43
Simple Stress and Strain Lecturer; Dr. Dawood S. Atrushi P P' P' r D d 1 2 d 1 2 max ave November 2014
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Simple Stress and Strain
Lecturer; Dr. Dawood S. Atrushi
115 You should also observe that the plates used to obtain a uniform stress distribution in the member of Fig. 2.55 must allow the mem-ber to freely expand laterally. Thus, the plates cannot be rigidly attached to the member; you must assume them to be just in contact with the member, and smooth enough not to impede the lateral expansion of the member. While such end conditions can actually be achieved for a member in compression, they cannot be physically realized in the case of a member in tension. It does not matter, however, whether or not an actual fixture can be realized and used to load a member so that the distribution of stresses in the member is uniform. The important thing is to imagine a model that will allow such a distribution of stresses, and to keep this model in mind so that you may later compare it with the actual loading conditions.
2.18 STRESS CONCENTRATIONSAs you saw in the preceding section, the stresses near the points of application of concentrated loads can reach values much larger than the average value of the stress in the member. When a structural member contains a discontinuity, such as a hole or a sudden change in cross section, high localized stresses can also occur near the dis-continuity. Figures 2.58 and 2.59 show the distribution of stresses in critical sections corresponding to two such situations. Figure 2.58 refers to a flat bar with a circular hole and shows the stress distribu-tion in a section passing through the center of the hole. Figure 2.59 refers to a flat bar consisting of two portions of different widths con-nected by fillets; it shows the stress distribution in the narrowest part of the connection, where the highest stresses occur. These results were obtained experimentally through the use of a photoelastic method. Fortunately for the engineer who has to design a given member and cannot afford to carry out such an analy-sis, the results obtained are independent of the size of the member and of the material used; they depend only upon the ratios of the geometric parameters involved, i.e., upon the ratio ryd in the case of a circular hole, and upon the ratios ryd and Dyd in the case of fillets. Furthermore, the designer is more interested in the maximum value of the stress in a given section, than in the actual distribution of stresses in that section, since the main concern is to determine whether the allowable stress will be exceeded under a given loading, and not where this value will be exceeded. For this reason, one defines the ratio
K 5
smax
save (2.48)
of the maximum stress over the average stress computed in the critical (narrowest) section of the discontinuity. This ratio is referred to as the stress-concentration factor of the given discontinuity. Stress- concentration factors can be computed once and for all in terms of the ratios of the geometric parameters involved, and the results obtained can be expressed in the form of tables or of graphs, as
2.18 Stress Concentrations
PP'
P'
rD
d12
d12
!max
!ave
Fig. 2.58 Stress distribution near circular hole in flat bar under axial loading.
PP'
P'
!max
!ave
dD
r
Fig. 2.59 Stress distribution near fillets in flat bar under axial loading.
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November 2014
Content
¢ Elastic deformations of axially loaded members
¢ Statically determinate and indeterminate members loaded axially
¢ Average shear Stress ¢ Stress concentrations ¢ Allowable stress ¢ IS Units
November, 14 2 Strenght of Materials I - DAT
Elastic Deformation of Axially Loaded Members
Consider a bar with ¢ gradually varying cross section along
its length, and ¢ subjected to concentrated loads at its
right end and also a variable external load distributed along its length
November, 14 Strenght of Materials I - DAT 3
Mechanical deformation
November, 14 Strenght of Materials I - DAT 4
Lecture Notes of Mechanics of Solids, Chapter 2 8
2.7 ELASTIC DEFORMATION OF AXIALLY LOADED MEMBER (SI&4
th p.120-127;3
rd p.122-129)
Now we are going to find the elastic deformation of a member subjected to axial loads. Let’s
consider a generalized bar shown in Fig. 2.7, which has a gradually varying cross-sectional
are along its length L. For a more general case, the bar is subjected to concentrated loads at its
right end and also a variable external load distributed along its length (such as a distributed
load could be for example, to represent the weight of a vertical bar or friction forces acting on
bar surface). Here we wish to find the relative displacement G of one end with respect to the other.
x
P(x)
dx
G
dx (original length)
P(x)+dP
dG (elongation of dx)
P
FBD
a
a View a-a
A(x)
L
Fig. 2.7 Thermal and mechanical deformation
We pick a differential element of length dx and cross-sectional area A(x). FBD can be drawn
as middle of Fig. 2.7. Assume that resultant internal axial force is represented as P(x). The
load P(x) will deform the element into the shape indicated by the dashed outline.
The average stress in the cross-sectional area would be � �)(
)(
xAxPx V
The average strain in the cross-sectional area would be � �dxdx GH
Provided these quantities do not exceed the proportional limit, we can relate them using
Hook’s law, i.e. H V E
Therefore
� � ¸¹·
¨©§
dxdxE
xAxP G)(
)(
Re-organize the equation, we have
� �dxxExA
xPd)(
)( G
For the entire length L of the bar, we must integrate this expression to find the required end
displacement
� �³ L
dxxExA
xP
0)(
)(G (2.16)
Where: G = displacement between two points L = distance between the points
P(x) = Internal axial force distribution
A(x) = Cross-sectional area
E(x) = Young’s modulus
Generalized bar with gradually varying cross-sectional along its length
Lecture Notes of Mechanics of Solids, Chapter 2 8
2.7 ELASTIC DEFORMATION OF AXIALLY LOADED MEMBER (SI&4
th p.120-127;3
rd p.122-129)
Now we are going to find the elastic deformation of a member subjected to axial loads. Let’s
consider a generalized bar shown in Fig. 2.7, which has a gradually varying cross-sectional
are along its length L. For a more general case, the bar is subjected to concentrated loads at its
right end and also a variable external load distributed along its length (such as a distributed
load could be for example, to represent the weight of a vertical bar or friction forces acting on
bar surface). Here we wish to find the relative displacement G of one end with respect to the other.
x
P(x)
dx
G
dx (original length)
P(x)+dP
dG (elongation of dx)
P
FBD
a
a View a-a
A(x)
L
Fig. 2.7 Thermal and mechanical deformation
We pick a differential element of length dx and cross-sectional area A(x). FBD can be drawn
as middle of Fig. 2.7. Assume that resultant internal axial force is represented as P(x). The
load P(x) will deform the element into the shape indicated by the dashed outline.
The average stress in the cross-sectional area would be � �)(
)(
xAxPx V
The average strain in the cross-sectional area would be � �dxdx GH
Provided these quantities do not exceed the proportional limit, we can relate them using
Hook’s law, i.e. H V E
Therefore
� � ¸¹·
¨©§
dxdxE
xAxP G)(
)(
Re-organize the equation, we have
� �dxxExA
xPd)(
)( G
For the entire length L of the bar, we must integrate this expression to find the required end
displacement
� �³ L
dxxExA
xP
0)(
)(G (2.16)
Where: G = displacement between two points L = distance between the points
P(x) = Internal axial force distribution
A(x) = Cross-sectional area
E(x) = Young’s modulus
¢ The average stress in the cross-sectional area;
November, 14 Strenght of Materials I - DAT 5
Lecture Notes of Mechanics of Solids, Chapter 2 8
2.7 ELASTIC DEFORMATION OF AXIALLY LOADED MEMBER (SI&4
th p.120-127;3
rd p.122-129)
Now we are going to find the elastic deformation of a member subjected to axial loads. Let’s
consider a generalized bar shown in Fig. 2.7, which has a gradually varying cross-sectional
are along its length L. For a more general case, the bar is subjected to concentrated loads at its
right end and also a variable external load distributed along its length (such as a distributed
load could be for example, to represent the weight of a vertical bar or friction forces acting on
bar surface). Here we wish to find the relative displacement G of one end with respect to the other.
x
P(x)
dx
G
dx (original length)
P(x)+dP
dG (elongation of dx)
P
FBD
a
a View a-a
A(x)
L
Fig. 2.7 Thermal and mechanical deformation
We pick a differential element of length dx and cross-sectional area A(x). FBD can be drawn
as middle of Fig. 2.7. Assume that resultant internal axial force is represented as P(x). The
load P(x) will deform the element into the shape indicated by the dashed outline.
The average stress in the cross-sectional area would be � �)(
)(
xAxPx V
The average strain in the cross-sectional area would be � �dxdx GH
Provided these quantities do not exceed the proportional limit, we can relate them using
Hook’s law, i.e. H V E
Therefore
� � ¸¹·
¨©§
dxdxE
xAxP G)(
)(
Re-organize the equation, we have
� �dxxExA
xPd)(
)( G
For the entire length L of the bar, we must integrate this expression to find the required end
displacement
� �³ L
dxxExA
xP
0)(
)(G (2.16)
Where: G = displacement between two points L = distance between the points
P(x) = Internal axial force distribution
A(x) = Cross-sectional area
E(x) = Young’s modulus
Lecture Notes of Mechanics of Solids, Chapter 2 8
2.7 ELASTIC DEFORMATION OF AXIALLY LOADED MEMBER (SI&4
th p.120-127;3
rd p.122-129)
Now we are going to find the elastic deformation of a member subjected to axial loads. Let’s
consider a generalized bar shown in Fig. 2.7, which has a gradually varying cross-sectional
are along its length L. For a more general case, the bar is subjected to concentrated loads at its
right end and also a variable external load distributed along its length (such as a distributed
load could be for example, to represent the weight of a vertical bar or friction forces acting on
bar surface). Here we wish to find the relative displacement G of one end with respect to the other.
x
P(x)
dx
G
dx (original length)
P(x)+dP
dG (elongation of dx)
P
FBD
a
a View a-a
A(x)
L
Fig. 2.7 Thermal and mechanical deformation
We pick a differential element of length dx and cross-sectional area A(x). FBD can be drawn
as middle of Fig. 2.7. Assume that resultant internal axial force is represented as P(x). The
load P(x) will deform the element into the shape indicated by the dashed outline.
The average stress in the cross-sectional area would be � �)(
)(
xAxPx V
The average strain in the cross-sectional area would be � �dxdx GH
Provided these quantities do not exceed the proportional limit, we can relate them using
Hook’s law, i.e. H V E
Therefore
� � ¸¹·
¨©§
dxdxE
xAxP G)(
)(
Re-organize the equation, we have
� �dxxExA
xPd)(
)( G
For the entire length L of the bar, we must integrate this expression to find the required end
displacement
� �³ L
dxxExA
xP
0)(
)(G (2.16)
Where: G = displacement between two points L = distance between the points
P(x) = Internal axial force distribution
A(x) = Cross-sectional area
E(x) = Young’s modulus
¢ The average strain in the cross-sectional area;
Using Hook’s law, i.e. σ = Eε;
November, 14 Strenght of Materials I - DAT 6
Lecture Notes of Mechanics of Solids, Chapter 2 8
2.7 ELASTIC DEFORMATION OF AXIALLY LOADED MEMBER (SI&4
th p.120-127;3
rd p.122-129)
Now we are going to find the elastic deformation of a member subjected to axial loads. Let’s
consider a generalized bar shown in Fig. 2.7, which has a gradually varying cross-sectional
are along its length L. For a more general case, the bar is subjected to concentrated loads at its
right end and also a variable external load distributed along its length (such as a distributed
load could be for example, to represent the weight of a vertical bar or friction forces acting on
bar surface). Here we wish to find the relative displacement G of one end with respect to the other.
x
P(x)
dx
G
dx (original length)
P(x)+dP
dG (elongation of dx)
P
FBD
a
a View a-a
A(x)
L
Fig. 2.7 Thermal and mechanical deformation
We pick a differential element of length dx and cross-sectional area A(x). FBD can be drawn
as middle of Fig. 2.7. Assume that resultant internal axial force is represented as P(x). The
load P(x) will deform the element into the shape indicated by the dashed outline.
The average stress in the cross-sectional area would be � �)(
)(
xAxPx V
The average strain in the cross-sectional area would be � �dxdx GH
Provided these quantities do not exceed the proportional limit, we can relate them using
Hook’s law, i.e. H V E
Therefore
� � ¸¹·
¨©§
dxdxE
xAxP G)(
)(
Re-organize the equation, we have
� �dxxExA
xPd)(
)( G
For the entire length L of the bar, we must integrate this expression to find the required end
displacement
� �³ L
dxxExA
xP
0)(
)(G (2.16)
Where: G = displacement between two points L = distance between the points
P(x) = Internal axial force distribution
A(x) = Cross-sectional area
E(x) = Young’s modulus
Lecture Notes of Mechanics of Solids, Chapter 2 8
2.7 ELASTIC DEFORMATION OF AXIALLY LOADED MEMBER (SI&4
th p.120-127;3
rd p.122-129)
Now we are going to find the elastic deformation of a member subjected to axial loads. Let’s
consider a generalized bar shown in Fig. 2.7, which has a gradually varying cross-sectional
are along its length L. For a more general case, the bar is subjected to concentrated loads at its
right end and also a variable external load distributed along its length (such as a distributed
load could be for example, to represent the weight of a vertical bar or friction forces acting on
bar surface). Here we wish to find the relative displacement G of one end with respect to the other.
x
P(x)
dx
G
dx (original length)
P(x)+dP
dG (elongation of dx)
P
FBD
a
a View a-a
A(x)
L
Fig. 2.7 Thermal and mechanical deformation
We pick a differential element of length dx and cross-sectional area A(x). FBD can be drawn
as middle of Fig. 2.7. Assume that resultant internal axial force is represented as P(x). The
load P(x) will deform the element into the shape indicated by the dashed outline.
