This PPT contain the basic topic about the strength of the material. Such as stress, strain, energy, principle of super position and various other topic of solid mechanics.

1. Mechanics of
solid

2. STRESS
Stress is defined as force per unit area.
𝜎 =
𝐹
𝐴
Normal stress
Compressive and tensile stresses constitute normal stress because the
force vector from which they are calculated is normal (perpendicular) to
the area on which it acts.
Tensile stress
Compressive stress
Shear stress
Shearing Stress is a type of stress that acts coplanar with a cross
section of material. Mathematically
𝜏 =
𝐹
𝐴

3. STRAIN
A change in shape due to the application of force is known as
deformation. Strain is the amount of deformation experienced by the
body in the direction of force applied, divided by the initial dimensions
of the body.
𝐿
∆𝐿 + 𝐿
The deformation that is easiest to observe is the stretching of a bar
made of a single material when a tensile force is applied to it. Suppose
that a force pulling on a bar of initial length L stretches it by an amount
ΔL, as in Fig. This quantity is known as longitudinal strain and is usually
denoted ε, that is
𝑆𝑡𝑟𝑎𝑖𝑛 𝜀 =
∆𝐿
𝐿
Shear strain
Shear strain can mathematically written as
𝛾 =
𝐹
𝐺
G is called modulas of rigidity
Poisson’s ratio
The ratio of lateral strain to longitudinal strain is as poisson’s ratio.
Mathematically
𝑚 = −
𝜀𝑙𝑎𝑡
𝜀𝑙𝑜𝑛𝑔

4. RELATION BETWEEN STRESS AND STRAIN
Hooke’s law; modulus of elasticity
the stress is directly proportional to the strain (within elastic limit) , and
we can write
𝜎 ∝ 𝜀 → 𝜎 = 𝐸𝜀
𝐸 =
𝜎
𝜀
=
𝐹𝐿
𝐴∆𝐿
The coefficient E is called the modulus of elasticity of the material
involved, or also Young’s modulus.
Stress-strain diagram
to distinguish some common characteristics among the stress strain diagrams of
various groups of materials and to divide materials into two broad categories on
the basis of these characteristics, namely, the ductile materials and the brittle
materials.
• Ductile materials, which comprise structural steel, as well as many alloys of
other metals, are characterized by their ability to yield at normal
temperatures.
• As the specimen is subjected to an increasing load, its length first increases
linearly with the load and at a very slow rate. Thus, the initial portion of the
stress-strain diagram is a straight line with a steep slope.
• after a critical value of the stress has been reached, the specimen undergoes
a large deformation with a relatively small increase in the applied load.
• The elongation of the specimen after it has started to yield can be 200 times
as large as its deformation before yield.
• After a certain maximum value of the load has been reached, the diameter of
a portion of the specimen begins to decrease. This phenomenon is known as
necking.

5. RELATION BETWEEN STRESS AND STRAIN
Stress-strain diagram
to distinguish some common characteristics among the stress strain diagrams of
various groups of materials and to divide materials into two broad categories on
the basis of these characteristics, namely, the ductile materials and the brittle
materials.
• Brittle materials, which comprise cast iron, glass, and stone, are characterized
by the fact that rupture occurs without any noticeable prior change in the
rate of elongation.
• Thus, for brittle materials, there is no difference between the ultimate
strength and the breaking strength.
• Also, the strain at the time of rupture is much smaller for brittle than for
ductile materials.

6. STRAIN ENARGY AND IMPACT LOADING
Strain energy when a load is applied gradually
Strain Energy is defined as the internal work done to deform a body by the
action of externally applied forces. Let’s say we have an axially loaded bar on
which load is applied gradually which increases from 0 to P and due to this load
the length of the bar increases by ΔL.
We can write, work done due to force P
𝑑𝑈 = 𝑃∆𝐿
Total work
𝑈 =
0
𝐿
𝑃∆𝐿
In the case of a linear and elastic deformation, the portion of the load-
deformation diagram involved can be represented by a straight line of equation
𝑃 = 𝑘𝐿
𝑈 =
0
𝐿
𝑘𝐿∆𝐿 = 𝑘𝐿
𝐿
2
=
𝑃𝐿
2
It can also interpreted as
work done = average load × deformation
𝑈 =
𝑃
2
∆𝐿 = 𝜎. 𝐴. 𝜀.
𝐿
2
= 𝑉. 𝜎.
𝜀
2
= 𝑉. 𝜎.
𝜎
𝐸
.
1
2
→ 𝑼 =
𝝈𝟐𝑽
𝟐𝑬
Strain Energy stored in a body when a load is applied suddenly
Work done = Strain energy
𝑃 × ∆𝐿 =
𝜎2. 𝐴. 𝐿
2𝐸
𝑃 × 𝜀𝐿 = 𝑃 ×
𝜎
𝐸
. 𝐿 =
𝜎2
. 𝐴. 𝐿
2𝐸
→ 𝜎 = 2
𝑃
𝐴
Then σ =
𝑃
𝐴
for gradually applied load σ =
2𝑃
𝐴
for suddenly applied
load
The stress applied by sudden load is twice the stress of gradual load when the
load is same. So the strain energy will be
𝑼 =
𝟒𝝈𝟐𝑽
𝟐𝑬
The strain energy stored in a body due to suddenly applied load
Compared to when it is applied gradually is four times.

7. STRAIN ENARGY AND IMPACT LOADING
Strain Energy stored in a body when a load is applied with impact
We know,
Work done = force × distance travelled
= 𝑃 ℎ + ∆𝐿
We can write
→ 𝑃 ℎ + ∆𝐿 =
𝜎2
. 𝐴. 𝐿
2𝐸
𝑃ℎ + 𝑃∆𝐿 =
𝜎2
. 𝐴. 𝐿
2𝐸
𝑃ℎ + 𝑃. 𝜀. 𝐿 = 𝑃ℎ + 𝑃.
𝜎
𝐸
. 𝐿 =
𝜎2. 𝐴. 𝐿
2𝐸
𝜎2
2
−
𝜎𝑃
𝐴
−
𝑃ℎ𝐸
𝐴𝐿
= 0
This is a quadratic equation
𝜎 =
𝑃
𝐴
+ 𝑃2 𝐴2 + 4 1 2 𝑃ℎ𝐸 𝐴𝐿
1
= 𝑃 𝐴 + 𝑃2 𝐴2 + 4 1 2 𝑃ℎ𝐸 𝐴𝐿
𝜎 = 𝑃 𝐴 + 𝑃 𝐴 1 + 2 ℎ𝐸𝐴 𝑃𝐿
𝜎 = 𝑃 𝐴 1 + 1 + 2 ℎ𝐸𝐴 𝑃𝐿
When ΔL is very small as compared to h, then
𝑃ℎ =
𝜎2
. 𝐴. 𝐿
2𝐸
𝜎 =
2𝑃ℎ𝐸
𝐴𝐿
h
P
∆𝐿

