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CHAPTER 1
STRESS AND STRAIN
STRESS AND STRAIN
Axial loading
Normal stress
Shear stress
Bearing stress
Strain
Stress-strain relations
Safety factor
Deformation of axial members
Statically indeterminate axial loading
Thermal stress in axial members
Stress And Strain
Mechanics of material is a study of the relationship between the external loads applied to a deformable body and the intensity of internal forces acting within the body.
Stress = the intensity of the internal force on a specific plane (area) passing through a point.
Strain = describe the deformation by changes in length of line segments and the changes in the angles between them
Normal Stress and Normal Strain
Normal Stress, the intensity of force, or force per unit area, acting
normal to A
A positive sign will be used to indicate a tensile stress (member in tension)
A negative sign will be used to indicate a compressive stress (member in compression)
= P / A
(a)
(b)
(c)
Stress ( ) = Force (P)
Cross Section (A)
Unit: Nm -²
Example 1.1:
Two solid cylindrical rods AB and BC are welded
together at B and loaded as shown. Knowing that
d1=50mm and d2=30mm, find average normal stress at
the midsection of (a) rod AB, (b) rod BC.
Solution:
2
2AB
ABAB m/MN7.35
4
m05.0
N70000
A
P
2
2/44.42
4
03.0
30000mMN
m
N
A
P
BC
BCBC
Two solid cylindrical rods AB and BC are
welded together at B and loaded as shown.
Determine the average normal stress at the
midsection of (a) rod AB, and (b) rod BC.
Example 1.2
Example 1.3:
The 80 kg lamp is supported by two rods AB and BC as shown in
the figure. If AB has a diameter of 10 mm and BC has a diameter of
8 mm, determine the average normal stress in each rod.
Solution:
Internal loading:
Average normal stress:
Example 1.4:
Member AC as shown in the figure is subjected to a vertical force of
3 kN. Determine the position x of this force so that the average
compressive stress at the smooth support C is equal to the average
tensile stress in the tie rod AB. The rod has a cross-sectional area of
400 mm2 and the contact area at C is 650 mm2.
Solution:
Internal loading:
Average normal stress:
Normal strain, is the elongation or contraction of a line
segment per unit of length
strain can be expressed as a
percentage strain
L
L
lengthoriginal
lengthchange instrain
Normal strain,
%100L
Lstrain
Determine the corresponding strain for a bar of length
L=0.600m and uniform cross section which undergoes
a deformation =15010-6m.
Example 1.5:
66
6
150 10 m250 10 m m
L 0 600m
250 10 250
/.
@
Example 1.6:
A force acting on the grip of the lever arm as shown in the figure
causes the arm to rotate clockwise through an angle of =0.002 rad.
Determine the average normal strain developed in the wire BC.
Solution:
Since =0.002 rad is small, the stretch in the wire CB is:
Example 1.7:
The plate is deformed into the dashed shape as shown in the figure.
If in this deformed shape horizontal lines on the plate remain
horizontal and do not change their length, determine the average
normal strain along the side AB .
Solution:
Line AB, coincident with the y-axis,
becomes line AB’ after deformation. The
length of this line is:-
The average normal strain for AB is
therefore:-
The negative sign indicates the strain
causes a contraction of AB
Example 1.8:
A short post constructed from a hollow circular tube of aluminium supports a compressive load of 240 kN. The inner and outer diameters of the tube are d1=90mm and d2=130mm, respectively, and its length is 1m. The shortening of the post due to the load is measured as 0.55 mm. Determine the compressive stress and strain in the post. (Disregard the weight of the post itself, and assume that the post does not buckle under the load.)
Solution:
232221
22 m1091.6m09.0m13.0
4dd
4A
MPa7.34m91.6
N000,240
A
P
2
55010550mm000,1
mm55.0
L
6
Assuming that the compressive load acts at the center of the hollow
tube, we can use the equation =P/A to calculate the normal stress.
The force P equals to 240kN and the cross-sectional area is:-
Therefore, the compressive stress in the post is:-
The compressive strain is:-
Stress-Strain Relations
Tensile test is a experiment to determine the load-
deformation behavior of the material.
Data from tensile test can be plot into stress and
strain diagram.
