124 MECHANICAL ENGINEERING PRINCIPLESFigure 10.5, wherea>b. Determine the
maximum resulting torque acting on the shaft
and then draw the torque diagram.aA
T 1 T 2
T bc
BxFigure 10.5From equilibrium considerations, (see Figure 10.5),
clockwise torque=the sum of the
anticlockwise torquesT=T 1 +T 2 ( 10. 11 )LetθC= the angle of twist at the pointC.
From equation (10.6),
T
J=Gθ
Lfrom which, T=
GθJ
LTherefore T 1 =
GθCJ
a( 10. 12 )and T 2 =
GθCJ
b( 10. 13 )Dividing equation (10.12) by equation (10.13) gives:
T 1
T 2=b
aor T 1 =
bT 2
a( 10. 14 )Substituting equation (10.14) into equation (10.11)
gives:
T=bT 2
a+T 2 =(
1 +b
a)
T 2orT 2 =
T
(
1 +b
a)=T
(
a+b
a)=Ta
(a+b)( 10. 15 )However,a+b= L
Therefore T 2 =
Ta
L( 10. 16 )Substituting equation (10.16) into equation (10.14)
gives:T 1 =b(
Ta
L)a=bT
L( 10. 17 )Asa>b,T 2 >T 1 ; therefore, maximumtorque=T 2 =Ta
LThe torque diagram is shown in Figure 10.6.T−Ta/L −Ta/LTb/L Tb/L0 0Figure 10.6 Torque diagramNow try the following exerciseExercise 50 Further problems on the
twisting of shafts- A shaft of uniform solid circular section
is subjected to a torque of 1500 N m.
Determine the maximum shear stress in
the shaft and its resulting angle of twist, if
the shaft’s diameter is 0.06 m, the shaft’s
length is 1.2 m, and the rigidity modulus,
G = 77 × 109 N/m^2. What power can
this shaft transmit if it is rotated at 400
rev/min? [35.4 MPa, 1.05°, 62.83 kW] - If the shaft in Problem 1 were replaced
by a similar hollow one of wall thickness
10 mm, but of the same outer diameter,
what would be the maximum shearing
stress in the shaft and the resulting angle
of twist? What power can this shaft
transmit if it rotated at 500 rev/min.
[44.1 MPa, 1.31°, 78.5 kW] - A boat’s propeller shaft transmits 50 hp
at 100 rev/min. Neglecting transmission
losses, determine the minimum diame-
ter of a solid circular section phosphor
bronze shaft, when the maximum permis-
sible shear stress in the shaft is limited to
40 MPa. What will be the resulting angle