2-Axial Loading.pdf

CHAPTER 2

AXIAL LOADING

FACULTY OF MECHANICAL ENGINEERING

DIVISION OF ENGINEERING MECHANICS

V{tÑàxÜ EV{tÑàxÜ E

Axial Loading

MEC411 – MECHANICS OF MATERIALS Ch 2 - 1

1. Ferdinand P. Beer, E. Russell Johnston,Jr, John T. Dewolf, David F. Mazurek “ Mechanics of Materials”

5th Edition in SI units

2. R.C.Hibbeler “ Mechanics of Materials “ Seventh Edition

Materials for this chapter are taken from :

CHAPTER 2

AXIAL LOADING



Suitability of a structure or machine may depend on the deformations in the

structure as well as the stresses induced under loading. Statics analyses

Introduction

alone are not sufficient.

Considering structures as deformable allows determination of member forces

and reactions which are statically indeterminate.

Determination of the stress distribution within a member also requires

consideration of deformations in the member.


Chapter 2 is concerned with deformation of a structural member under axial

loading. Later chapters will deal with torsional and pure bending loads.

CHAPTER 2

AXIAL LOADING



1. Saint-Venant’s Principle states that both

localized deformation and stress tend to

“even out” at a distance sufficiently

Saint-Venant’s Principle

“even out” at a distance sufficiently

removed from these regions.


CHAPTER 2

AXIAL LOADING



Normal Strain under Axial Loading


stress

normal strain

P

A

L

σ

δε

= =

= =L

A

P

A

P

δε

σ

=

==2

2

LL

A

P

δδε

σ

==

=

2

2

CHAPTER 2

AXIAL LOADING



Stress-Strain Test

rutlandplastics.co.uk


Tensile Test Machines & Specimensdeeshaimpex.com

www.tensilkut.com

CHAPTER 2

AXIAL LOADING



Stress-Strain Test: Ductile Materials

rupture

necking


Test specimen with tensile load -

elongation occurs until “necking”

and rupture.

CHAPTER 2

AXIAL LOADING



Stress-Strain Test: Brittle Materials

Pottery, glass, and cast iron


Pottery, glass, and cast iron

Material that breaks suddenly under

stress at a point just beyond its elastic

limit

Brittle materials may also break

suddenly when given a sharp knock

CHAPTER 2

AXIAL LOADING



AE

P

EE ===

σεεσ

From Hooke’s Law:

Deformations of Members under Axial Loading

From the definition of strain:

L

δε =

Equating and solving for the deformation,

AE

PL=δ


AE

With variations in loading, cross-section or

material properties,

∑=i ii

ii

EA

LPδ

CHAPTER 2

AXIAL LOADING



Example 2.1

STEPS:30 k�

45 k�75 k�

Determine the deformation of the

200

100mm 50mm

E GPa

D d

=

= =

STEPS:

Divide the rod into components at the

load application points.

Apply a free-body analysis on each

component to determine the internal

force

Evaluate the total of the component

120 mm 200 mm120 mm


Determine the deformation of the

steel rod shown under the given

loads.

Evaluate the total of the component

deflections.

CHAPTER 2

AXIAL LOADING



SOLUTION:

Divide the rod into three

components:

Apply free-body analysis to each component to

determine internal forces,

130kN

15kN

P

P

=

= −30 k�

A B CD

2

3

15kN

60kN

P

P

= −

=

Evaluate total deflection,

( ) ( ) ( )

1 1 2 2 3 3

1 2 3

3 3 3

1

30 10 0.12 15 10 0.12 60 10 0.21

i i

ii i

PL PL P L P L

A E E A A Aδ

= = + +

× − × ×= + +

× × × ×

∑P

2

30 k�

45 k�75 k�30 k�

30 k�

30 k�

45 k�

45 k�75 k�

P

P1

D


1 2

-3 2

1 2

120mm

7.854 10 m

L L

A A

= =

= = ×

3

3 2

3

200mm

1.96 10 m

L

A −

=

= ×

9 3 3 3

5

200 10 7.854 10 7.854 10 1.96 10

3.1758 10 m

− − −

−

= + + × × × ×

= ×

53.1758 10 m δ −= ×

P3

CHAPTER 2

AXIAL LOADING



The rigid bar BDE is supported by two links AB

Example 2.2

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


For the 30-kN force shown, determine the

deflection a) of B, b) of D, and c) of E.

