scoresofsteel.orgscoresofsteel.org/tatasteel/SB_Tata Steel_Mechanical... · Web viewMechanics of Solids Images.ppt, slide 121 Author Sudipa Sarkar Created Date 04/06/2018 02:06:00

Storyboard_V02
01
Welcome to the fourth module of Mechanics of Solids, in which you will understand about Stress and Strain.
Mechanics of Solids
2. Centre of Gravity and Moment of Inertia
3. Friction
7. Application: Columns and Struts
Show a classroom room with ‘Mechanical Curriculum’ written on the board/ wall/projector screen. Introduce a female character (Indian) who looks like a Professor. She is wearing a pair of spectacles. She will be talking and the text will appear inside speech blurb.
Text is already present on the projector screen. Animate the fourth module name to make it bigger or highlight the fourth module name with the images given here.
02
By the end of this module, you will be able to
· Define and calculate simple stress, different types of stresses and strain.
· Explain the relationship between stress and strain.
· Establish relationships among the elastic constants.
· Explain principal stresses and principal planes.
· Derive the expressions of normal stress and shear stress on an inclined plane in complex loaded element.
· Derive the expressions of principal stresses and principal planes.
· Derive the expressions of maximum shear stresses and their planes.
· Determine principal stresses and principal planes, and maximum shear stresses and their planes with the help of Mohr’s circle.
· Determine normal stress and shear stress on an inclined plane in complex loaded element with the help of Mohr’s circle.
· Define and derive thermal stresses.
· Analyse stresses by strain energy method for different load conditions.
Objective
· Define and calculate simple stress, different types of stresses and strain.
· Explain the relationship between stress and strain.
· Establish relationships among the elastic constants.
· Explain principal stresses and principal planes.
· Derive the expressions of normal stress and shear stress on an inclined plane in complex loaded element.
· Derive the expressions of principal stresses and principal planes.
· Derive the expressions of maximum shear stresses and their planes.
· Determine principal stresses and principal planes, and maximum shear stresses and their planes with the help of Mohr’s circle.
· Determine normal stress and shear stress on an inclined plane in complex loaded element with the help of Mohr’s circle.
· Define and derive thermal stresses.
· Analyse stresses by strain energy method for different load conditions.
Suitable animated image to depict Objective with bulleted text.
03
In this module, you will understand about simple stress and strain, principal stress and principal plane, thermal stress, and strain energy and impact loading.
Click each to learn about different types of stress and strain.
Introduction
Thermal Stress
Strain Energy and Impact Loading
Click each to learn about different types of stress and strain.
Show the image and animate the arrows with slight up and down continuous movement.
Make four clickable.
03.1.1
Let us get started with simple stress. Click the terms related to stress to know more about them.
· What is Stress
· Units of Stress
· Factor of Safety
Click the terms related to stress to know more about them.
When user clicks ‘Simple Stress and Strain’ sync text with VO.
Make four clickable.
PPT slide 2 & 3
Stress
When a member is subjected to loads, it develops resisting forces which results in the generation of stress.
To find the resisting forces developed, a section plane may be passed through the member and equilibrium of any one part may be considered. Each part is in equilibrium under the action of applied forces and internal resisting forces. The resisting forces may be conveniently split into normal and parallel to the section plane.
Stress can be normal or shear stress. Different types of normal and shear stresses are shown here.
The intensity of resisting force normal to the sectional plane is called intensity of normal stress. In practice, intensity of stress is called as ’stress’ only.
where, R is normal resisting force and
A is the area.
The resisting force parallel to the plane is called shear resistance. The intensity of resisting force parallel to the sectional plane is called shear stress (q).
where, Q is shear resistance and A is the area. Thus, stress at any point may be defined as resistance developed per unit area.
Stress
When a member is subjected to loads, it develops resisting forces which results in the generation of stress.
To find the resisting forces developed, a section plane may be passed through the member and equilibrium of any one part may be considered. Each part is in equilibrium under the action of applied forces and internal resisting forces. The resisting forces may be conveniently split into normal and parallel to the section plane.
Stress can be normal or shear stress. Different types of normal and shear stresses are shown here.
The intensity of resisting force normal to the sectional plane is called intensity of normal stress. In practice, intensity of stress is called as ’stress’ only.
where, R is normal resisting force and
A is the area.
The resisting force parallel to the plane is called shearing resistance. The intensity of resisting force parallel to the sectional plane is called shear stress (q).
where, Q is shear resistance and A is the area. Thus, stress at any point may be defined as resistance developed per unit area.
On clicking ‘Stress’, show the first image. The second and the third image will appear with the third paragraph and the fourth image will appear with the fourth paragraph.
In each case animate the arrows in their direction as up and down or back and forth movement.
In the second image, first ‘Stress’ will appear. Two lines will appear as two divisions followed by the text ‘Normal’ and ‘Shear’. Subsequently each division lines will appear followed by the text.
The first image will remain there throughout and the second and third image will disappear after the explanation.
03.1.1.2
PPT slide 4
Units of Stress
When Newton is taken as the unit of force and millimeter as unit of area, the unit of stress will be N/mm2.
The other derived units used in practice are kN/mm2, N/m2, kN/m2 or MN/m2. A stress of one N/m2 is known as Pascal and is represented by Pa.
1 MPa = 1 MN/ m2 = 1 × 106 N/(1000 mm)2 = 1 N/mm2
Unit of stress is used as MPa or N/mm2.
Units of Stress
When Newton is taken as the unit of force and millimeter as unit of area, the unit of stress will be N/mm2.
The other derived units used in practice are kN/mm2, N/m2, kN/m2 or MN/m2. A stress of one N/m2 is known as Pascal and is represented by Pa.
1 MPa = 1 MN/ m2 = 1 × 106 N/(1000 mm)2 = 1 N/mm2
Unit of stress is used as MPa or N/mm2.
On clicking ‘Units of Stress’, show text synced with VO.
03.1.1.3
Axial Stress (or Normal stress or Direct Stress)
A stress that tends to change the length of a body is called axial stress. 
Compressive stress is the axial stress which tries to shorten the length of the specimen.
Tensile stress is the axial stress which tries to elongate the length of the specimen.
Axial Stress (or Normal stress or Direct Stress)
A stress that tends to change the length of a body is called axial stress. 
Compressive stress is the axial stress which tries to shorten the length of the specimen.
Tensile stress is the axial stress which tries to elongate the length of the specimen.
On clicking ‘Axial Stress’ show the first and second image with the 1st paragraph. The first image disappears and the third image will appear next to the second image. The text for compressive and tensile stress can appear as pop-up text on the third image.
Sync text with VO.
PPT slide 6
Factor of Safety
Ultimate stress is the stress that a material can withstand while being stretched or pulled before breaking.
The maximum stress to which any member is designed is much less than the ultimate stress and this stress is called ‘working stress’. The ratio of ultimate stress to working stress is called factor of safety.
Factor of Safety =
Factor of safety for various materials depends upon their reliability. The following values are commonly taken in practice:
· For steel – 1.85
· For concrete – 3
Factor of Safety
Ultimate stress is the stress that a material can withstand while being stretched or pulled before breaking.
The maximum stress to which any member is designed is much less than the ultimate stress and this stress is called ‘working stress’. The ratio of ultimate stress to working stress is called factor of safety.
Factor of Safety =
Factor of safety for various materials depends up on their reliability. The following values are commonly taken in practice:
· For steel – 1.85
· For concrete – 3
· For timber – 4 to 6
On clicking ‘Factor of Safety’ show the image and text synced with VO.
03.1.2
Strain
Let us now understand about strain.The change in length per unit length is known as linear strain.
