Strength of Materials - Lecture Notes / Mehmet Zor

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DEFORMATION

3.1 General Information and the Concept of Strain( ε ),

3.2 Extension-Shortening Calculations in Rods

3.3 Hyperstatic Problems in Axial Loading

3.4 Thermal Loadings

3.5 General Hooke's Relations and Examples

3.

(tvid- 3a , tvid- 3b, tvid- 3c)

(tvid : turkish video)

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• In a solid system, not only the stresses but also the total deformations must not exceed a certain limit so that the functionality of the system besides its strength is preserved. For this reason, deformations are the most important comparison criterion after stresses in strength and must be taken into account at the design stage.

3 Deformation

• Total deformation calculations in a system are closely related to strain calculations and, like stresses, vary with the type of loading. Deformation calculations for different loadings such as bending, torsion and tensile are different from each other and each is a separate subject of strength

• In this section, the situations that cause axial loading (tensile-compressive loading) on the bars will be examined, and in this context, total elongation/shortening calculations in the elastic region, thermal loading, and hyperstatic situations will be explained.In the last part, Hooke's relations between stresses and strains, which are valid in the elastic region, will be explained and examples will be solved.

(Deformation calculations in the plastic region are more advanced topics and are generally taught at the graduate level.)

3.1 General Information

3.1.1 Importance of Deformation Calculations

Figure 3.1

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3.1.2 Concept of Strain (Unit Elongation) ( ε )

• The most important concept in deformation is the magnitude we call unit elongation,
• denoted by ε (read as « ipsilon »); It is also called strain.

• In its simplest definition, ε is the amount of extension (or shortening) of an object per unit length.

• The total elongation of a rod of length L₀ subjected to tension is ΔL , ..>> unit elongation:

• The index of ε changes depending on the direction in which the unit extension is ( could be ε _{x ,} ε _y or ε _z ).

3.1.3 Uniaxial Loading : If an object is loaded in only one direction (one of the x, y or z axes), we call this loading uniaxial loading.

​

• Loading rod-shaped elements only in the direction of their length (axis) is called axial loading. Since axial loading is in the direction of an axis, it is also uniaxial loading.

• This loading occurs in the form of tension or compression.
• The cross-sectional area or material of the rod may vary from region to region. These types of rods are called stepped rods.
• Now, we will learn to calculate the total amount of elongation or shortening in rod-shaped elements subjected to axial loading..>>.

(3.1)

3 Deformation

Figure 3.2

Figure 3.3

Figure 3.4

1.11.5 Tensile test and stress-strain diagram:

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that best characterizes the mechanical behavior of materials .

The part that connects to the chin

3 Deformation

Figure 3.5

Figure 3.6

Tensile test diagrams for different materials:

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Some Ductile Materials

A Brittle Material

Low

Carbon Steel

Aluminum Alloy

In brittle materials, the plastic zone is almost non-existent and is neglected.

3 Deformation

Figure 3.7

Figure 3.8

Figure 3.9

3.2.1 Calculation of elastic total elongation ( δ ) in rods under axial loading (Tension-Compression Situations) :

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From equation (3.1)

strain :

Let’s consider, an isotropic rod with cross section A subjected to axial tensile load . How much does this rod extend in the elastic region under the influence of the pulling force P ( δ =?)

Equations 3.3 give us the amount of elastic total elongation in a rod of constant cross section subjected to axial loading.

The trick is to know that the force in the equation is the internal force in the cut section, that is, P = F_internal = F_int.

This information will be much more useful in solving stepped bars.

(3.2a)

(3.3a)

(3.3b)

Or in more understandable terms:

(3.2b)

Stress:

Total elastic elongation

3 Deformation

Figure 3.10

Figure 3.11

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If at least one of the cross-sectional area (A), material (E) or axial force (P) values changes along the rod axis, it is called a stepped rod. The rod is divided into zones. Elongation amounts are calculated separately for each region and added together. The most important point here is to correctly determine the internal force in each region. This will be better understood with the examples to be solved.

