Properties of Metals, Stress-Strain & Elastic Constants - 2 | Civil Engineering SSC JE (Technical) - Civil Engineering (CE) PDF Download

Properties of Metals, Stress - Strain and Elastic Constants
 

ELONGATION OF BARS
1. A bar of uniform cross-sectional area

2. Non-uniform bar

3. Tapering bar of circular cross-section whose diameter changes from ‘d_1’ 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 ‘a’ to ‘b’.

5. Bars joined together parallel to each other A structural member composed of two or more elements of different materials to form a parallel arrangement and subjected to axial loading is termed as a compound bar. Such a section is also known as composite section

•  Composite Young’s modulus is given by

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 l = Unit - weight of the material = W/AL
L = Length of the specimen OR

W = Total weight of specimen
A = Cross-sectional area.

• Thus, the deformation of the bar under its own weight is half the deformation, if the body is subjected to the direct load equal to the weight of the body.

(b) For bar of tapering section (Conical bar) :

 OR  Deflection of prismatic bar

7. Bar of Uniform Strength :

• It is possible to maintain uniform stress at all the sections by increasing the area from the lower level to higher level.
• For a bar to have constant strength, the stress at any section, due to external load and the weight of the portion below it should be constant i.e.

 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

• When a material undergoes change in temperature, its length also changes and if the materials is free to expand or contract, no stresses are developed.

Free expansion in length is given by
ΔL = LaT Where,
L = Length of specimen
α = Coefficient of thermal change
T = Change in temperature

•  If the free expansion or contraction is prevented then stresses will be developed, which is given by σ = αTE
• If the ends of the rod yield by an amount ‘δ ’, in this case the total amount of expansion checked is given by ΔL' = L α T – δ
• It may be noted that thermal stress or temperature stress does not depend upon the cross sectional area.

 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 (α_c) is greater than coefficient of expansion of steel (α_s) hence free expansion of copper is more than that of steel.

• Let us take actual expansion of composite bar be ‘δ’ then L α_c T > δ > L α_s T. Therefore steel is in tension and copper is in compression when the temperature is raised.
• Let the σ_s and σ_c are stresses in steel and copper. Then for equilibrium of system, Total tension in steel = total compression in copper σ_sA_s = σ_cA_c ...(i) 

Also for composite section Actual expansion of steel = Actual expansion of copper, this gives following condition

Where E_s and E_c 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 = σ_sA_s = σ_cA_c ...(i) 
Total extension = 0 Therefore

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 ( Î_v )

• For rectangular bar having, length = l, width = b, and thickness = d. The volumetric strain is given by.

 
or Î_v = Î_x +Î_y +Î_z

• For cylindrical rod, having diameter = d.

Volumetric strain ‘ Î_v’ is given by
Î_v = 2Î_d +Î l = (2× Diametral strain) + Longitudinal strain

• For spherical body, having diameter = d.

 Diametral strain

• If a rectangular block is subjected to three normal stresses sx, sy and sz on all its faces, mutually perpendicular to each other in x, y and z directions respectively. Then volumetric strain of the specimen is given by

Î_v = Î_x +Î_y +Î_z

Under hydrogtatic loading

3. Shear Modulus

• When a body is subjected to shearing stresses, the shape of the body gets distorted. The measurement of this distortion is done by angle of distortion. Under pure shear the shape of the body gets distorted but the volume remains same.

Modulus of rigidity.

Shear modulus is also known as modulus of rigidity.

•  If under the shear, the shear strain is f then the linear strain in the diagonal of the specimen is given by
   linear strain of diagonal is half of the shear strain in the body..
• A shear stress in a given direction cannot exist without a balancing shear stress of equal intensity in a direction at right angles to it. This stress is called complimentary shear stress.

RELATIONSHIP BETWEEN ELASTIC CONSTANTS

Young’s modulus, E = Linear Stress / Linear Strain

Modulus of rigidity, G = Shear stress/ Shear strain

Bulk modulus, K = Direct Stress/ volumetric Strain

Poisson’s ratio, μ = 1/m = Lateral strain/ Linear Strain

 STRAIN ENERGY
The strain energy is equal to the work done by the load provided no energy is added or subtracted in the form of heat. It is denoted by U. Thus,

stress strain volume = (1/2)
For a prismatic bar,   Δ = PL/ AE

• The unit of strain Energy in SI units is Joule (J) which is equal to 1 N-m.
• Resilience is also known as strain energy density. It represents the ability of the material to absorb energy within elastic limit.

It is denoted by ‘u’. Thus,

When the stress (σ) is equal to proof stress (σ_f) at the elastic limit, the corresponding resilience is known as proof resilience.  Thus,

• Modulus of Resilience : - Proof Resilience / Volume =

• Strain energy of a prismatic bar with varying section is given by

• Strain energy of a prismatic bar hanging under its own weight is given by

• Strain energy of a hanging elastic body due to more than one load cannot be found merely by adding the energies obtained from individual loads. This is because of the fact that strain in such a case is not a linear function but is a quadratic function of the loads.