Transcript Static Machine Forces - University of Dayton

Section 1
Static Machine Forces
 Determines the forces in a machine while
links are stationary, or at constant velocity.
 Often used when the acceleration of the
machine components is small (<0.25g).
Forces
 Push or pull action.
 Vector quantity.
– Magnitude
– Direction
– Point of application
F = 100 lbs
300
Force Components
 It is often convenient to resolve a force into
orthogonal components.
F2
f
2
F
 Use right triangle to determine the
components
– F1=F sinf
– F2=F cosf
F1
1
Moments
 Twisting action produced by a force.
 Computed relative to a point.
d
F
 MA=F d
– F = Force
A
– A = Reference point
– d = perpendicular distance between force and
reference point.
 Vector
MA = 250 in lbs, cw
Problem 1-2
A force is applied to a box wrench as shown.
Determine the moment, relative to the center
of the nut, when f = 300, and b = 700.
b
8 in.
f
25 lb
Problem 1-4
A force is applied to the control lever as
shown. Determine the moment, relative to the
pivot block, when b = 300.
b
60 lb
12 in
18 in
Free-Body Diagrams
 Isolate the component(s) that must be
studied.
 Draw the component as if it were floating
freely.
 Replace all supports with the appropriate
force and/or couples (moments).
Free-Body Diagrams
D
E
F
C
B
A
FBD
Entire engine hoist
D
E
F
250 lbs
C
Bx
By
Ay
Free-Body Diagrams
FBD
Link DEF
D
Dx
Dy
E
F
CE
250 lbs
FBD
D
Base y
D
Dx
FBD
Cylinder CE
CE
C
CE
Bx
By
Ay
CE
Reaction Forces
 As a general rule, if the nature of the
contact prevents motion in a certain
direction, there must be a supporting force
in that direction.
F
Two-Force Members
 A component that is acted upon by only two
forces is known as a two-force member.
 A two force member will always be in
either tension or compression.
Problem 1-12
Draw free-body diagrams of the links for the
pliers when 10 lbs is applied to the handles
1.25 in
3.75 in
10 lbs
Problem 1-13
Draw free body diagrams of all the
components for the lift. The weight of the
crate and platform are 1200 lb and 400 lbs.
The weight of all other links is considered
80”
insignificant
18”
6”
18”
10”
24”
24”
24”
Static Equilibrium
 A machine component is in static
equilibrium when the combination of all
forces is zero.
 In addition, the net effect of all moments,
about any arbitrary point, must also result in
zero.
SFx = 0
SFy = 0
SMi = 0
Hints to Solve Statics Problems
 Draw free-body diagrams large and properly
labeled all forces.
 Identify any two-force members that are
contained in the machine.
 The shortest method of solution occurs when
the first FBD incorporates both given
information and what is required. However, a
free body diagram can, at most, be used to
determine three unknown forces.
Hints (con’t)
 If the first FBD fails to give a full solution,
draw several other free-body diagrams. Each
additional FBD generates three additional
independent equations. However it may
generate additional unknown forces. Keep track
of the total number of unknown forces and
independent equations.
 A FBD of the entire machine may be useful.
 Remember that internal forces between
connecting components are equal and opposite.
Problem 1-12
Determine the force onto the nut when 10 lbs
is applied to the handles of the pliers.
1.25 in
3.75 in
10 lbs
Problem 1-14
The clamp has a rated load of 1500 lb.
Determine the compressive force this creates
in the threaded rod, AB
2 in
5 in
3 in
1500 lb
5 in
2 in
4 in
Problem 1-13
Determine the cylinder force to maintain the
position of the lift. The weight of the crate and
platform are 1200 lb and 400 lbs. The weight
of all other links is considered insignificant
80”
45”
18”
6”
18”
10”
24”
24”
24”
Problem 1-25
Determine the force from the hydraulic
cylinder to keep the platform in the position
500 lb
24”
shown.
20.6”
16”
37.90
28”
75.60
30”
28”
37.90
12.9”
36”
36”
10”
Problem 1-16
Determine the force required by the hydraulic
cylinder to maintain position of the bucket.
5m
2.4 m
1200 N
2.0 m
0.9 m
1.2 m