3.2.2 Total elongation for stepped bars

If you cannot decide whether the internal force F_int in a region is tension or compression and you constantly confuse its sign, try this way: Always place the F_int value in the tensile direction on the separation surface of the cut you made. Calculate F_int from the static equilibrium of the right or left part of the cut and use the result exactly in the δ equation, along with its sign. This is a solution that will never let you down. This practical information was used in Example 3.1.

Note: We know from statics that when first placing a force, its direction can be chosen arbitrarily. If the sign turns out to be minus (-), it means that the direction we chose is opposite. However, it is used exactly in the equations with its sign without changing the direction. Or if we change its sign, we need to change the direction as well, but this is not preferred.

3.2.3 In order not to confuse the direction or sign of the internal force, a practical information:

(3.4)

3 Deformation

Figure 3.12

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Example 3.1. ⁽ⁿ⁾

Find the total deformation (total elongation amount) of the stepped steel bar in the figure for the given loading condition. Elasticity Modulus: E=200GPa

We divide the stepped bar into regions. When E, A or P changes, the region will change. In this case, a total of 3 regions will be formed as shown in the figure. Since it is the same material, E is the same in all regions.

Lengths of the regions: :

Cross-sectional areas of the regions:

Solution:

From Static Equilibrium, let's find the reaction force on wall A. :

3. Deformation / Total Elongation in Strain / Axial Loading

Figure 3.13

Figure 3.14

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We find the internal forces by cutting from each region. (We place each internal force as a tensile force to avoid confusing their signs.)

Remember: If we take the right part of the I-I segment for the 1st region, we still need to find the same result, which is proven below:

Region 1 (Section I-I, left part)

Region 3 (III-III section, right part)

Region 2 (right part of section II-II)

Although the internal force F was selected as tension, its sign was negative. We used it exactly as it is, with the sign, in the last equation without changing its direction.

(Section I-I right part)

Total Elongation:

3 Deformation / Total Elongation in Strain / Axial Loading

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The forces shown in the figure were applied through the symmetry axis of the stepped rod, consisting of steel and aluminum parts, fixed to wall A. Accordingly, calculate the value and direction of the displacement of end D?

E_steel =210GPa, E _{alum .} = 70GPa

Answer: -0.077mm

Example 3.2.

3. Deformation / Total Elongation in Strain / Axial Loading

Figure 3.15

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Example 3.3: Horizontal bars BC and DE, connected to the rigid bar AF with pins, are made of steel ( E_steel = 200GPa) and each has a width of 12 mm and a thickness of 6 mm. According to this; a-) Calculate the horizontal displacement of point A under the effect of the horizontal force P = 2.4 kN applied from point A.

Solution:

Since the horizontal rods receive force only from 2 points, they are two force members and carry force in the direction of their axes (otherwise they cannot be in equilibrium). ( To remember the concept of two force member, examine the topic of frames in statics course.)

3. Deformation / Hyperstatic Problems in Strain / Axial Loading

Direction selection of forces (reminder from statics): If the axis of a force is certain, its direction is chosen arbitrarily for the first time. When the same force is placed for the 2nd or 3rd time, its direction must be chosen depending on the first placement. For example: In element BC, we placed the F_CB force arbitrarily to the right at point C for the first time. Again, in the CB element, at point B, the F_CB force must be placed to the left (Because only then BC can remain in equlibrium). When placing the F_CB force at point C on the AF vertical bar, its direction must be to the left. (Because due to the action-reaction principle, AF and CB elements create opposing forces at the common contact point). When we calculate F_CB , if its sign is negative, it is interpreted as being opposite to the direction we chose, but this does not change the intensity of the force. It is used in other operations along with its sign without changing its direction. (See static lecture notes for more details.)

Figure 3.16

Figure 3.17

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. . . (1)

. . . (2)

. . . (3)

From the equilibrium of the AF vertical bar,

from similarity:

From equations (1) and (3):

From similarity:

Since AF is a rigid rod, its final state (A'F) remains linear.

