strength of material

1. Chapter- : Simple Stresses And Strains
Prepared by:
Trambadiya Dhruv Ashokbhai
Subject : (Mechanics Of Solids)
B.E.– 3rd semester

2. Topic:
* Stresses ,Strains & Elasticity,
* Hooke’s law, stress-strain diagram,
* Prismatic and non- prismatic bars,
* Composite and compound bars,
* Modulus Of Elasticity(E),
* Stress Strain Diagram,
* Stresses in bars of uniformly tapering circular section,
* Deformation of a body due to self weight: (Bar of uniform
section),
* Deformation of a body to self weight:(Bar of linearly varying
cross-section),
* Deformation of uniformly taping rectangular section,
* Bars With Cross-Section Varying In steps
* Stresses in Composite Bars

3. STRESS, STRAIN & ELASTICITY:
* Elasticity:
the ability of an object or material to resume its normal shape
after being stretched or compressed.
* Elastic Material:
If a martial fully recovers its original shap and size after
removal of external force.
* Elasto-plastic Material:
If the deformation exists partially even after the removal of
external force.

4. * Plastic Material:
If a material does not regain its original shape &size at all,
or it remains deformation even after removal of external
force.
* Ductile material:
Till now we have studied that can be drawn into wires is
called ductil.
* Brittle material:
If a material does not undergo any deformation on
application of force and fails due to rupture is called brittle
material.

5. * Stress (6):
It has been noted above that external force applied to a body
in equilibrium is reacted by internal forces set up within the
material. If, therefore, a bar is subjected to a uniform tension
or compression:
Tensile Stress due to tensile force
A tensile force makes the body longer
Compressive Stress due to compressive force
A Compression force makes the body shorter.

6. * Resistance offered by the material per unit cross- sectional
area is called STRESS
* Tensile and compressive forces are called DIRECT FORCES
* Stress is the force per unit area upon which it acts.
Note: Most of engineering fields used kPa, MPa, GPa.
Stress = σ = =

7. 7
* Strain (ε):
In each case, a force ‘F’ produces a deformation ‘x’ . In
engineering, we usually change this force into stress and the
deformation into strain and we define these as follows:
Strain is the deformation per unit of the original length.
Strain, ε = = Change in length/ Original length
(ε is called as Epsilon)
Strain has no unit’s since it is a ratio of length to length
L DL

8. Elastic materials-Hooke's law:
A material is said to be elastic if it returns to its original,
unloaded dimensions when load is removed.
A particular form of elasticity which applies to a large
range of engineering materials, at least over part of their load
range, produces deformations which are proportional to the
loads producing them.
Since loads are proportional to the stresses they produce
and deformations are proportional to the strains.
stress (σ ) α strain (ε )

9. Below the yield stress
Stress α Strain (ie) σ α ε
σ = E ε
Where E is a constant called as
Youngs Modulus or Modulus of
Elasticity

10. MODULUS OF ELASTICITY (E):
* Elastic materials always spring back into shape when
released. They also obey Young’s LAW.
* This is the law of spring which states that deformation is
directly proportional to the force. F/x = stiffness = kN/m
* E=



11. STRESS STRAIN DIAGRAM:

12. Strain
Stress
Stress- Strain Curve for Mild Steel (Ductile Material):
Plastic state
Of material
Elastic State
Of material
Yield stress
Point
E = modulus of
elasticity
Ultimate stress point
Breaking stress point

13. Prismatic and Non- prismatic bars:
* A bar having constant cross section throughout its length
is called prismatic bar.
* A bar which does not have a constant cross-section
throughout the length is called non-prismatic bar.

14. Composite and compound bars:
* If cross-section of bar is made up of tow or more different
materials, it is called composite bar.
* When a bar is made up of two or more materials along the
length of bar however, any cross section of the bar is made of
only one material is called compound bar.

