Unit 1
Simple Stresses and Strains
Materials Under Load
2
Bodies subjected to Load
• All bodies are deformable
• Subjected to loads, there is change is shape
and size
• Loads may act in all three cardinal directions
• Straining actions may be tension, compression
or shear
3
General Stress System
• On a three dimensional body, straining actions
can be reduced to three normal stresses in
cardinal directions and three shear stresses
• Normal stresses are - x, y, z
• Shear stresses - xy,xz, yx, yz, zx, zy
• In many elements, this can be reduced to two
dimensions or one dimension
4
Stress Systems
• In a beam, the stress system can be reduced
to a plane stress system of two dimensions
• A beam has bending stresses, tensile and
compressive, and shear stress
• The stress system of an element of a beam is
thus x, xy
5
Materials Under Load
6
Stress Systems
• A truss member is subjected to tension or
compression only
• Truss member is subjected to uniaxial stress
• The basic stress conditions are thus tension,
compression and shear
7
Materials Under Load
8
Normal Stress
• Normal stress acts normal to the cross section
of the member
• A member subjected to tension or
compression
• A member subjected to tension
• To find the stress, we take a section some
where in the middle
9
Stress
10
Tensile stress
• Separate the member at the section
• The member is in equilibrium after the
straining action
• So each part also is in equilibrium
• Equilibrium of each part requires a force equal
to P – the stress resultant
• Stress is uniformly distributed at X-X
• Near the load P, stress is not uniform
11
Stress
12
Tensile stress
• The tensile stress x area gives a force P equal
and opposite to P
• Normal stress is designated by 
• Normal stress  = P/A
• Unit of normal stress is load/area is N/mm²,
kN/m²
• Giga Pascal (GPa) is 109 N/m²
13
Tensile Stress
14
Compressive Stress
• Compressive force (thrust) represented by
two forces acting towards each other
• Thrust tends to reduce the length
• The stress resultant at section x-x are two
forces acting away from each other
• Compressive stress is normal to the cross
section
• Symbol for normal stress is .
15
Compressive Stress
16
Normal Stress
• A thin cylinder subjected to pressure has
stresses along the circumference called hoop
or circumferential stress
• A cylinder has also stress along its length
called longitudinal stress
• Both these are normal stresses
• Shear stress acts tangential to the cross
section; symbol is ; Average shear stress =
load by area; has the same units as normal
stress
17
Normal Stresses & shear stress
18
Strain
• Tensile force elongates a member;
compressive force causes decrease in length
• Strain is elongation/shortening per unit length
• Strain is a ratio and has no unit
• Shear strain is measured as the angle 
• Tan  = L/L; tan =  in radians
19
Modulii
• Strain is represented by the symbol ;
• For most material ratio of stress to strain is
constant up to a certain magnitude of stress
• Ratio of stress to strain is called modulus
• For normal stress and strain, this ratio is called
Young’s modulus of Elasticity with symbol E
20
Modulii
• The ratio of shear stress to shear strain
• (/) is called modulus of rigidity; symbols G,
N or C
• E for some materials (GPa) – Steel – 210;
Aluminium – 70; brass – 95
• Modulus of rigidity (GPa) – Steel – 85;
Aluminium – 25; brass – 35
• Modulii have the same units as stress
21
Elongation/Shortening
• Stress = ; Strain = ; E = /
• Elongation /shortening (L) =  L =  L/E
•  = P/A; L = PL/AE
• For a tapering round bar, Elongation
• L = 4PL/(πEd1d2)
• For a tapering bar of rectangular section
• L = PL log(w2/w1)/[Et(w2-w1)]
22
Tapering sections
[Add figures 3.9(b) and 3.10(b) in this frame]
23
Normal, Tensile, Compressive, &
Tangential Strain
24
Composite sections
• Two or more material elements rigidly joined
to act together
• P = P1 + P2
• Second condition of equal strains
• P1L/A1E1 = P2L/A2E2; 1/E1= 2/E2
• Modular ratio m = E1/E2; 1 = 2 m
25
Composite Sections
26
Stresses due to Temperature Change
27
Stress due to temperature
• Thermal stresses due to restraining
expansion/contraction due to change in
temperature
• Bar of single material, fully restrained
• Stress = E t
• Partial yielding by ‘a’;
• Stress = E (Lt – a)/L
28
Thermal Stress in Bar of Single
Material
29
Temperature stress in composite bars
• Two bars rigidly joined together, subjected to
temperature change
• No external restraint
• Thermal expansion coefficients 1, 2
• 1< 2; Free expansion of 1 = L1T
• Free expansion of 2 = L2T
• Material 1 expands more than L1T and in tension
• Material 2 expands less than L2T and in compression
30
Thermal Stress in Bar of Composite
Material
31
Shear stress and Strain