Example of test specimen
- note the dog-bone geometry
28
Universal Testing Machine - equipment used to
subject a specimen to tension, compression,
bending, etc. loads and measure its response
29
Stress-Strain Relations:
Universal Testing Machine
Stress-Strain Relations: Stress-Strain
Diagrams
30
A number of important mechanical properties of materials that be
deduced from stress-strain diagram are illustrated in figure above.
Point O-A = linear relationship between stress and strain
Point A = proportional limit (PL)
The ratio of stress to strain in this linear region of stress-strain
diagram is called Young Modulus or the Modulus of Elasticity.
• At point A-B, specimen begins yielding.
Point B = yield point
Point B-C = specimen continues to elongate without any increase in stress. Its refer as perfectly plastic zone
Point C = stress begins to increase
Point C-D = refer as the zone of strain hardening
Point D = ultimate stress/strength ; specimen begins to neck-down
Point E = fracture stress
< PL
Unit: MPa
31
Stress-Strain Relations: Stress-Strain
Diagrams (cont.)
Point O to A
Point C to D
Point D to E
At point E
Normal or engineering stress can be determine by dividing the applied load by the specimen original cross sectional area.
True stress is calculated using the actual cross sectional area at the instant the load is measured.
31
Stress-Strain Relations: Stress-Strain
Diagrams (cont.)
• Some of the materials like aluminum (ductile), does not have clearly yield point such as structural steel. Therefore, stress value called the offset yield stress, YL is used in lieu of a yield point stress.
As illustrated, the offset yield stress is determine by;
Drawing a straight line that best fits the data in initial (linear) portion of the stress-strain diagram
Second line is then drawn parallel to the original line but offset by specified amount of strain
The intersection of this second line with the stress-strain curve determine the offset yield stress.
Commonly used offset value is 0.002/0.2%32
Stress-Strain Relations: Stress-Strain
Diagrams (cont.)
• Brittle material such as ceramic and glass have low
tensile stress value but high in compressive stress.
Stress-strain diagram for brittle material.
33
Stress-Strain Relations: Stress-Strain
Diagrams (cont.)
Elasticity and Plasticity
Elasticity refers to the property of a material such that it returns
to its original dimensions after unloading
Any material which deforms when subjected to load and
returns to its original dimensions when unloaded is said to be
elastic.
If the stress is proportional to the strain, the material is said to
be linear elastic, otherwise it is non-linear elastic.
Beyond the elastic limit, some residual strain or permanent
strains will remain in the material upon unloading
The residual elongation corresponding to the permanent strain
is called the permanent set
34
• The amount of strain which is recovered upon unloading is
called the elastic recovery.
35
Poisson's Ratio,
When an elastic, homogenous and isotropic material is subjected
to uniform tension, it stretches axially but contracts laterally along
its entire length.
Similarly, if the material is subjected to axial compression, it
shortens axially but bulges out laterally (sideways).
The ratio of lateral strain to axial strain is a constant known as the
Poisson's ratio,
strainaxial
strainlateralv
36
Axial elongation and lateral contraction of a
bar in tension:
(a) before loading, and (b) after loading
Example 1.9:
A steel pipe of length L=1.2 m, outside diameter d2=150mm, and inside diameter d1=110mm is compressed by an axial force P=620kN. The material has modulus of elasticity E=200 GPa and Poisson’s ratio =0.30.
Determine the following quantities for the pipe:-
(a) the shortening,
(b) the lateral strain, ’
(c) the increase d2 in the outer diameter and the increase d1 in the inner diameter
(d) the increase t in the wall thickness
(e) the increase V in the volume of the material, and
(f) the dilatation e
Solution:
The cross-sectional area A and longitudinal stress are determined as
follows:
The axial strain may be found from Hooke’s Law:-
(a) The Shortening,
MPa9.75m10168.8
N620000
A
P
m10168.8m11.0m15.04
dd4
A
23
232221
22
5.379or105.379
MPa10200
MPa9.75
E,E 6
3
mm455.0mm102.1105.379LL,L
L 36
Solution (cont.):
(b) The lateral strain is obtained from Poisson’s Ratio
The positive sign for ’ indicates an increase in the lateral dimensions, as
expected for compression.
