e14 - applied mechanics: staticsbiomechanics.stanford.edu/e14/e14_s10.pdf · 2011. 5. 2. · 5....

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

e14 - applied mechanics: statics

2syllabus

e14 - applied mechanics: statics

fourth homework

due

3midterm teaching evaluation

today‘s objectivestoday‘s objectives





65. 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

75.3 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


tips and tricks

• apply MO = 0 first! direction solution of one unknown• consider reorienting the x,y system along the members• if a force is negative, it is pointing in the opposite direction• use MA = 0 as control after solving for all unknowns

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


example 5.13


example 5.13

155.5 free body diagram

procedure for drawing a FBD

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




pin-like

roller-like

W

BH

A

BV














support reactions - example 04

ball-and-socket joint



journal bearing



single smooth pin


example 5.14


example 5.14

285.6 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


what‘s common? what‘s different?what‘s most difficult? why?



WWb

A

Bv

Bh

W

A

Bv

Bh

W

A

W

A





W

A BA

B

W

W

A B

A B A B A B

W WW

A

BWW



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

applied loadingreactions @support & contactweight of the body

III. label forces & give dimensions











W

d1

Ax MAy


equilibrium analysis

WAx M

Ay

d1

MO = 0 :

Fx = 0 :

Fy = 0 :

Ax = 0

Ay - W = 0 Ay = W

M - W d1 = 0 M =W d1

the weight induces a moment at the fixed support
















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

e14 - applied mechanics: staticsbiomechanics.stanford.edu/e14/e14_s10.pdf · 2011. 5. 2. · 5....

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