Structures II, MACE, University of Manchester

Topic 5 Stress Analysis

Stress state under general loading conditions

By definition, a normal stress is perpendicular to the applied area, a shear stress is parallel to
the area. A force applied obliquely to a surface can be decomposed into two components:
normal and parallel respectively to the surface, yielding in normal and shear stresses.

Consider a body subjected to several loads P1, P2 etc (Fig. 1a). For a given point Q within the
body, a section is made passing through this point, using a plane parallel to the yz plane.

y P2 y P2 y
∆ Fy τxy
P3 ∆A
∆ Fz
Q ∆ Fx Q σx
τxz

P1 P4 P1

x x x
z z z
Figure 1a Figure 1b Figure 1c

The portion of the body to the left of the section is subjected to some of the original loads, and
normal and shear forces distributed over the section (Fig. 1b). The normal and shear forces
acting on a small area ∆A surrounding point Q are denoted by ∆Fx, ∆Fy and ∆Fz, the three
stress components σx, τxy, τxz at point Q on this surface (Fig. 1c) are then defined as:

∆Fx ∆Fy ∆Fz

σ x = lim τ xy = lim τ xz = lim (1)
∆A→0 ∆A ∆A→0 ∆A ∆A→0 ∆A

The above analysis may also be carried out by considering other two sections, parallel to the
xz and xy plane, respectively. This leads to six more stress components σy, τyx, τyz and σz, τzx,
τzy at point Q. All the stress components are shown on a cubic stress element in Fig. 2a.
P2 σy′
σy τy′x′
P3 τyz τyx τx′y′
y τy′z′
τxy σx′
τzy Q y′ Q
Q τz′y′ τx′z′
σx
τxz σz′ τz′x′
σz τzx
P1 P4
x′
z x
z′
Figure 2a Figure 2b

However, only six components are independent among the nine components, because τyx =

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τxy, τzx = τxz and τzy = τyz. (Please prove it yourself!).

The stress state at a given material point in a structure is fully determined by the six stress
components: normal stresses σx, σy and σz, and shear stresses τxy, τxz and τyz. Stresses acting
on inclined planes can be expressed by the above six stress components. For example, all the
stress components σx′, σy′, σz′, τx′y′, τx′z′ and τy′z′ on the stress element associated with a new
coordinate system x′y′z′ in Fig. 2b (by rotating the coordinate system xyz) can be expressed in
terms of σx, σy, σz, τxy, τxz and τyz in Fig. 2a.

It is import to know the relationship amongst these stresses as this plays a fundamental role in
analysing the failure of materials. The stresses on an inclined plane or surface at a point vary
according to the orientation of the axes. Larger stresses may occur on other planes. One of
the tools we use when we need to find the relationships between the stresses is the stress
transformation, which also leads to the development of Mohr’s circle.

Example 1 Two members of material of uniform 25°

cross section 50×80mm are glued together along a
P
plane a-a, that forms an angle of 25° with the vertical.
Knowing that the allowable stresses for the glued
a
joint are σ = 800 kPa and τ = 600 kPa. Determine
the largest axial load P that can be applied.

Solution
Axial normal stress σ x = P A
θ
Normal and shear stresses on the inclined plane
σ n = Pn Aθ = P cos θ ( A cos θ ) =σ xcos θ cos θ
τ nt = Pt Aθ = − P sin θ ( A cosθ ) = −σ xsin θ cosθ Pn
Applying the allowable stress P P
Pt
σ n = 0.821σ x= σ = 800kPa σ x= 974.4kPa
τ nt = 0.383σ x= τ = 600kPa σ x= 1566.6kPa
σn
The largest load P τnt
P = 974.4kPa × 50mm × 80mm = 3.9kN

Conventions:

A surface or plane is defined by its normal. A positive face of an element means its outward
normal points to the positive direction of one of the coordinate axes.

The subscript of the normal stress indicates along which axis the stress points to. For the
shear stress, the first subscript denotes the face on which it is exerted. The second subscript
indicates its direction (i.e. along which axis).

Normal stress is positive in tension, negative in compression. A shear stress is positive if it is

exerted on a positive face of an element and points to the positive direction of one of the
coordinate axes, or if it acts on a negative face of an element and is in the negative direction
of coordinate axis. Otherwise the shear stress is negative.

