Data : Power transmitted by shafts = 1500KW Rotational speed of driving pulley = 2000rpm Pre stress belt ratio = 2.05 Gear pressure angle = 20 deg Safety Factor SF or ns = 1.3 Yield strength of shaft material = y=400 MPa Ultimate tensile strength of shaft material = uts= 1000 MPa Material of keys AISI 1020 cold drawn = y = 350 MPa A/A student data = 64 Schematical illustration of assembly 1. General calculations for shaft 1 Calculate angular velocity for shaft 1 n Rn = s Rs => 2000 150 = s 150 => s = 2000rpm Calculate the shaft 1 input torque Torque = power/ angular velocity Tq1 = 1500KW/209.44 = 7162Nm Calculate the belt tension Belt ratio: 2.05 Pulley 2 radius: 0.15m T1 = (2.05 T2 )Tq1 = (2.05 T2R) - (T2 R) 7162 = (2.05T2 0.15)-(0.15 T2 )T2 = 7162/0.1575 T2= 45.47kN And T1= 93.22N Calculate the tangential and radial forces of gear 1 Radius of gear 1= Rgear1=0.16m The tangential force is given by: Ft = torque/ Rgear1 = 7162Nm/0.16m => Ft = 44.76kNThe radial force is given by: Fr = Fttan20o = (44.76kN)tan20o => Fr = 16.29kN

2. Shaft 1 forces and reactions

Free body diagram shaft1Calculating the reactions on z-x plane By taking moments at point A (44.76kN 50mm) + (R2z 300mm) = (138.69kN 420mm)R2z = 186.7kNBy summation of forces z-x plane fz=0 R1z + 138.69kN = 44.76kN + 186.7kN R1z = 92.77kN Calculating the reactions on y-x plane By taking moments at point A 16.29 kN 50mm = R2y 300mm R2y = 2.715kN By summation of forces y-x plane fy=0 16.29kN 2.715kN - R1y = 0 R1y = 13.57kN 3. Bending moment and torque diagrams for shaft 1 A B C D mm130 290mm 150mm R1z=1236,7 N Ft=550,5 N R2z=3084,9 N T1+T2=2398,63 N Mb=-160,38 Nm Mc=-359,77 Nm

Moment diagram in z-x plane The bending moment at B and C in Z-X plane are given by: Mb = (-R1z 0.05m) = (-92.77kN 0.05m) = -4.64kNm Mc = (-R1z 0.3m) + (Ft 0.25m) = -27.831kNm + 11.19kNm = -16.64kNm A B C D 130mm mm290 150mm R1y=138,35 N Fr=200,37 N R2y=62,02 N Mb=-18 Nm

Moment diagram in y-X plane The bending moment at B and C in Y-X are given by: Mb = (-R1y 0.05) = -13.57kN 0.05 = -0.6785kNm Mc= (-R1y 0.3) + (Fr 0.25) = (-13575N 0.3) + (16.29kN 0.25) = 0Nm A B C D mm130 290mm mm150 Tx(Nm) X(m) Tq=66,06

Torque diagram The resultant moment at b is Mb = Mbz2 + Mby2 = (-4.64)2 + (-0.6785)2 = 4.69kNm The moment at point C is: Mc = Mcz2 + Mcy2 = (-16.64)2 + (0)2 = 16.64kNm As seen from the bending moment diargams the maximum moment occurs at point C at the bearing and has a value of 16.64kNm The torque is constant 7162Nm between points B and D. The critical point of the shaft is at point B. Mx = 16.64kNm Torque = 7162Nm

4. Determine the smallest safe diameter Calculation of the endurance limit e for shaft 1 Data: ns = 1.3 , y = 400pa , Mc = 16.64kNm , Tc = 7162Nm , uts = 1000Mpa e=Ka*Kb*Kc*Kd*Ke*Kf*Kg* e e=0.504 uts = 0.504 1000 = 504Mpa Ka=surface factor (hot rolled steel) Ka = a utsb = 57.7 1000-0,718 = 0.4047 Kb=size factor Kb = (d/7.62)-0,1133 = (47/7.62)-0,1133 = 0.8134 Kc=reliability, 90% Kc = 0.897 Kd=temperature factor Kd= 1 Ke=duty cycle Ke= 1 Kf=fatigue stress Kf= 0.63 Kg=various Kg= 1 e = 0.4047 0.8134 0.897 1 1 0.63 1 504MPa e = 93.756Mpa The smallest safe diameter for shaft 1 is given by

