T/ Mohammed. A. Mustafa

Consider a bar rigidly clamped at one end and twisted at the other

end by a torque (twisting moment) T = Fd applied in a plane

perpendicular to the axis of the bar, as shown in Fig. 1. Such a bar is

in torsion. An alternative representation of the torque is the curved

arrow shown in the figure.

Fig. 1 Torque applied to a circular shaft

Introduction

Polar Moment of Inertia

J =𝜋

32(𝐷4 − 𝑑4)

J =𝜋

32𝑑4

For a hollow circular shaft of outer

diameter 𝑫 with a concentric circular

hole of diameter 𝒅 the polar moment of

inertia of the cross-sectional area is

given by

For a solid circular shaft of diameter 𝒅the polar moment of inertia of the

cross-sectional area is given by

A mathematical property of the geometry of the cross section which occurs in the study of

the stresses set up in a circular shaft subject to torsion is the polar moment of inertia J,

defined in a statics course

Polar Moment of Inertia

Torsional Shearing Stress

So, for either a solid or a hollow circular shaft subject to a twisting moment

𝑻 the torsional shearing stress τ at a distance r from the center of the

shaft is written as

The maximum shear stress is found by replacing ρ by the radius r of the shaft:

Distribution of shear stress

along the radius of a circular

shaft.

Shearing Strain

The ratio of the shear stress 𝜏 to the shear strain 𝛾 is called the shear modulus

Again the units of G are the same as those of shear stress, since the shear strain is dimensionless

𝐺 =𝜏

𝛾𝜏 = shear stress

𝛾 = shear strain

G = shear modulus

𝜃 =𝑇𝐿

𝐺𝐽𝜃 = angle of twist

𝐿 = length of the shaft

𝛾 = tan ∝ ≈ ∝ 𝛾 =𝑟𝜃

𝐿

SOLVED PROBLEMS

If a twisting moment of 1100 N.m is impressed upon a 4.4 cm diameter shaft, what is

the maximum shearing stress developed? Also, what is the angle of twist in a 150

cm length of the shaft? The material is steel for which G = 85 GPa.

SOLUTION

Step one: The polar moment of inertia of the cross-sectional area is

J =𝜋

32(𝑑4) =

𝜋

32× 0.0444 = 𝟑. 𝟔𝟖 × 𝟏𝟎−𝟕𝑚4

Step two: The maximum shear stress is developed at the outer fibers where r = 0.022 m:

𝜏𝑚𝑎𝑥 =𝑇𝑟

𝐽=

1100 × 0.022

3.68 × 10−7= 𝟔𝟓. 𝟖 × 𝟏𝟎𝟔 𝑷𝒂 𝒐𝒓 𝟔𝟓. 𝟖𝑴𝑷𝒂

Step three: The angle of twist q in a 1.5 m length of the shaft is

𝜃 =𝑇𝐿

𝐺𝐽=

1100 × 1.5

85 × 109(3.68 × 10−7)= 𝟎. 𝟎𝟓𝟐𝟕 𝐫𝐚𝐝 𝐨𝐫

Example (1):

A hollow 3 m long steel shaft must transmit a torque of 25 kN.m. The total angle of

twist in this length is not to exceed 2.5° and the allowable shearing stress is 90

MPa. Determine the inside and outside diameters of the shaft if G = 85 GPa.

Example (2):

SOLUTION

Step one: The angle of twist is θ = TL/GJ. Thus, in the 3-m length we have

2.5°𝜋 𝑟𝑎𝑑

180°=

(25000)(3)

85 × 109(𝜋/32)(𝐷4 − 𝑑4)∴ 𝐷4 − 𝑑4 = 206 × 10−6 𝑚4 → (𝟏)

Step two: The maximum shearing stress occurs at the outer fibers where r = D/2

90 × 106 =(25000)(𝐷/2)

(𝜋/32)(𝐷4 − 𝑑4)∴ 𝐷4 − 𝑑4 = (1415𝐷 × 10−6)𝑚4 → (𝟐)

Comparison of the right-hand sides of equations (1) and (2) indicates that

206 × 10−6 = (1415𝐷 × 10−6)

and thus D = 0.146 m or 146 mm. Substitution of this value into either of the equations then gives d = 0.126 m or 126 mm.

Consider two solid circular shafts connected by 5-cm- and 25-cm-pitch-diameter

gears as in Fig. (a). Find the angular rotation of D, the right end of one shaft, with

respect to A, the left end of the other, caused by the torque of 280 N.m applied at

D. The left shaft is steel for which G = 80 Gpa and the right is brass for which G =

33 GPa.

Example (4):

SOLUTION

Step one: A free-body diagram of the right shaft CD [Fig. (b)] reveals that a tangential force F

must act on the smaller gear. For equilibrium,

Fr = T 0.025 F = 280 ∴ F = 11200 NStep two: The angle of twist of the right shaft is

𝜃1 =𝑇𝐿

𝐺𝐽=

280 × 1.00

(33 × 109)𝜋 × 0.034/32= 𝟎. 𝟏𝟎𝟔𝟕𝐫𝐚𝐝

𝜃2 =𝑇𝐿

𝐺𝐽=

1400 × 1.20

(80 × 109)𝜋 × 0.064/32= 𝟎. 𝟎𝟏𝟔𝟓𝐫𝐚𝐝

𝜃 = 5𝜃2 + 𝜃1 = 5 × 0.0165 + 0.1067 = 𝟎. 𝟏𝟖𝟗 𝒓𝒂𝒅 𝒂𝒓 𝟏𝟎. 𝟖°

Step three: A free-body diagram of the left shaft AB is shown in Fig. (c). The force F is equal

and opposite to that acting on the small gear C. This force F acts 12.5 cm from the center line

of the left shaft; hence it imparts a torque of 0.125(11200) = 1400 N.m to the shaft AB.

Because of this torque there is a rotation of end B with respect to end A given by the angle

𝜃2, where

Step four: This angle of rotation 𝜃2 induces a rigid-body rotation of the entire shaft CD

because of the gears. In fact, the rotation of CD will be in the same ratio to that of AB as the

ratio of the pitch diameters, or 5:1. Thus a rigid-body rotation of 5(0.0165) rad is imparted to

shaft CD. Superposed on this rigid body movement of CD is the angular displacement of D

with respect to C, previously denoted by 𝜃1. Hence the resultant angle of twist of D with

respect to A is

Thank You