53
Allowable Stress
allow fail
P P
S F
= .
allow fail
S F
σ σ
= .
allow fail
S F
τ τ
= .
6. Design of Simple Connections
allow
P A
σ
=
allow
V A
τ
=
• Cross-Sectional Area of a Tension Member
P P
a a
P
τ
allow
allow
P A
σ
=
54
σ
b allow
P
V = P P
• Cross-Sectional Area of a Connector Subjected to Shear
allow
P A
σ
=
P P
ττττ
allow
• Required Area to Resist Bearing
allow b
P A
σ
=
55
• Required Area to Resist Shear by Axial Load
P
d P
l
allow
π τ
=
P
d Uniform shear stress τ
allow
56
Example 12a
The control arm is subjected to the loading shown. a Determine the shear stress for the steel at all pin b Determine normal stress in rod AB, plate D and E. The
thickness of plate D and E is 10 mm.
pin C 25 kN
13 kN 3
4 5
75 mm 50 mm
200 mm A
B
C D
E
φ
= 10 mm
φ
= 20 mm
D E
φ
= 25 mm
13 kN 50 mm
25 kN 50 mm
φ
= 25 mm
57
25 kN 13 kN
3 4
5
75 mm 50 mm
200 mm
• Reaction C
125 .
87 .
36 sin
25 075
. 13
2 .
= −
+
AB
F +
ΣM
C
= 0: F
AB
= 4.5 kN , ←
ΣF
y
= 0: +
C
y
+ 13 - 25sin36.87
o
= 0 C
y
= 2 kN, ↑ ΣF
x
= 0: +
-4.5 - C
x
+ 25cos36.87
o
= 0 C
x
= 15.5 kN, ←
kN 63
. 15
2 5
. 15
2 2
= +
= C
25 kN 13 kN
75 mm 50 mm
200 mm F
AB
C
y
C
x
A B
C
C
x
C
y
C 36.87
o
58
25 kN 13 kN
75 mm 50 mm
200 mm 4.5 kN
a Shear stress
15.63 kN 15.63 kN
Pin C Double shear
7.815 kN 7.815 kN
Pin D and E Single shear
pin C 25 kN
13 kN 3
4 5
75 mm 50 mm
200 mm A
B
C D
E
= 35 mm
φ
= 20 mm
φ
= 25 mm
φ
= 25 mm
⇐ =
= =
MPa 88
. 24
02 .
4 kN
815 .
7
2
π A
V τ
C C
C
⇐ =
= =
MPa 48
. 26
025 .
4 kN
13
2
π A
V τ
D D
D
⇐ =
= =
MPa 93
. 50
025 .
4 kN
25
2
π A
V τ
E E
E
59
25 kN 13 kN
75 mm 50 mm
200 mm 4.5 kN
b Normal stress
15.63 kN
Cale AB Plate D
φ
rod
= 10 mm
A
C D
E
σ
⇐ =
= =
MPa 7
. 56
010 .
4 kN
5 .
4
2
π σ
AB AB
A P
D E
φ
= 25 mm
13 kN 50 mm
25 kN 50 mm
φ = 25 mm
13 kN
σ
D
13 kN
13 kN t = 0.01 m
0.05 m ⇐
= =
=
MPa 26
01 .
05 .
kN 13
A P
σ
D
60
25 kN 13 kN
75 mm 50 mm
200 mm 4.5 kN
15.63 kN
φ
rod
= 35 mm
A
C D
E
Plate E
⇐ =
− =
=
MPa 100
01 .
025 .
05 .
kN 25
A P
E
σ
D E
φ
= 25 mm
13 kN 50 mm
25 kN 50 mm
φ
= 25 mm
25 kN 50 mm
25 kN 25 kN
50 mm σ
E
0. 05
m
0. 02
5 m
t = .0
1 m
61
Example 12b
The control arm is subjected to the loading shown. Determine the required diameter of the steel pin at C if the allowable shear stress for the steel is
τ
allow
= 55 MPa. Note in the figure that the pin is subjected to double shear.
22 kN 13 kN
3 4
5
75 mm 50 mm
200 mm A
B
C
62
• Internal Shear Force
125 .
87 .
36 sin
22 075
. 13
2 .