3. Deformation / Hyperstatic Problems in Strain / Axial Loading

Example 3.4

AB and CD bars are connected to a ceiling at their upper ends, and to the BDE rigid element at their lower ends with pins. The cross-sectional area of the AB bar , which is aluminum ( E = 70 GPa ) , is 500 mm² . The cross-sectional area of the CD steel bar (E = 200 GPa ) is 600 mm² . Calculate the displacements occurring at points B, D and E with the effect of 30kN force.

Solution :

Free Body Diagram and Equilibrium Equations for rijid element BDE.

Static calculations:

(since its sign is negative, F _AB is in the opposite direction to the selected direction)

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AB and CD

are two force elements.

Continuation of the Solution..>>

3. Deformation / Hyperstatic Problems in Strain / Axial Loading

Figure 3.18

Figure 3.19

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Since point A is fixed, the shortening in rod AB is equal to the displacement of point B.

Since point C is fixed, the extension in the rod DC is equal to the displacement of point D.

Since it is rigid, the BDE bar remains linear after deformation. (B'D'E' is also linear.)

Triangle similarities

14

II-II cut

3. Deformation / Hyperstatic Problems in Strain / Axial Loading

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Systems in which unknown forces cannot be calculated only with static equilibrium equations are called statically indeterminate (hyperstatic) systems, and the problems in which these are examined are called hyperstatic problems. In hyperstatic systems, the number of unknown forces is greater than the number of static equilibrium equations, and additional equations are needed to find all forces. These additional equations are obtained from the deformations of the system. In this section, hyperstatic systems under axial loading (tension-compression situations) will be examined through examples. �(In future topics, hyperstatic problems related to torsion and bending will also be discussed)

3.3 Statically Indeterminate ( Hyperstatic ) Problems in Axial Loading

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3.2.4 Superposition Principle: In elastic loading, the sum of the effects of more than one load is equal to the sum of the effects when the loads are applied separately. This is called the principle of superposition .

The total deflection of point C on the beam seen above will be equal to the sum of the deflections when the forces are applied separately.

3. Deformation / Hyperstatic Problems in Strain / Axial Loading

Figure 3.21

Figure 3.22

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Strength of Materials - Lecture Notes / Mehmet Zor

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Example 3.5

Find the reaction forces at points A and B for the loading case and boundary conditions given in the stepped bar in the figure.

Solution : As a Static Equlibrium equation only;

• We remove the ground B and replace it with the force R_B .
• According to the superposition method, we can first apply forces of 300kN and 600 kN and then R_B,
• When there are 300kN and 600kN, the elongation is δL.
• When there is only R_B the shortening is δ_R
• As a result, the total elongation: δ_R + δ_L = 0

For calculations, we will first find the internal force in each region..>>

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3. Deformation / Hyperstatic Problems in Strain / Axial Loading

Figure 3.23

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1 st region( Bottom part)

2 nd region

3rd region

4th region

When wall B is removed and 300kN and 600kN are applied at the same time

3. Deformation / Hyperstatic Problems in Strain / Axial Loading

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From equation (1) :

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Only when R _B is applied

5th region

6th region

3. Deformation / Hyperstatic Problems in Strain / Axial Loading

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α : It is the amount of change in the length of a rod of 1 unit length when its temperature is changed by 1 °C.

3.4.1 Thermal Elongation and Thermal Expansion Coefficient ( α )

3.4 Thermal Loads:

• If there is no restriction in solid objects, expansion occurs as a result of temperature increase or contraction occurs as a result of temperature decrease.
• When the temperature of rods with at least one free end is increased, thermal extensions will occur in their length.
• If these thermal extensions are prevented, stresses also occur, which are called thermal stresses.
• Our aim now is to calculate the deformations and stresses that occur in the rod elements as a result of temperature changes (thermal loading).

(Unit: 1/ ^o C )

1 ⁰ C temperature increase

1 unit length

Elongation : α

Δ T ⁰ C temperature increase

Elongation : m=?