8. STRAIN
Once s is obtained, deformation Δl or strain energy can be calculated:
𝑼 =
𝑺𝟐
𝑽
𝟐𝑬
If Δl is very small as compared to h, then,
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝑃ℎ
𝑆2𝑉
2𝐸
= 𝑃ℎ → 𝑆 =
2𝑃ℎ𝐸
𝑉
=
2𝑃ℎ𝐸
𝐴𝐿
Strain Energy stored in a body due to shear stress
We know
Work done = Force × Displacement
𝑤 =
𝑃
2
× ∅ℎ =
1
2
𝑞𝐴 ∅ℎ
𝑤 =
1
2
𝑞ℎ𝐿 ∅ℎ
𝑤 =
1
2
. 𝑞. ∅. ℎ. ℎ. 𝐿
𝑤 =
1
2
𝑞∅𝑉 =
1
2
. 𝑞.
𝑞
𝐺
. 𝑉 = 𝑞2
.
𝑉
2𝐺
𝒘 = 𝒒𝟐.
𝑽
𝟐𝑮
The strain energy stored in a body subjected to shear stress
𝑼 =
𝟏
𝟐
.
𝑽𝝉𝟐
𝑮
∅ ∅
ℎ
𝐿

9. DEFORMATION
A change in shape due to the application of force is known as
deformation.
Deformation due to axial loading
∆𝐿 =
𝐹𝐿
𝐴𝐸
or ∆𝐿 =
𝑃𝐿
𝐴𝐸
Deformation due to own weight
We know,
∆𝐿 =
𝑃𝐿
𝐸𝐴
Specific weight of the bar =
𝑊
𝐴𝐿
Load below small section =
𝑊
𝐴𝐿
× 𝐴𝑥 =
𝑊𝑥
𝐿
Deformation in the small section
=
𝑊𝑥𝑑𝑥
𝐸𝐴𝐿
Deformation of the bar
∆𝐿 =
𝑊
𝐸𝐴
0
𝐿
𝑥𝑑𝑥 =
𝑊
𝐸𝐴𝐿
.
𝑥2
2 0
𝐿
=
𝑊
𝐸𝐴𝐿
.
𝐿2
2
∆𝑳 =
𝟏
𝟐
.
𝑾𝑳
𝑨𝑬
Deformation of uniformly tappering circular bar
We know
∆𝐿 =
𝑃𝐿
(
𝜋
4
)𝑑2𝐸
Change in diameter with respect to length
𝑘 =
𝑑1 − 𝑑2
𝐿
Diameter in the small section
= 𝑑1 − 𝑑1 − 𝑑2 𝐿 𝑥 = 𝑑1 − 𝑘𝑥
Deformation in small section
=
𝑃𝑑𝑥
𝜋
4
𝑑1 − 𝑘𝑥 2𝐸
L
x
dx
d1
d2
dx
L
x
d

10. DEFORMATION
Deformation of the bar
∆𝐿 =
4𝑃
𝜋𝐸
0
𝐿
𝑑𝑥
𝑑1 − 𝑘𝑥 2
= −
4𝑃
𝜋𝑘𝐸
𝑑1
𝑑2
𝑑𝑧
𝑧2
=
4𝑃
𝜋𝐸𝑘
1
𝑧 𝑑1
𝑑2
=
4𝑃
𝜋𝐸
𝑑1 − 𝑑2
𝐿
×
1
𝑑2
−
1
𝑑1
=
4𝑃
𝜋𝐸
𝑑1 − 𝑑2
𝐿
×
𝑑1 − 𝑑2
𝑑1𝑑2
So the deformation of tapering circular bar
∆𝑷 =
𝟒𝑷𝑳
𝝅𝑬𝒅𝟏𝒅𝟐
Deformation of uniformly tappering flat bar
Change in width with respect to length
𝑎 − 𝑏
𝐿
= 𝑘
Width at the small section
𝑎 − 𝑘𝑥
From the definition of deformation at small section
=
𝑃𝑑𝑥
𝐸 𝑎 − 𝑘𝑥 𝑡
Deformation for bar
∆𝐿 =
𝑃
𝐸𝑡
0
𝐿
𝑑𝑥
𝑎 − 𝑘𝑥
∆𝐿 = −
𝑃
𝐸𝑡
𝑎 − 𝑏
𝐿 𝑎
𝑏
𝑑𝑧
𝑧
∆𝐿 = −
𝑃
𝐸𝑡
𝑎 − 𝑏
𝐿
ln
𝑏
𝑎
=
𝑷𝑳
𝒕𝑬 𝒂 − 𝒃
𝐥𝐧
𝒂
𝒃
a
b
t

11. STRESS
Stress on composite bar
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. In case of composite bars, the following
points should be kept in view:
• The extension or contraction of the bar being equal, the strain i.e.
deformation per unit length is also equal. ∆𝐿1 = ∆𝐿2
• The total external load on the bar is equal to the sum of the loads
carried by different materials. 𝑃 = 𝑃1 + 𝑃2
𝐴1
𝐸1
𝑃1
∆𝐿1
𝐴2
𝐸2
𝑃2
∆𝐿2
∆𝐿1 = ∆𝐿2
𝑃1𝐿
𝐸1𝐴1
=
𝑃2𝐿
𝐸2𝐴2
→ 𝑃1 =
𝑃2𝐸1𝐴1
𝐸2𝐴2
From the assumption
𝑃 = 𝑃1 + 𝑃2
𝑃 =
𝑃2𝐸1𝐴1
𝐸2𝐴2
+ 𝑃2 = 𝑃2 1 +
𝐸1𝐴1
𝐸2𝐴2
𝑷 = 𝑷𝟐 𝟏 +
𝑬𝟏𝑨𝟏
𝑬𝟐𝑨𝟐
𝑷 = 𝑷𝟏 𝟏 +
𝑬𝟐𝑨𝟐
𝑬𝟏𝑨𝟏
Again
𝑃1𝐿
𝐸1𝐴1
=
𝑃2𝐿
𝐸2𝐴2
We know
𝜎 =
𝑃
𝐴
Putting these value
𝜎1
𝐸1
=
𝜎2
𝐸2
→ 𝜎1 =
𝐸1
𝐸2
𝜎2
The ratio E1 / E2 is known as modular ratio of the two materials.