= −
−
AB
F +
ΣM
C
= 0: F
AB
= 13.125 kN
ΣF
y
= 0: +
C
y
- 13 - 22sin36.87
o
= 0 C
y
= 26.2 kN ΣF
x
= 0: +
-13.125 - C
x
+ 22cos36.87
o
= 0 C
x
= 4.47 kN
kN 58
. 26
2 .
26 47
. 4
2 2
= +
= C
22 kN 13 kN
75 mm 50 mm
200 mm F
AB
C
y
C
x
22 kN 13 kN
3 4
5
75 mm 50 mm
200 mm A
B
C
C
x
C
y
C
63
• Required Area
2 2
6 6
3
mm 242
m 10
6 .
241 10
55 10
29 .
13 =
× =
× ×
= =
− allow
τ V
A
2 2
mm 242
2 =
d π
d = 17.55 mm Use d = 20 mm
22 kN 13 kN
3 4
5
75 mm 50 mm
200 mm A
B
C
22 kN 13 kN
75 mm 50 mm
200 mm 13.125 kN
26.58 kN
26.58 kN
13.29 kN 13.29 kN
64
Example 13a
The two members are pinned together at B as shown below. Top views of the pin connections at A and B are also given. If the pins have an allowable shear stress
of
τ
allow
= 86 MPa , the allowable tensile stress of rod CB is
σ
t allow
= 112 MPa and the allowable bearing stress is
σ
b allow
= 150 MPa, determine to the smallest diameter of pins A and B ,the diameter of rod CB and the thickness of AB
necessary to support the load.
13 kN
A B
C
1 m 0.6 m
3 4
5 Top view
A
B
65
• Internal Force
13 kN
A B
F
BC
1 m 0.6 m
36.87
o
A
y
A
x
+ Σ M
A
= 0: 6
. 1
87 .
36 sin
1 13
= +
−
BC
F
F
BC
= 13.54 kN Σ
F
y
= 0: +
87 .
36 sin
54 .
13 13
= +
−
y
A
A
y
= 4.88 kN
Σ F
x
= 0: +
87 .
36 cos
54 .
13 =
+ −
x
A
A
x
= 10.83 kN
66
• Diameter of the Pins
13 kN
A B
F
BC
= 13.54 kN
1 m 0.6 m
36.87
o
A
y
= 4.88 kN
A
x
= 10.83 kN
Top view A
B 11.88 kN
Pin A
11.88 kN V
A
= 5.94 kN V
A
= 5.94 kN
2 2
3
mm 07
. 69
m kN
10 86
kN 94
. 5
= ×
= =
allow A
A
τ V
A
2 2
mm 07
. 69
4 =
A
d π
d
A
= 9.38 mm, Use d
A
= 10 mm
Pin B
F
BC
= 13.54 kN
V
B
= 13.54 kN
2 2
3
mm 4
. 157
m kN
10 86
kN 54
. 13
= ×
= =
allow B
B
V A
τ
2 2
mm 4
. 157
4 =
B
d
π
d
B
= 14.16 mm, Use d
B
= 15 mm
67
• Diameter of Rod
13.54 kN
13.54 kN
2 2
3
mm 9
. 120
m kN
10 112
kN 54
. 13
= ×
= =
allow t
BC
σ P
A
2 2
mm 9
. 120
4 =
BC
d π
d
BC
= 12.4 mm, Use d
BC
= 15 mm 13 kN
A B
F
BC
= 13.54 kN
1 m 0.6 m
36.87
o
A
y
= 4.88 kN
A
x
= 10.83 kN 11.88 kN
68
t
AB
φ
B
= 15 mm 20 kN
A B
13.54 kN 11.88 kN
t
AB
φ
A
= 10 mm
A B
σ
b A
σ
b B
m 00792
. 010
. 10
88 .
11 10
150
3 6
= ×
= ×
=
AB AB
allow b
t t
A P
σ
• Thickness
13 kN
A B
F
BC
= 13.54 kN
1 m 0.6 m
36.87
o
A
y
= 4.88 kN
A
x
= 10.83 kN 11.88 kN
m 00602
. 015
. 10
54 .
