ME2112 (Part 1) L2 - 5

Example 2.1 At a particular point A on the

surface of an aircraft making a landing, the
stresses have the magnitudes and directions
depicted on the stress element in Figure. Find
(a) The normal and shear stresses acting on an
inclined plane parallel to line a-a.
(b) The normal and shear stresses acting on an
inclined plane parallel to line b-b. In each
case, sketch the results on a properly
oriented element.

x y
y x  x  x
 y y
300 1200
a
b
0 450 600 x
45 x
b 300
a  y
(a)
y (b)

ME2112 (Part 1) L2 - 6

Solution: (a) Apply Eqs. (2.1) – (2.3) for   450 ,  x  10 MPa,  y  5 MPa
and  xy  6 MPa, we obtain
x y x  y
 x  cos 2   xy sin 2

2 2
10  5 10  5
  cos 900  6 sin 900  3.5 MPa
2 2
x y x  y
 y  cos 2   xy sin 2

2 2
10  5 10  5
  cos 900  6 sin 900
2 2
 8.5 MPa

x  y
 xy   sin 2   xy cos 2
2
10  5
 sin 900  6 cos 900  7.5 MPa
2
ME2112 (Part 1) L2 - 7

(b) As   300  900  1200 , from Eqs. (2.1) to (2.3), we have

x y x  y
 x   cos 2   xy sin 2
2 2
10  5 10  5
  cos 2400  6 sin 2400  3.95 MPa
2 2

x y x  y x y
 y   cos 2   xy sin 2  x
2 2 300 1200
10  5 10  5 b
  cos 2400  6 sin 2400 600 x
2 2 b 300
 y
 1.05 MPa y (b)

x  y
 xy   sin 2   xy cos 2
2
10  5
 sin 2400  6 cos 2400  9.5 MPa
2

ME2112 (Part 1) L2 - 8

Example 2.2 At a point on the surface of a machine the material is in biaxial stress with
 x  120 M Pa and  y   40 M Pa as shown in the Figure (a). Figure (b) shows an
inclined plane aa cut through the same point in the material but oriented at angle θ.
Determine all values of the θ between  90 0 and 90 0 such that no normal stresses act on
plane aa. For each angle θ, sketch a stress element having plane aa as one of its sides
and show all stresses acting on the element.

40 MPa
a

θ
x
120 MPa
a

(b)
(a)
ME2112 (Part 1) L2 - 9

Solution:  x  120 MPa,  y  40 MPa,  xy  0

a
Find angle θ
  y  x  y θ
 x  x  cos 2   xy sin 2
2 2
120  40 120  40
  cos 2  40  80 cos 2 a
2 2
For  x  0 , we obtain 0  40  80 cos 2 y

40 MPa
Hence, cos 2  (1/ 2)    600

Case 1:  x =0,   600

x
120 MPa
 x   y    x   y
 y   x   y   x  120  40  0  80

ME2112 (Part 1) L2 - 10

 x  y
 xy   sin 2   xy cos 2 y
x
2 80 MPa
120  40 y 600
 sin1200  0  69.3 MPa
2
x

Case 2:  x =0,   600

69.3 MPa
 x   y    x   y
 y   x   y   x  120  40  0  80
y
69.3 MPa y
 x  y
 xy   sin 2   xy cos 2 80 MPa
2
x
120  40
 sin(1200 )  0  69.3 MPa 600
2
x
ME2112 (Part 1) L2 - 21

Unlike the case of the principal planes on which no sharing stress occur, the max. shear stress planes are
usually not free of normal stress. The normal stresses are determined by introducing the values of 2s
into Eqs. (2.1) and (2.2):
( x   y ) / 2
 x  y  x  y sin 2 s 
 x   cos 2 s   xy sin 2 s 2
2 2   x  y 
   xy
2

 x  y  x  y  xy  xy ( x   y ) / 2  x   y  2 
   
2 2   x  y 
2
  x  y 
2 2  xy
  2
  2 cos 2 s 
    2
  x  y 
xy xy
 2   2 
   xy
2

 2 
 x  y  x  y
 y   cos 2 s   xy sin 2 s
2 2
 x  y  x  y  xy  xy ( x   y ) / 2  x  y
   