CHAPTER 2

AXIAL LOADING



Displacement of B:

( )( )( )( )

m3.0N1060926-

3

××

×−=

=AE

PLBδ

Free body: Bar BDE

Solution

( )( )m10514

Pa1070m10500

6

926-

−×−=

××=

↑= mm 514.0BδDisplacement of D:

( )( )3×

=AE

PLDδ( )

tensionF

F

M

CD

CD

B

kN90

m2.0m6.0kN300

0

+=

×+×−=

=∑


( )( )( )( )

m10300

Pa10200m10600

m4.0N1090

6

926-

3

−×=

××

×=

↓= mm 300.0Dδ

( )

ncompressioF

F

tensionF

AB

AB

CD

kN60

m2.0m4.0kN300

0M

kN90

D

−=

×−×−=

=

+=

∑

CHAPTER 2

AXIAL LOADING



Displacement of D:

( )200 mm0.514 mm

BB BH

DD HD

x

′=′

−( )200 mm0.514 mm

0.300 mm

73.7 mm

x

x

x

−=

=

( )400 73.7 mmE

EE HE

DD HD

δ

′=′

+=


↓= mm 928.1Eδ

( )400 73.7 mm

0.300 mm 73.7 mm

1.928 mm

E

E

δ

δ

+=

=

CHAPTER 2

AXIAL LOADING



Statically Indeterminate Problems

Structures for which internal forces and reactions

cannot be determined from statics alone are said to

be statically indeterminate.

A structure will be statically indeterminate whenever

it is held by more supports than are required to

maintain its equilibrium.


maintain its equilibrium.

CHAPTER 2

AXIAL LOADING



Redundant reactions are replaced with

unknown loads which along with the other

Statically Indeterminate Problems

loads must produce compatible

deformations.

Deformations due to actual loads and

redundant reactions are determined

separately and then added or superposed.


0=+= RL δδδ

CHAPTER 2

AXIAL LOADING



Example 2.3

Determine the reactions at A and B for theDetermine the reactions at A and B for the

steel bar and loading shown, assuming a

close fit at both supports before the loads are

applied.


CHAPTER 2

AXIAL LOADING



STEPS:

Consider the reaction at B as redundant,

release the bar from that support, and solve

Example 2.3

Require that the displacements due to the

loads and due to the redundant reaction be

Solve for the displacement at B due to the

redundant reaction at B.

release the bar from that support, and solve

for the displacement at B due to the applied

loads.


Solve for the reaction at A due to applied

loads and the reaction found at B.

loads and due to the redundant reaction be

compatible, i.e., require that their sum be

zero.

CHAPTER 2

AXIAL LOADING



Solve for the displacement at B due to the applied loads with

the redundant constraint released,

AAAA

PPPP

2626

34

3321

m10250m10400

N10900N106000

×==×==

×=×===

−−

EEA

LP

LLLL

AAAA

i ii

ii9

L

4321

2643

2621

10125.1

m 150.0

m10250m10400

×=∑=

====

×==×== −−

δ

Solve for the displacement at B due to the redundant

constraint,


( )∑

×−==

==

×=×=

−==

−−

i

B

ii

iiR

B

E

R

EA

LPδ

LL

AA

RPP

3

21

262

261

21

1095.1

m 300.0

m10250m10400

CHAPTER 2

AXIAL LOADING



Require that the displacements due to the loads and due to

the redundant reaction be compatible,

( )0

39 ××

=+= RL δδδ

( )

kN 577N10577

01095.110125.1

3

39

=×=

=×

−×

=

B

B

R

E

R

Eδ

Find the reaction at A due to the loads and the reaction at B

kN323

kN577kN600kN 3000

=

∑ +−−== Ay

R

RF


kN323=AR

kN577

kN323

=

=

B

A

R

R

CHAPTER 2

AXIAL LOADING



Example 2.4

The A-36 steel rod shown has a diameter of 5 mm.