13 10
150
3 6
= ×
= ×
=
AB AB
allow b
t t
A P
σ
mm 8
=
AB
t use
comparison By
69
Example 13b
The two members are pinned together at B as shown below. Top views of the pin connections at A and B are also given. If the pins have an allowable shear stress
of
τ
allow
= 86 MPa, the allowable tensile stress of rod CB is
σ
t allow
= 112 MPa and the allowable bearing stress is
σ
b allow
= 150 MPa, determine to the maximum load P that the beam can supported.
P
A B
C
1 m 0.6 m
3 4
5 Top view
A
B
φ
= 15 mm
φ
= 10 mm
t = 8 mm
φ
= 15 mm
t = 8 mm
70
• Internal Force P
A B
F
BC
1 m 0.6 m
36.87
o
A
y
A
x
+ Σ
M
A
= 0: 6
. 1
87 .
36 sin
1 =
+ −
BC
F P
F
BC
= 1.042P T 87
. 36
sin 042
. 1
; =
+ −
= Σ
↑ +
P P
A F
y y
A
y
= 0.375P 87
. 36
cos 042
. 1
; =
+ −
= Σ
→
+
P A
F
x x
A
x
= 0.834 kN
71
Pin A double shear
0.914P V
A
= 0.457P V
A
= 0.457P
kN 78
. 14
01 .
4 457
. kPa
10 86
2 3
= =
× =
P π
P A
V τ
A A
allow
Pin B single shear
F
BC
= 1.042P
V
B
= 1.042P
kN 58
. 14
015 .
4 042
. 1
kPa 10
86
2 3
= =
× =
P π
P A
V τ
B B
allow
P
A B
F
BC
= 1.042P
1 m 0.6 m
36.87
o
A
y
= 0.375P
0.834P
Top view A
B 0.914P
φ
= 10 mm
φ
= 15 mm
72
1.042P
1.042P
Rod AB P
A B
F
BC
= 1.042P
1 m 0.6 m
36.87
o
A
y
= 0.375P
0.834P 0.914P
kN 19
015 .
4 042
. 1
kPa 10
112
2 3
= =
× =
P π
P A
P σ
BC allow
φ
rod
= 15 mm
73
t
AB
= 10 mm
φ
B
= 15 mm 20 kN
A B
F
BC
= 1.042P 0.914P
t
AB
= 10 mm
φ
A
= 10 mm
A B
σ
b A
σ
b B
kN 41
. 16
010 .
010 .
914 .
10 150
6
= =
× =
P P
A P
σ
allow b
kN 60
. 21
01 .
015 .
042 .
1 10
150
6
= =
× =
P P
A P
σ
allow b
P
A B
F
BC
= 1.042P
1 m 0.6 m
36.87
o
A
y
= 0.375P
0.834P
Top view A
B 0.914P
φ
= 10 mm
φ
= 15 mm
t = 10 mm t = 10 mm
⇐ =
kN 58
. 14
P all
comparison By
74
Example 14
The suspender rod is supported at its end by a fixed-connected circular disk as shown. If the rod passes through a 40-mm diameter hole, determine the minimum
required diameter of the rod and the minimum thickness of the disk needed to support the 20 kN load. The allowable normal stress for the rod is σ
allow
= 60 MPa, and the allowable shear stress for the disk is τ
allow
= 35 MPa.
d 40 mm
t
20 kN 20 kN
40 mm
τ
allow
75
d 40 mm
t
20 kN 20 kN
40 mm
τ
allow
• Diameter of Rod
2 3
m kN
10 60
kN 20
× =
=
allow
σ P
A
2 3
2
m kN
10 60
kN 20
4 ×
= =
d π
A
d = 0.0206 m = 20.6 mm
• Thickness of Disk
2 3
m kN
10 35
kN 20
× =
=
allow
τ V
A
2 3
m kN
10 35
kN 20
m 02
. 2
× =
t π
t = 4.55x10
-3
m = 4.55 mm
76
Example 15
An axial load on the shaft shown is resisted by the collar at C, which is attached to the shaft and located on the right side of the bearing at B. Determine the largest
value of P for the two axial forces at E and F so that the stress in the collar does not exceed an allowable bearing stress at C of σ
b allow
= 75 MPa and allowable shearing stress of the adhesive at C of τ
allow
= 100 MPa , and the average normal stress in the shaft does not exceed an allowable tensile stress of σ
t allow
= 55 MPa.
