When a material is subjected to an external force, a resisting force is set up within the component. The internal resistance force per unit area acting on a material or intensity of the forces distributed over a given section is called the stress at a point.
Stress
1 kPa = 10^{3} Pa = 10^{3} N/ m^{2} (kPa = Kilo Pascal)
1 MPa = 10^{6} Pa = 10^{6}N/ m^{2} = 1 N/mm^{2 } (MPa = Mega Pascal)
1 GPa = 10^{9} Pa = 10^{9} N/ m^{2} (GPa = Giga Pascal)
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If σ> 0 the stress is tensile. i.e. The fibres of the component tend to elongate due to the external force. The beam is subjected to an external force tensile F and tensile stress distribution due to the force is shown in the figure. Tensile Stress
If σ< 0 the stress is compressive. i.e. The fibres of the component tend to shorten due to the external force. A member subjected to an external compressive force P and compressive stress distribution due to the force is shown in the given figure.
When forces are transmitted from one part of a body to other, the stresses developed in a plane parallel to the applied force are the shear stress. Shear stress acts parallel to plane of interest. Forces P is applied transversely to the member AB as shown. The corresponding internal forces act in the plane of section C and are called shearing forces. The corresponding average shear stress
(T) = P/Area
The displacement per unit length (dimensionless) is known as strain.
The elongation per unit length as shown in the figure is known as tensile strain.
ε_{t }= ΔL/L_{0}
It is engineering strain or conventional strain. Here we divide the elongation to original length not actual length (L_{0} +ΔL)
Tensile Strain
If the applied force is compressive then the reduction of length per unit length is known as compressive strain. It is negative. Then ε_{c} = (ΔL)/L_{0}
When a force P is applied tangentially to the element shown. Its edge displaced to dotted line. Where E is the lateral displacement of the upper face of the element relative to the lower face and L is the distance between these faces.
Shear StrainThen the shear strain is : formula
The true stress is defined as the ratio of the load to the cross section area at any instant.
True stress
Where σ and ε is the engineering stress and engineering strain respectively.
True strain
or engineering strain (ε) = e^{ε}_{t }1
The volume of the specimen is assumed to be constant during plastic deformation.
[ ∵ A_{o}L_{o }= AL ] It is valid till the neck formation.
Comparison of engineering and the true stressstrain curves shown below
stress vs strain
σ_{T }= L(ε_{T})^{n}
Where K is the strength coefficient
n is the strain hardening exponent
n = 0 perfectly plastic solid
n = 1 elastic solid
For most metals, 0.1< n < 0.5
Relations
The coefficient E is called the modulus of elasticity i.e. its resistance to elastic strain. The coefficient G is called the shear modulus of elasticity or modulus of rigidity.
A relationship similar to that for length changes holds for threedimensional (volume) change. For volumetric strain(ε_{v})
the relationship is (ε_{v}) = (VV_{0})/V_{0 }or (ε_{v}) = ΔV/V_{0 }= P/K
modulus
[Intext Question]
Poisson's ratio
(Under unidirectional stress in xdirection)
Ratios of various material
Ratio
Elongation for a bar
[Intext Question]
Elongation of a composite body
[Intext question]
Elongation of a tapered body
Elongation due to self weight
Structural members or machines must be designed such that the working stresses are less than the ultimate strength of the material.
Working stress
(n) = σ_{y }or σ_{p }or σ_{ult} / σ_{w}
ε = α(ΔT)
σ_{t} = αE(ΔT)
where, E = Modulus of elasticity
[Intext Question]
A brass rod placed within a steel tube of exactly same length. The assembly is making in such a way that elongation of the combination will be same. To calculate the stress induced in the brass rod, steel tube when the combination is raised by t^{ο}C then the following analogy have to do.
Where, δ = Expansion of the compound bar = AD in the above figure.
δ_{st}= Free expansion of the steel tube due to temperature rise t^{ο}C = α_{s}Lt = AB in the above figure.
δ_{sf} = Expansion of the steel tube due to internal force developed by the unequal expansion = BD in the above figure.
δ_{bt} = Free expansion of the brass rod due to temperature rise t^{ο}C = α_{b}Lt = AC in the above figure.
δ_{bt}= Compression of the brass rod due to internal force developed by the unequal expansion. = BD in the above figure.
And in the equilibrium equation
Tensile force in the steel tube = Compressive force in the brass rod
Where, σ_{s}= Tensile stress developed in the steel tube.
σ_{b}= Compressive stress developed in the brass rod.
A_{s}= Cross section area of the steel tube.
A_{b}= Cross section area of the brass rod.
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1. What is stress and how does it relate to mechanical engineering? 
2. What is the difference between tensile stress and compressive stress? 
3. How does Hook's law relate to stress and strain? 
4. What is Poisson's ratio and what does it tell us about materials? 
5. How does true stress and true strain differ from conventional stress and strain measurements? 

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