RecapIn the previous class we have studied the designing ofshafts on the basis of -- Torsional Load, -- Bending Stress and -- Combined Torsional and Bending Stress separately.Practiced the designing of Shafts by solving Several Problems duly considering the stress induced in it when they are subjected to external load.

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ObjectivesOn the completion of this period, you would be able to knowLoading systems on shaftsDesign of axleFormulae on axleProblems on axle

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loads on the shaft Fig:1

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ShaftFig:2

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Axle An axle is a central shaft for a rotating wheel or gear.

In some cases the axle may be fixed in position with a bearing or bushing sitting inside the hole in the wheel or gear to allow the wheel or gear to rotate around the axle.

In other cases the wheel or gear may be fixed to the axle, with bearings or bushings provided at the mounting points where the axle is supported.

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Rail axleFig:3

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Axle housingFig:4

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Fig:5

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Design of axleIn axle , if the loading is bending the following simple bending eq. is used

Or

The term [ I/Y] is called section modulus and is denoted by Z Where M =bending Moment, N-mm

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Design of axle Contd I = Moment of inertia of cross section about axis of rotation ,mm4 for solid shaft of diameter d

for hollow shaft of external diameter D, internal diameter d,

= bending stress (tensile or compressive) ,N/mm2 Y= distance from neutrl axis to the outer most fibre,mm Z = section modulus, mm3

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Design of axle ContdFinally after substituting I and Y value M becomes

for solid shafts (axle)

for hallow shafts (axle)

where

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A pair of wheels of a railway wagon carries a load of 50 KN on each axle box, acting at a distance of 100 mm outside the wheel base. The gauge of the rails is 1.4 m. find the diameter of the axle between the wheels, if the stress is not to exceed 100 MPaFig:6Problem :

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W = 50 KN=50000 N L = 100mm F b = 100 MPa= 100 N/mm2 Bending moment (B.M.) at A=Ma=0 B.M. at C = Mc=50000100=5106 N-mm B.M. at D = Md=(50000)(1400+100) - (5000014000)=5000N-mm

Solution:

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Solution Contd B.M. at B= Mb = (50000) 1600-500001500-50000100=0

Maximum B.M. at C and D=5106N/mm

D= diameter of axle

Max B.M.

5106 = 100 d3 Therefore d= 79.8=80mm

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Problem: Compute the diameter of the shaft shown in the fig.,. Take =60 N/mm2Fig.7

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Solution: W= 70Kn = 70000NTaking moments about A,Rb1800=70600Rb=23.33KN=23330 N Ra =70-23.33=46.67 KN

70 kNRbRa6001200Loading diagramBending diagramMmaxFig.8

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Solution ContdMax: bending moment occurs at point cMax: bending moment Mc=233301.2 = 27996 N-mMax: B.M= 27996 N-mMax: B.M. substituting all values D=168 mm

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SummaryAn axle is a central shaft for a rotating wheel or gear. Even shafts are subjected to all kinds of loads , axles are subjected to only bending moment are dealt.

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QuizAxle is a a) counter shaft in stationary wheel b) central shaft for rotating wheel c) both a& b d) none of above

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Short questionswith a sketch list out various kinds of loads acting on the shaftsDefine axle.

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Essay Question (Problem)A uniform solid shaft ABCDE is supported in bearings at A and D, and rotates at 900 rev/min (rpm).50 kW are delivered to the shaft through the 560 mm diameter belt pulley at C. 30 kW are extracted via the 280 mm diameter pulley at B, and 20 kW at the 210 mm diameter pulley at E. At each pulley, the two belt strands are parallel and the ratio of tensions in them is 3:1. Determine the minimum acceptable shaft diameter if the design stress of 100 MPa incorporates an allowance for fatigue. [ 40 mm diameter]

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ESSAY QUESTION( PROBLEM) contd..Fig.9

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