This document discusses torsion and formulas for calculating stresses and deformations in circular bars subjected to torsion. It includes: - Definitions of torsion as the twisting of a structural member by couples that produce rotation about its longitudinal axis. - Formulas for shear strain, stress, angle of twist, and torsional rigidity in linearly elastic circular bars under pure torsion. - The maximum shear stress occurs at the outer surface and varies linearly with the radial distance from the center. - Limitations that the bar must have a circular cross-section and the material must be linearly elastic. An example problem is given to calculate the maximum shear stress and angle of twist for a given

1. Lecturer;
Dr. Dawood S. Atrushi
January 2015
Torsion

2. Chapter 3 Torsion
of a structural member,
y couples that produce
gitudinal axis
T2 = P2 d2
Introduction
Torsion: twisting of a structural
member, when it is loaded by couples
that produce rotation about its
longitudinal axis.
January, 2015Torsion - DAT2
T1 = P1 d1
T2 = P2 d2

3. The couples T1, T2
are called;
¢ Torques,
¢ Twisting couples
or
¢ Twisting
moments.
Unit of T is:
N-m, lb-ft
January, 2015Torsion - DAT3
3.1 Introduction
Torsion : twisting of a structural member,
when it is loaded by couples that produce
otation about its longitudinal axis
T1 = P1 d1 T2 = P2 d2
the couples T1, T2 are called
orques, twisting couples or twisting
moments
unit of T : N-m, lb-ft
in this chapter, we will develop formulas
or the stresses and deformations produced
n circular bars subjected to torsion, such as
drive shafts, thin-walled members

4. Torsional Deformation of a
Circular Bar
Consider a bar or shaft of circular cross
section twisted by a couple T, assume
the left-hand end is fixed and the right-
hand end will rotate a small angle Φ,
called angle of twist.
January, 2015Torsion - DAT4

5. If every cross section has the same
radius and subjected to the same
torque, the angle Φ(x) will vary linearly
between ends.
January, 2015Torsion - DAT5
It is assumed
1. plane section remains plane
2. radii remaining straight and the cross
sections remaining plane and circular
3. if Φ is small, neither the length L nor
its radius will change

6. Consider an element of the bar dx, on
its outer surface we choose an small
element abcd,
January, 2015Torsion - DAT6
3. if is small, neither the length L nor its radius will cha
consider an element of the bar dx, on its outer surface we c
ll element abcd,

7. January, 2015Torsion - DAT7
During twisting the element rotate a
small angle dΦ, the element is in a
state of pure shear, and deformed into
ab'c’d.
during twisting the element rotate

8. Its shear strain ϒmax is;
January, 2015Torsion - DAT8
dΦ/dx represents the rate of change of
the angle of twist Φ, denote
θ= dΦ/dx as the angle of twist per unit
length or the rate of twist, then
!
ϒmax = r θ
2
during twisting the element rotate a small angle d
a state of pure shear, and deformed into ab'c'd, its she
b b' r d
max = CC = CC
a b dx

9. January, 2015Torsion - DAT9
In general, Φ and θ are function of x, in
the special case of pure torsion, θ is
constant along the length (every cross
section is subjected to the same torque)
max = r
in general, and are function of x, in the
pure torsion, is constant along the length (every
subjected to the same torque)
r
= C then max = CC
L L
and the shear strain inside the bar can be obtained
= = C max
r
for a circular tube, it can be obtained
max = r
in general, and are function of x, in the
pure torsion, is constant along the length (every
subjected to the same torque)
r
= C then max = CC
L L
and the shear strain inside the bar can be obtained
= = C max
r
for a circular tube, it can be obtained
And the shear strain inside the bar can
be obtained
in general, and are functio
pure torsion, is constant along th
subjected to the same torque)
= C then ma
L
and the shear strain inside the bar can
= = C max
r
for a circular tube, it can be obtained

10. January, 2015Torsion - DAT10
For a circular tube, it can be obtained:
L
side the bar can be obtained
C max
r
an be obtained
s are based only upon geometric concepts, they are
and the shear strain inside the bar can be obta
= = C max
r
for a circular tube, it can be obtained
r1
min = C max
r2
the above relationships are based only upon g

