Post on 09-Feb-2016
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Axially loaded member
Axial load and normal stress under equilibrium load, Elastic Deformation
1
Saint-Venant’s Principle
2
• 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
Saint-Venant’s Principle
3
• General rule: min. distance is at least equal to largest dimension of loaded x-section. 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
Elastic Deformation of an Axially Loaded Member
4
AE
PLDeformation can be calculated using
SIGN CONVENTION
ve When the member is tension
ve When the member is compression
Elastic Deformation of an Axially Loaded Member
AE
PLtotTotal deformation:
The Above Figure:
AE
LP
AE
LP
AE
LP CDCDBCBCABABtot
CDBCABtot
Elastic Deformation of an Axially Loaded Member
AE
LkN
AE
LkN
AE
LkN CDBCABtot
)7()3()5(
1) Internal Forces
2) Displacement calculation
Example 1
7
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
Internal axial load:
BACBDCA
Displacement of point A
Displacement of point B relative to C
CBCB /
mm
EA
LP
EA
LP
EA
LP
A
A
BA
BABA
CB
CBCB
DC
DCDCA
BACBDCA
61.0
)10)(210(600
)1000()10(75
)10)(210(1200
)750()10(35
)10)(210(1200
)500()10(453
3
3
3
3
3
Displacement of point A
Displacement of point B relative to C
3
3
/
/
)10)(210(1200
)750()10(35
CB
CBCB
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.
PAB=30 kNPAB=30 kN
mm
AB
06737.0
))10(126)(4/)75((
)1250)()10(30(32
3
PAB=30 kNP=20 kN
PBC=10 kN
P=5 kNPCD=-5 kN
Tension
Tension
mm
BC
07578.0
))10(126)(4/)50((
)1875)()10(10(32
3
Compression
mm
CD
12126.0
))10(126)(4/)25((
)1500)()10(5(32
3
mmtot 02189.012126.007578.006737.0
Total deformation
1) Internal Forces 2) Displacement calculation
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.
Lecture 1 14
F= 30kN
FDC
FAB
NF
NF
M
AB
DC
A
3
33
)10(20
)10(10900
)300()10(30
0
1) Equilibrium Equation
2) Based on FAB and FDC, FDC is predicted to deform more.
DC
AB
mm
DC
16667.0
))10(120)(600(
)1200)()10(10(3
3
mm
AB
33333.0
))10(120)(900(
)1800)()10(20(3
3
3) Angle of tilt
o
DCAB
01061.0
)900
(tan 1
Example 2
15
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.
m 003056.010200005.0
6.010809
3
/
AE
PLBC
16
m 001143.0001143.0
107010400
4.0108096
3
AE
PLB
Find the displacement of end C with respect to end B.
Displacement of end B with respect to the fixed end A,
Since both displacements are to the right,
mm 20.4m 0042.0/BCC B