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Stress Transformation

To make the analysis easier, let’s start with the case of plane stress, which mean stresses in
2D. If stresses are in xy plane, the plane stress is defined by the condition: σz = τxz = τyz = 0.
σx, σy and τxy may have nonzero values. Plane stress exists in thin plate with external forces
acting in the mid-plane of the plate. It also occurs on the free surface of a structural element.

y y
σy
σy

τxy
τxy
σx
σx O σx x
O
x τxy

z σy

Figure 3 plane stress representation in (a) 3D stress element, (b) 2D stress element

Let us have a square stress element in xy y′ y

coordinate system. When the stresses in an σy
inclined surface are needed associated with the τxy
θ
x′y′ system, which is achieved by rotating xy C τxy
system an angle of θ counter clockwise, we take
a wedge-shaped stress element ABC with two σx σx
τxy A B
sides of the wedge being parallel to x and y axes,
and the third side parallel to y′ (or perpendicular τxy x′
to x′). Note that we haven’t defined the value of σy θ
θ, thus this represents any arbitrary orientation
the x′y′ system may have. (a) O x

y′ y′
C C
τx′y′ dA τx′y′
x′ σx′ x′
σx dAcosθ σx θ
σx′ dA

τxy dAcosθ θ τxy θ

A B A B
τxy dAsinθ τxy
σy dAsinθ σy
(c) (b)
Figure 4 (a) stresses acting on the square element, (b) stresses acting on the wedge element,
and (c) forces acting on the wedge element (free-body diagram and the area of the inclined
face BC is set to be dA)

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The force acting on the left-hand and bottom faces of the wedge element in Fig. 4c can be
resolved into orthogonal components acting in the x′ and y′ directions. Two equations of
equilibrium can then be obtained by summing forces in those directions as follows:

σ x ' dA − σ x dA cos θ cos θ − τ xy dA cos θ sin θ − σ y dA sin θ sin θ − τ xy dA sin θ cos θ = 0

τ x ' y ' dA + σ x dA cos θ sin θ −τ xydA cos θ cos θ − σ y dA sin θ cos θ + τ xy dA sin θ sin θ = 0
Solving the above equations and after simple manipulation of trigonometry, we can obtain
σ x +σ y σ x −σ y
σ x' = + cos 2θ + τ xy sin 2θ (2a)
2 2
σ x −σ y
τ x' y ' = − sin 2θ + τ xy cos 2θ (2b)
2
These equations are usually called the stress transformation equations for plane stress
because they transform the stress components from one set of axes to another.
The transformation equations were derived solely from equilibrium. They are applicable to
stresses in any kind of material, whether linear or nonlinear, elastic or plastic.
By substituting θ +90° for θ, the normal stress σy′ can be obtained from equation (2a)
σ x +σ y σ x −σ y
σ y' = − cos 2θ − τ xy sin 2θ (2c)
2 2
One may also notice that

σx′ + σy′ = σx + σy (3)

This is called stress invariant, meaning the sum of the normal stresses on any two mutually
perpendicular planes remains the same. And you may use it to check your calculations.

Example 2 For the given state of plane-stress shown on the 80MPa

right, determine the normal and shear stresses after the element
shown has been rotated through 25° clockwise. Show the results
on a sketch of a properly oriented element. 55MPa
Solution
The stresses acting on the original element are: 40MPa
σx = 55 MPa σy = −80 MPa τxy = −40 MPa.
Stress element rotation angle is clockwise, therefore, θ = −25°
Substituting into the transformation equations (2), we get
σ x +σ y σ x −σ y y
σ x' = + cos 2θ + τ xy sin 2θ = 61.5MPa 86.5MPa
2 2
σ x −σ y
τ x' y ' = − sin 2θ + τ xy cos 2θ = 26.0 MPa 26.0MPa
2
Also the normal stress acting the y’ face is x
σ +σ y σx −σ y θ=25°
σ y' = x − cos 2θ − τ xy sin 2θ = −86.5MPa
2 2 61.5MPa
This stress can be checked by verified by σx’ + σy’ = σx + σy.
The stresses acting on the inclined element are shown on the right above.