12. STRESS
2m 2m
1.5m
25 N
Diameter 2 cm for all bar
𝐸𝑠 = 2.05 𝐺𝑁 𝑚2
𝐸𝑐 = 1.1 𝐺𝑁 𝑚2
We know,
∆𝐿𝑠 = ∆𝐿𝑐 = ∆𝐿𝑠
𝑃 = 𝑃𝑠 + 𝑃𝑐 + 𝑃𝑠 = 2𝑃𝑠 + 𝑃𝑐
So,
∆𝐿𝑠 = ∆𝐿𝑐 = ∆𝐿𝑠
𝑃𝑠𝐿𝑠
𝐸𝑠𝐴𝑠
=
𝑃𝑐𝐿𝑐
𝐸𝑐𝐴𝑐
=
𝑃𝑠𝐿𝑠
𝐸𝑠𝐴𝑠
𝑃𝑠𝐿𝑠
𝐸𝑠𝐴𝑠
=
𝑃𝑐𝐿𝑐
𝐸𝑐𝐴𝑐
𝑃𝑠 =
𝑃𝑐𝐿𝑐
𝐸𝑐𝐴𝑐
×
𝐸𝑠𝐴𝑠
𝐿𝑠
𝐴𝑐 =
𝜋
4
0.02 2 = 3.1415 × 10−4 𝑚2
𝐴𝑠 =
𝜋
4
0.02 2 = 3.1415 × 10−4 𝑚2
𝐴𝑐 = 𝐴𝑠
S C S
We get
𝑃 = 2𝑃𝑠 + 𝑃𝑐
𝑃 = 2
𝑃𝑐𝐿𝑐
𝐸𝑐𝐴𝑐
×
𝐸𝑠𝐴𝑠
𝐿𝑠
+ 𝑃𝑐
25 = 2
𝑃𝑐1.5
1.1 × 109
×
2.05 × 109
2
+ 𝑃𝑐
25 = 2.79𝑃𝑐 + 𝑃𝑐
𝑃𝑐 = 6.59 𝑁
Now
𝑃 = 2𝑃𝑠 + 𝑃𝑐
𝑃𝑠 =
25 − 6.59
2
= 9.21 𝑁
Stress
𝑆𝑠 =
𝑃𝑠
𝐴𝑠
=
9.21
3.1415 × 10−4
= 29.317 𝑘𝑁/𝑚2

13. STRESS
Thermal stresses in simple bar
The thermal stresses or strains, in a simple bar, may be found out as
discussed below:
• Calculate the amount of deformation due to change of temperature
with the assumption that bar is free to expand or contract.
• Calculate the load (or force) required to bring the deformed bar to
the original length.
• Calculate the stress and strain in the bar caused by this load.
∆𝐿
𝜎
𝐿
Let
L = Original length of the body
T = Increase in temperature
⍺ = Coefficient of linear expansion
We know that increase in length due
increase in temperature
∆𝐿 = 𝐿. 𝛼. 𝑡
If the ends of the bar are fixed to rigid supports, so that its expansion
is prevented, then compressive strain induced in the bar.
𝜺 =
∆𝑳
𝑳
=
𝑳. 𝜶. 𝒕
𝑳
= 𝜶. 𝒕
Stress
𝝈 = 𝑬. 𝜺 = 𝜶. 𝒕. 𝑬
If the supports yeild by Δ then
𝜺 =
∆𝑳 − ∆
𝑳
=
𝑳. 𝜶. 𝒕 − ∆
𝑳
= 𝜶. 𝒕 −
∆
𝑳
Stress
𝝈 = 𝑬 𝜶. 𝒕 −
∆
𝑳

14. STRESS
Thermal stresses in bars of circular Tapering bar
Let
L = Original length of the body
T = Increase in temperature
⍺ = Coefficient of linear expansion
We know that increase in length due
increase in temperature
∆𝐿 = 𝐿. 𝛼. 𝑡
P = Load required to bring the bar to the
orginal length
We know that decrase in length due to load P
∆𝐿 =
4𝑃𝐿
𝜋. 𝐸. 𝑑1. 𝑑2
𝐿. 𝛼. 𝑡 =
4𝑃𝐿
𝜋. 𝐸. 𝑑1. 𝑑2
𝑃 =
𝛼. 𝑡. 𝜋. 𝐸. 𝑑1. 𝑑2
4
Max stress
𝜎𝑚𝑎𝑥 =
𝑃
𝐴
=
𝛼. 𝑡. 𝜋. 𝐸. 𝑑1. 𝑑2
4
𝜋
4
𝑑2
2
=
𝜶. 𝒕. 𝑬. 𝒅𝟏
𝒅𝟐
If we substitute 𝑑1 = 𝑑2 we get 𝝈 = 𝜶. 𝒕. 𝑬
𝑑1
𝑑2
𝐿
𝐿1 𝐿2
Thermal stresses in bars of varying section
We know that as a result of the increase in
temperature, the bar will tend to expand. But since
it is fixed at its ends, therefore it will cause some
compressive stress in the body. Moreover, as the
thermal stress is shared equally by both the
portions, therefore
𝜎1𝐴1 = 𝜎2𝐴2
Total deformation of the body
∆𝐿 = ∆𝐿1 + ∆𝐿2 =
𝜎1𝐿1
𝐸1
+
𝜎2𝐿2
𝐸2

15. NUMERICAL
1. A rod 2m long at a temperature of 10℃. Find the expansion of
the rod when the temparature rise to 80℃ . If the expansion is
prevented. Find the stress induceed in the material.
𝛼 = 0.000012℃; 𝐸 = 1.0 × 105
𝑀𝑁/𝑚2
∆𝐿 = 𝛼. 𝑡. 𝐿 = 0.000012 80 − 10 2 = 1.68 × 10−3
𝑚
Stress induced
𝜎 = 𝐸. 𝑡. 𝛼 = 1.0 × 1011
× 70 × 0.000012 = 8.4 × 107
𝑁 𝑚2
2. Steel rod of 3cm diameter and 5m long is connected to two bricks.
And the rod is maintained at temperature of 95℃. Determine the
stress and pull excerted when temperature falls to 30℃.
i. The ends do not yield
ii. The ends yeild 0.12 cm
Stress
𝜎 = 𝛼. 𝐸. 𝑡 = 3 × 10−2
× 2 × 1011
× 95 − 30 × 11 × 10−6
= 4.29 × 106
𝑁 𝑚2
Stress when ends yield
𝜎 = 𝐸. 𝛼. 𝑡 −
∆
𝐿
= 2 × 1011
× 11 × 10−6
× 65 − 1.2 × 10−3
3 × 10−2
𝜎 = 7.85 × 109 𝑁/𝑚2

