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Fakultas TeknikJurusan Teknik SipilUniversitas Brawijaya Malang
Principal Plane
At any point in a strained material, there are three planes, mutuallyperpendicular to each other, which carry direct stresses only and no shearstress. It shows that out of these three direct stresses one will bemaximum the other minimum and the third an intermediate between thetwo. These particular planes, which have no shear stress are known asprincipal planes
Principal Stress
The magnitude of direct stress, across a principal plane, is known asprincipal stress
Method for Stresses on Oblique Section
• Analytical method• Graphical 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
Θ = angle which the oblique section makes with normal cross section EF
cossec 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.
cosPPn
Normal stress across the section EF,
2coscoscossec
cosp
A
P
A
P
Area
Forcepn
Tangential / shear stress across the section EF,
2sin
2cossin
sec
sin p
A
P
A
P
Area
Forcept
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:
21
22sin
2
ppppMax t
Resultant stress:
22
tnR ppp
A tension member is formed by connecting with glue two wooden scantingeach 7,5 x 15 cm at their end which are cut at an angle of 60°. The member issubjected to a pull P. Calculate safe value of P if permissible normal andshear stress in glue are 14 kg/cm2 and 7 kg/cm2 respectively.
Solution :
2cm 5,112155,7 Area
Angle of the joint with the normal
Θ = 90° - 60 ° = 30 °
kgP
cmkg
pp
pp
cmkg
pp
pp
cmkg
pcm
kgp
t
n
tn
1,18195,11217,16
17,1630cos30sin7
cossin
67,1830cos14
cos
7 14
2
22
2
22
Analytical Method for Stresses on an Oblique Section of a Body Subjected to a Direct Stresses in Two Mutually
Perpendicular Direction
Consider rectangular body ABCD of uniform cross-sectional area and
unit thickness subjected to mutually perpendicular principal tensile
stresses on the face AB, CD, and AD, BC. Also consider an oblique
section EF on which we are required to find out the stresses.
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
sincos
sincos
21
21
GFpBCpP
PPP
n
n
Tensile force tangential to plane EF
cossin
cossin
21
21
GFpBCpP
PPP
n
t
Normal Stress across the section EF
2cos22
2cos12
2cos12
sincos
sin/
sin
cos/
cos
sincos
sincos
2121
21
2
2
2
1
21
21
21
ppppp
ppp
ppp
GF
GFp
BC
BCpp
EF
GFp
EF
BCpp
EF
GFpBCp
EF
Pp
n
n
n
n
n
nn
Tangential Stress across the section EF
2sin2
cossin
cossincossin
sin/
cos
cos/
sin
cossin
cossin
2121
21
21
21
21
ppppp
ppp
GF
GFp
BC
BCpp
EF
GFp
EF
BCpp
EF
GFpBCp
EF
Pp
t
t
t
t
tt
Resultant stress:
22
tnR ppp
A point in a strained material is subjected to two mutuallyperpendicular tensile stress of 2000 kg/cm2 and 1000 kg/cm2.Determine the intensities of normal and resultant stress on a planeinclined at 30° to the axis of the minor stress.
Example :
Solution :
2
2121
22
021
175060cos2
10002000
2
10002000
2cos22
1000
30 2000
cmkg
p
ppppp
cmkg
p
cmkg
p
n
n
Tangential stress
2
21
43360sin2
10002000
2sin2
cmkg
p
ppp
t
t
Resultant stress
22222
8,10824331750cm
kgppp tnR
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
Consider rectangular body ABCD of uniform cross-sectional area and
unit thickness subjected to tensile stress in one plane accompanied by
a shear stress across the face AD, and BC. Also consider an oblique
section EF on which we are required to find out the stresses.
Let,
p = tensile stress on the face AD and BC
q = tensile stress across the face AD and BCΘ = angle which the oblique section makes with normal cross section EF
From the geometry , we find that the horizontal force acting on AD
ADpP1
Vertical force acting on AD
ADqP2
Horizontal force acting on GF
GFqP3
Normal Force across the section EF
cossincos 321 PPPPn
Tangential Force across the section EF
sincossin 321 PPPPt
Normal Stress across the section EF
2sin2cos12
cossin2cos
cossincossincos
sin/
cos
cos/
sin
cos/
cos
cossincos
cossincos
2
2
321
321
qp
p
qpp
qqpp
GF
GFq
AD
ADq
AD
ADpp
EF
P
EF
P
EF
Pp
EF
PPP
EF
Pp
n
n
n
n
n
nn
Tangential Stress across the section EF
2cos2sin2
1
sincos2sin2
1
sincoscossin
sin/
sin
cos/
cos
cos/
sin
sincossin
sincossin
22
22
321
321
qpp
qpp
qqpp
GF
GFq
AD
ADq
AD
ADpp
EF
P
EF
P
EF
Pp
EF
PPP
EF
Pp
t
t
t
t
t
tt
The planes of maximum and minimum normal stress maybe found out by equating the tangential stress to zero
p
qqp
qp
22tan2cos2sin
2
1
02cos2sin2
1
222
222
221
221
42 os
4
22sin
42 os
4
22sin
qp
pc
qp
q
qp
pc
qp
q
There are two principal planes, at right angle to each other. Their inclination with the normal cross section being θ1 and θ2 such that:
Values of Principal Stresses maybe found out by substituting the above value of 2θ1 and 2θ2 in equation:
2
2
1
22
1
22
2
22
2
1
1
1
22
42
1
2
4
2
422
2sin2
2cos
2
2sin2cos12
qpp
p
qpp
p
qp
q
qp
ppp
qpp
p
qp
p
n
n
n
n
n
And,
2
2
2
22
2
22
2
22
2
2
2
2
22
42
1
2
4
2
422
2sin2
2cos
2
2sin2cos12
qpp
p
qpp
p
qp
q
qp
ppp
qpp
p
qp
p
n
n
n
n
n
A point in a strained material is subjected to a compressive stressof 800 kg/cm2 and a shear stress of 560 kg/cm2. Determine themaximum and minimum intensities of direct stress.
