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10. ENERGY METHODS
DEFORMATION WORK
AND STRAIN ENERGY
10.1
(tvid- 10.1)
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Energy Methods/ Deformation Work and Strain Energy
k
k’
An object under the influence of external loads and in static balance does not move but changes shape. In this case, acting external loads (forces, moments) do deformation work.
Fk
F
10.1.1.b) If there is more than one force, how is the deformation work calculated?
We will try to understand the answers to these questions through 2 examples..>>
k
k’
.k’’
m
m: The final position of point k under the influence of all forces.
k'': projection of point m in the Fk direction
(a)
Figure 10.1.1
(b)
Figure 10.1.2
k: Application point of constant force Fk ,
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Energy Methods/ Deformation Work and Strain Energy
Ayşe comes first and sits at point c1 on an elastic beam.
Work done by weight WA :
Later, while Ayşe is sitting, Berrin comes and sits at point c2.
Work done by weight WB :
(Because Berrin hasn't settled down yet. There is no WB force at c2 at this moment.)
(Because when Berrin sat down, Ayşe was already sitting. WA force was present in C1 .)
From the instant Ayşe starts to sit, the force and deflection at point c1 increase linearly, and the force and deflection reach their maximum values at the last position of the sitting. For this reason, the F-ΔL diagram appears as follows. (This is a general rule. Try to notice this situation while sitting on a flexible floor or chair.)
Total Deformation Work:
before
Ayşe
Example-1: Deformation work when there are 2 forces in the same direction:
If Berrin had sat first,then Ayşe, or if they had both sat at the same time, the total Ud would not have changed.(Superposition method). This rule will be used in the future.
Berrin
Ayşe
after
WA
F
WA
F
WB
F
Figure 10.1.2
Figure 10.1.3
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Example -2 : Deformation when more than 2 forces are present in different directions
F
k
F
If first F,
then F1, and last F2
are applied to the object:
: The amount of deformation of k in the F direction due to the force F
: Additional deformation amount of k in the F direction due to other forces.
The deformation work of force F is:
The deformation work of force F1 is:
The deformation work of force F2 is:
(Contribution of F1 and F2)
(Contribution of F2)
If there were n forces:
Total Deformation Work:
+
(Contribution of F and F2)
(10.1.1.a)
(10.1.1.b)
(10.1.1.c)
(10.1.1.d)
10.1.2 Deformation Work of external moments:
(10.1.2.a)
Similar to equation 10.1.1.a, we first apply the Mk bending moment and then other loads. In this case, the deformation work of Mk is:
Deformation Work of External Torsional Moment for Circular Sections:
k
(10.1.2.b)
Energy Methods/ Deformation Work and Strain Energy
Figure 10.1.5
Figure 10.1.4
Figure 10.1.6
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1- Using strain energy density (ai) per unit volume or
2- Using internal force and internal moments.
10.1.3 Strain Energy (Ui) :
10.1.3.a- Calculation of Strain Energy Density (ai) and Total Deformation Energy (Ui)
The strain energy stored (absorbed) in unit volume is called Strain Energy Density and is denoted by ai.
We consider a cube element with edges of 1 unit (1 unit volume) and fixed at its back surface.
Internal normal force in x direction:
Strain in x direction :
From Hooke's equation for uniaxial loading:
or
(10.1.4.b)
(10.1.4.a)
Strain Energy (Ui) can be calculated by the following 2 methods:
(from equ. 10.1.1.a)
Tip 10.1.a: Total Strain Work (Ud) is stored in the object as strain energy (Ui).
(10.1.3)
If we squeeze an elastic ball, the total deformation work (Ud) of the squeezing forces is stored in the ball as strain energy (Ui), and this energy continues to remain on the ball as long as it remains in the squeezing position.
Now we will examine these 2 methods one by one:
Energy Methods/ Deformation Work and Strain Energy
Figure 10.1.7
Figure 10.1.8
(a)
(b)
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Hooke relation:
or
Or the same result can be obtained from the deformation work of a constant shear force Sxy.
Shear Force:
Elongation:
(from equ. 10.1.1.a)
or
or
By similar calculations
By similar calculations
or
or
(10.1.5.a)
(10.1.5.b)
(10.1.6.a)
(10.1.6.b)
(10.1.7.b)
(10.1.7.a)
(10.1.8.a)
(10.1.9.a)
(10.1.8.b)
(10.1.9.b)
25.03.2023
Energy Methods/ Deformation Work and Strain Energy
Figure 10.1.9
(a)
(b)
(c)
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(10.1.10)
(from equations 10.1.4.b,..10.1.9.b)
(10.1.11)
If we substitute the general Hooke relations into equation 10.1.11, :
(10.1.12.a)
Strain energy density in terms of Principal stresses
If we write equation 10.1.12.a in 1-2-3 principal axes:
Strain energy density in terms of Cartesian stress components
Strain energy density in terms of Principal strains
If we write equation 10.1.12.b in 1-2-3 principal axes:
(10.1.12.b)
(10.1.13.a)
Strain energy density in terms of Cartesian stress and strain components:
(10.1.13.b)
(shear strains are zero in the principal axes)
(shear stresses are zero on the principal axes)
If we obtain the stresses in terms of strains from the general Hooke relations and substitute them in equation 10.1.12.a:
Strain energy density in terms of Cartesian strains components
Question: The superposition method cannot be applied to equations (10.1.4.a-10.1.9.a) and they cannot be collected on top of each other. What do you think is the reason for this?
