Question

1.5 in. 1.25 in.

Two steel shafts, AB and CD, ara bolted together at the flange. Because the bolts are improperly sized, one shaft can rotate 1.25° before the shafts begin to rotate together. Find tha maximum shear stress for AB and CD when a torque of 510 kip*ft is applied to flange B

Answer 1

Assume Modulus of Rigidity for steel as 11. 2x 100 psi do G=11.2x106 psi tot Calculate the polar moment of inertia of shaft AB (AB: 012518) JAB = 70 - 17 x 225 m)4 = 0·24 24 Calculate the polar moment of inentia of Shaft CD Jco 7 De = I x (1.5 in)t = 0.497 194 Calculate the angle of twist of B wat A. 2 PAB= TABLAB and G JAB Here TAB is the toque applied at B wat A, LAB is the lergth of shaft, Gis the modulus of regedity. DAR - TAB (aft x 12h ) (11.2x100 psi) x 0.24194 Recoroarge the equation for Tosque applied TAB TAB= 11 2000 QAB ~(1) Calculate the angle of twist of c wat D Qoca Teoko - Tco ( 346 x 124 ) GJco (11.2x 106 ) + (0.407 int ) Rearrange the equation for Torque applied to TO

Tcp = 154622.2 doc (2) where the Given that one shaft can rotate 1.250 before the shafts begin to rotate together one So for correction shaft AB will rotate 1.250. de pirt too the rotation for connecting the shaft dos a la5° = (1-25 * ) = 0,0018 rad So the Tonque required for correction can be Calculated using (1) and QB TAB = 112000 dAB Substitute & Be for DAB TAB= 112000 (0.0218) = 2443.46 lbin = 2.443 kipsih Here Tao is the correction torque. CONCEPT So, the at flange Shaft AB correction remaining B only and CD. of the value cuate Since shaft of the applied torque Stress in the only after the then only shatt begin to rotate together. So, causes Calculate the value of torque (Tr) which rotation of shatt AB and CD Tr. To - The good C

How To is the torque applied at flange B. Substitute 2.443 14ps.in for TAB and 510 kip.ft for TB 79 - Sto meipft sem ) - 0.443 Kips in - 6117.554 kips.in Calculate the rotation caused by torque Th on the Shaft AB and CD. eodo slutilatie т? - The Teo" is the Here TAB torque acting on AB due to notation and Tot is the torque acting on CD due to rotation Substitute 112000 dr for Tags and 1546 22.20² for Tom 6119.557 Kipk-in = (12000 + 184622.2) 42 dr = 0.0229 Red Calculate the respective torque Taps and Tor TAB = 11 2000 (0.0228) at = 2569.8 Abin = 2.569 Kips.in Tror = 154622.2 (0.0228) = 3.547 kips.in

Maximum shear stress can now be calculated using the respective tonques acting on shatts due to rotation $9. TAB = 16TAB" = 6 TAB² x RAB Td3 JAB A3 Here is the radius of shadet AB Substitute 21569 kips.in for Tang? (1.25in) for RAB and 0.24 in 4 for JAB TAB = 2.569 Kips in x 1.25 -6.69 ksi 0.24 int - Tohxrco Tcp- J co Substitute 3.547 kips.in for Tamm (1.5 in) for nos and 0.497 for Ico To= 3.547 kipsin olisin = 5.350 ksi 0.497 int Therefore the manimum shear stress for AB is 16.69ksi. and for CD is s. 362 kesi.