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!