11. Circular Bars of Linearly
Elastic Materials
Shear stress τ in the bar of a linear
elastic material is
January, 2015Torsion - DAT11
nships are based only upon geometric concepts, they are
ar of any material, elastic or inelastic, linear or nonlinear
f Linearly Elastic Materials
in the bar of a
l is
s of elasticity
valid for a circular bar of any material, elastic o
3.3 Circular Bars of Linearly Elastic Materia
shear stress in the bar of a
linear elastic material is
= G
G : shear modulus of elasticity
G : shear modulus of elasticity

12. January, 2015Torsion - DAT12
With the geometric relation of the shear
strain, it is obtainedwith the geometric relation of the shear strain, it is obtain
max = G r
= G = C max
r
and in circular bar vary linear with the radial dist
the center, the maximum values max and max occur at the out
the shear stress acting on the plane of the
cross section are accompanied by shear
stresses of the same magnitude acting on
on of the shear strain, it is obtained
C max
r
with the geometric relation of the shear strain, it is o
max = G r
= G = C max
r
and in circular bar vary linear with the radial
the center, the maximum values max and max occur at the
the shear stress acting on the plane of the
with the geometric relation of the shear strain, it is
max = G r
= G = C max
r
and in circular bar vary linear with the radia
the center, the maximum values max and max occur at th
the shear stress acting on the plane of the

13. January, 2015Torsion - DAT13
¢ τ and ϒ in circular bar vary linear
with the radial distance ρ from the
center, the maximum values τmax and
ϒmax occur at the outer surface.
¢ The shear stress acting
on the plane of the
cross section are
accompanied by shear
stresses of the same
magnitude acting on
longitudinal plane of the
bar
= G = C max
r
and in circular bar vary linear with the radial distance fr
ter, the maximum values max and max occur at the outer surface
shear stress acting on the plane of the
section are accompanied by shear
s of the same magnitude acting on
dinal plane of the bar
he material is weaker in shear on
dinal plane than on cross-sectional
nes, as in the case of a circular bar made of wood, the first crack

14. January, 2015Torsion - DAT14
¢ If the material is weaker in shear on
longitudinal plane than on cross-
sectional planes, as in the case of a
circular bar made of wood, the first crack
due to twisting will appear on the surface
in longitudinal direction a rectangular
element with sides at 45 o to the axis of
the shaft will be subjected to tensile and
compressive stresses.
ccompanied by shear
magnitude acting on
the bar
s weaker in shear on
an on cross-sectional
ase of a circular bar made of wood, the first crack due
r on the surface in longitudinal direction
ment with sides at 45 o
to
t will be subjected to
ve stresses

15. January, 2015Torsion - DAT
Consider a bar subjected to pure
torsion, the shear force acting on an
element dA is τdA, the moment of this
force about the axis of barisτρ dA:
The Torsion Formula
planes, as in the case of a circular bar made of wood, the first crack due
wisting will appear on the surface in longitudinal direction
a rectangular element with sides at 45 o
to
axis of the shaft will be subjected to
sile and compressive stresses
e Torsion Formula
consider a bar subjected to pure torsion,
shear force acting on an element dA
dA, the moment of this force about
axis of bar is dA
dM = dA
dM = τρ dA

16. January, 2015Torsion - DAT16
Equation of moment equilibriumequation of moment equilibrium
T = dM = dA = G 2
dA = G
A A A
= G Ip [ = G ]
in which Ip = 2
dA is the polar moment of iner
A
r4
d4
Ip = CC = CC for circular cross section
2 32
the above relation can be written
T
= CC
A = G 2
dA = G 2
dA
A A
G ]
s the polar moment of inertia
for circular cross section
equation of moment equilibrium
T = dM = dA = G 2
dA = G
A A A A
= G Ip [ = G ]
in which Ip = 2
dA is the polar moment of inertia
A
r4
d4
Ip = CC = CC for circular cross section
2 32
the above relation can be written
T
= CC
equation of moment equilibrium
T = dM = dA = G 2
dA = G
A A A A
= G Ip [ = G ]
in which Ip = 2
dA is the polar moment of inertia
A
r4
d4
Ip = CC = CC for circular cross section
2 32
the above relation can be written
T
equation of moment equilibrium
T = dM = dA = G 2
dA = G 2
A A A A
= G Ip [ = G ]
in which Ip = 2
dA is the polar moment of inertia
A
r4
d4
Ip = CC = CC for circular cross section
2 32
the above relation can be written
T

