Third Edition LECTURE RODS: AXIAL LOADING AND DEFORMATION .pdf · rod or the member is homogeneous...

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• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering • A. J. Clark School of Engineering •Department of Civil and Environmental Engineering • A. J. Clark School of Engineering •Department of Civil and Environmental Engineering Third Edition LECTURE 3 2.8 Chapter by Dr. Ibrahim A. Assakkaf SPRING 2003 ENES 220 – Mechanics of Materials Department of Civil and Environmental Engineering University of Maryland, College Park RODS: AXIAL LOADING AND DEFORMATION LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 1 ENES 220 ©Assakkaf Deformations of Members under Axial Loading Uniform Member – Consider a homogeneous rod as shown in the figure of the next viewgraph. – If the resultant axial stress σ = P/A dos not exceed the proportional limit of the material, then Hooke’s law can applied, that is elasticity of modulus strain axial stress, axial = = = = E E ε σ ε σ

Transcript of Third Edition LECTURE RODS: AXIAL LOADING AND DEFORMATION .pdf · rod or the member is homogeneous...

Page 1: Third Edition LECTURE RODS: AXIAL LOADING AND DEFORMATION .pdf · rod or the member is homogeneous ... – The total deformation δof the rod or bar is ... LECTURE 3. RODS: AXIAL

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• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering

Third EditionLECTURE

32.8

Chapter

byDr. Ibrahim A. Assakkaf

SPRING 2003ENES 220 – Mechanics of Materials

Department of Civil and Environmental EngineeringUniversity of Maryland, College Park

RODS: AXIAL LOADING AND DEFORMATION

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 1ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Uniform Member– Consider a homogeneous rod as shown in

the figure of the next viewgraph.– If the resultant axial stress σ = P/A dos not

exceed the proportional limit of the material, then Hooke’s law can applied, that is

elasticity of modulusstrain axial stress, axial

===

=

E

Eεσ

εσ

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 2ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Uniform Member

P

δ

L

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 3ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Uniform Member– From Hooke’s law, it follows that

EAPL

LE

AP

L

EAP

E

=⇒=

=

=

=

δδ

δε

ε

εσ

, therefore,But

law)s(Hooke'

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 4ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Uniform Member– The deflection (deformation),δ, of the

uniform member subjected to axial loading P is given by

EAPL

=δ (1)

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 5ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Multiple Loads/Sizes– The expression for the deflection of the

previous equation may be used only if the rod or the member is homogeneous (constant E) and has a uniform cross sectional area A, and is loaded at its ends.

– If the member is loaded at other points, or if it consists of several portions of various cross sections, and materials, then

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 6ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Multiple Loads/Sizes– It needs to be divided into components

which satisfy individually the required conditions for application of the formula.

– Denoting respectively by Pi, Li, Ai, and Ei, the internal force, length, cross-sectional area, and modulus of elasticity corresponding to component i,then

∑∑==

==n

i ii

iin

ii AE

LP11

δδ

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 7ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Multiple Loads/Sizes

E1 E2 E3

L1 L2 L3

∑∑==

==n

i ii

iin

ii AE

LP11

δδ

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 8ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Multiple Loads/Sizes– The deformation of of various parts of a rod

or uniform member can be given by

∑∑==

==n

i ii

iin

ii AE

LP11

δδ

E1 E2 E3

L1 L2 L3

(2)

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 9ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 4– Determine the deformation of the steel rod

shown under the given loads. Assume that the modulus of elasticity for all parts is

– 29×106 psi

75 kips45 kips

12 in 12 in 16 in

30 kips

Area = 0.9 in2

Area = 0.3 in2

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 10ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 4 (cont’d)– Analysis of internal forces

75 kips45 kips 30 kips

1 2 3

30 kipsP3

kips 30

030 ;0

3

3

=∴

=+−=+→ ∑P

PFx

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 11ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 4 (cont’d)– Analysis of internal forces

75 kips45 kips 30 kips

1 2 3

30 kipsP2

kips 1503045 ;0

2

2

−=∴

=+−−=+→ ∑P

PFx

45 kips

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 12ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 4 (cont’d)– Analysis of internal forces

75 kips45 kips 30 kips

1 2 3

75 kips45 kips 30 kipsP1

kips 600304575 ;0

1

1

=∴

=+−+−=+→ ∑P

PFx

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 13ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 4 (cont’d)– Deflection

• Input parameters

lb 30,000kips 30lb 15,000kips 15

lb 60,000kips 60forces, internal of analysis From

in 3.0 in 9.0 in 9.0

in 16 in 12 in 12

3

2

1

23

22

21

321

==−=−=

==

===

===

PPP

AAALLL

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 14ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 4 (cont’d)– Carrying the values into the deformation

formula:

in 0759.03.0

)16(000,309.0

)12(000,159.0

)12(000,601029

1

1

6

3

33

2

22

1

113

1

=

+

−+

×

++== ∑

=

δ

δALP

ALP

ALP

EAELP

i ii

ii

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 15ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Relative DeformationA A

L

δA

δBB

P

DC

AEPL

ABAB =−= δδδ /

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 16ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Relative Deformation– If the load P is applied at B, each of the

three bars will deform.– Since the bars AC and AD are attached to

the fixed supports at C and D, their common deformation is measured by the displacement δA at point A.

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 17ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Relative Deformation– On the other hand, since both ends of bars

AB move, the deformation of AB is measured by the difference between the displacements δA and δB of points A and B.

