Fakultas Teknik
Jurusan Teknik Sipil
Principal Plane
At any point in a strained material, there are three planes, mutually perpendicular to each other, which carry direct stresses only and no shear stress. It shows that out of these three direct stresses one will be maximum the other minimum and the third an intermediate between the two. These particular planes, which have no shear stress are known as principal planes
Principal Stress
The magnitude of direct stress, across a principal plane, is known as principal stress
Method for Stresses on Oblique Section • Analytical method
Analytical Method for Stresses on an Oblique Section of a Body Subjected to a Direct Stresses in One Plane
Consider rectangular body ABCD of uniform cross-sectional area and unit thickness subjected to a principal tensile stress.
Let,
P = tensile force
A = cross sectional area
cos
sec
A
P
A
P
A
P
The intensity of tensile stress across the section EF:
The magnitude of tensile stress on section EF will be less than p, because the resisting section has a bigger area. But this stress is neither normal nor shear stress for section EF. Since the failure of this body will occur either by tension or by shear , it is therefore essential to know the normal and tangential stresses across the section EF.
cos
P
P
n
Normal stress across the section EF,
2cos
cos
cos
sec
cos
p
A
P
A
P
Area
Force
Tangential / shear stress across the section EF,
2
sin
2
cos
sin
sec
sin
p
A
P
A
P
Area
Force
p
t
Normal stress across the section EF will be maximum when cos2θ = 1 or θ
= 0°
Shear stress across the section EF will be maximum when sin 2θ = 1 or θ = 45° and 135 °
Maximum tangential stress:
2
1
2
2
sin
2
p
p
p
p
Max
t
Resultant stress:
2 2
t n
R
p
p
Solution :
2
cm
5
,
112
15
5
,
7
Area
Angle of the joint with the normal
Θ = 90° - 60 ° = 30 °
kg
P
cm
kg
p
p
p
p
cm
kg
p
p
p
p
cm
kg
p
cm
kg
p
t n
t n
1
,
1819
5
,
112
17
,
16
17
,
16
30
cos
30
sin
7
cos
sin
67
,
18
30
cos
14
cos
7
14
2 2
2 2
2 2
Analytical Method for Stresses on an Oblique Section of a Body Subjected to a Direct Stresses in Two Mutually
Perpendicular Direction
Let,
p1 = major tensile stress on the face AD and BC p2 = minor tensile stress on the face AB and CD
P1 = tensile force on the section EF (such that P1 = p1 x BC) P2 = tensile force on the section EF (such that P2 = p2 x GF)
Θ = angle which the oblique section makes with normal cross section EG
Tensile force perpendicular to plane EF
sin
cos
sin
cos
2 1
2 1
GF
p
BC
p
P
P
P
P
n n
Tensile force tangential to plane EF
cos
sin
cos
sin
2 1
2 1
GF
p
BC
p
P
P
P
P
n t
Tangential Stress across the section EF
2
sin
2
cos
sin
cos
sin
cos
sin
sin
/
cos
cos
/
sin
cos
sin
cos
sin
2 1
2 1
2 1
2 1
2 1
2 1
p
p
p
p
p
p
p
p
GF
GF
p
BC
BC
p
p
EF
GF
p
EF
BC
p
p
EF
GF
p
BC
p
EF
P
p
t t t t
t t
Resultant stress:
2 2
t n
R
p
p
A point in a strained material is subjected to two mutually perpendicular tensile stress of 2000 kg/cm2 and 1000 kg/cm2. Determine the intensities of normal and resultant stress on a plane inclined at 30° to the axis of the minor stress.
