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تاريخ التسجيل : 21/09/2008
 |  موضوع: سلسلة محاظرات  2008-09-28, 2:41 pm | |
| 4. Relationship between Stress and StrainRelationship between Stress and Strain are derived on the basis of the elastic behaviour of material bodies.
A standard mild steel specimen is subjected to a gradually increasing pull by Universal Testing Machine. The stress-strain curve obtained is as shown below. A -Elastic Limit
B - Upper Yield Stress
C - Lower Yield Stress
D -Ultimate Stress
E -Breaking Stress | | | 4.1. Elasticity and Elastic Limit
Elasticity of a body is the property of the body by virtue of which the body regains its original size and shape when the deformation force is removed. Most materials are elastic in nature to a lesser or greater extend, even though perfectly elastic materials are very rare.
The maximum stress upto which a material can exhibit the property of elasticity is called the elastic limit. If the deformation forces applied causes the stress in the material to exceed the elastic limit, there will be a permanent set in it. That is the body will not regain its original shape and size even after the removal of the deforming force completely. There will be some residual strain left in it. | | | <table cellSpacing=0 cellPadding=0 width=336 border=0><tr><td vAlign=top>
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Yield stress
When a specimen is loaded beyond the elastic limit the stress increases and reach a point at which the material starts yielding this stress is called yield stress.
Ultimate stress
Ultimate load is defined as maximum load which can be placed prior to the breaking of the specimen. Stress corresponding to the ultimate load is known as ultimate stress.
Working stress
Working stress= Yield stress/Factor of safety.
| | | 4.2. Hooke’s Law
Hooke’s law states that stress is proportional to strain upto elastic limit. If I is the stress induced in a material and e the corresponding strain, then according to Hooke’s law,
p / e = E, a constant.
This constant E is called the modulus of elasticity or Young’s Modulus, (named after the English scientist Thomas Young).
It has later been established that Hooke’s law is valid only upto a stress called the limit of proportionality which is slightly less than the elastic limit.
| | | 4.3. Elastic Constants
Elastic constants are used to express the relationship between stresses and strains. Hooke’s law , is stress/strain = a constant, within a certain limit. This means that any stress/corresponding strain = a constant, within certain limit. It follows that there can be three different types of such constants. (which we may call the elastic constants or elastic modulae) corresponding to three distinct types of stresses and strains. These are given below.
(i)Modulus of Elasticity or Young’s Modulus (E)
Modulus of Elasticity is the ratio of direct stress to corresponding linear strain within elastic limit. If p is any direct stress below the elastic limit and e the corresponding linear strain, then E = p / e.
(ii)Modulus of Rigidity or Shear Modulus (G)
Modulus of Rigidity is the ratio of shear stress to shear strain within elastic limit. It is denoted by N,C or G. if q is the shear stress within elastic limit and f the corresponding shear strain, then G = q / f.
(iii) Bulk Modulus (K)
Bulk Modulus is the ratio of volumetric stress to volumetric strain within the elastic limit. If pv is the volumetric stress within elastic limit and ev the corresponding volumetric strain, we have K = pv / ev. |
5. Poisson’s Ratio
Any direct stress is accompanied by a strain in its own direction and called linear strain and an opposite kind strain in every direction at right angles to it, lateral strain. This lateral strain bears a constant ratio with the linear strain. This ratio is called the Poisson’s ratio and is denoted by (1/m) or µ.
Poisson’s Ratio = Lateral Strain / Linear Strain.
Value of the Poisson’s ratio for most materials lies between 0.25 and 0.33. | | | 6. Complementary Strain
Consider a rectangular element ABCD of a body subjected to simple shear of intensity q as shown. Let t be the thickness of the element. Total force on face AB is , FAB = stress X area = q X AB X t. Total force on face CD is, FCD = q X CD Xt = q XAB Xt.
FAB and FCD being equal and opposite, constitute a couple whose moment is given by,
M =FAB X BC = q XAB X BC X t
Since the element is in equilibrium within the body, there must be a balancing couple which can be formed only by another shear stress of some intensity q’ on the faces BC and DA. This shear stress is called the complementary stress.
FBC = q’ X BC X t
FDA = q’ X DA X t = q’ X BC Xt
The couple formed by these two forces is M’ = FBC X AB = q’ X BC X t
For equilibrium, M’ = M.
Therefore q’ = q
This enables us to make the following statement.
In a state of simple shear a shear stress of any intensity along a plane is always accompanied by a complementary shear stress of same intensity along a plane at right angles to the plane. | | | <table cellSpacing=0 cellPadding=0 width=336 border=0><tr><td vAlign=top>
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</TD></TR></TABLE> | | | 7. Direct Stresses Developed Due to Simple Shear.
Consider a square element of side a and thickness t in a state of simple shear as shown in figure. It is clear that the shear stress on the forces of element causes it to elongate in the direction of the diagonal BD. Therefore a tensile stress of same intensity pt is induced in the elements along BD. ie, across the plane of the diagonal AC. The triangular portion ABC of the element is in equilibrium under the action of the following.  (1)FAC = Normal force on face AC = pt X AC X t = pt X v2 aXt (2)FAB = Tangential force on face AB = q X BC X t = q aXt (3)FBC = Tangential force on face BC = q X BC X t = q aXt
For equilibrium in the direction normal to AC,
FAC – FAB cos45 – FBC cos45 = 0
Pt X v2 at – q at X 1/v2 - q at X 1/v2 = 0
v2 pt – 2 q /v2 = 0
pt = q
It can also be seen that the shear stress on the faces of the element causes it to foreshorten in the direction of the diagonal BD. Therefore a compressive stress pc is induced in the element in the direction AC, ie across the plane of the diagonal BD. It can also be shown that pc = q.
It can thus be concluded that simple shear of any intensity gives rise to direct stresses of same intensity along the two planes inclined at 45º to the shearing plane. The stress along one of these planes being tensile and that along the other being compressive. | | |
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