UTM SKAA2223 - Stress Transformation
UTM SKAA2223 - Stress Transformation
UTM SKAA2223 - Stress Transformation
Materials and Structures
TOPIC 4
STRESS TRANSFORMATION
Lecturer:
Dr. Shek Poi Ngian
TOPIC 4
STRESS TRANSFORMATION
Analysis of Stress
• For this topic, the stresses to be considered
are not on the perpendicular and parallel
planes only but also on other inclined planes.
a b y
P P
x
z
a b
N=Pcos
P P
V=Psin
Analysis of Stress
P cos P
N cos 2 x cos 2
A A
cos
P sin P 1
sin cos x sin 2
A A 2
cos
Analysis of Stress
- On an element, there are 6 components of
stress :
x , y , z , xy , yz , and zx
x
z
y
y a xy
y’ x’
x x
x
a
All directions shown (axes
and on element) are taken y
as positive.
Stress Transformation
‐ Stresses on an element can be transformed using 2
methods:
i) Equations
ii) Mohr circle
i) EQUATIONS
- Consider an element rotated an amount of
about the z axis.
Stress Transformation
- Stresses on an inclined plane will be yielded and
can be expressed in term of x , y , xy and .
y’ x’
x’y’ x’
Sign Convention:
x x
Positive if counter-clockwise and A
usually taken from the vertical surface A cos xy A sin
(x plane) to the intended plane
xy (A sin) cos
Positive if tension or in the direction of
positive axis y
Positive if in the direction of positive
axis
Stress Transformation
Normal Stress:
x'
x y
x y
cos 2 xy sin 2
2 2
y'
x y
x y
cos 2 xy sin 2
2 2
Shear Stress:
sin 2
x y
cos 2
x' y' xy
2
Stress Transformation
In‐Plane Principle Stress:
2 xy
tan 2 p
x y
xy
2
2
2
Stress Transformation
Maximum In‐Plane Shear Stress:
tan 2 s
x y
2 xy
2
x y
max xy 2
2
Stress Transformation
Average Normal Stress:
When substituting the values for 2θs into the
equation for normal stress (σx’), there is also a
normal stress on the plane of maximum in-plane
shear stress, which can be determined by:
avg
x y
2
Stress Transformation
ii) Mohr circle
x y
2 2
x y
x '
x ' y '
2
xy
2
2 2
x x
xy
Y
max
y
min 2p(min)
max
(1) (2) yx
o
(x+y)/2 2p(max)
(+ve)
2s(max)
X yx
Y(+y, -yx)
(+ve) y
Example 1
The state of plane stress at a point on a body is shown
on the element in the Figure. Represent this stress
state in terms of the principal stresses.
90MPa
60MPa
x 20MPa
xy
y
Example 1 (cont.)
Solution
90MPa
60MPa
x = – 20MPa (Compression)
y = 90MPa (Tension) 20MPa
τxy = 60MPa (Clockwise)
xy
y
Example 1 (cont.)
Principal Stresses
xy
2
2
2
2
60
2 2
max and min 35 81.4
max 35 81.4 116MPa
min 35 81.4 46.4 MPa
Example 1 (cont.)
Orientation of Element
2 xy
tan 2 p
x y
2 60
tan 2 p
20 90
2 p 2 47.49
p 2 23.7
2 p1 47.49 180 132.51
p1 66.3
Example 1 (cont.)
The principal plane on which each normal stress acts
can be determined by applying:
x'
x y
x y
cos 2 xy sin 2
2 2
x'
20 90 20 90
cos(47.49) 60 sin(47.49)
2 2
x ' 46.4MPa
Hence, σmin = ‐46.4MPa acts on the
plane defined by θp2 = ‐23.7°,
whereas σmax = 116MPa acts on
the plane defined by θp1 = 66.3°.
Example 1 (cont.)
By replacing the θp1 and θp2 into the equation for shear
stress:
x' y'
x y
sin 2 xy cos 2
2
x' y'
20 90
sin( 47.49) 60 cos(47.49)
2
x' y' 0
No shear stress acts on this element.
Example 2
The state of plane stress at a point on a body is
represented on the element shown in the Figure.
Represent this stress state in terms of the maximum in‐
plane shear stress and associated average normal
stress. 90MPa
60MPa
x 20MPa
xy
y
Example 2 (cont.)
Maximum In‐Plane Shear Stresses
2
x y
max xy 2
2
2
20 90
max 60
2
2
max 81.4 MPa
Example 2 (cont.)
Orientation of Element
tan 2 s
x y
2 xy
tan 2 s
20 90
2 60
2 s 2 42.5
s 2 21.3
2 s1 180 42.5 222.5
s1 111.3
Worked Example
The proper direction of τmax on the element can be
determined by applying the equation:
x' y'
x y
sin 2 xy cos 2
2
x' y'
20 90
sin 42.5 60 cos 42.5
2
x' y' 81.4 MPa
Example 2 (cont.)
Average Normal Stress. Besides the maximum shear
stress, as calculated above, the element is also
subjected to an average normal stress determined
from the equation:
avg
x y
2
avg
20 90
2
avg 35MPa
Example 3
The state of plane stress at a point on a body is
represented on the element shown in the Figure.
