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