When changes are taking place in longitudinal direction, changes also take place in lateral direction. The nature of these changes in lateral direction are exactly opposite to that of changes in longitudinal direction i.e., if extension is taking place in longitudinal direction, the shortening of lateral dimension takes place and if shortening is taking place in longitudinal direction extension takes place in lateral directions.
The lateral strain may be defined as changes in the lateral dimension per unit lateral dimension.
Strain
The change in length per unit length is known as linear strain.
When changes are taking place in longitudinal direction, changes also take place in lateral direction. The nature of these changes in lateral direction are exactly opposite to that of changes in longitudinal direction i.e., if extension is taking place in longitudinal direction, the shortening of lateral dimension takes place and if shortening is taking place in longitudinal direction, extension takes place in lateral directions.
The lateral strain may be defined as changes in the lateral dimension per unit lateral dimension.
Show the first image. The second will appear when we are talking about shortening in the second paragraph followed by the third image. All the three images will remain on screen. Sync text with VO.
03.1.3
Stress Strain Relation: Hooke’s Law
It states that stress is directly proportional to strain within proportional limit.
p e
or p = E e, where ‘E’ is a proportionality constant, known as Young’s modulus or Modulus of elasticity.
Since elastic limit is very close to proportional limit, Hooke’s law is often stated to be valid up to elastic limit.
Stress Strain Relation: Hooke’s Law
It states that stress is directly proportional to strain within proportional limit.
p e
or p = E e, where ‘E’ is a proportionality constant, known as Young’s modulus or Modulus of elasticity.
Since elastic limit is very close to proportional limit, Hooke’s law is often stated to be valid up to elastic limit.
Show text and image synced with VO. The right hand side text part of the image should appear synced with VO.
03.1.4
Let us understand few situations related to the stress and strain phenomena such as:
· Axially loaded uniform cross-section bar
· Axially loaded stepped bar
Situations related to stress and strain phenomena:
· Axially loaded uniform cross-section bar
· Axially loaded stepped bar
The lady professor appears.
Stress p = P/A
Strain e =L
Axially loaded uniform cross-section bar
Consider the bars shown here.
Stress p = P/A
Strain e =L
From Hooke’s Law we have, modulus of elasticity
When user clicks ‘Axially loaded uniform cross-section bar’ show the image. Animate the arrows appearing next to P in their direction with continuous up and down movement. Sync text with VO.
03.1.4.2a
PPT slide 15 & 16
Axially loaded stepped bar
A typical bar with cross-sections varying in steps and subjected to axial load is as shown here.
The forces are acting on the cross-sections of the three portions. It is obvious that to maintain equilibrium the load acting on each portion is Ponly.
Hence, total change in length of the bar Δ = Δ1 + Δ2 + Δ3 =+
Click the example button to solve a problem.
Axially loaded stepped bar
The forces acting on the cross-sections of the three portions. It is obvious that to maintain equilibrium the load acting on each portion is Ponly.
Hence total change in length of the bar Δ = Δ1 + Δ2 + Δ3 =+
Problem
Click the example button to solve a problem.
On clicking ‘Axially loaded stepped bar’, the first image appears and stays there. Show the second image next to/below the first one.
Make a clickable button, ‘Problem’.
03.1.4.2b
PPT slide 17 & 18
The bar shown here is tested in universal testing machine. It is observed that at a load of 40 kN the total extension of the bar is 0.280 mm. Determine the Young’s modulus ofthe material.
Let us solve:
Problem
The bar shown here is tested in universal testing machine. It is observed that at a load of 40 kN the total extension of the bar is 0.280 mm. Determine the Young’s modulus ofthe material.
Let us solve:
On clicking ‘Problem’, show professor, saying ‘Let us solve’.
show image and text synced with VO. The text can appear on a board or a page of spiral bound notebook.
03.1.4.3
Bar with continuously varying cross-section
When the cross-section varies continuously, an elemental length of the bar should be considered and general expression for elongation of the elemental length derived. Then the general expression should be integrated over the entire length to get the total extension.
Let us find out the total extension in the case of:
· Prismatic bar
Click each to work out the derivation.
Bar with continuously varying cross-section
When the cross-section varies continuously, an elemental length of the bar should be considered and general expression for elongation of the elemental length derived. Then the general expression should be integrated over the entire length to get the total extension.
The total extension of:
Click each to work out the derivation.
When user clicks ‘Bar with continuously varying cross-section’ sync text with VO.
Make two clickable.
Prismatic bar
Consider an elemental length dx at a distance x from the larger end. The rate of change of breadth is (b1 - b2)/L.Hence, the width at section x is
where,
Extension of element = Pdx/AE
Prismatic bar
Consider an elemental length dx at a distance x from the larger end. The rate of change of breadth is (b1 - b2)/L.Hence, the width at section x is
where,
Extension of element = Pdx/AE
Total extension of the bar
When the user clicks ‘Prismatic bar’ show the images (both should remain on screen) and sync text with VO. For the first image, make the label appear one by one.
03.1.4.3.2
Uniformly varying circular section bar
A tapering rod has diameter d1 at one end and it tapers uniformly to a diameter d2 at the other end in a length L as shown here.
If modulus of elasticity of the material is E, let us find its change in length when subjected to an axial force P.
Change in diameter in length L is (d1 – d2)
since the rate of change of diameter, k = (d1 – d2) /L
Consider an elemental length of bar dx at a distance x from the larger end. The diameter of the bar at this section is d = (d1-kx).
Cross-sectional area A = d2/4
Extension of the entire bar
Nice to Know: For bar of uniform diameter, extension is PL/(d2E/4) and for tapering rod, it is PL/(d1.d2E/4).
Uniformly varying circular section bar
A tapering rod has diameter d1 at one end and it tapers uniformly to a diameter d2 at the other end in a length L as shown here.
If modulus of elasticity of the material is E, let us find its change in length when subjected to an axial force P.
Change in diameter in length L is (d1 – d2)
since the rate of change of diameter, k = (d1 – d2) /L
Consider an elemental length of bar dx at a distance x from the larger end. The diameter of the bar at this section is d = (d1-kx).
Cross-sectional area A = d2/4
Extension of the entire bar
Nice to Know
For bar of uniform diameter, extension is PL/(d2E/4) and for tapering rod, it is PL/(d1.d2E/4).
When user clicks ‘Uniformly varying circular section bar’,show the image and sync text with VO. Animate the arrows of the image in their direction.
Show a ‘Nice to Know’ button as clickable. On clicking the same, the text given below ‘Nice to Know’ appears
03.1.4.4
In the case of bar subjected to varying loads, you will look at three different types of bar which are:
· Stepped bar
· Solid conical bar
· Stepped bar
· Solid conical bar
Click each to learn more.
When user clicks ‘Bar subjected to varying loads’, show three clickable.
03.1.4.4.1
Since the algebraic sum of forces in X direction = 0
-P1+ P2 – P3 + P4 = 0 or P1+ P3= P2 + P4
Let us consider equilibrium of each part of stepped bar one by one, keeping only the left section force intact while achieving right section force by applying condition of equilibrium of the part.
To keep equilibrium of this part, let us apply another load P1 at section B.
Now consider the second part of stepped bar.
We had added force P1 at section B in first part. Let us add force P1 in the opposite direction at section B in the second part to nullify the effect of addition of this force in the first part.
To keep the second part in equilibrium, section C should have equal and opposite force acting on section B this part.
Let us consider the last part of stepped bar.
We had added force P1 – P2 at section C in second part. Let us add force P1 – P2 in opposite direction at section C in third part to nullify effect of addition of this force in second part.
To keep third part in equilibrium, section D should have equal and opposite force acting on section C this part.
You can verify the correctness of this process equating section D force in stepped bar to that in the third part and checking the condition of equilibrium of stepped bar.
Here, force P4 acts on section D of stepped bar. Thus, P1+P3-P2 = P4
or P1+P3 = P2 +P4 which is the condition of equilibrium of stepped bar as achieved in beginning.
