MECHANICS OF MATERIALS
CHAPTER
7 Transformations of Stress and Strain
MECHANICS OF MATERIALS
7 - 2
Transformations of Stress and Strain
IntroductionTransformation of Plane StressPrincipal StressesMaximum Shearing StressSample Problem 1Sample Problem 2Mohr’s Circle for Plane StressSample Problem 3Sample Problem 4Transformation of Plane StrainStresses in Thin-Walled Pressure Vessels
MECHANICS OF MATERIALS
7 - 3
Introduction• General state of stress at a point represented by
6 components,
),, :(Note
stresses shearing,,
stresses normal,,
xzzxzyyzyxxy
zxyzxy
zyx
• If axis are rotated, the same state of stress is represented by a different set of components, i.e., the stress components get transformed.
• First we consider transformation of stress components, due to rotation of coordinate axes. Then we consider a similar transformation of strain components.
),, :(Notestresses shearing,,
stresses normal,,
''''''''''''
''''''
'''
zxxzyzzyxyyx
xzzyyx
zyx
MECHANICS OF MATERIALS
7 - 4
Plane Stress
• Plane Stress - state of stress in which two faces of the cubic element are free of stress. Example,
.0nonzero, are,, xy zyzxzyx
• For example, state of plane stress occurs in thin plate subjected to forces acting in midplane of plate.
• Another example of plane stress is on free surface, i.e., unloaded point on surface.
MECHANICS OF MATERIALS
7 - 5
Transformation of Plane Stress
sinsincossin
coscossincos0
cossinsinsin
sincoscoscos0
AA
AAAF
AA
AAAF
xyy
xyxyxy
xyy
xyxxx
• Consider equilibrium of prismatic element with faces perpendicular to x, y, and x’ axes.
cos 2 sin 22 2
cos 2 sin 22 2
sin 2 cos 22
x y x yx xy
x y x yy xy
x yx y xy
• Solving for transformed stress components,
MECHANICS OF MATERIALS
7 - 6
Principal Stresses• Eliminating q , this yields equation of a
circle,
22
222
22
where
xyyxyx
ave
yxavex
R
R
• Principal stresses occur on principal planes on which there exist zero shearing stresses.
o
''
22
minmax,
90by separated angles twodefines :Note
0from,2
2tan
22
yxyx
xyp
xyyxyx
MECHANICS OF MATERIALS
7 - 7
Maximum Shearing StressMaximum shearing stress occurs for avex
2
:are
45by fromoffset and 90by separated angles twodefines :Note
from,2
2tan
22
''
o
o
ave'
minmax22
max
yxaveyx
p
xxy
yxs
xyyx
stressesnormalingCorrespond
R
MECHANICS OF MATERIALS
7 - 8
Mohr’s Circle for Plane Stress
• Used to graphically find principal stresses and planes and maximum shear stresses and planes
• For known plot points X and Yand construct circle centered at C.
xyyx ,,
22
22 xyyxyx
ave R
• Principal stresses obtained at A and B.
yx
xyp
ave R
22tan
minmax,
Direction of rotation of Ox to Oa (ie., in physical plane) is same as CX to CA(ie., in Mohr plane)
MECHANICS OF MATERIALS
7 - 9
Mohr’s Circle for Plane Stress
• From Mohr’s circle we can find state of stress at other axes orientations.
• For state of stress at angle q with respect to the xy axes, construct a new diametral line X’Y’ at angle 2q with respect to XY.
• Coordinates of X’, Y’ are the transformed normal and shear stresses.
MECHANICS OF MATERIALSProof of Mohr’s Circle construction
7 - 10
2sin2cos2sin2
2cos
)2sin2cos2cos2(sin)22sin(
2sin2cos22
)22cos(
2sin2cos22
2sin2cos2
)2sin2sin2cos2(cos)22cos(
''
'
'
RRRR
RR
R
RRR
RR
yxxy
yxxy
pppyx
xyyxyx
pavey
xyyxyxxyyx
ave
ppavepavex
So we get same formulae as before
MECHANICS OF MATERIALS
7 - 11
Mohr’s Circle for Plane Stress• Mohr’s circle for centric axial loading:
0, xyyx AP
(plane)element45on2
0
max A
Pxyyx
• Mohr’s circle for torsional loading:
JTc
xyyx 0(plane)element45on
0;
0
max xyyx JTc
MECHANICS OF MATERIALSPoints to note when drawing stress block.