The smallest safe diameter for shaft1 is d=133mm

5. General calculations for shaft 2 Calculate angular velocity for shaft 2 nRn = sRs => 2000160 = s100 => s = 3200rpm Calculate the shaft 2 input torque Torque = power/ angular velocity Tq1=1500kW/335.1 rad/s = 4476.28 Nm Calculate the tangential and radial forces of gear 2 The tangential and radial forces are equal and opposite to the ones on gear 2 Ft=44.76kN Fr=16.29kN

6. Shaft 2 forces and reactions

Free body diagram shaft 2 Calculating the reactions on z-x plane By taking moments at point B: (-44.76kN 0.13) + (R2z 0.26) = 0 R2z = 22.38kN By summation of forces z-x plane: fz=0 -+ = 0 R1z = 44.76 - 22.38 R1z = 22.38kN Calculating the reactions on y-x plane By taking moments at point B: -16.29 x 0.13 = R2y x 0.26R2y = 8.145kN By summation of forces y-x plane:fy=0 -16.29 + 8.145 + R1y = 0 R1y= 8.145kN 7. Bending moment and torque diagrams for shaft 2 The bending moment at B and C in Z-X plane are given by: Mb = (-44.76 x 0.13) + (22.38 x 0.26) = 0Nm Mc = R1z x 0.13 = 22.38 x 0.13 = 22.38kNm A B C D 130mm mm130 130mm Mz(Nm) R1z=275,25 N Ft=550,5 N Mc=35,8 Nm X(m) R2z=275,25N

Moment diagram in Z-X plane The bending moment at B and C in Y-X are given by: Mb = 8.145 x 0.26 16.29 x 0.13 = 0Nm Mc = R1y x 0.13 = 8.145kNm A B C D 1mm30 130mm mm130 Mz(Nm) R1y100,19 N Ft200,37 N Mc=13,025 Nm X(m) R2y=100,18 N

Moment diagram in y-X plane

A B C D mm130 mm130 mm130 Tx(Nm) R1z=275,25 N Tq44,04 Nm X(m) R2z=275,25N

Torque diagramThe resultant moment at b is: Mb = Mbz2 + Mby2 = (0)2 + (0)2 = 0Nm The moment at point C is: Mc = Mcz2 + Mcy2 = (22.38)2 + (8.145)2 = 23.82kNm As seen from the bending moment diargams the maximum moment occurs at point C at the gear and has a value of 23.82kNm The torque is constant 4476.28Nm between points A and C. The critical point of the shaft is at point C. Mx = 23.82kNm Torque = 4476.28Nm

8. Determine the smallest safe diameter Calculation of the endurance limit e for shaft 1 Data: ns = 1.3 , y = 400pa , Mc = 23.82kNm , Tc = 4476.28Nm , uts = 1000Mpa e=Ka*Kb*Kc*Kd*Ke*Kf*Kg* e e=0.504 uts = 0.504 1000 = 504Mpa Ka=surface factor (hot rolled steel) Ka = a utsb = 57.7 1000-0,718 = 0.4047 Kb=size factor Kb = (d/7.62)-0,1133 = (47/7.62)-0,1133 = 0.8134 Kc=reliability, 90% Kc = 0.897 Kd=temperature factor Kd= 1 Ke=duty cycle Ke= 1 Kf=fatigue stress Kf= 0.63 Kg=various Kg= 1 e = 0.4047 0.8134 0.897 1 1 0.63 1 504MPa e = 93.756Mpa The smallest safe diameter for shaft 1 is given by

The smallest safe diameter for shaft2 is d=149mm

9. Calculations of the keys and keyways Keys are used to secure the pulleys and gears on the shafts. They are used to transmit the torque from the shafts to the rotating elements. The size of the keys depends on the shaft diameter and is taken form the British Standard Metric Keyways for Square and Rectangular Parallel Keys table. They can fail from shear and from bearing. Shear stress calculation Tdesign=P/As P=T/0.5d=2T/d As=b*l , Tdesign=2T/dbl To avoid failure due to shear Tdesign 0.4Sy/ns Bearing stress calculation Failure due to compressive or bearing stress The compression or bearing area of the keys is Ac=l*h/2 , design=P/Ac=2T/0.5*dlh=4T/dlh To avoid failure due to compressive or bearing stress: design 0.9*Sy/ns

