Torsional and bending stresses in machine parts

1. Contents
1 Notations 2
2 Torsional Shear Stress (τ) 3
3 Shafts in Series and Parallel 4
4 Bending Stress in Straight Beams (σ) 4
5 Bending Stress in Curved Beams (σb) 7
6 Principal Stresses and Principal Planes 9
7 Determination of Principal Stresses for a Member Subjected to Bi-axial Stress 9
8 Application of Principal Stresses in Designing Machine Members 9
9 Theories of Failure Under Static Load 10
10 Maximum Principal or Normal Stress Theory (Rankines Theory) 10
11 Maximum Shear Stress Theory (Guests or Trescas Theory) 10
12 Maximum Principal Strain Theory (Saint Venants Theory) 10
13 MaximumStrain Energy Theory (Haighs Theory) 11
14 Maximum Distortion Energy Theory (Hencky and Von Mises Theory) 11
15 Eccentric Loading - Direct and Bending Stresses Combined 11
16 Shear Stresses in Beams 13
17 Examples 14
17.1 Torsional Shear Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
17.2 Shafts in Series and Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
17.3 Bending Stress in Straight Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
17.4 Bending Stress in Curved Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
17.5 Determination of Principal Stresses for a Member Subjected to Bi-axial Stress . . . . . . . . . . . . . . . . . . . . . . 28
17.6 Theories of Failure Under Static Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
17.7 Eccentric Loading - Direct and Bending Stresses Combined . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
17.8 Shear Stresses in Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
18 References 48
19 Contacts 48

2. 1 Notations
• τ = Torsional shear stress induced at the outer surface of the shaft or maximum shear stress.
• r = Radius of the shaft.
• T = Torque or twisting moment.
• J = Second moment of area of the section about its polar axis or polar moment of inertia.
• C = Modulus of rigidity for the shaft material.
• l = Length of the shaft.
• θ = Angle of twist in radians on a length l.
• T = Torque transmitted in N-m.
• w = Angular speed in rad/s.
• M = Bending moment acting at the given section about the centroidal axis.
• σ = Bending stress.
• I = Second moment of inertia of the cross-section about the neutral axis.
• y = Distance from the neutral axis to the extreme fiber (for straight beams).
= Distance from the neutral axis to the fiber under consideration. It is positive for the distances towards
the center of curvature and negative for the distances away from the center of curvature(for curved beams).
• E = Young’s modulus of the material of the beam.
• R = Radius of curvature of the centroidal axis of the beam.
• Z = Section modulus.
• Rn = Radius of curvature of the neutral axis.
• e = Distance from the centroidal axis to the neutral axis = RRn.
• A = Area of cross-section.
• σb = Bending stress.
• yi = Distance from the neutral axis to the inside fiber = RnRi.
• Ri = Radius of curvature of the inside fiber.
• yo = Distance from the neutral axis to the outside fiber = RoRn.
• Ro = Radius of curvature of the outside fiber.
• σc = Compressive stress.
• σt = Tensile stress due to direct load and bending.
• τ = Shear stress due to torsion.
• σyt = Yield point stress in tension as determined from simple tension test.
• σu = Ultimate stress.
• τmax = Maximum shear stress in a bi-axial stress system.
• τyt = Shear stress at yield point as determined from simple tension test.
• F.S. = Factor of safety.
• σt1 and σt2 = Maximum and minimum principal stresses in a bi-axial stress system.
• = Strain at yield point as determined from simple tension test.
• 1/m = Poissons ratio.
• E = Youngs modulus
• yc and yt = Distances of the extreme fibers on the compressive and tensile sides, from the neutral axis
respectively.
• F = Vertical shear force acting on the section.
• b = Width of the section under consideration.
• y = Distance between the C.G. of the area and the neutral axis.

