ENME400 Machine Design
Studio Problem 3 – Shaft Design for Failures Resulting from Static Loading
Description
The objective of this studio problem is to determine the diameter of shafts that were analyzed in Studio
Problem 2. Students will determine the diameter of the shaft considering static loading case with a
factor of safety of 3. Use low-ca
on, cold drawn steel including AISI XXXXXXXXXXsteels. For this studio
problem, assume uniform shaft diameter for each shaft.
• Determine the diameter of each shaft using the maximum-shear-stress theory.
• Determine the diameter of each shaft using the distortion-energy theory.
• Compare and discuss the results
• Document each student’s contribution to this assignment
Objective
Students will develop an appropriate understanding of using failure theories for static loading to design
a shaft.
Submission
Teams will submit their typed submissions through Canvas as one single PDF file.
Ru
ic
35 points for the diameter of the shafts using the maximum-shear-stress theory.
35 points for the diameter of the shafts using the distortion-energy theory.
10 points for results comparison and discussion
10 student’s contribution documentation
10 points for professionalism and well-articulated submission (clear, concise, coherent, complete)
https:/doc-0c-4k-prod-00-apps-viewer.googleusercontent.com/viewer2/prod-00/pdf/f0ku32248sojq4g47dghhvnd846t9dd4/7ipefn0imqvk66rlg28avnqp5fp28hlh/ XXXXXXXXXX/3/1 XXXXXXXXXX/APznzaYHLqOrFeUzSX-6wtTdKx3QVY1JDkt7Zx7P9fhFzl_BKnrLKyJPYE8Wdi6pfYQATKhmB6anhItYp6wF-tclAMqYB7VWkm6ZjJLMUKKq3HcWhYltmbZbWOno6qTQ34kxg897sYJ5YZKJMnP8DxvKJF58VA3VMdNWUOV7fcvgKBBrXfXrP8vkyHoHg6Mu1fsvGbsy89dx5UfT8LY8sYhnC0zfzop6om-0nRUpZzmzq8IsSAO-v7oRozwZ7b83ySFL8ADrDGjXwpi2wXbln9XB32LQStA50SGtZ-Zvcc8L3Rp0K20JGZjbNcEe3T1_ivEg-nahdpMZ3585bgl2D3L-OiE4WiaSllPgVDsI9Z3iFimtIM3pogQAbocAP298kLSWrI4VyuzTXNmj0GMCYTpqIPLey7ElPuLbBmJE0wOfdiRw1uQTQs4=?authuser=0&nonce=nlbnecpen99h8&user=1 XXXXXXXXXX&hash=s4kbjs4aiauq75cujop4ntl4o5a87fi7
Team 12
ENME XXXXXXXXXX
Studio Problem 2 - Free Body Diagram and Stress Calculations
Chase Fisher, Asres Lante, Shawn Park, Jevin Smith, Zereab Walelign, Alex Wilson
Instructor: Dr. M. Fazelpou
Fe
uary 25, 2024
Minimum Required Torque Calculations
Givens:
???????? ?? ?â„Ž???, ?
?
= 0. 025 ? ??????? ????????, ?
0
= 0 ?/?
????, ? = 50?? ????? ????????, ?
?
= 0. 1 ?/? ∆? = 2 ?
??????? ?â„Ž??? ????????, ?
?
= 0. 1 ?
Calculations:
Minimum Acceleration:
?
?
2 = ?
0
2 + 2?
?
(∆?) => ?
?
=
?
?
2−?
0
2
2∆? =
(0.1?/?)2−(0?/?)2
2*(2?) = XXXXXXXXXX?/?
2
Force Calculation:
? = ??
?
= 50 ?? * XXXXXXXXXX ?/?2 = 0. 125 ?
Torque:
on each wheelτ = ? * ?
?
= ? * ?
?
2 = XXXXXXXXXX ?) * (1 ?)/2 = XXXXXXXXXX ??
Free Body Diagram of Shafts
Differential to Wheel Shaft
- This shaft is the same on both sides of the differential
Gea
ox Output Shaft to Differential
- This shaft will have 2 gears, however force is only ever applied to one; for the purposes
of this assignment it will only include the 0.1 m/s gear.
Gea
ox LayShaft
- This shaft will have 3 gears, however force is only ever applied to 2 gears: the gear from
the input shaft and the gear selected within the output shaft.
Gea
ox Input Shaft from Moto
Reaction Forces and Torques on Shafts
(Using the found minimum torque where the wheels are located.)
Givens:
??????? ??????, Ï„ = XXXXXXXXXX ?? ?â„Ž??? ????????, ?
?
= 0. 025 ?
???? ?â„Ž??? ?????â„Ž, ?
?
= 0. 3 ? ??? ??â„Ž?? ?â„Ž????, ?
?
= 0. 1 ?
???? ????????, ?
?
= 0. 05 ?
Calculations:
- Force from torque on shaft:
? = Ï„/?
?
= XXXXXXXXXX??/(0. 025?/2) = 0. 5 ? = ?
????, ???
- Differential Input
?
????, ??
= 2?
????, ???
= 2 * 0. 5 ? = 1 ?
- Torque on Gea
ox to Differential Shaft:
Ï„ = ?
?
* ?
????, ??
= 0. 025 ? * 1 ?/2 = XXXXXXXXXX ??
- Force from Layshaft Gear on shaft:
Σ?
?â„Ž???
= Ï„ = ?
???
* ?
?
=> ?
???
= 0.0125 ??0.05 ? / 2 = 0. 5 ?
- Torque on the Layshaft: Ï„ = ?
???
* ?
?
= 0. 5? * 0. 025 ? = XXXXXXXXXX ??
- Because the 2 layshaft gear forces cancel out, use for the motor calc reaction and?
???
torque.
- Torque on Moto
Input Shaft:
Ï„
?????
= ?
???
* ?
?
= 0. 5? * 0. 025 ? = XXXXXXXXXX ??
- Force from the motor into the shaft:
Σ?
?
= Ï„
?????
= ?
?????
* ?
?
=> ?
?????
= Ï„
?????
?
?
= 0.0125 ??0.025? / 2 = 1 ?
Location and Magnitude of the Greatest Stresses
Maximum Shear Stress on Shaft = Ï„
???
= ???
?
???????
= XXXXXXXXXX ?
Radius of Shaft = XXXXXXXXXXm
Polar Moment of Inertia of Shaft = ? = π?
4
32 = 3. 83 * 10
−8 ?4
Max Stress = Ï„
???
= XXXXXXXXXX ???
Shear and Bending for Each Shaft
Axle Shafts
Gea
ox Output Shaft to Differential
Gea
ox LayShaft
Gea
ox Input Shaft from Moto
Student Contributions
- Chase Fisher - Minimum torque on driving shaft calculations
- Asres Lante - Reaction force calculations
- Shawn Park - Finalized edits
- Jevin Smith - Free Body Diagrams, Shear force and Bending Diagrams, Reaction Forces,
Formatting
- Zereab Walelign - Location and Magnitude of Greatest Stresses calculation, shear and
ending on the shafts
- Alex Wilson - Minimum torque on driving shaft calculations, format and compile work,
free body diagrams of gea
ox and motor shafts, shear diagrams for layshaft, output shaft
and input shaft, reaction force calculations for layshaft, input and output shaft.
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For the shaft design, we need to find normal stress and shear shears at a critical point.
Then, use normal stress and shear shears to calculate principal stresses, Sigma 1 > Sigma
2 > Sigma 3.
Tau_max = Sigma 1- Sigma 3
Then, find safety factor.