Δ T ⁰ C temperature increase

1 unit length

L length

(3.4)

3 Deformation / Thermal Loading

Δ T

Figure 3.24

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3.4.2 Thermal Stresses

Thermal stresses occur in objects whose thermal expansion is prevented. We increase the temperature of a rod placed between two fixed walls by DT. Since the extension of this rod is prevented, reaction forces P occur at both ends due to the walls. Due to these forces, compressive stresses occur in the rod. Because they occur due to temperature difference, these stresses are called thermal stresses. Now we will try to understand how these P forces and thermal stresses are calculated:

Δ T

Δ T

=

+

(Thermal Stress)

The total extension of the rod is zero. According to the principle of superposition , wall B is removed, allowing free extension. Then, the reaction force P occurring on wall B is applied and the rod shortens. The sum of this thermal extension ( δ_Τ ) and the shortening (δ_P) resulting from P becomes equal to zero (δ_Τ + δ_P = 0)

(3.5)

3 Deformation / Thermal Loading

Figure 3.25

(a)

(b)

(c)

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Example 3.6 ⁽ⁿ⁾

In the stepped rod in the figure, find a) the compressive forces and stresses occurring in the rods after the temperature increase of 110°C, b) the change in the length of the aluminum rod.

(1)

When the left wall is removed, the entire stepped bar lengthens by δ_T due to the temperature difference ΔT. It shortens by δ_P due to the force P appearing on the left wall. The result will be δ_T + δ_P = 0.6mm.

Solution:

Let's find the internal forces from sections I-I and II-II:

3 Deformation / Thermal Loading

From static equilibrium, Internal forces for both sections:

Alum.

Figure 3.26

Figure 3.27

(a)

(b)

(c)

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(2)

3 Deformation / Thermal Loading

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Sample Question 3.7 : The temperature of the stepped rod, consisting of cylindrical brass and aluminum parts, placed between the fixed walls A and C, is increased by 45 ^o C. Material properties are given in the table, and the diameters shown in the figure are d ₁ = 60 mm and d ₂ = 40 mm , respectively . According to this,

a-) Calculate the maximum stresses occurring in the brass and aluminum parts.

b-) What would the stresses be if there was a 0.5 mm gap between wall C and the aluminum?

3 Deformation / Thermal Loading

Figure 3.28

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How much can we heat the system in the figure without exceeding the safety limits?

Example (Question) 3.8*

(Answer: Δ T= 41 ^o C )

3.5 Hooke’s Laws: These are the relations between stress and strain in the elastic region. Hooke relations depend on the material properties we call elasticity modulus (E) and Poisson ratio ( ν ). We learned that in materials whose tensile diagram is linear, the slope of the elastic region is tan θ = E. Now we need to first understand the other material property, poisson ratio…>>

3 Deformation / Thermal Loading

Copper

Aluminum

Figure 3.29

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Here we can calculate 𝜀_𝑦 and 𝜀_𝑧 values with the Poisson ratio, which is a material property.

Poisson's ratio:

Attention : Formula (3.6 is valid only in case of x-axis (uniaxial) loading. If there were forces in the y or z directions in addition to P, this formula could not be used.

Hooke's equation for uniaxial loading .)

If the rod in the figure is subjected to P load in the x direction, only stress occurs in the x direction and no stress occurs in the y and z directions.

Realize it will happen.

,

,

(3.7ac)

Poisson's ratio is the material property that gives the deformation effect of a load in other directions.

3.5.1 Poisson Ratio ( ν ):

3 Deformation / Hooke’s Laws

(3.6)

Figure 3.30

(a)

(b)

(c)

(d)

3.5.2 Hooke's laws in general axial loading (Stress-Strain relations)

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The same terms in each loading are summed. Unit elongation values are obtained when all normal stresses exist at the same time.