3. 2 Torsional Shear Stress (τ)
When a machine member is subjected to the action of two equal and opposite couples acting in parallel planes
(or torque or twisting moment), then the machine member is said to be subjected to torsion. The stress set
up by torsion is known as torsional shear stress. It is zero at the centroidal axis and maximum at the outer
surface.
τ
r
=
T
J
=
Cθ
l
Figure 1: Torsional shear stress.
Assumptions:
1. The material of the shaft is uniform throughout.
2. The twist along the length of the shaft is uniform.
3. The normal cross-sections of the shaft, which were plane and circular before twist, remain plane and circular
after twist.
4. All diameters of the normal cross-section which were straight before twist, remain straight with their
magnitude unchanged, after twist.
5. The maximum shear stress induced in the shaft due to the twisting moment does not exceed its elastic
limit value.
Notes:
1. Since the torsional shear stress on any cross-section normal to the axis is directly proportional to the
distance from the center of the axis, therefore the torsional shear stress at a distance x from the center of
the shaft is given by
τx
x
=
τ
r
2.
T
J
=
τ
r
⇒ T = τ
J
r
For a solid shaft of diameter (d), the polar moment of inertia,
J = JXX + JY Y =
π
64
d4
+
π
64
d4
=
π
32
d4
∴ T = τ
π
32
d4 2
d
=
π
16
τ d3
In case of a hollow shaft with external diameter (do) and internal diameter (di), the polar moment of
inertia,
J =
π
32
d4
o − d4
i and r =
do
2
∴ T = τ
π
32
d4
o − d4
i
2
d
=
π
16
τ
d4
o − d4
i
do
=
π
16
τ d3
o 1 − k4
3. The expression (C x J) is called torsional rigidity of the shaft.
4. The strength of the shaft means the maximum torque transmitted by it. Therefore, in order to design a
shaft for strength, the above equations are used. The power transmitted by the shaft (in watts) is given by
P =
2πNT
60
= Tw

4. 3 Shafts in Series and Parallel
When two shafts of different diameters are connected together to form one shaft, it is then known as composite
shaft. Total angle of twist,
θ = θ1 + θ2 =
Tl1
C1J1
+
Tl2
C2J2
If the shafts are made of the same material, then C1 = C2 = C.
θ =
Tl1
CJ1
+
Tl2
CJ2
=
T
C
l1
J1
+
l2
J2
Figure 2: Shafts in series and parallel.
In such cases, the angle of twist is same for both the shafts, i.e.
θ1 = θ2 ⇒
T1l1
C1J1
=
T2l2
C2J2
⇒
T1
T2
=
l2
l1
C1
C2
J1
J2
T = T1 + T2
If the shafts are made of the same material, then C1 = C2.
∴
T1
T2
=
l2
l1
J1
J2
4 Bending Stress in Straight Beams (σ)
The following assumptions are usually made while deriving the bending formula.
1. The material of the beam is perfectly homogeneous (i.e. of the same material throughout) and isotropic
(i.e. of equal elastic properties in all directions).
2. The material of the beam obeys Hookes law.
3. The transverse sections (i.e. BC or GH) which were plane before bending, remain plane after bending
also.
4. Each layer of the beam is free to expand or contract, independently, of the layer, above or below it.
5. The Young’s modulus (E) is the same in tension and compression.
6. The loads are applied in the plane of bending.
Figure 3: Shafts in series and parallel.

5. M
I
=
σ
y
=
E
R
⇒ σ = y
E
R
σ =
M
I
y =
M
I/y
=
M
Z
Notes:
1. The neutral axis of a section always passes through its centroid.
2. In case of symmetrical sections such as circular, square or rectangular, the neutral axis passes through
its geometrical center and the distance of extreme fiber from the neutral axis is y = d/2, where d is the
diameter in case of circular section or depth in case of square or rectangular section.
3. In case of unsymmetrical sections such as L-section or T-section, the neutral axis does not pass through its
geometrical center. In such cases, first of all the centroid of the section is calculated and then the distance
of the extreme fibers for both lower and upper side of the section is obtained. Out of these two values, the
bigger value is used in bending equation.

7. 5 Bending Stress in Curved Beams (σb)
We have seen in the previous article that for the straight beams, the neutral axis of the section coincides with
its centroidal axis and the stress distribution in the beam is linear. But in case of curved beams, the neutral
axis of the cross-section is shifted towards the center of curvature of the beam causing a non-linear (hyperbolic)
distribution of stress. The application of curved beam principle is used in crane hooks, chain links and frames
of punches, presses, planers etc.
Figure 4: Bending stress in a curved beam.
σb =
M
Ae
y
Rn − y
Notes:
1. The bending stress in the curved beam is zero at a point other than at the centroidal axis.
2. If the section is symmetrical such as a circle, rectangle, I-beam with equal flanges, then the maximum
bending stress will always occur at the inside fiber.
3. If the section is unsymmetrical, then the maximum bending stress may occur at either the inside fiber or
the outside fiber. The maximum bending stress at the inside fiber is given by
σbi =
Myi
AeRi
The maximum bending stress at the outside fiber is given by
σbo =
Myo
AeRo
It may be noted that the bending stress at the inside fiber is tensile while the bending stress at the outside
fiber is compressive.
4. If the section has an axial load in addition to bending, then the axial or direct stress (σd) must be added
algebraically to the bending stress, in order to obtain the resultant stress on the section. Resultant stress,
σ = σd ± σb

9. 6 Principal Stresses and Principal Planes
It may be noted that out of these three direct stresses, one will be maximum and the other will be minimum.
These perpendicular planes which have no shear stress are known as principal planes and the direct stresses
along these planes are known as principal stresses.The planes on which the maximum shear stress act are
known as planes of maximum shear.
7 Determination of Principal Stresses for a Member Subjected to Bi-axial Stress
Maximum principal (or normal) stress,
σt1 =
σ1 + σ2
2
+
1
2
(σ1 − σ2)2
+ 4τ2
and minimum principal (or normal) stress,
σt1 =
σ1 + σ2
2
−
1
2
(σ1 − σ2)2
+ 4τ2
and the inclination of principal planes,
tan 2θ =
2τ
σ1 − σ2
The planes of maximum shear stress are at right angles to each other and are inclined at 45o
to the principal
planes. The maximum shear stress is given by one-half the algebraic difference between the principal
stresses, i.e.
τmax =
σt1 − σt2
2
=
1
2
(σ1 − σ2)2
+ 4τ2
Notes:
1. When a member is subjected to direct stress in one plane accompanied by a simple shear stress , then the
principal stresses are obtained by substituting σ2 = 0
σt1 =
σ1
2
+
1
2
(σ1)2
+ 4τ2
σt2 =
σ1
2
−
1
2
(σ1)2
+ 4τ2
τmax =
σt1
2
=
1
2
(σ1)2
+ 4τ2
2. In the above expression of σt2, the value of 1
2
(σ1)2
+ 4τ2 is more than σ1
2
Therefore the nature of σt2 will
be opposite to that of σt1, i.e. if σt1 is tensile then σt2 will be compressive and vice-versa.
8 Application of Principal Stresses in Designing Machine Members
There are many cases in practice, in which machine members are subjected to combined stresses due to simul-
taneous action of either tensile or compressive stresses combined with shear stresses. In many shafts such as
propeller shafts, C-frames etc., there are direct tensile or compressive stresses due to the external force and
shear stress due to torsion, which acts normal to direct tensile or com-pressive stresses. The shafts like crank
shafts, are subjected simultaneously to torsion and bending. In such cases, the maximum principal stresses, due
to the combination of tensile or compressive stresses with shear stresses may be obtained.
The results obtained in the previous article may be written as follows:
1. Maximum tensile stress,
σt(max) =
σt
2
+
1
2
(σt)2
+ 4τ2
2. Maximum compressive stress,
σc(max) =
σc
2
+
1
2
(σc)2
+ 4τ2
3. Maximum shear stress,
τmax =
1
2
(σt)2
+ 4τ2

10. 9 Theories of Failure Under Static Load
The principal theories of failure for a member subjected to bi-axial stress are as follows:
1. Maximum principal (or normal) stress theory (also known as Rankines theory).
2. Maximum shear stress theory (also known as Guests or Trescas theory).
3. Maximum principal (or normal) strain theory (also known as Saint Venant theory).
4. Maximum strain energy theory (also known as Haighs theory).
5. Maximum distortion energy theory (also known as Hencky and Von Mises theory).
Since ductile materials usually fail by yielding i.e. when permanent deformations occur in the material and
brittle materials fail by fracture, therefore the limiting strength for these two classes of materials is normally
measured by different mechanical properties. For ductile materials, the limiting strength is the stress at yield
point as determined from simple tension test and it is, assumed to be equal in tension or compression. For
brittle materials, the limiting strength is the ultimate stress in tension or compression.
10 Maximum Principal or Normal Stress Theory (Rankines Theory)
According to this theory, the failure or yielding occurs at a point in a member when the maximum principal or
normal stress in a bi-axial stress system reaches the limiting strength of the material in a simple tension test.
Since the limiting strength for ductile materials is yield point stress and for brittle materials (which do not
have well defined yield point) the limiting strength is ultimate stress, therefore according to the above theory,
taking factor of safety (F.S.) into consideration, the maximum principal or normal stress (σt1) in a bi-axial
stress system is given by
σt1 =
σyt
F.S.
, for ductile materials
=
σu
F.S.
, for brittle materials
Since the maximum principal or normal stress theory is based on failure in tension or compression and ignores
the possibility of failure due to shearing stress, therefore it is not used for ductile materials. However, for brittle
materials which are relatively strong in shear but weak in tension or compression, this theory is generally used.
11 Maximum Shear Stress Theory (Guests or Trescas Theory)
According to this theory, the failure or yielding occurs at a point in a member when the maximum shear stress
in a bi-axial stress system reaches a value equal to the shear stress at yield point in a simple tension test.
Mathematically,
τmax =
τyt
F.S.
Since the shear stress at yield point in a simple tension test is equal to one-half the yield stress in tension,
therefore the equation may be written as
τmax =
τyt
2 x F.S.
12 Maximum Principal Strain Theory (Saint Venants Theory)
According to this theory, the failure or yielding occurs at a point in a member when the maximum principal
(or normal) strain in a bi-axial stress system reaches the limiting value of strain (i.e. strain at yield point) as
determined from a simple tensile test. The maximum principal (or normal) strain in a bi-axial stress system is
given by
max =
σt1
E
−
σt2
mE
∴ According to the above theory,
max =
σt1
E
−
σt2
mE
= =
σyt
E x F.S.
σt1 −
σt2
m
=
σyt
F.S.
This theory is not used, in general, because it only gives reliable results in particular cases.

11. 13 MaximumStrain Energy Theory (Haighs Theory)
According to this theory, the failure or yielding occurs at a point in a member when the strain energy per unit
volume in a bi-axial stress system reaches the limiting strain energy (i.e. strain energy at the yield point ) per
unit volume as determined from simple tension test.
We know that strain energy per unit volume in a bi-axial stress system,
U1 =
1
2E
(σt1)2
+ (σt2)2
−
2σt1σt2
m
and limiting strain energy per unit volume for yielding as determined from simple tension test,
U2 =
1
2E
σyt
F.S.
2
According to the above theory,
U1 = U2
∴
1
2E
(σt1)2
+ (σt2)2
−
2σt1σt2
m
=
1
2E
σyt
F.S.
2
(σt1)2
+ (σt2)2
−
2σt1σt2
m
=
σyt
F.S.
2
This theory may be used for ductile materials.
14 Maximum Distortion Energy Theory (Hencky and Von Mises Theory)
According to this theory, the failure or yielding occurs at a point in a member when the distortion strain energy
(also called shear strain energy) per unit volume in a bi-axial stress system reaches the limiting distortion energy
(i.e. distortion energy at yield point) per unit volume as determined from a simple tension test. Mathematically,
the maximum distortion energy theory for yielding is expressed as
(σt1)2
+ (σt2)2
− 2σt1σt2 =
σyt
F.S.
2
This theory is mostly used for ductile materials in place of maximum strain energy theory.
Note: The maximum distortion energy is the difference between the total strain energy and the strain energy
due to uniform stress.
15 Eccentric Loading - Direct and Bending Stresses Combined
An external load, whose line of action is parallel but does not coincide with the centroidal axis of the machine
component, is known as an eccentric load. The distance between the centroidal axis of the machine component
and the eccentric load is called eccentricity and is generally denoted by e. The examples of eccentric loading,
from the subject point of view, are C-clamps, punching machines, brackets, offset connecting links etc.
The magnitude of this direct compressive stress is given by
σo =
P1
A
or
P
A
The forces P1 and P2 will form a couple equal to P x e which will cause bending stress. This bending stress
is compressive at the edge AB and tensile at the edge CD. The magnitude of bending stress at the edge AB is
given by
σb =
P e yc
I
(compressive)
and bending stress at the edge CD,
σb =
P e yt
I
(tensile)
According to the principle of superposition, the maximum or the resultant compressive stress at the edge AB,
σc =
P e yc
I
+
P
A
=
M
Z
+
P
A
= σb + σo

