CVEN214- Lecture 4 Axial Loads -Dr. Wael Alnahhal

7/25/2019 CVEN214- Lecture 4 Axial Loads -Dr. Wael Alnahhal

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COLLEGE OF ENGINEERING

DEPARTMENT OF CIVIL &ARCHITECTURALENGINEERING

CVEN 214: STRENGTH OF MATERIALS

WAEL I. ALNAHHAL, Ph. D., P. Eng

Spring, 2015

Chapter 4: AXIAL LOADS


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ELASTIC DEFORMATION OF AN AXIALLY

LOADED MEMBER

Provided these quantities do not exceed the proportional

limit, we can relate them using Hookes Law, i.e. = E

( )

( ) dx

d

xA

xP== and

( )( )

( )( )

( )( )

=

=

=

L

ExA

dxxP

ExA

dxxPd

dx

dE

xA

xP

0


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EXAMPLE 1

The assembly shown in Fig. 47a consists of an aluminumtubeAB having a cross-sectional area of 400 mm2. A steel

rod having a diameter of 10 mm is attached to a rigid collarand passes through the tube. If a tensile load of 80 kN isapplied to the rod, determine the displacement of the end Cof the rod. Take Est= 200 GPa, Eal= 70 GPa.


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EXAMPLE 1 (cont.)

Find the displacement of end C with respect to end B.

Displacement of end B with respect to the fixed endA,

Since both displacements are to the right,

Solutions

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

==

==

m001143.0001143.0107010400

4.0108096

3

AE

PLB

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

+=+

== m003056.010200005.0

6.010809

3

/

AE

PLBC

==+= mm20.4m0042.0/BCCC


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EXAMPLE 4.2


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EXAMPLE 4.2(CONTINUED)


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PRINCIPLE OF SUPERPOSITION

It can be used to simply problems having complicated

loadings. This is done by dividing the loading into

components, then algebraically adding the results.

It is applicable provided the material obeys Hookes Law

and the deformation is small.

If P = P1 + P2 and d d1 d2, then the deflection at

location x is sum of two cases, x= x1+ x2


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COMPATIBILITY CONDITIONS

When the force equilibrium condition alone cannot

determine the solution, the structural member is called

statically indeterminate.

In this case, compatibility conditions at the constraint

locations shall be used to obtain the solution.


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EXAMPLE 4.4


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EXAMPLE 4.4 (CONTINUED)


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EXAMPLE 4.5


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FORCE METHOD OF ANALYSIS

It is also possible to solve statically indeterminate problem

by writing the compatibility equation using the superposition

of the forces acting on the free body diagram.


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EXAMPLE 4.9

The A-36 steel rod shown in Fig. 417a has a diameter of 10

mm. It is fixed to the wall atA, and before it is loaded there is

a gap between the wall at B and the rod of 0.2 mm.

Determine the reactions atA and Neglect the size of the

collar at C. Take Est = 200GPa.


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EXAMPLE 4.9 (cont.)

Using the principle of superposition,

From Eq. 4-2,

Substituting into Eq. 1, we get

Solutions

( )[ ]( )

( ) ( )[ ] ( )( )

( ) ( )[ ] ( ) BBABBB

ACP

FF

AE

LF

AE

PL

9

92

3

92

3

103944.7610200005.0

2.1

105093.010200005.0

4.01020

===

===

( ) ( )10002.0 BP =+

( ) ( )( ) (Ans)kN05.41005.4

103944.76105093.00002.0

3

93

==

=

B

B

F

F


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EXAMPLE 4.9 (cont.)

From the free-body diagram,

Solutions

( )

(Ans)kN0.16

005.420

0

=

=+

=+

A

A

x

F

F

F


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THERMAL STRESS

Ordinarily, the expansion or contraction Tis linearly

related to the temperature increase or decreaseT thatoccurs.

If the change in temperature varies throughout the length of

the member, i.e.T =T (x), or if varies along the length,then

TLT =

= linear coefficient of thermal expansion, property of the material

= algebraic change in temperature of the member= original length of the member

= algebraic change in length of the member

TT

T

dxTT =


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EXAMPLE 4.10


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STRESS CONCENTRATION

The stress concentration factor Kis a ratio of the

maximum stress to the average stress acting at thesmallest cross section; i.e.

avg

K

max=


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STRESS CONCENTRATION (cont.)

K is independent of the material properties

K depends only on the specimens geometry and the typeof discontinuity


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INELASTIC AXIAL DEFORMATION

When a material is stressed beyond the elastic range, it

starts to yield and thereby causes permanent deformation.Among various inelastic behavior, the common cases

exhibit elastoplasticor elastic-perfectly-plastic behavior.


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RESIDUAL STRESS

After an axially loaded member is stressed beyond yield

stress, it will create residual stress in the member when theloads are removed.

Consider the stress history of a prismatic member made

from an elastoplastic material.

Path OA: Member is loaded to reach yield stress Y Path AC: Member deforms plastically

Path CD: Unloading but permanent strain 0 remains


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EXAMPLE 4-14

The rod shown in Fig. 430a has a radius of 5 mm and ismade of an elastic perfectly plastic material for which Y=

420 MPa and E = 70 GPa, Fig. 430c. If a force of P = 60 kNis applied to the rod and then removed, determine theresidual stress in the rod.


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EXAMPLE 3 (cont.)

An elastic analysis will produce FA= 45kN and FB= 15kN. This results in

a stress of

The maximum possible force developed inAC is

From the equilibrium of the rod,

Solutions

( ) ( )

( )

( )tensionMPa191

005.0

15

MPa420ncompressioMPa573005.0

45

2

2

==

=>==

CB

YAC

( ) ( ) ( ) kN0.33005.010420 23 === AF YYA

kN0.273360 ==BF


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EXAMPLE 3 (cont.)

The stress in each segment of the rod is therefore

Since CB responds elastically,

Solutions

( )

( ) ( ) (OK)MPa420tensionMPa344

005.0

27

ncompressioMPa420

2


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EXAMPLE 3 (cont.)

Here the yield strain is

The residual stress in each member is

This residual stress is the same for

both segments, which is to be

expected.

Solutions

( )( )

006.01070

104209

6

===E

YY

( )

( ) (Ans)MPa153191344

(Ans)MPa153573420

==

=+=

rCB

rAC


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EXAMPLE 4.14


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CVEN214- Lecture 4 Axial Loads -Dr. Wael Alnahhal

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