7 - 12
plane.on which acts stress principalwhich identify not do they However,).,,(act they on which planes principal theand ve),–or ve( sense/sign
and magnitude their yield stresses principalfor Formulae
minmax p
plane.on which acts stress principalhich identify w torelations
ation transformin the angles put the , angles associated and, stresses principal findyou once So
p
minmax
MECHANICS OF MATERIALSPoints to note when drawing stress block.
7 - 13
. of ve)-or ve( sense/sign theyieldnot do they However, act.hey on which t planes and of
magnitudeonly yield stressshear maximumfor Formulae
max
smax
. of ve)-or ve( sense/signcorrect find to relationsation transformin theit put , findyou once So
max
s
sense/signcorrect with , , associated find torelations
rmation in transfo themuse and and findjust can you So
maxmaxmin
ps
MECHANICS OF MATERIALSPoints to note when drawing stress block.
7 - 14
act.hey on which t
, planes theand,,, of sense and magnitudescorrect gives which circleMohr usecan you y,Alternatel
spmaxminmax
MECHANICS OF MATERIALS
7 - 15
Example1
For state of plane stress shown, find (a) principal planes, (b) principal stresses, (c) maximum shearing stress and corresponding normal stress.
MECHANICS OF MATERIALS
7 - 16
Example 1
• Find element orientation for principal stresses:
1.233,1.532
333.11050
40222tan
p
yx
xyp
6.116,6.26p
• Find principal stresses:
22
22
minmax,
403020
22
xy
yxyx
MPa30MPa70
min
max
MPa10
MPa40MPa50
y
xyx
MECHANICS OF MATERIALS
7 - 17
Example 1
MPa10MPa40MPa50
y
xyx
21050
2''
yxaveyx
• Find corresponding normal stress:
MPa20
• Find maximum shearing stress:
22
22
max
4030
2
xy
yx
MPa50max
45 ps
6.71,4.18s
MECHANICS OF MATERIALS
7- 18
Example 2
Horizontal force P = 600 N magnitude applied to end D of lever ABD. Find (a) normal and shearing stresses on element at H having sides parallel to x and y axes, (b) principal planes and principal stresses at H.
MECHANICS OF MATERIALS
7- 19
Example 2
Find equivalent force-couple system at center of transverse section passing through H.
Nm150m25.0N600
Nm270m45.0N600N600
xMTP
Find normal and shearing stresses at H.
421
441
m015.0m015.0Nm270
m015.0m015.0Nm150
JTcI
Mc
xy
y
MPa9.50MPa6.560 yyx
Note: tyz due to bending is zero at H
MECHANICS OF MATERIALS
7- 20
Example 2Find principal stresses and planes.
119,0.612
8.16.560
9.50222tan
p
yx
xyp
5.59,5.30p
22
22
minmax,
9.502
6.5602
6.560
22
xy
yxyx
MPa9.29MPa5.86
min
max
MECHANICS OF MATERIALS
7 - 21
Example 3
For state of plane stress shown, (a) construct Mohr’s circle, find (b) principal planes, (c) principal stresses, (d) maximum shearing stress and corresponding normal stress.