2 2   x  y 
2
  x  y 
2 2
    2
xy     2
xy
 2   2 

x y
 avg     (2.8c)
2

ME2112 (Part 1) L2 - 22

Example 2.3 At a particular point in a machine frame, the material is in a plane stress
condition with  x  7 MPa,  y  2 MPa,  xy  5 MPa acting on an element as shown in Fig.
(a). Find (a) The principal stresses. (b) The maximum shear stresses and the associated
normal stresses. Sketch the results found in (a) and (b) on properly oriented elements.

y  y  2 MPa

 xy  5 MPa

 x  7 MPa

(a)
ME2112 (Part 1) L2 - 23
x y
Solution: (a) The largest normal stresses. Using Eqs. (2.6a) and (2.6b)
121.7 0
2 x
x y   y   xy y
 1, 2    x    xy2 tan 2 p 
2  2  ( x   y ) / 2

2
72 72
     5  4.5  5.59
2
y
2  2  x
y 76.7 0
 1  10.09 MPa,  2  1.09 MPa x

1 2(5)
 p  tan 1  31.70 and 121.70
2 72
 ( x   y ) / 2
tan 2 s 
(b) Maximum shear stress. Using Eq. (2.7) & (2.8a)  xy

 72
2
 y  1
 max   x    xy2  s  tan 1     13.3 and 76.7
0 0
The plane of max. positive shear
 2 
2  25  stress  max is defined by angle  s1
2
x y 7  2  s1   p  450  31.70  450  13.30
72
    5  5.59 MPa      4.5 MPa
2

 2  2 2  s   p  450  31.70  450  76.70

2

ME2112 (Part 1) L2 - 24

Example 2.4 A shear wall in a reinforced concrete building is subjected to a vertical uniform
load of q and a horizontal force H, as shown in the Fig. (a). (The force H represents the
effects of wind and earthquake loads.) As consequence of these loads, the stresses at point A
on the surface of the wall have the values shown in Fig. (b) (compressive stress equals to 10
MPa and shear stress equals to 2 MPa. (a) Determine the principal stresses and show them on
a sketch of a properly oriented element. (b) Determine the maximum shear stresses and
associated normal stresses and show them on a sketch of a properly oriented element.
q
y 10 MPa
H
A 2 MPa

A 2 MPa

(a) (b)
ME2112 (Part 1) L2 - 25

Solution:  x  0 MPa,  y  10 MPa,  xy  2 MPa

(a) Principal stresses
2 xy 2(2)
tan 2 p    0.4
 x  y 0  (10)

2 p  21.800 and  p =  10.900

y 10 MPa
2 p  180  21.80 =158.20 and  p =79.10
0 0 0 0
2 MPa

A
x  y  x  y 2 MPa

 x   cos 2   xy sin 2
2 2 x

0  10 0  10
For 2 p  21.800 :  x   cos(21.800 )  (2) sin(21.809 )  0.4 MPa
2 2
0  10 0  10
For 2 p =158.200 :  x   cos(158.200 )  (2) sin(158.209 )  10.4 MPa
2 2
  y  x  y  x  y  x  y
 x  x   90 0   y  
cos 2   xy sin 2    cos 2   xy sin 2
2 2 2 2

ME2112 (Part 1) L2 - 26

Hence,  1  0.4 MPa and  p  10.900 y y

5 MPa y
y
 2  10.4 MPa and  p =79.10 0 10.4 MPa 5 MPa
34.100
0
(b) Maximum shear stresses. 79.10 A
A x x
2 10.900
  x  y  x
 max      xy
2 0.4 MPa 5.4 MPa 55.90
 2 
x
2
 0  (10)  Max. positive
    (2)  5.4 MPa
2
shear stress
 s   p  450  10.90  450  55.90
 2 
1

 s   p  450  10.90  450  34.10

2

x  y 0  10 124.10 0 y
    avg    5 MPa 5 MPa
2 2 5 MPa
34.100
A Max. positive
1  0  (10)  x shear stress
 s  tan 1     34.1 and 124.1
0 0

2  2  ( 2)   s   p  450  79.10  450  34.10

1
5.4 MPa
 s   p  450  79.10  450  124.10
2
ME2112 (Part 1) L2 - 31

READING QUIZ
1) Which of the following statement is incorrect?

a) The principal stresses represent the maximum and minimum normal

stress at the point.

b) When the state of stress is represented by the principal stresses, no

shear stress will act on the element.

c) When the state of stress is represented in terms of the maximum in-plane

shear stress, no normal stress will act on the element.

d) For the state of stress at a point, the maximum in-plane shear stress
usually associated with the average normal stress.

ME2112 (Part 1) L2 - 32

READING QUIZ
2) Which of the following statement is untrue?
a) In 2D state of stress, the orientation of the element representing the
maximum in-plane shear stress can be obtained by rotating the
element 45° from the element representing the principle stresses.
b) In 2D state of stress, the sum of the normal stresses on two
perpendicular planes is dependent on θ, angle of rotation of element
with respect to the horizontal direction.
c) If the in-plane principal stresses are of opposite sign, then the
maximum in-plane shear stress is given by  max  ( 1   2 ) / 2
d) The radius of Mohr’s circle in 2D state of stress is equal to maximum
in-plane shear stress.
ME2112 (Part 1) L3 - 15

Example 2.5 The state of plane stress at a point (in a pressure vessel) is represented
by the figure shown. Determine the stresses on an element oriented at 300 counter-
clockwise from the position shown. Illustrate your answer on a diagram.

12 MPa

300
8 MPa x
x
6 MPa

ME2112 (Part 1) L3 - 16

Solution: From Eqs. (2.1) to (2.3), we have y

 y x y 12 MPa
 x  x  cos 2   xy sin 2
2 2
x y x y 8 MPa
 y   cos 2   xy sin 2 x
2 2 6 MPa
x y  x  8 MPa
 xy   sin 2   xy cos 2
2  y  12 MPa
Substituting   300 , we obtain  xy  6 MPa

 x  ( 8  12) / 2  [( 8  12) / 2] cos 600  ( 6) sin 600  8.2 MPa

 y  ( 8  12) / 2  [( 8  12) / 2] cos 600  ( 6) sin 600  12.2 MPa
 xy  [( 8  12) / 2] sin 600  ( 6) cos 600  5.66 MPa
To construct the Mohr’s circle
 avg  ( x   y ) / 2  ( 8  12) / 2  2 MPa (Center of circle)
2
  y 
2
  8  12 
R   x    xy2     ( 6)  11.66 MPa (Radius of circle)
2

 2   2 
ME2112 (Part 1) L3 - 17

 avg  2 MPa, R  11.66 MPa Question: Where should point A be ?


12  y  12 MPa
8 B
Clockwise,
10
+v shear y
A  x  8 MPa
x 8
 y
A B’ x
R  11.66 6 2  xy  6 MPa
4
6
Point A? 2
600
-8 -6 -4 -2 0 2C 4 6 8 10 12 14 
Point A?  y  xy  x
-2
6 B’ 300
A’ -4
B A’ x
 x -6 y
12
-8
-10

-12

ME2112 (Part 1) L3 - 18

From the Mohr’s circle,

 x  8.2 MPa 
12
 y  12.2 MPa Clockwise,
8
+v shear 10
 xy  5.66 MPa 12.2
x 8
 y
A B’
 y  12 MPa R  11.66 6 2
4
B 6
5.66
y 2
A  x  8 MPa
0
60
-8 -6 -4 -2 0 2C 4 6 8 10 12 14
x 
-2
5.66 6
 xy  6 MPa -4
 x A’
B
-6 y
12
-ve shear stress at 8.22 -8
this point -10
12.2 5.66 8.2
-12
B’ 300
y
x
A’ x
-ve shear stress on Mohr’s circle, hence anticlockwise direction
on the element
ME2112 (Part 1) L4 - 11

Vessels for pressurized storage such as oil refinery tanks and gas tanks etc.

ME2112 (Part 1) L4 - 12

Example 2.6
A hollow pressurized sphere having
radius R = 150 mm and wall
thickness t = 13 mm is lowered in
a lake. The compressed air in the
tank is at a pressure of 140 kPa
(gauge pressure when the tank is
out of the water).

At what depth D0, will the wall of the

tank be subjected to a compressive
stress of 700 kPa?
ME2112 (Part 1) L4 - 13

Solution. Characteristic parameter t/R = 13mm/150mm = 0.087

(a) In air, pi = 140 kPa, pe = 0
Under water, pi is maintained, pe increases
R
 1   2    ( pi  pe ) , pe  D0
2t
  9.81 kN/m3 (weight density of water)

For compressive stress of 700 kPa

R
( pi  D0 )  700 kPa
2t
Substitute numerical values:

150 103
(140 103  9.81103 D0 ) 3
 700 103
2  13 10
 D0 = 26.64 m

ME2112 (Part 1) L4 - 14

Discussion 1: What can we learn from this example?

Fuselage of aircraft

r
Tangential stress:  t  ( pi  pe )
t
On ground: pi  pe
r
High in sky: pi is maintained,

pe decreases
ME2112 (Part 1) L4 - 15

Discussion 2: t

The maximum tangential stress in the wall of

the tank (with an open top) due to water
pressure occurs
(A) at top of the standpipe
(B) at the middle of the standpipe
(C) at the base of the standpipe
The axial stress in the wall of the tank due to
the water pressure is
(A) > 0
(B) = 0
(C) < 0

ME2112 (Part 1) L4 - 16

Example 2.7 A closed-end cylindrical tank is constructed with a helical weld that makes an
angle   50 about the longitudinal axis and supported by two cradles as shown in the Fig.
0

(a). The vessel has inner radius r  2 m, wall thickness t  10 mm, and is subjected to an
internal pressure of p  500 kPa. Find (a) The tangential and axial stresses. (b) The normal
and shear stresses acting perpendicular and parallel to the weld, respectively.

 y  xy

 x

t

a
ME2112 (Part 1) L4 - 17

Solution: (a) The tangential and longitudinal stresses are given by Eqs. (2.11) and (2.12).
pr 500(103 )( 2) 
t    100 MPa,  a  t  50 MPa
t 0.01 2
The stresses are indicated in Fig. (b).
(b) An element oriented at   900  500  400 has its sides parallel and perpendicular to the
weld as shown in Fig. (c). Observing that there are both tangential and axial stresses, we
draw Mohr’s circle as shown in the Fig. (d).

y
x
0
50 40 0

ME2112 (Part 1) L4 - 18

So, a counterclockwise angle of 2  800 on the circle locates point Bcorresponding to the stresses
on the x face. The average normal stress and radius of circle are
1 1
 avg  (100  50)  75 MPa, R  (100  50)  25 MPa
2 2
Hence, tensile stress:
The coordinates of point B are then
 x  70.7 MPa
 x   avg  R cos 2  75  25 cos 80  70.7 MPa
0

 y  79.3 MPa
 xy  R sin 2  25 sin 800  24.6 MPa Shear stress:
Similarly,  y   avg  R cos 2  75  25 cos 80  79.3 MPa
0  xy  24.6 MPa

Normal stress in y’ axis

A
B
y

t
a 800

A1 D E

B1

x
(d)
ME2112 (Part 1) L4 - 27

Example 2.8 The strain components at a point in a machine member are given by
 x  900  ,  y  100  ,  xy  600 Using Mohr’s circle, determine the principal strains and the
maximum shearing strains.   10  6

Solution: Center of circle:

y  y  100

x  y 900  ( 100) B
 avg    400   xy   x  900
2 2
600  A

x
Radius of circle:

2 2
    y    xy 
2 2
 900  100   600 
R   x           583 
 2   2   2   2 

ME2112 (Part 1) L4 - 28


2 (½) Max. shearing strain
y  y  100
B(-100,300)
R  583 B
(983, 0)
(-183, 0) C x  xy   x  900
2  y 0 2 p 1  600  A
A(900,-300) x

400
On plane A,  x  900 ,on plane A the shear
direction is anticlockwise, hence it is indicated
as a –ve shear strain on the Mohr’s circle
From the Mohr’s circle   xy / 2  600 / 2  300 
 1  983  ,  2   183 
On plane B,  y  100  , on plane B the shear
Maximum shearing strain direction is clockwise, hence it is indicated as a
 max  1166  max / 2  583   max  2  583 +ve shear strain on the Mohr’s circle
 xy / 2  600 / 2  300