2001, W. E. Haisler
1
Chapter 11: Axial Deformation and Stress of Bars
Axial Deformation and Stress of Bars (Chapter 11)
y
ux (x)
A
w
h
x
cross section (A)
may be any shape
z
L
L is large
compared to h and w
The 4 governing equations that must be satisfied are:
 xx
Static Equilibrium (from COLM):
0
x
u x
Stress-Strain:  xx  E xx
Kinematics:  xx 
x
Boundary Conditions: Depends on the problem
2001, W. E. Haisler
2
Chapter 11: Axial Deformation and Stress of Bars
Example 1. Consider an elastic axial bar as follows:
Axial force F applied to cross-section at right end (x=L)
Left end fixed so displacement B.C. of ux(x  0)  0
Beam is prismatic (constant cross-sectional area of A)
y
fixed from
motion at
x=0
ux(x)
x
z
A
F
L
Determine: ux(x),  xx ( x) and ux(x=L).
F
Equilibrium at x:  xx 
; Displace B.C.: ux(x  0)  0
A
u x
Kinematics Relation (strain-displacement):  xx 
x
2001, W. E. Haisler
3
Chapter 11: Axial Deformation and Stress of Bars
Constitutive Relation:  xx  E xx
Combine last three equations to obtain:
u x
F
  xx   xx / E  ( ) / E
x
A
u x
F
Thus
 ( ) / E or dux [( F )/ E]dx .
A
x
A
Integrate from 0 to x' to obtain the axial displacement:
u x ( x ')  
x'
0
F
F x'
F
dx 
dx  ( ) x ' C

EA 0
EA
AE
Apply the boundary condition: ux(x  0)  0
=>
C=0
2001, W. E. Haisler
4
Chapter 11: Axial Deformation and Stress of Bars
F
Solution for axial bar in tension: u x ( x)  ( ) x
EA
Strain is given by:  xx
u x F


x EA
Stress is given by:  xx  E xx
F

A
and displacement at end,  end  u x ( x  L)  FL
EA
Note that  end is the elongation of a bar of length L with
cross-sectional area A and Young’s modulus E and
subjected to a tensile force of F as shown below.
F
L
L 
F
2001, W. E. Haisler
5
Chapter 11: Axial Deformation and Stress of Bars
Consider a truss structure as done
in ENGR 211. By the method of
joints, the force in bar EF is
FAE  2.628kips . If the bar has a
cross-sectional area of 2 sq. in.,
then the stress is given by  xx
A
B
C
D
E
8 ft
4 kips
I
H
8 ft
G
8 ft
F
8 ft
8 ft
F
  1,314 psi . Suppose the
A
truss is made of steel. The elongation of the truss memeber
(2,268lb)(11.314 ft 12ftin )
 0.00531 in . In
is then   FL 
6
2
EA
29 x10 psi (2in )
this case, the bar is in compression and the negative
indicates that the member has shortened.
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
6
Example 2: Elastic bar constructed of two different bars, fixed
between two walls and loaded with a force P applied at point B:
E1
A1
E 2 A2
P
B
L1
L
2
Determine: axial force and stress in each bar, axial displacement
at point B.
For this problem, we have 4 relationships to satisfy:
1. Equilibrium of horizontal forces at any point
2. Kinematics (strain/displacements in horizontal direction)
3. Stress-Strain (material properties)
4. Boundary Conditions
2001, W. E. Haisler
7
Chapter 11: Axial Deformation and Stress of Bars
Use a “free-body diagram” to determine equilibrium of the
forces acting on a segment of the bar. Assume forces in the
bars are P1 and P2 (positive). Take a free-body of the beam
cross-section at point B (where the axial load P is applied):
xx
P (applied load)
xx
1
(stress in bar 2)
2


P
P1
P2 (force inbar 2)

Each stress can be written in terms of an equivalent force over an area
so that: P1 
A  xx dA   xx A1
1
1
1
and P2    xx2 dA   xx2 A2
A2
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
8
Equilibrium of the free body at B in terms of forces requires that
 Fhorizontal  0  P  P2  P1
or P  P1  P2
(1)
NOTE:
Any problem which can not be solved for the internal forces
by force equilibrium alone requires additional equations
(defining displacements) in order to complete the solution.
Such a problem is called statically indeterminate.
The above problem is thus statically indeterminate!
We know from boundary conditions that the bar’s total
elongation between the two fixed walls is zero. First,
calculate the deformation (elongation) of bars 1 and 2:
2001, W. E. Haisler
E1
9
Chapter 11: Axial Deformation and Stress of Bars
A1
A1

