Axial load and normal stress under equilibrium load, Elastic
Deformation
1

• Localized deformation occurs at each end, and the deformations decrease as measurements are taken further away from the ends
• At section c-c
, stress reaches almost uniform value as compared to a-a
, b-b
• c-c is sufficiently far enough away from P so that localized deformation “vanishes”, i.e., minimum distance
2

• General rule: min. distance is at least equal to largest dimension of loaded xsection. For the bar, the min. distance is equal to width of bar
• This behavior discovered by Barré de
Saint-Venant in 1855, this the name of the principle
• Saint-Venant Principle states that localized effects caused by any load acting on the body, will dissipate/smooth out within regions that are sufficiently removed from location of load
• Thus, no need to study stress distributions at that points near application loads or support reactions
3

Elastic Deformation of an Axially
Loaded Member
Deformation can be calculated using  
PL
AE
SIGN CONVENTION
   ve
When the member is tension
   ve
When the member is compression
4

Elastic Deformation of an Axially
Loaded Member
Total deformation:
The Above Figure:
 tot
  PL
AE
 tot
 
AB
 
BC
 
CD
 tot

P
AB
L
AB
AE

P
BC
L
BC
AE

P
CD
L
CD
AE

Elastic Deformation of an Axially
Loaded Member
1) Internal Forces
2) Displacement calculation
  tot

( 5 kN ) L
AB
AE

(

3 kN ) L
BC
AE

(

7 kN ) L
CD
AE

Composite A-36 steel bar shown made from two segments AB and BD. Area
A
= 600 mm 2 and
A
BD
= 1200 mm 2 .
AB
Determine the vertical displacement of end A and displacement of
B relative to
C
.
A-36 Steel: E = 210 GPa
7

Internal axial load:

Displacement of point A

A
 
DC
 
CB
 
BA
Displacement of point B relative to C
 
B / C

CB

Displacement of point A

A
 
DC
 
CB
 
BA

A

A

A



P
DC
L
1200 (
DC

210

0 .
61 mm
P
)(
CB
10
L
)
3
CB
A
DC
E A
CB
E

45 ( 10 )
3
( 500 )

P
BA
A
L
BA
BA
E

35 ( 10 )
3
( 750 )
1200 ( 210 )( 10 )
3

75 ( 10 )
3
( 1000 )
600 ( 210 )( 10 )
3
Displacement of point B relative to C

B / C

B / C
 
CB

35 ( 10 )
3
( 750 )
1200 ( 210 )( 10 )
3

The copper shaft is subjected to the axial loads shown. Determine the displacement of end A with respect to end D. The diameter of each segment are d
126 GPa.
AB
= 75 mm, d
BC
= 50 mm and d
CD
= 25 mm. Take E cu
=

P
AB
=30 kN
1) Internal Forces
P
AB
=30 kN
P
AB
=30 kN
P
CD
=-5 kN
P=20 kN
P=5 kN
2) Displacement calculation

AB
Tension


( 30 ( 10 )
3
)( 1250 )
(

( 75 )
2
/ 4 )( 126 ( 10 )
3
)

0 .
06737 mm
Tension
P
BC
=10 kN 
BC


( 10 ( 10 )
3
)( 1875 )
(

( 50 )
2
/ 4 )( 126 ( 10 )
3
)

0 .
07578 mm
Compression

CD



( 5 ( 10 )
3
)( 1500 )
(

( 25 )
2
/ 4 )( 126 ( 10 )
3
)

0 .
12126 mm
Total deformation
 tot

0 .
06737

0 .
07578

0 .
12126

0 .
02189 mm

The assembly consist of three titanium (Ti-6A1-4V) rods and a rigid bar AC. The cross sectional area for each rod is given in the figure. If a force 30kN is applied to the ring F, determine the angle of tilt of bar
AC.

1) Equilibrium Equation

M
A

0
F
DC
F
AB

30 ( 10 )
3
(
900

20 ( 10 ) 3 N
300 )

10 ( 10 )
3
N
F
DC
F
AB
2) Based on F
AB and F
DC
, F
DC is predicted to deform more.

DC


( 10 ( 10 )
3
)( 1200 )
( 600 )( 120 ( 10 )
3
)

0 .
16667 mm

DC

AB


( 20 ( 10 )
3
)( 1800 )
( 900 )( 120 ( 10 )
3
)

0 .
33333 mm
3) Angle of tilt

AB
 
 tan

1
(

AB
 
900
DC
0 .
01061 o
Lecture 1
)
F= 30kN
14

The assembly shown in the figure consists of an aluminum tube
AB having a cross-sectional area of 400 mm 2 . A steel rod having a diameter of 10 mm is attached to a rigid collar and passes through the tube. If a tensile load of 80 kN is applied to the rod, determine the displacement of the end C of the rod.
Take E st
= 200 GPa, E al
= 70 GPa.
15

Find the displacement of end C with respect to end B.

C / B

PL
AE





80 10
0 .
005
 
 
  
200
    
0 .
003056 m

Displacement of end B with respect to the fixed end A,

B

PL
AE
 


400
80
10
10

 
  
70
    
0 .
001143

0 .
001143 m

Since both displacements are to the right,

C
 
B
 
C / B

0 .
0042 m

4 .
20 mm

16