The average stress in the cross-sectional area would be � �)(
)(
xAxPx V
The average strain in the cross-sectional area would be � �dxdx GH
Provided these quantities do not exceed the proportional limit, we can relate them using
Hook’s law, i.e. H V E
Therefore
� � ¸¹·
¨©§
dxdxE
xAxP G)(
)(
Re-organize the equation, we have
� �dxxExA
xPd)(
)( G
For the entire length L of the bar, we must integrate this expression to find the required end
displacement
� �³ L
dxxExA
xP
0)(
)(G (2.16)
Where: G = displacement between two points L = distance between the points
P(x) = Internal axial force distribution
A(x) = Cross-sectional area
E(x) = Young’s modulus
Lecture Notes of Mechanics of Solids, Chapter 2 8
2.7 ELASTIC DEFORMATION OF AXIALLY LOADED MEMBER (SI&4
th p.120-127;3
rd p.122-129)
Now we are going to find the elastic deformation of a member subjected to axial loads. Let’s
consider a generalized bar shown in Fig. 2.7, which has a gradually varying cross-sectional
are along its length L. For a more general case, the bar is subjected to concentrated loads at its
right end and also a variable external load distributed along its length (such as a distributed
load could be for example, to represent the weight of a vertical bar or friction forces acting on
bar surface). Here we wish to find the relative displacement G of one end with respect to the other.
x
P(x)
dx
G
dx (original length)
P(x)+dP
dG (elongation of dx)
P
FBD
a
a View a-a
A(x)
L
Fig. 2.7 Thermal and mechanical deformation
We pick a differential element of length dx and cross-sectional area A(x). FBD can be drawn
as middle of Fig. 2.7. Assume that resultant internal axial force is represented as P(x). The
load P(x) will deform the element into the shape indicated by the dashed outline.
The average stress in the cross-sectional area would be � �)(
)(
xAxPx V
The average strain in the cross-sectional area would be � �dxdx GH
Provided these quantities do not exceed the proportional limit, we can relate them using
Hook’s law, i.e. H V E
Therefore
� � ¸¹·
¨©§
dxdxE
xAxP G)(
)(
Re-organize the equation, we have
� �dxxExA
xPd)(
)( G
For the entire length L of the bar, we must integrate this expression to find the required end
displacement
� �³ L
dxxExA
xP
0)(
)(G (2.16)
Where: G = displacement between two points L = distance between the points
P(x) = Internal axial force distribution
A(x) = Cross-sectional area
E(x) = Young’s modulus
Where δ = displacement between two points L = distance between the points P(x) = Internal axial force distribution A(x) = Cross-sectional area E(x) = Young’s modulus
For the entire length L of the bar;
Constant Load and Cross-Sectional Area For P(x) = P = constant (no axially distributed load) A(x) = A = constant (uniform area) E(x) = E = constant (homogeneous material) the displacement is;
November, 14 Strenght of Materials I - DAT 7
Lecture Notes of Mechanics of Solids, Chapter 2 9
Constant Load and Cross-Sectional Area In many engineering cases, the structural members experience a constant load and have a constant cross-sectional area and made of one homogenous material, i.e.
P(x) = P = constant (no axially distributed load) A(x) = A = constant (uniform area) E(x) = E = constant (homogeneous material)
From Eq. (2.16), we have
EAPL
G (2.17)
Multi-Segment Bar If the bar is subjected to several different axial forces or cross-sectional areas or Young’s moduli, the above equation can be used for each segment. The total displacement can be computed from algebraic addition as
¦ i ii
ii
EALP
G (2.18)
Example 2.4: The composite bar shown in the figure is made of two segments, AB and BC, having cross-sectional areas of AAB=200mm2 and ABC=100mm2. Their Young’s moduli are EAB=100GPa and EBC=210GPa respectively. Find the total displacement at the right end.
Step 1 FBDs for Segments AB and BC. Assume the internal forces are in tension. Step 2 Equilibriums Internal force in AB
00 12 ��� �o� ¦� FFPF ABx kNPAB 30� ?
(Opposite to our assumption of tension, so Segment AB is in compression) Internal force in BC
00 1 �� �o� ¦� FPF BCx kNFPBC 101 ? (in tension)
Step 3 Compute the total deformation by using Eq. (2.18)
69
3
69
3
10100102102.41010
102001010041030
�� uuuuu
�uuuuu�
� � BCBC
BCBC
ABAB
ABABBCABAC AE
LPAELP
GGG
mmmAC 4004.0002.0006.0 � � �� G (towards left)
4m 4.2m
F1=10kNF2=40kN
A BC
E1A1 E2A2
F1=10kNF2=40kN
CE1A1 E2A2
B
PAB
F1=10kN
C
PBC
FBD1
FBD2
Multi-Segment Bar
¢ If the bar is subjected to several different axial forces or cross-sectional areas or Young’s moduli, the equation can be used for each segment. The total displacement can be computed from algebraic addition as;
November, 14 Strenght of Materials I - DAT 8
Lecture Notes of Mechanics of Solids, Chapter 2 9
Constant Load and Cross-Sectional Area In many engineering cases, the structural members experience a constant load and have a constant cross-sectional area and made of one homogenous material, i.e.
P(x) = P = constant (no axially distributed load) A(x) = A = constant (uniform area) E(x) = E = constant (homogeneous material)
From Eq. (2.16), we have
EAPL
G (2.17)
Multi-Segment Bar If the bar is subjected to several different axial forces or cross-sectional areas or Young’s moduli, the above equation can be used for each segment. The total displacement can be computed from algebraic addition as
¦ i ii
ii
EALP
G (2.18)
Example 2.4: The composite bar shown in the figure is made of two segments, AB and BC, having cross-sectional areas of AAB=200mm2 and ABC=100mm2. Their Young’s moduli are EAB=100GPa and EBC=210GPa respectively. Find the total displacement at the right end.
Step 1 FBDs for Segments AB and BC. Assume the internal forces are in tension. Step 2 Equilibriums Internal force in AB
00 12 ��� �o� ¦� FFPF ABx kNPAB 30� ?
(Opposite to our assumption of tension, so Segment AB is in compression) Internal force in BC
00 1 �� �o� ¦� FPF BCx kNFPBC 101 ? (in tension)
Step 3 Compute the total deformation by using Eq. (2.18)
69
3
69
3
10100102102.41010
102001010041030
�� uuuuu
�uuuuu�
� � BCBC
BCBC
ABAB
ABABBCABAC AE
LPAELP
GGG
mmmAC 4004.0002.0006.0 � � �� G (towards left)
4m 4.2m
F1=10kNF2=40kN
A BC
E1A1 E2A2
F1=10kNF2=40kN
CE1A1 E2A2
B
PAB
F1=10kN
C
PBC
FBD1
FBD2
Example 4 The composite bar shown in the figure is made of two segments, AB and BC, having cross-sectional areas of AAB=200 mm2 and ABC=100 mm2. Their Young’s moduli are EAB=100 GPa and EBC=210 GPa respectively. Find the total displacement at the right end.
November, 14 Strenght of Materials I - DAT 9
Lecture Notes of Mechanics of Solids, Chapter 2 9
Constant Load and Cross-Sectional Area In many engineering cases, the structural members experience a constant load and have a constant cross-sectional area and made of one homogenous material, i.e.
P(x) = P = constant (no axially distributed load) A(x) = A = constant (uniform area) E(x) = E = constant (homogeneous material)
From Eq. (2.16), we have
EAPL
G (2.17)
Multi-Segment Bar If the bar is subjected to several different axial forces or cross-sectional areas or Young’s moduli, the above equation can be used for each segment. The total displacement can be computed from algebraic addition as
¦ i ii
ii
EALP
G (2.18)
Example 2.4: The composite bar shown in the figure is made of two segments, AB and BC, having cross-sectional areas of AAB=200mm2 and ABC=100mm2. Their Young’s moduli are EAB=100GPa and EBC=210GPa respectively. Find the total displacement at the right end.
Step 1 FBDs for Segments AB and BC. Assume the internal forces are in tension. Step 2 Equilibriums Internal force in AB
00 12 ��� �o� ¦� FFPF ABx kNPAB 30� ?
(Opposite to our assumption of tension, so Segment AB is in compression) Internal force in BC
00 1 �� �o� ¦� FPF BCx kNFPBC 101 ? (in tension)
Step 3 Compute the total deformation by using Eq. (2.18)
69
3
69
3
10100102102.41010
102001010041030
�� uuuuu
�uuuuu�
� � BCBC
BCBC
ABAB
ABABBCABAC AE
LPAELP
GGG
mmmAC 4004.0002.0006.0 � � �� G (towards left)
4m 4.2m
F1=10kNF2=40kN
A BC
E1A1 E2A2
F1=10kNF2=40kN
CE1A1 E2A2
B
PAB
F1=10kN
C
PBC
FBD1
FBD2
November, 14 Strenght of Materials I - DAT 10
Lecture Notes of Mechanics of Solids, Chapter 2 9
Constant Load and Cross-Sectional Area In many engineering cases, the structural members experience a constant load and have a constant cross-sectional area and made of one homogenous material, i.e.
P(x) = P = constant (no axially distributed load) A(x) = A = constant (uniform area) E(x) = E = constant (homogeneous material)
From Eq. (2.16), we have
EAPL
G (2.17)
Multi-Segment Bar If the bar is subjected to several different axial forces or cross-sectional areas or Young’s moduli, the above equation can be used for each segment. The total displacement can be computed from algebraic addition as
¦ i ii
ii
EALP
G (2.18)
Example 2.4: The composite bar shown in the figure is made of two segments, AB and BC, having cross-sectional areas of AAB=200mm2 and ABC=100mm2. Their Young’s moduli are EAB=100GPa and EBC=210GPa respectively. Find the total displacement at the right end.
Step 1 FBDs for Segments AB and BC. Assume the internal forces are in tension. Step 2 Equilibriums Internal force in AB
00 12 ��� �o� ¦� FFPF ABx kNPAB 30� ?
(Opposite to our assumption of tension, so Segment AB is in compression) Internal force in BC
00 1 �� �o� ¦� FPF BCx kNFPBC 101 ? (in tension)
Step 3 Compute the total deformation by using Eq. (2.18)
69
3
69
3
10100102102.41010
102001010041030
�� uuuuu
�uuuuu�
� � BCBC
BCBC
ABAB
ABABBCABAC AE
LPAELP
GGG
mmmAC 4004.0002.0006.0 � � �� G (towards left)
4m 4.2m
F1=10kNF2=40kN
A BC
E1A1 E2A2
F1=10kNF2=40kN
CE1A1 E2A2
B
PAB
F1=10kN
C
PBC
FBD1
FBD2
Solution Example 4
Step 1, Free body diagram
Step 2, Equilibriums Internal force in the whole bar, ABC;
November, 14 Strenght of Materials I - DAT 11
Solution Example 4
Lecture Notes of Mechanics of Solids, Chapter 2 9
Constant Load and Cross-Sectional Area In many engineering cases, the structural members experience a constant load and have a constant cross-sectional area and made of one homogenous material, i.e.
P(x) = P = constant (no axially distributed load) A(x) = A = constant (uniform area) E(x) = E = constant (homogeneous material)
From Eq. (2.16), we have
EAPL
G (2.17)
Multi-Segment Bar If the bar is subjected to several different axial forces or cross-sectional areas or Young’s moduli, the above equation can be used for each segment. The total displacement can be computed from algebraic addition as
¦ i ii
ii
EALP
G (2.18)
Example 2.4: The composite bar shown in the figure is made of two segments, AB and BC, having cross-sectional areas of AAB=200mm2 and ABC=100mm2. Their Young’s moduli are EAB=100GPa and EBC=210GPa respectively. Find the total displacement at the right end.
Step 1 FBDs for Segments AB and BC. Assume the internal forces are in tension. Step 2 Equilibriums Internal force in AB
00 12 ��� �o� ¦� FFPF ABx kNPAB 30� ?
(Opposite to our assumption of tension, so Segment AB is in compression) Internal force in BC
00 1 �� �o� ¦� FPF BCx kNFPBC 101 ? (in tension)
Step 3 Compute the total deformation by using Eq. (2.18)
69
3
69
3
10100102102.41010
102001010041030
�� uuuuu
�uuuuu�
� � BCBC
BCBC
ABAB
ABABBCABAC AE
LPAELP
GGG
mmmAC 4004.0002.0006.0 � � �� G (towards left)
4m 4.2m
F1=10kNF2=40kN
A BC
E1A1 E2A2
F1=10kNF2=40kN
CE1A1 E2A2
B
PAB
F1=10kN
C
PBC
FBD1
FBD2
Lecture Notes of Mechanics of Solids, Chapter 2 9
Constant Load and Cross-Sectional Area In many engineering cases, the structural members experience a constant load and have a constant cross-sectional area and made of one homogenous material, i.e.
P(x) = P = constant (no axially distributed load) A(x) = A = constant (uniform area) E(x) = E = constant (homogeneous material)
From Eq. (2.16), we have
EAPL
G (2.17)
Multi-Segment Bar If the bar is subjected to several different axial forces or cross-sectional areas or Young’s moduli, the above equation can be used for each segment. The total displacement can be computed from algebraic addition as
¦ i ii
ii
EALP
G (2.18)
Example 2.4: The composite bar shown in the figure is made of two segments, AB and BC, having cross-sectional areas of AAB=200mm2 and ABC=100mm2. Their Young’s moduli are EAB=100GPa and EBC=210GPa respectively. Find the total displacement at the right end.
Step 1 FBDs for Segments AB and BC. Assume the internal forces are in tension. Step 2 Equilibriums Internal force in AB
00 12 ��� �o� ¦� FFPF ABx kNPAB 30� ?
