Chapter 1
PROPERTIES OF METALS, STRESS- STRAIN AND ELASTIC CONSTANTS
PROPERTIES OF METALS
1. Ductility
• Ductility is the property by virtue of which material can be stretched to a reduced section under the action of tensile force. Large deformations are thus possible in ductile materials before the absolute failure or rupture takes place, some of the examples are mild steel, aluminium, copper, manganese, lead, nickel, brass,
bronze, monal metal etc.
2. Brittleness
3. Malleability
4. Toughness
Modulus of toughness
where,
Sty is the tensile yield strength,
Stu is the tensile ultimate strength,
εy is the strain at yield,
εu is the ultimate strain (total strain at failure), and
E is the elastic modulus.Modulus of resilience (u)=
5. Hardness
Based upon this there are two methods of hardness measurement :
(i) Scratch hardness - Commonly measured by Mohr's test.
(ii) Indentation hardness (abrasion) measured by deformation.
6. Fatigue
The behaviour of material under variable loads (dynamic loads) is referred to as “fatigue”. In recent past several failures of structures have been noted due to fatigue.
Factors affecting fatigue are :
(i) Loading conditions
(ii) Frequency of loading
(iii) Corrosion
(iv) Temperature
(v) Stress concentration
7. Creep & Stress Relaxation
STRESS-STRAIN CURVE
Simple tension test for mild steel :
Important Points to be noted :
1. Strain that occurs before the yield point is called elastic strain and that which occurs after yield point with no increase in stress is called plastic strain. For mild steel, Plastic Strain = 10 to 15 times of elastic strain
2. Ideal curve for tension is shown, however actual behaviour is different and indicates apparently reduced yield stress in compression for mild steel. The divergence between tension and compression results it is explained by Bauchinger and is called “Bauchinger effect”.
3. The stress is defined as the ratio of load to original area (Ao) It is also known as ‘engineering stress’ or ‘stress’ or ‘nominal stress’.
Engg. stress ‘or’ Nominal stress =
On the contrary when changing area is taken into account, the ratio of load to actual area (A) is called ‘true stress’. True stress = (P/A)
SIMPLE STRESS AND STRAIN
(i) Direct stresses or normal stresses which may be tensile or compressive.
(ii) Shear or Tangential stresses.
(iii) Transverse or Bending stresses.
(iv) Torsional or Twisting stresses.
• For direct stresses, if area under consideration is original area, then it is known as
engineering stress or nominal stress or simply stress. But, if the area taken is actual
area, then stress is known as true stress.
Nominal Stress =
True Stress = (P/A)
here,
Ao = Original area of specimen
A = Actual area of specimen
• When a prismatic bar is subjected to axial load, it undergoes a change in length. This change is length is usually called deformation. The deformation per unit length of the bar is termed as strain. Since strain is deformation per unit length, it is a dimensionless quantity. But sometimes in practice, strain is recorded in forms such as mm/m or μm/m etc.
Tension and Compression Monotonic Stress-strain Diagrams
Tensile stress-strain diagram for differents steels
1. Hooke’s Law
Stress ∝ Strain
( Stress/Strain)= E(const.)
Where, E = Young’s modulus of elasticity.
2. Proof Stress
Stress-strain curve for ‘aluminium’
ELONGATION OF BARS
1. A bar of uniform cross-sectional area
2. Non-uniform bar
dL = (P/E)[L1/A1 + L2/A2+ _ _ _ _ _ ]
3. Tapering bar of circular cross-section whose diameter changes from ‘d1’ at one end to ‘d2’ at the other end.
where P is the load applied & E is the Young’s modulus of elasticity. L is the length of the specimen.
4. Tapering bar of rectangular cross-section having uniform thickness ‘t’ but width of the bar varies from ‘b1’ to 'b2'
5. Bars joined together parallel to each other
Where A1 and E1 are properties of one bar and A2 and E2 are properties of other bar.
6. Elongation due to self weight
(a) For bar of uniform section (Prismatic bar): -
Where λ = Unit - weight of the material
L= Length of the specimen
W = Total weight of specimen
A = Cross-sectional area.