P1
L
1
Free Body #1

A2
P
P1
P2 = for ce in
bar 2
E
2
A
2
P
2
L2
B
Free Body #2
Free Body #3
#1 and 3 are not really free-body diagrams!!!
P1L1
P2 L2
elongation of bar 1 = 1 
; elongation of bar 2 =  2 
A1E1
A2 E2
Displacement B.C.
total elongation = 0 =  +
1
2
P1L1 P2 L2
0

A1E1 A2 E2
We now have two equations to solve for P1 and P2 :
(2)
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
 1
 L
 1
 A1E1
10
1 
 P1   P 

L2     
  P2   0 
A2 E2 
Solution of the equations for P1 and P2 gives:
P
P1 
 A2 E2 L1 
1
 A E L 
 1 1 2
and
P
P2 
 A1E1L2 
1
 A E L 
 2 2 1
Stresses are:
P1
 xx 
1
A1
P2
 xx 
2
A2
Note: for P to right,  xx is tensile (+) and  xx is compressive (-).
1
2
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
11
Displacement at point B is
P1L1
P( L1L2 )
 B  1 

A1E1 A1E1L2  A2 E2 L1
Note that for P to the right (positive P) ,  B is positive (to the
right) as expected.
Special Cases: L1  L2  L
1. A1  A2 , E1  E2 :  P1  P / 2 (tension), P2   P / 2 (comp)
2. A1  A  2 A2 (bar 1 has larger area), E1  E2 :
 P1  (2 / 3) P , P2  (1/ 3) P (bar 1 carries more load)
 xx  (2 / 3) P / A,  xx  (2 / 3) P / A (same!)
1
2
3. A1  A2 , E1  2 E2 (bar 1 has higher E, i.e., is "stiffer"):
 P1  (2 / 3) P , P1  (1/ 3) P (bar 1 carries more load)
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
12
General Solution Procedure for Bars
1. Determine number of free-body diagrams required.
Generally need a cut in each bar. However, if applied load or
bar properties (E, A, etc.) change along the bar, then you need
a cut on either side of where the change occurs.
2. Draw free-body diagrams labeling all unknown forces, Pi .
Must be consistent by assuming + directions; must be
consistent with equal and opposite forces on adjacent freebody diagrams where a cut was made).
3. Write appropriate COLM and COAM equations.
4. Determine if problem is statically determinate or
indeterminate.
5. If indeterminate, determine appropriate displacement
kinematic boundary conditions for the problem
(elongations sum to zero, elongations opposite, one
elongation multiple of another, etc.).
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
13
6. Determine elongations  i in terms of unknown (assumed)
internal forces. Remember positive direction of elongation is
in same direction as assumed member force Pi .
7. Evaluate kinematic B.C. in terms of displacements (now in
terms of unknown internal forces).
8. Solve equations obtained in steps 3 and 7 for the unknown
internal forces.
9. If appropriate, plot member force for all bars, i.e., the
structure (particularly when bars in series; will help you
understand what is going on).
Evaluate elongations, strains and stresses in each bar (by
substituting the internal forces). If you followed the "+"=tension
sign convention rigorously, then + elongation means stretching
and - means compression, + stress means tension, etc.
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
14
Example 3: Two elastic bars in parallel.
A1,E1,L1
A 2 ,E 2 ,L 2
1
2
P
A rigid horizontal bar is
attached to the bottom of
each vertical bar.
Horizontal bar remains
horizontal when pulled
downwards.
Determine: force, stress and deflection in each vertical bar.
As before, we have four governing equations: Equilibrium,
Stress/strain, Kinematics and Boundary Conditions.
2001, W. E. Haisler
15
Chapter 11: Axial Deformation and Stress of Bars
Use a “free-body diagram” to determine equilibrium of the
forces acting in each bar. Assume force in the bars are P1
and P2 (positive). Take a free-body by cutting each bar
below its fixed point:
P1
P2
A1,E1,L1
L1
A 2 ,E 2 ,L 2
1
1
L2
2
2
2
1
P
P1
P2
2001, W. E. Haisler
16
Chapter 11: Axial Deformation and Stress of Bars
Equilibrium of the free body in terms of forces requires that
 Fvertical  0  P1  P2  P
or
P  P1  P2
(1)
Elongation of each bar is:
P1L1
elongation of bar 1 = 1 
A1E1
P2 L2
elongation of bar 2 =  2 
A2 E2
Displacement B.C.:
 =
1
2
or
P1L1 P2 L2

A1E1 A2 E2
(2)
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
17
We now have two equations and two unknowns (P1 and P2).
Writing the two equations in matrix notation:
 1
 L
 1
 A1E1
1 
 P1   P 