(Opposite to our assumption of tension, so Segment AB is in compression) Internal force in BC
00 1 �� �o� ¦� FPF BCx kNFPBC 101 ? (in tension)
Step 3 Compute the total deformation by using Eq. (2.18)
69
3
69
3
10100102102.41010
102001010041030
�� uuuuu
�uuuuu�
� � BCBC
BCBC
ABAB
ABABBCABAC AE
LPAELP
GGG
mmmAC 4004.0002.0006.0 � � �� G (towards left)
4m 4.2m
F1=10kNF2=40kN
A BC
E1A1 E2A2
F1=10kNF2=40kN
CE1A1 E2A2
B
PAB
F1=10kN
C
PBC
FBD1
FBD2
Internal force in the segment BC;
November, 14 Strenght of Materials I - DAT 12
Step 3, Deformation The total deformation;
Lecture Notes of Mechanics of Solids, Chapter 2 9
Constant Load and Cross-Sectional Area In many engineering cases, the structural members experience a constant load and have a constant cross-sectional area and made of one homogenous material, i.e.
P(x) = P = constant (no axially distributed load) A(x) = A = constant (uniform area) E(x) = E = constant (homogeneous material)
From Eq. (2.16), we have
EAPL
G (2.17)
Multi-Segment Bar If the bar is subjected to several different axial forces or cross-sectional areas or Young’s moduli, the above equation can be used for each segment. The total displacement can be computed from algebraic addition as
¦ i ii
ii
EALP
G (2.18)
Example 2.4: The composite bar shown in the figure is made of two segments, AB and BC, having cross-sectional areas of AAB=200mm2 and ABC=100mm2. Their Young’s moduli are EAB=100GPa and EBC=210GPa respectively. Find the total displacement at the right end.
Step 1 FBDs for Segments AB and BC. Assume the internal forces are in tension. Step 2 Equilibriums Internal force in AB
00 12 ��� �o� ¦� FFPF ABx kNPAB 30� ?
(Opposite to our assumption of tension, so Segment AB is in compression) Internal force in BC
00 1 �� �o� ¦� FPF BCx kNFPBC 101 ? (in tension)
Step 3 Compute the total deformation by using Eq. (2.18)
69
3
69
3
10100102102.41010
102001010041030
�� uuuuu
�uuuuu�
� � BCBC
BCBC
ABAB
ABABBCABAC AE
LPAELP
GGG
mmmAC 4004.0002.0006.0 � � �� G (towards left)
4m 4.2m
F1=10kNF2=40kN
A BC
E1A1 E2A2
F1=10kNF2=40kN
CE1A1 E2A2
B
PAB
F1=10kN
C
PBC
FBD1
FBD2
Lecture Notes of Mechanics of Solids, Chapter 2 9
Constant Load and Cross-Sectional Area In many engineering cases, the structural members experience a constant load and have a constant cross-sectional area and made of one homogenous material, i.e.
P(x) = P = constant (no axially distributed load) A(x) = A = constant (uniform area) E(x) = E = constant (homogeneous material)
From Eq. (2.16), we have
EAPL
G (2.17)
Multi-Segment Bar If the bar is subjected to several different axial forces or cross-sectional areas or Young’s moduli, the above equation can be used for each segment. The total displacement can be computed from algebraic addition as
¦ i ii
ii
EALP
G (2.18)
Example 2.4: The composite bar shown in the figure is made of two segments, AB and BC, having cross-sectional areas of AAB=200mm2 and ABC=100mm2. Their Young’s moduli are EAB=100GPa and EBC=210GPa respectively. Find the total displacement at the right end.
Step 1 FBDs for Segments AB and BC. Assume the internal forces are in tension. Step 2 Equilibriums Internal force in AB
00 12 ��� �o� ¦� FFPF ABx kNPAB 30� ?
(Opposite to our assumption of tension, so Segment AB is in compression) Internal force in BC
00 1 �� �o� ¦� FPF BCx kNFPBC 101 ? (in tension)
Step 3 Compute the total deformation by using Eq. (2.18)
69
3
69
3
10100102102.41010
102001010041030
�� uuuuu
�uuuuu�
� � BCBC
BCBC
ABAB
ABABBCABAC AE
LPAELP
GGG
mmmAC 4004.0002.0006.0 � � �� G (towards left)
4m 4.2m
F1=10kNF2=40kN
A BC
E1A1 E2A2
F1=10kNF2=40kN
CE1A1 E2A2
B
PAB
F1=10kN
C
PBC
FBD1
FBD2
Lecture Notes of Mechanics of Solids, Chapter 2 9
Constant Load and Cross-Sectional Area In many engineering cases, the structural members experience a constant load and have a constant cross-sectional area and made of one homogenous material, i.e.
P(x) = P = constant (no axially distributed load) A(x) = A = constant (uniform area) E(x) = E = constant (homogeneous material)
From Eq. (2.16), we have
EAPL
G (2.17)
Multi-Segment Bar If the bar is subjected to several different axial forces or cross-sectional areas or Young’s moduli, the above equation can be used for each segment. The total displacement can be computed from algebraic addition as
¦ i ii
ii
EALP
G (2.18)
Example 2.4: The composite bar shown in the figure is made of two segments, AB and BC, having cross-sectional areas of AAB=200mm2 and ABC=100mm2. Their Young’s moduli are EAB=100GPa and EBC=210GPa respectively. Find the total displacement at the right end.
Step 1 FBDs for Segments AB and BC. Assume the internal forces are in tension. Step 2 Equilibriums Internal force in AB
00 12 ��� �o� ¦� FFPF ABx kNPAB 30� ?
(Opposite to our assumption of tension, so Segment AB is in compression) Internal force in BC
00 1 �� �o� ¦� FPF BCx kNFPBC 101 ? (in tension)
Step 3 Compute the total deformation by using Eq. (2.18)
69
3
69
3
10100102102.41010
102001010041030
�� uuuuu
�uuuuu�
� � BCBC
BCBC
ABAB
ABABBCABAC AE
LPAELP
GGG
mmmAC 4004.0002.0006.0 � � �� G (towards left)
4m 4.2m
F1=10kNF2=40kN
A BC
E1A1 E2A2
F1=10kNF2=40kN
CE1A1 E2A2
B
PAB
F1=10kN
C
PBC
FBD1
FBD2
Statically Determinate Members Loaded Axially Statically determinate ¢ System with the same
number of unknown reactions as equations of statics.
¢ Known reactions can be determined strictly from equilibrium equations.
November, 14 Strenght of Materials I - DAT 13
Lecture Notes of Mechanics of Solids, Chapter 2 10
2.8 STATICALLY INDETERMINATE MEMBERS LOADED AXIALLY (SI&4th Ed p. 134-139; 3rd Ed p. 137-142) Statically Determinate and Indeterminate When a bar is supported at one end and subjected to an axial force P at the other end as shown in Fig. 2.8(a), there is only one unknown reaction force FA. By using the equations of statics, the unknown reaction can easily be determined. So such a system with the same number of unknown reactions as equations of statics is called statically determinate. – i.e. known reactions can be determined strictly from equilibrium equations.
FA
P(a) Statically determinate
P
(b) Statically indeterminate
FA
FB
C
B
AA
B
LAC
LCB
L
Fig. 2.8 Statically determinate and indeterminate structures
If the bar is also restricted at the free end as shown in Fig. 2.8(b), it has 2 unknown reactions FA and FB, one known force P and one equation of statics as:
00 �� n� ¦ PFFF BAy PFF BA �? (2.19)
It cannot be solved if do not introduce more other conditions. If the system has more unknown forces than equations of statics it is called statically indeterminate. Compatibility Conditions What we need is an additional equation that specifies how the structure is displaced due to the applied loading. Such an equation is usually termed the compatibility equation. Since there are 2 unknown and only 1 equation of statics herein, what we need is an additional equation that specifies how the structure is displaced due to the applied loading. Such an equation is usually termed the compatibility equation or kinematic conditions. In order to determine the compatibility for this example we need to determine how point C is going to move, and how much point B moves in relation to point A. Now, since both ends of the bar are fully fixed, then the total change in length between A and B must be zero. Basically the amount that length AC elongates CB contracts as shown in Fig. 2.9, so the equation can be written as:
0 G�G CBAC (2.20)
Statically Indeterminate Members Loaded Axially Statically indeterminate
The system that has more unknown forces than equations of statics.
The reactions can not be determined only from equilibrium equations.
November, 14 Strenght of Materials I - DAT 14
Lecture Notes of Mechanics of Solids, Chapter 2 10
2.8 STATICALLY INDETERMINATE MEMBERS LOADED AXIALLY (SI&4th Ed p. 134-139; 3rd Ed p. 137-142) Statically Determinate and Indeterminate When a bar is supported at one end and subjected to an axial force P at the other end as shown in Fig. 2.8(a), there is only one unknown reaction force FA. By using the equations of statics, the unknown reaction can easily be determined. So such a system with the same number of unknown reactions as equations of statics is called statically determinate. – i.e. known reactions can be determined strictly from equilibrium equations.
FA
P(a) Statically determinate
P
(b) Statically indeterminate
FA
FB
C
B
AA
B
LAC
LCB
L
Fig. 2.8 Statically determinate and indeterminate structures
If the bar is also restricted at the free end as shown in Fig. 2.8(b), it has 2 unknown reactions FA and FB, one known force P and one equation of statics as:
00 �� n� ¦ PFFF BAy PFF BA �? (2.19)
It cannot be solved if do not introduce more other conditions. If the system has more unknown forces than equations of statics it is called statically indeterminate. Compatibility Conditions What we need is an additional equation that specifies how the structure is displaced due to the applied loading. Such an equation is usually termed the compatibility equation. Since there are 2 unknown and only 1 equation of statics herein, what we need is an additional equation that specifies how the structure is displaced due to the applied loading. Such an equation is usually termed the compatibility equation or kinematic conditions. In order to determine the compatibility for this example we need to determine how point C is going to move, and how much point B moves in relation to point A. Now, since both ends of the bar are fully fixed, then the total change in length between A and B must be zero. Basically the amount that length AC elongates CB contracts as shown in Fig. 2.9, so the equation can be written as:
0 G�G CBAC (2.20)
Compatibility Conditions
¢ What we need is an additional equation that specifies how the structure is displaced due to the applied loading. Such an equation is usually termed the compatibility equation.
November, 14 Strenght of Materials I - DAT 15
Lecture Notes of Mechanics of Solids, Chapter 2 10
2.8 STATICALLY INDETERMINATE MEMBERS LOADED AXIALLY (SI&4th Ed p. 134-139; 3rd Ed p. 137-142) Statically Determinate and Indeterminate When a bar is supported at one end and subjected to an axial force P at the other end as shown in Fig. 2.8(a), there is only one unknown reaction force FA. By using the equations of statics, the unknown reaction can easily be determined. So such a system with the same number of unknown reactions as equations of statics is called statically determinate. – i.e. known reactions can be determined strictly from equilibrium equations.
FA
P(a) Statically determinate
P
(b) Statically indeterminate
FA
FB
C
B
AA
B
LAC
LCB
L
Fig. 2.8 Statically determinate and indeterminate structures
If the bar is also restricted at the free end as shown in Fig. 2.8(b), it has 2 unknown reactions FA and FB, one known force P and one equation of statics as:
00 �� n� ¦ PFFF BAy PFF BA �? (2.19)
It cannot be solved if do not introduce more other conditions. If the system has more unknown forces than equations of statics it is called statically indeterminate. Compatibility Conditions What we need is an additional equation that specifies how the structure is displaced due to the applied loading. Such an equation is usually termed the compatibility equation. Since there are 2 unknown and only 1 equation of statics herein, what we need is an additional equation that specifies how the structure is displaced due to the applied loading. Such an equation is usually termed the compatibility equation or kinematic conditions. In order to determine the compatibility for this example we need to determine how point C is going to move, and how much point B moves in relation to point A. Now, since both ends of the bar are fully fixed, then the total change in length between A and B must be zero. Basically the amount that length AC elongates CB contracts as shown in Fig. 2.9, so the equation can be written as:
0 G�G CBAC (2.20)
Compatibility Condition
November, 14 Strenght of Materials I - DAT 16
Lecture Notes of Mechanics of Solids, Chapter 2 11
FA
C
B
A
G=0C’
G =GAC +GCB=0
FA
GAC
C
C’PAC
FA
FBD
B
C
C’Elongated
Contracted
GCB
FB
PCB
FBD
Fig. 2.9 Compatibility condition
Let’s respectively look at the free body diagram for segment AC and CB as in Fig. 2.9. (indeed FBD can be in any level of structural system or structural members). For segment AC,
00 � n� ¦ ACAyPFF AAC FP ? Tension (+)
ACAC
ACA
ACAC
ACACAC EA
LFEALP
G elongation (+) (2.21)
For segment CB,
00 � n� ¦ CBBy PFF BCB FP � ? Compression (�)
CBCB
CBB
CBCB
CBCBCB EA
LFEALP
� � G contraction (�) (2.22)
Compatibility condition:
0 ¸̧¹
·¨̈©
§�� �
CBCB
CBB
ACAC
ACACBAC EA
LFEALF
GGG (2.23)
Combining Compatibility equation (2.23) with the equation of statics (2.19), we now can solve for the two unknowns FA and FB as,
°¯
°®
�
�
PFFEALF
EALF
BA
CBCB
CBB
ACAC
ACA 0 (2.24)
i.e. P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
ACAC
AC
B
� P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
CBCB
CB
A
�
If ConstEAEA CBCBACAC , we have
PL
LF AC
B and PL
LF CB
A (2.25)
The amount that length AC elongates CB contracts
¢ Now, since both ends of the bar are fully fixed, then the total change in length between A and B must be zero.