Example :
Solution :
2 2
1 2
1
2 2
0 2
1
1750
60
cos
2
1000
2000
2
1000
2000
2
cos
2
2
1000
30
2000
cm
kg
p
p
p
p
p
p
cm
kg
p
cm
kg
p
n n
Tangential stress
2 2
1
433
60
sin
2
1000
2000
2
sin
2
cm
kg
p
p
p
p
t t
Resultant stress
2 2
2 2
2
8
,
1082
433
1750
cm
kg
p
p
Analytical Method for Stresses on an Oblique Section of a Body Subjected to a Direct Stresses in One Plane Accompanied by a
Simple Shear Stress
Let,
p = tensile stress on the face AD and BC
q = tensile stress across the face AD and BC
Θ = angle which the oblique section makes with normal cross section EF
From the geometry , we find that the horizontal force acting on AD
p
AD
P
1Vertical force acting on AD
q
AD
P
2Horizontal force acting on GF
q
GF
P
3Normal Force across the section EF
sin
cos
cos
2 31
P
P
P
P
n
Tangential Force across the section EF
cos
sin
sin
2 31
P
P
P
The planes of maximum and minimum normal stress maybe found out by equating the tangential stress to zero
p
q
q
p
q
p
2
2
tan
2
cos
2
sin
2
1
0
2
cos
2
sin
2
1
2 2
2 2
2 2
2 2
1 2
2 1
4
2
os
4
2
2
sin
4
2
os
4
2
2
sin
q
p
p
c
q
p
q
q
p
p
c
q
p
q
Values of Principal Stresses maybe found out by substituting the above value of 2θ1 and 2θ2 in equation:
2 2
1
2 2
1
2 2
2
2 2
2
1 1 1
2
2
4
2
1
2
4
2
4
2
2
2
sin
2
2
cos
2
2
sin
2
cos
1
2
q
p
p
p
q
p
p
p
q
p
q
q
p
p
p
p
q
p
p
p
q
p
p
n n n n n
And,
2 2
2
2 2
2
2 2
2
2 2
2
2 2 2
2
2
4
2
1
2
4
2
4
2
2
2
sin
2
2
cos
2
2
sin
2
cos
1
2
q
p
p
p
q
p
p
p
q
p
q
q
p
p
p
p
q
p
p
p
q
p
p
n n n n n
A point in a strained material is subjected to a compressive stress of 800 kg/cm2 and a shear stress of 560 kg/cm2. Determine the maximum and minimum intensities of direct stress.
Example :
Solution :
'
14
27
'
28
54
2
4
,
1
800
560
2
2
2
tan
stress
e
compressiv
to
normal
with the
makes
plane
plane
principal
which the
angle
560
800
0 0
2 2
P
q
cm
kg
q
Minimum intensity of direct stress
2 2
2 1
2 2
1
2
,
288
560
2
800
2
800
2
2
cm
kg
p
q
p
p
p
n n
2 2
2
2
2 2
2
2
,
1088
2
800
2
800
2
2
cm
kg
q
p
q
p
p
p
n n
Analytical Method for Stresses on an Oblique Section of a Body Subjected to a Direct Stresses in Two Mutually Perpendicular
Direction Accompanied by a Simple Shear Stress
Let,
p1 = tensile stress on the face AD and BC p2 = tensile stress on the face AB and CD q = shear stress across the face AD and BC
Θ = angle which the oblique section makes with normal cross section EG
From the geometry , we find that the horizontal force acting on AD
q
AD
P
1Vertical force acting on AD:
P
2
q
AD
Horizontal force acting on GF:P
3
q
GF
Vertical force acting on GF:
P
4
p
2
GF
sin
cos
sin
cos
2 3 41
P
P
P
P
P
n
Tangential Force across the section EF
cos
sin
cos
sin
2 3 41
P
P
P
P
P
t
sin
2
cos
2
0
2
1
2
1
p
q
p
1 2
21 2
1
2
2
1
2
tan
2
cos
2
sin
2
1
p
p
q
p
p
q
q
p
p
There are two principal planes at the right angles to each other. Their inclination with the normal cross-section being θ1 and θ2 such that:
2 2 21
2 1
2
2 2
2 1
2
2 2
2 1
2 1
1
2 2
2 1
1
4
2
os
4
2
2
sin
4
2
os
4
2
2
sin
q
p
p
p
p
c
q
p
p
q
q
p
p
p
p
c
q
p
p
q
Values of Principal Stresses maybe found out by substituting the above value of 2θ1 and 2θ2 in equation:
The principal stress pn1 will be maximum whereas the stress pn2 will be minimum. The planes of maximum shear now be found out. These planes are at right angles of each other and are inclined at 45° to principal planes. The maximum shear stress will be given by the relation:
2
A point is subjected to a tensile stress of 60 N/mm2 and a compressive stress of 40 N/mm2, acting on two mutually perpendicular planes and a shear stress of 10 N/mm2 on these planes. Determine the principal stresses as well as maximum shear stress. Also find out the value of maximum shear stress.
Example :
Solution :
stress
principal
minor
stress
principal
major
10
stress
Shear
40
stress
Minor
60
2 1
2
2 2
2 1
n n
p
p
mm
N
q
mm
N
p
mm
N
2 2 2 1 2 2 2 1 2 1 1
61
10
2
40
60
2
40
60
2
2
mm
N
p
q
p
p
p
p
p
n n
on)
(compressi
41
10
2
40
60
2
40
60
2
2
2 2 2 2 2 2 2 1 2 1 2mm
N
p
q
p
p
p
p
p
n n
2 2 151
2
41
61
2
max
mm
N
p
p
p
t
n
n
A little
knowledge
that
acts
is
worth
infinitely
more
than
much