16. PRINCIPLE OF SUPER POSITION
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.

17. 50 kN
P kN 120 kN 40 kN 60 kN
0.5 m
0.5 m
0.5 m 0.5 m
50 kN 50 kN
60 kN
60 kN
20 kN
20 kN
140 kN
140 kN
From the above figure
𝑃 = 90 𝑘𝑁
𝐸 = 2 × 105
𝑁 𝑚𝑚2
= 2 × 1011
𝑁 𝑚2
We know
∆𝐿1 =
𝑃1𝐿1
𝐴1𝐸
=
50000 × 0.5
𝜋
4
0.1 2 × 2 × 1011
= 1.6 × 10−5 𝑚
∆𝐿2 =
𝑃2𝐿2
𝐴2𝐸
=
140000 × 0.5
𝜋
4
0.2 2 × 2 × 1011
= 1.1 × 10−5
𝑚
∆𝐿3 =
4𝑃𝐿
𝜋𝐸𝑑1𝑑2
=
4 × 0.5 × 20000
𝜋 × 2 × 1011 × 0.2 × 0.4
= 7.96 × 10−7
𝑚
∆𝐿4 =
𝑃4𝐿4
𝐴4𝐸
=
0.5 × 60000
𝜋
4
0.4 2 × 2 × 1011
= 1.19 × 10−6
𝑚
Now
∆𝐿 = ∆𝐿1 + ∆𝐿2 + ∆𝐿3 + ∆𝐿4
= 1.6 × 10−5 + 1.1 × 10−5 + 7.96 × 10−7 + 1.19 × 10−6
= 2.8989 × 10−5
𝑚
10 cm
20 cm
40 cm
20 cm
40 cm
PRINCIPLE OF SUPER POSITION

18. ELASTICITY
Elasticity, ability of a deformed material body to return to its original
shape and size when the forces causing the deformation are removed.
Hooke's law
The modern theory of elasticity generalizes Hooke's law to say that
the strain (deformation) of an elastic object or material is proportional
to the stress applied to it. Mathematically
𝐸 =
𝐹
𝐴
∆𝐿
𝐿
=
𝐹𝐿
𝐴∆𝐿
E is called elastic constant or young modulus
𝐿
∆𝐿 + 𝐿
Relation between K, G and E
The relation among different elastic modulas/constant
• 𝐾 =
𝑚𝐸
3(𝑚−2)
• 𝐺 =
𝑚𝐸
2(1+𝑚)
Here
• m = Poisson’s ratio
• E = Elasticity
• G = Rigidity modulus
• K = Bulk modulus

19. THIN CYLINDRICAL AND SPHERICAL SHELLS
In engineering field, we daily come across vessels of cylindrical and
spherical shapes containing fluids such as tanks, boilers, compressed air
receivers etc. Generally, the walls of such vessels are very thin as
compared to their diameters. These vessels, when empty, are subjected
to atmospheric pressure internally as well as externally. In such a case,
the resultant pressure on the walls of the shell is zero. But whenever a
vessel is subjected to internal pressure (due to steam, compressed air
etc.) its walls are subjected to tensile stresses.
In general, if the thickness of the wall of a shell is less than 1/10th to
1/15th of its diameter, it is known as a thin shell.
Failure of a cylindrical shell due to an internal pressure
if these stresses exceed the permissible limit, the cylinder is likely to
fail in any one of the following two ways as shown in Fig.
1. It may split up into two troughs
2. It may split up into two cylinder
Stresses in thin cylindrical shells
A little consideration will show that the walls of the cylindrical shell
will be subjected to the following two types of tensile stresses:
1. Circumferential stress or Hoop stress
2. Longitudinal stress.

20. THIN CYLINDRICAL AND SPHERICAL SHELLS
Circumferential stress
Consider a thin cylindrical shell subjected to an internal pressure as
shown in Fig. We know that as a result of the internal pressure, the
cylinder has a tendency to split up into two troughs as shown in the
figure. Let
• L = Length of the shell,
• d = Diameter of the shell,
• t = Thickness of the shell and
• p = Intensity of internal pressure.
Total pressure along the diameter (say X-X axis) of the shell,
𝑃 = Intensity of the pressure × area = 𝑝. 𝑑𝑥. 𝐿
and circumferential stress in the shell,
𝜎𝑐 =
Total pressure
Resisting section
=
𝑝. 𝑑. 𝐿
2𝑡𝐿
=
𝑝𝑑
2𝑡
Longitudinal stress
Consider the same cylindrical shell, subjected to the same internal
pressure as shown in Fig. We know that as a result of the internal
pressure, the cylinder also has a tendency to split into two pieces as
shown in the figure.
Total pressure along its length (say Y-Y axis) of the shell
𝑃 = Intensity of the pressure × area = 𝑝.
𝜋
4
. 𝑑2
Longitudinal stress in the shell
𝜎𝑙 =
Total pressure
Resisting section
=
𝑝.
𝜋
4
. 𝑑2
𝜋. 𝑡. 𝑑
=
𝑝𝑑
4𝑡
=
1
2
𝜎𝑐

21. THIN CYLINDRICAL AND SPHERICAL SHELLS
Change in dimension due to internal pressure
We know,
𝜎𝑐 =
𝑝𝑑
2𝑡
; 𝜎𝑙 =
𝑝𝑑
4𝑡
; poison′
s ratio =
1
m
;
ΔL = Change in length; Δd = Change in diameter
Change in length
ΔL = 𝜀𝐿. 𝐿 → εL =
ΔL
L
=
𝜎𝐿
𝐸
−
1
𝑚
.
𝜎𝑐
𝐸
=
1
𝐸
𝑝𝑑
4𝑡
−
1
𝑚
.
𝑝𝑑
2𝑡
ΔL
L
=
1
𝐸
.
𝑝𝑑
2𝑡
1
2
−
1
𝑚
𝚫𝐋 =
𝒑𝒅𝑳
𝟐𝒕𝑬
𝟏
𝟐
−
𝟏
𝒎
𝜺𝑳 =
𝒑𝒅
𝟐𝒕𝑬
𝟏
𝟐
−
𝟏
𝒎
Change in diameter
Δd = 𝜀𝑑. 𝑑 → εd =
Δd
d
=
σc
E
−
1
𝑚
.
𝜎𝐿
𝐸
=
𝑝𝑑
2𝑡𝐸
−
1
𝑚
.
𝑝𝑑
4𝑡𝐸
Δd
d
=
𝑝𝑑
2𝑡𝐸
1 −
1
2𝑚
𝚫𝐝 =
𝒑𝒅𝟐
𝟐𝒕𝑬
𝟏 −
𝟏
𝟐𝒎
𝜺𝒅 =
𝒑𝒅
𝟐𝒕𝑬
𝟏 −
𝟏
𝟐𝒎
Change in volume
ΔV = 𝑉 + ΔV − V
∆𝑉 =
𝜋
4
𝑑 + Δ𝑑 2
𝐿 + ∆𝐿 −
𝜋
4
𝑑2
. 𝐿
∆𝑉 =
𝜋
4
𝑑2
∆𝐿 + 2∆𝑑. 𝑑. 𝐿
∆𝑉
𝑉
=
𝜋
4
𝑑2
∆𝐿 + 2∆𝑑. 𝑑. 𝐿
𝜋/4 𝑑2. 𝐿
=
∆𝐿
𝐿
+ 2
∆𝑑
𝑑
= 𝜀𝐿 + 2𝜀𝑐 . 𝑉

