MEMS Project 2 Assignment. Design of a Shaft to Transmit Torque Between Two Pulleys

Transcription

1 MEMS 029 Project 2 Assignment Design of a Shaft to Transmit Torque Between Two Pulleys Date: February 5, 206 Instructor: Dr. Stephen Ludwick

2 Product Definition Shafts are incredibly important in order to transmit torque and rotation from one element to another. Usually secured in frictionless bearings that are mounted in pillow blocks, almost no energy is lost in transmitting the rotation. They are also useful because they allow the rotation of one gear to simultaneously control the rotation of multiple other gears. The functional requirements of a shaft and gear system include: The material must be strong enough to support the loads but must also be cheap Has a yield strength large enough to prevent against initial static yielding due to the applied forces Has an endurance strength large enough to ensure infinite life Has a factor of safety against failure from yielding of at least 2.0 Has a factor of safety against failure from fatigue of at least.25 Minimal friction loss in the bearings allowing for maximum transfer of torque between the gears Must be open at the ends of the bearings allowing for axial growth due to an increase in temperature The parts of the shaft assembly are of sizes that can be bought through specific vendors The pulleys and shaft behave as a rigid body Proposed Model Figure : Problem 3-70 from Shigley's Mechanical Engineering Design

3 Figure 2: Full assembly Figure 3: Drawing of assembly labeling parts

4 Table : Description of parts in drawing Free Body Diagrams Shaft Figure 4: Free body diagram of the shaft in xy- and xz-plane

5 Bearing Figure 5: Free body diagram of the bearing and pillow block Pulley Pulley A Pulley B Figure 6: Free body diagram of pulley A Figure 7: Free body diagram of pulley B

6 Technical Analysis To begin, the torques can be summed around pulley A which then allows the tensions in pulley B to be found. T A = T big R A T small R A = (4 in)[(300 lbf) (50 lbf)] = 000 lbf in From Figure, it can be seen that pulley A will cause the shaft to rotate counter clockwise around the x-axis. This means that the T will be the tight side of pulley B while T2 will be the loose side. It is stated in the problem that the belt tension on the loose side at B is 5 percent of the tension on the tight side. T B = R B (T tight T loose ) = R B (T T 2 ) = R B (T 0.5T ) = 0.85R B T T = 0 = T A T B T A = T B 0.85R B T = 000 lbf in T = 000 lbf in in = lbf in T 2 = 0.5T = 0.5(392.6 lbf in) = lbf in Now that the tensions at pulley B are known, the support reactions from the bearing and pillow block can be found. Looking at Figure 4 and ignoring the weights of the individual components, the forces and moments can be summed in order to solve for the support reactions. In order to be conservative, the weight of the pulleys and shaft were rounded up substantially from the weight calculated by the SolidWorks mass properties tool using AISI 020 steel. Later on in the analysis, the material can be changed as needed in order to meet the necessary strength requirement. The SolidWorks mass properties tool estimated the 6-inch pulley to be 3.4 lbs., the 8-inch pulley to be 5.65 lbs., and the axle to be 4.48 lbs. These can be rounded up to 5 lbs., 8 lbs., and 7 lbs. respectively. xy-plane M o = 0 = (350 lbf)(8 in) (8 lbf)(8 in) (7 lbf)( in) (5 lbf)(6 in) + R cy (22 in) R cy = 7.22 lbf F y = 0 = R oy + (350 lbf) (8 lbf) (7 lbf) (5 lbf) + R cy R oy = lbf xz-plane M o = 0 = ( lbf)(6 in) + R cz (22 in) R cz = lbf F y = 0 = R oz + ( lbf) + R cz R oz = lbf

7 R o =< 0, lbf, lbf > R c =< 0, 7.22 lbf, lbf > xy-plane xz-plane Figure 8: Shear and bending moment diagrams for xy-plane Figure 9: Shear and bending moment diagrams for xz-plane Looking at Figure 8 and 9, the total bending moments at A and B can be found using the Pythagorean Theorem. M A = ( 984 lbf in) 2 + ( lbf in) 2 = 966. lbf in M B = ( lbf in) 2 + ( lbf in) 2 = lbf in M max = M B = lbf point B (x = 6 in) With the maximum bending moment experienced by the beam, the normal and shear stress can be found under this loading condition. σ = M maxc I = M max ( d 2 ) 32M max 32( lbf in) = πd 3 = π( in) 3 = ksi 64 πd4

8 τ = Tr J = T ( d 2 ) 6T 6(000 lbf in) = = πd3 π( in) 3 = ksi 32 πd4 Now, using Mohr s Circle, the maximum stresses at any element can be found. σ,2 = σ 2 ± ( σ 2 ) 2 + τ 2 = 2.288ksi 2 = ksi,.6 ksi ksi ± ( 2 2 ) + (5.093 ksi) 2 τ max = ( σ 2 2 ) ksi + τ 2 = ( ) (5.093 ksi) 2 =.80 ksi With this information, we can check the safety for initial yielding with a factor of safety of 2.0 using the Distortional Energy Theory. η y = S y σ S y = η y σ 2 σ σ 2 + σ 2 2 = (2.0) (22.44 ksi) 2 (22.44 ksi)(.6 ksi) + (.6 ksi) 2 = ksi The yield strength for AISI 020 steel which is our current material is ksi; therefore, the material will yield under the loading conditions. The endurance strength must now be calculated to check if the shaft has infinite life. The factor of safety for fatigue will be taken to be.25. σ a = 2.28 ksi, σ m = 0 ksi, τ a = 0 ksi, τ m = ksi σ a = σ a 2 + 3τ a 2 = σ a = 2.28 ksi σ m = σ m 2 + 3τ m 2 = 3(5.093 ksi) 2 = 8.82 ksi S e = k a k b k c S e S e = 0.5S ut Choosing to have the shaft made with a machined or cold drawn surface finish with a one-inch diameter that is subjected to only bending: b k a = as ut, a = 2.7, b = k b = 0.879d 0.07, d = in

