%% Vertical shaft Calculation %% Force analysis (finding tangentail force on bevel % gear at the top of the vertical shaft T2=15.82 %Tangent force on the seccond bevel gear(Top bevel gear on vertical shaft) D2=.12 %meters R2=D2/2 Ftan2=T2/R2 %Tangential force on the second bevel gear. %% Axial force on the second bevel gear %degees to radians deg1=20 %pressure angle deg2=21.81 %cone pitch angle rad1=deg1*(3.14/180) rad2=deg2*(3.14/180) FV0F=Ftan2*tan(rad1)*sin(rad2) %% Tangential force acting on the third bevel gear % at the bottom of the vertical shaft) D3=.135 R3=D3/2 Ftan3=T2/R3 %tangential force on the third bevel gear. %% Radial force acting on the third bevel gear FH2F=Ftan3*tan(rad1)*sin(rad2) %% Axial force acting on the third bevel gear FV1F=Ftan3*tan(rad1)*cos(rad2) %% Moment about the third bevel gear from % the axial force FV1 acting at a distance of R3 % from the edge of the third bevel geat to the shaft M1F=R3*FV1F %% Force of grvity acting on the vertical shaft m=.253 g=9.81 Fg=m*g %% The thrust reactionary force in the side plane % for the bearing at the bottom of the vertical shaft %Sum of fores in the vertical direction FV2F=FV0F+Fg-FV1F %% Finding the radial reactionary force in the frontal plane % for the bearing at the bottom of the vertical shaft %Sum of the moments at bearing 1 is equal to 0. d1=.119 d2=.170 d3=.115 FH3F=(M1F+(-(Ftan2*d3)-(d1*FH2F)))/(d2) %Hroizontal radial reactionary force for the % bearing at the bottom of the vertical shaft. % FH3 is negative so the initial assumption % of the direction to the right was wrong. % IT will be pointing to the left. %% Finding the radial reactionary force in the frontal plane % for the bearing at the middle of the vertical shaft %Sum of forces in the x-direcction %FH1F+FH2F-FH3F-Ftan2=0 FH1F=-FH2F+(-FH3F)+Ftan2 %Radial force on the middile bearing %% Force analysis in the side plane %% Radial force on the second bevel gear at the top of % the vertical shaft FH1S=Ftan2*tan(rad1)*cos(rad2) %% Finding the radial reactionary forces on the bottom % bearing in the side plane %Sum of moment about the bearing in the middle d1=.119 d2=.170 d3=.115 %+(d1*Ftan3)+(d2*FH3S)+(d3*FH1S)=0 FH3S=(-(d3*FH1S)-(d1*Ftan3))/d2 %Negative so in opposite direction %Radial force on bottom bearing %in the side plane. %% Find the radial reaction force in the side plane for the bearing at % the middle of the vertical shaft %Sum of forces in the x-direction %-FH3S+Ftan3-FH1S+FH2S=0 FH2S=-(FH3S)-Ftan3+FH1S %% Resultant moments at the critical sections %Side plane d4=.051 d5=.089 MH1F=(d4*(-FH3F)) MH2F=M1F+(((-FH3F)-FH2F)*d1)+MH1F MH3F=MH2F+(((-FH3F)-FH2F-FH1F)*d5) %front plane MH1S=((-FH3S)*d4) MH2S=(((-FH3S)-Ftan3)*d1)+MH1S MH3S=MH2S+(((-FH3S)-Ftan3-FH2S)*d5) %Take the max moments from the frontal and side planes to get the %resultant moment % section 1 (Max horizontal and vertical moment)(Keyseat) %MH1F=23.76 %MH1S=16.55 Mres1=sqrt((MH1F^2)+(MH1S^2)) % section 2 (Max horizontal and vertical moment) %MH2F=37 %MH2S=20.31 Mres2=sqrt((MH2F^2)+(MH2S^2)) %Section 3(Max horizontal and vertical moment)(Step) %MH3F=18.18 %MH3S=6.17 Mres3=sqrt((MH3F^2)+(MH3S^2)) %% Formula for sizing a shaft in static loading %dmin1=(1/Ky)*(Load^1/3)*Kser %% Find the shaft yield strength factor (1/Ky) based on % energy distortion theory %SSy=(1/Ky)=[(10*Cs*K)/(Sy)]^1/3 %Size factor (Cs).Shaft will be smaller than 60mm so Cs=1. Cs=1 %% Ultimate strength and yield strength % of Marine grade stainless steel shaft %Rockwell Hardness 91B is equivalent to a tensile strength of %91000Psifor marine grade nitronic 50 stainless steel. %Yield strength of 55000 psi %% Yield strength of stainless steel grade 1702 %High carbon stainless steel grade 1702 can reach a yield strength of %2502795077 Pa with heat treatment. Sy=363000.12*6894.75729 %Yeild strength of shaft (Pa) %Sy=91000*6894.75729 %Stress concentration factor (K=2 means that %splines, keyseats are present. K=1.6 %Stress concentration factor at section 1 is K=1.6 %due to the step %Section 2, no geometry change s K=1 %Section 3, no geometry change so K=1.6 due to step FS=1.8 %safety factor B=(FS*10*Cs*K)/(Sy) SSy=B^(1/3) %Shaft yield strength factor %% Service factor %Service factor (Kser), Light transmissions Kser=1 %% Formula for Combined loading based on energy distortion theory % (Von Mises) %Load=sqrt([(Mres2-(FA3*dassume/8)]^2+(3/4)*(T2^2))^1/3 %Section 1: axial force=FV0F, Moment=Mres1, Torque=T2,keyseat %Section 2: Axial force=FV0F, Moment=Mres2, Torque=T2. %Section 3: Axial force=FV0F, Moment=Mres3, Torque=T2,step %The section with the largest minimum diameter is the critical section. %Assume a d =.015m dassume=.015 %meters A=(FV0F*dassume)/8 %This can be negleted if FA3 is small compared to %Mres but it's not so it can't be. Have to %assume a minimum