¢ The amount that length AC elongates CB contracts; so the equation can be written as:
δAC +δCB =0
November, 14 Strenght of Materials I - DAT 17
From the free body diagram for segment AC and CB;
November, 14 Strenght of Materials I - DAT 18
Lecture Notes of Mechanics of Solids, Chapter 2 11
FA
C
B
A
G=0C’
G =GAC +GCB=0
FA
GAC
C
C’PAC
FA
FBD
B
C
C’Elongated
Contracted
GCB
FB
PCB
FBD
Fig. 2.9 Compatibility condition
Let’s respectively look at the free body diagram for segment AC and CB as in Fig. 2.9. (indeed FBD can be in any level of structural system or structural members). For segment AC,
00 � n� ¦ ACAyPFF AAC FP ? Tension (+)
ACAC
ACA
ACAC
ACACAC EA
LFEALP
G elongation (+) (2.21)
For segment CB,
00 � n� ¦ CBBy PFF BCB FP � ? Compression (�)
CBCB
CBB
CBCB
CBCBCB EA
LFEALP
� � G contraction (�) (2.22)
Compatibility condition:
0 ¸̧¹
·¨̈©
§�� �
CBCB
CBB
ACAC
ACACBAC EA
LFEALF
GGG (2.23)
Combining Compatibility equation (2.23) with the equation of statics (2.19), we now can solve for the two unknowns FA and FB as,
°¯
°®
�
�
PFFEALF
EALF
BA
CBCB
CBB
ACAC
ACA 0 (2.24)
i.e. P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
ACAC
AC
B
� P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
CBCB
CB
A
�
If ConstEAEA CBCBACAC , we have
PL
LF AC
B and PL
LF CB
A (2.25)
Lecture Notes of Mechanics of Solids, Chapter 2 11
FA
C
B
A
G=0C’
G =GAC +GCB=0
FA
GAC
C
C’PAC
FA
FBD
B
C
C’Elongated
Contracted
GCB
FB
PCB
FBD
Fig. 2.9 Compatibility condition
Let’s respectively look at the free body diagram for segment AC and CB as in Fig. 2.9. (indeed FBD can be in any level of structural system or structural members). For segment AC,
00 � n� ¦ ACAyPFF AAC FP ? Tension (+)
ACAC
ACA
ACAC
ACACAC EA
LFEALP
G elongation (+) (2.21)
For segment CB,
00 � n� ¦ CBBy PFF BCB FP � ? Compression (�)
CBCB
CBB
CBCB
CBCBCB EA
LFEALP
� � G contraction (�) (2.22)
Compatibility condition:
0 ¸̧¹
·¨̈©
§�� �
CBCB
CBB
ACAC
ACACBAC EA
LFEALF
GGG (2.23)
Combining Compatibility equation (2.23) with the equation of statics (2.19), we now can solve for the two unknowns FA and FB as,
°¯
°®
�
�
PFFEALF
EALF
BA
CBCB
CBB
ACAC
ACA 0 (2.24)
i.e. P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
ACAC
AC
B
� P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
CBCB
CB
A
�
If ConstEAEA CBCBACAC , we have
PL
LF AC
B and PL
LF CB
A (2.25)
November, 14 Strenght of Materials I - DAT 19
Lecture Notes of Mechanics of Solids, Chapter 2 11
FA
C
B
A
G=0C’
G =GAC +GCB=0
FA
GAC
C
C’PAC
FA
FBD
B
C
C’Elongated
Contracted
GCB
FB
PCB
FBD
Fig. 2.9 Compatibility condition
Let’s respectively look at the free body diagram for segment AC and CB as in Fig. 2.9. (indeed FBD can be in any level of structural system or structural members). For segment AC,
00 � n� ¦ ACAyPFF AAC FP ? Tension (+)
ACAC
ACA
ACAC
ACACAC EA
LFEALP
G elongation (+) (2.21)
For segment CB,
00 � n� ¦ CBBy PFF BCB FP � ? Compression (�)
CBCB
CBB
CBCB
CBCBCB EA
LFEALP
� � G contraction (�) (2.22)
Compatibility condition:
0 ¸̧¹
·¨̈©
§�� �
CBCB
CBB
ACAC
ACACBAC EA
LFEALF
GGG (2.23)
Combining Compatibility equation (2.23) with the equation of statics (2.19), we now can solve for the two unknowns FA and FB as,
°¯
°®
�
�
PFFEALF
EALF
BA
CBCB
CBB
ACAC
ACA 0 (2.24)
i.e. P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
ACAC
AC
B
� P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
CBCB
CB
A
�
If ConstEAEA CBCBACAC , we have
PL
LF AC
B and PL
LF CB
A (2.25)
Lecture Notes of Mechanics of Solids, Chapter 2 11
FA
C
B
A
G=0C’
G =GAC +GCB=0
FA
GAC
C
C’PAC
FA
FBD
B
C
C’Elongated
Contracted
GCB
FB
PCB
FBD
Fig. 2.9 Compatibility condition
Let’s respectively look at the free body diagram for segment AC and CB as in Fig. 2.9. (indeed FBD can be in any level of structural system or structural members). For segment AC,
00 � n� ¦ ACAyPFF AAC FP ? Tension (+)
ACAC
ACA
ACAC
ACACAC EA
LFEALP
G elongation (+) (2.21)
For segment CB,
00 � n� ¦ CBBy PFF BCB FP � ? Compression (�)
CBCB
CBB
CBCB
CBCBCB EA
LFEALP
� � G contraction (�) (2.22)
Compatibility condition:
0 ¸̧¹
·¨̈©
§�� �
CBCB
CBB
ACAC
ACACBAC EA
LFEALF
GGG (2.23)
Combining Compatibility equation (2.23) with the equation of statics (2.19), we now can solve for the two unknowns FA and FB as,
°¯
°®
�
�
PFFEALF
EALF
BA
CBCB
CBB
ACAC
ACA 0 (2.24)
i.e. P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
ACAC
AC
B
� P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
CBCB
CB
A
�
If ConstEAEA CBCBACAC , we have
PL
LF AC
B and PL
LF CB
A (2.25)
Lecture Notes of Mechanics of Solids, Chapter 2 11
FA
C
B
A
G=0C’
G =GAC +GCB=0
FA
GAC
C
C’PAC
FA
FBD
B
C
C’Elongated
Contracted
GCB
FB
PCB
FBD
Fig. 2.9 Compatibility condition
Let’s respectively look at the free body diagram for segment AC and CB as in Fig. 2.9. (indeed FBD can be in any level of structural system or structural members). For segment AC,
00 � n� ¦ ACAyPFF AAC FP ? Tension (+)
ACAC
ACA
ACAC
ACACAC EA
LFEALP
G elongation (+) (2.21)
For segment CB,
00 � n� ¦ CBBy PFF BCB FP � ? Compression (�)
CBCB
CBB
CBCB
CBCBCB EA
LFEALP
� � G contraction (�) (2.22)
Compatibility condition:
0 ¸̧¹
·¨̈©
§�� �
CBCB
CBB
ACAC
ACACBAC EA
LFEALF
GGG (2.23)
Combining Compatibility equation (2.23) with the equation of statics (2.19), we now can solve for the two unknowns FA and FB as,
°¯
°®
�
�
PFFEALF
EALF
BA
CBCB
CBB
ACAC
ACA 0 (2.24)
i.e. P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
ACAC
AC
B
� P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
CBCB
CB
A
�
If ConstEAEA CBCBACAC , we have
PL
LF AC
B and PL
LF CB
A (2.25)
Lecture Notes of Mechanics of Solids, Chapter 2 11
FA
C
B
A
G=0C’
G =GAC +GCB=0
FA
GAC
C
C’PAC
FA
FBD
B
C
C’Elongated
Contracted
GCB
FB
PCB
FBD
Fig. 2.9 Compatibility condition
Let’s respectively look at the free body diagram for segment AC and CB as in Fig. 2.9. (indeed FBD can be in any level of structural system or structural members). For segment AC,
00 � n� ¦ ACAyPFF AAC FP ? Tension (+)
ACAC
ACA
ACAC
ACACAC EA
LFEALP
G elongation (+) (2.21)
For segment CB,
00 � n� ¦ CBBy PFF BCB FP � ? Compression (�)
CBCB
CBB
CBCB
CBCBCB EA
LFEALP
� � G contraction (�) (2.22)
Compatibility condition:
0 ¸̧¹
·¨̈©
§�� �
CBCB
CBB
ACAC
ACACBAC EA
LFEALF
GGG (2.23)
Combining Compatibility equation (2.23) with the equation of statics (2.19), we now can solve for the two unknowns FA and FB as,
°¯
°®
�
�
PFFEALF
EALF
BA
CBCB
CBB
ACAC
ACA 0 (2.24)
i.e. P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
ACAC
AC
B
� P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
CBCB
CB
A
�
If ConstEAEA CBCBACAC , we have
PL
LF AC
B and PL
LF CB
A (2.25)
Combining compatibility equation with the equation of statics, we now can solve for the two unknowns FA and FB as;
November, 14 Strenght of Materials I - DAT 20
Lecture Notes of Mechanics of Solids, Chapter 2 11
FA
C
B
A
G=0C’
G =GAC +GCB=0
FA
GAC
C
C’PAC
FA
FBD
B
C
C’Elongated
Contracted
GCB
FB
PCB
FBD
Fig. 2.9 Compatibility condition
Let’s respectively look at the free body diagram for segment AC and CB as in Fig. 2.9. (indeed FBD can be in any level of structural system or structural members). For segment AC,
00 � n� ¦ ACAyPFF AAC FP ? Tension (+)
ACAC
ACA
ACAC
ACACAC EA
LFEALP
G elongation (+) (2.21)
For segment CB,
00 � n� ¦ CBBy PFF BCB FP � ? Compression (�)
CBCB
CBB
CBCB
CBCBCB EA
LFEALP
� � G contraction (�) (2.22)
Compatibility condition:
0 ¸̧¹
·¨̈©
§�� �
CBCB
CBB
ACAC
ACACBAC EA
LFEALF
GGG (2.23)
Combining Compatibility equation (2.23) with the equation of statics (2.19), we now can solve for the two unknowns FA and FB as,
°¯
°®
�
�
PFFEALF
EALF
BA
CBCB
CBB
ACAC
ACA 0 (2.24)
i.e. P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
ACAC
AC
B
� P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
CBCB
CB
A
�
If ConstEAEA CBCBACAC , we have
PL
LF AC
B and PL
LF CB
A (2.25)
Lecture Notes of Mechanics of Solids, Chapter 2 11
FA
C
B
A
G=0C’
G =GAC +GCB=0
FA
GAC
C
C’PAC
FA
FBD
B
C
C’Elongated
Contracted
GCB
FB
PCB
FBD
Fig. 2.9 Compatibility condition
Let’s respectively look at the free body diagram for segment AC and CB as in Fig. 2.9. (indeed FBD can be in any level of structural system or structural members). For segment AC,
00 � n� ¦ ACAyPFF AAC FP ? Tension (+)
ACAC
ACA
ACAC
ACACAC EA
LFEALP
G elongation (+) (2.21)
For segment CB,
00 � n� ¦ CBBy PFF BCB FP � ? Compression (�)
CBCB
CBB
CBCB
CBCBCB EA
LFEALP
� � G contraction (�) (2.22)
Compatibility condition:
0 ¸̧¹
·¨̈©
§�� �
CBCB
CBB
ACAC
ACACBAC EA
LFEALF
GGG (2.23)
Combining Compatibility equation (2.23) with the equation of statics (2.19), we now can solve for the two unknowns FA and FB as,
°¯
°®
�
�
PFFEALF
EALF
BA
CBCB
CBB
ACAC
ACA 0 (2.24)
i.e. P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
ACAC
AC
B
� P
EAL
EAL
EAL
F
CBCB
CB
ACAC
AC
CBCB
CB
A
�
If ConstEAEA CBCBACAC , we have
PL
LF AC
B and PL
LF CB
A (2.25)
Example 5 Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average normal stress in both bars if increase the temperature from 10°C to 210°C.
November, 14 Strenght of Materials I - DAT 21
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
November, 14 Strenght of Materials I - DAT 22
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
November, 14 Strenght of Materials I - DAT
23
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
Total deformation of copper;
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
Similarly, we have;
November, 14 Strenght of Materials I - DAT 24
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V � Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
These two expanding bars should fill the gap, we prescribe a compatibility condition as;
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
Solve for F; F = 206.2 N
November, 14 Strenght of Materials I - DAT 25
The average normal stress can be computed as;
Lecture Notes of Mechanics of Solids, Chapter 2 12
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutments. Originally, there is a gap of 5mm between the ends as shown in the figure. Determine average
normal stress in both bars if increase the temperature from 10qC to 210qC.
Copper AluminumDcu=17 × 10-6
Ecu=110GPaDal=23×10-6
Eal=69GPa
T0=10oC
T=210oC
'T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
GT,cu
GF,cu
GT,Al
GF,Al
F
F
Thermally expanded GT,cu due to 'T
'T
Mechanical force push it back by GF,cu
Copper
Copper
Aluminum
Aluminum
d = 0.01m
'T
522 10857010
4
143
4
�u S
...dA m2 and CT o20010210 � '
Let’s firstly look at the copper bar. When the bar system is heated up from 10qC to 210qC, the
copper bar expand towards right by Cu,TG . After the copper bar touch to the aluminum bar, a
mechanical force F will develop, which will prevent the copper bar from expanding further.