9 k c = S e = (2.7)S ut 0.879( in) 0.07 () S ut =.87S ut Using the Goodman criteria for fatigue failure,.25 = σ a + σ m η f S e S ut 2.28 ksi 8.82 ksi = S.87S ut S ut = ksi ut This means that a steel with a yield strength of ksi or higher and an ultimate tensile strength of ksi or high is required in order to meet the requirements for the factor of safety. Looking on a steel can be picked out that fulfils the necessary requirements. An additional requirement that was assumed earlier is that the steel was machined or cold drawn. From these requirements, AISI 050 carbon steel is picked. Its main applications are for shafts and gears which makes it a perfect material for the application. Table 2: Properties of AISI 050 carbon steel Yield Strength, Sy Ultimate Tensile Strength, Sut Elastic Modulus, E Shear Modulus, G 84. ksi 00 ksi ksi 600 ksi Now that a material is chosen, a deflection analysis can be performed. The beam will be analyzed in both the xy and xz directions. The xz direction will be the first to be analyzed. EIy = M(x) EIy = 23x < x 6 in > EIy = 6.5x < x 6 in > 2 + C EIy = 20.5x < x 6 in > 3 + C x + C 2 [EIy] x=0 = 0 = C 2 [EIy] x=22 in = 0 C = 20.5(22 in)3 75.6(22 in 6 in) 3 = 984 in 2 22 in y(x) = EI ( 20.5x < x 6 in > x)

10 The minimum elastic modulus is chosen since it leads to the largest deflection. y(8 in) = (27557 ksi) ( 64 π( in)4 ) ( 20.5(8 in) in 2 (8 in) = in y(6 in) = The deflection at pulley A is solved for below. (27557 ksi) ( ( 20.5(6 in) π( in)4 ) < (6 in) 6 in > in 2 (6 in) = in EIy = M(x) EIy = 22.77x < x 8 in > 7 < x in > 5 < x 6 in > EIy = 06.38x < x 8 in > < x in > < x 6 in > 2 + C EIy = 35.46x < x 8 in > 3.7 < x in > < x 6 in > +C x + C 2 [EIy] x=0 = 0 = C 2 [EIy] x=22 in = 0 C = 35.46(22 in)3 57(4 in) 3 +.7( in) (6 in) 3 22 in = 032 in 2 y(x) = EI ( 35.46x < x 8 in > 3.7 < x in > < x 6 in y(8 in) = y(6 in) = > +032x) (27557 ksi) ( ( 35.46(8 in) (8 in)) = in 64 π( in)4 ) (27557 ksi) ( ( 35.46(6 in) < 6 in 8 in > π( in)4 ) < 6 in in > (6 in)) = in These deflections can be combines to find the total deflections at pulley A and pulley B. δ A = (0.046 in) 2 + (0.046 in) 2 = in

11 δ B = (0.047 in) 2 + (0.034 in) 2 = in The maximum deflection that the shaft experiences is inches and occurs at pulley A, 8 inches from the first bearing. The components will be assembled such as that in Figure 2 and will be assembled as follows. The bearings will be mounted inside of the pillow blocks. After this, the shaft will be inserted into one of the bearings. The ends of the shaft will be threaded so that a cap can be screwed onto the end. The pulleys will then be heated causing them to expand, and they will be placed in their desired locations and allowed to cool. This will cause them to shrink fit onto the shaft allowing for a fixed position. The other end of the shaft will be inserted into the other bearing, and the other cap will be screwed onto the end of the shaft allowing for the assembly to be stable. In Figure 0 below, the parts are shown with their needed tolerances. The shaft must be slightly smaller than the bearing so that it can fit through. The holes in the pulleys must be slightly smaller than the shaft. This allows for the pulleys to be heated causing the hole to expand to larger than the shaft. As the pulley then cools, the hole will shrink back to its original size allowing for the pulley to compress onto the shaft allowing for a shrink fit. Figure 0: Parts with needed tolerances (units are in inches)

12 Figure : Areas where the tolerances need to be applied Conclusions In order for the shaft to withstand the loads exerted by the applied forces on the pulleys, the material of the shaft had to be carefully chosen. The material had to have a yield strength large enough to prevent initial yielding and had to have an endurance and ultimate tensile strength large enough to prevent eventual fatigue failure. After analyzing the maximum normal stresses and shear stress that the shaft experienced and using these to determine the maximum strengths necessary with a factor of safety using the distortional energy theory, AISI 050 carbon steel was observed to be the material best suited for the job. The factor of safety against yielding for AISI 050 steel is 3.65, and the factor of safety against fatigue failure is.44 using Goodman criteria which satisfies our functional requirements. This shaft would not break and would have an infinite life. Perhaps one risk item that could be studied later on in a future stage of the project would be that the pulley would be exerting a constant normal force on all sides of the shaft since it was attached by shrink fit. This distribution may lead to changes in the stress state of the shaft which change our factors of safety. This would likely not be a large enough factor to make any significant changes, however.