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Mohr’s Circle for Plane Stresses

The transformation equation for plane stress can be represented in a graphical form by a plot
known as Mohr’s circle. This graphical representation is extremely useful because it enables
you to visualize the relationship between the normal and shear stresses acting on various
inclined planes at a point in a stressed body. It also provides a means for calculating
maximum normal stresses, maximum shear stresses, and stresses on inclined planes.

Equation (2) can be converted into

2 2
 σ +σ y  σ −σ y 
 σ x ' − x  + τ x2' y ' =  x  + τ xy2 (4a)
 2   2 
or
(σ x ' − σ ave )2 + τ x2' y ' = R 2 (4b)
With notation
2
σ x +σ y σ −σ y 
σ ave = R =  x  + τ xy2 (5)
2  2 

This is an equation of a circle in standard algebraic form. The coordinates are σx′ and τx′y′, the
radius is R, and the centre of the circle has coordinates σx′ = σave and τx′y′ = 0.

There are a variety of ways to construct Mohr’s circle, depending upon which stresses are
known and which are to be found. Here comes one procedure to draw Mohr’s circle with
stresses known in the xy system:

Y (σy, −τxy, θ=90°)

S2 D (σx′, τx′y′, θ)

R
2θ
B (σ2, 0) A (σ1, 0)
O C σx′
2θP

X(σx, τxy, θ=0)

σave S1(σave, τ 12)

τx′y′

Figure 5 Construction of Mohr’s circle

1) Draw a set of coordinate axes with σx′ as abscissa (positive to the right) and τx′y′
ordinate (positive downwards).
2) Locate point X (σx′ =σx and τx′y′ =τxy), representing the stress condition on the x face of
the stress element. Note that point X on the circle corresponds to θ =0.

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3) Locate point Y (σx′ =σy and τx′y′ =−τxy), representing the stress condition on the y face
of the stress element. Note that point Y on the circle corresponds to θ =90°.
4) Locate the centre C of the circle at the point σx′ =σave =0.5×(σx + σy) and τx′y′ =0.
5) Draw a line from point X to point Y. This line is the diameter of the circle and passes
through the centre C.
6) Using point C as the centre, draw Mohr’s circle through points X and Y.

Note σx′ is plotted positive to the right and τx′y′ positive downwards. The angle θ of the
inclined plane should be doubled on the circle. The angle 2θ is measured from point X and is
positive counter clockwise.

Principal Stresses and Maximum in-plane Shear Stress

It is clearly shown on the Mohr’s circle in Fig. 5 that there are two planes (represented by
points A and B) on which the normal stress reaches its maximum and minimum values and the
shear stress is zero. The planes on which there is no shear stress are called principal planes
and the corresponding normal stresses on the principal planes are called principal stresses.

From the Mohr’s circle, we have the formula for the principal stresses:

2
σ x +σ y σ x −σ y 
σ 1, 2 = σ ave ± R = ±   + τ xy2 (6a)
2  2 

The orientation of the two principal planes is defined by the angle θp in Fig. 5 which are
counter clockwise from the old xy system. From Mohr’s circles we get

2τ xy
tan 2θ p = (6b)
σ x −σ y

The two principal planes are perpendicular to each other. θp2 = θp1 + 90°.

Maximum in-plane shear stresses are the bottom and top of the circle at points S1 and S2.
There are the following relationships

2
σ1 − σ 2 σ x −σ y 
τ 12 = R = =   + τ xy2 and θ s = θ p ± 45 o (7)
2  2 

This means the plane of maximum in-plane shear stress is 45 degree to the principle planes.
Note that the normal stresses are usually not zero on the maximum in-plane shear stress
planes. Can you find out how much it is?

Stresses at inclined planes (point D on the circle) can also be obtained easily once the
principal angle θp is determined:

σ x ' = σ ave + R cos(2θ − 2θ p ) (8a)

τ x ' y ' = − R sin(2θ − 2θ p ) (8b)

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Example 3 An element in plane stress state is subjected to 30MPa

stresses σx = 60 MPa, σy = 30 MPa and τxy = 20 MPa, as shown
on the right. Using Mohr’s circle determine the following 20MPa
quantities: (a) the principal stresses, and (b) the maximum in-
plane shear stresses, (c) the stresses acting on an element 60MPa
inclined at an angle θ = 45°. Show all the results on sketches of
properly oriented elements.