It is attached to the fixed wall at A, and before it is

loaded there is a gap between the wall at and theloaded there is a gap between the wall at and the

rod of 1 mm. Determine the reactions at A and B’.

Consider the support at B’ as redundant

and using principle of superposition,

Solution


0.001 p Bδ δ+ → = −

CHAPTER 2

AXIAL LOADING



Thus,

( ) ( )320 10 0.4PL

Example 2.4

By substituting into compatibility equation,

( )60.001 0.002037 0.3056 10 F−= −( ) ( )

( ) ( )

( )( ) ( )

( )

9

9

6

20 10 0.4

0.0025 200 10

0.002037

1.2

0.0025 200 10

0.3056 10

ACP

BB ABB

B

PL

AE

FF L

AE

F

δπ

δπ

−

= =

=

= =

=

m

( )( )

6

3

0.001 0.002037 0.3056 10

3.39 10 3.39 kN (Ans)

B

B

F

F

−= −

= =

From the free-body diagram,


0;

20 3.39 0

16.6

kN (Ans)

x

A

A

F

F

F

+ → =

− + − =

=

∑

CHAPTER 2

AXIAL LOADING



Problems Involving Temperature Changes

A temperature change results in a change in length or

thermal strain. There is no stress associated with the

thermal strain unless the elongation is restrained by

the supports.

( ) ;

thermalexpansioncoef.

T P

PLT L

AEδ α δ

α

= ∆ =

=

Treat the additional support as redundant and apply

the principle of superposition.

The thermal deformation and the deformation from


( )

0

0

T P

PLT L

AE

δ δ δ

α

= + =

∆ + =

The thermal deformation and the deformation from

the redundant support must be compatible.

( )

( )

0T P

P AE T

PE T

A

δ δ δ

α

σ α

= + =

= ∆

= = ∆

CHAPTER 2

AXIAL LOADING



A steel rod is stretched between two rigid walls and carries a tensile load of 5000 N

at 20°C. If the allowable stress is not to exceed 130 MPa at -20°C, what is the

Example 2.5

at 20°C. If the allowable stress is not to exceed 130 MPa at -20°C, what is the

minimum diameter of the rod? Assume α = 11.7 µm/(m�°C) and E = 200 GPa.


CHAPTER 2

AXIAL LOADING



Apply the total deformation formula

and obtain the area of the rod.

Get the diameter considering

formula for area.

Example 2.5

and obtain the area of the rod.

( )

( )

( )( )( ) 5000

T st

L PLL T

E AE

PE T

A

δ δ δ

σα

σ α

= − +

= − ∆ +

= − ∆ +

formula for area.

21137.36

4

13.22 .

d

d mm Ans

π =

=


( )( )( )6

2

5000130 11.7 10 200000 20 20

137.36

A

A mm

−= − × − − +

=

CHAPTER 2

AXIAL LOADING



Example 2.6

The rigid bar is fixed to the top of the three

posts made of A-36 steel and 2014-T6posts made of A-36 steel and 2014-T6

aluminum. The posts each have a length of

250 mm when no load is applied to the bar,

and the temperature is T1 = 20°C.

Determine the force supported by each post

if the bar is subjected to a uniform distributed

load of 150 kN/m and the temperature is


raised to T2 = 80°C.

CHAPTER 2

AXIAL LOADING



From free-body diagram we have

0; yF+ ↑ =∑

Example 2.6

Solution

The final position of the top of each

post is equal to its displacement

caused by the temperature increase

The top of each post is displaced by

an equal amount and hence,

( )3

0;

2 90 10 0 (1)

y

st al

F

F F

+ ↑ =

+ − =

∑ caused by the temperature increase

and internal axial compressive force.