We assume that due to such a mechanical force, the copper bar is pressed back by Cu,FG . The
real total deformation of copper bar will be computed as
CuFCuTCu ,, GGG � (elongation +, Contraction �)
Similarly, we have
Al,FAl,TAl G�G G (elongation +, Contraction �)
Because these two expanding bars should fill the gap, we prescribe a compatibility condition as
0050.CuAl G�G (2.26)
From these two equations, we have
� � � � 0050.Al,FAl,TCu,FCu,T G�G�G�G
i.e. 0050.AELF
TLAELF
TLAlAl
AlAlAl
CuCu
CuCuCu ¸̧
¹
·¨̈©
§ u�'D�¸̧
¹
·¨̈©
§ u�'D
NAE
LAE
LTLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
1085.71069
8.0
1085.710110
4.0
005.08.020010234.02001017005.0
5959
66
uuu�
uuu
�uuu�uuu
�
�'�'
��
��DD
The average normal stress can be computed as
MPa..
.AF
63210857
22065
u V �
Average Shear Stress
Shear Stress The intensity or force per unit area acting tangentially to A is called Shear Stress, (tau), and it is expressed as;
November, 14 Strenght of Materials I - DAT 26
Lecture Notes of Mechanics of Solids, Chapter 2 13
o u u�u
uuuu
�uuu ¸̧¹
·¨̈©
§ u�' �
���
��
mmm
AELFTL
CuCu
CuCuCuCuFCuTCu
35.11035.11055.91036.1
1085.710110
4.02.2064.02001017
363
59
6
,,DGGG
m u u�u
uuuu
�uuu ¸̧¹
·¨̈©
§ u�' �
���
��
mmm
AELF
TLAlAl
AlAlAlAlFAlTAl
65.31065.31005.31068.3
1085.71069
8.02.2068.02001023
363
59
6
,,DGGG
2.9 AVERAGE SHEAR STRESS (SI&4th
Ed p. 32-39; 3rd
Ed p. 35-41)
The intensity or force per unit area acting tangentially to A is called Shear Stress, (tau), and it
is expressed as in Eq. (2.3) as:
AF
lim tA '
'W
' 0o (2.3)
In order to show how the shear stress can develop in a structural member, let’s take a block as
an example. The block is supported by two rigid bodies and an external force F is applied
vertically as shown in Fig. 2.10. If the force is large enough, it will cause the material of the
block to deform and fail along the vertical planes as shown.
A FBD of the unsupported center segment indicate that shear force V=F/2 must be applied at
each section to hold the segment in equilibrium.
F
F F
VV
A
B
C
D
Sectioned area A
Block
Rigid Rigid
Fig. 2.10 Average shear stress
020 � n� ¦ FVFy
2/FV ?
The average shear stress distributed over each sectioned area that develops the shear force is
defined by
AV
avg W (2.27)
Wavg = assume to be the same at each point over the section
V = Internal shear force
A = Area at the section
Let’s take a block supported by two rigid bodies and an external force F as an example;
Lecture Notes of Mechanics of Solids, Chapter 2 13
o u u�u
uuuu
�uuu ¸̧¹
·¨̈©
§ u�' �
���
��
mmm
AELFTL
CuCu
CuCuCuCuFCuTCu
35.11035.11055.91036.1
1085.710110
4.02.2064.02001017
363
59
6
,,DGGG
m u u�u
uuuu
�uuu ¸̧¹
·¨̈©
§ u�' �
���
��
mmm
AELF
TLAlAl
AlAlAlAlFAlTAl
65.31065.31005.31068.3
1085.71069
8.02.2068.02001023
363
59
6
,,DGGG
2.9 AVERAGE SHEAR STRESS (SI&4th
Ed p. 32-39; 3rd
Ed p. 35-41)
The intensity or force per unit area acting tangentially to A is called Shear Stress, (tau), and it
is expressed as in Eq. (2.3) as:
AF
lim tA '
'W
' 0o (2.3)
In order to show how the shear stress can develop in a structural member, let’s take a block as
an example. The block is supported by two rigid bodies and an external force F is applied
vertically as shown in Fig. 2.10. If the force is large enough, it will cause the material of the
block to deform and fail along the vertical planes as shown.
A FBD of the unsupported center segment indicate that shear force V=F/2 must be applied at
each section to hold the segment in equilibrium.
F
F F
VV
A
B
C
D
Sectioned area A
Block
Rigid Rigid
Fig. 2.10 Average shear stress
020 � n� ¦ FVFy
2/FV ?
The average shear stress distributed over each sectioned area that develops the shear force is
defined by
AV
avg W (2.27)
Wavg = assume to be the same at each point over the section
V = Internal shear force
A = Area at the section
November, 14 Strenght of Materials I - DAT 27
Lecture Notes of Mechanics of Solids, Chapter 2 13
o u u�u
uuuu
�uuu ¸̧¹
·¨̈©
§ u�' �
���
��
mmm
AELFTL
CuCu
CuCuCuCuFCuTCu
35.11035.11055.91036.1
1085.710110
4.02.2064.02001017
363
59
6
,,DGGG
m u u�u
uuuu
�uuu ¸̧¹
·¨̈©
§ u�' �
���
��
mmm
AELF
TLAlAl
AlAlAlAlFAlTAl
65.31065.31005.31068.3
1085.71069
8.02.2068.02001023
363
59
6
,,DGGG
2.9 AVERAGE SHEAR STRESS (SI&4th
Ed p. 32-39; 3rd
Ed p. 35-41)
The intensity or force per unit area acting tangentially to A is called Shear Stress, (tau), and it
is expressed as in Eq. (2.3) as:
AF
lim tA '
'W
' 0o (2.3)
In order to show how the shear stress can develop in a structural member, let’s take a block as
an example. The block is supported by two rigid bodies and an external force F is applied
vertically as shown in Fig. 2.10. If the force is large enough, it will cause the material of the
block to deform and fail along the vertical planes as shown.
A FBD of the unsupported center segment indicate that shear force V=F/2 must be applied at
each section to hold the segment in equilibrium.
F
F F
VV
A
B
C
D
Sectioned area A
Block
Rigid Rigid
Fig. 2.10 Average shear stress
020 � n� ¦ FVFy
2/FV ?
The average shear stress distributed over each sectioned area that develops the shear force is
defined by
AV
avg W (2.27)
Wavg = assume to be the same at each point over the section
V = Internal shear force
A = Area at the section
Lecture Notes of Mechanics of Solids, Chapter 2 13
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2.9 AVERAGE SHEAR STRESS (SI&4th
Ed p. 32-39; 3rd
Ed p. 35-41)
The intensity or force per unit area acting tangentially to A is called Shear Stress, (tau), and it
is expressed as in Eq. (2.3) as:
AF
lim tA '
'W
' 0o (2.3)
In order to show how the shear stress can develop in a structural member, let’s take a block as
an example. The block is supported by two rigid bodies and an external force F is applied
vertically as shown in Fig. 2.10. If the force is large enough, it will cause the material of the
block to deform and fail along the vertical planes as shown.
A FBD of the unsupported center segment indicate that shear force V=F/2 must be applied at
each section to hold the segment in equilibrium.
F
F F
VV
A
B
C
D
Sectioned area A
Block
Rigid Rigid
Fig. 2.10 Average shear stress
020 � n� ¦ FVFy
2/FV ?
The average shear stress distributed over each sectioned area that develops the shear force is
defined by
AV
avg W (2.27)
Wavg = assume to be the same at each point over the section
V = Internal shear force
A = Area at the section
Vertical equilibrium;
November, 14 Strenght of Materials I - DAT 28
Lecture Notes of Mechanics of Solids, Chapter 2 13
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�uuu ¸̧¹
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AELF
TLAlAl
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8.02.2068.02001023
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6
,,DGGG
2.9 AVERAGE SHEAR STRESS (SI&4th
Ed p. 32-39; 3rd
Ed p. 35-41)
The intensity or force per unit area acting tangentially to A is called Shear Stress, (tau), and it
is expressed as in Eq. (2.3) as:
AF
lim tA '
'W
' 0o (2.3)
In order to show how the shear stress can develop in a structural member, let’s take a block as
an example. The block is supported by two rigid bodies and an external force F is applied
vertically as shown in Fig. 2.10. If the force is large enough, it will cause the material of the
block to deform and fail along the vertical planes as shown.
A FBD of the unsupported center segment indicate that shear force V=F/2 must be applied at
each section to hold the segment in equilibrium.
F
F F
VV
A
B
C
D
Sectioned area A
Block
Rigid Rigid
Fig. 2.10 Average shear stress
020 � n� ¦ FVFy
2/FV ?
The average shear stress distributed over each sectioned area that develops the shear force is
defined by
AV
avg W (2.27)
Wavg = assume to be the same at each point over the section
V = Internal shear force
A = Area at the section
Average shear stress distributed over each section;
Lecture Notes of Mechanics of Solids, Chapter 2 13
o u u�u
uuuu
�uuu ¸̧¹
·¨̈©
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6
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2.9 AVERAGE SHEAR STRESS (SI&4th
Ed p. 32-39; 3rd
Ed p. 35-41)
The intensity or force per unit area acting tangentially to A is called Shear Stress, (tau), and it
is expressed as in Eq. (2.3) as:
AF
lim tA '
'W
' 0o (2.3)
In order to show how the shear stress can develop in a structural member, let’s take a block as
an example. The block is supported by two rigid bodies and an external force F is applied
vertically as shown in Fig. 2.10. If the force is large enough, it will cause the material of the
block to deform and fail along the vertical planes as shown.
A FBD of the unsupported center segment indicate that shear force V=F/2 must be applied at
each section to hold the segment in equilibrium.
F
F F
VV
A
B
C
D
Sectioned area A
Block
Rigid Rigid
Fig. 2.10 Average shear stress
020 � n� ¦ FVFy
2/FV ?
The average shear stress distributed over each sectioned area that develops the shear force is
defined by
AV
avg W (2.27)
Wavg = assume to be the same at each point over the section
V = Internal shear force
A = Area at the section
Assume to be uniform over the section Internal shear force Area at the section
Lecture Notes of Mechanics of Solids, Chapter 2 13
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�uuu ¸̧¹
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2.9 AVERAGE SHEAR STRESS (SI&4th
Ed p. 32-39; 3rd
Ed p. 35-41)
The intensity or force per unit area acting tangentially to A is called Shear Stress, (tau), and it
is expressed as in Eq. (2.3) as:
AF
lim tA '
'W
' 0o (2.3)
In order to show how the shear stress can develop in a structural member, let’s take a block as
an example. The block is supported by two rigid bodies and an external force F is applied
vertically as shown in Fig. 2.10. If the force is large enough, it will cause the material of the
block to deform and fail along the vertical planes as shown.
A FBD of the unsupported center segment indicate that shear force V=F/2 must be applied at
each section to hold the segment in equilibrium.
F
F F
VV
A
B
C
D
Sectioned area A
Block
Rigid Rigid
Fig. 2.10 Average shear stress
020 � n� ¦ FVFy
2/FV ?
The average shear stress distributed over each sectioned area that develops the shear force is
defined by
AV
avg W (2.27)
Wavg = assume to be the same at each point over the section
V = Internal shear force
A = Area at the section
Stress concentrations ¢ For a uniform cross-
sectional bar that is applied an axial force, both experiment and theory of elasticity find that the normal stress will be uniformly distributed over the cross-section.
November, 14 Strenght of Materials I - DAT 29
71.3 STRESSES IN THE MEMBERS OF A STRUCTUREWhile the results obtained in the preceding section represent a first and necessary step in the analysis of the given structure, they do not tell us whether the given load can be safely supported. Whether rod BC, for example, will break or not under this loading depends not only upon the value found for the internal force FBC, but also upon the cross-sectional area of the rod and the material of which the rod is made. Indeed, the internal force FBC actually represents the resul-tant of elementary forces distributed over the entire area A of the cross section (Fig. 1.7) and the average intensity of these distributed forces is equal to the force per unit area, FBCyA, in the section. Whether or not the rod will break under the given loading clearly depends upon the ability of the material to withstand the corre-sponding value FBCyA of the intensity of the distributed internal forces. It thus depends upon the force FBC, the cross-sectional area A, and the material of the rod. The force per unit area, or intensity of the forces distributed over a given section, is called the stress on that section and is denoted by the Greek letter s (sigma). The stress in a member of cross-sectional area A subjected to an axial load P (Fig. 1.8) is therefore obtained by dividing the magnitude P of the load by the area A:
s 5
PA
(1.5)
A positive sign will be used to indicate a tensile stress (member in tension) and a negative sign to indicate a compressive stress (mem-ber in compression). Since SI metric units are used in this discussion, with P ex-pressed in newtons (N) and A in square meters (m2), the stress s will be expressed in N/m2. This unit is called a pascal (Pa). How-ever, one finds that the pascal is an exceedingly small quantity and that, in practice, multiples of this unit must be used, namely, the kilopascal (kPa), the megapascal (MPa), and the gigapascal (GPa). We have
1 kPa 5 103 Pa 5 103 N/m2
1 MPa 5 106 Pa 5 106 N/m2
1 GPa 5 109 Pa 5 109 N/m2
When U.S. customary units are used, the force P is usually expressed in pounds (lb) or kilopounds (kip), and the cross-sectional area A in square inches (in2). The stress s will then be expressed in pounds per square inch (psi) or kilopounds per square inch (ksi).†
†The principal SI and U.S. customary units used in mechanics are listed in tables inside the front cover of this book. From the table on the right-hand side, we note that 1 psi is approximately equal to 7 kPa, and 1 ksi approximately equal to 7 MPa.
Fig. 1.7
A
FBCFBC A! !
Fig. 1.8 Member with an axial load.
(a) (b)
A
PA
P' P'
P
! !
1.3 Stresses in the Members of a Structure
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Stress Concentration l If we drill a hole for some reasons in the
component, the typical example is to build a connection with other structural elements.