Solution σy =30
Construction of Mohr’s circle. Y
D (θ=45°)
1) Set up the axes for Mohr’s circle. 2θ=90°
−τxy R
2) Locate point X (σx′=σx=60MPa, τx′y′=τxy=20MPa).
3) Locate point Y (σx′=σy=30MPa and τx′y′=−20MPa). F E A
O B C σx′
4) The abscissa of the centre C of Mohr’s circle is
τxy=20
σ x +σ y 2θ p
σ ave= OC = = 45 MPa X
2 σave S
5) The circle is drawn through point X and Y with
σx =60
centre at C. The radius of the circle is
τx′y′
2
σ x −σ y 
R = CE 2 + EX 2 =   + τ xy 2 = 25 MPa
 2 
(a) Principal stresses. They are represented by points A and B on the Mohr’s circle.
σ 1 = OC + CA = σ ave + R = 70 MPa y
σ 2 = OC − BC = σ ave − R = 20 MPa 20MPa
EX 2τ xy 70MPa
tan 2θ p1 = = = 1.3333
CE σ x − σ y 26.6°
2θ p = 53.13° θ p1 = θ p = 26.6° θ p 2 = 116.6° x

The principal stresses and planes are shown on the right.

(b) Maximum in-plane shear stress. It is represented by point S on Mohr’s circle.

τ 12 = R = 25 MPa
θ s1 = θ p1 − 45° = 26.6° − 45° = −18.4° y
45MPa
This angle is negative because it is measured clockwise on
the circle. 25MPa
x
The normal stresses acting on the plane of maximum in- θ=18.4°
plane shear stresses are equal to σave, which is the abscissa 45MPa
of the centre C of the circle (45MPa). These stresses are
shown on the right.

(c) Stresses on an element inclined at θ = 45°. The stresses acting on an element inclined at
an angle θ = 45° are given by the coordinates of point D, which is at an angle 2θ = 90° from

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point X. Therefore the coordinates of point D are y

σ x ' = σ ave + R cos(2θ − 2θ p1 ) = 65 MPa 25MPa 65MPa

τ x ' y′ = − R sin(2θ − 2θ p1 ) = −15 MPa θ=45°

To find σy′ replacing θ by θ +90°: x

15MPa
σ y ' = σ ave + R cos(2θ + 180° − 2θ p1 ) = 25 MPa
These stresses are shown on the right.

σy
Example 4 For the state of plane stress shown, it is known that
the direct and shear stress are directed as shown and σx = 110 MPa, τxy=?
σy = 60 MPa and σ2 = 35 MPa. Using Mohr’s circle, determine
the principal planes, the principal stresses and the maximum in- σx
plane shear stress.

Solution
Construction of Mohr’s circle. σy = 60
1) Set up the axes for Mohr’s circle. Y
2) Locate point B (σx′=σ2=35MPa, τx′y′=0)
3) The abscissa of the centre C of Mohr’s circle is −τxy
σx +σ y B F E A
σ ave= OC = = 85 MPa O C σx′
2
σ2=35 R τxy=?
4) Locate point C (σx′=σave=85MPa, τx′y′=0).
5) The radius of the circle is X
σave S
R = σ ave −σ 2= 50 MPa
6) The circle is drawn with radius R and centre at C.
σx = 110
τx′y′
The radius of the circle can also be obtained as
2
σ −σ y 
R = CE + EX =  x
2 2
 + τ xy2
 2 
2
 σ −σ y 
Therefore τ xy = ± R −  x
2
 = ±43.3 MPa
 2 
Principal directions
EX 2τ xy
tan 2θ p1 = = = 1.732
CE σ x − σ y
θ p1 = 30° , θ p 2 = 120°
The principal stress σ 1 =σ ave+ R = 135 MPa
Maximum in-plane shear stress is represented by point S on Mohr’s circle
τ 12 = R = 50 MPa

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Principal plane, principal stress and maximum shear stress for a general state of stress