(c) The increase in outer diameter equals the lateral strain times diameter:
(d) The increase in wall thickness is found in the same manner as the increases in
the diameters; thus,
Note that under compression, all three quantities increase.
66 109.113)105.379(30.0,
mm0125.0mm110109.113dd
mm0171.0mm150109.113dd
611
622
mm00228.0mm0125.0mm0171.02
1
2
ddt
,or
mm00228.0mm20109.113tt
12
6
Solution (cont.):
(e) The change in volume of the material is,
The volume change is negative, indicating a decrease in volume, as expected for
compression.
(f) Finally, the dilatation is,
3m10488.160.01105.379m2.1m10168.8
21AL21VV
6623
o
00015.0 0.60-110379.5-2-1e -6
Shear Stress
Shear stress, : The intensity of force, or force per unit
area, acting tangent to A.
Average shear stress distributed over each sectioned area
that develops shear force is defined by:
avg = Average shear stress at the section, which is assumes
to be the same at each point located on the section
V = Internal resultant shear force at the section determined
from the equitions of equilibrium
A = Area at the section
A
Vavg
41
Depending on the type of connection, a connecting element (bolt, rivet, pin) may be subjected to single shear or double shear as shown.
Rivet in Single Shear
4
d
P
A
V
2
42
Shear Stress (cont.)
22
2
)4
(2d
P
d
P
A
V
43
Rivet in Double Shear
Shear Stress (cont.)
Example 1.10:
The inclined member as shown in the figure is subjected to a
compressive force of 3000N. Determine the average shear stress
along the horizontal plane defined by EDB.
Solution:
The average shear stress acting on the
horizontal plane defined by EDB is:
Shear Stress (cont.)
A
F
A
Pave
Single Shear
A
F
A
P
2ave
Double Shear
Shear Strain
The effect of shear stress is to distort the shape of a body
by inducing shear strains
The shear strain, is a measure of the angular distortion of
the body.
L
Vx
L
x
44
Example 1.11:
The plate is deformed into the dashed shape as shown in the figure.
If in this deformed shape horizontal lines on the plate remain
horizontal and do not change their length, determine the average
shear strain in the plate relative to the x and y axis .
Solution:
The angle of BAC, 90 between the sides
of the plate, referenced from the x, y axes
changes to ’ due to the displacement of
B to B’. Since xy =/2-’ is shown in the
figure, thus,
Example 1.12:
A punch for making holes in steel plates is shown in the figure a.
Assume that a punch having a diameter of 20mm is used to
punch a hole in a 6.5mm plate as shown in the cross-sectional
view (figure b). If a force P=125kN is required, what is the
average shear stress in the plate and average compressive
stress in the punch?
Solution:
The average shear stress in the plate is,
The average compressive stress in the plate is,
MPa306
m0065.0m02.0
N10125
dt
P
A
P 3
savg
MPa398
4/m02.0
N10125
4/d
P
A
P
22punch
c
3
Bearing Stress
Bearing stress is also known as a contact stress
Definition: The bearing stress σb is the average compressive
normal stress acting over a contact surface. It is given by the
load divided by the projected contact area.
Bearing stress occurs in joints, connections, pins, etc.
– anywhere you have surfaces in contact.
Bearing stress in rivet and plat;
td
Pb
45
The connection shown in the figure consists of five
steel plates, each 2.5 mm thick, to be joined by a
single bolt. Determine the required diameter of the bolt
if the allowable bearing stress, σb, is 180.0 MPa and
the allowable shear stress, allow, is 45.0 MPa?
Example 1.13:
Solution:
Solution (cont.):
Shear Modulus
It also known as Shear Modulus of Elasticity or the Modulus
of Rigidity.
Value of shear modulus can be occur from the linear region
of shear stress-strain diagram.
The modulus young (E), poisson’s ratio() and the modulus
of rigidity (G) can be related as
G Unit : Pa
)1(2
EG
48
Example 1.14:
A bearing pad of the kind used tosupport machines an bridge girdersconsists of a linearly elastic material(usually an elastomer, such asrubber) capped by a steel plate asshown in the figure a. Assume thatthe thickness of the elastomer is h,the dimension of the plate are ab,and the pad is subjected to ahorizontal shear force V.
Obtain formulas for the averageshear stresses in the elastomer andthe horizontal displacement d of theplate (figure b).
