Translation of axially loaded members
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Transcript Translation of axially loaded members
Axially loaded member
Axial load and normal stress under equilibrium load, Elastic
Deformation
1
Saint-Venant’s Principle
•
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
Saint-Venant’s Principle
•
•
•
•
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 L
P L
PAB LAB
BC BC CD CD
AE
AE
AE
Elastic Deformation of an Axially
Loaded Member
1) Internal Forces
2) Displacement calculation
tot
( 3kN ) LBC
( 7 kN ) LCD
(5kN ) LAB
AE
AE
AE
Example 1
Composite A-36 steel bar shown made
from two segments AB and BD. Area AAB
= 600 mm2 and ABD = 1200 mm2.
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
PDC LDC
P L
P L
CB CB BA BA
ADC E
ACB E
ABA E
45(10) 3 (500)
35(10) 3 (750)
75(10) 3 (1000)
3
3
1200( 210)(10)
1200( 210)(10)
600( 210)(10) 3
0.61m m
Displacement of point B relative to C
B / C CB
B/C
35(10) 3 (750)
1200(210)(10) 3
Solve it!
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 dAB = 75 mm, dBC= 50 mm and dCD = 25 mm. Take Ecu =
126 GPa.
1) Internal Forces
PAB=30 kN
2) Displacement calculation
Tension
(30(10) 3 )(1250)
AB
( (75) 2 / 4)(126(10) 3 )
0.06737m m
PAB=30 kN
Tension
P=20 kN
PAB=30 kN
PBC=10 kN
PCD=-5 kN
P=5 kN
(10(10) 3 )(1875)
BC
( (50) 2 / 4)(126(10) 3 )
0.07578m m
Compression
(5(10) 3 )(1500)
CD
( (25) 2 / 4)(126(10) 3 )
0.12126m m
Total deformation
tot 0.06737 0.07578 0.12126 0.02189m m
Solve it!
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.
FDC
1) Equilibrium Equation
M
FDC
FAB
A
0
30(10) 3 (300)
10(10) 3 N
900
20(10) 3 N
FAB
F= 30kN
2) Based on FAB and FDC, FDC is predicted to deform more.
DC
(10(10) 3 )(1200)
DC
(600)(120(10) 3 )
0.16667m m
(20(10) 3 )(1800)
AB
(900)(120(10) 3 )
0.33333m m
3) Angle of tilt
t an1 (
AB DC
900
o
0.01061
Lecture 1
AB
)
14
Example 2
The assembly shown in the figure consists of an aluminum tube
AB having a cross-sectional area of 400 mm2. 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 Est = 200 GPa, Eal = 70 GPa.
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Find the displacement of end C with respect to end B.
C / B
PL
80 103 0.6
0.003056m
9
AE 0.005 20010
Displacement of end B with respect to the fixed end A,
PL
80 103 0.4
B
0.001143 0.001143m
6
9
AE 40010 70 10
Since both displacements are to the right,
C B C / B 0.0042m 4.20 mm
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