Axial LoadingPurposes
� We have seen normal stress and normal strain concepts in axially
loaded members,
� Here in this chapter, we will learn a method to calculate deformations in
these members,
� This method will be useful in finding internal forces and reactions when
equilibrium equations are not enough to determine reaction and internal
forces,
� Also, we will see how to calculate deformation due to thermal stress.
APPLICATIONS
Most concrete columns are reinforced with steel rods; and
these two materials work together in supporting the applied
load. Are both subjected to axial stress and how much?
Axial LoadingSaint-Venant Principle
� Notice that as the section is further
away from where the load is applied,
stress distributions, therefore also
deformations/strains, smoothens out,
and become almost uniform.
� In this case, average normal stress
formula becomes valid.
� This is called the Saint-Venant
principle.
Axial LoadSaint-Venant Principle
� The approximate distance where the uniform stresses occurs start around
the depth “d” where d is the largest dimension of the section.
d
d
Axial LoadingSaint-Venant Prensibi
� Saint Venant principle states that (Barre de Saint-Venant, 1855, French): If one
wants to analyze stress distribution in an element, it is not necessary to analyze
the stresses in the sections close to where the load is applied.
� It is sufficient to analyze stresses where uniform stress distributions take place
which occur at a certain distance (about distance d) from where the load is
applied.
� If one wants to analyze stresses at the vicinity of the load, then more
sophisticated theories, such as theory of elasticity, must be used.
Load Applied
Axial LoadingSaint-Venant Prensibi
STRAIN AND STRESS DISTRIBUTIONS:
SAINT-VENANT’S PRINCIPLE
• The smoothing out of the stress distribution is
an illustration of Saint-Venant’s principle.
Barre de Saint-Venant, a French engineer and
mathematician, observed that near loads, high
localized stresses may occur, but away from the
load at a distance equal to the width or depth of
the member, the localized effect disappears and
the value of the stress can be determined from
an elementary formula.
Such as:
Elastic Deformation in Axially Loaded Members
� Here we will develop a formula to calculate deformations in an axially loaded
member using stress-strain relationship and Hooke’s law.
� Let’s consider the axially loaded member with varying cross section:
� Note that there are concentrated forces at both ends of the bar, as well as
distributed axial load along the bar. Distributed axial force can be the self-weight
of the bar, or frictional forces.
� Based on the Saint-Venant principle let’s analyze a section where
smooth/average normal stress occurs. Let’s analyze a section with a length of dx
and a cross-sectional area of A(x):
� The force P(x) will be deforming the element of thickness of dx as shown with
the dotted line. The stress and the deformation of this element can be calculated
as follows:
( ) ve
( )
P x d
A x dx
δσ ε= =
Elastic Deformation in Axially Loaded Members
� If we assume that these values stay within the elastic limit/proportional limit, we
can combine these two quantities using the Hooke’s law:
( )=E
( )
( )
( )
E
P x d
A x dx
P x dxd
A x E
σ ε
δ
δ
=
=
� The total change in length of the bar can be calculated by integrating both sides:
0
( )
( )
LP x dx
A x Eδ = ∫
Elastic Deformation in Axially Loaded Members
Here δ is the relative displacement of one point on the bar to another point.
0
( )
( )
LP x dx
A x Eδ = ∫
� A special form of this formula can be found if the cross-sectional area and
axial load are assumed to be constant (non-varying).
PL
AEδ =
Elastic Deformation in Axially Loaded Members
� If the bar/element is subjected to different concentrated axial loading along the
bar then the expression for the axial deformation takes the form given below:
PL
AEδ =∑
Elastic Deformation in Axially Loaded Members
� In order to apply the formula given above, a positive sign convention must be
introduced: if the force elongates the bar it is assumed to be positive, and if the
force shortens the bar it is assumed to be negative.
PL
AEδ =∑
Elastic Deformation in Axially Loaded Members
PL
AEδ =∑
� Calculate the total axial deformation of the bar under the axial loading shown
above. Why do we have to use the deformation formula in the specific form
given below?
Axial Internal Force Diagram
Elastic Deformation in Axially Loaded Members
� A steel column is subjected to the axial forces shown below. The cross=sectional
areas of the different segments of the column are AAB = 1 in2 ve ABD = 2 in2. If the
elasticity modulus of the steel is E = 29(103) kip/in2 calculate total displacement
of point A on the column. Also, calculate the relative displacement of point B with
respect to point C (1 kip = 4448 kN ve 1 in = 2.54 cm and 1 ft = 30.48 cm).