Solution:
Assume that the shear stresses in elastomer are uniformly distributed
throughout its entire volume. Then the shear stress on any horizontal
plane through the elastomer equals the shear force V divided by the area
of plane (figure a):
ab
Vavg
From the equation of Hooke’s Law in shear;
,Ge
eee
avg
abG
V
G
ab
V
G
In which Ge is the shear modulus of the elastomer material.
eabG
Vtanhtanhd
The horizontal displacement d is equal to h tan ;
In most practical situations the shear strain is a small angle, and in
such cases we may replace tan by and obtain;
eabG
hVhd
Solution (cont.):
Stress on an inclined plane
50
sincos PVPF
• Resolve P into components normal and
tangential to the oblique section,
cossin
cos
sin
cos
cos
cos
00
2
00
A
P
A
P
A
V
A
P
A
P
A
F
• The average normal and shear stresses on
the oblique plane are
Stress on an inclined plane (cont.)
Allowable Stress/ Safety Factor
Applied load that is less than the load the member can fully support. (maximum load)
One method of specifying the allowable load for the designor analysis of a member is use a number called the Factorof Safety (FS).
Allowable-Stress Design
allow
fail
F
FFS FS > 1
FSor
FS
yieldallow
yieldallow
63
Deformations Under Axial Loading
AE
P
EE
• From Hooke’s Law:
• From the definition of strain:
L
• Equating and solving for the deformation,
AE
PL
• With variations in loading, cross-section or
material properties,
i ii
ii
EA
LP
Example 1.15:
Bar aluminium AB, bar tembaga BC, dan bar keluli CD adalah dihubungkan secara sempurna diantara satu sama lain dan diikat tegar pada dinding seperti yang ditunjukkan dalam rajah dibawah. Luas keratan rentas AB, BC dan CD masing-masing 200mm2, 150mm2, dan 100mm2. Tentukan jumlah pemanjangan yang berlaku keatas bar ABCD akibat daripada beban-beban yang dikenakan. Modulus keanjalan bagi aluminium, tembaga, dan keluli adalah masing-masing 70GPa, 110GPa, dan 200GPa.
12 kN10 kN 8 kN
A BC D
0.6m 0.2m0.4m
Solution:
kN12P
0PkN12
0F
CD
CD
x
kN4P
0PkN8kN12
0F
BC
BC
x
10 kN 8 kN
A B C D
0.6m 0.2m0.4m
12kN
12 kNPCD12kN8kNPBC
m102.1
)m/N10200)(m10100(
m2.0N12000
AE
PLL
4
292
CDCD
6
m10697.9
)m/N10110)(m10150(
m4.0N4000
AE
PLL
5
292
BCBC
6
Solution:
kN6P
0PkN10kN8kN12
0F
AB
AB
x
m10571.2
)m/N1070)(m10200(
m6.0N6000
AE
PLL
4
292
ABAB
6
10 kN 8 kN
C D
12kN
B
PAB
m10013.4m102.1m10697.9m10571.2
LLLL
554
CDBCABABCD
4
Example 1.16:
Bar tegar ABC dalam rajah di bawah berada dalam keadaan mengufuk ketika tiada beban dikenakan. Dengan beban sebanyak 10kN dikenakan di C, panjang serta luas keratan rentas rod BD masing-masing 1m dan 80mm2, dan modulus keanjalan rod BD adalah 200GPa, tentukan anjakan yang berlaku keatas titik C.
10kN
0.5m1mA B C
D
Solution:
N15000m0.1
)m5.1(N10000P
0)m0.1(P)m5.1(N10000
0M
rod
rod
A
m10375.9
)m/N10200)(m1080(
m1N15000
AE
PLL
4
2926rod
rod
AX
AY10kN
Prod
0.5m1m
B CA
+
1m 0.5m
LCLrod
m10406.1
10375.95.1
L5.1
5.11
L
3
4
rodC
Crod
Example 1.17
Determine the deformation of
the steel rod shown under the
given loads.
in. 618.0 in. 07.1
psi1029 6
dD
E
SOLUTION:
• Divide the rod into components at
the load application points.
• Apply a free-body analysis on each
component to determine the
internal force
• Evaluate the total of the component
deflections.
