Question Solved1 Answer Q1. (30 P) A steel shaft with an outside diameter of 20 mm is supported in bearings at its ends. Two pulleys are keyed to the shaft, and the pulleys carry belt tensions as shown. The yield strength of the steel is Sy=440 MPa. Determine the factors of safety predicted at points H and K by the maximum-shear-stress theory and by the maximum-distortion-energy theory. Ignore the shear stresses caused by the shear forces. (Note: Draw FBD, V and M diagrams and show all your calculations). 1400 N 200 N 90 y mm 160 mm 160 mm 160 mm 160 mm K D 1100 N 135 mm X 300 N

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Transcribed Image Text: Q1. (30 P) A steel shaft with an outside diameter of 20 mm is supported in bearings at its ends. Two pulleys are keyed to the shaft, and the pulleys carry belt tensions as shown. The yield strength of the steel is Sy=440 MPa. Determine the factors of safety predicted at points H and K by the maximum-shear-stress theory and by the maximum-distortion-energy theory. Ignore the shear stresses caused by the shear forces. (Note: Draw FBD, V and M diagrams and show all your calculations). 1400 N 200 N 90 y mm 160 mm 160 mm 160 mm 160 mm K D 1100 N 135 mm X 300 N
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Transcribed Image Text: Q1. (30 P) A steel shaft with an outside diameter of 20 mm is supported in bearings at its ends. Two pulleys are keyed to the shaft, and the pulleys carry belt tensions as shown. The yield strength of the steel is Sy=440 MPa. Determine the factors of safety predicted at points H and K by the maximum-shear-stress theory and by the maximum-distortion-energy theory. Ignore the shear stresses caused by the shear forces. (Note: Draw FBD, V and M diagrams and show all your calculations). 1400 N 200 N 90 y mm 160 mm 160 mm 160 mm 160 mm K D 1100 N 135 mm X 300 N
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Given Data:-{:[d=20mm],[S_(y)=440mpa.]:}Solution:-=> Free body Diagram:-plane x-yfor Static Equillibrium,{:[sumM_(A=0)(1600 xx160)-(R_(D)xx640)=0],[R_(D)=400N],[sum Fy=0quad1600-RA-R_(D)=0.],[R_(P)=1200N]:}{:[SFA=-RA=-1200N],[SFB=-1200+1600=400N],[SF_(C)=-1200+1600=400N=SF_(D)],[SFD=400N". "],[(BMD)/(BMA)=BMD=0". "],[BM_(B)=1200 xx160=192000N*mm],[{:BM_(H)=1200 xx320=11600 xx160)=128000N*mm],[BMC=1200 xx480=5210000Nxxmn=6400Nmm]:}{:[1400+200],[=1600N]:}Torve alagram(1600 xx360)TB=(1400-200)xx90//2=54000N-mmTt=(1100-300)xx135//2=54000N-mmPhare x-zFor Statk Equilibrium,Torque DiougnamBMDBMA:BM_(D)=0{:[BMB=(350 xx(60)=56000N-mm],[BMA=(350 xx320)=112000N-mm],[BMC=(350 xx480)=168000N-mm]:}Internal Loadinis at sectuon (HK)Bentunig moment, M=sqrt((128000)^(2)+(112000)^(2))=170082.33N-mm Torsional moment, T=54000N-mm.Sorens At Sectuon itk shoar load, V=sqrt(400^(2)+350^(2))=531.51NMaximum.Torsional shear Stren,Transverse shear stren, tau_("iran ")=(4)/(3)((v)/(t)) Trorsion =(T)/(J)R=(16+)/(pid^(3))=(16 xx54000 ... See the full answer