2001, W. E. Haisler
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Chapter 11: Example - Bar with Distributed Axial Load
Bar with Distributed Axial Load
25 lbf / in
100 in
x
100 in
5,000 lb
6
E=10x10 psi
A=2 in2
Lets do some free-body diagrams to determine the internal
axial force P in the two sections of the bar.
2001, W. E. Haisler
Chapter 11: Example - Bar with Distributed Axial Load
P(x)
FB #1 (for x>100”)
Cut bar at some point x
x
within the distributed
load. Apply COLM in x direction:
25 lbf / in
2
5,000 lbf
200“-x
0   Fx   P( x)  25(200  x)  5,000
Hence,
P( x)  25 x lbf
FB #2 (for x<100”)
Apply COLM in x
direction:
(for x>=100”)
P(x)
x
25 lbf / in
100”
200-x
0   Fx   P( x)  25(100)  5,000
Hence,
P( x)  2,500 lbf
(for x<=100”)
5,000 lbf
2001, W. E. Haisler
Chapter 11: Example - Bar with Distributed Axial Load
Plot the internal axial force, P(x)
P(x)
5,000 lbf
2,500
100”
200”
What do you think the stress is?
At x=0”:  xx  P / A  2,500 lbf / A
At x=100”:  xx  P / A  2,500 lbf / A
At x=200”:  xx  P / A  5,000 lbf / A
x
3
2001, W. E. Haisler
Chapter 11: Example - Bar with Distributed Axial Load
4
You could also get P( x) by integration. P( x) and distributed
load px ( x) are related by:
For 100  x  200", px  25 lbf / in .
dP
  px .
dx
Integrate Diff. Eq. to get: P( x)  25 x  C1.
B.C.: P(200")  5,000lbf  25lbf / in[200"]  C1.
Thus, C1  0 , and P( x)  25x lbf (100  x  200")
For 0  x  100", px  0 .
Integrate Diff. Eq. to get: P ( x)  C2 .
B.C. (using P( x)  25x above): P(100")  25[100"]  C2 .
Thus, C2  2,500 , and P( x)  2,500 lbf (0  x  100")
2001, W. E. Haisler
5
Chapter 11: Example - Bar with Distributed Axial Load
Distributed axial load ( p x ) and internal force (P) diagrams:
P( x)
px ( x)
100”
200”
x
5,000 lbf
2,500 lbf
25 lbf / in
100”
200”
x
Observations:
1. Where distributed axial load p x is 0, internal axial force
P is a constant.
2. Where distributed axial load p x is a non-zero constant
value, internal axial force P is a linear function of x.
3. This is because P is the integral of p x .
2001, W. E. Haisler
6
Chapter 11: Example - Bar with Distributed Axial Load
Now lets solve for the displacement u x ( x)
25 lbf / in
100 in
x
100 in
5,000 lb
6
E=10x10 psi
A=2 in2
Have to work as two sections. Why? Because one has
distributed load and the other does not. From the previous
free-body work, we also saw that the internal axial force
must be written separately for each section, i.e., one for
x<=100” and one for x>=100”.
Each section will have 2 Boundary Conditions: 1 on
displacement, u x , and 1 on internal axial force, P .
2001, W. E. Haisler
Chapter 11: Example - Bar with Distributed Axial Load
7
0  x  100" - Call this section 1 and the displacement u x1 ( x)
du x1
B.C.: 1) P(100")  AE
(100)  2,500 lbf (see prev. FB)
dx
2) u x1 (0")  0
Solve Diff. Eq. (note px  0 for section 1)
d 
du x1 
 AE
   px  0
dx 
dx 
du x1
Integrate to get:  AE
 C1
dx
Apply B.C.#1:
du x1 (100")
P(100")  AE
 2,500lbf  C1
dx
du x1
So, we have: AE
 2,500lbf
dx
 C1  2,500lbf
2001, W. E. Haisler
Chapter 11: Example - Bar with Distributed Axial Load
Integrate to get: 
1
u x1 ( x) 
(2,500 x  C2 )
AE
Apply B.C.#2:
1
u x1 (0)  0 
(2,500[0]  C2 )
AE
 C2  0
1
So, we have: u x1 ( x) 
(2,500 x) in
AE
Note: A  2in2 and E  10 x106 psi
Now do the next section.
( 0  x  100")
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2001, W. E. Haisler
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Chapter 11: Example - Bar with Distributed Axial Load
100  x  200" - Call this section 2 and the displacement u x 2 ( x)
du x 2
B.C.: 1) P(200")  AE
(200)  5,000 lbf (see prev. FB)
dx
1
2) u x 2 (100")  u x1 (100") 
(2,500[100"])  2.5 x105 / AE
AE
Solve Diff. Eq. (note px  25 lbf / in for section 2)
d 
du x 2 
 AE
   px  (25 lbf / in)  25 lbf / in
dx 
dx 
du x 2
Integrate to get:  AE
 25 x  C3
dx
Apply B.C.#1:
du x 2 (200")
P(200")  AE
 5,000lbf  25(200")  C3
dx
 C3  0
2001, W. E. Haisler
Chapter 11: Example - Bar with Distributed Axial Load
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du x 2
So, we have: AE
 25 x
dx
Integrate to get: 
1
u x 2 ( x) 
(25 x 2 / 2  C4 )
AE
Apply B.C.#2:
2.5 x105
1
u x 2 (100") 

(25[100]2 / 2  C4 )
AE
AE
 C4  1.25 x105
1
So, we have: u x 2 ( x) 
(12.5 x 2  1.25 x105 ) in (100  x  200")
AE
Note: A  2in2 and E  10 x106 psi
2001, W. E. Haisler
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Chapter 11: Example - Bar with Distributed Axial Load
Now lets determine strain and stress in each section from u x .
du x1 2,500
,
 xx1 

dx
AE
 xx 2
du x 2 25 x
,


dx
AE
Plot  xx A:
 xx A
2,500lbf
 xx1  E xx1 
A
 xx 2  E xx 2
25 x lbf

A
2,500 lbf
 xx 2 (100")  E xx 2 
A
5,000 lbf
 xx 2 (200")  E xx 2 
A
5,000 lbf
2,500
100”
200”
x
2001, W. E. Haisler
Chapter 11: Example - Bar with Distributed Axial Load
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If you substitute A  2in2 and E  10 x106 psi , you have the
following:
( 0  x  100")
ux1( x)  1.25 x104 x in
ux 2 ( x)  6.25 x107 x 2  6.25x103 in (100  x  200")
u x (200")  0.03125 in
du x1
 xx1 
 1.25 x104 in / in ,
dx
 xx 2
du x 2

 (1.25 x106 ) x in / in ,
dx
 xx (0")   xx (100")  1, 250 psi ,
 xx1  E xx1  1, 250 psi
 xx 2  E xx 2  12.5 x psi
 xx (200")  2,500 psi
2001, W. E. Haisler
Chapter 11: Example - Bar with Distributed Axial Load
13
linear
quadratic
constant
linear
Observations:
1. Where the displacement varies linearly with x, the
stress is a constant.
2. Where the displacement varies quadratically with x, the
stress varies linearly.
3. This is because strain is first derivative of displacement.