Having considered plane stress cases, let’s look at a general state of stress in 3D. An
infinitesimal tetrahedron ABCO is defined as shown below. A set of known stresses on the
three faces of the mutually perpendicular planes is given and the unknown principal planes
and principal stresses in the 3D coordinate xyz are sought.
y

n y
τxy B τyz
σn
σz Unit vector
m β
τxz α
σx γ l
O n
A x
τxz x
τxy z
C τyz
σy
z
(a) (b)
Figure 6

A unit normal n to the oblique plane defines its orientation with direction cosines l, m and n,
where l = cosα, m = cosβ and n = cosγ. If the area of ABC is dA, then we have areas of BOC,
AOC and AOB as ldA, mdA and ndA, respectively.

Suppose the oblique plane ABC is the principal plane. Because there is no shear stress on
principal plane, force equilibrium of the tetrahedron ABCO in three axes directions leads to
∑ Fx = 0 (σ n dA)l − σ x ldA − τ xy mdA − τ xz ndA = 0 i.e. (σ x − σ n )l + τ xy m + τ xz n = 0
∑ Fy = 0 (σ n dA)m − σ y mdA − τ yz ndA − τ xy ldA = 0 i.e. τ xy l + (σ y − σ n )m + τ yz n = 0 (9)
∑ Fz = 0 (σ n dA)n − σ z ndA − τ xz ldA − τ yz mdA = 0 i.e. τ xz l + τ yz m + (σ z − σ n )n = 0
For nontrivial l, m and n, the linear equations must have
σ x −σ n τ xy τ xz
τ xy σ y −σ n τ yz = 0 (10)
τ xz τ yz σ z −σ n
Expansion of the determinant gives the characteristic equation
σ n3 − I1σ n2 + I 2σ n − I 3 = 0 (11)
where
I1 = σ x + σ y + σ z
I 2 = σ xσ y + σ xσ z + σ yσ z − τ xy2 − τ xz2 − τ yz2 (12)
I 3 = σ xσ yσ z − σ xτ yz2 − σ yτ xz2 − σ zτ xy2 + 2τ xyτ xzτ yz
Equation (11) has three real roots. These roots are the eigenvalues of the determinant in (10)
and are the principal stresses of the problem. Principal stresses are orthogonal to each other.

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Note that the principal stresses do not change when the coordinate system changes, thus the
I’s in (12) remain constant, and are called the 1st, 2nd and 3rd stress invariants, accordingly. We
have used the 1st invariant already in the plane stress cases. As before, we can use I1 as one
way to check whether the calculated principal stresses are correct. The determination of
principal directions is demonstrated in the following example.

Example 5 Find the stress invariants, the principal stresses, and the corresponding
principal directions at a material point with a stress state as follows
σx = 10 MPa σy = 20 MPa σz = 20 MPa τxy = 20 MPa τxz = 30 MPa τyz = 0

Solution

Stress invariants
I 1 = σ x + σ y + σ z = 50
I 2 = σ xσ y + σ xσ z + σ y σ z − τ xy2 − τ xz2 − τ yz2 = −500
I 3 = σ xσ yσ z − σ xτ yz2 − σ yτ xz2 − σ zτ xy2 + 2τ xyτ xzτ yz = −22000
Characteristics equation
σ n3 − 50σ n2 − 500σ n + 22000 = 0 ⇒ (σ n − 20)(σ n2 − 30σ n − 1100) = 0
Three roots for three principal stresses
σ 1= 51.4 σ 2= 20 σ 3= −21.4

Principal directions
1) For σ1, substitute σn = σ1 = 51.4 into the second and third equation in (9)
20l + (20 − 51.4)m = 0 ( ∑ Fy = 0 )
30l + (20 − 51.4)n = 0 ( ∑ Fz = 0 )
l 2 + m2 + n2 = 1 (Unit vector)
l1 = 0.657 m1 = 0.418 n1 = 0.627

2) For σ2, substitute σn = σ2 = 20 into the first and second equation in (9)
(10 − 20)l + 20m + 30n = 0 ( ∑ Fx = 0 )
20l + (20 − 20)m = 0 ( ∑ Fy = 0 )
2 2 2
l + m + n =1 (Unit vector)
l 2= 0 m2 = 0.832 n2 = −0.555