Total Strain Energy:
(10.1.14)
Energy Methods/ Deformation Work and Strain Energy
Reminder: General Hooke’s Laws (stress-strain relations in the elastic region)
Figure 10.1.10
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10.1.5 Calculation of the total strain energy (Ui) resulting from Section Effects (Internal forces and internal moments):
or from the strain energy density
We found the same result.
10.1.5.a-) Strain Energy due to Internal Normal Force (N):(UN)
Equ.10.1.1.a:
(equ. 3.3a )
(Since there is no other force)
From equ. 10.1.4.b :
;
;
(10.1.15.a)
(10.1.15.b)
For L length piece in stepped bars :
For single rod of length L :
Now we will obtain the strain energy from the internal force and internal moment types (section effects) occurring inside the object.
A normal internal force N occurs in the bar section subjected to the tensile force P.
For this loading type:
Total Volume:
Energy Methods/ Deformation Work and Strain Energy
Figure 10.1.11
Figure 10.1.12
Figure 10.1.14
(a)
(b)
(c)
Figure 10.1.13
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Internal moments Mz arising on the surfaces of the differential part of length 𝑑𝑠 removed from a beam subject to simple bending cause a convexity of 𝑑𝜃.
10.1.5.b-) Strain Energy (UM) due to Simple Internal Bending Moment (Mz)
Equ. (9.2.1) :
Strain energy in the beam due to internal bending moment Mz :>>
From geometry, for small deformations:
(10.1.16)
or we find from the energy density as follows:
From equa. 10.1.4.b:
Normal stress:
We got the same result
( equ. 5.1.1 )..>>
Energy Methods/ Deformation Work and Strain Energy
Figure 10.1.15
(a)
(b)
Figure 10.1.16
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10.1.5.c-) Strain Energy due to Internal Shear Force (V) in Simple Bending with Shear: Uv
We want to find the deformation work and therefore the deformation energy caused by the internal shear force of length dx of a beam subjected to simple bending with shear. (Actually, there are bending moments and normal stresses in the section, but we did not show them in the figure because we only examined the effect of shear force.)
Strain energy density due to 𝜏 stress is from equation (10.1.7.b):
Strain energy density due to shear force V:
Total strain energy due to shear force V:
Shear stress at all points on the 𝑎−𝑐 line:
(Static moment of area 𝐴′ )
Symmetrical beam section
(10.1.17)
Energy Methods/ Deformation Work and Strain Energy
Figure 10.1.17
(a)
(b)
(c)
(d)
,
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10.1.5.d-) - Strain Energy due to Internal Torsional Moment (T) for Circular Sections: UT
or
It is found from the energy density as follows:
Total Strain Energy:
Angle of twist formed by the effect of the torsional moment T:
(Equation: 4.1.5)
(Area under the curve)
Differential deformation work and strain energy resulting from Tint moment in the section of length dx:
Angle of twist for the section of length dx:
(10.1.18b)
Strain energy density from equation 10.1.7.b:
Shear stress distribution occurring in torsion:
(equa. 4.1.6)
Total Deformation Work:
We found the same result.
(Polar moment of inertia)
Valid for circular sections.
As a result of the torsional moment T applied to a built-in shaft, a torsional internal moment Tint = T occurs at any cross-section..
Energy Methods/ Deformation Work and Strain Energy
(10.1.18a)
Figure 10.1.18
Figure 10.1.19
Figure 10.1.20
Figure 10.1.21
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10.1.6 When all cross-sectional effects are present simultaneously, Total Strain Energy: (𝑈𝑖 )
For torsional moment T in circular cross-section members
for V shear force in simple bending with shear
For bending moment M in simple bending
For N normal forces
(tension or compression)
Equations (10.1.20) are valid for isotropic materials and in the elastic loading region.
Energy Methods/ Deformation Work and Strain Energy
y
x
a-) In Rods with Linear Axis
b-) In Rods with Curvilinear Axis
Since the position of any cross-section is expressed relative to the orbital coordinate s, equation 10.1.20.a will become:
Figure 10.1.22
Figure 10.1.23
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Example 10.1.1
3
4
5
y
x
60mm
cross-section of beam
y
z
200mm
Solution:
First of all, we must determine the section effects (internal forces and internal moments) in any section.
3
y
x
4
5
The + direction of the internal influences in the left part of cut I-I is selected in this way. (See: topic 5.2.3)
Strain energy due to tensile force N:
Strain energy due to shear force V:
Energy Methods/ Deformation Work and Strain Energy
From equation (10.1.15.b) :
From equation (10.1.17) :
Figure 10.1.24
Figure 10.1.25
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Strain energy due to bending moment
Total Strain Energy:
60mm
y
z
200mm
When all values are substituted in the above equation
Numeric values :
From equ.10.1.19 :
Energy Methods/ Deformation Work and Strain Energy
From equation (10.1.16)
found.
, ν =0.25