• Construct Mohr’s circle
MPa504030
MPa40MPa302050
MPa202
10502
22
CXR
FXCF
yxave
MECHANICS OF MATERIALS
7 - 22
Example 3• Principal planes and stresses
5020max CAOCOA
MPa70max
5020max BCOCOB
MPa30max
1.53230402tan
p
p CFFX
6.26p
MECHANICS OF MATERIALS
7 - 23
Example 3
• Maximum shear stress
45ps
6.71s
Rmax
MPa 50max
ave
MPa 20
MECHANICS OF MATERIALS
7 - 24
Example 4
For state of plane stress shown, find (a) principal planes and the principal stresses, (b) stress components on element obtained by rotating given element counterclockwise through 30 degrees. • Construct Mohr’s circle
MPa524820
MPa802
601002
2222
FXCFR
yxave
MECHANICS OF MATERIALS
7 - 25
Example 4
• Principal planes and stresses
4.672
4.220482tan
p
p CFXF
clockwise7.33 p
5280max
CAOCOA5280
max
BCOCOA
MPa132max MPa28min
MECHANICS OF MATERIALS
7 - 26
Example 4
6.52sin52
6.52cos52806.52cos5280
6.524.6760180
XK
CLOCOLKCOCOK
yx
y
x
• Stress components after counterclockwise rotation by 30o
Points X’ and Y’ on Mohr’s circle, that correspond to stress components on rotated element, are obtained by rotating XY ccw through 602
MPa3.41
MPa6.111MPa4.48
yx
y
x
MECHANICS OF MATERIALS
7 - 27
Transformation of Plane Strain• Plane strain - deformations of the material
take place in parallel planes and are the same in each of those planes.
• Example: Long bar subjected to uniformly distributed transverse loads (ie., normal to z-axis). State of plane strain exists in any transverse section not located too close to the ends of the bar.
• Plane strain occurs in a plate subjected along its edges to a uniformly distributed (in z-direction) load and restrained from expanding or contracting laterally by smooth, rigid and fixed supports
0
:strain of components
x zyzxzxyy
MECHANICS OF MATERIALS
7 - 28
Transformation of Plane Strain• State of strain at point Q results in
different strain components with respect to the xy and x’y’ coordinate systems.
• We get strain transformation relations similar to those for stress transformation (see details in next two slides)
MECHANICS OF MATERIALSTransformation of Plane Strain
7 - 29
2sin2
2cos22
cossinsincos)(
'
22'
xyyxyxx
xyyxx
,sin/,cos/,)()()(Use
)(2)21()()21()()21()(
)2/cos()1()1(2
)1()()1()()1()(strains,inquadraticsneglectingrule,cosineUse
222
22'
2
22222'
2
sysxyxs
yxyxs
yxyxs
xyyxx
xyyx
yxx
2sin2
2cos22'
xyyxyxy
,in2/put 'x
MECHANICS OF MATERIALSTransformation of Plane Strain
7 - 30
2cos2
2sin22
'' xyyxyx
xyyxxOB
21
' )4/(,4/For
OyOxOB
yxOBxy
andofbisectoriswhere
2
2cos2
2sin22
)4/(
Thus,'.and'ofbisectoris'where
22system,inThus,
''
''''''
xyyxyxxOB
yxOByxOByx
OyOxOB
x'y'
''''
'''' 2but;;:Note yx
yxyyxx
MECHANICS OF MATERIALS
7 - 31
Mohr’s Circle for Plane Strain• Since strain transformation relations are of
same form as stress transformation, for plane problems, Mohr’s circle techniques apply.
• Abscissa for center C, and radius R , are22
222
xyyxyx
ave R
• Principal axes of strain and principal strains,
RR aveave
yx
xyp
minmax
2tan
22max 2 xyyxR
• Maximum in-plane shearing strain,
MECHANICS OF MATERIALSDrawing strain block.
7 - 32
MECHANICS OF MATERIALSDrawing strain block.
7 - 33
MECHANICS OF MATERIALS
7 - 34
Measurements of Strain: Strain Rosette
• Strain gages indicate normal strain through changes in resistance.
yxOBxy 2
• With a 45o rosette, ex and ey are measured directly. gxy is obtained indirectly with,
3332
32
3
2222
22
2
1112
12
1
cossinsincos
cossinsincos
cossinsincos
xyyx
xyyx
xyyx
• Normal and shearing strains may be obtained from normal strains in any three directions,
MECHANICS OF MATERIALS
7 - 35
Stresses in Thin-Walled Cylindrical Pressure Vessels
• Cylindrical vessel with principal stressessq=s1 = hoop stresssx = s2 = longitudinal stress
tpr
xrpxtFz
1
1 220
• Hoop stress:
21
2
22
22
20
tpr
rprtF
x
x
• Longitudinal stress:
MECHANICS OF MATERIALS
7- 36
Stresses in Thin-Walled Cylindrical Pressure Vessels
Maximum in-plane shearing stress:
tpr42
12)planeinmax(
Maximum out-of-plane shearing stress corresponds to a 45o rotation of the plane stress element around a longitudinal axis
tpr22max
Points A and B correspond to hoop stress, s1, and longitudinal stress, s2
Note: Plane stress Mohr’s circle is only ADBE. The remaining is due to 3-D state of stress with third principal stress being zero. 3-D state of stress is also used in finding maximum out-of-plane shearing stress by using third principal stress as zero. This is not covered in this course. Only maximum in-plane shearing stress is covered in this course.