November, 14 Strenght of Materials I - DAT 30
Lecture Notes of Mechanics of Solids, Chapter 2 14
2.10 STRESS CONCENTRATIONS (SI&4th Ed p. 156-161; 3rd Ed p. 159-163) For a uniform cross-sectional bar that is applied an axial force, both experiment and theory of elasticity find that the normal stress will be uniformly distributed over the cross-section Stress Concentration However, if we drill a hole for some reasons in the component, the typical example is to build a connection with other structural elements. For such a case, if we cut from the hole’s center plane, we find that the stress distribution is no longer uniform, as in Fig. 2.11(a). It may distribute over such a smaller area in highly uneven pattern. We call this phenomenon as Stress Concentration. Stress Concentration Factor In engineering practice, though, the actual stress distribution does not have to be determined. Instead, only the maximum stress at these sections must be known, and the member is then designed to resist this highest stress when the axial load is applied. The specific values of the maximum normal stress at the critical section can be determined by experimental methods or by advanced mathematical techniques using the theory of elasticity. The results of these investigations are usually reported in graphical form (as in Fig. 2.11(c)) in terms of Stress Concentration Factor K.
avg
KVV max (2.28)
in which Vavg=P/A’ is the assumed average stress as in Fig. 2.11(b). Provided K has been known from the figures or tables (as in Fig. 2.11(c)), and the average normal stress has been calculated from Vavg=P/A’, where A’ is the smallest cross-sectional area. Then from the above equation the maximum stress at the cross section can be computed as:
'APKK avgmax VV (2.29)
(a)
(b) (c)
P
P
A’
Stre
ss C
once
ntra
tion
Fact
or K
w 2r
r/w
Fig. 2.11 Stress concentration
Stress concentration occurs in the case that there is a sudden change in cross-sectional area. By observing Fig. 2.11(c), it is interesting to note that the bigger the ratio of change in the sectional area, the higher the stress concentration.
l In engineering practice, the actual stress distribution does not need. Instead, only the maximum stress at these sections must be known.
November, 14 Strenght of Materials I - DAT 31
115 You should also observe that the plates used to obtain a uniform stress distribution in the member of Fig. 2.55 must allow the mem-ber to freely expand laterally. Thus, the plates cannot be rigidly attached to the member; you must assume them to be just in contact with the member, and smooth enough not to impede the lateral expansion of the member. While such end conditions can actually be achieved for a member in compression, they cannot be physically realized in the case of a member in tension. It does not matter, however, whether or not an actual fixture can be realized and used to load a member so that the distribution of stresses in the member is uniform. The important thing is to imagine a model that will allow such a distribution of stresses, and to keep this model in mind so that you may later compare it with the actual loading conditions.
2.18 STRESS CONCENTRATIONSAs you saw in the preceding section, the stresses near the points of application of concentrated loads can reach values much larger than the average value of the stress in the member. When a structural member contains a discontinuity, such as a hole or a sudden change in cross section, high localized stresses can also occur near the dis-continuity. Figures 2.58 and 2.59 show the distribution of stresses in critical sections corresponding to two such situations. Figure 2.58 refers to a flat bar with a circular hole and shows the stress distribu-tion in a section passing through the center of the hole. Figure 2.59 refers to a flat bar consisting of two portions of different widths con-nected by fillets; it shows the stress distribution in the narrowest part of the connection, where the highest stresses occur. These results were obtained experimentally through the use of a photoelastic method. Fortunately for the engineer who has to design a given member and cannot afford to carry out such an analy-sis, the results obtained are independent of the size of the member and of the material used; they depend only upon the ratios of the geometric parameters involved, i.e., upon the ratio ryd in the case of a circular hole, and upon the ratios ryd and Dyd in the case of fillets. Furthermore, the designer is more interested in the maximum value of the stress in a given section, than in the actual distribution of stresses in that section, since the main concern is to determine whether the allowable stress will be exceeded under a given loading, and not where this value will be exceeded. For this reason, one defines the ratio
K 5
smax
save (2.48)
of the maximum stress over the average stress computed in the critical (narrowest) section of the discontinuity. This ratio is referred to as the stress-concentration factor of the given discontinuity. Stress- concentration factors can be computed once and for all in terms of the ratios of the geometric parameters involved, and the results obtained can be expressed in the form of tables or of graphs, as
2.18 Stress Concentrations
PP'
P'
rD
d12
d12
!max
!ave
Fig. 2.58 Stress distribution near circular hole in flat bar under axial loading.
PP'
P'
!max
!ave
dD
r
Fig. 2.59 Stress distribution near fillets in flat bar under axial loading.
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Stress Concentration If we drill a hole for some reasons in the component, the typical example is to build a connection with other structural elements.
November, 14 Strenght of Materials I - DAT 32
Lecture Notes of Mechanics of Solids, Chapter 2 14
2.10 STRESS CONCENTRATIONS (SI&4th Ed p. 156-161; 3rd Ed p. 159-163) For a uniform cross-sectional bar that is applied an axial force, both experiment and theory of elasticity find that the normal stress will be uniformly distributed over the cross-section Stress Concentration However, if we drill a hole for some reasons in the component, the typical example is to build a connection with other structural elements. For such a case, if we cut from the hole’s center plane, we find that the stress distribution is no longer uniform, as in Fig. 2.11(a). It may distribute over such a smaller area in highly uneven pattern. We call this phenomenon as Stress Concentration. Stress Concentration Factor In engineering practice, though, the actual stress distribution does not have to be determined. Instead, only the maximum stress at these sections must be known, and the member is then designed to resist this highest stress when the axial load is applied. The specific values of the maximum normal stress at the critical section can be determined by experimental methods or by advanced mathematical techniques using the theory of elasticity. The results of these investigations are usually reported in graphical form (as in Fig. 2.11(c)) in terms of Stress Concentration Factor K.
avg
KVV max (2.28)
in which Vavg=P/A’ is the assumed average stress as in Fig. 2.11(b). Provided K has been known from the figures or tables (as in Fig. 2.11(c)), and the average normal stress has been calculated from Vavg=P/A’, where A’ is the smallest cross-sectional area. Then from the above equation the maximum stress at the cross section can be computed as:
'APKK avgmax VV (2.29)
(a)
(b) (c)
P
P
A’
Stre
ss C
once
ntra
tion
Fact
or K
w 2r
r/w
Fig. 2.11 Stress concentration
Stress concentration occurs in the case that there is a sudden change in cross-sectional area. By observing Fig. 2.11(c), it is interesting to note that the bigger the ratio of change in the sectional area, the higher the stress concentration.
Lecture Notes of Mechanics of Solids, Chapter 2 14
2.10 STRESS CONCENTRATIONS (SI&4th Ed p. 156-161; 3rd Ed p. 159-163) For a uniform cross-sectional bar that is applied an axial force, both experiment and theory of elasticity find that the normal stress will be uniformly distributed over the cross-section Stress Concentration However, if we drill a hole for some reasons in the component, the typical example is to build a connection with other structural elements. For such a case, if we cut from the hole’s center plane, we find that the stress distribution is no longer uniform, as in Fig. 2.11(a). It may distribute over such a smaller area in highly uneven pattern. We call this phenomenon as Stress Concentration. Stress Concentration Factor In engineering practice, though, the actual stress distribution does not have to be determined. Instead, only the maximum stress at these sections must be known, and the member is then designed to resist this highest stress when the axial load is applied. The specific values of the maximum normal stress at the critical section can be determined by experimental methods or by advanced mathematical techniques using the theory of elasticity. The results of these investigations are usually reported in graphical form (as in Fig. 2.11(c)) in terms of Stress Concentration Factor K.
avg
KVV max (2.28)
in which Vavg=P/A’ is the assumed average stress as in Fig. 2.11(b). Provided K has been known from the figures or tables (as in Fig. 2.11(c)), and the average normal stress has been calculated from Vavg=P/A’, where A’ is the smallest cross-sectional area. Then from the above equation the maximum stress at the cross section can be computed as:
'APKK avgmax VV (2.29)
(a)
(b) (c)
P
P
A’
Stre
ss C
once
ntra
tion
Fact
or K
w 2r
r/w
Fig. 2.11 Stress concentration
Stress concentration occurs in the case that there is a sudden change in cross-sectional area. By observing Fig. 2.11(c), it is interesting to note that the bigger the ratio of change in the sectional area, the higher the stress concentration.
A´ is the smallest cross section
Stress Concentration Factor If we drill a hole for some reasons in the component, the typical example is to build a connection with other structural elements.
November, 14 Strenght of Materials I - DAT 33
Lecture Notes of Mechanics of Solids, Chapter 2 14
2.10 STRESS CONCENTRATIONS (SI&4th Ed p. 156-161; 3rd Ed p. 159-163) For a uniform cross-sectional bar that is applied an axial force, both experiment and theory of elasticity find that the normal stress will be uniformly distributed over the cross-section Stress Concentration However, if we drill a hole for some reasons in the component, the typical example is to build a connection with other structural elements. For such a case, if we cut from the hole’s center plane, we find that the stress distribution is no longer uniform, as in Fig. 2.11(a). It may distribute over such a smaller area in highly uneven pattern. We call this phenomenon as Stress Concentration. Stress Concentration Factor In engineering practice, though, the actual stress distribution does not have to be determined. Instead, only the maximum stress at these sections must be known, and the member is then designed to resist this highest stress when the axial load is applied. The specific values of the maximum normal stress at the critical section can be determined by experimental methods or by advanced mathematical techniques using the theory of elasticity. The results of these investigations are usually reported in graphical form (as in Fig. 2.11(c)) in terms of Stress Concentration Factor K.
avg
KVV max (2.28)
in which Vavg=P/A’ is the assumed average stress as in Fig. 2.11(b). Provided K has been known from the figures or tables (as in Fig. 2.11(c)), and the average normal stress has been calculated from Vavg=P/A’, where A’ is the smallest cross-sectional area. Then from the above equation the maximum stress at the cross section can be computed as:
'APKK avgmax VV (2.29)
(a)
(b) (c)
P
P
A’
Stre
ss C
once
ntra
tion
Fact
or K
w 2r
r/w
Fig. 2.11 Stress concentration
Stress concentration occurs in the case that there is a sudden change in cross-sectional area. By observing Fig. 2.11(c), it is interesting to note that the bigger the ratio of change in the sectional area, the higher the stress concentration.
Lecture Notes of Mechanics of Solids, Chapter 2 14
2.10 STRESS CONCENTRATIONS (SI&4th Ed p. 156-161; 3rd Ed p. 159-163) For a uniform cross-sectional bar that is applied an axial force, both experiment and theory of elasticity find that the normal stress will be uniformly distributed over the cross-section Stress Concentration However, if we drill a hole for some reasons in the component, the typical example is to build a connection with other structural elements. For such a case, if we cut from the hole’s center plane, we find that the stress distribution is no longer uniform, as in Fig. 2.11(a). It may distribute over such a smaller area in highly uneven pattern. We call this phenomenon as Stress Concentration. Stress Concentration Factor In engineering practice, though, the actual stress distribution does not have to be determined. Instead, only the maximum stress at these sections must be known, and the member is then designed to resist this highest stress when the axial load is applied. The specific values of the maximum normal stress at the critical section can be determined by experimental methods or by advanced mathematical techniques using the theory of elasticity. The results of these investigations are usually reported in graphical form (as in Fig. 2.11(c)) in terms of Stress Concentration Factor K.
avg
KVV max (2.28)
in which Vavg=P/A’ is the assumed average stress as in Fig. 2.11(b). Provided K has been known from the figures or tables (as in Fig. 2.11(c)), and the average normal stress has been calculated from Vavg=P/A’, where A’ is the smallest cross-sectional area. Then from the above equation the maximum stress at the cross section can be computed as:
'APKK avgmax VV (2.29)
(a)
(b) (c)
P
P
A’St
ress
Con
cent
ratio
n Fa
ctor
K
w 2r
r/w
Fig. 2.11 Stress concentration
Stress concentration occurs in the case that there is a sudden change in cross-sectional area. By observing Fig. 2.11(c), it is interesting to note that the bigger the ratio of change in the sectional area, the higher the stress concentration.
Allowable Stress
¢ When the stress (intensity of force) of an element exceeds some level, the structure will fail.
¢ We usually adopt allowable force or allowable stress to measure the threshold of safety in engineering.
November, 14 Strenght of Materials I - DAT 34
Uncertainties that we must take into account in engineering:
¢ The load for design may be different from the actual load.
¢ Size of structural member may not be very precise due to manufacturing and assembly.
¢ Various defects in material due to manufacturing processing.
November, 14 Strenght of Materials I - DAT 35
To consider such uncertainties a Factor of Safety, F.S., is usually introduced;
November, 14 Strenght of Materials I - DAT 36
Lecture Notes of Mechanics of Solids, Chapter 2 3
Moreover, there are following several reasons that we must take into account in engineering:
x The load for design may be different from the actual load.
x Size of structural member may not be very precise due to manufacturing and assembly.
x Various defects in material due to manufacturing processing.
One simple method to consider such uncertainties is to use a number called the Factor of Safety, F.S., which is a ratio of failure load Ffail (found from experimental testing) divided by
the allowable one Fallow
allow
fail
FF
.S.F (2.5)
If the applied load is linearly related to the stress developed in the member, as in the case of
using A/F V , then we can define the factor of safety as a ratio of the failure stress Vfail to
the allowable stress Vallow
allow
failSFVV
.. (2.6)
Usually, the factor of safety is chosen to be greater than 1 in order to avoid the potential
failure. This is dependent on the specific design case. For nuclear power plant, the factor of
safety for some of its components may be as high as 3. For an aircraft design, the higher the
F.S. (safer), the heavier the structure, therefore the higher in the operational cost. So we need
to balance the safety and cost.
