBC TT 707.0707.

))6(6)3(

663(

222

BDT BDBDBD TTT 667.667.333.

Example Problem SolutionSolution:

1-FBD2-Force (vectors)3-Eqn of Equil

z

x

y6m6mAx

C

B

D6m3m

1 kN(0,0,6)

(-3,6,0)

(6,0,0)

Ay

Az

TBC

TBD

5 unknowns

BDBDBDBD

BCBCBC

zyx

TTT

TT

AAA

667.667.333.

707.0707.

010000

T

T

A

F

0F

3 from ∑F , 2 from ∑M

0 xF0333.707. BDBCx TTA

0 yF1000667. BDy TA

0 zF0667.707. BDBCz TTA

0 BDBC TTAF

0 AxM

0AM

0 AyM

Solution:1-FBD2-Force (vectors)3-Eqn of Equil

z

x

y6m6mAx

C

B

D6m3m

1 kN(0,0,6)

(-3,6,0)

(6,0,0)

Ay

Az

TBC

TBD

5 unknowns

BDBDBDBD

BCBCBC

zyx

TTT

TT

AAA

667.667.333.

707.0707.

010000

T

T

A

F

3 from ∑F , 2 from ∑M

Example Problem Solution

010000600FrAB

0)()()( BDABBCABAB TrTrFr

BCBCBCAB TT 707.0707.600Tr

BDBDBDBDAB TTT 667.667.333.600Tr

006000

0243.40 BCT 024 BDBD TT

060004 BDTNTBD 1500

0)1500(2243.4 BCTNTBC 707

NT

NT

BC

BD

707

1500

Substituting TBD and TBC into ∑F equations:

0667.707.

0

1000667.

0

0333.707.

0

BDBCz

z

BDy

y

BDBCx

x

TTA

TA

TTA

F

F

F 0 xF

Solution:1-FBD2-Force (vectors)3-Eqn of Equil

z

x

y6m6mAx

C

B

D6m3m

1 kN(0,0,6)

(-3,6,0)

(6,0,0)

Ay

Az

TBC

TBD

Example Problem Solution

0)1500(333.)707(707. xA

0xA

0 yF

1000)1500(667. yA

0yA

0 zF0)1500(667.)707(707. zA

NAz 1500