Hooke's laws for axial loading

3 Deformation / Hooke’s Laws

Figure 3.31

(a)

(b)

(c)

(d)

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3.5.3 Important details and Hooke's laws in case of shearing:

G : shear ( stiffness ) modulus

Hooke's Equation in case of shearing in the xy plane :

1.Index (i) : Plane normal

2.Index (j) : Direction of stress

Let's remember the meaning of indices:

In the elastic region, there is a linear relationship between shear stress and shear strain angle in each plane, which is called Hooke's relation in shear. If we examine it for the x-y plane;

Similar relations can be written in other planes:

3 Deformation / Hooke's Laws

(3.8)

These stresses cause shear strain angles (γ_xy , γ_yz , γ_xz ).

Figure 3.32

(a)

(b)

(c)

3.5.4 To summarize: Hooke's relations in the most general case:

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Temperature effect:

(3.9e)

(3.9a)

(3.9b)

(3.9c)

(3.9d)

(3.9f)

General Hooke's Relations

( Valid in the elastic region for isotropic materials.)

3 Deformation / Hooke's Laws

Some conclusions to be obtained from Hooke's Equations:

• σ Normal stresses have no effect on γ deformations.
• Shear stresses (τ) have no effect on ε unit elongations.
• Δ T temperature change has no effect on γ deformations.

Figure 3.33

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3 Deformation / Hooke's Laws

Solution:

The pulling force P is balanced from the adhesion surface.

adhesion surface

Front View of Force Distribution

Stress and strain at a point

Figure 3.34

Example 3.10

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A circle with a diameter of d = 225 mm is drawn on the plate in the figure. Stresses of σ_x = 84 MPa and σ_z = 140 MPa occurred in the 18 mm thick plate due to the effect of plane forces. Since E = 70 GPa and ν = 1/3, find the changes in a) AB diameter, b) CD diameter, c) plate thickness.

Total elongations (Deformations);

Total elongation ( δ ) = unit elongation ( strain ε ) initial length in xo direction (L)

Solution:

3 Deformation / Hooke's Laws

Since AB .. is in the x direction

Since CD.. is in the z direction

Since the thickness t is in the y direction

We will reach the conclusion from the formula:

Figure 3.35

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(Since movement is limited in the y direction, no deformation occurs in that direction.

The initial and final lengths remain the same.)

Calculation of total elongations:

If we apply general Hooke relations to this system:

Solution:

3 Deformation / Hooke's Laws

(The movement of the object in the z direction is free. It changes shape freely upwards.

For this reason, no stress occurs. Don't forget! Stress occurs only if movement is blocked.)

Example 3.11 : A prismatic elastic element with dimensions of 10x20x30 mm ³ is placed in a fixed cavity, with its surfaces in contact. A compression force of F=-60kN, parallel to the x-axis, was applied to the outer surface of this element . Frictions can be neglected.

Accordingly, calculate the amount of change in the length of each side. E=100GPa, ν = 0.3

Figure 3.36

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Example (Question) 3.12 :

We press the pastry dough and shape it into a shape.

In this situation;

a-) In which of the x, y, z axes does stress occur?

b-) In which axes does deformation occur?

c-) Can stresses be calculated with F/A?

d-) Can deformations be calculated with Hooke's laws?

Discuss the reasons for the answers among yourselves.

Answers:

a-) z,

b-) x, y and z

c-) Yes,

d-) No

3. Deformation / Hooke's Laws

Figure 3.37

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Example 3.13

3. Deformation / Hooke's Laws

Figure 3.38

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3.6 Answered questions on the subject

3. Deformation / Questions with Answers

Question 3.14* In the figure, the stresses occurring in the plate are:

σ _x = 150 MPa , σ _z = 100 MPa .

Considering that E = 200 GPa and ν = 0.3 ;

Calculate the changes in a-) side length AB, b-) side length BC, c) length AC.

Answers: a-) 0.06mm, b-) 0.0206 mm, c-) 0.06mm

Figure 3.39

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Question 3.15*

Q.