With forces in each part known, you can easily find out stress and deformation of each part and movement of each section.
Stepped bar
Since the algebraic sum of forces in X direction = 0
-P1+ P2 – P3 + P4 = 0 or P1+ P3= P2 + P4
Let us consider equilibrium of each part of stepped bar one by one, keeping only the left section force intact while achieving right section force by applying condition of equilibrium of the part.
To keep equilibrium of this part, let us apply another load P1 at section B.
Now consider the second part of stepped bar.
We had added force P1 at section B in first part. Let us add force P1 in the opposite direction at section B in the second part to nullify the effect of addition of this force in the first part.
To keep the second part in equilibrium, section C should have equal and opposite force acting on section B this part.
Let us consider the last part of stepped bar.
We had added force P1 – P2 at section C in second part. Let us add force P1 – P2 in opposite direction at section C in third part to nullify effect of addition of this force in second part.
To keep third part in equilibrium, section D should have equal and opposite force acting on section C this part.
You can verify the correctness of this process equating section D force in stepped bar to that in the third part and checking the condition of equilibrium of stepped bar.
Here, force P4 acts on section D of stepped bar. Thus, P1+P3-P2 = P4
or P1+P3 = P2 +P4 which is the condition of equilibrium of stepped bar as achieved in beginning.
With forces in each part known, you can easily find out stress and deformation of each part and movement of each section.
When user clicks ‘Stepped bar’, show the first image. The image remains on screen. The next two images appear with each paragraph as shown. These two images disappear and then the next three images appear with each paragraph as shown. These three disappear and the last three appear with each paragraph.
03.1.4.4.2
A bar of uniform cross section A and length L is suspended from top. Find the expression for the extension of the bar due to self weight only if Young’s modulus is E and the unit weight of the material is .
Consider an element of length dx at a distance x from the free end. The load acting on this section is P = A.x.
Extension of elemental length = ==
Extension of bar =dx
Thus, this extension is half the extension of the bar if the load is equal to the self weight, applied at the end which would have been
Uniform section bar with self load
A bar of uniform cross section A and length L is suspended from top. Find the expression for the extension of the bar due to self weight only if Young’s modulus is E and the unit weight of the material is .
Consider an element of length dx at a distance x from the free end. The load acting on this section is P = A.x.
Extension of elemental length ===
Extension of bar =dx
Thus, this extension is half the extension of the bar if the load is equal to the self weight, applied at the end which would have been
When user clicks ‘Uniform section bar with self load’, show the first paragraph of the text as a problem statement. Show the first image. The label L and the double arrowed line next to it will appear synced with VO. Similarly, show dx and x, along with the line next to them synced with VO. Show the second and third image just after this. All the three images remain on screen.
03.1.4.4.3
Solid conical bar
A solid conical bar of uniformly varying diameter has diameter D at one end and zero at the other end. If length of the bar is L, modulus of elasticity E and unit weight , find the extension of the bar due to self weight only.
Consider an elemental length dx at a distance x from the free end. Let the diameter at this section be ‘d’.
Self weight acting on this element = times volume of portion below the section
= ) x
=
=
Solid conical bar
A solid conical bar of uniformly varying diameter has diameter D at one end and zero at the other end. If length of the bar is L, modulus of elasticity E and unit weight , find the extension of the bar due to self weight only.
Consider an elemental length dx at a distance x from the free end. Let the diameter at this section be ‘d’.
Self weight acting on this element = times volume of portion below the section
= ) x
=
=
When user clicks ‘Solid conical bar’, show the first paragraph of the text as a problem statement. show the first image and then the second image. In the second image, the label D, L, dx, x and d and the double arrowed line next to them will appear synced with VO. Both the images remain on screen.
03.1.5
Let us now understand some more concepts related to shear stress and shear strain such as:
· Shear Stress
· Simple Shear
· Shear Strain
Some more shear stress and shear strain phenomena:
· Shear Stress
· Simple Shear
· Shear Strain
Three clickable.
Rivet/bolt subjected to direct shear
Shear Stress
You can see a bar, subjected to direct shear force i.e., the force parallel to the cross-section of bar.
The section of a rivet/bolt subjected to direct shear is shown here.
Let Q be the shear force and q the shear stress acting on the section. Then, with usual assumptions that stresses are uniform we get,
For equilibrium,
Thus, the shear stress is equal to shear force per unit area.
Shear Stress
Let Q be the shear force and q the shear stress acting on the section. Then, with usual assumptions that stresses are uniform we get,
For equilibrium,
Thus, the shear stress is equal to shear force per unit area.
On clicking ‘Shear Stress’, the first two images appear one by one with the first paragraph. In the first image, the arrow with P appears first and then the arrow with Q. Similarly, for the second image Q appears followed by R.
The last three images appear one by one with the second paragraph. Q, R and the arrows will appear as explained.
03.1.5.2
Simple Shear
A material is said to be in a state of simple shear if it is subjected to only shearing stress.
Consider a bolt subjected to pure shear as shown here.
A rectangular element at this section is as shown in the figure.
Let the intensity of shear stress be qab and thickness of element be ‘t’. Consider the equilibrium of the element.
Vertical force on AB = qab × AB × t
This can be balanced by vertical downward force on CD.
qab × AB × t = qcd × CD × t
qab = qcd = q
q × AB × t × AD = q′ × AD × t × AB
We get q= q′.
Thus, if a section is subjected to pure shear, the state of stress in any element at that section is as explained here.
Consider a square element under simple shear. Since it is square element, AC = a√2.
Consider section along AC and let ‘p’ be the stress on this section. From equilibrium condition of system of forces in the direction normal to AC, we get
p × AC × t = q × CD × t cos 45° + q × AD × t sin 45°
pa √2t = qat 1/√2 + qat 1/√2= qat√2
p = q
Thus, simple shear gives rise to tensile and compressive stresses across planes inclined at 45° to the shearing planes, the intensity of direct stresses being of same magnitude as shearing stress.
Simple Shear
A material is said to be in a state of simple shear if it is subjected to only shearing stress.
Let the intensity of shear stress be qab and thickness of element be ‘t’. Consider the equilibrium of the element.
Vertical force on AB = qab × AB × t
This can be balanced by vertical downward force on CD.
qab × AB × t = qcd × CD × t
qab = qcd = q
q × AB × t × AD = q′ × AD × t × AB
We get q= q′.
Thus, if a section is subjected to pure shear, the state of stress in any element at that section is as explained here.
Consider a square element under simple shear. Since it is square element, AC = a√2.
Consider section along AC and let ‘p’ be the stress on this section. From equilibrium condition of system of forces in the direction normal to AC, we get
p × AC × t = q × CD × t cos 45° + q × AD × t sin 45°
pa √2t = qat 1/√2 + qat 1/√2= qat√2
p = q
Thus, simple shear gives rise to tensile and compressive stresses across planes inclined at 45° to the shearing planes, the intensity of direct stresses being of same magnitude as shearing stress.
On clicking ‘Simple Shear’, show images one by one, synced with the relevant paragraph text and VO. The 2nd and third images appear together with the third paragraph.
The first four images appear with the paragraphs and remain. These will disappear and the 5th image will appear. 6th, 7th and the 8th image appear after this along with relevant paragraph and these four images remains on screen together.
In each image arrows and labels will appear animated.
03.1.5.3
Shear Strain
Shear stress has a tendency to distort the element to position AB′C′D from the original position ABCD as shown in the figure here.
This deformation is expressed in terms of angular displacement and is called shear strain.
Shear strain = since angle is small)
Shear Strain
Shear stress has a tendency to distort the element to position AB′C′D from the original position ABCD as shown in the figure here.
This deformation is expressed in terms of angular displacement and is called shear strain.
