Unit 2 simple stresses and strain

MECHANICS OF STRUCTURE
Simple Stresses and Strains
Presented by
Mr. P. D. Akarte

Mechanical properties of material
 Introduction :
The Various type of structure and its various member
subjected to various type of load are made from various
compound and composite material. Fundamental concept
likes various types of stress , strain, elastic constant and
property of material are basically needed to understand
the subject in proper way.
 Mechanical Property of material
1. Strength 2. Stiffness 3. Elasticity
4. Plasticity 5. Malleability 6. Ductility
7. Brittleness 8. Hardness 9. Toughness
10. Fatigue 11. Creep

Type of Load
Type of load
Direct Load or
Axial Load
Tensile force or
tensile load
Compressive load
Eccentric
load Shear load

1. Axial or Direct Load : When a load whose line of
action is passing through axis of body
2. Eccentric Load : When load whose line of action is
not passing through axial of body

Shear Force or Load : A Force or load which act tangentially to the
plane of body
Stress : It is the internal resisting force per unit area
Unit ; N/mm2
Strain : It is the ratios of change length to the original length.
Unit : No Unit

Shear stress : ( q)
When two equal and opposite forces acts
tangentially on the body so the body tends
to slide across the section due to this
stress is developed is called as shear
stress.
Unit : N/mm2

Types of Strain
Longitudinal Strain : It is define as the ratio
of change in length to the original length.
Lateral Strain:
It the ratio of change in lateral dimension
to the original dimension
Volumetric strain: Volumetric strain is
defined as the ratio of change in volume
to original volume.

Elasticity : It is define as the property of the
material due to which the material regain its
original shape and size after removal of the
external load is called as elasticity.
Hooke’s law : It is that when a elastic
material is loaded within its elastic limit,
then stress produce is directly proportional
to strain.
Mathematically,

Modulus of Elasticity Or Young’s Modulus
( E )
Within elastic limit, the ratio of stress to strain
is a constant and this constant is called as
constant of proportionality.

Change in length or deformation of body
Suppose a bar of length L is subjected to force of P KN so
that deformation produce shown in below fig.
Where,
P = Force applied.
A = Cross section area.
E = Modulus of elasticity.
L = length of bar.

 Q.1 Calculate longitudinal stress developed in 2 cm
diameter bar undergo tensile force of 120 kN.
Solution:
Diameter d = 2 cm = 2 x 10 = 20 mm.
Tensile force P = 120 kN = 120 x 10³ N
we know,
Stress ( σ ) = P / A A = л /4 x 20²
= 120 x 10³/ 314.15
= -----------

Q.2 Determine the tensile force on a steel bar of
circular cross section 25 mm diameter, if strain
equal to 0.75 x 10-3 . Consider E for steel is 200
Gpa.
Given Data :
1. Diameter of steel bar : 25 mm
2. Strain ( e ) : 0.75 x 10-3
3. E= 200 Gpa = 200 x 103
4. To find : P = ?

 Step 1: To find stress in the bar
We know
σ = E x e
= 200 x 103 x 0.75 x 10-3
= ------ N/mm2
Step 2: To find Tensile force in the bar
σ = P/A
P = σ x A
=

Que: A steel rod of cross sectional area 20 x 20 mm and
length of 1.5m long subjected to a axial pull of 50 kN.
Calculate the Stress developed in rod. Also calculate change
in length. Take E= 200 Gpa.
Solution: -
1. Stress developed in rod = axial pull /area
= 50 x 1000 / 20 x 20
= 12.5 N/mm2
2.Change in length (∂l ) = Pl/AE
= 50 x 1000 x 1500/ 20 x 20 x 200 x 1000
=0.937 mm.

 Q.4 In a tension test a metallic rod of diameter
16 mm. produces an elongation of 48 mm when
subjected to 90 kN load. The length of the bar
is 150 mm. Find modulus of elasticity.
 Given Data
1. D = 16 mm
2. δl = 48 mm
3. P = 90 kN. = 90 x 10³ N
4. L = 150 mm

 Q.5 A steel rod of 25mm in diameter and 2 m
long is subjected to an axial pull of 45 kN. Find
i) the stress
ii) Strain
iii) Elongation.
Take E = 200 Gpa.
Given Data:
1. D = 25 mm
2. L = 2 m
3. P = 45 kN. = 45 x 10³ N.
4. E = 200 Gpa = 200 x 10 ³ Mpa

Q.6 A steel Flat 100 mm wide, 12 mm thick and 5 m
long carries an axial load of 20 kN Find the stress,
strain and change in length.
E = 2.1 x 105 Mpa

Q. 7 A wire 4 mm in diameter, 4 m long is subjected
to an axial pull of 1890 N is stretched by 3mm
under the axial pull. Find stress, strain induced.
Also find modulus of elasticity.
Given data:
1. d = 4 mm
2. L = 4 x 10³ mm.
3. P = 1890 N
4. δl = 3 mm .