A B
P C
E F
2P
60 mm 20 mm
80 mm
77
A B
P C
E F
2P
60 mm 20 mm
80 mm
3P
Position Load
3P 2P
• Axial Stress
3P P
2P
A P
allow t
=
σ
2 2
3
m 03
. 3
m kN
10 55
π P
= ×
P
1
= 51.8 kN
• Bearing Stress
A P
allow b
=
σ
] m
03 .
m 04
. [
3 m
kN 10
75
2 2
2 3
π π
P −
= ×
B
C 60 mm
80 mm
3P
P
2
= 55 kN
78
A B
P C
E F
2P
60 mm 20 mm
80 mm
3P
Position Load
3P 2P
The largest load that can be applied to the shaft is P = 51.8 kN.
shear allow
A P
=
τ
] 020
. m
04 .
2 [
3 m
kN 10
100
2 3
π P
= ×
P
3
= 55 kN
• Shearing Stress
B
C 60 mm
80 mm
3P
20 mm
• Axial Stress
• Shearing Stress • Bearing Stress
P
3
= 55 kN P
1
= 51.8 kN
P
2
= 55 kN
79
Example 16
The rigid bar AB shown supported by a steel rod AC having a diameter of 20 mm and an aluminum block having a cross-sectional area of 1800 mm
2
. The18-mm- diameter pins at A and C are subjected to single shear. If the failure stress for the
steel and aluminum is σ
st fail
= 680 Mpa and σ
al fail
= 70 MPa, respectively, and the failure shear stress for each pin is τ
fail
= 900 MPa, determine the largest load P that can be applied to the bar. Apply a factor of safety of F.S = 2.0.
P
Steel 0.75 m
2 m Aluminum
A B
C
80
+ ΣM
B
= 0: P1.25 m - F
AC
2 m = 0 --------- F
AC
= 0.625P 1.25 m
P
A B
F
AC
F
B
0.75 m
+ ΣM
A
= 0: P0.75 m - F
B
2 m = 0 ---------- F
B
= 0.375P P
Steel 0.75 m
2 m Aluminum
A B
C
• Rod AC
AC AC
fail st
allow st
A F
S F
= =
.
σ σ
2 3
m π
0.01 0.625
2 kPa
10 680
P =
×
P
1
= 171 kN
• Block B
B B
fail al
allow al
A F
S F
= =
.
σ σ
2 6
3
m 10
1800 0.375
2 kPa
10 70
−
× =
× P
P
2
= 168 kN
81
P
Steel 0.75 m
2 m Aluminum
A B
C
• Pin A or C
pin AC
fail allow
A F
S F
= =
.
τ τ
2 3
m π
0.009 0.625
2 kPa
10 900
P =
×
P
3
= 183 kN
Largest load P = 168 kN
1.25 m
P
A B
F
AC
= 0.625P
F
B
= 0.375P 0.75 m
• Pin A or C P
3
= 183 kN
Summary • Rod AC
P
1
= 171 kN
• Block B P
2
= 168 kN
1
STRAIN
Deformation Strain
Normal Strain Shear Strain
Cartesian Strain Components
2
Deformation
Deformed body θ
A B
C
Undeformed body
uA
A´ B´
C´ θ´
3
Strain
x y
z x
y
z
x y
1 ε
x
1 x
y
1
1 ε
y
• Normal Strain
4
x y
x y
2 π
x y
xy
γ
xy
γ ε
y
1
1 ε
x
• Shear Strain
5
Example 1
The slender rod shown subjected to an increase of temperature along its axis, which creates a normal strain in the rod of
ε
z
= 4010
-3
z
12
, where z is given in meters. Determine a the displacement of the end B of the rod due to the
temperature increase, and b the average normal strain in the rod.
200 mm z
dz A
B
6
200 mm z
dz A
B
a the displacement of the end B of the rod due to the temperature increase
dz z
dz ]
10 40
[
2 1
3 −
=
∫
−
= ∆
m 2
. 2
1 3
] 10
40 [
dz z
B ↓
= =
−
m z
00239 .
3 2
10 40
m 2
. 2
3 3
b the average normal strain in the rod.
mm mm
0119 .
mm 200
mm 39
. 2
= =
∆ ∆
− ∆
= s
s s
avg
ε
7
Example 2
A force acting on the grip of the lever arm shown causes the arm to rotate clockwise through an angle of θ = 0.002 rad. Determine the average normal
strain developed in the wire BC.