15. Deformation of a body due to self weight:
(Bar of uniform section)
Consider a bar of circular cross section and uniform
diameter throughout. Consider it to be suspended from a
rigid support and its top end, such that it is in a hanging in
a vertical position as shown in the figure.
Let,
A = Uniform cross sectional area of the bar
E = Young’s modulus for the bar
L = Length of the bar
ρ = Weight of the bar, per unit length, for the material of
the bar

16. * Consider an element of length ‘dy’ at a distance of ‘y’ from
the bottom of the bar being elongated due to the force ‘P’, at
section x-x, as shown in the figure.
Weight of the portion below x-x = P = W.X
Change in the length of the element
dX=
* For total change in the length of the bar, we need to
integrate along the length
* Total change in length =

17. On integrating, we get,
δL=
* This is the expression for the elongation of a uniform bar
under self weight.

18. Stresses in bars of uniformly tapering
circular section:
Consider a circular bar of
uniformly reaping section as shown in figure.
Let P = Force applied on bar (Tensile) (N)
L = Length of the bar (m)
d1 = Diameter of the bigger end of the bar (m), and
d2 = Diameter of the smaller end of the bar (m)
Now, consider a small element of length dx, of the
bar, at a distance x from the bigger end as shown
in figure

19. * Consider an infinitesimal element of thickness dx, diameter
Dx at a distance x from face with diameter D1.
Deformation of the element d(Δx)=
* Ax= x Dx²; Dx= D1 -
Let (D1-D2)/L=k; Then Dx= D1-kx
d(ΔLx)=
* Integrating from x=0 to x=L 4PL/(πED1D2)

20. * Let D1-kx=λ; then dx= -(d λ/k)
When x=0, λ=D1; When x=L, λ=D2
ΔLx =

21. Deformation of a body to self weight: (Bar
of linearly varying cross-section)
Let us consider small strip of thickness dy
at distance y from apex having radius ry
force on the section at distance y from te
apex
= y.W
But, ry = ro / L . y
Stress= =
Total downward deflection of apex

22. * Deformation due to self weight
dL= . dy =

23. * Width of the bar at sectionx-x ,
* Area of the sectionx-x=(a-K.x).t
Deformation of uniformly taping
rectangular section:

24. * Stress=
Extension of the small elemental length dx
=
Total elegation from 0 to L,
.dx
Then , put K = a-b / L

25. BARS WITH CROSS-SECTIONS
VARYING IN STEPS:
* A typical bar with cross-sections varying in steps and
subjected to axial load
* length of three portions L1, L2 and L3 and the respective
cross-sectional areas are A1, A2, A3
* E = Young’s modulus of the material
* P = applied axial load.

26. 26
Stress, strain and extension of each of these portions are:
Portio
n
Stress Strain Extension
1 σ1 = P/ A1 e1 = σ1 / E δ1 = P L1 / A1 E
2 σ2 = P/ A2 e2 = σ2 / E δ2 = P L2 / A2 E
3 σ3 = P/ A3 e3 = σ3 / E δ3 = P L3 / A3 E

27. Total Elongation: δ
δ1 + δ2 + δ3 = [P L1 / A1 E] + [P L2 / A2 E] + [P L3 / A3 E]

28. Stresses in Composite Bars:
28
* A composite bar may be defined as a bar made up of two or more
different materials, joined together, in such a manner that the system
extends or contracts as one unit, equally, when subjected to tension
or compression.
* The extension or contraction of
the bar is being equal
* The total external load on the bar is
equal to the sum of the loads carried
by different materials.

29. P1 = Load carried by bar 1,
A1 = Cross-sectional area of bar 1,
σ1 = Stress produced in bar 1,
E1 = Young's modulus of bar 1,
* P2, A2, σ2, E2 = Corresponding values of bar 2,
P = Total load on the composite bar,
l = Length of the composite bar, and
δl = Elongation of the composite bar.
* We know that P = P1 + P2
* Stress in bar 1,
* strain in bar 1,

30. 30
* Elongation in bar -1:
* Elongation in bar -2:
There fore,
δl1 = δl2
 The ratio E1 / E2 is known as modular ratio of the two materials

31. Thank you