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Static Force Analysis for Skid Loader – Virtual Work
A trunnion mount hydraulic cylinder actuates the arm of a skid steer loader as shown below. At this position, e = 40 inches, = 61.131°, e^ = -12 ips, ^ = -0.3625 rad/s.
Determine the force on the hydraulic cylinder required to lower an 800 lbf payload attached to point D by a cable. The payload moves with constant velocity at the position shown. You may neglect the effects of friction. The weight of the arm and cylinder are small compared to the payload. Show your work.
FCYLINDER __________________________
What corresponding hydraulic pressure would be required for a cylinder with a 3 inch DIA bore?
PCYLINDER __________________________
Is this value reasonable? Why?
If you include friction between the piston and cylinder wall, will it increase or decrease your computation for pressure.
increase decrease Why?
What value would you use for the coefficient of friction between the piston and cylinder wall?
_________________________________ Why?
Should your analysis be different if the cylinder were retracting at constant velocity instead of the payload moving at constant velocity?
yes no Why?
A
B
C
e
Not to scale AB = 36 inches AC = 42 inches AD = 96 inches = 16°
D
Payload
VD = AD ^ = 34.8 ips
power FC VC PVD 0
-FC ( e^ ) + P (cos ) VD = 0
2261.7 lbf up/left
from Newtonian solution FC = 2261.9 lbf
320 psi (^) A = D^2 / 4 = 7.069 in^2
OK, industrial hydraulics often go to 3000 psi
pressure pushes up friction force will be up opposing piston motion
0.1 lubricated
constant e^ means ^ will not be constant means velocity of the payload will not be constant, therefore must account for acceleration of payload mass
Static Force Analysis for Sewing Machine – Virtual Work
Determine crank torque T 12 required to maintain this sewing machine linkage in static equilibrium as shown below for applied load P = 10 N. Assume that friction and the weight of the links are negligible.
from velocity solution 2 = +8 rad/sec V = 64.71 cps downF
actual power, no friction, no springs T 12 2 VFP 0
assume T 12 is CCW ( +T 12 ) ( +8 rad/sec)
- ( -64.72 cm/sec ) ( +10 N ) = 0
T 12 = + 25.75 N.cm
A
D
B
C
E
G
F
AB = 1.60 cm BC = 3.57 cm DC = 2.24 cm CE = 1.60 cm DE = 2.74 cm EF = 3.81 cm AG = 1.42 cm DG = 3.81 cm
DCE = 90
CDE = 35.7
constant 2
P
T 12
Static Force Analysis for Pushups – Virtual Work
A person doing pushups can be modeled as a four bar linkage. The ground is the base link, the forearms are link 2, the upper arms are link 3, and the torso and legs are link 4 as shown below. The wrists are revolute O 2 , the elbows are revolute A, the shoulders are revolute B, and the toes are revolute O 4. Mass of the torso/legs is 180 lbm and the mass center is located at G 4. Assume that all muscular effort is provided by the triceps as torque T 32 across the elbows.
For an initial estimate, use the additional assumptions: a) 4 is constant at this position b) Weight of the arms is negligible compared to weight of the torso/legs. c) Friction is negligible at A, B, C and D. d) No muscular torque is generated at A, C and D.
Determine angular velocity across the elbows 2/3 for the position and velocity provided above.
Determine elbow torque T 32 for the position and velocity provided above.
Do the magnitude and direction for your answer seem reasonable? Why?
Rate the last four assumptions and state your reasoning.
b) constant 4 1=poor 2=acceptable for an approximation 3=very good
c) weight of arms is negligible 1=poor 2=acceptable for an approximation 3=very good
d) friction is negligible 1=poor 2=acceptable for an approximation 3=very good
e) no muscle force at A, C, D 1=poor 2=acceptable for an approximation 3=very good
Determine 4 of the torso/legs when the forearm is aligned with the upper arms ( 2 = 3 ).
A
D
B
C
2 ^4
B
C
4
3 2
G 4
AD = 52 inch AB = 12 in BC = 14 in CD = 57.7 in DG 4 = 39 in
2 = 0.5 rad/sec 3 = -1.435 rad/sec 4 = -0.387 rad/sec
2/3 = 2 - 3 = +1.935 rad/sec
VG4 = DG 4 4 = 15.093 ips UP, AG4T^ = DG 4 4 = 0, AG4N^ = DG 4 42 = 5.84 ips 2 = 0.015 G negligible
W 4 VG 4 T 32 2 / 3 0 -W 4 cos(180°- 4 ) VG4 + T 32 2/3 = 0 T 32 = +1351.2 in.lbf
W 4
2/3^ VG
T 32
from Newtonian solution T 32 = 1351.2 in.lbf CCW