 Step 1; To find the stress
σ = P/A A = л /4 x 4²
= 1890 / 12.56
Step 2 : To find the strain
e = δl / l
Step 3 : To find the Modulus of elasticity
E= σ / e

 Q.8 : A load of 6 kN is to be raised with help of
steel cable. Find the minimum diameter of steel
cable if stress is not exceed 110 N/mm2.
 Given Data :
 1. load P = 6 kN.
 2. stress σ = 110 N/mm2.
 3. To find d =?
We know
σ = P/A

We know
σ = P/A
A = P /σ
= 6 x 10³ / 110
A = л/ 4 x d²
54.54 = л/ 4 x d²
d =

 Q.9 A steel rod is subjected to an axial pull of 25
kN Find the maximum diameter if the stress is not
exceed to 100 N /mm2. The length of the rod is
2000 mm and take E = 2.1 x 105 Mpa.

Deformation of body or bars of
cross section subjected to axial load
 Total Elongation : δl = δl1 + δl2 + δl3

 Q. 1 A mild steel stepped bar 6 m length is 25 mm
in diameter of 2 m and 15 mm diameter for
remaining length. It is subjected to a tensile load of
70 kN Calculate the change in length if its modulus
if elasticity is 200 kN/mm2.

 Change in length
δl = δl1 + δl2 + δl3
= (Pl/AE) + (PL/ AE ) + (PL/AE)
=

 Q.2 A Bar as shown in fig. is axially loaded. If
the maximum stress induced in bar is 100 Mpa,
calculate the value of P. Take E = 102 GPa for
both part.

 Q.3 A brass bar having cross section area 1000
mm2 is subjected to axial forces of 50 kN as shown
in fig. Find the total change in length.
Take E = 1.05 x 105 N/ mm2

 Q.4 A bar having cross-section as given in fig.
No. 3 is subjected to a tensile load of 150 kN.
Calculate the change in length of each part along
with the total change in length if E= 2 x 105
N/m. ( W-19)

 Q.5 A bar ABCD of varying cross section is
subjected to an axial pull of 75 kN Part AB is
300 mm long, hollow circular with external
diameter 30 mm and internal diameter 26 mm.
Part BC is square in section of side 15 mm and
is 150 mm long. Part CD is 200 mm long and
solid circular section of diameter 20 mm.
determine deformation of each part and net
deformation.
 Take Eab = 200 Gpa, Ebc = 100 Gpa and
Ecd = 75 Gpa.

Q.6 A circular bar of 1000 mm length has cross –
section as given below :
First 200 mm has a diameter 10 mm, second 500
mm has a diameter 25 mm and the last 300 mm
has diameter of 15 mm. Determine the maximum
axial pull which the bar may be subjected if the
maximum stress is limited to 150 MPa. Find total
elongation of the bar. Take E = 200 GPa. (S-18)

Numerical on forced applied in
intermediate section
 Q.1 Find the total elongation of the bar
shown in fig. Take E = 200 kN/mm2.

 Solution:
 Given Data:
d1 = 30 mm. P1 = 50 kN = 50 x 103 N
d2 = 50 mm. P2 = 30 kN = 30 x 103 N
d2 = 15 mm. P3 = 10 kN = 10 x 103 N
L1 = 2 m = 2 x 103 mm
L1 = 2 m = 2 x 103 mm
L1 = 2 m = 2 x 103 mm

Total Change in length
= (Pl/AE) + (PL/ AE ) + (PL/AE)
=

Q.2 A Compound bar having steel rod of dia. 35 mm and
solid copper rod of dia. 20mm and aluminum square rod
of 10 mm is as shown in following figure. Find change
in length of bar. Take modulus of elasticity Es = 210
kN/mm2, Ec = 110 GPa and Eal = 70GPa. (S-19)

1. To find unknown force P
∑Fx=0
-30+P-5+10=0
P-25=0 P=25kN

 Q.3 A bar of uniform cross section area 100 mm2 is
subjected to axial forces as shown in fig . Calculate
the net change in length of the bar. Take E = 25 x
105 N/mm2.