22. NUMERICAL
1. Given that
𝑑 = 800 𝑚𝑚 = 0.8 𝑚; 𝑡 = 10 𝑚𝑚 = 0.01; 𝑃 = 2.5 𝑀𝑃𝑎 = 2.5 × 106
𝑁/𝑚2
𝜎𝑐 =
𝑃𝑑
2𝑡
=
2.5 × 109
× 0.8
2 × 0.01
= 1 × 108 𝑁/𝑚2
𝜎𝑙 = 0.5 × 𝜎𝑙 = 1 × 1011
× 0.5 = 5 × 107
𝑁/𝑚2
2. Given that
𝑑 = 40 𝑚𝑚 = 0.04 𝑚; 𝑡 = 5 𝑚𝑚 = 0.005; 𝜎𝑙 = 30 𝑀𝑃𝑎 = 30 × 109
𝑁/𝑚2
𝜎𝑙 =
𝑃𝑑
2𝑡
→ 30 × 106
=
𝑃 × 0.04
2 × 0.005
→ 𝑃 =
30 × 106
4
= 7.5 × 106
𝑁/𝑚2
3. Given that
𝑑 = 800 𝑚𝑚 = 0.8 𝑚; 𝑡 = 10 𝑚𝑚 = 0.01; 𝑃 = 2.5 𝑀𝑃𝑎 = 2.5 × 106
𝑁/𝑚2
𝐿 = 4𝑚;
1
𝑚
= 0.25; 𝐸 = 200 × 109
𝑁/𝑚2
Δd =
𝑃𝑑2
2𝐸𝑡
1 −
1
2𝑚
=
2.5 × 106
× 0.8 2
2 × 200 × 109 × 0.01
1 −
0.25
2
Δd = 3.5 × 10−4𝑚 = 0.35 𝑚𝑚
∆𝑙 =
𝑃𝑑𝑙
2𝑡𝐸
1
2
−
1
𝑚
=
2.5 × 106 × 4 × 0.8
2 × 200 × 109 × 0.01
0.5 − 0.25
∆𝑙 = 5 × 10−4
= 0.5 𝑚𝑚

23. THIN CYLINDRICAL AND SPHERICAL SHELLS
Thin spherical shell
Consider a thin spherical shell subjected to an internal pressure as
shown in Fig. We know that as a result of the internal pressure, the
shell is likely to be torn away along the centre of the sphere. Let,
• d = Diameter of the shell,
• t = Thickness of the shell and
• p = Intensity of internal pressure.
𝑃 = Intensity of the pressure × area = 𝑝.
𝜋
4
. 𝑑2
Stress in the shell,
𝜎 =
Total pressure
Resisting section
=
𝑝.
𝜋
4
. 𝑑2
𝜋. 𝑡. 𝑑
=
𝑝𝑑
4𝑡

24. TORTION OF CIRCULAR SHAFT
Shear stress in a circular shaft subjected to torsion
Consider a circular shaft fixed at one end subjected to a torque at
the other end as shown in fig.
We know,
Shear strain = Deformation per unit length
Φ =
𝐴𝐴′
𝐶𝐴
=
𝐴𝐴′
𝑙
=
𝑅. 𝜃
𝑙
Also
Shear stress = 𝜏; Modulas of rigidity = C
𝐶 =
𝜏
Φ
𝜏 = 𝐶. Φ = C.
𝑅. 𝜃
𝑙
𝜏
𝑅
= 𝐶.
𝜃
𝑙
If 𝜏𝑥 is the intensity of shear stress at a distance 𝑥 from the centre of
the shaft
𝜏𝑥
𝑥
=
𝜏
𝑅
= 𝐶.
𝜃
𝑙
Maximum torque transmitted by a circular solid shaft
We know, Shear stress at x
𝜏𝑥
𝑥
=
𝜏
𝑅
Shear stress at that section, 𝜏𝑥 =
𝜏
𝑅
. 𝑥
Turning force at that section
𝐹 = Stress × Area
𝐹 =
𝜏
𝑅
. 𝑥 2𝜋𝑥. 𝑑𝑥
𝐹 =
𝜏
𝑅
. 2𝜋𝑥2
. 𝑑𝑥

25. TORTION OF CIRCULAR SHAFT
𝐹 =
𝜏
𝑅
. 2𝜋𝑥2
. 𝑑𝑥
Turning moment at this section
𝑑𝑇 =
𝜏
𝑅
. 2𝜋𝑥2. 𝑑𝑥 . 𝑥
𝑑𝑇 =
𝜏
𝑅
. 2𝜋𝑥3
. 𝑑𝑥
0
𝑇
𝑑𝑇 =
0
𝑅
𝜏
𝑅
. 2𝜋𝑥3
. 𝑑𝑥
𝑇 =
2𝜋𝜏
𝑅
𝑥4
4 0
𝑅
=
2𝜋𝜏
𝑅
𝑅4
4
𝑇 =
𝜋𝜏𝑅3
2
=
𝜋
16
× 𝐷3
× 𝜏
Polar moment of inertia
Reletion between angle of twist and torque
𝑇 =
𝜋
16
× 𝜏 × 𝐷3
→ 𝜏 = 𝑇 ×
16
𝜋
×
1
𝐷3
Also
𝜏 = 𝐶.
𝑅. 𝜃
𝑙
Equiting the both equation
𝑇 ×
16
𝜋
×
1
𝐷3 = 𝐶.
𝑅. 𝜃
𝑙
𝑇 ×
16
𝜋
×
1
𝐷3
×
1
𝑅
= 𝐶.
𝜃
𝑙
𝑇32
𝜋𝐷4 =
𝐶𝜃
𝑙
→
𝑇
𝜋
32
𝐷4
=
𝐶𝜃
𝑙
We consider
𝐽 =
𝜋
32
𝐷4
This is called polar moment of inertia.
Then we get
𝑻
𝑱
=
𝑪𝜽
𝒍
=
𝝉
𝑹

