Problems on Torsion, weight ratio of solid to hollow shaft, weight saving %

2. Q1. Determine the diameter of a solid shaft which will transmit 300 kW at 250 r.p.m; The
maximum shear stress should not exceed 30 N/mm2 angle of twist should not be more than
1o in a shaft length of 2 m. Take modulus of rigidity = 1 x 105 N/mm2.

5. So, D= 107.5 mm is not recommended

6. Q2. A 500 mm long solid steel [G = 80 GPa] shaft is being designed to transmit a torque T =
20 Nm. The maximum shear stress in the shaft must not exceed 70 MPa, and the angle of
twist must not exceed 3° in the 500 mm length. Determine the minimum diameter d
required for the shaft.
𝑻
𝑱
=
𝝉
𝑹

7. 𝑻
𝑱
=
𝑮𝜽
𝑳
𝜋 ⅆ4
32
=
𝑇𝐿
𝐺𝜃

8. Q3. Determine the diameter of solid steel shaft which will transmit 90
kW at 160 r.p.m; Also determine the length of the shaft if the twist
must not exceed 1o over the entire length. The maximum shear stress
is limited to 60 N/mm2. Take the value of modulus of rigidity = 8 x 104
N/mm2

9. Q4. A hollow cylindrical steel shaft is 1.5 m long and has inner and outer diameters respectively equal to 40 and
60 mm (a) What is the largest torque that can be applied to the shaft if the shearing stress is not to exceed 120
MPa?
(b) What is the corresponding minimum value of the shearing stress in the shaft?
A hollow cylindrical steel shaft:
length = 1.5 m
inner diameter Di= 40 mm
outer diameters Do= 60 mm
(a)What is the largest torque= ?
shearing stress is not to exceed 120 MPa

10. (a) What is the largest torque= ?
shearing stress is not to exceed 120 MPa
Simple Torsion Equation =>
𝑻
𝑱
=
𝝉
𝑹
=
𝑮𝜽
𝑳
𝑻
𝑱
=
𝝉
𝑹
=> T= Zp 𝝉
T=
𝜋
16
𝐷0
4−𝐷ⅈ4
𝐷0
𝝉
=
𝜋
16
604
−404
60
∗ 120
= 4084070.45 N-mm
=4.08 kN-m
(b) What is the corresponding
minimum value of the shearing stress
in the shaft?
𝒒
𝒓
=
𝝉
𝑹
q=
𝝉
𝑹
. r
q=
𝟏𝟐𝟎
𝟑𝟎
∗ 𝟐𝟎
 q = 80 MPa

11. Q5. Two shafts of the same material and of same lengths are subjected
to the same torque, if the first shaft is of a solid circular section and the
second shaft is of hollow circular section, whose internal diameter is 2/3
of the outside diameter and the maximum shear stress developed in
each shaft is the same, compare the weights of the shafts.

15. Q6. A solid circular shaft and a hollow circular shaft whose inside
diameter is (
𝟑
𝟒
) of the outside diameter, are of the same material, of
equal lengths and are required to transmit a given torque. Compare the
weights of these two shafts if the maximum shear stress developed in
the two shafts are equal.

18. Q7. A solid cylindrical shaft is to transmit 300 kW power at 100 r.p.m.
(a) If the shear stress is not to exceed 80 N/mm2, find its diameter.
(b) (b) What percent saving in weight would be obtained if this shaft is
replaced by a hollow one whose internal diameter equals to 0.6 of
the external diameter, the length, the material and maximum shear
stress being the same

21. Q8. The final drive to a conveyor that feeds coal to a railroad car is a solid shaft loaded in
pure torsion and carrying 800 Nm of torque
i) if the acceptable shear stress is 39MPa, find its diameter
ii) What percentage saving in weight would be obtained if this shaft is replaced by a
hollow shaft whose outside diameter of 60 mm is available in the same material as
specified for the solid shaft. Compute what maximum inside diameter the hollow shaft
can have that would result in the same stress in the solid shaft. The length , material
are being the same.

22. Torque T= 800 Nm = 8*105 N-mm
Acceptable shear stress 𝜏= 39MPa
i) If Solid shaft finding diameter:
𝑻
𝑱
=
𝝉
𝑹
T= Zp 𝝉
Zp =
𝜋 d3
16
d3 = (
16𝑇
𝜋 𝜏
)
d3 = (
16 ∗8 ∗105
𝜋 ∗39
)
=> d= 47.09 mm