Chapter four - Philadelphia University. axial... · Chapter four Laith Batarseh Axial load Home t...

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Chapter four Chapter four

Laith Batarseh Laith Batarseh

Axial load Axial load

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4.1. Saint-Venant’s principle 4.1. Saint-Venant’s principle

Assume that we apply axial load P to the gridded rectangular bar as

shown in the figure. Assume also that the bar is under elastic

deformation.

As you can see, the deformation shown by

the grid is not uniform. This means that the

stress distribution is non uniform because of

the relationship between the stress and strain

(Hook’s law).

We can study this irregular stress

distribution by sectioning the bar from

different locations as shown in the figure

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Let the section a-a is near to the point of applying load and as we go

further from the point of load application, we pass through sections b-b

and c-c.

Due to the shape of deformation, the stress distribution is represented

in the next figure



It is obvious from the figure that as we move further from the

application point of the load, the stress distribution become more

uniform. In fact, there is a common distance from the point of load

appliance. This distance should be at least equal the largest dimension in

the cross-section area. For our case, this distance is the width of the bar.

After this distance, the stress distrubtion becomes uniform and we can

find the average stress using σavg = P/A.

You can also observe the none uniform deformation in the supporting

point.

This manner of behavior is called Saint-Venant’s principle.

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4.2. elastic deformation of an axially loaded member 4.2. elastic deformation of an axially loaded member

Assume the following non uniform cross sectional member:

To find the stress, we take infinitesimal element dx



As mentioned before, the stress and strain are related to this element

are found by using the next equation

Hook’s law states : σ = E.ε. Therefore:

dx

dδε

xA

xP and

L

ExA

dxxP

0

dx

dE

xA

xP

ExAdxxP

d Rearrange the previous equation:

Integrate to find δ:

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According to the previous derivation, to find the total deformation in

axially loaded member, we use this equation

For a constant load (i.e. P(x) = P) and cross-section area (i.e. A(x) =A)

the deflection in the member can be found simply:

Where L is the total length of the member

L

ExA

dxxP

0

AE

PL



When the member is subjected to several different axial loads (P)

AE

PL

The final thing is the sign convention. We can assume:

If the load and the elongation is tensile then the sign is positive

for both P and δ.

If the load and the elongation is compersion then the sign is

negative for both P and δ.

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4.3. principle of superposition 4.3. 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 Hooke’s 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, δx = δx1 + δx2


4.4. statically indeterminate axially loaded members 4.4. statically indeterminate axially loaded members

In some cases as the one shown in the figure, the number reaction unknowns is larger than the number of equilibrium equations. These cases are called statically indeterminate

In such cases, other equations are required to solve the unknowns. This condition is called compatibility or kinematic condition.

For example, in the case shown in the figure, the equilibrium vertical forces equation is

As you can see: one equation, two unknowns

0 PFF AB

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4.4. statically indeterminate axially loaded members 4.4. statically indeterminate axially loaded members

to find other equation (i.e. compatibility condition), the total deflection is zero or: the deflection at A equal the –deflection at point B. mathematically,

The final equation is the 2nd equation to solve the two unknowns

AC

BCBA

BCBACABA

L

LFF

AE

LF

AE

LF

00

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4.5. statically indeterminate: force method 4.5. statically indeterminate: force method

It is another way to solve statically indeterminate problem

Consider the statically indeterminate shown in fig.(a), the displacement

caused by the force P at point B is δp so the reaction (FB) must cause a

displacement δB equal to δp :

δp - δB =0


4.5. statically indeterminate: force method 4.5. statically indeterminate: force method

Substitute the deflection formula

This is the 1st equation. The 2nd equation is derived from the

equilibrium condition

L

LPF

AE

LF

AE

PL ACB

BACBP 00

L

LPF

PFL

PLPFF

CBA

AAC

BA

00

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4.6. thermal stresses 4.6. thermal stresses

When the temperature of a body changed: incerase or decrease, the

body will expand or contract according the change (or difference) in the

temperature.

The relation between the thermal deflection(δT) and temperature

difference (∆T) is given by:

δT = α.∆T.L

Where:

• δT is the thermal deflection

•α is the thermal expansion coefficient

•∆T the temperature change

•L is the length of the member

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4.6. thermal stresses 4.6. thermal stresses

In the case of free members (i.e. one end is free to deflect), the thermal

deflection take place without generating stresses. These cases are the

statically determinate members.

If the member is restricted by supports (i.e. statically indeterminate),

the thermal deflection will cause thermal stresses which must be

consider in the design problems.

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4.7. stress concentration 4.7. stress concentration

It was mentioned previously that the stress may be concentrated in the

points of force application. At this point we can derive an equation to

relate the stress in these points to the applied load.

Assume the following member. The actual stress distribution shown in

Fig(b) is derived from the deflection of the grid shown in Fig(a) while the

stress distribution

Shown In Fig(c) is

averaged stress

distribution

to the actual one

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Apply force equilibrium condition

In engineering practice, there is a relation between the maximum

stress and averaged stress represented by The stress concentration

factor K

A

dAP .

avg

K

max



K is independent of the material properties K depends only on the specimen’s geometry and the type of

discontinuity

If we know K and the average stress is calculated as: P/A where

A is the area of the smallest cross section, then the maximum

stress (σmax) can be determined easily .

A

PKmax

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To find K, use the following charts


4.8. inelastic axial deformation 4.8. 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 elastoplastic or elastic-perfectly-plastic behavior.

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4.9. residual stress 4.9. residual stress

After an axially loaded member is stressed beyond yield stress, it will create residual stress in the member when the loads are removed. Consider the stress history of a prismatic member made from an elastoplastic material. Path OA: Member is loaded to reach yield stress σY Path AC: Member deforms plastically Path CD: Unloading but permanent strain ε0 remains

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