26. TORTION OF CIRCULAR SHAFT
Maximum torque transmitted by a circular hollow shaft
𝑇 =
𝜋
16
× 𝜏 ×
𝐷4
− 𝑑4
𝐷
Polar moment of inertia
𝐽 =
𝜋
16
𝐷4
− 𝑑4
Power transmitted by a shaft
We have already discussed that the main purpose of a shaft is to
transmit power from one shaft to another.
𝑃 =
2𝜋𝑁𝑇
60
The unit will be watt
T = Torque
N = r. p. m

27. BENDING MOMENT AND SHEAR FORCE
Shear force
Shear force is the force that acts perpendicular to the longitudinal axis
of a structural member, such as a beam or a column, and tends to
cause the member to slide or shear along its cross section. In other
words, it is the force that is exerted on a section of a structure
perpendicular to the plane of the section. Shear force is usually
denoted by the symbol V and its unit is Newtons (N) or pounds (lb).
• When a force acts downwards on the left-hand side of the section,
the shear force is positive (+) and acts upwards on the right-hand
side of the section. Conversely, when a force acts upwards on the
left-hand side of the section, the shear force is negative (-) and acts
downwards on the right-hand side of the section.
• The shear at any given point of a beam is positive when the external
forces (loads and reactions) acting on the beam tend to shear off
the beam at that point as indicated in Fig.
• In other words, the sign convention of shear force is such that it is
positive when it tends to create a clockwise moment on the section,
and negative when it tends to create an anticlockwise moment on
the section.

28. BENDING MOMENT AND SHEAR FORCE
Bending moment
Bending moment, on the other hand, is the moment that tends to
cause a structural member to bend or deform around a neutral axis. It
is the algebraic sum of the moments about the neutral axis of all the
forces acting on one side of the section. Bending moment is usually
denoted by the symbol M and its unit is Newton-meters (Nm) or
pound-feet (lb-ft).
• The sign convention of bending moment is based on the direction
of the moment acting on a section of a structure. If the moment is
causing clockwise rotation on a section, it is considered negative,
and if it is causing counter clockwise rotation, it is considered
positive.
• The bending moment at any given point of a beam is positive when
the external forces acting on the beam tend to bend the beam at
that point as indicated in Fig.
• For example, if a beam is supported at both ends and a load is
applied at the midpoint of the beam, the bending moment will be
maximum at the midpoint and will be negative on the right side of
the section and positive on the left side of the section, as the
moment is causing clockwise rotation on the right side and counter
clockwise rotation on the left side.

29. BENDING MOMENT AND SHEAR FORCE
Shear force and bending moment diagram
A shear force diagram (SFD) shows the variation of shear force
along the length of a structural member. The diagram is plotted with
shear force on the y-axis and the length of the member on the x-axis.
The diagram shows positive values of shear force as upward or
rightward forces and negative values of shear force as downward or
leftward forces.
A bending moment diagram (BMD) shows the variation of bending
moment along the length of a structural member. The diagram is
plotted with bending moment on the y-axis and the length of the
member on the x-axis.
Simply supported beam with a point load at its mid point:
𝐴 𝐵
𝑊
𝐿
𝑅𝐴 =
𝑊
2
𝑅𝐵 =
𝑊
2
Shear force at A and B
Since the load is in the mid point, the
reaction force will be,
𝑅𝐴 = 𝑅𝐵 =
𝑊
2
Bending at poin C with respect to A
and B is same
𝑀𝑐 = 𝑅𝐴 × 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
𝑊
2
×
𝐿
2
=
𝑊𝐿
4
𝐶
𝐶
𝑊𝐿
4
Simply supported beam with a point load at any point:
𝐴 𝐵
𝑊
𝐿
𝑅𝐴 =
𝑊𝑏
𝐿
𝑅𝐵 =
𝑊𝑎
𝐿
𝐶
𝑏
𝑎
Reaction force at A
𝑅𝐴 × 𝐿 − 𝑊 × 𝑏 = 0; 𝑅𝐴 =
𝑊𝑏
𝐿
Reaction force at B
𝑅𝐵 × 𝐿 − 𝑊 × 𝑎 = 0; 𝑅𝐵 =
𝑊𝑎
𝐿
Bending moment at C
𝑀𝐶 = 𝑅𝐴 × 𝑎 =
𝑊𝑎𝑏
𝐿
; 𝑀𝐶 = 𝑅𝐵 × 𝑏 =
𝑊𝑎𝑏
𝐿
𝐶
𝑊𝑎𝑏
𝐿

30. BENDING MOMENT AND SHEAR FORCE
Reaction force at A
𝑅𝐴 × 2.5 − 2 × 1.5 − 4 × 1 = 0; 𝑅𝐴 =
7
2.5
= 2.8
Reaction force at B
𝑅𝐵 × 2.5 − 4 × 1.5 − 2 × 1 = 0; 𝑅𝐵 = 3.2
Bending moment at C
𝑀𝐶 = 𝑅𝐴 × 1 = 2.8 × 1 = 2.8
Bending moment at D
𝑀𝐷 = 𝑅𝐴 × 1.5 = 2.8 × 1.5 − 2 × 0.5 = 3.2
Shear force diagram
𝐹𝐴 = +𝑅𝐴 = +2.8;
𝐹𝐶 = 2.8 − 2 = 0.8;
𝐹𝐷 = 0.8 − 4 = −3.2;
𝐹𝐵 = −3.2;
Bending moment diagram
𝑀𝐴 = 𝑀𝐵 = 0;
𝑀𝐶 = 2.8;
𝑀𝐷 = 3.2