• Shear stress is found in riveted joints, keyed-
shafts, beams and rotating shafts
• Shear stress acts tangential to the surface
• Shear stress  shear strain ( = L/L)
•  = G 
32
Shear Stress & Strain
33
Complementary Shear Stress
• Shear stress on two opposite faces
• Accompanied by shear stress of opposite
nature on the other two opposite faces
• Shear stress is taken positive when the
moment of the forces about a point within the
element is clockwise
• Shear stress on opposite faces is
complementary shear stress
34
Complementary Shear Stress
35
State of Pure Shear
• When element is subjected to only shear
stress, it is state of pure shear
• The element elongates along one diagonal
and shortens along the other
• There is tensile stress along one diagonal and
compressive stress along the other
• Stresses along the diagonal is the same as that
acting on the sides
36
Stress & Strain along Diagonals
37
Poisson’s ratio
• Material stressed in one direction has stresses
of opposite nature in the other two directions
• Ratio of longitudinal strain to lateral strain is
Poisson’s ratio
• Symbol for Poisson’s ratio = 1/m, 
• Poisson’s ratio lies between o.25 & 0.34 and
cannot be more than 0.5
38
Lateral Strain & POISSON’s Ratio
39
Multi- axial Stresses & Generalized
Hooke’s Law
40
Generalized Hooke’s Law
• Three stresses x, y, z acting on element
• Strain x = [x - y - z]/E
• x + y + z = (1-2) [x + z + z]/E
• x = z = z = ; x = y = z
• E = (1 - 2) ; 1- 2 is positive;  <0.5
• Shear strain xy = xy/G; yz = yz/G; zx =
zx/G
41
Bulk Modulus
• Volumetric Strain v = V/V
• Bulk Modulus K = /v [x=y=z=]
• v = x + y +z
• x = x/E - [y + z]/E
• v = (x + y + z)(1 - 2)/E
• v = 3(1 - 2)/E [x = y = z = ]
42
Bulk Modulus
43
Elastic Constants
• Elastic constants – E, G, K and 
• Relationships between constants
• G = E / [2(1 + )]
• K = E / [3(1 - 2)]
• G = 3K (1- 2) / [2(1+ )]
• E = 9KG / (3K + G)
44
Relationship between Elastic
Constants
45
Mechanical Properties
• Elasticity –regaining shape and size on
removal of load
• Homogeneous – having same elastic
properties every where
• Isotropic – same elastic properties in all
directions at a point
• Plasticity – material undergoes deformation
but does not regain the shape and size on
removal of load
46
Mechanical Properties
• Ductility – helps to draw thin wires
• Brittleness – tendency to shatter on impact
• Toughness – ability to undergo large plastic
deformations
• Resilience – ability to recover size and shape
after deformation
• Hardness – ability to resist wear, cutting,
scratching or intendation
47
Mechanical Properties
• Fatigue – failure caused by repeated cycles of
loading even under low stress
• Creep – deformation under constant stress
over a long period of time
48
Testing of mild steel
• Standard specimen and test procedure as per
BIS specification
• Standard test specimen with gauge length
marked
• Testing done on universal testing machine
• Extensometers used to measure extension
over gauge length
• Specimen tested to failure
49
Stress Strain Diagram for Mild Steel
50
Stress – Strain diagram
• Stress – strain diagram has an initial straight
portion where stress  strain
• Slope of the straight line is E
• Proportional limit – stress up to which the
diagram is linear
• Elastic limit – stress up to which when
stressed, shape and size cab regained
• Yield point – stress beyond which plastic
deformation takes place
51
Stress-Strain diagram
• Ultimate stress – Maximum stress ordinate
• Breaking stress – stress at which material
breaks (lower than ultimate stress)
• As the area decreases when loaded, true
stress when plotted against strain shows a
different type of diagram
52
Stress-Strain diagram
• Percentage elongation – is the increase in
length expressed as a percentage of original
length
• Percent reduction in area – area measured at
neck. Difference in area expressed as a
percentage of original area
• Value of E for all types of steels is nearly the
same
53
Stress Strain Diagram for Mild Steel
54
Stress Strain Diagram for Mild
Steel
55
 -  diagrams
• Stress-strain diagram for other materials show
a lot of variation
• Different steels have an initial straight portion;
high strength steels do not show much plastic
deformation and hence brittle
• Concrete does not have a well-defined
straight part
56
Stress Strain Diagram for Other
Material
57
Stress Strain Diagram for Other
Material
58
Factor of Safety
• Stress allowable under working loads
• Yield stress / working stress is factor of safety
• Factor of safety is less for steel as it is a
factory manufactured product
• For site-made product like concrete under
not-so-controlled conditions, factor safety is
high
59

Strength of Materials - Unit 1 Full PPT.ppt

  • 1.
  • 2.