B
A L
2L C
8
B
A L
2L C
B´ 2L
B
A L
2L C
Lcos
θ
Lsin θ
θ
• Solution
001 .
2 002
. 2
2 =
= =
= −
=
θ θ
ε
L L
CB CB
CB
avg
Since θ is small, the stretch in wire CB, is using Eq.2-6. The average normal strain in the wire is then
, sin
θ θ
L L
≈
9
Example 3
The plate is deformed into the dashed shape shown. If in this deformed shape horizontal lines on the plate remain horizontal and do not change their length,
determine a the average normal strain along the side AB, and b the average shear strain in the plate relative to the x and y axes.
250 mm
300 mm 3 mm
2 mm B
C x
y
A
10
250 mm 3 mm
2 mm B
y
A B´
mm 018
. 248
3 2
250
2 2
= +
− =
AB
a the average normal strain along the side AB
250 250
018 .
248 −
= −
= AB
AB AB
avg AB
ε
mmmm 10
93 .
7
3 −
− =
b the average shear strain in the plate relative to the x and y axes.
250 mm
300 mm 3 mm
2 mm B
C x
y
A B´
γ
xy
θ´ rad
0121 .
2 250
3 tan
1
= −
=
− xy
γ 2
θ π
γ
− =
xy
11
Example 4
The plate shown is held in the rigid horizontal guides at its top and bottom, AD and BC. If its right side CD is given a uniform horizontal displacement of 2 mm,
determine a the average normal strain along the diagonal AC, and b the shear strain at E relative to the x, y axes.
150 mm
150 mm A
B C
D
E
2 mm x
y
12
A
B 75 mm
C´ D´
E´ 75 mm
76 mm 76 mm
a the average normal strain along the diagonal AC
m 21213
. 150
. 150
.
2 2
= +
= AC
m 21355
. 152
. 150
.
2 2
= +
= AC
150 mm
150 mm A
B C
D
E
2 mm x
y
mmmm 00669
. 21213
. 21213
. 21355
. =
− =
− =
AC AC
AC
avg AC
ε
θ
´
13
A
B 75 mm
C´ D´
E´ 75 mm
76 mm 76 mm
θ
´
b the shear strain at E relative to the x, y axes.
75 76
2 tan
=
θ
150 mm
150 mm A
B C
D
E
2 mm x
y
rad 58404
. 1
759 .
90 180
759 .
90 =
= =
o o
o
π θ
rad 0132
. 58404
. 1
2 −
= −
= π
γ
xy
1
M
ECHANICAL
P
ROPERTIES OF
M
ATERIALS
Simple Tension Test The Stress-Strain Diagram
Stress-Strain Behavior of Ductile and Brittle Materials
Hooke’s Law Strain Energy
Poisson’s Ratio The Shear Stress-Strain Diagram
Failure of Materials Due to Creep and Fatigue
2
400 350
300 250
200 150
100 50
εεεε
mmmm
σ σ
σ σ
MPa
0.10 0.20
0.30 Upper scale
0.40 0.00
0.0020 0.0030 0.0010
Lower scale
0.0040 0.0000
Stress Strain Relationship
L d
P P
3
necking strain
hardening yielding
elastic region
The Stress-Strain Diagram
εεεε σ
σ σ
σ
σ
pl
proportional limit elastic limit
σ
f
σ
u
σ
f
´
fracture stress ultimate stress
elastic behavior
plastic behavior true fracture stress
yield stress σ
Y
4
σ
Y
0.002 0.2 offset
300 250
200 150
100 50
ε
mmmm
σ
MPa
0.005 0.010
Offset yield strength for material with no yield points Offset Yield Stress
5
σ
y
u
= 230 MPa
σ
y
l
= 220 MPa A
P =
σ
L
δ ε
=
400 350
300 250
200 150
100 50
ε
mmmm
σ
MPa
0.10 0.20
0.30
Upper scale
0.40 0.00
0.0020 0.0030 0.0010
Lower scale
0.0040 0.0000
E = = 200x10
3
MPa= 200 GPa 0.001
σ
fail
= 295 MPa
σ
pl
= 200 MPa
σ
u
= 390 MPa
200 L
d
P P
6
400 350
300 250
200 150
100 50
ε
mmmm
σ
MPa
