1 mon/wed/fri, 12:50-2:05pm, 370-370 e14 - applied mechanics: statics 2 syllabus e14 - applied mechanics: statics fourth homework due 3 5. 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 4 5.3 equations of 2d equilibrium three alternative sets of eqns M O = 0 F x = 0 F y = 0 • two force & one moment equilibrium equations M B = 0 F = 0 • one force & two moment equilibrium equations M A = 0 M C = 0 • no force & three moment equilibrium equations M B = 0 M A = 0 line through A and B must be ! to x-axis A, B and C must not be on one line
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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
55.3 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
65.3 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
75.3 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
85.3 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
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
135.5 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
145.5 free body diagram
support reactions in 2d - memorize!
roller pin fixed
no motion # force / no rotation # moment
155.5 free body diagram
support reactions in 2d - memorize!
no motion # force / no rotation # moment
pin-like
roller-like
W
BH
A
BV
165.5 free body diagram
support reactions - example 04
ball-and-socket joint
175.5 free body diagram
support reactions - example 05
journal bearing
185.5 free body diagram
support reactions - example 08
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
205.6 equations of 3d equilibirum
example 5.15
215.6 equations of 3d equilibirum
example 5.15
225.6 equations of 3d equilibirum
example 5.15
235.7 constraints & statical determinacy
too much support
2d3 equations
3d6 equations
245.7 constraints & statical determinacy
too much support
2d3 equations
3d6 equations
2d5 unknowns
3d8 unknowns
W
F
N1
T3
M
T1
N3
T2
N2
statically indeterminate
255.7 constraints & statical determinacy
too much support
2d3 equations
2d8 unknowns
265.7 constraints & statical determinacy
not enough support
2d3 equations
2d3 equations
275.7 constraints & statical determinacy
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!
285.7 constraints & statical determinacy
not enough support
MA = 0 moment is not supported # system will rotate!
M
the system is improperly supported if all reaction forces areparallel