1 mon/wed/fri, 12:50-2:05pm, 370-370 e14 - applied mechanics: statics 2 syllabus e14 - applied mechanics: statics fourth homework due 3 midterm teaching evaluation today‘s objectives 4 midterm teaching evaluation today‘s objectives
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
4midterm teaching evaluation
today‘s objectivestoday‘s objectives
5midterm 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
85.3 equations of 2d equilibrium
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
105.4 two- and three-force members
three-force members
MO=0 requires that thethree forces form aconcurrent (meeting at acommon point O) or parallel(meeting at ") force system
115.4 two- and three-force members
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
125.4 two- and three-force members
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
135.4 two- and three-force members
example 5.13
145.4 two- and three-force members
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
165.5 free body diagram
procedure for drawing a FBD
175.5 free body diagram
support reactions in 2d - memorize!
roller pin fixed
no motion # force / no rotation # moment
185.5 free body diagram
support reactions in 2d - memorize!
no motion # force / no rotation # moment
pin-like
roller-like
W
BH
A
BV
195.5 free body diagram
no motion # force / no rotation # moment
support reactions in 3d - memorize!
205.5 free body diagram
no motion # force / no rotation # moment
support reactions in 3d - memorize!
215.5 free body diagram
support reactions in 3d - memorize!
no motion # force / no rotation # moment
225.5 free body diagram
support reactions in 3d - memorize!
no motion # force / no rotation # moment
235.5 free body diagram
support reactions - example 04
ball-and-socket joint
245.5 free body diagram
support reactions - example 05
journal bearing
255.5 free body diagram
support reactions - example 08
single smooth pin
265.5 free body diagram
example 5.14
275.5 free body diagram
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
295. equilibrium of a rigid body
what‘s common? what‘s different?what‘s most difficult? why?
305. equilibrium of a rigid body
what‘s common? what‘s different?what‘s most difficult? why?
WWb
A
Bv
Bh
W
A
Bv
Bh
W
A
W
A
315. equilibrium of a rigid body
what‘s common? what‘s different?what‘s most difficult? why?
325. equilibrium of a rigid body
what‘s common? what‘s different?what‘s most difficult? why?
W
A BA
B
W
W
A B
A B A B A B
W WW
A
BWW
335.3 equations of 2d equilibrium
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
345.3 equations of 2d equilibrium
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
355.3 equations of 2d equilibrium
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
Ax MAy
365.3 equations of 2d equilibrium
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
375.3 equations of 2d equilibrium
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
385.3 equations of 2d equilibrium
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
395.3 equations of 2d equilibrium
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
405.3 equations of 2d equilibrium
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