this is the next part of studio 2 which you helped me with last time in order XXXXXXXXXXso all the things you need for the last assignment are in order12209 and I just need you to do Maximum Shear...

this is the next part of studio 2 which you helped me with last time in order
12209

.. so all the things you need for the last assignment are in order

12209 and I just need you to do Maximum Shear Stress Theory for this which is the first question

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-carbon, cold drawn steel including AISI 1020-1050 steels. 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. Rubric 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/1710419250000/3/107778041916381461330/APznzaYHLqOrFeUzSX-6wtTdKx3QVY1JDkt7Zx7P9fhFzl_BKnrLKyJPYE8Wdi6pfYQATKhmB6anhItYp6wF-tclAMqYB7VWkm6ZjJLMUKKq3HcWhYltmbZbWOno6qTQ34kxg897sYJ5YZKJMnP8DxvKJF58VA3VMdNWUOV7fcvgKBBrXfXrP8vkyHoHg6Mu1fsvGbsy89dx5UfT8LY8sYhnC0zfzop6om-0nRUpZzmzq8IsSAO-v7oRozwZ7b83ySFL8ADrDGjXwpi2wXbln9XB32LQStA50SGtZ-Zvcc8L3Rp0K20JGZjbNcEe3T1_ivEg-nahdpMZ3585bgl2D3L-OiE4WiaSllPgVDsI9Z3iFimtIM3pogQAbocAP298kLSWrI4VyuzTXNmj0GMCYTpqIPLey7ElPuLbBmJE0wOfdiRw1uQTQs4=?authuser=0&nonce=nlbnecpen99h8&user=107778041916381461330&hash=s4kbjs4aiauq75cujop4ntl4o5a87fi7 Team 12 ENME 400 - 0102 Studio Problem 2 - Free Body Diagram and Stress Calculations Chase Fisher, Asres Lante, Shawn Park, Jevin Smith, Zereab Walelign, Alex Wilson Instructor: Dr. M. Fazelpour February 25, 2024 Minimum Required Torque Calculations Givens: ???????? ?? ?ℎ???, ? ? = 0. 025 ? ??????? ????????, ? 0 = 0 ?/? ????, ? = 50?? ????? ????????, ? ? = 0. 1 ?/? ∆? = 2 ? ??????? ?ℎ??? ????????, ? ? = 0. 1 ? Calculations: Minimum Acceleration: ? ? 2 = ? 0 2 + 2? ? (∆?) => ? ? = ? ? 2−? 0 2 2∆? = (0.1?/?)2−(0?/?)2 2*(2?) = 0. 0025?/? 2 Force Calculation: ? = ?? ? = 50 ?? * 0. 0025 ?/?2 = 0. 125 ? Torque: on each wheelτ = ? * ? ? = ? * ? ? /2 = (0. 125 ?) * (1 ?)/2 = 0. 00625 ?? Free Body Diagram of Shafts Differential to Wheel Shaft - This shaft is the same on both sides of the differential Gearbox 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. Gearbox 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. Gearbox Input Shaft from Motor Reaction Forces and Torques on Shafts (Using the found minimum torque where the wheels are located.) Givens: ??????? ??????, τ = 0. 00625 ?? ?ℎ??? ????????, ? ? = 0. 025 ? ???? ?ℎ??? ?????ℎ, ? ? = 0. 3 ? ??? ??ℎ?? ?ℎ????, ? ? = 0. 1 ? ???? ????????, ? ? = 0. 05 ? Calculations: - Force from torque on shaft: ? = τ/? ? = 0. 00625??/(0. 025?/2) = 0. 5 ? = ? ????, ??? - Differential Input ? ????, ?? = 2? ????, ??? = 2 * 0. 5 ? = 1 ? - Torque on Gearbox to Differential Shaft: τ = ? ? * ? ????, ?? = 0. 025 ? * 1 ?/2 = 0. 0125 ?? - Force from Layshaft Gear on shaft: Σ? ?ℎ??? = τ = ? ??? * ? ? => ? ??? = 0.0125 ??0.05 ? / 2 = 0. 5 ? - Torque on the Layshaft: τ = ? ??? * ? ? = 0. 5? * 0. 025 ? = 0. 0125 ?? - Because the 2 layshaft gear forces cancel out, use for the motor calc reaction and? ??? torque. - Torque on Motor/Input Shaft: τ ????? = ? ??? * ? ? = 0. 5? * 0. 025 ? = 0. 0125 ?? - Force from the motor into the shaft: Σ? ? = τ ????? = ? ????? * ? ? => ? ????? = τ ????? /? ? = 0.0125 ??0.025? / 2 = 1 ? Location and Magnitude of the Greatest Stresses Maximum Shear Stress on Shaft = τ ??? = ??? ? ??????? = 0. 003125 ? Radius of Shaft = 0.0125 m Polar Moment of Inertia of Shaft = ? = π? 4 32 = 3. 83 * 10 −8 ?4 Max Stress = τ ??? = 1019. 90 ??? Shear and Bending for Each Shaft Axle Shafts Gearbox Output Shaft to Differential Gearbox LayShaft Gearbox Input Shaft from Motor 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 bending on the shafts - Alex Wilson - Minimum torque on driving shaft calculations, format and compile work, free body diagrams of gearbox and motor shafts, shear diagrams for layshaft, output shaft and input shaft, reaction force calculations for layshaft, input and output shaft. \ i I ' I 1 sF.~~.a F,-=- - - - Zt1o (o. 1,.\ fo • o'l..,\ ,re•:<) :::....0="" -:b="" 0,3\~j°a="" -="" (="" m._="" d+~="" ld-bi)f;1="" s="~" o~~(m1:}="" -="" {),="" ~~.:rf\"="" 1~0="" -="" o·="" i!;,{="" 1="" l'v',(j="" -="." 0="" )="-t"> D . 3 (M _g - D. &""h eft\r\'L. _-- \!-t-~-"' Q !* t Ji .. I 5 ----~ -- J ---~ Zi\::-o O·UM_ro}- - F,j [ o,d'") =- 0 L-,, D • 2..--) r\l'r·a, - M it;} )l = --- C) /0 L o ·IS~ -~ 1 -=--0 ~o .- \IV\ I - --- - I - t,: ... rrJ tf :: 'ii ' x _ o ,- 01 1' 1.- 'f f L ':. lo , J & , I ~ p ( ( ,5 + ' N i. "t 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.
Answered 1 days AfterMar 14, 2024

Answer To: this is the next part of studio 2 which you helped me with last time in order XXXXXXXXXXso all the...

Dr Shweta answered on Mar 15 2024
SOLUTION.PDF