Calculation of the key and the keyway for pulley 2 on shaft 1 Shaft dia= d = 133mm Torque = T = 7162Nm Key yield strength y = 350Mpa Key size (mm)= 60 x 32 x 20 Keyway size (mm)= 60 x 32 x 16 (depth) (4,3 hub) A. Failure due to shear Tdesign = (2 x 7162) / ( 0.06 x 0.032 x 0.02) = 373.02Mpa ns = (0.4 x Sy ) / Tdesign = (0.4 x 350) / 373.02 = 0.37

B. Failure due to bearing design = P/Ac = (4 x 7162) / ( 0.06 x 0.032 x 0.02) = 746.04Mpa nS = (0.9 x Sy ) / design = (0.9 x 350) / 746.04 = 0.42 Calculation of the key and the keyway for gear 1 on shaft 1 Shaft dia= d = 133mm Torque= T = 7162Nm Key yield strength y = 350Mpa Key size (mm)= 60 x 32 x 20 Keyway size (mm)= 60 x 32 x 16 (depth) (4,3 hub) A. Failure due to shear Tdesign = (2 x 7162) / ( 0.06 x 0.032 x 0.02) = 373.02Mpa ns = (0.4 x Sy ) / Tdesign = (0.4 x 350) / 373.02 = 0.37

B. Failure due to bearing design = P/Ac = (4 x 7162) / ( 0.06 x 0.032 x 0.02) = 746.04Mpa nS = (0.9 x Sy ) / design = (0.9 x 350) / 746.04 = 0.42

Calculation of the key and the keyway for pulley3 on shaft 2 Shaft dia= d = 149mm Torque= T = 4476.25Nm Key yield strength y= 350Mpa Key size (mm)= 65 x 37 x 25 Keyway size (mm)= 65 x 37 x 20 (depth) (3,3 hub) A. Failure due to shear Tdesign = (2 x 4476.25) / ( 0.065 x 0.037 x 0.025) = 148.9Mpa ns = (0.4 x Sy ) / Tdesign = (0.4 x 350) / 148.9 = 0.94

B. Failure due to bearing design = P/Ac = (4 x 4476.25) / ( 0.065 x 0.037 x 0.025) = 297.8Mpa nS = (0.9 x Sy ) / design = (0.9 x 350) / 297.8 = 1.06 Calculation of the key and the keyway for gear 2 on shaft 2 Shaft dia= d = 149mm Torque= T = 4476.25Nm Key yield strength y= 350Mpa Key size (mm)= 65 x 37 x 25 Keyway size (mm)= 65 x 37 x 20 (depth) (3,3 hub) A. Failure due to shear Tdesign = (2 x 4476.25) / ( 0.065 x 0.037 x 0.025) = 148.9Mpa ns = (0.4 x Sy ) / Tdesign = (0.4 x 350) / 148.9 = 0.94

B. Failure due to bearing design = P/Ac = (4 x 4476.25) / ( 0.065 x 0.037 x 0.025) = 297.8Mpa nS = (0.9 x Sy ) / design = (0.9 x 350) / 297.8 = 1.06

10. Calculations of the critical speed of rotation for shaft 2 The calculations for the critical speed are based on the diameter of the shaft between points B and C. the maximum deflection is at point C. shaft diameter d = 149mm Yangs modulus of elasticity E = 210000 N/mm2 Find the resultant force at point C F= Ft2+Fr2 F= + = 47.85kN The second moment of area of the shaft for 149mm diameter is: Calculation of the maximum deflection at point C The shaft at boints B and D behaves like a simply supported beam. The maximum deflection is given by:

Calculation of the critical speed of rotation The critical speed is given by: The critical speed in RPM is given by: The critical speed of rotation for shaft 2 is 16.1 RPM So the critical rotational speed of shaft 2 is much smaller than the actual.11. Attachments