MECHANICS OF MATERIALS
7- 37
Stresses in Thin-Walled Spherical Pressure Vessels
Spherical pressure vessel:
tpr
rptr
2
)2(
21
22
Maximum out-of-plane shearing stress
tpr412
1max
Mohr’s circle for in-plane transformations reduces to a point
0constant
plane)-max(in
21
Note: Plane stress Mohr’s circle is only AB, i.e., circle with zero radius, i.e., a point. The dotted circle is due to 3-D state of stress with third principal stress being zero. 3-D state of stress is also used in finding maximum out-of-plane shearing stress, by using third principal stress as zero. This is not covered in this course. Only maximum in-plane shearing stress is covered in this course.
MECHANICS OF MATERIALS
8- 38
Principle Stresses in a Beam
For beam subjected to transverse loading
ItVQ
ItVQ
IMc
IMy
mxy
mx
Can the maximum normal stress within the cross-section be larger than
IMc
m
MECHANICS OF MATERIALS
8- 39
Principle Stresses in a Rectangular section Beam
MECHANICS OF MATERIALS
8- 40
Principle Stresses in a Beam
Cross-section shape results in large values of txy near the surface where sx is also large.
smax may be greater than sm (since tb is large)
MECHANICS OF MATERIALS
8- 41
Example 5
160-kN force applied at tip of W200x52 rolled-steel beam.
Neglect effects stress concentrations at fillets, determine whether normal stresses at section A-A’ satisfy sall =150 MPa
MECHANICS OF MATERIALS
8- 42
Example 5Determine shear and bending moment in
Section A-A’
kN160
m-kN60m375.0kN160
A
AVM
Calculate normal stress at top surface and at flange-web junction.
MPa9.102mm103mm4.90MPa2.117
MPa2.117m10512
mkN6036
cyσ
SM
bab
Aa
MECHANICS OF MATERIALS
8- 43
Example 5
Calculate principal stresses at flange-web junction
MPa 150MPa9.159
5.952
9.1022
9.102 22
2221
21
max
bbb
Design specification not satisfied.
Calculate shear stress at flange-web junction.
MPa5.95m0079.0m107.52
m106.248kN160
m106.248
mm106.2487.966.12204
46
36
36
33
ItQV
Q
Ab
MECHANICS OF MATERIALS
8- 44
Example 6
sall = 165 Mpa, tall = 100 Mpa. Select wide-flange beam to be used.
MECHANICS OF MATERIALS
8- 45
Example 6
Calculate required section modulus, select appropriate beam section.
section beam 920select W53
mm1959Pa10651Nm10323 3
6
3max
min
all
MS
Maximum shear and bending moment from SFD, BMD.
kN4.193
kN 9.54withkNm2.323
max
max
V
VM
Reactions at A and D.
kN5.1840
kN5.2650
AD
DA
RM
RM
323.2
MECHANICS OF MATERIALS
8- 46
Example 6Find maximum shearing stress.
Assume uniform shearing stress in web (conservative, see shallow parabolic variation, slide 10, shear stresses chapter)
MPa100MPa 6.35m 105436.6N 104.193
26-
3max
max
webA
V
Find maximum normal stress.
MPa1.10m106.4365
N 54900
MPa147mm5.266mm9.250MPa1.156
MPa1.156m1007.2
Nm323200
26b
36max
web
bab
a
AV
cyσ
SM
MPa165MPa1.147
MPa1.102MPa147
2MPa147 2
2
max
MECHANICS OF MATERIALS
8- 47
Design of Transmission Shaft
If power is transferred to and from shaft by gears or sprocket wheels, the shaft is subjected to transverse loading (due to forces in mating gears) as well as shear loading (due to torque from these forces).