– That is by relative displacement of B with respect to A, or

EAPL

ABAB =−= δδδ /(3)

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 18ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 5The rigid bar BDE is supported by two links AB and CD.

Link AB is made of aluminum (E = 70 GPa) and has a cross-sectional area of 500 mm2. Link CD is made of steel (E = 200 GPa) and has a cross-sectional area of (600 mm2).

For the 30-kN force shown, determine the deflection a) of B, b) of D, and c) of E.

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 19ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 5 (cont’d)

SOLUTION:

• Apply a free-body analysis to the bar BDE to find the forces exerted by links AB and DC.

• Evaluate the deformation of links ABand DC or the displacements of Band D.

• Work out the geometry to find the deflection at E given the deflections at B and D.

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 20ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 5 (cont’d)Free body: Bar BDE

( )

( )ncompressioF

F

tensionF

F

M

AB

AB

CD

CD

B

kN60

m2.0m4.0kN300

0M

kN90

m2.0m6.0kN300

0

D

−=

×−×−=

=

+=

×+×−=

=

SOLUTION:

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 21ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 5 (cont’d)

Displacement of B:

( )( )( )( )

m10514

Pa1070m10500m3.0N1060

6

926-

3

−×−=

××

×−=

=AEPL

↑= mm 514.0Bδ

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 22ENES 220 ©AssakkafDeformations of Members under

Axial LoadingExample 5 (cont’d)

Displacement of D:

( )( )( )( )

m10300

Pa10200m10600m4.0N1090

6

926-

3

−×=

××

×=

=AEPL

↓= mm 300.0Dδ

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 23ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 5 (cont’d)Displacement of D:

( )

mm7.73

mm 200mm 0.300mm 514.0

=

−=

=′′

xx

xHDBH

DDBB

( )

mm 928.1mm 7.73

mm7.73400mm 300.0

=

+=

=′′

E

E

HDHE

DDEE

δ

δ

↓= mm 928.1Eδ

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 24ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Nonuniform Deformation– For cases in which the axial force or the

cross-sectional area varies continuously along the length of the bar, then Eq. 1

– (PL / EA) is not valid.– Recall that in the case of variable cross

section, the strain depends on the position of point Q, where it is computed from

dxdδε =

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 25ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Nonuniform Deformation

xL

P

dxdδε =

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 26ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Nonuniform Deformation– Solving for dδ and substituting for ε

– But ε = σ / E, and σ = P/A. therefore

dxd εδ =

dxEAPdx

Edxd ===

σεδ (4)

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 27ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Nonuniform Deformation– The total deformation δ of the rod or bar is

obtained by integrating Eq. 4 over the length L as

∫=L

x

x dxEAP

0

δ (5)

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 28ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 6– Determine the deflection of point a of a

homogeneous circular cone of height h, density ρ, and modulus of elasticity E due to its own weight.

h

a

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 29ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 6 (cont’d)– Consider a slice of thickness dy– P = weight of above slice– = ρg (volume above)

hyδy r

( ) ydyEg

rE

yrg

EAPdyd

yrgP

331

31

2

2

2

ρπ

πρδ

πρ

=

==

=

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 30ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 6 (cont’d)

Egh

yEgydy

Eg

ydyEgd

hh

6

233

3

20

2

0

ρδ

ρρδ

ρδ

=

==

=

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 31ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Normal Stresses in Tapered Bar– Consider the following tapered bar with a

thickness t that is constant along the entire length of the bar.

h(x)h1 h2x

L

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 32ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Normal Stresses in Tapered Bar

– The following relationship gives the height h of tapered bar as a function of the location x

( )Lxhhhhx 121 −+= (6)

h(x)h1 h2x

L

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 33ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Normal Stresses in Tapered Bar

– The area Ax at any location x along the length of the bar is given by

( )

−+==

LxhhhtthA xx 121 (7)

h(x)h1 h2x

L

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 34ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Normal Stresses in Tapered Bar

– The normal stress σx as a function of x is given

( )

−+

==

Lxhhht

PAP

xx

121

σ(8)

h(x)h1 h2x

L

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 35ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 7– Determine the normal stress as a function of x

along the length of the tapered bar shown if– h1 = 2 in– h2 = 6 in– t = 3 in, and– L = 36 in– P = 5,000 lb h(x)h1 h2

x

L

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 36ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 7 (cont’d)– Applying Eq. 8, the normal stress as a

function of x is given by

( )

( )

x

xx

Lxhhht

PAP

x

x

xx

+=

+=

−+

=

−+

==

1815000

36

5000

36)2623

5000

121

σ

σ

σ

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 37ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Example 7 (cont’d)– Max σ = 833.3 psi

• At x =0– Min σ = 277.8 psi

• At x =36 in

h(x)h1 h2x

L

x (in) σ (psi)0 833.33 714.36 625.09 555.6

12 500.015 454.518 416.721 384.624 357.127 333.330 312.533 294.136 277.8

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 38ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Deflection of Tapered Bar– Consider the following tapered bar with a

thickness t that is constant along the entire length of the bar.

h(x)h1 h2x

L

dx

LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 39ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Deflection of Tapered Bar– Recall Eq. 5

– Substitute Eq. 7

– into Eq. 5, therefore

∫=L

x

x dxEAP

0

δ

( ) dxxhhLhEt

PL L

1

0 121∫ −+

=δ (9)

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LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 40ENES 220 ©Assakkaf

Deformations of Members under Axial Loading

Deflection of Tapered BarIntegrating Eq. 9, the deflection of a tapered bar is given by

( )[ ]LhhhhEt

PL12

12

ln1−

=δ (10)