Shear strain = since angle is small)
On clicking ‘Shearing Strain’, animate the first image. Initially show only the square ABCD. Show the blue lines with arrow. Then the dotted line appears with the angle and AB′C′D gets formed. Show the next two images after this.
Animate the last image where show the parallelepiped shape with height H, then Force Q with arrows is applied. The parallelepiped shifts with displacement D. After this show rest of the labels one by one.
03.1.6
Let us now learn how to find stresses and elongation in a composite or compound bar.
Bars made up of two or more materials are called composite or compound bars. They may have same length or different lengths. The ends of different materials of the bar are held together under loaded conditions.
Consider a member with two materials. Let the load shared by material 1 be P1 and that by material 2 be P2. Then
i. From equation of equilibrium of the forces, we get P = P1 + P2
ii. Since the ends are held securely, we get l1 = l2
where, l1 and l2 are the extension of the bars of material 1 and 2 respectively. Thus,
Using two relations for P1 and P2, their values can be found out. Then stresses in material 1 and 2 are p1 = P1 / (dand pPdSimilarly, extension of bar 1 and 2 is given byl1 = l2 = P1L1/A1E1
Stresses and Elongation in Composite or Compound Bars
Bars made up of two or more materials are called composite or compound bars. They may have same length or different lengths. The ends of different materials of the bar are held together under loaded conditions.
Consider a member with two materials. Let the load shared by material 1 be P1 and that by material 2 be P2. Then
i. From equation of equilibrium of the forces, we get P = P1 + P2
ii. Since the ends are held securely, we get l1 = l2
where, l1 and l2 are the extension of the bars of material 1 and 2 respectively. Thus,
Using two relations for P1 and P2, their values can be found out. Then stresses in material 1 and 2 are p1 = P1 / (dand pPdSimilarly, extension of bar 1 and 2 is given byl1 = l2 = P1L1/A1E1
Show the first image left side image and right side image appear one by one.
Show the second and third image along with the first image.
03.1.6
In order to understand stress, strain and volume change associated with a body subjected to forces acting in more than one direction, you must learn about Poisson’s Ratio and Volumetric Strain.
Click each to know more.
Stress, strain and volume change associated with a body subjected to forces acting in more than one direction
· Poisson’s Ratio
Show Professor character.
Make two clickable.
Poisson’s Ratio
When a material undergoes changes in length, it undergoes changes of opposite nature in lateral directions. For example, if a bar is subjected to direct tension in its axial direction it elongates and at the same time its sides contract.
If we define the ratio of change in axial direction to original length as linear strain and change in lateral direction to the original lateral dimension as lateral strain, it is found that within the elastic limit, there is a constant ratio between lateral strain and linear strain. This constant ratio is called
Poisson’s ratio. Thus,
Poisson’s ratio =
It is denoted by or For most of metals its value is between 0.25 and 0.33. Its value for steel is 0.3 and for concrete is 0.15.
Poisson’s Ratio
When a material undergoes changes in length, it undergoes changes of opposite nature in lateral directions. For example, if a bar is subjected to direct tension in its axial direction it elongates and at the same time its sides contract.
If we define the ratio of change in axial direction to original length as linear strain and change in lateral direction to the original lateral dimension as lateral strain, it is found that within the elastic limit, there is a constant ratio between lateral strain and linear strain. This constant ratio is called
Poisson’s ratio. Thus,
Poisson’s ratio =
It is denoted by or For most of metals its value is between 0.25 and 0.33. Its value for steel is 0.3 and for concrete is 0.15.
When user clicks ‘Poisson’s Ratio’ show the first image. Show the rectangle with continuous lines first. The arrows will appear on both the sides of the rectangle and then the dotted lines and L.
Show thesecond image with the second paragraph.
03.1.6.2
Volumetric Strain
When a member is subjected to stresses, it undergoes deformation in all directions. Hence, there will be change in volume. The ratio of the change in volume to original volume is called volumetric strain.
Thus, ev =
V = Original volume
It can be shown that volumetric strain is sum of strains in three mutually perpendicular directions.
ev= ex + ey + ez
For example, consider a bar of length L, breadth b and depth d as shown.
Now, V= Lbd
δV = δL bd + L δb d+ Lbδd
eV
ev= ex + ey + ez
Now, consider a circular rod of length L and diameter das shown here. Volume of the bar,
V = d2L
δV = 2dδd + d2δL (since V is a function of d and L)
ev= ex + ey + ez; since ey = ez =
In general for any shape, volumetric strain may be taken as sum of strains in three mutually perpendicular directions.
Volumetric Strain
When a member is subjected to stresses, it undergoes deformation in all directions. Hence, there will be change in volume. The ratio of the change in volume to original volume is called volumetric strain.
Thus, ev =
V = Original volume
It can be shown that volumetric strain is sum of strains in three mutually perpendicular directions.
ev= ex + ey + ez
For example, consider a bar of length L, breadth b and depth d as shown.
Now, V= Lbd
δV = δL bd + L δb d+ Lbδd
eV
ev= ex + ey + ez
Now, consider a circular rod of length L and diameter das shown here. Volume of the bar,
V = d2L
δV = 2dδd + d2δL (since V is a function of d and L)
ev= ex + ey + ez; since ey = ez =
In general for any shape, volumetric strain may be taken as sum of strains in three mutually perpendicular directions.
When user clicks ‘Volumetric Strain’ sync text with VO. The first image should appear along with the third paragraph as shown. This image will disappear when the second image appears with 5th paragraph. The last image appears with the last paragraph. In each case animate the labels and arrows.
03.1.7
Elastic constants are mechanical properties which help us to find out the deformation in a body subjected to forces.
Modulus of elasticity, modulus of rigidity and bulk modulus are the three elastic constants.
Click each to know more.
Elastic Constants
Elastic constants are mechanical properties which help us to find out the deformation in a body subjected to forces.
· Modulus of Elasticity
· Modulus of Rigidity
Make three clickable.
03.1.7.1
Modulus of elasticity (Young’s Modulus) ‘E’ has been already defined as the ratio of linear stress to linear strain within the elastic limit.
Modulus of Elasticity
Modulus of elasticity (Young’s Modulus) ‘E’ has been already defined as the ratio of linear stress to linear strain within the elastic limit.
When the user clicks ‘Modulus of Elasticity’ show text synced with VO.
03.1.7.2
Modulus of rigidity is defined as the ratio of shearing stress to shearing strain within the elastic limit and is usually denoted by letter G or N. Thus,
G =
q = Shearing stress
Modulus of Rigidity
Modulus of rigidity is defined as the ratio of shearing stress to shearing strain within the elastic limit and is usually denoted by letter G or N. Thus,
G =
q = Shearing stress
= Shearing strain
When the user clicks ‘Modulus of Rigidity’ show the first image as a cuboid first. The arrow with F appears and the cuboid shifts to the position as shown. Angle and x will appear alongside.
03.1.7.3
Bulk Modulus
When a body is subjected to identical stresses p in three mutually perpendicular directions as shown in the figure, the body undergoes uniform changes in three directions without undergoing distortion of shape.
The ratio of change in volume to original volume has been defined as volumetric strain (ev). The bulk modulus, K is defined as
where, p = identical pressure in three mutually perpendicular directions
ev = , Volumetric strain
ΔV = Change in volume
and V = Original volume
Thus, bulk modulus may be defined as the ratio of identical pressure ‘p’ acting in three mutually perpendicular directions to corresponding volumetric strain.
Bulk Modulus
When a body is subjected to identical stresses p in three mutually perpendicular directions, the body undergoes uniform changes in three directions without undergoing distortion of shape.