Normal stresses due to transverse loads may be large, should be included in determination of maximum shearing stress.
Shearing stresses due to transverse loads are usually small and their contribution to maximum shear stress may be neglected.
MECHANICS OF MATERIALS
8- 48
Design of a Transmission Shaft
Shaft section requirement,
all
TM
cJ
max
22
min
Maximum shearing stress,
22max
222
2
max
2 section,-crossannular or circular afor
22
TMJc
JI
JTc
IMc
mm
At any section,
JTc
MMMI
Mc
m
zym
222where
MECHANICS OF MATERIALS
8- 49
Example 7
Shaft rotates at 480 rpm, transmits 30 kW from motor to gears G and H; 20 kW is taken off at gear G and 10 kW at gear H. sall = 50 Mpa. Find smallest permissible diameter for shaft.
MECHANICS OF MATERIALS
8- 50
Example 7Find gear torques and corresponding tangential
forces. Use,
kN49.2mN199Hz82
kW10
kN63.6mN398Hz82
kW20
kN73.3m0.16
mN597
mN597Hz82
kW302
DD
CC
E
EE
E
FT
FT
rTF
fPT
Find reactions at A and B.
kN90.2kN80.2
kN22.6kN932.0
zy
zy
BB
AA
HHHDDD
GGGCCC
MMMEEE
TPPTTPPTTPPT
MECHANICS OF MATERIALS
8- 51
Example 7Identify critical shaft section from torque and bending
moment diagrams. D comes out as critical section.
mN1357
5973731160 222max
222
TMM zy
At C it is 1319 N.m. At E it is 1003 N.m
MECHANICS OF MATERIALS
8- 52
Example 7Find minimum allowable shaft diameter.
36
222
m1014.27MPa50
mN 1357
all
zy TMMcJ
mm 7.512 cd
m25.85m02585.0
m1014.272
363
c
ccJ
For solid circular shaft,
MECHANICS OF MATERIALS
8- 53
Stresses Under Combined Loadings
Wish to find stresses in slender structural members subjected to arbitrary loadings.
Pass section through points of interest. Determine force-couple system at centroid of section required to maintain equilibrium.
System of internal forces consist of three force components and three couple vectors.
Determine stress distribution by applying the superposition principle.
MECHANICS OF MATERIALS
8- 54
Stresses Under Combined LoadingsAxial force and bending moments yield
normal stresses.Shear forces and twisting couple yield
shearing stresses.
Find principal stresses, maximum shearing stress.
MECHANICS OF MATERIALS
8- 55
Example 8
Find principle stresses, principal planes, maximum shearing stress, at H.
MECHANICS OF MATERIALS
8- 56
Example 8Internal forces in Section EFG.
mkN3m100.0kN300
mkN5.8
m200.0kN75m130.0kN50
kN75kN50kN 30
zy
x
zx
MM
M
VPV
Section properties,
463121
463121
23
m10747.0m040.0m140.0
m1015.9m140.0m040.0
m106.5m140.0m040.0
z
x
I
I
A
MECHANICS OF MATERIALS
8- 57
Example 8Normal stress at H.
MPa66.0MPa2.233.8093.8m1015.9
m025.0mkN5.8m10747.0
m020.0mkN3m105.6
kN50
46
4623-
x
x
z
zy I
bMI
aMAP
Shearing stresses at H.
0MPa52.17
m040.0m1015.9m105.85kN75
m105.85
m0475.0m045.0m040.0
46
36
3611
yx
x
zyz tI
QV
yAQ
Note: 2-D (i.e., plane) state of stress only at H. In interior it is 3-D state of stress
MECHANICS OF MATERIALS
8- 58
Example 8Principal stresses and maximum, shearing
stress, principal planes.
98.13
96.2720.33
52.172tan
MPa4.74.370.33
MPa4.704.370.33
MPa4.3752.170.33
pp
min
max
22max
p
CDCY
ROC
ROC
R
98.13
MPa4.7
MPa4.70
MPa4.37
min
max
max
p