As shown in the figure below a cantilevered circular shaft d

As shown in the figure below, a cantilevered circular shaft (diameter 5 inches) has a welded bar of length 2 ft attached at the end. The bar is subjected to a load P in the horizontal direction. The shaft also has a downward distributed load lb/ft, and a torque T = 300 ft-lbs. Assume that the attached bar and the weld is much stronger than the shaft. The desired factor of safety N = 1.5 and . Note that (Hint: Don’t forget the load P causes a moment at the end of the beam).

(a). Determine the wall reactions MA, Ay, and Ax.

(b). Draw the shear and bending moment diagram for the circular shaft.

(c). Find the maximum moment in the shaft.

(d). Find the largest load P so that yielding does not occur using the maximum shear stress theory.

(e). Find the largest load P so that yielding does not occur using the octahedral shear stress theory.

yield point stress=20000 psi

Circular shaft

Solution

solution:

1)here for stattic equillibrium we have that

Fx=0

Ax-P=0

Ax=P lbf

Fy=0

Ay-w1*4=0

Ay=4w lbf

2)hence moment at A is given by

Ma=(w1*4)*(2+4/2)+2*P=16w1+2P lbf ft

3)here hear forces in cantilever beam is given by if naming is given as A at wall,D at 2\',C at 4\',E at 6\' and b at 8\'.

Fal=0 lbf

Far=4w1

Fdl=4w1

Fdr=4w1

Fcl=4w1-2w1=2w1

Fel=4w1-4w1=0

Feb=0 lbf

hence shear force diagram would be horizontal between A to D ,then inclined curve till point E and again horizontal till point B.

4)here bending moment in the shaft is given by as follows

Ma=-(16w1+2P)

Md=-(16w1+2P)

Mc=-(16w1+2P)+2w1=14w1+2P

Me=-(16w1+2P)+8w1=-(8w1+2P)

Mbl=-2P

Mbr=0

hence bending moment diagram would be horizontal between pointA to D,again second degree curve between point D to E and again inclined between point E to point B,then to zero line.

5)hence maximum bending moment would be at A which is equal to

Ma=-(16w1+2P)

6)here as per maximum shear stress theory

tmax=.5*Syt/Nf=.5*20000/1.5=(16/pi*d^3)(M^2+T^2)^.5

here on putting all value we get bending moment as

M=16w1+2P=163624.34 ft lb

where for w1=10000 lb/ ft

P=1812.17 lbf

7)here maximum octahedral shear stress theory or von mises theory we get

Sv=syt/Nf=20000/1.5=(32/pi*d^3)(M^2+.75T^2)^.5

on putting value we get moment as

M=16w1+2P=163624.41 lbf ft

here again for w1=10000 lb/ft,we get maximum value of P to avoid yeilding of shaft as follows

P=1812.20 lbf

8)hence it is seems that for given factor of safety maximum value of P that can applied over shaft is closely same.