The ratio of change in volume to original volume has been defined as volumetric strain (ev). The bulk modulus, K is defined as
where, p = identical pressure in three mutually perpendicular directions
ev = , Volumetric strain
V = Original volume
Thus, bulk modulus may be defined as the ratio of identical pressure ‘p’ acting in three mutually perpendicular directions to corresponding volumetric strain.
When the user clicks ‘Bulk Modulus’ show the first cuboid. Show force P with arrows on the cuboid. The cuboid shrinks to become smaller in size and dotted lines are to be shown as the original cuboid (2nd image).
03.1.8
Now that you know about the elastic constants, you can derive the relationship among them. Let us look at the
· Relationship between Modulus of Elasticity and Modulus of Rigidity
· Relationship between Modulus of Elasticity and Bulk Modulus
· Relationship among E, G and K
Click each to determine the relationships.
Relationship of Elastic Constants
· Relationship between Modulus of Elasticity and Bulk Modulus
· Relationship among E, G and K
Click each to determine the relationships.
Make three clickable.
03.1.8.1
Consider a square element ABCD of sides ‘a’ subjected to pure shear ‘q’ as shown in the Figure. AEC′D shown is the deformed shape due to shear q. Drop perpendicular BF to diagonal DE. Let φ be the shear strain and G modulus of rigidity.
Now, strain in diagonal BD = = =
Since angle of deformation is very small we can assume BEF = 45°, hence EF = BE cos45°
Strain in diagonal BD= =
= , since =
Now, we know that the pure shear gives rise to axial tensile stress q in the diagonal direction of DB and axial compression q at right angles to it. These two stresses cause tensile strain along the diagonal DB.
Tensile strain along the diagonal DB = + μ = (1 + μ)
From these equations we get,
x = (1 + μ)
Relationship between Modulus of Elasticity and Modulus of Rigidity
Consider a square element ABCD of sides ‘a’ subjected to pure shear ‘q’ as shown in the Figure. AEC′D shown is the deformed shape due to shear q. Drop perpendicular BF to diagonal DE. Let φ be the shear strain and G modulus of rigidity.
Now, strain in diagonal BD = = =
Since angle of deformation is very small we can assume BEF = 45°, hence EF = BE cos45°
Strain in diagonal BD= =
= , since =
Now, we know that the pure shear gives rise to axial tensile stress q in the diagonal direction of DB and axial compression q at right angles to it. These two stresses cause tensile strain along the diagonal DB.
Tensile strain along the diagonal DB = + μ = (1 + μ)
From these equations we get,
x = (1 + μ)
E = 2G (1 + μ)
When user clicks, ‘Relationship between Modulus of Elasticity and Modulus of Rigidity’, animate the first image. Initially, show a square ABCD and a, q with their arrows. The dotted lines will appear with angle φ to form AEC′D and then the diagonal ED synced with VO. The second image appears next to the first one.
03.1.8.2
Cubic element subjected to stresses p in the three mutually perpendicular direction x, y and z
Relationship between Modulus of Elasticity and Bulk Modulus
Consider a cubic element subjected to stresses p in the three mutually perpendicular direction x, y and z as shown in the figure.
Now the stress p in x direction causes tensile strain in x direction while the stress p in y and in z direction cause compressive strains μ in x direction.
Hence, ex = - μ - μ = (1 - 2μ)
Similarly, ey = (1 - 2μ)
and ez= (1 - 2μ)
= (1 - 2μ)
K = =
Relationship between Modulus of Elasticity and Bulk Modulus
The stress p in x direction causes tensile strain in x direction while the stress p in y and in z direction cause compressive strains μ in x direction.
Hence, ex = - μ - μ = (1 - 2μ)
Similarly, ey = (1 - 2μ)
and ez= (1 - 2μ)
= (1 - 2μ)
K = =
or E = 3K (1 – 2μ)
When user clicks, ‘Relationship between Modulus of Elasticity and Bulk Modulus’, show the first image. At first, show only the cube then all the ‘P’s with animated arrows. Subsequently, all the ‘p’s appear. ex, ey, ez will appear synced with text and VO.
03.1.8.3
We know, E = 2G (1 + μ) …………… (a)
E = 3K (1 – 2μ) ………….. (b)
By eliminating μ between these two equations, we can get the relationship between E, G and K, free from the term μ.
From equation (a), μ = – 1
Substituting it in equation (b), we get
E = 3K (1 – 2 ( – 1))
= 3K (1 - + 2) = 3K (3 - )
= 9K -
=
We know, E = 2G (1 + μ) …………… (a)
E = 3K (1 – 2μ) ………….. (b)
By eliminating μ between these two equations, we can get the relationship between E, G and K, free from the term μ.
From equation (a), μ = – 1
Substituting it in equation (b), we get
E = 3K (1 – 2 ( – 1))
= 3K (1 - + 2) = 3K (3 - )
= 9K -
=
or =
When user clicks, ‘Relationship among E, G and K’, sync text with VO.
03.2.1
You have understood about simple stress and it is clear that a structural member need not be always under simple stress. It may be subjected to direct stresses in different directions and may be subjected to shear stresses also.
A beam is usually subjected to axial stresses due to bending moment and shear stresses due to shear force.
A boiler is subjected to longitudinal and hoop stresses due to the pressure inside it. The stresses are mutually perpendicular to each other.
In this section of the module, our discussion is limited to two-dimensional stress system. First, general expression for stresses on a plane inclined at a selected axis is discussed. Then the terms principal planes and stresses are explained and the expressions to get them are presented.
Click each concept to learn about them.
Principal Stress and Principal Plane
Structural member need not be always under simple stress. It may be subjected to direct stresses in different directions and may be subjected to shear stresses also.
A beam is usually subjected to axial stresses due to bending moment and shear stresses due to shear force.
A boiler is subjected to longitudinal and hoop stresses due to the pressure inside it. The stresses are mutually perpendicular to each other.
In this section of the module, our discussion is limited to two-dimensional stress system.
· Stresses in inclined planes
Click each concept to learn about them.
When user clicks ‘Principal Stress and Principal Plane’, show text, synced with VO. Show images synced with the relevant paragraph.
Make four clickable.
PPT slide 4-7
Stresses in inclined planes
Consider the element in a structural member which is under a general two-dimensional state of stress as shown in the the figure.
Note the sign convention:
+ve face +ve direction is +ve stress for px and py. In other words, tensile stresses are +ve. Shearing stress as shown in the figure is +ve stress.
We are interested in finding the state of stress on plane DE which makes anticlockwise angle θ with the plane of stress px, in other words with y axis.
For simplicity, let us consider thickness of the element as unity. We are interested in finding normal and tangential stresses acting on the plane DE.
Let normal stress be pn and tangential stress pt as shown in the second figure. Since the system is in equilibrium,
Σ Forces normal to DE = 0 gives
pn× DE × 1 = px × CD × 1 × cos θ + q × CD × 1 × sin θ + py × CE × 1 × sin θ + q × CE × 1 × cos θ
∴pn = pxcos θ + py sin θ + q sin θ + q cos θ
Since, = cos θ and = sin θ, we get
pn = px cos2 θ + pYsin2 θ + q cos θ sin θ + q sin θ cos θ
= px+ py + 2q sin θcos θ
Since,θ= and θ=
Thus, …. (4.1)
Similarly, from the equilibrium condition of forces tangential to plane DE, we get
pt x DE x 1 = px × CD × 1 × sin θ – q × CD × 1 × cos θ – py × CE × 1× cos θ + q × CE × 1× sinθ
pt= px sin θ – py cos θ - q cos θ + q sin θ
But = cos θ and = sin θ
pt= px sin θ cos θ – py sin θ cos θ - qcos2θ+ q sin2θ
= (px– py) sin θ cos θ -q (cos2θ- sin2θ)
Since 2 sin cos = sin 2
and cos2θ- sin2θ = cos 2
Thus,
Thus, in case of a general two-dimension stress system, the normal and tangential stresses acting on a plane making anticlockwise angle θ with the plane of px (y direction) are as given by equations 4.1 and 4.2.