3. Deformation / Questions with Answers

Figure 3.40

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The prismatic element with dimensions of 10x30x40cm3 was placed between the walls as shown in the figure and was loaded with two forces parallel to the x and y axes. As a result of this loading, the 10cm thickness parallel to the z axis increased to 10 + 4x10^-4 cm. The Poisson ratio of the element's material is known as ν = 0.2. According to this; a-) Calculate the modulus of elasticity, b-) Calculate the total elongation of the other edges.

Answers: a-) 10GPa, b-) -32x10 ^-3 mm, -24x10 ^-3 mm

Question 3.16

3. Deformation / Questions with Answers

Figure 3.41

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3. Deformation / Questions with Answers

Figure 3.42

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3. Deformation / Questions with Answers

Figure 3.43

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A brass plate was placed between two aluminum plates and glued, and the composite plate in the figure is obtained. Aluminum and Brass plates each have dimensions of 250mmx30mmx5mm. Calculate the normal stresses occurring in the aluminum and brass plates due to the axial compression force P = -30kN applied to the composite plate. E _{alum .} =70GPa, E _brass = 105GPa,

Answer : σ_{alum .} = -57.1 MPa , σ_brass = -85.7MPa

3. Deformation / Questions with Answers

Example (Question) 3.19*

Figure 3.44

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Example (Question) 3.20* The rigid bar AB is placed freely on the bars AC and BD, whose elastic modulus are 200 GPa and 100 GPa, respectively. A vertical force of 100 kN is applied to the rod AB from point E. Find how much point E will be displaced under the effect of this force.

Answer: 0.534mm

3. Deformation / Questions with Answers

Figure 3.45

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Example (Question) 3.21) The 25 mm diameter steel shaft is placed inside an aluminum tube. When a pressure force of P = 160kN is applied to this system a-) Calculate the stresses occurring in steel and aluminum.

b-) Calculate the amount of change in the length of each.

( E_steel = 200GPa, E_{alum .} = 70GPa)

Answer : a-) σ_steel = -116.3 MPa , σ_{alum .} = -40.7MPa, b-)-0.145mm

3. Deformation / Questions with Answers

Figure 3.46

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Example (Question) 3.22 The stepped bar, made of steel and brass, is placed between two fixed walls at both ends, subjected to axial forces of 60kN and 40kN at points B and D, respectively. Accordingly, find the maximum stresses and their locations in steel and brass.( E _steel = 200GPa and E _brass = 105GPa) (Answers: R _A =67.8kN, R _E =32.2kN , σ _AB = 53.98MPa, σ _DE = -52.65MPa)

3. Deformation / Questions with Answers

Figure 3.47

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A 25mm diameter brass shaft is placed inside a 60mm diameter aluminum tube. If the temperature of the system is increased from 15 ^o C to 195 ^o C, find the normal stresses in Aluminum and Brass.

Answer: σ _alum = -8.15 MPa , σ _brass = 38.82 MPa

3. Deformation / Questions with Answers

Example (Question) 3.23*

E = 105 GPa

α = 20.9 x 10 ^-6 / ^o C

σ _yield = 180 MPa

E = 70 GPa

α = 23.6 x 10 ^-6 / ^o C

σ _yield = 150 MPa

Brass:

Aluminum:

Figure 3.48

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3. Deformation / Questions with Answers

At the same time, the temperature of the system is increased. According to this; a-) How many degrees should the temperature of the system be increased to close the 0.3mm gap? ( ΔT = ? ) b-) In this case, calculate the changes in the diameters of the Copper and Aluminum parts.

The circular cross-section stepped bar in the figure consists of Copper and Aluminum parts.

A compression force of P = 2kN is applied from the free end A.

Example (Question) 3.24*

Aluminum:

E = 70 GPa

α = 23 x 10 ^-6 / ^o C

σ _safety = 50 MPa

τ _safety = 25 MPa

• = 0.27

G = 27GPa

​

Copper:

E = 100 GPa

α = 27 x 10 ^-6 / ^o C

σ _safety = 100 MPa

τ _safety = 50 MPa

• = 0.3

G = 38GPa

​

Figure 3.49

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Example (Question) 3.25*

3. Deformation / Questions with Answers

Figure 3.50