0.10 0.20
0.30
Upper scale
0.40 0.00
0.0020 0.0030 0.0010
Lower scale
0.0040 0.0000
σ
pl
= 180 MPa
σ
y
= 250 MPa
σ
u
= 360 MPa
σ
fail
= 310 MPa
E = = 200x10
3
MPa = 200 GPa 0.0005
100
7
Stress-Strain Behavior of Ductile and Brittle Materials
0.02 0.04
0.06 100
200 300
400
σ
MPa
ε
mmmm
σ−ε σ−ε
σ−ε σ−ε
diagrams for a methacrylate plastic
Brittle material
Ductile material
8
• Elongation
100 L
L L
elongation Percent
f
− =
100 A
A A
area of
reduction Percent
f
− =
9
• Temperature Effects:
0.02 0.04
0.06 100
200 300
400
σ
MPa
ε
mmmm
σ−ε σ−ε
σ−ε σ−ε diagrams for a methacrylate plastic
10
o
C
40
o
C
70
o
C Brittle to Ductile
Ductile to Brittle
10
σ
ε
Hooke’s Law
σ
pl
ε
pl
= =
pl pl
E
ε σ
Constant
11
- Apply load to failure • Elastic and Plastic Behavior of Materials
1
σ
ε
PL EL
Failure
12
- Apply and release load
σ
ε
PL EL
a Load is less than proportional limit
1
2
σ
ε
PL EL
b Load is more than proportional limit, but less than elastic limit
1
2
13
2 3
1
2 3
4
Mechanical hysteresis
1
σ
ε
PL EL
c Load is more than elastic limit, and reaply
σ
ε
PL EL
d Repeated load is more than elastic limit loading
14
2
σ
ε
PL EL
Repeated loading n times
3
4
mechanical hysteresis
1
σ
εεεε
O
1
Apply load once
- Comparison
15
E E
elastic region
plastic region
O´
unload
ε σ
A
O load
A´
B
permanent set
elastic region
plastic region
elastic recovery
ε σ
O´ A´
mechanical hysteresis
16
u
t
u
r
Strain Energy
• Modulus of Resilience
σ
pl
ε
pl
ε σ
Modulus of resilience u
r
• Modulus of Toughness
ε σ
Modulus of toughness u
t
E pl
u
pl pl
r 2
2 1
2 1
σ ε
σ
= =
17
Modulus of resiliency u
r
= Area under curve proportional limit u
r
= 120.001180 MPa = 90 kPa = 90 kNm
2
= 90 kN•mm
3
= 90 kJm
3
Upper scale Lower scale
400 350
300 250
200 150
100 50
ε
mmmm
σ
MPa
0.10 0.20
0.30 0.40
0.00 0.0020 0.0030
0.0010 0.0040
0.0000
Energy per unit volume = 90 kNm
2
1 m
3
= 90 kN•m = 90 kJ σ
σ σ
σ
pl
= 180 MPa
Modulus of resiliency • Modulus of Resilience
18
400 350
300 250
200 150
100 50
ε
mmmm
σ
MPa
0.10 0.20
0.30 0.40
0.00 0.0020 0.0030
0.0010
Upper scale Lower scale
0.0040 0.0000
Modulus of hyper-resiliency EL
2 1
3
19
400 350
300 250
200 150
100 50
εεεε mmmm σ
σ σ
σ MPa
0.10 0.20
0.30
Upper scale
0.40 0.00
0.0020 0.0030 0.0010
Lower scale
0.0040 0.0000
Failure
Modulus of toughness • Modulus of Toughness
20
Example 1
A tension test for a steel alloy results in the stress-strain diagram shown. Calculate the modulus of elasticity and the yield strength based on a 0.2 offset.
Identify on the graph the proportional limit, elastic limit, ultimate stress and the fracture stress.
400 350
300 250
200 150
100 50
ε
mmmm
σ
MPa
0.10 0.20
0.30
Upper scale
0.40 0.00
0.0020 0.0030 0.0010
Lower scale
0.0040 0.0000
21
400 350
300 250
200 150
100 50
ε
mmmm
σ
MPa
0.10 0.20
0.30
Upper scale
0.40 0.00
0.0020 0.0030 0.0010
Lower scale
0.0040 0.0000
σ σ
σ σ
y
E
• Modulus of Elasticity
GPa mm
mm MPa
E 200
0005 .