Example :
Solution :
'1427'28542
4,1800
560222tan
stress ecompressiv tonormal with themakes
plane plane principal which theangle
560
800
00
2
2
P
q
cmkg
q
cmkg
p
Minimum intensity of direct stress
22
2
1
2
2
1
2,2885602
800
2
800
22
cmkg
p
qpp
p
n
n
22
2
2
2
2
2
2,10882
800
2
800
22
cmkg
qp
qpp
p
n
n
Maximum intensity of direct stress
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
Consider rectangular body ABCD of uniform cross-sectional area and
unit thickness subjected to tensile stress and shear stress. Consider an
oblique section EF on which we are required to find out the stresses.
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 BCΘ = angle which the oblique section makes with normal cross section EG
From the geometry , we find that the horizontal force acting on AD
ADqP1
Vertical force acting on AD: ADqP2
Horizontal force acting on GF: GFqP3
Vertical force acting on GF: GFpP 24
sincossincos 4321 PPPPPn
Tangential Force across the section EF
cossincossin 4321 PPPPPt
Normal Force across the section EF
Normal Stress across the section EF
2sin2cos22
2
2cos
22sin
2
2cos
2
2cos12
2sin2cos12
sincossin2cos
sincossincossincos
sin/
sin
sin/
cos
cos/
sin
cos/
cos
sincossincos
sincossincos
2121
2211
21
2
2
2
1
2
2
2
1
21
4321
4321
qpppp
p
ppq
ppp
pq
pp
pqpp
pqqpp
GF
GFp
GF
GFq
AD
ADq
AD
ADpp
EF
P
EF
P
EF
P
EF
Pp
EF
PPPP
EF
Pp
n
n
n
n
n
n
n
nn
Tangential Stress across the section EF
2cos2sin2
1
cossincossin
cossinsincoscossin
sin/
cos
sin/
sin
cos/
cos
cos/
sin
cossincossin
cossincossin
21
22
21
2
22
1
21
4321
4321
qppp
qppp
pqqpp
GF
GFp
GF
GFq
AD
ADq
AD
ADpp
EF
P
EF
P
EF
P
EF
Pp
EF
PPPP
EF
Pp
t
t
t
t
t
tt
02cos2sin2
121 qpp
2121
21
2
2
12tan
2cos2sin2
1
pp
q
pp
q
qpp
Maximum and minimum normal stress may be found out by
equating the tangential stress to zero
There are two principal planes at the right angles to each other. Theirinclination with the normal cross-section being θ1 and θ2 such that:
22
21
212
22
21
2
22
21
211
22
21
1
42 os
4
22sin
42 os
4
22sin
qpp
ppc
qpp
q
qpp
ppc
qpp
q
Values of Principal Stresses maybe found out by substituting the above value of 2θ1 and 2θ2 in equation:
2
2
21211
22
2121
1
22
21
2
22
21
2
21211
22
21
22
21
2121211
21211
22
42
1
2
4
2
422
4
2
422
2sin2cos22
qpppp
p
qpppp
p
qpp
q
qpp
ppppp
qpp
qpp
ppppppp
qpppp
p
n
n
n
n
n
2
2
21212
22
2121
2
22
21
2
22
21
2
21212
22
21
22
21
2121212
21212
22
42
1
2
4
2
422
4
2
422
2sin2cos22
qpppp
p
qpppp
p
qpp
q
qpp
ppppp
qpp
qpp
ppppppp
qpppp
p
n
n
n
n
n
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:
2max 21 nn
t
ppp
A point is subjected to a tensile stress of 60 N/mm2 and acompressive stress of 40 N/mm2, acting on two mutuallyperpendicular planes and a shear stress of 10 N/mm2 on theseplanes. Determine the principal stresses as well as maximum shearstress. Also find out the value of maximum shear stress.
Example :
Solution :
stress principalminor
stress principalmajor
10 stressShear
40 stressMinor
60
2
1
2
22
21
n
n
p
p
mmNq
mmNp
mmNp
22
2
1
2
2
21211
61102
4060
2
4060
22
mmNp
qpppp
p
n
n
on)(compressi41102
4060
2
4060
22
22
2
2
2
2
21212
mmNp
qpppp
p
n
n
221 51
2
4161
2max
mmNpp
p nnt
Maximum shear stress:
A little knowledge thatacts is worth infinitely more than much knowledge that is idle.