  • 3.
    Bodies subjected toLoad • All bodies are deformable • Subjected to loads, there is change is shape and size • Loads may act in all three cardinal directions • Straining actions may be tension, compression or shear 3
  • 4.
    General Stress System •On a three dimensional body, straining actions can be reduced to three normal stresses in cardinal directions and three shear stresses • Normal stresses are - x, y, z • Shear stresses - xy,xz, yx, yz, zx, zy • In many elements, this can be reduced to two dimensions or one dimension 4
  • 5.
    Stress Systems • Ina beam, the stress system can be reduced to a plane stress system of two dimensions • A beam has bending stresses, tensile and compressive, and shear stress • The stress system of an element of a beam is thus x, xy 5
  • 6.
  • 7.
    Stress Systems • Atruss member is subjected to tension or compression only • Truss member is subjected to uniaxial stress • The basic stress conditions are thus tension, compression and shear 7
  • 8.
  • 9.
    Normal Stress • Normalstress acts normal to the cross section of the member • A member subjected to tension or compression • A member subjected to tension • To find the stress, we take a section some where in the middle 9
  • 10.
  • 11.
    Tensile stress • Separatethe member at the section • The member is in equilibrium after the straining action • So each part also is in equilibrium • Equilibrium of each part requires a force equal to P – the stress resultant • Stress is uniformly distributed at X-X • Near the load P, stress is not uniform 11
  • 12.
  • 13.
    Tensile stress • Thetensile stress x area gives a force P equal and opposite to P • Normal stress is designated by  • Normal stress  = P/A • Unit of normal stress is load/area is N/mm², kN/m² • Giga Pascal (GPa) is 109 N/m² 13
  • 14.
  • 15.
    Compressive Stress • Compressiveforce (thrust) represented by two forces acting towards each other • Thrust tends to reduce the length • The stress resultant at section x-x are two forces acting away from each other • Compressive stress is normal to the cross section • Symbol for normal stress is . 15
  • 16.
  • 17.
    Normal Stress • Athin cylinder subjected to pressure has stresses along the circumference called hoop or circumferential stress • A cylinder has also stress along its length called longitudinal stress • Both these are normal stresses • Shear stress acts tangential to the cross section; symbol is ; Average shear stress = load by area; has the same units as normal stress 17
  • 18.
    Normal Stresses &shear stress 18
  • 19.
    Strain • Tensile forceelongates a member; compressive force causes decrease in length • Strain is elongation/shortening per unit length • Strain is a ratio and has no unit • Shear strain is measured as the angle  • Tan  = L/L; tan =  in radians 19
  • 20.
    Modulii • Strain isrepresented by the symbol ; • For most material ratio of stress to strain is constant up to a certain magnitude of stress • Ratio of stress to strain is called modulus • For normal stress and strain, this ratio is called Young’s modulus of Elasticity with symbol E 20
  • 21.
    Modulii • The ratioof shear stress to shear strain • (/) is called modulus of rigidity; symbols G, N or C • E for some materials (GPa) – Steel – 210; Aluminium – 70; brass – 95 • Modulus of rigidity (GPa) – Steel – 85; Aluminium – 25; brass – 35 • Modulii have the same units as stress 21
  • 22.
    Elongation/Shortening • Stress =; Strain = ; E = / • Elongation /shortening (L) =  L =  L/E •  = P/A; L = PL/AE • For a tapering round bar, Elongation • L = 4PL/(πEd1d2) • For a tapering bar of rectangular section • L = PL log(w2/w1)/[Et(w2-w1)] 22
  • 23.
    Tapering sections [Add figures3.9(b) and 3.10(b) in this frame] 23
  • 24.
    Normal, Tensile, Compressive,& Tangential Strain 24
  • 25.
    Composite sections • Twoor more material elements rigidly joined to act together • P = P1 + P2 • Second condition of equal strains • P1L/A1E1 = P2L/A2E2; 1/E1= 2/E2 • Modular ratio m = E1/E2; 1 = 2 m 25
  • 26.
  • 27.
    Stresses due toTemperature Change 27
  • 28.
    Stress due totemperature • Thermal stresses due to restraining expansion/contraction due to change in temperature • Bar of single material, fully restrained • Stress = E t • Partial yielding by ‘a’; • Stress = E (Lt – a)/L 28
  • 29.
    Thermal Stress inBar of Single Material 29
  • 30.
    Temperature stress incomposite bars • Two bars rigidly joined together, subjected to temperature change • No external restraint • Thermal expansion coefficients 1, 2 • 1< 2; Free expansion of 1 = L1T • Free expansion of 2 = L2T • Material 1 expands more than L1T and in tension • Material 2 expands less than L2T and in compression 30
  • 31.
    Thermal Stress inBar of Composite Material 31
  • 32.
