e14 - applied mechanics: staticsbiomechanics.stanford.edu/e14/e14_s11.pdfe14 - applied mechanics:...

1mon/wed/fri, 12:50-2:05pm, 370-370

e14 - applied mechanics: statics

2syllabus

e14 - applied mechanics: statics

fourth homework

due

35. equilibrium of a rigid body

• to develop equations of equilibrium for a rigid body• to introduce the concept of a free body diagram for a rigid body• to show how to solve rigid body equilibrium problems

today‘s objectives

45.3 equations of 2d equilibrium

three alternative sets of eqns

MO = 0 Fx = 0 Fy = 0• two force & one moment equilibrium equations

MB = 0 F = 0• one force & two moment equilibrium equations

MA = 0

MC = 0• no force & three moment equilibrium equations

MB = 0 MA = 0

line through A and B must be ! to x-axis

A, B and C must not be on one line


procedure for drawing a FBD

I. isolate the system of interestII. identify all forces & moments

applied loadingreactions @support & contactweight of the body

III. label forces & give dimensions











W

d1

AH2

H1

d2


equilibrium analysis

MO = 0 :

Fx = 0 :

Fy = 0 :

H2 = -H1

A - W = 0 A = W

H1d2 - Wd1=0 H1 = W d1/d2

for d1= 2d2, both arms have to support twice the weight

W

d1

AH2

H1

d2O

H1 + H2 =0

tension

com-pression

95.4 two- and three-force members

• pin-connected at both ends• weightless• no extra forces acting on it

two-force members

FAB and FBA are equal andopposite FAB = FBA,, resultingfrom Fx=0 and Fy=0, andlie along the same line ofaction, resulting from MO=0


three-force members

MO=0 requires that thethree forces form aconcurrent (meeting at acommon point O) or parallel(meeting at ") force system


three-force members

MO=0 requires that the three forces form a concurrentforce system, their lines of action interset at a common point

W

F1F2


three-force members

MO=0 requires that the three forces form a concurrentforce system, their lines of action interset at a common point

W

F2

F1100N

800N

600N

450Ngraphic solution using vector addition

W

F1F2

135.5 free body diagram


I. isolate the system ofinterest

II. identify all forces &momentsapplied loadingreactions

support, contact forcesweight of the body

III. label each force & givedimensions

Fhr

Fhl

Ffr

Ffl

W


support reactions in 2d - memorize!

roller pin fixed

no motion # force / no rotation # moment


support reactions in 2d - memorize!

no motion # force / no rotation # moment

pin-like

roller-like

W

BH

A

BV


support reactions - example 04

ball-and-socket joint



journal bearing



single smooth pin

195.6 equations of 3d equilibirum

force and moment equilibrium

(MR)O = MO = 0

• forces sum up to zero

• moments at point O sum up to zero

FR = F = 0 Fz = 0 Fx = 0 Fy = 0

Mz = 0 Mx = 0 My = 0


example 5.15


example 5.15


example 5.15

235.7 constraints & statical determinacy

too much support

2d3 equations

3d6 equations


too much support

2d3 equations

3d6 equations

2d5 unknowns

3d8 unknowns

W

F

N1

T3

M

T1

N3

T2

N2

statically indeterminate


too much support

2d3 equations

2d8 unknowns


not enough support

2d3 equations

2d3 equations


not enough support

2d3 equations

2d2 unknowns

2d3 equations

2d3 unknowns

MA = 0 moment is not supported # system will rotate!

Fy = 0 force is not supported # system will translate!


not enough support

MA = 0 moment is not supported # system will rotate!

M

the system is improperly supported if all reaction forces areparallel

F1 F2 F3

2d3 equations

2d3 unknowns

296. structural analysis

equilibrium of structural systems

306. structural analysis

equilibrium of structural systems

W1

W2

F

N

F

N T

FH

FVM

e14 - applied mechanics: staticsbiomechanics.stanford.edu/e14/e14_s11.pdfe14 - applied mechanics:...

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