Question

Problem 4- The Fig. below shows a shaft carrying a pulley A and a gear B and supported in two bearings C and D. The shaft transmits 30 kW at 200 r.p.m. The tangential force Ft on the gear B acts vertically upwards as shown. The pulley delivers the power through a belt to another pulley of equal diameter vertically below the pulley A. The ratio of tensions T1/T2 is equal to 2.5. The gear and the pulley weigh 1000 N and 2500 N respectively. The permissible shear stress for the material of the shaft may be taken as 60 MPa. Assuming the weight of the shaft to be negligible in comparison with the other loads, determine its diameter. Take shock and fatigue factors for bending and torsion as Km - 2 and Ke - 1.5 respectively. Note that:- for the gear the Ft = Torque/radius of the gear Pulley B Gear А R 1250 c D 750 750 350 1250 Attach File

static and material strength

Answer 1

The detailed step-by-step calculations for the given problem is described in the images below:

Page 1 30 KW= * Criven Data : - Power, pa 30x103 W RPM , N=200 s.p.m. Ratio of tensions, T 7 7 7 42 - 2.5 = 1000 N - Weight of the Grear 'B', WB - Weight of the Pulley 'A', WA - Permissible shear stress for shaft, = 2500N T = 60mla Permissible - shock and fatigue factor for bending, km = 2 -Shock and fatigue factor for torsion, kt = 1.5 Diameter of Pulley A, DA= 1250 mm. Diameter of crear 'B', DB = 750 mm. - Ft A Pulley $750 E Gear А B - B с - 가 250 1ooo 3 50 1250 * SK (All Dimensions are in mm. )

Page 2 - Now, the torque transmitted by the shaft, T = Рx6o 30 x 103x60 27N 27 X 200 1432.39 Nm 1432.394103 N.mm こ Now, Let T, and T = Tensions in the right side and slack side of the belt on Pulley Å - since, the torque on the pulley is same as that of shaft. Therefore, (T, - Tz). RAFT (T. -T.) = 14322.39 x 10 3 = 14 32. 39 x 10 3 625 2291.83 N CT, -T2) Now, it is given that, I 2.5 T,=2.5T2 Tz from equation - 2.5T2 - T2 =2291.83 = 2291,83 1.5 1. Ta 1527.89 N and T = 2.58 1527.89 1 in T, = 3819.72N -

Page 3 i Total vertical load acting downward on the shaft at 'A', = T, + Tz + WA = 3819.72 +1527.89 +2500 T - = 7847.61 N • Assuming that the torque on the Grear 'B' is same as that of the shaft, therefore the tangential force acting vertically upward on the creas 'B', 1432.39 X103 - 3819.71N RB since, the weight of creca 'B' (Wg = 1000 N) acts vertically downward, therefore the total vertical lood acting upward on the shaft ct 'B', = F - WB t 375 - 3819, 71 - 1000 = 2819.71 N Now, let us find the reactions ct the bearings c'and 'D'. Let, Rc and Ro Reactions et c' and 'p' respectively, 2819.71 CN) A с B Rc 7847.61 CN) 250 TODO * 350 (vertical Loading Diagram)

Page 4 - Taking moments about 'D', we get, (Rc x1000) = (7847.61 X1250 ) +(2819.71x350) = 98,09,512.5 + 9,86,898,5 Rc こ 10,796.411 N 1 Now, for the equilibrium of the shaft, Rp + 7847.61= Rc+2819.71 - RO 5768.511 N - Now, it is clear that, Bending Moments at '' and 'B' =0 (shaft ends.) Bending moment at c' 7847.61 X 250 = 1961.9X163 N.mm - Bending moment at 'b' = 2819.71 × 350 = 986.8998103 N.mm we can see that the bending moment is mabimum at c. : Mcheimum Bending moment =m= =mc - 1967.9 xlo nomm Now, the equivalent twisting moment is given by Te = √ (kmxm ) 2 + (KXT) ² =J (2x1961.9x103) + (1.5 x 1432.39 X 103)2 . Te = 4473.55 X 103 Nimm

Page - 5 Now, we know that, the equivalent twisting moment is also given by Te 7 ? permissible x diette) .:(ashart) 3 16 X Te TXT permissible こ 16 = 16 X 4473.55 X 10- TX60 d shaft こ 72.41 mm ~ 73 mm - Ang