# Strength of material Book RK bansal
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STRESSES AND STRAINS
(SIMPLE & PRINCIPAL)
Stress is proportional to strain within its elastic limit. This law is known as Hooke‟s law.
The material will not return to original shape ifthe applied stress is more than E.
ζ αε Stress - ζ Linear Strain – ε
Therefore, ζ = Eε Where E Modulus of Elasticity or Young‟s Modulus.
Stresses are three types tensile, compressive, and shear stress. Moment and
torsion will produced any of these stresses.
Strain is nothing but deformation (change in length, breadth, height, diameter,
therefore area or volume) of the body or material due to load. Therefore strain is change in
dimension to the original dimension. It may be length or volume.
TYPES OF STRESSES : Only two basic stresses exists : (1) normal stress and (2) shear stress. Other stresses either are similar to these basic stresses or are a combination of this e.g. bending stress is a combination tensile, compressive and shear stresses. Torsional stress, as encountered in twisting of a shaft is a shearing stress. Let us define the normal stresses and shear stresses in the following sections.
Normal stresses : We have defined stress as force per unit area. If the stresses are normal to the areas concerned, then these are termed as normal stresses. The normal stresses aregenerally denoted by a Greek letter (ζ)
This is also known as uniaxial state of stress, because the stresses acts only in one direction
however, such a state rarely exists, therefore we have biaxial and triaxial state of stresses
where either the two mutually perpendicular normal stresses acts or three mutually
perpendicular normal stresses acts.
Tensile or compressive Stresses:
The normal stresses can be either tensile or compressive whether the stresses acts out of
the area or into the area
Shear Stresses:
Let us consider now the situation, where the cross – sectional area of a block of material is
subject to a distribution of forces which are parallel, rather than normal, to the area
concerned. Such forces are associated with a shearing of the material, and are referred to as
shear forces. The resulting stress is known as shear stress
Principle of Superposition
The principle of superposition states that when there are numbers of loads are acting together
on an elastic material, the resultant strain will be the sum of individual strains caused by each
load acting separately.
Types of problem
Both ends are free (to expand or shrink) determinate structure: Total change in length is equal to algebraic sum of change in length of each
section of its load P, length L, Area A, and Young‟s modulus E.
These parameters may
vary from section to section. The material is free to expand and shrink.
δL =δ 1+ δ2+ δ3+ …..+ δn
Both ends are fixed (cannot expand or shrink) indeterminate structure:
Total change in length is zero because the ends are fixed which will not allow the
sections to expand or shrink. Load or stress is produced by expansion or shrinkage of the
section istaken bythe ends. Therefore ends carry some load or stress.
Using principle of superposition the reactions at the end of each section is found
from free body diagram. Equate the direction of force in free body diagram to force
applied for each section,
PAB- PBc= P1
PBC + PCD = P2Equations –(A)
The equation shows that the section AB and BC is under tension and CD under
compression. The direction of load in each section can be chosen as we desire, but if the final
result is negative then the direction chosen is incorrect but the answer is correct. So in other
wordstensile force is actually a compressive force vice versa.
Sum of change in length of each section due to expansion is equal to sum of
change in length of each section due to compression. The load P, length L, Area A, and
Young‟s modulus E parameters may vary from section to section.
Expansion section = Compression section
δ1+ δ2+....+ δn= δ3+ δ4+ …..+ δn Equations – (B)
Composite Material of Equal length
Reinforced Columns, Supporting load, Suspended load, Composite structure of
equal length (example pipe inside a pipe) these problems can be solved with the following
expression.
The change length is same for all materials in that structure. Example in
reinforced concrete column (RCC), steel and concrete length change equally, similarly for
supporting load, suspended load, and composite structure of equal length. Therefore to solve these
problems use the following expressions
