MCDESG-125 Design Problems
Assignment 4B
Statics – Find Reactions
MCDESG-125 Page 1
Name: _______________________________________ Team ___________________
Item
ATV Lift Lift table Score Co
ections
1 Top Pad Table /10
2 Idler Lift Arm Idler Frame /
3 Active Lift Arm Active Frame /20
4 Base Assembly Base Assembly /20 Individual Score: (1-6)
5 Cylinder Link Pair - /
6 Bicycle Stopping Reactions /40
7 Free Body Diagram Pipe (Length) /20 Team Score: (7-9)
8 Maximum Deflection /40
9 Maximum Stress /40
Total Score
Total Percent %
Multiplied by 10.00% %
On Time: 0.20% %
Added to Grade %
Hand in ONE cover sheet each with Reaction Calculations
Only one Pipe Support is necessary on one person’s cover sheet
Individual scores apply to the ATV Lift/Lift Table & Bicycle Stopping calculations;
everyone gets the same score for the Pipe Support calculations
MCDESG-125 Design Problems
Assignment 4B
Statics – Find Reactions
MCDESG-125 Page 2
MCDESG-125 Design Problems
Assignment 4B
Statics – Find Reactions
MCDESG-125 Page 3
ATV Lift: (if that is the project that you chose)
Use your Free Body Diagrams from last week. Find the
eactions at the Top Pad. Use the regular analytical (fingers
& a calculator) MA = 0, etc. Be sure to perform a check.
Transfer the “actions” to the reactions on the Idler
Lift Arm and the Active Lift Arm.
Show the forces on the Idler Lift Arm. There
isn’t any calculation here as it is a Link, this is
simply for completeness. Show it in its actual
orientation.
Calculate the forces on the Active Lift Arm. First
transfer the “action” from the Top Pad. The
magnitude of the force due to the piston is as yet
unknown, but its direction is known. The direction
of this force is the angle of the Cylinder Link Pair
as defined in the Layout Drawing.
MCDESG-125 Design Problems
Assignment 4B
Statics – Find Reactions
MCDESG-125 Page 4
Finally, find the reactions at the wheels on
the Base Assembly. (The reactions at the
wheels are vertical.)
The force at Pin C is transfe
ed from
the Idler Lift Arm. The force at the
other Pin D is transfe
ed from the
Active Arm just solved. The load
from the Piston is actually at a fixed
support and should be a force and a
moment. This can
e sidestepped (see
ight): draw a line
through pins C & D
and run the line of
action of the
Cylinder Link Pair
through this line.
Draw rectangles
around the forces to
find the vertical
components.
Complete the
calculation with
MG = 0, etc. Be
sure to perform an
independent check
at the end.
What should the
eactions at the
wheels add up to?
Every force and moment
combination can be replaced
y a force alone in a certain
position and line of action.
MCDESG-125 Design Problems
Assignment 4B
Statics – Find Reactions
MCDESG-125 Page 5
Lift Table: (if that is the project that you chose)
Calculate the two reactions
from the Frames to the 500lb
load on the Table.
Transfer the force from the pin at the front of the Table to the Idler
Frame
o Calculate the remaining two forces on the Idler Frame.
Transfer the forces from the roller beneath the Table and
the force at the center of the Idler Frame to the Active
Frame and solve for the force in the Piston and the
eaction at the pin where the Frame attaches to the
Base. Remember that you know the direction of the force
at the piston from your layout drawing. Use this angle to four
(4) sig. fig.
Finally, solve for the reactions at the wheels by transfe
ing forces
to the Base Assembly. What should the reactions at the wheels total
to? Be sure to perform and independent check at the end.
MCDESG-125 Design Problems
Assignment 4B
Statics – Find Reactions
MCDESG-125 Page 6
Bicycle Stopping Reactions: (Individual)
See attached Bicycle drawings. The center of gravity and weight is
shown for the rider and the bicycle itself, as well as relevant
dimensions.
Assuming the bicycles are moving at a constant velocity in a straight line, find the reactions at
the front and rear wheels. Assume that the rider weighs 160lb in each case:
o The “ordinary” weighs 40lb
o The “safety” weighs 21lb
o The “recumbent” weighs 36lb
Use SM = 0; take moments about where the front wheel touches the roadway and then about
where the rear wheel touches the roadway. Then use SFy to check your answers.
OVERWRITE THE ATTACHED AUTOCAD TO MAKE YOUR FBD FOR EACH CASE.
Note that the first one already is an FBD
If each bike is travelling at 18mph and needs to stop in 40ft, find the deceleration. Use equations
from the previous assignment. Yes, the value is the same in each case, so only calculate once!
Once something is being accelerated/decelerated, there is a force involved:
F = ma, where m = W/g and g = 32.2 ft/s² (acceleration due to gravity)
and ‘a’ is acceleration calculated above.
E.g. If a = 10ft/sec² (it doesn’t!) for the rider: FR = (160/ XXXXXXXXXX) = 49.69lb
and for the bike: FB = (40/ XXXXXXXXXX) = 12.42l
Then:
Σ?? = −? XXXXXXXXXX XXXXXXXXXX) − XXXXXXXXXX) − XXXXXXXXXX) =
0
∴ ? =
XXXXXXXXXX50) − XXXXXXXXXX) − XXXXXXXXXX)
39.75
= −33.49??
Σ?? = ?(39.75) − XXXXXXXXXX) − XXXXXXXXXX) − XXXXXXXXXX) − XXXXXXXXXX) = 0
∴ ? =
XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX)
39.75
= 233.49??
. . . and check:
Σ?? = −33.49 − 40 − XXXXXXXXXX = 0
What does it mean when a reaction is negative? (R = -33.49lb)
Be sure to find reactions when
NOT
aking first and then
eactions when
aking.
MCDESG-125 Design Problems
Assignment 4B
Statics – Find Reactions
MCDESG-125 Page 7
Now solve for reactions at the wheels for the “Safety” and the “Recumbent” using the
deceleration that you calculated for the 40’ stopping distance. State if anyone is going over the
handlebars.
MCDESG-125 Design Problems
Assignment 4B
Statics – Find Reactions
MCDESG-125 Page 8
MCDESG-125 Design Problems
Assignment 4B
Statics – Find Reactions
MCDESG-125 Page 9
Pipe Support: (Group – attach one submission to one cover sheet)
Draw a Free Body Diagram of the pipe between two supports. What kind of loading is this? Are
the supports “simple”?
Go to Appendix A-25 in “Applied Strength of Materials” by Mott or Appendix A14-3 in “Machine
Elements in Mechanical Design” by Mott and look for an appropriate FBD. The load on the pipe
is its own weight - is that a point load or distributed?
Go to the Appendix where you found the FBD. Use the deflection formula to confirm that the
pipe does not droop more than 1.00". Note that you need “E”, Young’s Modulus and “I”, the
cross-sectional Moment of Inertia. Use E = 30 x 106 psi; look up (or calculate) I in a table of Pipe
Properties. Note that you will have to calculate the deflection for 18" Sch 80 and 6" Sch 40. The
lighter pipe may be more flexible.
Calculate the stress and confirm that it is less than 10,000psi. Note that the formulas given are
only for M, Moment. Use either:
? =
??
?
?? ? =
?
?