31. BENDING MOMENT AND SHEAR FORCE
Simply suppoted beam with uniformly distributed load
𝐴 𝐵
𝑤 = 𝑢𝑛𝑖𝑡 𝑙𝑜𝑎𝑑
𝐿
𝑅𝐴 =
𝑤𝐿
2
𝑅𝐵 =
𝑤𝐿
2
𝐶
𝐶
𝑤𝐿2
8
𝑥
Reaction force at A and B
𝑅𝐴 = 𝑅𝐵 =
𝑤𝐿
2
Shear force diagram
𝐹𝐴 = +𝑅𝐴 = +
𝑤𝐿
2
;
𝐹𝐶 =
𝑤𝐿
2
−
𝑤𝐿
2
= 0;
𝐹𝐵 = −
𝑤𝐿
2
;
𝐹𝑥 = −𝑅𝐵 + 𝑤𝑥 = 𝑤𝑥 −
𝑤𝐿
2
Bending moment diagram
𝑀𝐴 = 𝑀𝐵 = 0;
𝑀𝐶 = 𝑅𝐴.
𝐿
2
=
𝑤𝐿2
4
−
𝑤𝐿
2
.
𝐿
4
=
𝑤𝐿2
8
;
𝑀𝑥 = 𝑅𝐵. 𝑥 =
𝑤𝐿𝑥
2
− 𝑤𝑥.
𝑥
2
=
𝑤𝐿𝑥
2
−
𝑤𝑥2
2
𝐴 𝐵
𝑤 = 1000 𝑘𝑔/𝑚
4𝑚
𝐶 𝐸
𝑅𝐴 = 𝑅𝐵 =
1000 × 4
2
= 2000
SFD
𝐹𝐴 = 2000;
𝐹𝑐 = 2000 − 1000 × 2 = 0:
𝐹𝐵 = −2000;
BMD
𝑀𝐴 = 𝑀𝐵 = 0
𝑀𝐷 = 2000 × 1 − 1000 × 1 ×
1
2
= 1500;
𝑀𝐶 = 2000 × 2 − 1000 × 2 ×
2
2
= 2000;
𝐷
1 1
𝐶
𝑤𝐿2
8

32. BENDING MOMENT AND SHEAR FORCE
𝐴 𝐵
𝑤 = 500 𝑘𝑔/𝑚
4
𝐶 𝐸
𝐷
2 2
1000 𝑁 1000 𝑁
Reaction force at A and B
𝑅𝐴 × 8 = 500 × 4 × 4; 𝑅𝐴 =
4000
8
= 1000
𝑅𝐵 = 1000 𝑘𝑔
SFD
𝐹𝐴 = +1000;
𝐹𝐷 = +1000;
𝐹𝐶 = 1000 − 500 × 2 = 0;
𝐹𝐸 = −500;
𝐹𝐵 = −2000;
BMD
𝑀𝐴 = 𝑀𝐵 = 0;
𝑀𝐷 = 1000 × 2 = 2000;
𝑀𝐶 = 1000 × 4 − 500 × 2 ×
2
2
= 3000;
𝑀𝐸 = 1000 × 6 − 500 × 4 ×
4
2
= 2000;
𝐶 𝐸
𝐷
𝐴 𝐵
𝑤 = 500 𝑘𝑔/𝑚
4
𝐶
𝐸
𝐷
2 2
1000 𝑁 1000 𝑁
Reaction force at B (taking moment at
A)
𝑅𝐵 × 4 + 1000 × 2
= 500 × 4 × 2 + 1000 × 6
𝑅𝐵 =
8000
4
= 2000 𝑘𝑔
𝑅𝐴 = 𝑅𝐵 = 2000 𝑘𝑔
SFD
𝐹𝑐 = −1000;
𝐹𝐴 = −1000 + 2000 = 1000
𝐹𝐸 = 1000 − 500 × 2 = 0
𝐹𝐵 = 0 − 500 × 2 + 2000 = 1000
𝐹𝐷 = 1000 − 1000 = 0
BMD
𝑀𝑐 = 0; 𝑀𝐴 = −1000 × 2 = −2000
𝑀𝐸 = −1000 × 4 + 2000 × 2 − 500 × 2 × 1
= −1000
𝑀𝐵 = −1000 × 2 = −2000

33. Reaction force at B taking moment at A
𝑅𝐵 × 5 = 0.5 × 2 × 4 + 1.5 × 2 × 2 + 0.5 × 1 × 0.5 = 10.25
𝑅𝐵 = 2.05 𝑘𝑁
𝑅𝐴 = 0.5 × 1 + 1.5 × 2 + 0.5 × 2 − 2.05 = 2.45 𝑘𝑁
SFD
𝐹𝐴 = 2.45 𝑘𝑁;
𝐹𝐶 = +2.45 − 0.5 × 1 = 1.95 𝑘𝑁;
𝐹𝐷 = 1.95 − 1.5 × 2 = −1.05 𝑘𝑁;
𝐹𝐵 = −2.05
BMD
𝑀𝐴 = 0;
𝑀𝐶 = 2.45 × 1 − 0.5 × 1 × 0.5 = 2.2
𝑀𝐷 = 2.05 × 2 − 0.5 × 2 × 1 = 3.1
𝑀𝐵 = 0
The maximum bending moment is at M, x distance from c where shear
force changes sign.
1.95
𝑥
=
1.05
2 − 𝑥
→ 𝑥 = 1.3 𝑚
𝑀𝑀 = 2.45 × 2.3 − 0.5 × 1 × (0.5 + 1.3) − 1.5 × 1.3 × 0.65 = 3.47

34. Reaction force at A by taking moment at B
𝑅𝐴 × 5 = 10 × 2 × 3
𝑅𝐴 = 12 𝑘𝑁
Reaction force at A
𝑅𝐵 = 10 × 2 − 12 = 8 𝑘𝑁
Reaction force at A by taking moment at B
𝑅𝐴 × 4 = 4 × 2.5 + 2 × 1 × 2
𝑅𝐴 = 3.5 𝑘𝑁
Reaction force at A
𝑅𝐵 = (2 × 1 + 4) − 3.5 = 2.5 𝑘𝑁
Reaction force at A by taking moment at B
𝑅𝐴 × 6 = 4 × 1.5 × 5.25 + 2 × 3 × 1.5 + (5 × 1.5)
𝑅𝐴 = 8 𝑘𝑁
Reaction force at A
𝑅𝐵 = (4 × 1.5 + 5 × 3 + 5) − 8 = 18 𝑘𝑁

35. Reaction force at A by taking moment at B
𝑅𝐴 × 3 = 4.5 × 3 × 1.5 − 4.5 × 1 × 0.5
𝑅𝐴 = 6 𝑘𝑁
Reaction force at A
𝑅𝐵 = 4.5 × 4 − 6 = 12 𝑘𝑁