Determine the stress components acting on the inclined
plane a‐a.
90MPa
60MPa a
x 20MPa
20°
xy
a
y
Example 3 (cont.)
Solution
Normal Stress:
x'
20 90 20 90
cos(70) 60 sin(70)
2 2
x ' 2.57 MPa (Compressive Stress) 70°
20°
Shear Stress:
x' y'
20 90
sin(70) 60 cos(70) 110°
2
x' y' 31.16 MPa (Opposite Direction)
Example 3 (cont.)
70°
x 20° 20MPa
60MPa
y
110°
90MPa
Example 4
The state of plane stress at a point on a body is
represented on the element shown in the Figure.
Determine the principal stresses acting at this point.
90MPa
60MPa
x 20MPa
xy
y
Example 4 (cont.)
Construction of the Circle.
x = – 20MPa (Compression)
y = 90MPa (Tension)
τxy = 60MPa (Clockwise)
The Centre of Circle is at
avg
x y
20 90
35MPa
2 2
The Radius of the Circle is
x y 20 90
2 2
R
2
xy 2
2 60 2
81.4 MPa
Example 4 (cont.)
2θpmax = 132.6°
60MPa
35MPa
20MPa
tan 2θp2 = 60 / 55
(-20, 60) 2θp2 = tan-1 1.09
2θp2 = 47.49°
θp2 = 23.74 °
Ψ = 80° – 33.69°
Example 4 (cont.)
Ψ = 46.31 ° (20, - 30)
At Point A,
A
2θsmax =
x’ = 20 – 36.06sin46.31° Ψ= 33.69°
x’ = - 6.07MPa 46.31° - 30MPa
x’y’ = 36.06cos46.31° σmin = - 2θpmax =
16.06MPa 56.31°
x’y’ = 24.91MPa
σmax = 56.06MPa
20MPa 20MPa
tan 2θs1 = 20 / - 30
tan 2θp2 = - 30 / 20
2θs1 = tan-1 0.67
2θp2 = tan-1 - 1.5
2θs1 = - 33.69°
2θp2 = - 56.31°
θs1 = - 16.85 ° θp2 = - 28.15 °
Ψ = 80° – 33.69°
Example 4 (cont.)
Ψ = 46.31 ° (20, - 30)
At Point A,
A (-6.07, -24.91)
2θsmax =
x’ = 20 – 36.06sin46.31° Ψ= 33.69°
x’ = - 6.07MPa 46.31° - 30MPa
x’y’ = 36.06cos46.31° σmin = - 2θpmax =
16.06MPa 56.31°
x’y’ = 24.91MPa
tan 2θp2 = - 30 / 20
σmax = 56.06MPa
2θp2 = tan-1 - 1.5
2θp2 = - 56.31°
20MPa 20MPa
θp2 = - 28.15 °
tan 2θs1 = 20 / - 30
2θs1 = tan-1 0.67
2θs1 = - 33.69° B (46.06, 24.91)
θs1 = - 16.85 °
Example 5
The state of plane stress at a point on a body is
represented on the element shown in the Figure.
Determine the maximum in‐plane shear stresses and
the orientation of the element upon which they act.
90MPa
60MPa
x 20MPa
xy
y
Example 5 (cont.)
tan 2θs1 = 55 / 60
2θs1 = tan-1 1.09
2θs1 = 42.5°
τmax = - 81.4MPa
θs1 = 21.3 °
2θs1 = 42.5°
35MPa
20MPa
(-20, 60) τmax = 81.4MPa
Example 6
The state of plane stress at a point on a body is
represented on the element shown in the Figure.
Represent this state of stress on an element oriented
30° counterclockwise from the position shown.
90MPa
60MPa
x 20MPa
xy
y
Example 6 (cont.)
tan 2θ = 55 / 60
2θ = tan-1 1.09 B
2θ = 42.5°
θ = 21.3 °
Ψ = 60° – 42.5°
Ψ = 17.5 °
At Point A,
x’ = 35 + 81.4sin17.5°
x’ = 59.48MPa
42.5°
x’y’ = 81.4cos17.5° x’y’
Ψ = 17.5°
x’y’ = 77.63MPa (-20, 60)
x’
A
20MPa 35MPa
Example 6 (cont.)
B x’
At Point B,
x’ = 81.4sin17.5° - 35
Ψ = 17.5°
x’ = -10.52MPa x’y’
x’y’ = - 81.4cos17.5°
x’y’ = - 77.63MPa
42.5°
x’y’
Ψ = 17.5°
(-20, 60)
x’
A
20MPa 35MPa
Example 6 (cont.)
30°
References
1. Hibbeler, R.C., Mechanics Of Materials, 8th Edition in SI
units, Prentice Hall, 2011.
2. Gere dan Timoshenko, Mechanics of Materials, 3rd Edition,
Chapman & Hall.
3. Yusof Ahmad, ‘Mekanik Bahan dan Struktur’ Penerbit UTM
2001