    Shear stress andStrain • Shear stress is found in riveted joints, keyed- shafts, beams and rotating shafts • Shear stress acts tangential to the surface • Shear stress  shear strain ( = L/L) •  = G  32
  • 33.
    Shear Stress &Strain 33
  • 34.
    Complementary Shear Stress •Shear stress on two opposite faces • Accompanied by shear stress of opposite nature on the other two opposite faces • Shear stress is taken positive when the moment of the forces about a point within the element is clockwise • Shear stress on opposite faces is complementary shear stress 34
  • 35.
  • 36.
    State of PureShear • When element is subjected to only shear stress, it is state of pure shear • The element elongates along one diagonal and shortens along the other • There is tensile stress along one diagonal and compressive stress along the other • Stresses along the diagonal is the same as that acting on the sides 36
  • 37.
    Stress & Strainalong Diagonals 37
  • 38.
    Poisson’s ratio • Materialstressed in one direction has stresses of opposite nature in the other two directions • Ratio of longitudinal strain to lateral strain is Poisson’s ratio • Symbol for Poisson’s ratio = 1/m,  • Poisson’s ratio lies between o.25 & 0.34 and cannot be more than 0.5 38
  • 39.
    Lateral Strain &POISSON’s Ratio 39
  • 40.
    Multi- axial Stresses& Generalized Hooke’s Law 40
  • 41.
    Generalized Hooke’s Law •Three stresses x, y, z acting on element • Strain x = [x - y - z]/E • x + y + z = (1-2) [x + z + z]/E • x = z = z = ; x = y = z • E = (1 - 2) ; 1- 2 is positive;  <0.5 • Shear strain xy = xy/G; yz = yz/G; zx = zx/G 41
  • 42.
    Bulk Modulus • VolumetricStrain v = V/V • Bulk Modulus K = /v [x=y=z=] • v = x + y +z • x = x/E - [y + z]/E • v = (x + y + z)(1 - 2)/E • v = 3(1 - 2)/E [x = y = z = ] 42
  • 43.
  • 44.
    Elastic Constants • Elasticconstants – E, G, K and  • Relationships between constants • G = E / [2(1 + )] • K = E / [3(1 - 2)] • G = 3K (1- 2) / [2(1+ )] • E = 9KG / (3K + G) 44
  • 45.
  • 46.
    Mechanical Properties • Elasticity–regaining shape and size on removal of load • Homogeneous – having same elastic properties every where • Isotropic – same elastic properties in all directions at a point • Plasticity – material undergoes deformation but does not regain the shape and size on removal of load 46
  • 47.
    Mechanical Properties • Ductility– helps to draw thin wires • Brittleness – tendency to shatter on impact • Toughness – ability to undergo large plastic deformations • Resilience – ability to recover size and shape after deformation • Hardness – ability to resist wear, cutting, scratching or intendation 47
  • 48.
    Mechanical Properties • Fatigue– failure caused by repeated cycles of loading even under low stress • Creep – deformation under constant stress over a long period of time 48
  • 49.
    Testing of mildsteel • Standard specimen and test procedure as per BIS specification • Standard test specimen with gauge length marked • Testing done on universal testing machine • Extensometers used to measure extension over gauge length • Specimen tested to failure 49
  • 50.
    Stress Strain Diagramfor Mild Steel 50
  • 51.
    Stress – Straindiagram • Stress – strain diagram has an initial straight portion where stress  strain • Slope of the straight line is E • Proportional limit – stress up to which the diagram is linear • Elastic limit – stress up to which when stressed, shape and size cab regained • Yield point – stress beyond which plastic deformation takes place 51
  • 52.
    Stress-Strain diagram • Ultimatestress – Maximum stress ordinate • Breaking stress – stress at which material breaks (lower than ultimate stress) • As the area decreases when loaded, true stress when plotted against strain shows a different type of diagram 52
  • 53.
    Stress-Strain diagram • Percentageelongation – is the increase in length expressed as a percentage of original length • Percent reduction in area – area measured at neck. Difference in area expressed as a percentage of original area • Value of E for all types of steels is nearly the same 53
  • 54.
    Stress Strain Diagramfor Mild Steel 54
  • 55.
    Stress Strain Diagramfor Mild Steel 55
  • 56.
     - diagrams • Stress-strain diagram for other materials show a lot of variation • Different steels have an initial straight portion; high strength steels do not show much plastic deformation and hence brittle • Concrete does not have a well-defined straight part 56
  • 57.
    Stress Strain Diagramfor Other Material 57
  • 58.
    Stress Strain Diagramfor Other Material 58
  • 59.
    Factor of Safety •Stress allowable under working loads • Yield stress / working stress is factor of safety • Factor of safety is less for steel as it is a factory manufactured product • For site-made product like concrete under not-so-controlled conditions, factor safety is high 59