Stresses in inclined planes
Consider the element in a structural member which is under a general two-dimensional state of stress.
Note the sign convention:
+ve face +ve direction is +ve stress for px and py. In other words, tensile stresses are +ve. Shearing stress as shown in the figure is +ve stress.
We are interested in finding the state of stress on plane DE which makes anticlockwise angle θ with the plane of stress px, in other words with y axis.
For simplicity, let us consider thickness of the element as unity. We are interested in finding normal and tangential stresses acting on the plane DE.
Let normal stress be pn and tangential stress pt as shown in the second figure. Since the system is in equilibrium,
Σ Forces normal to DE = 0 gives
pn× DE × 1 = px × CD × 1 × cos θ + q × CD × 1 × sin θ + py × CE × 1 × sin θ + q × CE × 1 × cos θ
∴pn = pxcos θ + py sin θ + q sin θ + q cos θ
Since, = cos θ and = sin θ, we get
pn = px cos2 θ + pYsin2 θ + q cos θ sin θ + q sin θ cos θ
= px+ py + 2q sin θcos θ
Since,θ= and θ=
Thus, …. (4.1)
Similarly, from the equilibrium condition of forces tangential to plane DE, we get
pt x DE x1 = px × CD × 1 × sin θ – q × CD × 1 × cos θ – py × CE × 1 × cos θ + q × CE × 1 × sinθ
pt= px sin θ – py cos θ - q cos θ + q sin θ
But = cos θ and = sin θ
pt= px sin θ cos θ – py sin θ cos θ -qcos2θ+ q sin2θ
= (px– py) sin θ cos θ -q (cos2θ- sin2θ)
Since 2 sin cos = sin 2
and cos2θ- sin2θ = cos 2
Thus,
Thus, in case of a general two-dimension stress system, the normal and tangential stresses acting on a plane making anticlockwise angle θ with the plane of px (y direction) are as given by equations 4.1 and 4.2.
On clicking ‘Stresses in inclined plane’, show the images with relevant paragraph. Animate the images to form gradually. For the second image sync the appearance of pn and pt with text and VO.
03.2.1.1b
Angle of Obliquity of Resultant Stress
The resultant of the stresses on the given inclined plane can be found as
which is inclined at to the plane.
tan pn / pt
In other words, the resultant is inclined at to the plane of px.
Angle of Obliquity of Resultant Stress
The resultant of the stresses on the given inclined plane can be found as
which is inclined at to the plane.
tan pn / pt
In other words, the resultant is inclined at to the plane of px.
Continue the second image from previous page (with only the parts as shown here). The part with green lines will appear animated.
03.2.1.2
PPT slide 8
The planes on which shearing stresses are zero are called principal planes.
To locate principal plane pt=0, from equation 4.2,
There are two values for 2θ which differ by 180° for which equation 4.3can be satisfied.
Let 2θ1 and 2θ2 be the solution. Referring to the figure, we find
Similarly,
2θ1 and 2θ2 differ by 180°. Hence, we can say θ1 and θ2 differ by 90°. Thus, the directions of principal planes to the plane of pnare given by equation 4.4. Another principal plane is at right angles to it.
Principal plane
The planes on which shearing stresses are zero are called principal planes.
To locate principal plane pt=0, from equation 4.2,
There are two values for 2θ which differ by 180° for which equation 4.3 can be satisfied.
Let 2θ1 and 2θ2 be the solution. Referring to the second figure, we find
Similarly,
2θ1 and 2θ2 differ by 180°. Hence, we can say θ1 and θ2 differ by 90°. Thus, directions of principal planes to the plane of pn are given by equation 4.4. Another principal plane is at right angles to it.
On clicking ‘Principal plane’, show images according to the paragraphs and text synced with VO. Animate the labels and arrows.
03.2.1.3
Principal stresses
Principal stresses are the normal stresses on principal planes. Hence, the values of principal stresses may be obtained by substituting θ1 and θ2 values for θ in the expression for pn.
Denoting the values as p1 and p2, we get, p1 = pn at θ = θ1
and, p2 = pn at θ = θ2
Thus, the principal stresses are given by
It can be proved that principal stresses are maximum and minimum stresses also. To find extreme value of normal stress pn,
Thus, the principal planes are the planes of maximum/minimum normal stresses also. Plane corresponding to θ1 gives the maximum value while plane corresponding to θ2 gives minimum normal stress.
Principal stresses
Principal stresses are the normal stresses on principal planes. Hence, the values of principal stresses may be obtained by substituting θ1 and θ2 values for θ in the expression for pn.
Denoting the values as p1 and p2, we get, p1 = pn at θ = θ1
and, p2 = pn at θ = θ2
Thus, the principal stresses are given by
It can be proved that principal stresses are maximum and minimum stresses also. To find extreme value of normal stress pn,
Thus, the principal planes are the planes of maximum/minimum normal stresses also. Plane corresponding to θ1 gives the maximum value while plane corresponding to θ2 gives minimum normal stress.
On clicking ‘Principal stresses’, show the image and sync text with VO.
03.2.1.4
Plane for maximum shear stress
For maximum shear
tan 2θ × tan 2θ′ = –1
Hence, 2θ′ and 2θ values differ by 90°. In other word planes of extreme shearing stresses are at 45° to the principal planes.
Value of maximum shearing stress may be obtained by substituting θ′ for θ in equation for pt (equation4.2).
Now,
For maximum shear
tan 2θ × tan 2θ′ = –1
Hence, 2θ′ and 2θ values differ by 90°. In other word planes of extreme shearing stresses are at 45° to the principal planes.
Value of maximum shearing stress may be obtained by substituting θ′ for θ in equation for pt (equation 4.2).
Now,
From the figure, we get
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03.2.2
PPT slide 18
The direction of principal planes and the planes of maximum shearing stresses may be indicated as shown in the figures.
The direction of principal planes and the planes of maximum shearing stresses
Show the images. Animate them by showing arrow movement and making labels and angles appear one by one.
03.2.3
So far, you have learnt analytical method of finding principal stresses and principle plain as well as normal stress and shear stress on an inclined plain.
Mohr’s Circle is the graphical method to find principal stresses, principal planes, maximum shear stresses and their planes in a complex stress system. It can also give normal stress and shear stress on an inclined plane within a member loaded with complex stress system.
Let us learn how to draw a Mohr’s Circle.
1. If value of px and py both are positive, choose centre point ‘O’ at left end of paper. If one of the values of normal stress is negative (means it is compressive stress), choose centre point at middle of paper. Draw a horizontal line from point ‘O’.
2. Select a proper scale and mark points ‘A’ and ‘B’ on horizontal line from ‘O’ keeping their distance equal to px and py respectively. Then, OA = px and OB = py.
3. Mark centre of Mohr’s circle ‘C’ at midpoint of AB.
4. From ‘A’ and ‘B’, draw lines AD and BE perpendicular to line OA such that AD = BE = q. Meet points DE intersecting at centre ‘C’.
5. With centre ‘C’ and radius CD (or CE), draw a circle. This is Mohr’s circle.
6. Draw perpendicular lines CF and CI on horizontal line from ‘O’ from centre ‘C’.
Mohr’s Circle
Mohr’s Circle is the graphical method to find principal stresses, principal planes, maximum shear stresses and their planes in a complex stress system. It can also give normal stress and shear stress on an inclined plane within a member loaded with complex stress system.
Steps to draw a Mohr’s Circle:
1. If value of px and py both are positive, choose centre point ‘O’ at left end of paper. If one of the values of normal stress is negative (means it is compressive stress), choose centre point at middle of paper. Draw a horizontal line from point ‘O’.
