TWISTING OF SHAFTS 123and θ= 0 .0543 rad
= 0 .0543 rad×360 °
2 πradi.e. angle of twist,θ= 3. 11 °
Problem 2. If the shaft in Problem 1 were
replaced by a hollow tube of the same
external diameter, but of wall thickness
0.005 m, what would be the maximum shear
stress in the shaft due to the same applied
torque, and the resulting twist of the shaft.
The material properties of the shaft may be
assumed to be the same as that of Problem 1.Internal shaft diameter,
D 1 =D 2 − 2 ×wall thickness= 0. 0525 − 2 × 0. 005i.e. D 1 = 0 .0425 m
The polar second moment of area for a hollow tube,
J= 32 π
(
D^42 −D^41)
from Problem 12, page 93.Hence, J=
π
32(
0. 05254 − 0. 04254)= 4. 255 × 10 −^7 m^4From equation (10.6),
τ
r=T
JHence,maximum shear stress,
τˆ=Tr
J=1424 N m×0. 0525
2m4. 255 × 10 −^7 m^4
= 87 .85 MPaFrom equation (10.6),
τ
r=Gθ
Lfrom which,θ=
τL
Gr=87. 85 × 106N
m^2×2m70 × 109N
m^2× 0 .02625 mi.e. θ= 0 .0956 rad
= 0 .0956 rad×360 °
2 πradi.e. angle of twist,θ= 5. 48 °
Problem 3. What would be the maximum
shear stress and resulting angle of twist on
the shaft of Problem 2, if the wall thickness
were 10 mm, instead of 5 mm?Internal shaft diameter,D 1 =D 2 − 2 ×wall thickness= 0. 0525 − 2 × 10 × 10 −^3i.e. D 1 = 0 .0325 mThe polar second moment of area for a hollow tube,J=π
32(
D 24 −D 14)Hence, J=π
32(
0. 05254 − 0. 03254)= 6. 36 × 10 −^7 m^4From equation (10.6),τ
r=T
J
Hence,maximum shear stress,τˆ=Tr
J=1424 N m×0. 0525
2m6. 36 × 10 −^7 m^4
= 58 .75 MPaFrom equation (10.6),τ
r=Gθ
Lfrom which, θ=τL
Gr=58. 75 × 106N
m^2×2m70 × 109N
m^2× 0 .02625 mi.e. θ= 0 .06395 rad= 0 .06395 rad×360 °
2 πrad
i.e. angle of twist,θ= 3. 66 °N.B. From the calculations in Problems 1 to 3, it
can be seen that a hollow shaft is structurally more
efficient than a solid section shaft.Problem 4. A shaft of uniform circular
section is fixed at its ends and it is subjected
to an intermediate torqueT, as shown in