17. January, 2015Torsion - DAT17
The above relation can be written
GIp: torsional rigidity
the above relation can be written
T
= CC
G Ip
G Ip : torsional rigidity
the angle of twist can be expressed as
T L
= L = CC is measured in r
G Ip
L
torsional flexibility f = CC
The angle of twist Φ can be expressed as
T
= CC
G Ip
G Ip : torsional rigidity
the angle of twist can be expressed as
T L
= L = CC is measured in radians
G Ip
L
torsional flexibility f = CC
G Ip
G Ip
torsional stiffness k = CC
L
and the shear stress is
Torsional flexibility;
G Ip
G Ip : torsional rigidity
the angle of twist can be expressed as
T L
= L = CC is measured
G Ip
L
torsional flexibility f = CC
G Ip
G Ip
torsional stiffness k = CC
L
and the shear stress is
Torsional stiffness;
G Ip
G Ip : torsional rigidity
the angle of twist can be expressed as
T L
= L = CC is measur
G Ip
L
torsional flexibility f = CC
G Ip
G Ip
torsional stiffness k = CC
L
and the shear stress is

18. January, 2015Torsion - DAT18
The shear stress is;
The maximum shear stress τmax at ρ = r is;
5
torsional stiffness k = CC
L
and the shear stress is
T T
= G = G CC = CC
G Ip Ip
the maximum shear stress max at = r is
T r 16 T
max = CC = CC
Ip d3
for a circular tube

19. T r 16 T
= CC = CC
Ip d3
ular tube
= (r2
4
- r1
4
) / 2 = (d2
4
- d1
4
) / 32
ow tube is very thin
(r2
2
+ r1
2
) (r2 + r1) (r2 - r1) / 2
(2r2
) (2r) (t) = 2 r3
t = d3
t / 4 January, 2015Torsion - DAT19
For a circular tube
If the hollow tube is very thin
for a circular tube
Ip = (r2
4
- r1
4
) / 2 = (d2
4
- d1
4
) / 32
if the hollow tube is very thin
Ip (r2
2
+ r1
2
) (r2 + r1) (r2 - r1) / 2
= (2r2
) (2r) (t) = 2 r3
t = d3
t
limitations
1. bar have circular cross section (either solid or holl
p
for a circular tube
Ip = (r2
4
- r1
4
) / 2 = (d2
4
- d1
4
) / 32
if the hollow tube is very thin
Ip (r2
2
+ r1
2
) (r2 + r1) (r2 - r1) / 2
= (2r2
) (2r) (t) = 2 r3
t = d3
t / 4
limitations
1. bar have circular cross section (either solid or hollow)
2. material is linear elastic
for a circular tube
Ip = (r2
4
- r1
4
) / 2 = (d2
4
- d1
4
) / 32
if the hollow tube is very thin
Ip (r2
2
+ r1
2
) (r2 + r1) (r2 - r1) / 2
= (2r2
) (2r) (t) = 2 r3
t = d3
t / 4
limitations
1. bar have circular cross section (either solid or hollow
2. material is linear elastic
= CC = CC
Ip d3
r tube
(r2
4
- r1
4
) / 2 = (d2
4
- d1
4
) / 32
w tube is very thin
(r2
2
+ r1
2
) (r2 + r1) (r2 - r1) / 2
(2r2
) (2r) (t) = 2 r3
t = d3
t / 4
circular cross section (either solid or hollow)
16 T
CC
d3
= (d2
4
- d1
4
) / 32
thin
(r2 + r1) (r2 - r1) / 2
) = 2 r3
t = d3
t / 4

20. January, 2015Torsion - DAT20
Limitations
1. bar have circular cross section (either
solid or hollow)
2. material is linear elastic
5
and the shear stress is
T T
= G = G CC = CC
G Ip Ip
the maximum shear stress max at = r is
The shear stress; = G =
the maximum shear stres
Note that the above equations cannot be used for
bars of noncircular shapes, because their cross
sections do not remain plane and their maximum
stresses are not located at the farthest distances
from the midpoint.

21. ions do not remain plane and their maximum
rthest distances from the midpoint
section
G = 80 GPa
= ?
5o
, T = ?
16 x 340 N-M
Example 1
A solid bar of circular cross section d =
40 mm, L = 1.3 m, and G = 80 GPa.
a) for T = 340 N-m, find τmax and Φ.
b) for τallowable = 42 MPa and Φallowable =
2.5o, find T.
January, 2015Torsion - DAT21
1.3 m
40 mm

22. Solution
January, 2015Torsion - DAT22
6
a solid bar of circular cross section
d = 40 mm, L = 1.3 m, G = 80 GPa
(a) T = 340 N-m, max, = ?
(b) all = 42 MPa, all = 2.5o
, T = ?
(a) 16 T 16 x 340 N-M
max = CC = CCCCCCC = 27.1
d3
(0.04 m)3
Ip = d4
/ 32 = 2.51 x 10-7
m4
T L 340 N-m x 1.3 m
= CC = CCCCCCCCCC = 0.
G Ip 80 GPa x 2.51 x 10-7
m4
a) The maximum shear stress τmax ;
6
r cross section
3 m, G = 80 GPa
max, = ?
all = 2.5o
, T = ?
16 x 340 N-M
= CCCCCCC = 27.1 MPa
(0.04 m)3
2 = 2.51 x 10-7
m4
340 N-m x 1.3 m
CCCCCCCCCC = 0.02198 rad = 1.26o
80 GPa x 2.51 x 10-7
m4
The angle of twist Φ;
6
Example 3-1
a solid bar of circular cross section
d = 40 mm, L = 1.3 m, G = 80 GPa
(a) T = 340 N-m, max, = ?
(b) all = 42 MPa, all = 2.5o
, T = ?
(a) 16 T 16 x 340 N-M
max = CC = CCCCCCC = 27.1
d3
(0.04 m)3
Ip = d4
/ 32 = 2.51 x 10-7
m4
T L 340 N-m x 1.3 m
= CC = CCCCCCCCCC = 0
G Ip 80 GPa x 2.51 x 10-7
m4
arthest distances from the midpoint
section
G = 80 GPa
= ?
5o
, T = ?
16 x 340 N-M
CCCCCCC = 27.1 MPa
(0.04 m)3
2.51 x 10-7
m4
340 N-m x 1.3 m
CCCCCCCC = 0.02198 rad = 1.26o
GPa x 2.51 x 10-7
m4

23. January, 2015Torsion - DAT23
b) Due to τallowable = 42 MPa
(b) due to all = 42 MPa
T1 = d3
all / 16 = (0.04 m)3
x 42 MPa / 16 = 528
due to all = 2.5o =
2.5 x rad / 180o
= 0.0
T2 = G Ip all / L = 80 GPa x 2.51 x 10-7
m4
x
= 674 N-m
thus Tall = min [T1, T2] = 528 N-m
Example 3-2
a steel shaft of either solid bar or circular tube
b) Due to Φallowable = 2.5o = 2.5 x PI/180 = 0.04363
rad;
(b) due to all = 42 MPa
T1 = d3
all / 16 = (0.04 m)3
x 42 MPa / 16 = 528 N-m
due to all = 2.5o =
2.5 x rad / 180o
= 0.0436
T2 = G Ip all / L = 80 GPa x 2.51 x 10-7
m4
x 0.0
= 674 N-m
thus Tall = min [T1, T2] = 528 N-m
Example 3-2
a steel shaft of either solid bar or circular tube
e to all = 42 MPa
T1 = d3
all / 16 = (0.04 m)3
x 42 MPa / 16 = 528 N-m
ue to all = 2.5o =
2.5 x rad / 180o
= 0.04363 rad
T2 = G Ip all / L = 80 GPa x 2.51 x 10-7
m4
x 0.04363 / 1.3 m
= 674 N-m
hus Tall = min [T1, T2] = 528 N-m
3-2
(b) due to all = 42 MPa
T1 = d3
all / 16 = (0.04 m)3
x 42 MPa / 16 = 528 N-m
due to all = 2.5o =
2.5 x rad / 180o
= 0.0436
T2 = G Ip all / L = 80 GPa x 2.51 x 10-7
m4
x 0.0
= 674 N-m
thus Tall = min [T1, T2] = 528 N-m
Example 3-2
m)3
x 42 MPa / 16 = 528 N-m
x rad / 180o
= 0.04363 rad
GPa x 2.51 x 10-7
m4
x 0.04363 / 1.3 m
= 528 N-mThus;

24. Example 2
A steel shaft of either solid bar or circular
tube is exposed to T = 1200 N-m. τall = 40
MPa, θ=0.75o/m, G= 78 Gpa.
January, 2015Torsion - DAT24
p all / L = 80 GPa x 2.51 x 10-7
m4
x 0.04363 / 1.3 m
N-m
= min [T1, T2] = 528 N-m
ther solid bar or circular tube
m, all = 40 MPa
m G = 78 GPa
0 of the solid bar
a) Determine do of
the solid bar
b) For the hollow
shaft t=d2/10,
determine d2
c) Determine d2/do,
Whollow/Wsolid

25. Solution
January, 2015Torsion - DAT25
a) For the solid shaft, due to τmax = 40 MPa;
(a) determine d0 of the solid bar
(b) for the hollow shaft, t = d2 / 10, determine d2
(c) determine d2 / d0, Whollow / Wsolid
(a) for the solid shaft, due to all = 40 MPa
d0
3
= 16 T / all = 16 x 1200 / 40 = 152.8
d0 = 0.0535 m = 53.5 mm
due to all = 0.75o
/ m = 0.75 x rad / 180o
/ m = 0
Ip = T / G all = 1200 / 78 x 109
x 0.01309 = 117
d0
4
= 32 Ip / = 32 x 117.5 x 10-8
/ = 1197
d0 = 0.0588 m = 58.8 mm
thus, we choose d0 = 58.8 mm [in practical design, d
olid bar
d2 / 10, determine d2
hollow / Wsolid
o all = 40 MPa
l = 16 x 1200 / 40 = 152.8 x 10-6
m3
= 53.5 mm
m = 0.75 x rad / 180o
/ m = 0.01309 rad / m
1200 / 78 x 109
x 0.01309 = 117.5 x 10-8
m4
= 32 x 117.5 x 10-8
/ = 1197 x 10-8
m4
= 58.8 mm
58.8 mm [in practical design, d0 = 60 mm]
all
(a) determine d0 of the solid bar
(b) for the hollow shaft, t = d2 / 10, determine d2
(c) determine d2 / d0, Whollow / Wsolid
(a) for the solid shaft, due to all = 40 MPa
d0
3
= 16 T / all = 16 x 1200 / 40 = 152.
d0 = 0.0535 m = 53.5 mm
due to all = 0.75o
/ m = 0.75 x rad / 180o
/ m = 0
Ip = T / G all = 1200 / 78 x 109
x 0.01309 = 11
d0
4
= 32 Ip / = 32 x 117.5 x 10-8
/ = 119
d0 = 0.0588 m = 58.8 mm
thus, we choose d0 = 58.8 mm [in practical design,
all = 0.75o
/ m G = 78 GPa
(a) determine d0 of the solid bar
(b) for the hollow shaft, t = d2 / 10, determine d2
(c) determine d2 / d0, Whollow / Wsolid
(a) for the solid shaft, due to all = 40 MPa
d0
3
= 16 T / all = 16 x 1200 / 40 = 152.8
d0 = 0.0535 m = 53.5 mm
due to all = 0.75o
/ m = 0.75 x rad / 180o
/ m = 0
Ip = T / G all = 1200 / 78 x 109
x 0.01309 = 11
d0
4
= 32 Ip / = 32 x 117.5 x 10-8
/ = 119
d0 = 0.0588 m = 58.8 mm
thus, we choose d0 = 58.8 mm [in practical design,
40 MPa
= 78 GPa
olid bar
d2 / 10, determine d2
hollow / Wsolid
o all = 40 MPa
l = 16 x 1200 / 40 = 152.8 x 10-6
m3
= 53.5 mm
m = 0.75 x rad / 180o
/ m = 0.01309 rad / m
1200 / 78 x 109
x 0.01309 = 117.5 x 10-8
m4
= 32 x 117.5 x 10-8
/ = 1197 x 10-8
m4
= 58.8 mm
T = 1200 N-m, all = 40 MPa
all = 0.75o
/ m G = 78 GPa
(a) determine d0 of the solid bar
(b) for the hollow shaft, t = d2 / 10, determine d2
(c) determine d2 / d0, Whollow / Wsolid
(a) for the solid shaft, due to all = 40 MPa
d0
3
= 16 T / all = 16 x 1200 / 40 = 15
d0 = 0.0535 m = 53.5 mm
due to all = 0.75o
/ m = 0.75 x rad / 180o
/ m =
Ip = T / G all = 1200 / 78 x 109
x 0.01309 = 1
d0
4
= 32 Ip / = 32 x 117.5 x 10-8
/ = 11
d0 = 0.0588 m = 58.8 mm
= 40 MPa
= 78 GPa
e solid bar
= d2 / 10, determine d2
Whollow / Wsolid
to all = 40 MPa
all = 16 x 1200 / 40 = 152.8 x 10-6
m3
m = 53.5 mm
/ m = 0.75 x rad / 180o
/ m = 0.01309 rad / m
= 1200 / 78 x 109
x 0.01309 = 117.5 x 10-8
m4
= 32 x 117.5 x 10-8
/ = 1197 x 10-8
m4

26. January, 2015Torsion - DAT26
7
due to all = 0.75o
/ m = 0.75 x rad / 180o
/ m = 0
Ip = T / G all = 1200 / 78 x 109
x 0.01309 = 117
d0
4
= 32 Ip / = 32 x 117.5 x 10-8
/ = 1197
d0 = 0.0588 m = 58.8 mm
thus, we choose d0 = 58.8 mm [in practical design, d
(b) for the hollow shaft
d1 = d2 - 2 t = d2 - 0.2 d2 = 0.8 d
(a) for the solid shaft, due to all = 40 MPa
d0
3
= 16 T / all = 16 x 1200 / 40 = 152.8 x
d0 = 0.0535 m = 53.5 mm
due to all = 0.75o
/ m = 0.75 x rad / 180o
/ m = 0.0
Ip = T / G all = 1200 / 78 x 109
x 0.01309 = 117.5
d0
4
= 32 Ip / = 32 x 117.5 x 10-8
/ = 1197 x
d0 = 0.0588 m = 58.8 mm
thus, we choose d0 = 58.8 mm [in practical design, d0
(b) for the hollow shaft
d1 = d2 - 2 t = d2 - 0.2 d2 = 0.8 d2
7
d0 = 0.0535 m = 53.5 mm
due to all = 0.75o
/ m = 0.75 x rad / 180o
/ m = 0
Ip = T / G all = 1200 / 78 x 109
x 0.01309 = 117
d0
4
= 32 Ip / = 32 x 117.5 x 10-8
/ = 1197
d0 = 0.0588 m = 58.8 mm
thus, we choose d0 = 58.8 mm [in practical design,
(b) for the hollow shaft
d1 = d2 - 2 t = d2 - 0.2 d2 = 0.8
7
d0 = 0.0535 m = 53.5 mm
due to all = 0.75o
/ m = 0.75 x rad / 180o
/ m =
Ip = T / G all = 1200 / 78 x 109
x 0.01309 = 11
d0
4
= 32 Ip / = 32 x 117.5 x 10-8
/ = 119
d0 = 0.0588 m = 58.8 mm
thus, we choose d0 = 58.8 mm [in practical design
(b) for the hollow shaft
d1 = d2 - 2 t = d2 - 0.2 d2 = 0.8
7
/ all = 16 x 1200 / 40 = 152.8 x 10 m
535 m = 53.5 mm
.75o
/ m = 0.75 x rad / 180o
/ m = 0.01309 rad / m
all = 1200 / 78 x 109
x 0.01309 = 117.5 x 10-8
m4
p / = 32 x 117.5 x 10-8
/ = 1197 x 10-8
m4
588 m = 58.8 mm
d0 = 58.8 mm [in practical design, d0 = 60 mm]
aft
- 2 t = d2 - 0.2 d2 = 0.8 d2
b) For the hollow shaft
7
d0
3
= 16 T / all = 16 x 1200 / 40 = 152.8 x 10
d0 = 0.0535 m = 53.5 mm
due to all = 0.75o
/ m = 0.75 x rad / 180o
/ m = 0.0130
Ip = T / G all = 1200 / 78 x 109
x 0.01309 = 117.5 x
d0
4
= 32 Ip / = 32 x 117.5 x 10-8
/ = 1197 x 10
d0 = 0.0588 m = 58.8 mm
thus, we choose d0 = 58.8 mm [in practical design, d0 = 6
(b) for the hollow shaft
d1 = d2 - 2 t = d2 - 0.2 d2 = 0.8 d2
Ip = (d2
4
- d1
4
) / 32 = [d2
4
- (0.8d2)4
] / 32 = 0
due to all = 40 MPa
Ip = 0.05796 d2
4
= T r / all = 1200 (d2/2) / 4
4
- (0.8d2)4
] / 32 = 0.05796 d2
4