( ) ( ) ( )

( ) ( ) ( )

st st stT F

al al alT F

δ δ δ

δ δ δ

+ ↓ = − +

+ ↓ = − +


( ) (2)st alδ δ+ ↓ =

CHAPTER 2

AXIAL LOADING



( ) ( ) ( ) ( )FalTstFstTst δδδδ +−=+−

Applying Eq. 2 gives

With reference from the material properties, we have

( )[ ]( )( ) ( )( ) ( )[ ] ( )[ ]( )( ) ( )

( ) ( )[ ]( ) (3) 109.165216.1

101.7303.0

25.025.020801023

1020002.0

25.025.020801012

3

92

6

92

6

−=

+−−=+−− −−

alst

alst

FF

FF

ππ

Solving Eqs. 1 and 3 simultaneously yields


Solving Eqs. 1 and 3 simultaneously yields

(Ans) kN 123 and kN 4.16 =−= alst FF

CHAPTER 2

AXIAL LOADING



Supplementary Problem 2

1. An aluminium tube is fastened between a steel rod and a bronze rod as shown. Axial

loads are applied at the positions indicated. Find the value of P that will not exceed a

maximum overall deformation of 2 mm or a stress in the steel of 140 MPa, in the

aluminium of 80 MPa, or in bronze of 120 MPa. Assume that the assembly is suitablyaluminium of 80 MPa, or in bronze of 120 MPa. Assume that the assembly is suitably

braced to prevent buckling.

D C1.2 m

Aluminium

A = 600 mm2

E = 70 GPa

Steel

A = 300 mm2

E = 200 GPa

Bronze

A = 450 mm2

E = 83 GPa

0.6 m 1.0 m 0.8 m

P 4P

2P3P


2. The assembly consists of three titanium (Ti-6A1-

4V) rods and a rigid bar AC. The cross-sectional

area of each rod is given in the figure. If a force

of 30 kN is applied to the ring F, determine the

horizontal displacement of point F.A

D C

B

E

1.2 m

0.3 m

ACD = 600 mm2

1.8 m

AAB = 900 mm2

F

0.6 m

0.3 m

AEF = 1200 mm2

CHAPTER 2

AXIAL LOADING




3. The composite bar in Fig. 2.3 is stress-free

before the axial loads P1 and P2 are

applied. Assuming that the walls are rigid,

500 mm 250 mm 350 mm

P1 P

2applied. Assuming that the walls are rigid,

calculate the stress in each material if P1 =

150 kN and P2 = 90 kN.Aluminium

A = 900 mm2

E = 70 GPa

Steel

A = 2000 mm2

E = 200 GPa

Bronze

A = 1200 mm2

E = 83 GPa

4. A rigid block of mass M is supported by

three symmetrically spaced rods as shown.

Each copper rod has an area of 900 mm2;


Each copper rod has an area of 900 mm2;

E = 120 GPa; and the allowable stress is 70

MPa. The steel rod has an area of 1200

mm2; E = 200 GPa; and the allowable

stress is 140 MPa. Determine the largest

mass M which can be supported.

CHAPTER 2

AXIAL LOADING



5. The assembly has the diameters and

material make-up indicated. If it fits securely

between its fixed supports when the

1.2 m 1.8 m 0.9 m

DA


between its fixed supports when the

temperature is T1 = 20°C, determine the

average normal stress in each material

when the temperature reaches T2 = 40°C.

6. The two circular rod segments, one of

aluminum and the other of copper, are fixed

to the rigid walls such that there is a gap of

0.2 mm between them when T = 15°C.

Stainless Steel

E = 200 GPa

Ø = 100 mm

Bronze

E = 83 GPa

Ø = 200 mm

D

Aluminium

E = 70 GPa

Ø = 300 mm

CB

A

Copper Aluminium

0.2 mm


0.2 mm between them when T1 = 15°C.

What larger temperature T2 is required in

order to just close the gap? Each rod has a

diameter of 30 mm . Determine the average

normal stress in each rod if T2 = 95°C.

Aluminium

E = 70 GPa

α = 24 (10-6)/oC

Copper Aluminium

100 mm 200 mm

Copper

E = 126 GPa

α =17(10-6)/oC

CHAPTER 2

AXIAL LOADING




7. Four steel bars jointly support a mass of 15

Mg as shown in Fig. 2.7. Each bar has a

60o

C DBA

cross-sectional area of 600 mm2. Find the

load carried by each bar after a

temperature rise of 50°C. Assume α = 11.7

µm/(m�°C) and E = 200 GPa.15

Mg

15o15o

60o


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