SOLUTION:
• Divide the rod into three
components:
221
21
in 9.0
in. 12
AA
LL
23
3
in 3.0
in. 16
A
L
• Apply free-body analysis to each
component to determine internal forces,
lb1030
lb1015
lb1060
33
32
31
P
P
P
• Evaluate total deflection,
in.109.75
3.0
161030
9.0
121015
9.0
121060
1029
1
1
3
333
6
3
33
2
22
1
11
A
LP
A
LP
A
LP
EEA
LP
i ii
ii
in. 109.75 3
Example 1.18:
The rigid bar BDE is supported by two
links AB and CD.
Link AB is made of aluminum (E = 70
GPa) and has a cross-sectional area of
500 mm2. Link CD is made of steel (E
= 200 GPa) and has a cross-sectional
area of (600 mm2).
For the 30-kN force shown, determine
the deflection a) of B, b) of D, and c)
of E.
SOLUTION:
• Apply a free-body analysis to
the bar BDE to find the forces
exerted by links AB and DC.
• Evaluate the deformation of
links AB and DC or the
displacements of B and D.
• Work out the geometry to find
the deflection at E given the
deflections at B and D.
Displacement of B:
m10514
Pa1070m10500
m3.0N1060
6
926-
3
AE
PLB
mm 514.0B
Displacement of D:
m10300
Pa10200m10600
m4.0N1090
6
926-
3
AE
PLD
mm 300.0D
Free body: Bar BDE
ncompressioF
F
tensionF
F
M
AB
AB
CD
CD
B
kN60
m2.0m4.0kN300
0M
kN90
m2.0m6.0kN300
0
D
SOLUTION:
Displacement of E:
mm 7.73
mm 200
mm 0.300
mm 514.0
x
x
x
HD
BH
DD
BB
mm 928.1E
mm 928.1
mm 7.73
mm7.73400
mm 300.0
E
E
HD
HE
DD
EE
Static Indeterminacy
• Structures for which internal forces and reactions cannot be determined from statics alone are said to be statically indeterminate.
A structure will be statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium.
Redundant reactions are replaced with unknown loads which along with the other loads must produce compatible deformations
• Deformations due to actual loads and redundant reactions are determined separately and then added or superposed.
0 RL
65
Determine the reactions at A and B for the steel bar and
loading shown, assuming a close fit at both supports
before the loads are applied.
66
Example 1.19
Solution:
BA
BA
y
RR
RR
F
900
0600300
0
PAD
AAD
ADA
y
RP
PR
F
0
0
PDC
300
0300
0
ADC
DCA
y
RP
PR
F
PCK
300
0300
0
ACK
CKA
y
RP
PR
F
Solution:
PKB
900
0600300
0
AKB
KBA
y
RP
PR
F
kNR
RRRR
E
AE
PL
AE
PL
AE
PL
AE
PL
A
AAAA
KBCKDCAD
AB
KBCKDCADAB
323
010400
15.0900
10400
15.0300
10250
15.0300
10250
15.01
0
0
6666
kNRR AB 577323900900
Two cylindrical rods, CD made of steel (E = 200 GPa) and AC made
of aluminum (E = 72 GPa), are joined at C and restrained by rigid
supports at A and D. Determine
(a) The reactions at A and D,
(b) the deflection of point C.
67
Example 1.20
Solution:
AABAAB
x
RPRP
F
,0
0PAB
PBC
31080
080
0
ABC
ABC
x
RP
RP
F
PCD
310140
06080
0
ACD
ACD
x
RP
RP
F
Solution:
Solution:
Thermal Stresses
• A temperature change results in a change in length or thermal strain. There is no stress associated with the thermal strain unless the elongation is restrained by the supports.
coef.expansion thermal
AE
PLLT PT
• Treat the additional support as redundant and apply the principle of superposition.
0
0
AE
PLLT
PT
• The thermal deformation and the deformation from
the redundant support must be compatible.
TEA
P
TAEPPT
0
At room temperature (21°C) a 0.5 mm gap exists between the
ends of the rods shown. At a later time when the temperature has
reached 160°C, determine:
(a) The normal stress in the aluminum rod,
(b) the change in length of the aluminum rod.
Example 1.21
Solution:
Solution: