The shaft in the figure below is supported on journal bearings that do not offer any resistance t...
The shaft in the figure below is supported on journal bearings that do not offer any resistance to axial load. The yield strength of the material is Ơ,-300 MPa and the safety factor is FS-2.5 1) 2) 3) 4) Determine the reaction at the supports. Draw the shear force, bending and torsion moment diagrams Derive an expression for the components of the stress tensor at a cross section of the shaft Neglect the shear stress due to the transverse shear forces and determine to the nearest millimeter, the smallest diameter of the shaft that will support the loading strength. Maximum Shear Stress Theory (MSST) Distortion Energy Theory (DET) a. b. 30P 250 CA30 150 N 100 mm 100-N 50皿 SON 250 zR
The shaft in the figure below is supported on journal bearings that do not offer any resistance to axial load. The yield strength of the material is Ơ,-300 MPa and the safety factor is FS-2.5 1) 2) 3) 4) Determine the reaction at the supports. Draw the shear force, bending and torsion moment diagrams Derive an expression for the components of the stress tensor at a cross section of the shaft Neglect the shear stress due to the transverse shear forces and determine to the nearest millimeter, the smallest diameter of the shaft that will support the loading strength. Maximum Shear Stress Theory (MSST) Distortion Energy Theory (DET) a. b. 30P 250 CA30 150 N 100 mm 100-N 50皿 SON 250 zR
SOLUTION Given: To design shagt 300 Ma FS 2.5 Gy = 300 MPa FS = 2.5 The belt is inclind in both pulleys Firsty, we have to calculate component of loads in z and 150mm 30% İOON -auus 50 by = -50 Sir(30') -150 Sire") 250m D 500 mm 150 mn, B by=-100 N VLD 10ON 175N LOON (down wards ulley Fez = -100 cos(30)-250 cos130.) A18-75N mm (356.25N) 29687.5 N-mm 39062.5 N 73-2 N 303-IN -75 N 175 N (downwards) : Now, vertcal loodny diugam (VD) rs aam (VLD) RAv and Rov te to be 13532 5 N-mm drawn HeMD cal cul ヲ RAv+ RBV-275 .(i) 4600.75 N- → 100X250 + 175x750 -RBVX'ODO-O 39332 S N-m 3262631 N-mm bom en (i) RAV = 118.75 N Now we have to draw vertcal berdrny mo ment diagpam Resuteant BMD MIDE 11675%250-2968 7.5 MC,-156.25%250 39062.5 45000 N-mm N-mm om. N-m
Now, horizontal ng diegpom (HLD) RAH +173.21-303-11 +RBH → 17 3.21 x 250-303.", 750 + RBH X1000 RBH 184.03 N Gon (i); drauw in horzental bending moment dingsym MpH 54 13 x 250 3 532.5 N-mm McH -184.03 x 250-46007.5 N-mm Calculaton pesultant bendng moment at D and C 39332-5 N-mm = -4600 .75 ) 2 +(39062-5)2 Now cal cu lating Tozque at D c T. Tc = (250-100)XLOO = 15000 N-mm caloldion oTar ) Takin Kb-1, Te-93542035 67 m Thas, critical cross secton is the setion at tbe point wherc pulley c is tnstalled |(Tenax = (Te)c = 42035.67 N-mm
NoW こ300MPa FSこ2-5 (a) Maximum shear Sbress theony (MSST) may _ 으 의 °es.309 : 60HN FS 2.5 Let d' be diameto of shaft 16(Te) = max 6016x 4 2095.67 = (3573-19)⅓ = 15-268mm d 15-288 mm (b) Distortion ene y theay(DET) 0.577 x 30O69.24 MP FS L6 x 420 95.67 69 24 z dご (5096.353) 14.575 d14.575 mm