100 =
=
• Yield Strength
σ
y
= 250 MPa
σ σ
σ σ
pl
• Proportional Limit
σ
pl
= 180 MPa σ
σ σ
σ
EL
• Elastic Limit
σ
pl
= 220 MPa σ
σ σ
σ
u
• Ultimate Stress
σ
u
= 365 MPa
• Fracture Stress
σ
f
= 310 MPa σ
σ σ
σ
fail
22
Example 2
An aluminum specimen shown has a diameter of d = 25 mm, a gauge length of
L = 250 mm and is subjected to an axial load of 294.5 kN. If a portion of the
stress-strain diagram for the material is shown, determine the approximate elongation of the rod when the load is applied. If the load is removed, does the
rod return to its original length? Also, compute the modulus of resilience both before and after the load application.
0.01 0.02
0.04 150
300 450
σ
MPa
ε
mmmm 0.03
750 600
L d
294.5 kN 294.5 kN
23
0.01 0.02
0.04 150
300 450
σ
MPa
ε
mmmm 0.03
750 600
F
O 294.5 kN
294.5 kN
L = 250 mm
294.5 kN 294.5 kN
L = 250 mm
d = 25 mm
• The Load is Applied
- Normal Stress
MPa m
N A
P 600
025 .
4 10
5 .
294
2 3
= ×
= =
π σ
- The Strain
ε
= 0.023 mmmm
- The Elongation
δ
=
ε
L = 0.023 mmmm250 mm = 5.75 mm
D
0.023 5.75 mm2
5.75 mm2 A
B
24
294.5 kN 294.5 kN
• The Load is Removed
0.01 0.02
0.04 150
300 450
σ
MPa
ε
mmmm 0.03
750 600
F
O D
0.023 A
B L
= 250 mm d
= 25 mm
- Permanent Strain
CD MPa
GPa mm
mm MPa
E 600
. 75
006 .
450 =
= =
CD = 0.008 mmmm The permanent strain, ε
OC
= 0.023 - CD ε
OC
= 0.023 - 0.008 = 0.015 mmmm
- Normal Stress
MPa m
N A
P 600
025 .
4 10
5 .
294
2 3
= ×
= =
π σ
- The Permanent Elongation
δ
=
ε
L = 0.015 mmmm250 mm = 3.75 mm
ε
pl
=0.006 ε
OC
σ
pl
=
E E
CD C
294.5 kN 294.5 kN
L = 250 mm
3.75 mm2 3.75 mm2
25
u
r initial
u
r final
0.01 0.02
0.04 150
300 450
σ
MPa
ε
mmmm 0.03
750 600
F
O D
0.023 A
B L
= 250 mm d
= 25 mm 294.5 kN
294.5 kN
- Normal Stress
MPa m
N A
P 600
025 .
4 10
5 .
294
2 3
= ×
= =
π σ
ε
pl
=0.006 σ
pl
=
CD = 0.008 C
• Modulus of Resilience
- Modulus of Resilience
3
40 .
2 008
. 600
2 1
2 1
m MJ
mm mm
MPa u
pl pl
final r
= =
=
ε σ
3
35 .
1 006
. 450
2 1
2 1
m MJ
mm mm
MPa u
pl pl
initial r
= =
=
ε σ
26
Example 3
An aluminum rod shown has a circular cross section and is subjected to an axial load of 10 kN. If a portion of the stress-strain diagram for the material is shown,
determine the approximate elongation of the rod when the load is applied. If the load is removed, does the rod return to its original length? Take E
al
= 70 GPa.
600 mm 400 mm
20 mm 15 mm
A B
C 10 kN
10 kN
50 60
40
O
ε
mmmm
σ
MPa
0.02 0.08
0.10 0.04
0.06 0.12
30 20
10
27
50 60
40
O
ε
mmmm
σ
MPa
0.02 0.08 0.10
0.04 0.06
0.12 30
20 10
• The Load is Applied
600 mm 400 mm
20 mm 15 mm
A B
C 10 kN
10 kN
MPa m
kN A
P
AB
83 .
31 01
. 10
2
= =
=
π σ
MPa m
kN A
P
BC
6 .
56 0075
. 10
2
= =
=
π σ
mm mm
Pa Pa
E
al AB
AB
0004547 .