L2     
  P2   0 

A2 E2 
Note: for simple problems, you can use Cramer’s rule of
determinants to solve the system of equations. Otherwise,
you Maple, EES, calculator, etc..
Solve for P1 and P2 gives:
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
1
L2
PL2
0
A2 E2
P
A2 E2
P1 


L2
L1
A2 E2 L1
1
1

1
A2 E2 A1E1
A1E1L2
L1
L2
A1E1 A2 E2
P
1
P
L1
0
A1E1
P
P2 

A1E1L2
1
1
1
A2 E2 L1
L1
L2
A1E1 A2 E2
18
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
19
Stress in each bar is
 xx
1
P1
P / A1


A1 1  A2 E2 L1
A1E1L2
and the deflection is:
P1L1
1 

A1E1
Special Cases:
L1  L2  L for all cases below!
 xx
2
P2
P / A2


A2 1  A1E1L2
A2 E2 L1
PL1
 2

A2 E2 L1 
A1E1 1 

A
E
L
1 1 2 

2001, W. E. Haisler
20
Chapter 11: Axial Deformation and Stress of Bars
1. A1  A2 , E1  E2 :  P1  P / 2 (tension), P2  P / 2 (tension)
2. A1  A  2 A2 (bar 1 has larger area), E1  E  E2 :
 P1  (2 / 3) P , P2  (1/ 3) P (bar 1 carries more load)
 xx  (2 / 3) P / A,  xx  (2/ 3) P / A (stresses same!)
1
2
 P1 is twice P2 , but A1is 1/2 of A2 , so stress is same
3. A1  A  A2 , E1  E  2 E2 (bar 1 has higher E, i.e., "stiffer"):
 P1  (2 / 3) P , P2  (1/ 3) P (bar 1 carries more load)
 xx  (2 / 3) P / A,  xx  (1/ 3) P / A (stress not same)
1
2
Since 1   2 (kinematic BC) and L1  L2 , strain is same
in each bar; but E1 is twice E2 , so  xx is twice  xx .
1
2
2001, W. E. Haisler
21
Chapter 11: Axial Deformation and Stress of Bars
Class Exercise: The horizontal bar is rigid and pinned at it's left
end. The horizontal bar rests on two vertical bars as shown and
has a 1 Kip load at its right end. For each vertical bar, determine:
1) force, 2) stress, 3) displacement at it's top and 4) axial strain.
10 in
10 in
10 in
2
8 in
1
8 in
1 Kip
E1  30 x106 psi
A1  1 in 2
E2  20 x106 psi
A2  2 in 2
You can assume that motion is "small" so that the vertical bars
remain vertical when loaded; also ends are rounded so they carry
only axial forces.
2001, W. E. Haisler
22
Chapter 11: Axial Deformation and Stress of Bars
Example 4: Uniaxial elastic bar subjected to a uniform
temperature increase of T:
A
E=Young's modulus
 =coefficient of
E
thermal expansion
L
The bar is fixed between two walls and has a constant crosssection A.
Determine: axial strain and stress in the bar and the force on the
wall.
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
23
For this problem, we have 4 governing equations to satisfy:
 xx
1. Equilibrium:
0
x
2. Stress-Strain:  xx 
elastic
E xx
total
elastic
thermal
thermal
  xx
  xx
 T ,
3. Kinematics:  xx
,  xx
total u x
= axial strain measured/observed
 xx 
x
(e.g., by a strain gage)
4. Boundary Conditions: u( x  0)  0 & u( x  L)  0
total u x
or  xx 
0
x
2001, W. E. Haisler
24
Chapter 11: Axial Deformation and Stress of Bars
Solution
u
x =0 (zero because of B.C.)
total 
1)  xx
x
2) Combine stress-strain, kinematics, and boundary
condition to obtain
0
elastic
total
thermal
total
 xx  E xx
 E ( xx
  xx
)  E (  xx
3)  xx  P / A