3) For σ3, substitute σn = σ3 = −21.4 into the second and third equation in (9)
20l + (20 + 21.4)m = 0 ( ∑ Fy = 0 )
30l + (20 + 21.4)n = 0 ( ∑ Fz = 0 )
2 2 2
l + m + n =1 (Unit vector)
l 3 = −0.754 m3 = 0.364 n3 = 0.546

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Mohr’s circle for a general state of stress

For the general case, consider a stress element oriented in the principal directions as shown in
Fig. 7 (σ1, σ2 and σ3 are principal stresses in the present case). If an inclined plane parallel to
the axis 3 is cut through the element, the stresses on the inclined plane are normal stress σ and
shear stress τ both acting in the 1-2 plane. We can use the stress transformation equations
derived for plane stress to find σ and τ on the inclined plane, because they are found from
equations of force equilibrium in 1-2 plane, and are independent of σ3.

σ2
σ
σ1 τ σ3
σ1

σ3 σ3
σ2 Figure 7

Therefore, in each plane perpendicular to one of the principal stresses, we can draw a Mohr’s
circle. Three Mohr’s circles can be put together on a single diagram as in Fig. 8.

σ3 σ2 σ1
Ο σ

τ12
τ23

τ13
τ Figure 8

It can be proved that the stresses on any inclined planes are located in the shaded area formed
by the three Mohr’s circles in Fig. 8. The maximum tensile or compressive normal stresses
and maximum shear stress can then be obtained.

The maximum and minimum principal stresses (algebraic) are

σ max = Max[σ 1 , σ 2 , σ 3 ] (13)

σ min = Min[σ 1 , σ 2 , σ 3 ] (14)

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The maximum tensile normal stress is

t
σ max = Max[σ 1 , σ 2 , σ 3 ] if t
σ max >0 (15)
The maximum compressive normal stress is
c
σ max = Min[σ 1 , σ 2 , σ 3 ] if c
σ max <0 (16)
and the maximum shear stress is
1
τ max = (σ max − σ min ) (17)
2
Note that there are three maximum in-plane (in 1-2, 2-3 and 1-3 planes) shear stresses:
1 1 1
τ 12 = σ1 − σ 2 , τ 23 = σ2 −σ3 , τ 13 = σ1 − σ 3 (18)
2 2 2
Only the largest maximum in-plane shear stress above is the maximum shear stress τmax.

Special attention needs to be drawn for the cases of plane stress when calculating maximum
shear stress. The shear stress obtained below
1
τ 12 = (σ 1 − σ 2 )
2
is the maximum in-plane shear stress in the 1-2 plane as stated before. It may or may not be
the maximum shear stress, depending on the values of σ1 and σ2.

By definition, the third principle stress is zero in plane stress state. For the Mohr’s circle
shown below on the left – the dark circle, where both principal stresses are positive, the actual
smallest principle stress in reality, σmin, is zero in the 3D space. Therefore, we would have
τ max = σ 1 2 , and not (σ 1 − σ 2 ) 2 . Similarly, when both principle stresses in plane stress are
negative (so the dark circle would be on the negative domain of σ axis), the largest principle
stress, σmax, is zero, thus τ max = σ 2 2 , and not (σ 1 − σ 2 ) 2 .

In contrast, see the diagram below on the right. What if the two principal stresses have
opposite signs for the plane stress cases?

σ2 σ1 σ2 σ1 σ2 σ1
σ σ σ

τmax=σ1/2 τmax=|σ2|/2 τmax=(σ1−σ2)/2

τ τ τ

Figure 9

Rule of thumb for plane stresses: when calculate the maximum shear stresses, always look for
the maximum and minimum principals stresses in a three dimensional state.