(b) For bar of tapering section (Conical bar) : Deflection of prismatic bar
OR
7. Bar of Uniform Strength :
Where A1 is the area at upper section where the bar is fixed with immovable support. A2 is area at lower section which is freely hanging.
TEMPERATURE STRESSES
Free expansion in length is given by
ΔL = LαT
Where, L = Length of specimen
α = Coefficient of thermal change
T = Change in temperature
which is given by
σ = αTE
ΔL' = L α T – δ
TEMPERATURE STRESSES IN COMPOSITE SECTIONS
(a) When ends of a compound bar are free to expand :
If the temperature is increased then the metal with greater value of a will be in compression and the other metal will be in tension. On the other hand if the temperature is decreased then the nature of stresses will be change i.e., metal with greater value of a will be in tension and the other metal will be in compression.
Example:- In copper and steel composite section, since coeff. of expansion of copper (αc) is greater than coefficient of expansion of steel (αs) hence free expansion of copper is more than that of steel.
Let us take actual expansion of composite bar be ‘δ’ then L αc T > δ > L αs T.
Therefore steel is in tension and copper is in compression when the temperature is raised.
Total tension in steel = total compression in copper
σs As = σc Ac ...(i)
Also for composite section
Actual expansion of steel = Actual expansion of copper, this gives following condition
Where Es and Ec are modulus of elasticity of steel and copper respectively.
Note : When ends of a compound bar are restrained, then on increasing the temperature metal with greater value of a will be in compression and the other metal will be in tension and vice versa.
(b) When two bars are connected end to end between two fixed supports
• If the temperature is increased both the bars will undergo change in length. But due to the supports, the change in length is restrained. hence both the bars will experience compressive stresses. If the temperature is decreased then both the bars will experience tensile stresses.
Total compressive force = σs As = σc Ac ...(i)
Total extension = 0
Therefore
σsLs/Es + σcLc/Es + ΔT(αsLs + αcLc) = 0
ELASTIC CONSTANTS
1. Poisson’s ratio (μ)
The value of m lies between 3 and 4 for most of the metals.
μ = 0.05 to 0.1 for glass
= 0.1 to 0.2 for concrete
= 0.25 to 0.42 for metals
= 0.5 for pure rubber (for perfectly plastic body)
2. Volumetric strain (Îv)
or
Let,
L = Length of the cylindrical rod
d = Diameter of the cylindrical rod
Volume of the cylindrical rod, V = (Π/4) d2. L
ΔL= Change in length of the cylindrical rod
Δd= Change in diameter of the cylindrical rod
ΔV= Change in volume of the cylindrical rod
Let us determine the final dimensions of the cylindrical rod
Final length of the cylindrical rod = L+ ΔL
Final diameter of the cylindrical rod = d - Δd
Final volume of the cylindrical rod = (Π/4) (d - Δd) 2. (L+ ΔL)
Change in volume of the cylindrical rod = Final volume – initial volume
Change in volume of the cylindrical rod, ΔV = (Π/4) [(d2.L – 2.Δd.L.d + ΔL. d2)-(d2.L)]
ΔV= (Π/4) [ΔL. d2– 2.Δd.L.d]
Volumetric strain also known as bulk strain will be determined as following
Ԑv= Change in volume /original volume
Ԑv= dV/V
Ԑv= [(Π/4) (ΔL. d2– 2.Δd.L.d)]/ [(Π/4) d2.L]
Ԑv= [(ΔL. d2– 2.Δd.L.d)/ (d2.L]
Ԑv= (ΔL/L) – 2(Δd/d)
3. Shear Modulus
Modulus of rigidity.
Shear modulus is also known as modulus of rigidity.
i.e. linear strain of diagonal is half of the shear strain in the body.
RELATIONSHIP BETWEEN ELASTIC CONSTANTS
Young’s modulus,
Modulus of rigidity,
Bulk modulus,
Poisson’s ratio,
Relationships:
STRAIN ENERGY
For a prismatic bar,
U = (p2/2E)[L1/A1 + L2/A2 + _ _ _ _ ]
where 'g' is the weight of bar per unit volume.