Question

1. Consider the stepped (note change in diameter at bearing B) shaft with gear and belt drive as shown below. The shaft is rotating and can be modeled as simply supported at locations A and B 2000N (+ z direction) 2000N (- z direction) 600N 10 mm 100 mm rsimm 150 mm Diameter 400 mm 15 mm 150 mm B 75 mm Fp (- z direction) a. Determine the force on the belt, Fp b. Identify the location on the shaft most likely to fail and calculate the equivalent mean and alternating stresses at that point. c. If the shaft is machined from normalized 1020 steel, find the safety factor for infinite life. If infinite life is not possible, find the number of cycles to failure. Assume the load are such that the stresses increase proportionally

Answer 1

DOON R000 N 2000N 16mm 100mm Dia =400 mm 10 mm 8 Gear 斤 150 mm 150mm 75 mm ay Torque transmitted = 2000x 100-2x100 N-mm { by gear Also, this torque is only Hronsmited to belb 2x10 (2000-Fr ) 200 Fp 1000 N b Free -Body diagram of shoft in X-Y plane 60ON RA 45x103 N-mm 8MD inx- plane Free-Body diagram of shaft in zx plane 2000+p 30001N 2000N RAxx 300-2000 x150-3000x15-o RAx® I 750 N N-m 1 225x10 Kat KBK t 3000= 2000 N-mm RB,--2750 사

Resultant Bending Moment at c" Resu Itant Bending Moment at'e 225 x103 N-mm Since torque is same at section c and B2xio N-mm 50) (2625X10266 3X N-nn Magnitude oF BM roiln decide the roeaker section, hve have to ch ack stresses at both location as Br is having stress concentration. So even the Bending moment is less stress concentration can result in eriica Stress es At Extreme Fibres, if Bending stress at Bc = 32 Mc 32x266-33x1o-= 808.79 MPa Theoritical Bending stress at B 32MB 32x 225xt0 2291.331 MPa tHence B is the section most lihety to fail From Databooh for bending Stress at s" due to stress cancentration 157x 2291 83 due to Bending = 3598.175 MPa Torsional shear stress at 6xS1013.591 MFa From Data book , fortorsion Torsional shear stress at B' -4x1018-59に1426.027 M a Since the shaft is rotating the flbres subiected to completely reversed bending stress and a constant torsional shear stress 3598.175 Ma Trnar :Trnin= 142 6.027 MPa tm-1426027 MPa

pplying Distortion- Energy theory Mean equivalent sness (one):-1 Gitan = 2 Amplitude equivalent stress (6%) /0a2 tat-3598,i7SMa For 1020 steel Normalisea? t3tm = 2469.951 MPa Syt 345 MPa Sut= 440 MPa. For steels endurance limit (se). ossut=220MPa Corrected endurance lirnit (Se Ka Kb Ke kp Se 220MPa. Since Amplitude stress is greater than endurance stress ,hece it is a finite life problem Also the amplitude stress is even- higher thanthe Sut of materfal Hence the shaft can't sustain even one cyde It will fail