Example - 1
Some of the relationships between the
SI and Anglo-saxon units:
• 1 in2 = 6.4516 cm2 (6.452×10-4 m2)
• 29×103 kip/in2 = ~200 GPa
• 1 ft = 12 in = 30.48 cm
Example – 1 (cont.’ed)
�Internal forces can be calculated by the section method:
Axial Internal ForceDiagram
Example – 1 (cont.’ed)
� By also taking into account the positive sign assumption, the total displacement
of point A can be calculated as
� Similarly, the displacement of point B relative to point C
can be calculated as
Here note that point B is moving away from point C, since
the result is positive.
B
C
Example - 2
Aluminum (area of 400 mm2)
Steel (diameter of 10 mm)
The assembly shown below consists of an aluminum tube AB havinga cross-sectional area of 400 mm2. A steel rod having a diameter of10 mm is attached to a rigid collar and passes through the tube. If atensile load of 80 kN is applied to the rod, determine thedisplacement of the end C of the rod. Take Est = 200 GPa, Eal = 70 GPa.
� Internal Force: With reference to the free-body-diagrams given below, internal
forces in the elements can be found as shown below
Example – 2 (cont.’ed)
� Notice that the tube is under compression and the bar is under tension loads.
� Displacements: Let’s first calculate the displacement of the end C with respect
to the end B (notice that the end B displaces as well)
Example – 2 (cont.’ed)
� Point B displaces with respect to point A which is fixed (it does not move, fixed to
a rigid support)
Positive sign indicates that C moves to the right with respect to B because the
bar elongates.
Example – 2 (cont.’ed)
� Since both points B and C displace to the right, the absolute displacement of
point C with respect to the fixed point A can be calculated as follows:
Example - 3
PRINCIPLE OF SUPERPOSITION
• It can be used to simplify problems having complicated loadings. This is
done by dividing the loading into components, then algebraically
adding the results (deformations).
• It is applicable provided that 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 (an
arbitrary point along the beam) is sum of two cases, δx = δx1 + δx2
• If the deformation is not small, then the following happens
• In this case, the superposition principle does not hold!
( ) 1 2P d Pd Pd≠ +
COMPATIBILITY CONDITIONS
• When the equilibrium equations alone cannot determine the
solution (internal forces/support reactions), the structural
member is called statically indeterminate.
• In this case, compatibility conditions at the constraint
locations shall be used to obtain the solution.
• For example, the stresses and elongations in the 3 steel wires
are different, but their displacement at the common joint A
must be the same.
Axially Loaded Statically Indeterminate Elements
� Consider the bar shown below where the bar is supported at both ends. In order
to find the support forces, equations of equilibrium are not enough.
0; 0B A
F F F P= + − =∑The reactions FB and FA can not be determined with only one equation.
Therefore the bar is said to be statically indeterminate.
(1)+
� The necessary extra equation can be written by taking into account the
deformations. This extra condition is called the compatibility or the kinematic
conditions.
� For this problem, the compatibility condition can be written as follows:
/ 0A B
δ =
This condition can be expressed using force-
displacement equation:
0A AC B BCF L F L
AE AE− =
If we take AE to be constant eq.’s (1) and (2) can
be solved together to calculate support reactions:
(2)
CB AC
A B
L LF P F P
L L= =
Since the results are found to be
positive, the assumed directions for
the support forces were correct.
A
B
Since ends A and B are
supported/fixed, they can not move
away from each other.
Axially Loaded Statically Indeterminate Elements
Force Method for Axially Loaded Members
Superposition Principle
0 = δP - δB (Compatibility Equation)
Equation of equilibrium applied along the vertical
direction:
If eq.’s (1) and (2) are solved together
FB can be found by substituting FA in equation (2).
Superposition
Example - 4
Örnek – 4 (devam)
Example – 4 (cont.’ed)
A
Example – 4 (cont.’ed)
Example – 4Hwk: Solve the
problem using the Force Method
Example - 5
Example – 5 (cont.’ed)
Free-body-diagram of the rigid bar
Example – 5 (cont.’ed)
Notice that the compatibility
condition is based on the
assumption that the bar is a
rigid bar.
(3)
Example – 5 (cont.’ed)
(3)
Notice that the necessary
third equation comes from
the compatibility condition!