The value of F.S. can be found in design codes and engineering handbook. More often, we
use Eq. (2.6) to compute the allowable stress:
..SFfail
allow
VV (2.7)
Example 2.2: In Example 2.1, if the maximum allowable stress for copper is
VCu,allow=50MPa, please determine the minimum size of the wire/cable from the material
strength point of view.
Mathematically, allowCudmg
AF
,24
VS
V d
Therefore: mmmgdallowCu
469.410469.44 3
,
u t �
SV
Obviously, the lower the allowable stress, the bigger the cable size. Stress is an indication of
structural strength and elemental size.
In engineering, there are two significant problems associated with stress as follows.
Problem (1) Stress Analysis: for a specific structure, we can determine the stress level.
With the stress level, we then justify the safety and reliability of a structural
member, i.e. known size A and load F, to determine stress level: AF V
Problem (2) Engineering Design: Inversely, we can design a structural member based on the
allowable stress so that it can satisfy the safety requirements, i.e. known material’s
allowable stress V allow and load F, to design the element size: allowFA Vt
Lecture Notes of Mechanics of Solids, Chapter 2 3
Moreover, there are following several reasons that we must take into account in engineering:
x The load for design may be different from the actual load.
x Size of structural member may not be very precise due to manufacturing and assembly.
x Various defects in material due to manufacturing processing.
One simple method to consider such uncertainties is to use a number called the Factor of Safety, F.S., which is a ratio of failure load Ffail (found from experimental testing) divided by
the allowable one Fallow
allow
fail
FF
.S.F (2.5)
If the applied load is linearly related to the stress developed in the member, as in the case of
using A/F V , then we can define the factor of safety as a ratio of the failure stress Vfail to
the allowable stress Vallow
allow
failSFVV
.. (2.6)
Usually, the factor of safety is chosen to be greater than 1 in order to avoid the potential
failure. This is dependent on the specific design case. For nuclear power plant, the factor of
safety for some of its components may be as high as 3. For an aircraft design, the higher the
F.S. (safer), the heavier the structure, therefore the higher in the operational cost. So we need
to balance the safety and cost.
The value of F.S. can be found in design codes and engineering handbook. More often, we
use Eq. (2.6) to compute the allowable stress:
..SFfail
allow
VV (2.7)
Example 2.2: In Example 2.1, if the maximum allowable stress for copper is
VCu,allow=50MPa, please determine the minimum size of the wire/cable from the material
strength point of view.
Mathematically, allowCudmg
AF
,24
VS
V d
Therefore: mmmgdallowCu
469.410469.44 3
,
u t �
SV
Obviously, the lower the allowable stress, the bigger the cable size. Stress is an indication of
structural strength and elemental size.
In engineering, there are two significant problems associated with stress as follows.
Problem (1) Stress Analysis: for a specific structure, we can determine the stress level.
With the stress level, we then justify the safety and reliability of a structural
member, i.e. known size A and load F, to determine stress level: AF V
Problem (2) Engineering Design: Inversely, we can design a structural member based on the
allowable stress so that it can satisfy the safety requirements, i.e. known material’s
allowable stress V allow and load F, to design the element size: allowFA Vt
Lecture Notes of Mechanics of Solids, Chapter 2 3
Moreover, there are following several reasons that we must take into account in engineering:
x The load for design may be different from the actual load.
x Size of structural member may not be very precise due to manufacturing and assembly.
x Various defects in material due to manufacturing processing.
One simple method to consider such uncertainties is to use a number called the Factor of Safety, F.S., which is a ratio of failure load Ffail (found from experimental testing) divided by
the allowable one Fallow
allow
fail
FF
.S.F (2.5)
If the applied load is linearly related to the stress developed in the member, as in the case of
using A/F V , then we can define the factor of safety as a ratio of the failure stress Vfail to
the allowable stress Vallow
allow
failSFVV
.. (2.6)
Usually, the factor of safety is chosen to be greater than 1 in order to avoid the potential
failure. This is dependent on the specific design case. For nuclear power plant, the factor of
safety for some of its components may be as high as 3. For an aircraft design, the higher the
F.S. (safer), the heavier the structure, therefore the higher in the operational cost. So we need
to balance the safety and cost.
The value of F.S. can be found in design codes and engineering handbook. More often, we
use Eq. (2.6) to compute the allowable stress:
..SFfail
allow
VV (2.7)
Example 2.2: In Example 2.1, if the maximum allowable stress for copper is
VCu,allow=50MPa, please determine the minimum size of the wire/cable from the material
strength point of view.
Mathematically, allowCudmg
AF
,24
VS
V d
Therefore: mmmgdallowCu
469.410469.44 3
,
u t �
SV
Obviously, the lower the allowable stress, the bigger the cable size. Stress is an indication of
structural strength and elemental size.
In engineering, there are two significant problems associated with stress as follows.
Problem (1) Stress Analysis: for a specific structure, we can determine the stress level.
With the stress level, we then justify the safety and reliability of a structural
member, i.e. known size A and load F, to determine stress level: AF V
Problem (2) Engineering Design: Inversely, we can design a structural member based on the
allowable stress so that it can satisfy the safety requirements, i.e. known material’s
allowable stress V allow and load F, to design the element size: allowFA Vt
Fallow : Allowable load Ffail : Failure load
σallow : Allowable stress
The factor of safety is chosen to be greater than 1
Example 6
¢ If the maximum allowable stress for copper in Example 1 is σCu,allow=50 MPa, please determine the minimum size of the wire/cable from the material strength point of view.
November, 14 Strenght of Materials I - DAT 37
Lecture Notes of Mechanics of Solids, Chapter 2 2
apparently, the stress may not reflect the real mechanical status at this point. In other words, we need to consider both magnitude and direction of the force.
Now let’s resolve the force 'F in normal and tangential direction of the acting area as Fig. 2.1b). The intensity of the force or force per unit area acting normally to section A is called Normal Stress, V (sigma), and it is expressed as:
AF
lim nA '
'V
' 0o (2.2)
If this stress “pulls” on the area it is referred as Tensile Stress and defined as Positive. If it “pushes” on the area it is called Compressive Stress and defined as Negative.
The intensity or force per unit area acting tangentially to A is called Shear Stress, W (tau), and it is expressed as:
AF
lim tA '
'W
' 0o (2.3)
Average Normal Stress (SI&4th p. 24-31, 3rd p. 27-34) To begin, we only look at beams that carry tensile or compressive loads and which are long and slender. Such beams can then be assumed to carry a constant stress, and Eq. (2.2) can be simplified to:
AF
V (2.4)
We call this either Average Normal Stress or Uniform Uniaxial Stress. Units of Stress The units in the SI system is the Newton per square meter or Pascal, i.e. : Pa = N/m2. In engineering, Pa seems too small, so we usually use:
Kilo Pascal KPa (=Pau103) e.g. 20,000Pa=20kPa Mega Pascal MPa (=Pau106) e.g. 20,000,000Pa=20MPa Giga Pascal GPa (=Pau109) e.g. 20,000,000,000Pa=20GPa
Example 2.1: An 80 kg lamp is supported by a single electrical copper cable of diameter d = 3.15 mm. What is the stress carried by the cable. To determine the stress in the wire/cable as Eq. (2.4), we need the cross sectional area A of the cable and the applied internal force F:
2622
10793.7400315.0
4mdA �u
u SS
N.mgF 7848980 u
so MPa..A
F 6100107937
7846
u
�V
Allowable Stress (SI&4th p. 48-49, 3rd p. 51-52) From Example 2.1, we may concern whether or not 80kg would be too heavy, or say 100.6MPa stress would be too high for the wire/cable, from the safety point of view. Indeed, stress is one of most important indicators of structural strength. When the stress (intensity of force) of an element exceeds some level, the structure will fail. For convenience, we usually adopt allowable force or allowable stress to measure the threshold of safety in engineering.
80kgF
a a
Sectiona-a
A
d
FBD
F
Solution
November, 14
Strenght of Materials I - DAT
38
Lecture Notes of Mechanics of Solids, Chapter 2 3
Moreover, there are following several reasons that we must take into account in engineering:
x The load for design may be different from the actual load.
x Size of structural member may not be very precise due to manufacturing and assembly.
x Various defects in material due to manufacturing processing.
One simple method to consider such uncertainties is to use a number called the Factor of Safety, F.S., which is a ratio of failure load Ffail (found from experimental testing) divided by
the allowable one Fallow
allow
fail
FF
.S.F (2.5)
If the applied load is linearly related to the stress developed in the member, as in the case of
using A/F V , then we can define the factor of safety as a ratio of the failure stress Vfail to
the allowable stress Vallow
allow
failSFVV
.. (2.6)
Usually, the factor of safety is chosen to be greater than 1 in order to avoid the potential
failure. This is dependent on the specific design case. For nuclear power plant, the factor of
safety for some of its components may be as high as 3. For an aircraft design, the higher the
F.S. (safer), the heavier the structure, therefore the higher in the operational cost. So we need
to balance the safety and cost.
The value of F.S. can be found in design codes and engineering handbook. More often, we
use Eq. (2.6) to compute the allowable stress:
..SFfail
allow
VV (2.7)
Example 2.2: In Example 2.1, if the maximum allowable stress for copper is
VCu,allow=50MPa, please determine the minimum size of the wire/cable from the material
strength point of view.
Mathematically, allowCudmg
AF
,24
VS
V d
Therefore: mmmgdallowCu
469.410469.44 3
,
u t �
SV
Obviously, the lower the allowable stress, the bigger the cable size. Stress is an indication of
structural strength and elemental size.
In engineering, there are two significant problems associated with stress as follows.
Problem (1) Stress Analysis: for a specific structure, we can determine the stress level.
With the stress level, we then justify the safety and reliability of a structural
member, i.e. known size A and load F, to determine stress level: AF V
Problem (2) Engineering Design: Inversely, we can design a structural member based on the
allowable stress so that it can satisfy the safety requirements, i.e. known material’s
allowable stress V allow and load F, to design the element size: allowFA Vt
Lecture Notes of Mechanics of Solids, Chapter 2 3
Moreover, there are following several reasons that we must take into account in engineering:
x The load for design may be different from the actual load.
x Size of structural member may not be very precise due to manufacturing and assembly.
x Various defects in material due to manufacturing processing.
One simple method to consider such uncertainties is to use a number called the Factor of Safety, F.S., which is a ratio of failure load Ffail (found from experimental testing) divided by
the allowable one Fallow
allow
fail
FF
.S.F (2.5)
If the applied load is linearly related to the stress developed in the member, as in the case of
using A/F V , then we can define the factor of safety as a ratio of the failure stress Vfail to
the allowable stress Vallow
allow
failSFVV
.. (2.6)
Usually, the factor of safety is chosen to be greater than 1 in order to avoid the potential
failure. This is dependent on the specific design case. For nuclear power plant, the factor of
safety for some of its components may be as high as 3. For an aircraft design, the higher the
F.S. (safer), the heavier the structure, therefore the higher in the operational cost. So we need
to balance the safety and cost.
The value of F.S. can be found in design codes and engineering handbook. More often, we
use Eq. (2.6) to compute the allowable stress:
..SFfail
allow
VV (2.7)
Example 2.2: In Example 2.1, if the maximum allowable stress for copper is
VCu,allow=50MPa, please determine the minimum size of the wire/cable from the material
strength point of view.
Mathematically, allowCudmg
AF
,24
VS
V d
Therefore: mmmgdallowCu
469.410469.44 3
,
u t �
SV
Obviously, the lower the allowable stress, the bigger the cable size. Stress is an indication of
structural strength and elemental size.
In engineering, there are two significant problems associated with stress as follows.
Problem (1) Stress Analysis: for a specific structure, we can determine the stress level.
With the stress level, we then justify the safety and reliability of a structural
member, i.e. known size A and load F, to determine stress level: AF V
Problem (2) Engineering Design: Inversely, we can design a structural member based on the
allowable stress so that it can satisfy the safety requirements, i.e. known material’s
allowable stress V allow and load F, to design the element size: allowFA Vt
Lecture Notes of Mechanics of Solids, Chapter 2 3
Moreover, there are following several reasons that we must take into account in engineering:
x The load for design may be different from the actual load.
x Size of structural member may not be very precise due to manufacturing and assembly.
x Various defects in material due to manufacturing processing.
One simple method to consider such uncertainties is to use a number called the Factor of Safety, F.S., which is a ratio of failure load Ffail (found from experimental testing) divided by
the allowable one Fallow
allow
fail
FF
.S.F (2.5)
If the applied load is linearly related to the stress developed in the member, as in the case of
using A/F V , then we can define the factor of safety as a ratio of the failure stress Vfail to
the allowable stress Vallow
allow
failSFVV
.. (2.6)
Usually, the factor of safety is chosen to be greater than 1 in order to avoid the potential
failure. This is dependent on the specific design case. For nuclear power plant, the factor of
safety for some of its components may be as high as 3. For an aircraft design, the higher the
F.S. (safer), the heavier the structure, therefore the higher in the operational cost. So we need
to balance the safety and cost.
The value of F.S. can be found in design codes and engineering handbook. More often, we
use Eq. (2.6) to compute the allowable stress:
..SFfail
allow
VV (2.7)
Example 2.2: In Example 2.1, if the maximum allowable stress for copper is
VCu,allow=50MPa, please determine the minimum size of the wire/cable from the material
strength point of view.
Mathematically, allowCudmg
AF
,24
VS
V d
Therefore: mmmgdallowCu
469.410469.44 3
,
u t �
SV
Obviously, the lower the allowable stress, the bigger the cable size. Stress is an indication of
structural strength and elemental size.