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Example 6 For the state of stress shown, determine the maximum shear stress when
(a) σz = 0, (b) σz = −45 MPa (c) σz = 165 MPa

σy = 20
y Y

20MPa
−τxy
75MPa E A
B O C σ
σz 100MPa σave R τxy=75

z x
X
σx = 100

Solution τ

The third principal stress is σz

The stress components in xy plane are
σx = 100 MPa, σy = 20 MPa, τxy = 75 MPa
Draw Mohr’s circle in xy plane.
2
σ x +σ y σ x −σ y 
σ ave= OC = = 60 MPa 2
R = CE + EX 2
=   + τ xy2 = 85 MPa
2  2 
Principal stresses:
σ 1 = σ ave + R = 145 MPa σ 2 = σ ave − R = −25 MPa
Maximum shear stress:
(a) σz = σ3 = 0, then τ max = (σ max − σ min ) 2 = (σ 1 − σ 2 ) 2 = 85 MPa
(b) σz = σ3 = −45, then τ max = (σ max − σ min ) 2 = (σ 1 − σ 3 ) 2 = 95 MPa
(c) σz = σ3 = 165, then τ max = (σ max − σ min ) 2 = (σ 3 − σ 2 ) 2 = 95 MPa

Example 7 For the state of stress shown, determine two values of σy for which the
maximum shear stress is 75MPa.
σ y

y Y

σy −τxy

40MPa A
B O C σ
70MPa σave R τxy=40

z x S
X
σx = 70
τ

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Solution

The third principal stress is σ3 = σz =0.

Draw Mohr’s circle in xy plane.
2
σx +σ y σx −σ y 
σ ave=
2
= 35 + 0.5σ y R =   + τ xy
2
= (35 − 0.5σ y )2 + 402
 2 
Principal stresses:
σ 1 = σ ave + R σ 2 = σ ave − R
There are two possibilities: (i) σ1> σ2 > 0 (ii) σ1> 0 and σ2 < 0

(i) If σ1> σ2 > 0, then

τ max = (σ 1 − σ 3 ) 2 = (σ ave + R ) 2
2 × 75 = (35 + 0.5σ y ) + (35 − 0.5σ ) 2
y + 40 2
σ y = 130
Check: when σy = 130, σ1>=150>0 and σ2 =50 > 0, satisfy the assumption, ∴ solution.

(ii) If σ1> 0 and σ2 < 0, then

τ max = (σ 1 − σ 2 ) 2 = R
75 = (35 − 0.5σ ) y
2
+ 40 2
σ y = −56.9 or σ y = 197
Check: when σy = −56.9, σ1> 0 and σ2 < 0, satisfy the assumption, ∴ solution.
Check: when σy = 197, σ1> 0 and σ2 > 0, against the assumption, ∴ not solution.

Stresses in Pressurized Thin Wall Vessels

Thin-wall pressurized vessels have wide applications in engineering sections, such as gas
tanks and aircraft fuselages under internal pressure and submarines under external pressure.
Usually they are in the cylindrical or spherical shapes. The assumption that only thin walls
are being used means that the bending resistance of the walls is very low and that all the
stresses in the walls can be considered to be uniform through the thickness and tangential to
the surface of the vessel. Stresses in thin wall pressure vessels can be calculated directly,
using the equilibrium equations.

Stresses in a Cylindrical Pressure Vessel

Consider the cylindrical vessel (shown in Fig. 11) of inside radius r, wall thickness t and
gauge pressure p (where gauge pressure is the difference between inside pressure and outside
pressure). The ratio t/r should be less than about 0.1 in order the vessel to be counted as thin
wall structure.

The stresses on a small element of the wall are defined as hoop stress (σh around the
circumference) and axial stress (σa along the axis). The element is in plane stress state. The
internal and/or external pressure actually exerts a normal stress in the direction through

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Structures II, MACE, University of Manchester

thickness of the wall but usually this stress is orders of magnitude smaller than those in the
hoop and axial directions. It will therefore be neglected as an engineering approximation.
σh

p
σa σa
r

t
σh
Figure 10

Consider the equilibrium of half of a pressure cylinder split down its axial length as shown in Fig. 11a.
There should be axial stresses at both ends but they would not contribute to the equilibrium in the
direct of interest, which is perpendicular to the axial axis. They are therefore not shown in Fig. 11a.

y L y
2πrtσa
σh tL
t
t
z r
r
z 2prL
x
2
pπr
σh tL x

Figure 11(a) Figure 11(b)

Force equilibrium gives

∑F z = 2σ h tL − 2 prL = 0
This results in
pr
σh = (19)
t
The axial stress σa depends on the type of end constraints applied to the cylinder. The most
common case, generally referred to as closed-ended, is where the ends are closed by some
form of end caps, which can be flat, hemispherical or some other curved shapes. Since the end
caps are attached to the cylinder, the axial force created by the internal pressure on these caps
is transferred to the cylinder walls. It should be noted that the present formulae is not
applicable in the regions of the cylinder close to the end caps, where the stress distribution is
highly nonlinear and more complicated methods for the analysis are required. Now consider
the equilibrium of a pressure cylinder split perpendicular to its axis as shown in Fig. 11b.