diameter. %% Calculate the load at the critical sections % based on energy distortion theory (Von Mises) Load1=sqrt([(Mres3-(FV0F*dassume/8))]^2+(3/4)*(T2^2)) %% Place all factors in the formula for minimum diameter %dmin1=(1/Ky)*(Load^1/3)*Kser dmin1=(SSy)*(Load1)^1/3*Kser %% Critical Minimum diameter occurs at section 2 where the % middle bear is. %Minimum diameter at section 1 is 14.9 mm under static loading. %minimum critical diameter is 22.4 mm under static loading %Minimum diameter at section 3 is 11.6 mm under static loading %% The resultant moment will act at section 2 % The critical element will be situated %at the furthest edge from the bending plane and it will experience %maximum shear torsional stress, axial bending stress %and axial ompressive stress. %% Dynamic loading analysis (Fatigue) %% Formula for sizing a shaft in dynamic loading %dmin2=(1/Kf)*(Load^1/3)*Kser %% Find the fatigue strength factor based on distortion energy theory % (Von Mises) %SSf=(1/Kf)=[(10*Cs*K)/(Se)]^1/3 %% Calculate the fatigue strength factor %Now we know that the critical section is section 2 so dynamic %analysis will occur at this section. %Size factor Cs=1 %Shaft diameter lower than 60mm, Cs=1 %Stress conentration factor K=1 %K=1.6 at section 1 (keyseat) %K=1 at section 2 %K=1.6 at section 3 (step) %Service factor Kser=1 %Light transmission %Safety factor FS2=1.8 %Section 1 can only reach a safety factor of 1.6 %Section 2 can %Section 3 can only reach a safety factor of 1.4 Sul=450000*6894.75729 %Psi to Pa Se=.50*Sul %Surface factor Sf=1 %Surface factor is .8 beause if we're going to machine it. %This decreases the endurance limit. C=(FS2*10*Cs*K)/(Se*Sf) SSf=C^(1/3) %Shaft fatigue strength factor %% Calculate the load at the critical section % based on energy distortion theory (Von Mises) Load2=sqrt([(Mres2-(FV0F*dassume/8))]^2+(3/4)*(T2^2)) %% Place all factors in the formula for minimum diameter %dmin2=(1/Kf)*(Load2^1/3)*Kser dmin2=(SSf)*(Load2^1/3)*Kser %Originally the nitric 50 stainless steel bar was going to be used %but it gave large minimum diameters so high carbon stainless steel was used %as a replacment. The high carbon stainless steel %has a higher yield strength and tensile strength %than nitric 50 so it was able to decrease the minimum diameter. %The minimum diameter at section 3 based the dynamic loading % is 13.6 mm. Compared to 11.6 mm from static loading. %The minimum diameter at section 2 is based on the dynamic loading %which is 26.2mm. Compared to 24.4 mm for static loading. %The minimum diameter at section 1 is based on the dynamic loading %which is 17.5 mm. Compared to 14.9 mm for static loading. %the minimum diameters have a factor of safety of 1.8. %% Recommended Factor of safety of 1.8 %% Finding the resultant force on the bottom bearing % (FB1resN) FB1resN=sqrt((FH3F^2)+(FH3S^2)) %Resultant radial force %(Newtons) FB1reslb=FB1resN*0.224809 %Resultant radial force in pounds %% Finding the equivalent force for the bearing at the bottom ratio=abs(FV2F/FB1resN) %Thrust force FV2F/Fb1resN < .35 % therefore the equivalent force (Fe) on the bearing is %equal to the resultant radial force (FB1resN). %Fe=FV2F/FB1resN %% Finding the design life required %Hours in 10 years 87600 %Can't attain 1o years, only 3700 hours due to load capacity of %bearings on the market. hours=8500 rpm=180 minperhour=60 deslife=hours*minperhour*rpm %design life in cycles (revs) %This is the max design life that the bottom bearing can have %due to the rated dynamic capacity of the bearings on the market. %% Finding the dynamic capacity rating for the bottom bearing C1=FB1reslb*((deslife/10^6)^1/3) C1a = C1*4.44822 %(Newtons) %the maximum capacity found on the market was (1950 lbs) which %is higher than the calculated at 1932 at a dia of 15 mm. %The bottom bearing will have to be changed every 7000 hours %assuming that it is used regularly. %440C stainless steel bearing with a dia of 15 mm was chosen. %% Finding the resultant force on the bearing at the middle of % the vertical shaft % (FB2resN) FB2resN=sqrt((FH1F^2)+(FH2S^2)) %resultant radial force on middle %bearing (Newtons) FB2reslb=FB2resN*0.224809 %% Finding the dynamic capacity rating for the bearing at the % middle of the vertical shaft %% Finding the design life required for the middle bearing hours=4820 rpm=180 minperhour=60 deslife=hours*minperhour*rpm %design life in cycles (revs) %% Dynamic rated load capacity for the bearing at % the middle of the vertical shsft C2=FB2reslb*((deslife/10^6)^1/3) %Rated capacity pounds C2b=C2*4.44822 %(Newtons) %Themax dynamic load capacity found was %1600 lbs which is higher than the calculated 1594 lbs. %This will require the bearing to be replaced every %4820 hours if regular use is assumed. %440C Stainless steel ball bearing with a diaof 15 mm.