 T )   ET
P   xx A  ( ET ) A   ETA
(bar is in compression)
P is the force in the bar; hence force on wall is also P.
2001, W. E. Haisler
25
Chapter 11: Axial Deformation and Stress of Bars
Suppose we have an aluminum bar of area A=0.1 in2 with
E=10x106 psi and =6x10-6 (in/in)/F
and thermal loading of T=250F (250F temperature rise
above the reference temperature)
Then
 xx   ET  (10 x106 psi)(6 x106 in / in / F )(250 F )  15,000 psi
and
P   xx A  15,000 psi(0.1in 2 )  1,500lbf
(compression)
2001, W. E. Haisler
26
Chapter 11: Axial Deformation and Stress of Bars
Bar with Distributed Axial Load
Consider a bar with a distributed axial load p x ( x ) [units of
force/length].
px ( x)
x
Determine the differential equation relating the internal
force, P, to the applied distributed load, p x ( x ) . Note that the
internal force P will be a function of x. First, consider a
free-body of a differential section at any point x:
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
P( x)
px ( x)
x
dx
27
P ( x  dx )
x  dx
For equilibrium in x direction:
P(x  dx)  P(x)  pxdx  0 .
P
 px  0
Divide by dx and take limit to obtain
x
 ux
Recall P   xx A  ( E xx ) A  EA
. Substitute P into
x

 ux
( EA
)  px  0 .
equilibrium to obtain the governing ODE
x
x
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
28
Note that in obtained the last solution, we have combined
the required governing equations:
 xx
1. Static Equilibrium (from COLM):
  gx  0
x
2. Constitutive (Stress-Strain):  xx  E xx
u x
3. Kinematics (Strain-Displacement):  xx 
x
and obtained a single second-order ODE

 ux
( EA
)  px  0
x
x
In solving this ODE, we will need two boundary conditions;
one for the axial displacement ( u x ), and one for internal force
(P, or stress  xx ).
2001, W. E. Haisler
29
Chapter 11: Axial Deformation and Stress of Bars
Two similar, but different, examples of axial loading:
Case 1:
A=2 in2
x
100 in
F
20,000lb
3
u x ( x) 
x 7

10
x
2
EA
10 psi (2in )
F=20,000 lb = 20 Kips
E=10x10 6 psi
 u x (100")  0.1"
F 20,000lb
 xx  
 10,000 psi
2
A
2in
u x
7
3


E


E

10
psi
(10
in / in)  10,000 psi
or,
xx
xx
x
Note: stress is constant (not a function of x)!
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
p x  200lb / in
Case 2:
Note: total load
on beam is
x
30
6
E=10x10 psi
2
A=2 in
100 in
(200lb / in)(100in)  20,000lb (same as case 1, but now
distributed over length).
 ux

( EA
)   p x  200lb / in
Governing ODE:
x
x
d ( EA
du x
dx
)  200dx
 integrate  EA
du x
dx
 200 x  C1
Need to apply a boundary condition to obtain C1. Recall that
 xx
internal axial force P
du x
 P   xx A  E xx A  E


A
A
A
dx
2001, W. E. Haisler
31
Chapter 11: Axial Deformation and Stress of Bars
B.C. for P (internal axial force):
P(x)
Do a free body of the bar at any x:
x
p x  200lb / in
100” - x
0   Fx   P( x)  200(100  x)  P( x)  200lb / in(100" x)
For x=100", P(100")=0
du x
Applying B.C. for P (note: P   xx A  EA
):
dx
P ( x  100")  0  EA
du x
dx
x 100"
 200(100")  C1
 C1  20,000lb
2001, W. E. Haisler
32
Chapter 11: Axial Deformation and Stress of Bars
Substituting C1 into the reduced ODE:
EA
du x
dx
 200 x  20,000 
du x
1