10 70
10 83
. 31
9 6
= ×
× =
=
σ ε
ε
BC
= 0.045 mmmm
The elongation of the rod is δ = Σ
ε
L =
ε
AB
L
AB
+
ε
BC
L
BC
= 0.0004547600 mm + 0.045400 mm = 18.3 mm
31.83 56.6
28
50 60
40
O
ε
mmmm
σ
MPa
0.02 0.08 0.10
0.04 0.06
0.12 30
20 10
• The Load is Removed
600 mm 400 mm
20 mm 15 mm
A B
C
MPa m
kN A
P
AB
83 .
31 01
. 10
2
= =
=
π σ
MPa m
kN A
P
BC
6 .
56 0075
. 10
2
= =
=
π σ
The elongation of the rod is δ´ = Σ
ε
L = 0 +
ε
OG
L
BC
= 0.0442400 mm = 17.7 mm
parallel
G ε
BC
= 0.045 mmmm mm
mm Pa
Pa E
al BC
rec
000808 .
10 70
10 6
. 56
9 6
= ×
× =
=
σ ε
ε
OG
The permanent strain, ε
OG
= 0.0450 - 0.000808 = 0.0442 mmmm 56.6
10 kN 10 kN
σ
pl
= 31.83
29
L
Tension r
Poisson’s Ratio
L
Compression Original Shape
r
P P
Final Shape δ´
δ2 δ2
Original Shape Final Shape
δ2 δ2
P P
δ´
r and
L
lat long
δ ε
δ ε
= =
long lat
ε ε
ν
− =
30
P
z x
y t
P
L b
z y
z x
ε ε
ε ε
ν
− =
− =
Assumption:
• Homogeneous
• Isotropic • Elastic
31
Example 4
A bar made of A-36 steel has the dimensions shown. If an axial force of P = 80 kN is applied to the bar, determine the change in its length and the change in the
dimensions of its cross section after applying the load. The material behaves elastically. Take E = 200 GPa and ν
st
= 0.32.
P = 80 kN
z x
y 50 mm
P = 80 kN
1.5 m 100 mm
32
MPa m
m kN
A P
z
16 05
. 1
. 80
= =
=
σ
z
Pa Pa
ε
10 16
10 200
6 9
=
δ
z
= ε
z
L
z
= [8010
-6
1.5 m] = 120 µm
P = 80 kN
z x
y 50 mm
P = 80 kN
1.5 m 100 mm
• The change in the bar’s length δ δ
δ δ
- z direction
z z
st
E
ε σ
=
µ ε
80 10
80
6
= =
−
mm mm
z
- x and y direction
z y
z x
long lat
st
Ratio s
Poisson
ε ε
ε ε
ε ε
ν
− =
− =
− =
:
ε
x
= ε
y
= - ν
st
ε
z
= -0.32[8010
-6
] = -25.6 µ
δ
x
= -ε
x
L
x
= -[25.610
-6
0.1 m = -2.56 µm
δ
y
= -ε
y
L
y
= -[25.610
-6
0.05 m = -1.28 µm
33
x y
The Shear Stress-Strain Diagram
x y
τ
xy
π2 - γ
xy
γ
xy
2 γ
xy
2
τ
= Gγ
τ
γ
τ
pl
γ
pl
τ
u
γ
u
τ
f
γ
f
G 1
2
ν
+ =
E G
34
Example 5
A specimen of titanium alloy is tested in torsion and the shear stress-strain diagram is shown.
a Determine the shear modulus G, the proportional limit, and the ultimate shear stress.
b Determine the distance d that the top of a block of this material, shown, could be displaced horizontally by a shear force V of 135 MN .
300
200
O
γ
rad
τ
MPa
0.73 100
400
0.54 0.008
50 mm 100 mm
75 mm
γ d
V
35
300
200
O
γ
rad
τ
MPa
0.73 100
400
- Proportional limit ;
τ
pl
= 270 MPa
- Ultimate shear stress; τ
u
= 370 MPa
γ
pl
= 0.008
G
a The shear modulus, proportional limit, and the ultimate shear stress.
GPa rad
MPa G
75 .
33 008
. 270
= =
- Shear Modulus ;
τ
u
= 370
τ
pl
= 270
36
50 mm 100 mm
75 mm
1.35 MN