12. and the maximum or resultant tensile stress at the edge CD,
σt =
P e yt
I
−
P
A
=
M
Z
−
P
A
= σb − σo
Figure 5: Eccentric loading.
Notes:
1. When the member is subjected to a tensile load,
then the above equations may be used by inter-
changing the subscripts c and t.
2. When the direct stress σo is greater than or equal
to bending stress σb , then the compressive stress
shall be present all over the cross-section.
3. When the direct stress σo is less than the bending
stress σb, then the tensile stress will occur in the
left hand portion of the cross-section and com-
pressive stress on the right hand portion of the
cross-section.
In case the eccentric load acts with eccentricity
about two axes, then the total stress at the ex-
treme fiber
=
P
A
±
P ex x
IXX
±
P ey y
IY Y
Figure 6: Eccentric load with eccentricity about two axes.

13. 16 Shear Stresses in Beams
It has been observed that the effect of the shear stress, as compared to the bending stress, is quite negligible
and is of not much importance. But, sometimes, the shear stress at a section is of much importance in the
design. It may be noted that the shear stress in a beam is not uniformly distributed over the cross-section but
varies from zero at the outer fibers to a maximum at the neutral surface.
Figure 7: Shear stress in rectangular, circular and I-section beam.
The shear stress at any section acts in a plane at right angle to the plane of the bending stress and its value is
given by
τ −
F
Ib
A y
The following values of maximum shear stress for different cross-section of beams may be noted :
1. For a beam of rectangular section, the shear stress at a distance y from neutral axis is given by
I =
bh3
12
∴ τ =
F
2I
h2
4
− y2
=
3F
2bh
h2
− 4y2
and maximum shear stress,
y =
h
2
∴ τmax =
3F
2bh
τ(average) =
F
Area
=
F
bh
∴ τmax = 1.5τ(average)

14. 2. or a beam of circular section, the shear stress at a distance y from neutral axis is given by
τ =
F
3I
d2
4
− y2
=
16F
3πd4
d2
− 4y2
and the maximum shear stress,
y =
h
2
∴ τmax =
4F
3π
4
d2
τ(average) =
F
Area
=
F
π
4
d2
∴ τmax =
4
3
τ(average)
3. For a beam of I-section, the maximum shear stress occurs at the neutral axis and is given by
τmax =
F
Ib
B
8
H2
− h2
+
bh2
8
Shear stress at the joint of the web and the flange
=
F
8I
H2
− h2
and shear stress at the junction of the top of the web and bottom of the flange
=
F
8I
B
b
H2
− h2
17 Examples
17.1 Torsional Shear Stress

18. 17.2 Shafts in Series and Parallel

20. 17.3 Bending Stress in Straight Beams

22. 17.4 Bending Stress in Curved Beams

28. 17.5 Determination of Principal Stresses for a Member Subjected to Bi-axial Stress

32. 17.6 Theories of Failure Under Static Load

37. 17.7 Eccentric Loading - Direct and Bending Stresses Combined

47. 17.8 Shear Stresses in Beams

48. 18 References
1. R.S. KHURMI, J.K. GUPTA, A Textbook Of Machine Design
19 Contacts
mohamed.atyya94@eng-st.cu.edu.eg