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28. The shaft shown in Figure P5-28 is supported by bear- ings at each end, which have bores of 20.0 mm. Design the shaft to carry the given load if it is steady and the shaft is stationary. Make the dimension a as large as pos- sible while keeping the stress safe. Determine the required d = 20mm D= ? -a R = ? Length not to scale 5.4 kN d 20mm -125 mm. -250 mm- FIGURE P5-28 (Problems 28, 29, and 30) Problem 5-30 INPUT DATA: Bore Sizes, d= Applied force = Max fillet allowed- Metal Required= Design Factor, N = Length of bore to middle = Full length = RESULTS: Manufacturing Process: Tensile Strength, S_u = Endurance Strength, S_n - Material Factor, C_m= Stress Factor, C_st = 20 5.4 2 SAE 1137 cold-drawn steel 3 125 250 =Stress cold-drawn 676 MPa 252 MPa 1 Unitless 1 Unitless 0.81 Unitless 0.813070766 Unitless 95.76123077 MPa Reliability Factor, C_R= Size Factor, C_S= S'_sm= S_su 507 MPa Estimated Actual Endurnace limit S_n' = 165.9640048 k_t= 1.6 Q= 23113.18372 49.0090887 D= Size Factor when d=20mm, C_S= Estimated Actual Endurnace limit S_n' = S'_sm when d=20 mm = k_t= Max Moment = d = 0.899294106 Unitless 183.563913 105.9163778 MPa 1.8 13288.83958 N*mm 4.921481481 mm mm KN mm mm mm FORMULAS: diameter in the middle portion. The maximum fillet per- missible is 2.0 mm. Use SAE 1137 cold-drawn steel. Use a design factor of 3. 29. Repeat Problem 28, except using a rotating shaft. 30. Repeat Problem 28, except using a shaft that is rotating and transmitting a torque of 150 N m from the left bear- ing to the middle of the shaft. Also, there is a profile key- seat at the middle under the load. Appendix 3 Figure 5-11 Wrought steel has C_m-1.0 C_st 0 for bending stress Table 5-3 C_S=(D/7.62)^-0.11 0.577*S_n'=S'_sm S_su 0.75*S_u S_n'=(S_n)*(C_m)*(C_st)*(C_R)"(C_S) Figure A18-1 Iforgor ((Q*16)/pi)^1/3=D C_S=(D/7.62)^-0.11 S_n'=(S_n)*(C_m)*(C_st)*(C_R)*(C_S) 0.577*S_n' I forgor again d=Moment/Force

28. The shaft shown in Figure P5-28 is supported by bear- ings at each end, which have bores of 20.0 mm. Design the shaft to carry the given load if it is steady and the shaft is stationary. Make the dimension a as large as pos- sible while keeping the stress safe. Determine the required d = 20mm D= ? -a R = ? Length not to scale 5.4 kN d 20mm -125 mm. -250 mm- FIGURE P5-28 (Problems 28, 29, and 30) Problem 5-30 INPUT DATA: Bore Sizes, d= Applied force = Max fillet allowed- Metal Required= Design Factor, N = Length of bore to middle = Full length = RESULTS: Manufacturing Process: Tensile Strength, S_u = Endurance Strength, S_n - Material Factor, C_m= Stress Factor, C_st = 20 5.4 2 SAE 1137 cold-drawn steel 3 125 250 =Stress cold-drawn 676 MPa 252 MPa 1 Unitless 1 Unitless 0.81 Unitless 0.813070766 Unitless 95.76123077 MPa Reliability Factor, C_R= Size Factor, C_S= S'_sm= S_su 507 MPa Estimated Actual Endurnace limit S_n' = 165.9640048 k_t= 1.6 Q= 23113.18372 49.0090887 D= Size Factor when d=20mm, C_S= Estimated Actual Endurnace limit S_n' = S'_sm when d=20 mm = k_t= Max Moment = d = 0.899294106 Unitless 183.563913 105.9163778 MPa 1.8 13288.83958 N*mm 4.921481481 mm mm KN mm mm mm FORMULAS: diameter in the middle portion. The maximum fillet per- missible is 2.0 mm. Use SAE 1137 cold-drawn steel. Use a design factor of 3. 29. Repeat Problem 28, except using a rotating shaft. 30. Repeat Problem 28, except using a shaft that is rotating and transmitting a torque of 150 N m from the left bear- ing to the middle of the shaft. Also, there is a profile key- seat at the middle under the load. Appendix 3 Figure 5-11 Wrought steel has C_m-1.0 C_st 0 for bending stress Table 5-3 C_S=(D/7.62)^-0.11 0.577*S_n'=S'_sm S_su 0.75*S_u S_n'=(S_n)*(C_m)*(C_st)*(C_R)"(C_S) Figure A18-1 Iforgor ((Q*16)/pi)^1/3=D C_S=(D/7.62)^-0.11 S_n'=(S_n)*(C_m)*(C_st)*(C_R)*(C_S) 0.577*S_n' I forgor again d=Moment/Force

Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter6: Stresses In Beams (advanced Topics)
Section: Chapter Questions
Problem 6.3.1P: Repeat Problem 6.2-17 but now use a transformed-section approach.
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Repeat Problem 28, except using a shaft that is rotating
and transmitting a torque of 150 N * m from the left bearing to the middle of the shaft. Also, there is a profile keyseat at the middle under the load. 