 200 x  20,000 
dx EA


1
Integrate above to obtain: u x 
100 x 2  20,000 x  C2
EA
B.C. for u x : bar is fixed at left end, so u x ( x  0)  0


1
u x ( x  0)  0 
100(0)2  20,000(0)  C2  C2  0
EA
Hence,

1
ux 
100 x 2  20,000 x
EA

2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
33
Substituting EA values gives final displacement solution:
u x ( x)  103 x  0.5(105 ) x 2
Axial displacement at x=100" (end):
u x ( x  100")  0.1" .05"  0.05in
Stress is given by:
 xx  E xx
u x
E
 107 psi (103  105 x)
x
 xx ( x  0)  10,000 psi
 xx ( x  50)  5,000 psi
 xx ( x  100)  0
Axial stress varies with x!!
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
34
Note on these similar but very different cases:
1. Concentrated load of 20,000 lb at end
2. Distributed load of 200 lb/in (but total load of 20,000 lb)
Displacements completely different:
Case 1: displacement varies linearly with x (0.1" at end)
Case 2: displacement varies quadraticly with x (0.05" at
end).
Stresses are completely different:
Case 1: stress is constant (10,000 psi)
Case 2: stress varies linearly with x (10,000 psi at left
end, to 0 at right end)
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
35
An Alternate Approach for Bars with Axial Distributed Load
The 4 governing equations that must be satisfied for 1-D are:
 xx
1. Static Equilibrium (from COLM):
  gx  0
x
2. Constitutive (Stress-Strain):  xx  E xx
u x
3. Kinematics (Strain-Displacement):  xx 
x
4. Boundary Conditions: Depends on the problem
In COLM, recall that  g x is the body force per unit volume
applied to the system. Consider a prismatic bar with crossapplied force
section area, A, and length, L. Then  g x 
. If
AL
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
36
we multiply by A, then we can define the applied force per
unit length of the bar as p x   g x A. Note that p x  p x ( x ) .
px
A
x
L
Now take COLM and multiply  g x by A/A:
 xx
 xx
 (  g x A) / A  0 
 px / A  0
x
x
COLM can now be written as:
 xx
1a. Static Equilibrium (from COLM):
 px / A  0
x
2001, W. E. Haisler
37
Chapter 11: Axial Deformation and Stress of Bars
Consider the problem where p x and A are both constant (not
function of x). Let p x  p
p x  constant  p
A
x
L
 xx
1. Apply COLM:
 p / A  0.
x
Integrate wrt to x to obtain:  xx  ( p / A) x  C1.
Need B.C. for  xx . At x=L,  xx ( x  L)  0 . Get this from
p
free body: 0   Fx   xx A  p ( L  x)
 xx ( x)
Thus:  xx  p ( L  x) / A   xx ( L)  0
x
L-x
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
Apply B.C. to stress solution:
 xx ( x  L)  0  ( p / A) L  C1  C1  ( p / A) L
Solution for stress becomes:
 xx  ( p / A) x  ( p / A) L
or
p
 xx ( x)  ( L  x)
A
2. Apply Constitutive Equation:  xx  E xx
 xx  E xx
p
 ( L  x)
A
or
 xx
p

( L  x)
AE
u x
3. Apply Kinematics:  xx 
x
u x
p
 xx 

( L  x) . Integrate wrt to x to obtain:
x AE
38
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
39
p
u x ( x) 
( Lx  x 2 / 2)  C2
AE
Need B.C. for u x ( x) . Bar is fixed at x=0, thus u x ( x  0)  0 .
Applying the B.C. to displacement solution:
p
u x ( x  0)  0 
[ L(0)  (0) 2 / 2]  C2  C2  0
AE
Hence, solution for axial displacement is
p
u x ( x) 
( Lx  x 2 / 2)
AE
Recall, the above solution for stress, strain and displacement
is only valid for p x =constant= p . If p x or A is a function of
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
40
x, then in Step 1/1a (solution of COLM equation), use
p x  p x ( x ) and A  A( x).
Note that this alternate approach to solving problems with a
distributed axial load is really no different then the first
approach. In the alternate approach, we solve the governing
equations in succession:
 xx
1. Static Equilibrium (from COLM):
 px / A  0
x
2. Constitutive (Stress-Strain):  xx  E xx
u x
3. Kinematics (Strain-Displacement):  xx 
x
In the first approach, we combined all equations to get a
u

( EA x )  p x  0 .
second order ODE to solve:
x
x
2001, W. E. Haisler
41
Chapter 11: Axial Deformation and Stress of Bars
Class Exercise: Distributed load of 200 lb/in over length of bar
and 10,000 lb point force at end:
p x  200lb / in
x
100 in
10,000 lb
6
E=10x10 psi
A=2 in2
Determine:
a) u x ( x), and u x (50"), u x (100")
b)  xx ( x), and  xx (0),  xx (50"),  xx (100")
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
42
Other Cases
1. Assume a long bar which is slightly tapered with an end
load F, and whose cross-sectional area is a function of x:
A=A(x).
x
L
A=A(x)
F
Stress-Strain:  xx   xx / E
F
Equilibrium at x:  xx ( x) 
A( x)
Displacement B.C.: ux(x  0)  0
u
Kinematics:  xx  x
x
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
43
Combine all constitutive and kinematics equations to obtain
 ux
F
  xx   xx / E  ( ) / E
x
A
or
F
du x 
dx
AE
Integrate from 0 to x':
u x ( x ')  u x (0)  
x'
0
Displace B.C.: ux(0)  0
so
F
F x' 1
dx  
dx
E 0 A
AE
F x' 1
u x ( x ')  
dx
E 0 A
In order to complete the solution, must have a specific case
for A=A(x).
2001, W. E. Haisler
Chapter 11: Axial Deformation and Stress of Bars
2. Non-constant, axial force at any cross-section, F=F(x).
x
L
F
A=A(x)
Same as above except leave F inside of the integral.
u x ( x ')  
x'
0
F
1 x' F
dx  
dx
E 0 A
AE
F can be a function of x if gravity acts in the x direction.
44