Question

EXERCISE 1 A cast iron shaft ASTM-25 with an ultimate tensile strength of 25,000 psi, is subjected to a torsional load which is completely reversed. The load is to be applied an indefinite number of cycles. The shaft is 2” in diameter and is joined to a 3" diameter shaft whit a 0.5" radius fillet. The factor of safety is to be 2. What is the maximum torque that can be applied to the shaft? Solve by two ways: 1) Using Soderberg's equation 2) Using Ss=Tc/

Answer 1

Criven: Sut = 25,000 psi (ultimate tensile strength) 0.5 in zin 3in subjected to completely reversed toorsional load Minoch shaft diameter = 2 inch mason shaft diameter = 3 inch radius of fillet =0.5 inch Factor of safety = 2 Endurance limiti (for infinite life) Se = 0.4 Sut foor cast iron Se = 0.4 X 25,000 Se = 10,000 psi modified endurance limit I Se = = sex kaxkb xkexkd x ke

where, Ka= size factor ） Ka~ 0.81 foor ain dia shaft Kb = loading factor Ky = 0.577 for torsional loading Ke= Surface factor Ke=1 foor Polished surface. Kg = temperature factor Kd = 0.82 For T2450°C Ke= Reliability factor Ke=1 for 50% reliability. sel = 10,000 X 0.81* 0.577 X 1 X 0.82 x 1 Se' = 3832.43 Psi torque Given shaft to completely reversed subjected -T lb-in ie Tmin (minimum Torque) Tmax (maximum Torque) = +T lb-in and (a) = Tmax - Tmin TET) – TET Alternating Torque 2 Ta -T mean Torque (Tm) = Tmaxt min 2 TT = 2. Alternating stress (W) = Mean stress (m)=0 16 Ta Td3 16T Td² Ta =

Foo mla = 15 2 } Kz 1:19 8 0:5 = 0.25 2 Ford Sut = 25000 PSI } q~0.82 r=0.5 in - Kp = 1+ q (1-KL) = 1+ 0.82 (I-1.19) H 0.156 = 1.156 fatigue stress concentration factor Now, from Soderberg Equation : Ке 1 Ym Syt FS x KF 0+ 16 T Td3 11 FiS x KF Se 16 T Tx 23 x 3832.43 2X 1.156 T= TT X 2?x 3932: 43 2 x 1-156 x 16 T= 2603.8 lb-in 2) From Ss = TC J where Is = shear stress T= Torque c= distance from the center of the shaft to outer layer

C=_=_=lin J= Polar moment of inertia = TT 32 7 7 2 24 Food Fatigue failure Ss = sel Ке isle = TxC J Ке TX1 = 3832:43 1:156 T= 5207.6 lb-in A 04 32