 

 

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28. The shaft shown in Figure P5-28 is supported by bear- ings at each end, which have bores of 20.0 mm. Design the shaft to carry the given load if it is steady and the shaft is stationary. Make the dimension a as large as pos- sible while keeping the stress safe. Determine the required d = 20mm D= ? -a R = ? Length not to scale 5.4 kN d 20mm -125 mm. -250 mm- FIGURE P5-28 (Problems 28, 29, and 30)
Transcribed Image Text:28. The shaft shown in Figure P5-28 is supported by bear- ings at each end, which have bores of 20.0 mm. Design the shaft to carry the given load if it is steady and the shaft is stationary. Make the dimension a as large as pos- sible while keeping the stress safe. Determine the required d = 20mm D= ? -a R = ? Length not to scale 5.4 kN d 20mm -125 mm. -250 mm- FIGURE P5-28 (Problems 28, 29, and 30)
Problem 5-30 INPUT DATA: Bore Sizes, d= Applied force = Max fillet allowed- Metal Required= Design Factor, N = Length of bore to middle = Full length = RESULTS: Manufacturing Process: Tensile Strength, S_u = Endurance Strength, S_n - Material Factor, C_m= Stress Factor, C_st = 20 5.4 2 SAE 1137 cold-drawn steel 3 125 250 =Stress cold-drawn 676 MPa 252 MPa 1 Unitless 1 Unitless 0.81 Unitless 0.813070766 Unitless 95.76123077 MPa Reliability Factor, C_R= Size Factor, C_S= S'_sm= S_su 507 MPa Estimated Actual Endurnace limit S_n' = 165.9640048 k_t= 1.6 Q= 23113.18372 49.0090887 D= Size Factor when d=20mm, C_S= Estimated Actual Endurnace limit S_n' = S'_sm when d=20 mm = k_t= Max Moment = d = 0.899294106 Unitless 183.563913 105.9163778 MPa 1.8 13288.83958 N*mm 4.921481481 mm mm KN mm mm mm FORMULAS: diameter in the middle portion. The maximum fillet per- missible is 2.0 mm. Use SAE 1137 cold-drawn steel. Use a design factor of 3. 29. Repeat Problem 28, except using a rotating shaft. 30. Repeat Problem 28, except using a shaft that is rotating and transmitting a torque of 150 N m from the left bear- ing to the middle of the shaft. Also, there is a profile key- seat at the middle under the load. Appendix 3 Figure 5-11 Wrought steel has C_m-1.0 C_st 0 for bending stress Table 5-3 C_S=(D/7.62)^-0.11 0.577*S_n'=S'_sm S_su 0.75*S_u S_n'=(S_n)*(C_m)*(C_st)*(C_R)"(C_S) Figure A18-1 Iforgor ((Q*16)/pi)^1/3=D C_S=(D/7.62)^-0.11 S_n'=(S_n)*(C_m)*(C_st)*(C_R)*(C_S) 0.577*S_n' I forgor again d=Moment/Force
Transcribed Image Text:Problem 5-30 INPUT DATA: Bore Sizes, d= Applied force = Max fillet allowed- Metal Required= Design Factor, N = Length of bore to middle = Full length = RESULTS: Manufacturing Process: Tensile Strength, S_u = Endurance Strength, S_n - Material Factor, C_m= Stress Factor, C_st = 20 5.4 2 SAE 1137 cold-drawn steel 3 125 250 =Stress cold-drawn 676 MPa 252 MPa 1 Unitless 1 Unitless 0.81 Unitless 0.813070766 Unitless 95.76123077 MPa Reliability Factor, C_R= Size Factor, C_S= S'_sm= S_su 507 MPa Estimated Actual Endurnace limit S_n' = 165.9640048 k_t= 1.6 Q= 23113.18372 49.0090887 D= Size Factor when d=20mm, C_S= Estimated Actual Endurnace limit S_n' = S'_sm when d=20 mm = k_t= Max Moment = d = 0.899294106 Unitless 183.563913 105.9163778 MPa 1.8 13288.83958 N*mm 4.921481481 mm mm KN mm mm mm FORMULAS: diameter in the middle portion. The maximum fillet per- missible is 2.0 mm. Use SAE 1137 cold-drawn steel. Use a design factor of 3. 29. Repeat Problem 28, except using a rotating shaft. 30. Repeat Problem 28, except using a shaft that is rotating and transmitting a torque of 150 N m from the left bear- ing to the middle of the shaft. Also, there is a profile key- seat at the middle under the load. Appendix 3 Figure 5-11 Wrought steel has C_m-1.0 C_st 0 for bending stress Table 5-3 C_S=(D/7.62)^-0.11 0.577*S_n'=S'_sm S_su 0.75*S_u S_n'=(S_n)*(C_m)*(C_st)*(C_R)"(C_S) Figure A18-1 Iforgor ((Q*16)/pi)^1/3=D C_S=(D/7.62)^-0.11 S_n'=(S_n)*(C_m)*(C_st)*(C_R)*(C_S) 0.577*S_n' I forgor again d=Moment/Force
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