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Chapter four Chapter four
Laith Batarseh Laith Batarseh
Axial load Axial load
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Axial load Axial load
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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Axial load Axial load
4.1. Saint-Venant’s principle 4.1. Saint-Venant’s principle
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
Axial load Axial load
4.1. Saint-Venant’s principle 4.1. Saint-Venant’s principle
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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Axial load Axial load
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
Axial load Axial load
4.2. elastic deformation of an axially loaded member 4.2. elastic deformation of an axially loaded member
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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Axial load Axial load
4.2. elastic deformation of an axially loaded member 4.2. elastic deformation of an axially loaded member
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
Axial load Axial load
4.2. elastic deformation of an axially loaded member 4.2. elastic deformation of an axially loaded member
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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Axial load Axial load
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
Axial load Axial load
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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Axial load Axial load
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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Axial load Axial load
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
Axial load Axial load
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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Axial load Axial load
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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Axial load Axial load
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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Axial load Axial load
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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Axial load Axial load
4.7. stress concentration 4.7. stress concentration
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
Axial load Axial load
4.7. stress concentration 4.7. stress concentration
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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Axial load Axial load
4.7. stress concentration 4.7. stress concentration
To find K, use the following charts
Axial load Axial load
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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Axial load Axial load
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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