A workpiece weighing 50 lb is clamped to a work table in readiness for a machini

Question

A workpiece weighing 50 lb is clamped to a work table in readiness for a machining operation. Table and clamps together weigh 250 lb. The milling cutter applies a downward load of p = 25 lb. Support is by two paralled slide bars of diameter "D", each having two slide bearings placed 20 in apart. E for the bar material is 29 times 10^6 lb/in^2. If the vertical downward displacement of the table under the symmetrical loading is limited to 10^-3 in and a conservative design factor of 2 is used, what is the minimum value for "D"?

Explanation / Answer

solution:

1)here for given arrangement total load supported by two bars is given by

W=weigh of clamp,table+weigh of workpiece+cutting force

W=250+50+25=325 lb

as two bars are symmetrically sharing load and this loading on each bar at centre of bar will create maximum displacement of bar

force on each bar at center=F=W/2=162.5 lb

2)where maximum displacement will be at centre of bar at distance of l=40 in and it is given as

dyallowable=F*L^3/(48*E*I)

F=162.5 lb

L=80 in

E=29*10^6 lb/in2

I=(pi/64)*D^4

dy=vertical limiting displacement is given by

dylimiting=1*10^-3 in

design factor=dylimiting/dyallowable

dyallowable=1*10^-3/2=0.5*10^-3 in

on putting all value we get that

dyallowable=F*L^3/(48*E*I)

(.5*10^-3)=(162.5*80^3)/(48*29*10^6*(pi/64)*D^4)

on solving we get that

D=7.0248 in

3)in this way minimum diameter of bars which are subjected to bending in simply supported case of beam as D=7.0248 in and this diameter is same for both the bars.

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