2. Select a proper scale and mark points ‘A’ and ‘B’ on horizontal line from ‘O’ keeping their distance equal to px and py respectively. Then, OA = px and OB = py.
3. Mark centre of Mohr’s circle ‘C’ at midpoint of AB.
4. From ‘A’ and ‘B’, draw lines AD and BE perpendicular to line OA such that AD = BE = q. Meet points DE intersecting at centre ‘C’.
5. With centre ‘C’ and radius CD (or CE), draw a circle. This is Mohr’s circle.
6. Draw perpendicular lines CF and CI on horizontal line from ‘O’ from centre ‘C’.
Show the first image. The square appears first followed by the arrows along with labels.
The second image should be animated as the steps mentioned.
03.2.4
From Mohr’s circle you can find Principal stresses, principal planes, maximum shear stresses and planes of maximum shear stresses
Mohr’s circle intersects horizontal line from ‘O’ at point G and H such that:
· Major principal stress, p1 = OG
· Minor principal stress, p2 = OH
· Inclination of Major principal stress from a vertical plane in complex stressed system = θ1 = ACD
· Inclination of Minor principal stress from a vertical plane in complex stressed system = θ2 = DCH = 90o + θ1 = 90o + ACD
· Maximum shear stress, qmax= CF = CI
Finding Principal stresses, principal planes, maximum shear stresses and planes of maximum shear stresses
Mohr’s circle intersects horizontal line from ‘O’ at point G and H such that:
· Major principal stress, p1 = OG
· Minor principal stress, p2 = OH
· Inclination of Major principal stress from a vertical plane in complex stressed system = θ1 = ACD
· Inclination of Minor principal stress from a vertical plane in complex stressed system = θ2 = DCH = 90o + θ1 = 90o + ACD
· Maximum shear stress, qmax= CF = CI
Continue the earlier image. Draw the lines, angles and labels as mentioned in the text.
03.2.5
From Mohr’s circle you can also find normal stress, shear stress and resultant stress on an inclined plane within complex stress system
Let inclined plane make an angle θ with vertical.
1. Draw radius CJ at an angle 2θ with radius CD such that obtuse angle
DCJ = 2θ.
2. From point J, draw perpendicular on horizontal line from O intersecting it at K.
3. By measurement, find normal stress on inclined plane, pn = OK and shear stress on inclined plane, qt = BJ
Nice to Know:
Mohr’s circle gives approximate result if scale is used to find the value of stresses. However, the exact value of stresses can also be found out using trigonometry for getting the length of lines and angles involved.
Finding normal stress, shear stress and resultant stress on an inclined plane within complex stress system
Let inclined plane make an angle θ with vertical.
1. Draw radius CJ at an angle 2θ with radius CD such that obtuse angle
DCJ = 2θ.
2. From point J, draw perpendicular on horizontal line from O intersecting it at K.
3. By measurement, find normal stress on inclined plane, pn = OK and shear stress on inclined plane, qt = BJ
Nice to Know
Mohr’s circle gives approximate result if scale is used to find the value of stresses. However, the exact value of stresses can also be found out using trigonometry for getting the length of lines and angles involved.
Show the first image and then draw the angle θ with vertical as shown in the second image. Simultaneously, show the third image a and draw the lines angles and labels as mentioned in text to finally make it look like the fourth image.
03.3.1
Stresses are also induced without the application of load, for example thermal stress.
When a body is heated or cooled and its deformation is restrained, thermal stresses are induced in the body.
Let us understand how to determine thermal stresses in a body.
Thermal Stress
Stresses are also induced without the application of load, for example thermal stress.
When a body is heated or cooled and its deformation is restrained, thermal stresses are induced in the body.
When user clicks ‘Thermal Stress’, show Professor explaining. Sync text with VO.
03.3.2
Every material expands when temperature rises and contracts when temperature falls. It is established experimentally that the change in length is directly proportional to the length of the member L and change in temperature t. Thus,
tL= α tL
The constant of proportionality α is called coefficient of thermal expansion and is defined as change in unit length of material due to unit change in temperature.
Table shows coefficient of thermal expansion for some of the commonly used engineering materials.
Thermal Stresses
Every material expands when temperature rises and contracts when temperature falls. It is established experimentally that the change in length is directly proportional to the length of the member L and change in temperature t. Thus,
tL= α tL
The constant of proportionality α is called coefficient of thermal expansion and is defined as change in unit length of material due to unit change in temperature.
Table shows coefficient of thermal expansion for some of the commonly used engineering materials.
Sync text with VO. The table appears with last paragraph.
03.3.3
Thermal stresses occur depending on the expansion allowed.
If the expansion of the member is freely permitted, as shown in Figure, no thermal stresses are induced in the material.
If the free expansion is prevented fully or partially the stresses are induced in the bar, by the support forces. Referring to the figure shown, if free expansion is permitted the bar would have expanded by
= α tL
Since the support is not permitting it, the support force P develops to keep it at the original position. Magnitude of this force is such that contraction is equal to free expansion, i.e.
or p = E t,
which is the temperature stress. It is compressive in nature in this case.
Consider the case shown in Figure in which free expansion is prevented partially.
In this case free expansion = α tL
Expansion prevented = α tL – δ
The expansion is prevented by developing compressive force P at supports
Thermal Stresses
If the expansion of the member is freely permitted, as shown in Figure, no thermal stresses are induced in the material.
If the free expansion is prevented fully or partially the stresses are induced in the bar, by the support forces. Referring to the figure shown, if free expansion is permitted the bar would have expanded by
= α tL
Since the support is not permitting it, the support force P develops to keep it at the original position. Magnitude of this force is such that contraction is equal to free expansion, i.e.
or p = E t,
which is the temperature stress. It is compressive in nature in this case.
Consider the case shown in Figure in which free expansion is prevented partially.
In this case free expansion = α tL
Expansion prevented = α tL – δ
The expansion is prevented by developing compressive force P at supports
03.3.4
Thermal Stresses in Compound Bars
When temperature rises, the two materials of the compound bar experience different free expansion. Since they are prevented from separating, the two bars will have common position. This is possible only by extension of the bar which has less free expansion and contraction of the bar which has more free expansion. Thus, one bar develops tensile force and another develops the compressive force.
Consider the compound bar as shown here. Let α1, α2 be coefficient of thermal expansion and E1, E2 be moduli of elasticity of the two materials respectively. If rise in temperature is ‘t’,
Free expansion of bar 1 = α1 tL
Free expansion of bar 2 = α2 tL
Let α1> α2. Hence, the positions of the two bars, if the free expansions are permitted, are at AA and BB as shown in the figure.
Since the two bars are rigidly connected at the ends, the final position of the end will be somewhere between AA and BB, say at CC. It means that bar 1 will experience compressive force P1 which will contract it by 1 and bar 2 will experience tensile force P2 which will expand it by 2.
The equilibrium of horizontal forces gives,
P1 = P2, say P
αtL - 1 = αtL + 2
1 + 2 = αtL - αtL = (α- αtL
If the cross-sectional areas of the bars are A1 and A2, we get
= (α- αtL
From the above equation force P can be found and hence the stresses p1 and p2 can be determined.
Thermal Stresses in Compound Bars
When temperature rises, the two materials of the compound bar experience different free expansion. Since they are prevented from separating, the two bars will have common position. This is possible only by extension of the bar which has less free expansion and contraction of the bar which has more free expansion. Thus, one bar develops tensile force and another develops the compressive force.
Consider the compound bar as shown here. Let α1, α2 be coefficient of thermal expansion and E1, E2 be moduli of elasticity of the two materials respectively. If rise in temperature is ‘t’,
Free expansion of bar 1 = α1 tL
Free expansion of bar 2 = α2 tL
Let α1> α2. Hence, the positions of the two bars, if the free expansions are permitted, are at AA and BB as shown in the figure.
Since the two bars are rigidly connected at the ends, the final position of the end will be somewhere between AA and BB, say at CC. It means that bar 1 will experience compressive force P1 which will contract it by 1 and bar 2 will experience tensile force P2 which will expand it by 2.
The equilibrium of horizontal forces gives,
P1 = P2, say P
αtL - 1 = αtL + 2
1 + 2 = αtL - αtL = (α- αtL
If the cross-sectional areas of the bars are A1 and A2, we get
= (α- αtL
From the above equation force P can be found and hence the stresses p1 and p2 can be determined.
Show images according synced with the text paragraphs and VO.
03.4.1
Strain Energy and Impact Loading
When the external forces are applied on a body, the resisting force develops gradually. It develops fully when complete deformation takes place. Figure shows the development of the resisting force with deformation.
As the resisting force moves with deformation, work is done by this force which is stored as energy. When the external force is removed, this stored energy springs the material back to the original position. This energy which is stored in a body due to straining of the body is called strain energy.
Consider a bar of length L. The cross sectional area A is subjected to axial load P. Let the resistance developed be R.
When deformation is zero, resistance R = 0.
When deformation is then= eL, R = P.
Work done by the resisting force = average resistance x = [(0 + P) / 2] x eL
= P e L = p A e L = p e V (Since AL = Volume V) (Since AL = Volume V)
But strain energy = Work done by internal force R
Strain Energy = p e V =stress x strain x volume
Writing for strain e, we get, strain energy = p V = p2
Strain energy per unit volume is defined as Resilience and hence
Resilience =p2
The maximum strain energy which can be stored by a body without undergoing permanent deformation is called Proof Resilience. Hence, proof resilience is equal to strain energy in the body corresponding to the stress at elastic limit.
Proof Resilience = pe2, where pe is the stress at elastic limit.
Strain Energy and Impact Loading
When the external forces are applied on a body, the resisting force develops gradually. It develops fully when complete deformation takes place.
As the resisting force moves with deformation, work is done by this force which is stored as energy. When the external force is removed, this stored energy springs the material back to the original position. This energy which is stored in a body due to straining of the body is called strain energy.
Consider a bar of length L. The cross sectional area A is subjected to axial load P. Let the resistance developed be R.
When deformation is zero, resistance R= 0.
When deformation is then= eL, R = P.
Work done by the resisting force = average resistance x = [(0 + P) / 2] x eL
= P e L = p A e L = p e V (Since AL = Volume V) (Since AL = Volume V)
But strain energy = Work done by internal force R
Strain Energy = p e V =stress x strain x volume
Writing for strain e, we get, strain energy = p V = p2
Strain energy per unit volume is defined as Resilience and hence
Resilience =p2
The maximum strain energy which can be stored by a body without undergoing permanent deformation is called Proof Resilience. Hence, proof resilience is equal to strain energy in the body corresponding to the stress at elastic limit.
Proof Resilience = pe2, where pe is the stress at elastic limit.
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03.4.2
Stress analysis by energy method
Stress analysis due to various types of loads can be done by the strain energy method. In this method, strain energy is equated to the work done by the loads.
This procedure is illustrated for various load cases such as:
· Gradually applied load
· Suddenly applied load
· Shock load
Stress analysis by energy method
Stress analysis due to various types of loads can be done by the strain energy method.
In this method, strain energy is equated to the work done by the loads.
This procedure is illustrated for various load cases such as:
· Gradually applied load
· Suddenly applied load
· Shock load
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Gradually applied load
Consider a bar of length L, cross sectional area A subjected to load P. If the load P is gradually applied, the load increases from 0 to P as extension increases from 0 to gradually.
Hence work done by load = paverage x = [(0 + P)/2] x = P e L
Equating it to strain energy, we getp eV=P e Lp A L=P L
Hence,
p =
Gradually applied load
Consider a bar of length L, cross sectional area A subjected to load P. If the load P is gradually applied, the load increases from 0 to P as extension increases from 0 to gradually.
Hence work done by load = paverage x = [(0 + P)/2] x = P e L
Equating it to strain energy, we getp eV=P e Lp A L=P L
Hence,
p =
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03.4.2.2
Breaking of sling when load is just about to touch the unloading surface.
An example of suddenly applied load
Suddenly applied load
Due to sudden application of loads, there will be instantaneous high stresses and strains. The suddenly applied load P means, P is acting even when A = 0 and acts throughout the period of extension. Hence,
Work done by load= P x = P e L
The resisting force (internal force) develops gradually from zero to maximum value R and hence strain energy stored =p e A L
Equating the work done by the load to the strain energy, we get
P e L = p e A L
Hence, stress
p =
Thus, stress developed due to applications of a load suddenly is twice that due to same load being applied gradually.
Suddenly applied load
Due to sudden application of loads, there will be instantaneous high stresses and strains. The suddenly applied load P means, P is acting even when A = 0 and acts throughout the period of extension. Hence,
Work done by load= P x = P e L
The resisting force (internal force) develops gradually from zero to maximum value R and hence strain energy stored =p e A L
Equating the work done by the load to the strain energy, we get
P e L = p e A L
Hence, stress
p =
Thus, stress developed due to applications of a load suddenly is twice that due to same load being applied gradually.
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03.4.2.3
Impact load (Freely falling load)
In this type of loading, the load falls from a height, say h, before it acts on the body. Figure shows such a loading case on a bar of length L, in which load W falls from a height h before applying load on the bar.
After striking the collar, load W acts through a further distance of .
Work done by load = W (h + ) = W (h+ eL) = W (h + L)
Strain energy stored =p2 AL
Equating the work done by the load to the strain energy, we get
p2 AL = W (h + L) p2 = (h + L) = + p p2 - p - = 0
By solving above quadratic equation for positive root as applicable for maximum instantaneous stress,
p = [ + {( )2 + 4 }1/2] p = [ + + 4 x ()2}1/2]
Hence,
p = [1 + (1 + )1/2]
Thus, in this case, the instantaneous stress produced by W is much more than the suddenly applied case (p = 2W).
In most of the cases, compared to height L, the extension is too small. Hence, if is neglected in comparison to h, work done by load = W.hand
By equating strain energy to this, we get
p2 AL = W h
Impact load (Freely falling load)
In this type of loading, the load falls from a height, say h, before it acts on the body.
After striking the collar, load W acts through a further distance of .
Work done by load = W (h + ) = W (h + eL) = W (h + L)
Strain energy stored = p2 AL
Equating the work done by the load to the strain energy, we get
p2 AL = W (h + L) p2 = (h + L) = + p p2 - p - = 0
By solving above quadratic equation for positive root as applicable for maximum instantaneous stress,
p = [ + {( )2 + 4 }1/2] p = [ + + 4 x ()2}1/2]
Hence,
p = [1 + (1 + )1/2]
Thus, in this case, the instantaneous stress produced by W is much more than the suddenly applied case (p = 2W).
In most of the cases, compared to height L, the extension is too small. Hence, if is neglected in comparison to h, work done by load = W.hand
By equating strain energy to this, we get
p2 AL = W h
p = ( )1/2
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03.4.2.4
Shock load
Shock load is measured in terms of energy. It is an externally applied energy. If U units of shock load is applied to a bar, the stress produced in the bar can be obtained by equating strain energy to the shock energy. Thus, instantaneous stress produced in such a case is given by
p2 AL= U,
Hence, p = ( )1/2
Shock load
Shock load is measured in terms of energy. It is an externally applied energy. If U units of shock load is applied to a bar, the stress produced in the bar can be obtained by equating strain energy to the shock energy. Thus, instantaneous stress produced in such a case is given by
p2 AL= U,
Hence, p = ( )1/2