27. January, 2015Torsion - DAT27
Ip = (d2
4
- d1
4
) / 32 = [d2
4
- (0.8d2)4
] / 32 = 0.05796 d2
4
due to all = 40 MPa
Ip = 0.05796 d2
4
= T r / all = 1200 (d2/2) / 40
d2
3
= 258.8 x 10-6
m3
d2 = 0.0637 m = 63.7 mm
due to all = 0.75o
/ m = 0.01309 rad / m
all = 0.01309 = T / G Ip = 1200 / 78 x 109
x 0.05796 d2
4
d2
4
= 2028 x 10-8
m4
d2 = 0.0671 m = 67.1 mm
thus, we choose d0 = 67.1 mm [in practical design, d0 = 70 mm]
(c) the ratios of hollow and solid bar are

28. January, 2015Torsion - DAT28
c) The ratios of hollow and solid bar are
thus, we choose d0 = 67.1 mm [in practical design,
(c) the ratios of hollow and solid bar are
d2 / d0 = 67.1 / 58.8 = 1.14
Whollow Ahollow (d2
2
- d1
2
)/4
CCC = CCC = CCCCCC = 0
Wsolid Asolid d0
2
/4
the hollow shaft has 14% greater in diameter but 53% le
Example 3-3
a hollow shaft and a solid shaft has same
material, same length, same outer radius R,
and r = 0.6 R for the hollow shaft
d2 = 0.0671 m = 67.1 mm
thus, we choose d0 = 67.1 mm [in practical design, d
(c) the ratios of hollow and solid bar are
d2 / d0 = 67.1 / 58.8 = 1.14
Whollow Ahollow (d2
2
- d1
2
)/4
CCC = CCC = CCCCCC = 0.4
Wsolid Asolid d0
2
/4
the hollow shaft has 14% greater in diameter but 53% les
Example 3-3
a hollow shaft and a solid shaft has same
material, same length, same outer radius R,
10-8
m4
= 67.1 mm
67.1 mm [in practical design, d0 = 70 mm]
solid bar are
/ 58.8 = 1.14
ollow (d2
2
- d1
2
)/4
CC = CCCCCC = 0.47
olid d0
2
/4
% greater in diameter but 53% less in weightThe hollow shaft has a 14% greater diameter
but 53% less weight.

29. max R / L
x R / L
T1
Non-uniform Torsion
1) Constant torque through each
segment
January, 2015Torsion - DAT29
SH is 36% greater than SS
3.4 Nonuniform Torsion
(1) constant torque through each segment
TCD = - T1 - T2 + T3
TBC = - T1 - T2 TAB = - T1
n n Ti Li
= i = CC
i=1 i=1
Gi Ipi
(2) constant torque with continuously
varying cross section
SH is 36% greater than S
3.4 Nonuniform Torsion
(1) constant torque through each segm
TCD = - T1 - T2 + T3
TBC = - T1 - T2 TAB =
n n Ti Li
= i = CC
i=1 i=1
Gi Ipi
(2) constant torque with continuousl
varying cross section
SS = TS / WS = 0.5 max R / L
SH is 36% greater than SS
3.4 Nonuniform Torsion
(1) constant torque through each segment
TCD = - T1 - T2 + T3
TBC = - T1 - T2 TAB = - T1
n n Ti Li
= i = CC
i=1 i=1
Gi Ipi
(2) constant torque with continuously
varying cross section

30. 2) Constant torque with continuously
varying cross section
January, 2015Torsion - DAT30
9
2 AB 1
n Ti Li
CC
i=1
Gi Ipi
ith continuously
ion
T dx
d = CCC
G Ip(x)
L L T dx
= d = CCC
0 0
G Ip(x)
(3) continuously varying cross section and
T dx
d = CCC
G Ip(x)
L L T dx
= d = CCC
0 0
G Ip(x)
(3) continuously varying cross section and

31. 3) Continuously varying cross section
and continuously varying torque
January, 2015Torsion - DAT31
)
L T dx
= CCC
0
G Ip(x)
ying cross section and
ing torque
L T(x) dx
= CCC
0
G Ip(x)
CDE, d = 30 mm
T dx
d = CCC
G Ip(x)
L L T dx
= d = CCC
0 0
G Ip(x)
(3) continuously varying cross section and
continuously varying torque
L L T(x) dx
= d = CCC
0 0
G Ip(x)
T dx
d = CCC
G Ip(x)
L L T dx
= d = CCC
0 0
G Ip(x)
(3) continuously varying cross section and
continuously varying torque
L L T(x) dx
= d = CCC
0 0
G Ip(x)