36. Cantilever with point load at free end
BENDING MOMENT AND SHEAR FORCE
We know that shear force at any
section X from a distance x from
free end, is equal to the total
unbalance vertical force
𝐹𝑥 = 𝑤
𝑀𝑥 = −𝑤. 𝑥
0.5 0.5 0.5
1000 1000 1000
SFD
𝐹𝐵 = 1000; 𝐹𝑐 = 2000; 𝐹𝐷 = 3000
BMD
𝑀𝐵 = 0;
𝑀𝐶 = −1000 × 0.5 = −500
𝑀𝐷 = −1000 × 1 − 1000 × 0.5 = −1500
𝑀𝐴 = −1000 × 1.5 − 1000 × 1 − 1000 × 0.5 = −3000
B
C
D

37. COLUMNS AND STRUTS
Introduction
A structural member, subjected to an axial compressive force, is
called a strut. As per definition, a strut may be horizontal, inclined or
even vertical. But a vertical strut, used in buildings or frames, is called a
column.
Failure of a column or struts
t has been observed, that when a column or a strut is subjected to
some compressive force, then the compressive stress induced,
𝜎 =
𝑃
𝐴
• A little consideration will show, that if the force or load is gradually
increased the column will reach a stage, when it will be subjected to
the ultimate crushing stress. Beyond this stage, the column will fail
by crushing. The load corresponding to the crushing stress, is called
crushing load.
It has also been experienced that sometimes, a compression member
does not fail entirely by crushing, but also by bending i.e., buckling.
This happens in the case of long columns. It has also been observed
that all the short columns fail due to their crushing. But,
• If a long column is subjected to a compressive load, it is subjected to
a compressive stress. If the load is gradually increased, column will
reach a stage, when it will start buckling. The load, at which the
column just buckles is called buckling load, critical load or crippling
load and the column is said to have developed an elastic instability.
• A little consideration will show that for a long column, the value of
buckling load will be less than the crushing load. Moreover, the value
of buckling load is low for long columns and relatively high for short
columns.

38. EULER’S COLUMN THEORY
Assumption
1. Initially the column is perfectly straight and the load applied is truly
axial.
2. The cross-section of the column is uniform throughout its length.
3. The column material is perfectly elastic, homogeneous and isotropic
and thus obeys Hooke's law.
4. The length of column is very large as compared to its cross-
sectional dimensions.
5. The shortening of column, due to direct compression (being very
small) is neglected.
6. The failure of column occurs due to buckling alone.
Sign convention
Types of end conditions of column
1. Both ends hinged,
2. Both ends fixed,
3. One end is fixed and the other hinged, and
4. One end is fixed and the other free.

39. COLUMN WITH BOTH ENDS HINGED
Consider a long column AB with a length of l is
hinged at its both ends. Its carrying a critical load P
at B. As a result of loading the column deflect into a
curve AX1B as shown in Fig.
Now consider a secttion X at a distance x from A.
𝑃 = Load on the column
𝑦 = Deflection of the column at X
Moment due to critical load
𝑀 = −𝑃. 𝑦
𝐸𝐼
𝑑2
𝑦
𝑑𝑥2
= −𝑃. 𝑦
𝐸𝐼
𝑑2𝑦
𝑑𝑥2
+ 𝑃. 𝑦 = 0
𝑑2𝑦
𝑑𝑥2 +
𝑃. 𝑦
𝐸𝐼
= 0
The general solution for above differential equation,
𝑦 = 𝐶1. cos 𝑥.
𝑃
𝐸𝐼
+ 𝐶2 sin 𝑥.
𝑃
𝐸𝐼
We know, when 𝑥 = 0, 𝑦 = 0, then C1 = 0
When 𝑥 = 𝑙, 𝑦 = 0 then
sin 𝑙.
𝑃
𝐸𝐼
= 0
sin 𝑙.
𝑃
𝐸𝐼
= sin 0 = sin 𝜋 = sin 2𝜋
Taking the least significant value
𝑙.
𝑃
𝐸𝐼
= 𝜋 → 𝑃 =
𝝅𝟐𝑬𝑰
𝒍𝟐

40. EULER’S FORMULA AND EQUIVALENT LENGTH
In the previous articles, we have derived the relations for the crippling
load under various end conditions. Sometimes, all these cases are
represented by a general equation called Euler's formula,
𝑃𝐸 =
𝜋2
𝐸𝐼
𝐿𝑒
2
Where 𝐿𝑒 equivalent length. The equivalent length for all the condition
are given below.
Conditions
Relation between
𝒍 and 𝑳𝒆
Formula
Both end hinged 𝐿𝑒 = 𝑙 𝑃 =
𝜋2𝐸𝐼
𝑙2
Both end fixed 𝐿𝑒 =
𝑙
2
𝑃 =
4𝜋2𝐸𝐼
𝑙2
One end hinged,
other end fixed
𝐿𝑒 =
𝑙
2
𝑃 =
2𝜋2𝐸𝐼
𝑙2
One end fixed, other
end free
𝐿𝑒 = 2𝑙 𝑃 =
𝜋2𝐸𝐼
4𝑙2
Slenderness ratio
We know that,
𝑃𝐸 =
𝜋2𝐸𝐼
𝐿𝑒
2
Now substituting 𝐼 = 𝐴𝑘2 where 𝐴 is cross sectional area and 𝑘 is the
least redius of gyration.
𝑃𝐸 =
𝜋2
𝐸(𝐴𝑘2
)
𝐿𝑒
2 = 𝑃𝐸 =
𝜋2
𝐴
𝐿𝑒
𝑘
2
Where
𝐿𝑒
𝑘
is called the slenderness ratio.

41. PRINCIPAL STRESSES AND STRAIN
Principal planes
Principal planes are the planes within the material such that the
resultant stresses across them are completely normal stresses or planes
across which the shear stresses is zero.
Principal stress
Principal stresses are those stresses which are acting on the principal
planes.
• The plane carrying the maximum normal stress is called the
major principal plane and the stress acting on it is called major
principal stress.
• The plane carrying minimum normal stress is known as minor
principal plane and the stress acting on it is called as minor
principal stress.
Normal stress and shear stress on obli

42. • আখলাক অর্ থকক?
• ভাললা আখলাক এর উধারণ দাও? ৫টি
• সৃটির সসবা কক?
• সৃটির সসবায় উধাহরন দাও? ৫টি
• "রাহমাতুকিল আলামীন" অর্ থকক?
• তুকম ককভালব সদশ সক ভাললাবাসলব কলখ?
• ক্ষমা কার গুণ?
• মহানবী (স) সকান শহলর দাওয়াত কদলত সেলয় লাকিত হলয়কিললন?
• ক্ষমার গুরুত্ব কলখ?
• সততা কক?
• সতামালদর মলধে সলব থ
াত্তম সসই বোক্তি,

43. • আখলাক অর্ থকক?