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 ... In this case, average normal stress ... Note that there are concentrated forces at both ends of the bar, as well as distributed axial ...
Axial Loading Purposes 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.
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 Loading Saint-Venant Principle Notice that as the section is further away from where the load is applied, stress
deformations/strains, smoothens out, and become almost uniform. In this case, average normal stress formula becomes valid. This
Axial Load Saint-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
Axial Loading Saint-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.
Axial Loading Saint-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.
Elastic Deformation in Axially Loaded Members Based
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:
P ( x) dδ ve ε = σ= A( x) dx
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:
σ = Eε P( x) dδ =E A( x) dx P ( x)dx dδ = A( x) E The total change in length of the bar can be calculated by integrating both sides: L
P ( x)dx δ =∫ 0 A( x ) E
Elastic Deformation in Axially Loaded Members L
P ( x)dx δ =∫ A( x) E 0 Here δ is the relative displacement of one point on the bar to another point. 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 δ= AE
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 δ =∑ AE
Elastic Deformation in Axially Loaded Members PL δ =∑ AE 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.
Elastic Deformation in Axially Loaded Members
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?
PL δ =∑ AE Axial Internal Force Diagram
Example - 1 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).
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 Force Diagram
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.
Example - 2 The assembly shown below consists of an aluminum tube AB having a cross-sectional area of 400 mm2. A steel rod having a diameter of 10 mm is attached to a rigid collar and passes through the tube. If a tensile load of 80 kN is applied to the rod, determine the displacement of the end C of the rod. Take Est = 200 GPa, Eal = 70 GPa.
Steel (diameter of 10 mm) Aluminum (area of 400 mm2)
Example – 2 (cont.’ed) Internal Force: With reference to the free-body-diagrams given below, internal forces in the elements can be found as shown below
Notice that the tube is under compression and the bar is under tension loads.
Example – 2 (cont.’ed) 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)
Positive sign indicates that C moves to the right with respect to B because the bar elongates. Point B displaces with respect to point A which is fixed (it does not move, fixed to a rigid support)
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
P ( d ) ≠ Pd1 + Pd 2 •
In this case, the superposition principle does not hold!
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.
∑ F = 0;
FB +FA − P = 0
The reactions FB and FA can not be determined with only one equation. Therefore the bar is said to be statically indeterminate.
Axially Loaded Statically Indeterminate Elements 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:
δ A/ B = 0 A
Since ends A and B are supported/fixed, they can not move away from each other.
This condition can be expressed using forcedisplacement equation:
FA LAC FB LBC − =0 AE AE B
If we take AE to be constant eq.’s (1) and (2) can be solved together to calculate support reactions:
LCB LAC FA = P FB = P L L
Since the results are found to be positive, the assumed directions for the support forces were correct.
Force Method for Axially Loaded Members Superposition Principle 0 = δP - δB
Equation of equilibrium applied along the vertical direction:
Superposition If eq.’s (1) and (2) are solved together
FB can be found by substituting FA in equation (2).
Example - 4
Örnek – 4 (devam)
Example – 4 (cont.’ed)
Example – 4 (cont.’ed)
Example – 4 Hwk: 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.
Example – 5 (cont.’ed) (3)
Notice that the necessary third equation comes from the compatibility condition!