In engineering, there are two significant problems associated with stress as follows.
Problem (1) Stress Analysis: for a specific structure, we can determine the stress level.
With the stress level, we then justify the safety and reliability of a structural
member, i.e. known size A and load F, to determine stress level: AF V
Problem (2) Engineering Design: Inversely, we can design a structural member based on the
allowable stress so that it can satisfy the safety requirements, i.e. known material’s
allowable stress V allow and load F, to design the element size: allowFA Vt
Lecture Notes of Mechanics of Solids, Chapter 2 3
Moreover, there are following several reasons that we must take into account in engineering:
x The load for design may be different from the actual load.
x Size of structural member may not be very precise due to manufacturing and assembly.
x Various defects in material due to manufacturing processing.
One simple method to consider such uncertainties is to use a number called the Factor of Safety, F.S., which is a ratio of failure load Ffail (found from experimental testing) divided by
the allowable one Fallow
allow
fail
FF
.S.F (2.5)
If the applied load is linearly related to the stress developed in the member, as in the case of
using A/F V , then we can define the factor of safety as a ratio of the failure stress Vfail to
the allowable stress Vallow
allow
failSFVV
.. (2.6)
Usually, the factor of safety is chosen to be greater than 1 in order to avoid the potential
failure. This is dependent on the specific design case. For nuclear power plant, the factor of
safety for some of its components may be as high as 3. For an aircraft design, the higher the
F.S. (safer), the heavier the structure, therefore the higher in the operational cost. So we need
to balance the safety and cost.
The value of F.S. can be found in design codes and engineering handbook. More often, we
use Eq. (2.6) to compute the allowable stress:
..SFfail
allow
VV (2.7)
Example 2.2: In Example 2.1, if the maximum allowable stress for copper is
VCu,allow=50MPa, please determine the minimum size of the wire/cable from the material
strength point of view.
Mathematically, allowCudmg
AF
,24
VS
V d
Therefore: mmmgdallowCu
469.410469.44 3
,
u t �
SV
Obviously, the lower the allowable stress, the bigger the cable size. Stress is an indication of
structural strength and elemental size.
In engineering, there are two significant problems associated with stress as follows.
Problem (1) Stress Analysis: for a specific structure, we can determine the stress level.
With the stress level, we then justify the safety and reliability of a structural
member, i.e. known size A and load F, to determine stress level: AF V
Problem (2) Engineering Design: Inversely, we can design a structural member based on the
allowable stress so that it can satisfy the safety requirements, i.e. known material’s
allowable stress V allow and load F, to design the element size: allowFA Vt
IS UNITS from
MECHANICS OF MATERIALS
Ferdinand P. Beer E. Russell Johnston, Jr.
John T. Dewolf David F. Mazurek
November, 14 Strenght of Materials I - DAT 39
SI Prefixes
November, 14 Strenght of Materials I - DAT 40
ISBN: 0073380288Author: Beer, Johnston, Dewolf, and MazurekTitle: MECHANICS OF MATERIALS
Front endsheetsColor: 4Pages: 2, 3
U.S. Customary Units and Their SI Equivalents
Quantity U.S. Customary Units SI Equivalent
Acceleration ft/s2 0.3048 m/s2
in./s2 0.0254 m/s2
Area ft2 0.0929 m2
in2 645.2 mm2
Energy ft ? lb 1.356 JForce kip 4.448 kN lb 4.448 N oz 0.2780 NImpulse lb ? s 4.448 N ? sLength ft 0.3048 m in. 25.40 mm mi 1.609 kmMass oz mass 28.35 g lb mass 0.4536 kg slug 14.59 kg ton 907.2 kgMoment of a force lb ? ft 1.356 N ? m lb ? in. 0.1130 N ? mMoment of inertia Of an area in4 0.4162 3 106 mm4
Of a mass lb ? ft ? s2 1.356 kg ? m2
Power ft ? lb/s 1.356 W hp 745.7 WPressure or stress lb/ft2 47.88 Pa lb/in2 (psi) 6.895 kPaVelocity ft/s 0.3048 m/s in./s 0.0254 m/s mi/h (mph) 0.4470 m/s mi/h (mph) 1.609 km/hVolume, solids ft3 0.02832 m3
in3 16.39 cm3
Liquids gal 3.785 L qt 0.9464 LWork ft ? lb 1.356 J
SI Prefixes
Multiplication Factor Prefix † Symbol
1 000 000 000 000 5 1012 tera T 1 000 000 000 5 109 giga G 1 000 000 5 106 mega M 1 000 5 103 kilo k 100 5 102 hecto‡ h 10 5 101 deka ‡ da 0.1 5 1021 deci ‡ d 0.01 5 1022 centi ‡ c 0.001 5 1023 milli m 0.000 001 5 1026 micro m 0.000 000 001 5 1029 nano n 0.000 000 000 001 5 10212 pico p 0.000 000 000 000 001 5 10215 femto f 0.000 000 000 000 000 001 5 10218 atto a
† The first syllable of every prefix is accented so that the prefix will retain its identity. Thus, the preferred pronunciation of kilometer places the accent on the first syllable, not the second.
‡ The use of these prefixes should be avoided, except for the measurement of areas and vol-umes and for the nontechnical use of centimeter, as for body and clothing measurements.
Principal SI Units Used in Mechanics
Quantity Unit Symbol Formula
Acceleration Meter per second squared p m/s2
Angle Radian rad †Angular acceleration Radian per second squared p rad/s2
Angular velocity Radian per second p rad/sArea Square meter p m2
Density Kilogram per cubic meter p kg/m3
Energy Joule J N ? mForce Newton N kg ? m/s2
Frequency Hertz Hz s21
Impulse Newton-second p kg ? m/sLength Meter m ‡ Mass Kilogram kg ‡ Moment of a force Newton-meter p N ? m Power Watt W J/sPressure Pascal Pa N/m2
Stress Pascal Pa N/m2
Time Second s ‡ Velocity Meter per second p m/sVolume, solids Cubic meter p m3
Liquids Liter L 1023 m3
Work Joule J N ? m
† Supplementary unit (1 revolution 5 2p rad 5 3608). ‡ Base unit.
bee80288_ifc.indd Page 1 10/26/10 4:39:07 PM user-f499bee80288_ifc.indd Page 1 10/26/10 4:39:07 PM user-f499 /Volumes/201/MHDQ251/bee80288_disk1of1/0073380288/bee80288_pagefiles/Volumes/201/MHDQ251/bee80288_disk1of1/0073380288/bee80288_pagefiles
Principal SI Units Used in Mechanics
November, 14 Strenght of Materials I - DAT 41
ISBN: 0073380288Author: Beer, Johnston, Dewolf, and MazurekTitle: MECHANICS OF MATERIALS
Front endsheetsColor: 4Pages: 2, 3
U.S. Customary Units and Their SI Equivalents
Quantity U.S. Customary Units SI Equivalent
Acceleration ft/s2 0.3048 m/s2
in./s2 0.0254 m/s2
Area ft2 0.0929 m2
in2 645.2 mm2
Energy ft ? lb 1.356 JForce kip 4.448 kN lb 4.448 N oz 0.2780 NImpulse lb ? s 4.448 N ? sLength ft 0.3048 m in. 25.40 mm mi 1.609 kmMass oz mass 28.35 g lb mass 0.4536 kg slug 14.59 kg ton 907.2 kgMoment of a force lb ? ft 1.356 N ? m lb ? in. 0.1130 N ? mMoment of inertia Of an area in4 0.4162 3 106 mm4
Of a mass lb ? ft ? s2 1.356 kg ? m2
Power ft ? lb/s 1.356 W hp 745.7 WPressure or stress lb/ft2 47.88 Pa lb/in2 (psi) 6.895 kPaVelocity ft/s 0.3048 m/s in./s 0.0254 m/s mi/h (mph) 0.4470 m/s mi/h (mph) 1.609 km/hVolume, solids ft3 0.02832 m3
in3 16.39 cm3
Liquids gal 3.785 L qt 0.9464 LWork ft ? lb 1.356 J
SI Prefixes
Multiplication Factor Prefix † Symbol
1 000 000 000 000 5 1012 tera T 1 000 000 000 5 109 giga G 1 000 000 5 106 mega M 1 000 5 103 kilo k 100 5 102 hecto‡ h 10 5 101 deka ‡ da 0.1 5 1021 deci ‡ d 0.01 5 1022 centi ‡ c 0.001 5 1023 milli m 0.000 001 5 1026 micro m 0.000 000 001 5 1029 nano n 0.000 000 000 001 5 10212 pico p 0.000 000 000 000 001 5 10215 femto f 0.000 000 000 000 000 001 5 10218 atto a
† The first syllable of every prefix is accented so that the prefix will retain its identity. Thus, the preferred pronunciation of kilometer places the accent on the first syllable, not the second.
‡ The use of these prefixes should be avoided, except for the measurement of areas and vol-umes and for the nontechnical use of centimeter, as for body and clothing measurements.
Principal SI Units Used in Mechanics
Quantity Unit Symbol Formula
Acceleration Meter per second squared p m/s2
Angle Radian rad †Angular acceleration Radian per second squared p rad/s2
Angular velocity Radian per second p rad/sArea Square meter p m2
Density Kilogram per cubic meter p kg/m3
Energy Joule J N ? mForce Newton N kg ? m/s2
Frequency Hertz Hz s21
Impulse Newton-second p kg ? m/sLength Meter m ‡ Mass Kilogram kg ‡ Moment of a force Newton-meter p N ? m Power Watt W J/sPressure Pascal Pa N/m2
Stress Pascal Pa N/m2
Time Second s ‡ Velocity Meter per second p m/sVolume, solids Cubic meter p m3
Liquids Liter L 1023 m3
Work Joule J N ? m
† Supplementary unit (1 revolution 5 2p rad 5 3608). ‡ Base unit.
bee80288_ifc.indd Page 1 10/26/10 4:39:07 PM user-f499bee80288_ifc.indd Page 1 10/26/10 4:39:07 PM user-f499 /Volumes/201/MHDQ251/bee80288_disk1of1/0073380288/bee80288_pagefiles/Volumes/201/MHDQ251/bee80288_disk1of1/0073380288/bee80288_pagefiles
U.S
. Cus
tom
ary
Uni
ts a
nd T
heir
SI
Equi
vale
nts
November, 14 Strenght of Materials I - DAT 42
ISBN: 0073380288Author: Beer, Johnston, Dewolf, and MazurekTitle: MECHANICS OF MATERIALS
Front endsheetsColor: 4Pages: 2, 3
U.S. Customary Units and Their SI Equivalents
Quantity U.S. Customary Units SI Equivalent
Acceleration ft/s2 0.3048 m/s2
in./s2 0.0254 m/s2
Area ft2 0.0929 m2
in2 645.2 mm2
Energy ft ? lb 1.356 JForce kip 4.448 kN lb 4.448 N oz 0.2780 NImpulse lb ? s 4.448 N ? sLength ft 0.3048 m in. 25.40 mm mi 1.609 kmMass oz mass 28.35 g lb mass 0.4536 kg slug 14.59 kg ton 907.2 kgMoment of a force lb ? ft 1.356 N ? m lb ? in. 0.1130 N ? mMoment of inertia Of an area in4 0.4162 3 106 mm4
Of a mass lb ? ft ? s2 1.356 kg ? m2
Power ft ? lb/s 1.356 W hp 745.7 WPressure or stress lb/ft2 47.88 Pa lb/in2 (psi) 6.895 kPaVelocity ft/s 0.3048 m/s in./s 0.0254 m/s mi/h (mph) 0.4470 m/s mi/h (mph) 1.609 km/hVolume, solids ft3 0.02832 m3
in3 16.39 cm3
Liquids gal 3.785 L qt 0.9464 LWork ft ? lb 1.356 J
SI Prefixes
Multiplication Factor Prefix † Symbol
1 000 000 000 000 5 1012 tera T 1 000 000 000 5 109 giga G 1 000 000 5 106 mega M 1 000 5 103 kilo k 100 5 102 hecto‡ h 10 5 101 deka ‡ da 0.1 5 1021 deci ‡ d 0.01 5 1022 centi ‡ c 0.001 5 1023 milli m 0.000 001 5 1026 micro m 0.000 000 001 5 1029 nano n 0.000 000 000 001 5 10212 pico p 0.000 000 000 000 001 5 10215 femto f 0.000 000 000 000 000 001 5 10218 atto a
† The first syllable of every prefix is accented so that the prefix will retain its identity. Thus, the preferred pronunciation of kilometer places the accent on the first syllable, not the second.
‡ The use of these prefixes should be avoided, except for the measurement of areas and vol-umes and for the nontechnical use of centimeter, as for body and clothing measurements.
Principal SI Units Used in Mechanics
Quantity Unit Symbol Formula
Acceleration Meter per second squared p m/s2
Angle Radian rad †Angular acceleration Radian per second squared p rad/s2
Angular velocity Radian per second p rad/sArea Square meter p m2
Density Kilogram per cubic meter p kg/m3
Energy Joule J N ? mForce Newton N kg ? m/s2
Frequency Hertz Hz s21
Impulse Newton-second p kg ? m/sLength Meter m ‡ Mass Kilogram kg ‡ Moment of a force Newton-meter p N ? m Power Watt W J/sPressure Pascal Pa N/m2
Stress Pascal Pa N/m2
Time Second s ‡ Velocity Meter per second p m/sVolume, solids Cubic meter p m3
Liquids Liter L 1023 m3
Work Joule J N ? m
† Supplementary unit (1 revolution 5 2p rad 5 3608). ‡ Base unit.
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Thank You
November, 14 Strenght of Materials I - DAT 43