∑F x = σ a 2πrt − pπr 2 = 0
Hence
pr
σa = (20)
2t
It means that if the pressure is higher than the safety limit the cylinder will crack in the axial

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direction due to σh, not in the circumferential direction. The construction of a cylinder should
have stronger connections along the axial direction.

Stresses in a Spherical Pressure Vessel

Fig. 12a shows thin-walled spherical vessel with an internal pressure p, a constant wall
thickness t and radius r (t/r < 0.1). For a sphere there are symmetry of geometry and loading
about each and every plane passing through its centre. Due to symmetry, stress at any location
in any direction tangential to the spherical surface is same.

pπr2

Figure 12(a) Figure 12(b)

Fig. 12b shows the free body diagram of the sphere, formed by cutting it along any plane
through its centre. Symmetry ensures that there is no shear stress acting on the edges of the
wall. For equilibrium of the forces
2πrtσ = pπr 2
Therefore
pr
σ= (21)
2t

Example 8 A thin-walled cylindrical bottle has hemispherical ends with a diameter of

150mm, a parallel length of 760mm and constant thickness. It is used to contain oil at a gauge
pressure of 35MPa. Determine the required thickness if the vessel material can withstand a
maximum tensile stress of 345MPa.

Solution
pr pr
Stress in the cylindrical part: σh = and σa =
t 2t
pr
Stress in the hemispherical ends: σ=
2t
35 × 10 6 × 0.075
The highest stress is σh: σh = ≤ 345 × 10 6 t ≥ 7.61 mm
t

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Tutorial Questions (set 5)

5.1 A single horizontal force of 900N is applied to end D of lever ABD. Knowing portion
AB has a diameter of 36mm: (a) determine the normal and shear stresses at point H,
(b) draw Mohr’s circle, (c) determine the principal planes and principal stresses at H.
Answer: (a) 0MPa, 58.9MPa, 53.1MPa; (c) 59.5o, –30.5o, 90.2MPa, −31.3MPa

y
200mm
B 150mm
540mm
300mm 2.5kN H
D
36 mm
120mm H
P 280N⋅m
A
z
Q5.1 x 2.5kN Q5.2

5.2 The axle of a car is acted upon by the forces and couple shown. Knowing that the
diameter of the axis is 32mm: (a) determine the normal and shear stresses at point H,
(b) draw Mohr’s circle, (c) determine the principal planes and principal stresses at
point H; (d) the maximum shear stress at the point.
Answer: (a) −116.57MPa, 0MPa, −43.52MPa;
(c) −71.6°, 18.4°, 14.42MPa, −131MPa; (d) 72.74MPa

5.3 For the plane stress shown in Q5.3, determine the value of the shear stress for which
the maximum shear stress is (a) 60MPa, (b) 78MPa.
Answer: (a) 40MPa, (b) 72MPa

40MPa 48MPa

52MPa

100MPa
τxy σz
126MPa
Q5.3 Q5.4

5.4 For the state of stress shown in Q5.4, determine two values of σz for which the
maximum shear stress is 80MPa.
Answer: −8MPa and 182MPa

5.5 Find (a) the stress invariants, (b) the principal stresses, and (c) the corresponding
principal directions at a material point with a state of stress as follows:
σx = 20 MPa σy = 40 MPa σz = 15 MPa
τxy = 0 MPa τxz = 10MPa τyz = 20MPa
Answer: (a) 75, 1200, 0; (b) 51.86MPa, 23.14 MPa, 0 MPa;
(c) (0.1581, 0.8493, 0.5036); (0.8990, −0.3348, 0.2823);
(0.4082, 0.4082, −0.8165)

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