32. Example 3
In a solid shaft ABCDE;
d = 30 mm;
T1 = 275 N-m
T2 = 175 N-m
T3 = 80 N-m
G = 80 GPa
L1 = 500 mm
Determine τmax in each pat and ΦBD.
January, 2015Torsion - DAT32
L T(x) dx
d = CCC
0
G Ip(x)
ft ABCDE, d = 30 mm
m T2 = 450 N-m
m G = 80 GPa
m L2 = 400 mm
x in each part and BD
T2 - T1 = 175 N-m

33. Solution
January, 2015Torsion - DAT33
T3 = 175 N-m G = 80 GPa
L1 = 500 mm L2 = 400 mm
determine max in each part and BD
TCD = T2 - T1 = 175 N-m
TBC = - T1 = - 275 N-m
16 TBC 16 x 275 x 103
BC = CCC = CCCCCC = 51.9 MPa
d3
303
16 TCD 16 x 175 x 103
CD = CCC = CCCCCC = 33 MPa
d3
303
BD = BC + CD
4 4
T3 = 175 N-m G = 80 GPa
L1 = 500 mm L2 = 400 mm
determine max in each part and BD
TCD = T2 - T1 = 175 N-m
TBC = - T1 = - 275 N-m
16 TBC 16 x 275 x 103
BC = CCC = CCCCCC = 51.9 MPa
d3
303
16 TCD 16 x 175 x 103
CD = CCC = CCCCCC = 33 MPa
d3
303
BD = BC + CD
d4
304
T3 = 175 N-m G = 80 GPa
L1 = 500 mm L2 = 400 mm
determine max in each part and BD
TCD = T2 - T1 = 175 N-m
TBC = - T1 = - 275 N-m
16 TBC 16 x 275 x 103
BC = CCC = CCCCCC = 51.9 MPa
d3
303
16 TCD 16 x 175 x 103
CD = CCC = CCCCCC = 33 MPa
d3
303
BD = BC + CD
4 4
T3 = 175 N-m G = 80 GPa
L1 = 500 mm L2 = 400 mm
determine max in each part and BD
TCD = T2 - T1 = 175 N-m
TBC = - T1 = - 275 N-m
16 TBC 16 x 275 x 103
BC = CCC = CCCCCC = 51.9 MPa
d3
303
16 TCD 16 x 175 x 103
CD = CCC = CCCCCC = 33 MPa
d3
303
BD = BC + CD
d4
304

34. January, 2015Torsion - DAT34
10
CD = CCC = CCCCCC = 33 MPa
d3
303
BD = BC + CD
d4
304
Ip = CC = CCC = 79,520 mm2
32 32
10
CD
CD = CCC = CCCCCC = 33 MPa
d3
303
BD = BC + CD
d4
304
Ip = CC = CCC = 79,520 mm2
32 32
TBC L1 - 275 x 103
x 500
BC = CCC = CCCCCCCC = - 0.0216 rad
G Ip 80 x 103
x 79,520
TCD L2 175 x 103
x 400
CD = CCC = CCCCCCCC = 0.011 rad
G Ip 80 x 103
x 79,520
BD = BC + CD = - 0.0216 + 0.011 = - 0.0106 rad = - 0.61o
ample 3-5
TBC L1 - 275 x 103
x 500
BC = CCC = CCCCCCCC = - 0.0216 rad
G Ip 80 x 103
x 79,520
TCD L2 175 x 103
x 400
CD = CCC = CCCCCCCC = 0.011 rad
G Ip 80 x 103
x 79,520
BD = BC + CD = - 0.0216 + 0.011 = - 0.0106 rad = - 0.61o
ample 3-5
TBC L1 - 275 x 103
x 500
BC = CCC = CCCCCCCC = - 0.0216 rad
G Ip 80 x 103
x 79,520
TCD L2 175 x 103
x 400
CD = CCC = CCCCCCCC = 0.011 rad
G Ip 80 x 103
x 79,520
BD = BC + CD = - 0.0216 + 0.011 = - 0.0106 rad = - 0.61o
ample 3-5