1) To calculate, P
∑Fx=0
1-P + 4 -2 =0
P = 5 - 2
P = 3kN
Total Change in length
= (Pl/AE) + (PL/ AE ) + (PL/AE)
=

 Q.4 A Composite bar comprising of
aluminum and steel as show in fig. Find
the value of P if net deformation produced
in the bar is 2 mm. Take Es = 2 x 105
N/mm2 and Eal = 7 x 104 N/mm2.

 Q.5 Determine the magnitude of P for
equilibrium and total elongation of the
bar shown in fig. Take E = 210 Gpa.
Also calculate minimum stress
induced.

 Q. 6 A compound bar having steel rod of
dia of 35 mm and copper rod of 20 mm
and aluminum square rod of 10 mm is
as shown in fig. Find the change in
length of bar take modulus of elasticity
Es = 210 Gpa, Ec = 110 Gpa and Eal =
70 Gpa. ( S-19)

Stress Strain Curve for Mild Steel

A : Proportionality limit
B : Elastic limit.
C & D : Yield point.
E : Ultimate stress
F : Breaking or failure point.

Stress Strain Curve for HYSD Bar
 2 : Proportionality Limit
 4: Breaking point.

Factor of safety
 It is the ratio of maximum stress to the
working stress.
FOS = Max stress / working stress
 It is based on the yield point stress.
For ductile material
FOS = Yield point stress/ working Stress

Composite Section
 A structural member consist of two of
more dissimilar material joined
together to act as a unit is called as
composite section

Modular Ratio
 It is the ratio of young’s modulus of
elasticity of two different material in
construction by composite material.
m = (E1 / E2 )

Stresses in Composite
material
 Procedure to Solve Numerical
1) Total Load
P = P1 + P2
P = σ 1 A1 + σ2 A2
2) Compare Elongation for both material
δl1 = δl2
Make relation between stresses of both material
using modular ration
m = (E1 / E2 )

 Q.1. A steel cube 40 mm inside diameter and 4
mm metal thickness is filled with concrete.
Determine the stress in the each material due
to an axial thrust of 100 kN.
Take Es = 2.1 x 105 N/mm2
and Ec = 0.14 x 105 N/mm2.
d
D

1 ) Total Load
P = P1 + P2
P = σ 1 A1 + σ2 A2
As= л/4 ( D2 - d2 )
Ac= л/4 x d2
2) Compare Elongation of both material
δls = δlc
(Pl/AE) s = (Pl/AE)c
σs / Es = σc / Ec ------------ σ = P /A
σs / 2.1 x 105 = σc / 0.14 x 105
σs = ----------- σc

Total Load
P = P1 + P2
100 x 103 = σs As + σc Ac

 Q.2 A composite Bar consisting of steel rod 30
mm in diameter enclosed in a copper tube of
external diameter 60 mm and internal diameter
30 mm. It is subjected to compressive axial load
of 110 kN Calculate stresses developed in steel
and copper. Take Es = 210 kN /mm2. Ec = 105
kN / mm2.
Given:
Es = 210 kN /mm2.
= 210 x 103 N / mm2
Ec = 105 kN / mm2.
= 105 x 103 N / mm2

1 ) Total Load
P = P1 + P2
P = σ 1 A1 + σ2 A2
Ac= л/4 ( D2 - d2 )
As= л/4 x d2
δls = δlc
σs / Es = σc / Ec ------------ σ = P /A
σs / 210x 103 = σc / 105 x 103
σs = ----------- σc

Total Load
P = P1 + P2
110 x 103 = σs As + σc Ac

 Q.3 A RCC column 400 x 400 mm is
reinforced with 4 bars of 20mm diameter.
Determine the stresses induced in steel
and concrete. It is subjected to an axial
load of 500 kN and modular ratio is 13.
 Given :
P = 500 kN = 500 x 103 N
Modular ratio m = 13
Es / Ec = 13
Es = 13 Ec

1 ) Total Load
P = Ps + Pc
P = σ s As + σc Ac
As= n л/4 x d2
As= 4 x л/4 x 202
Ag= 400 x400
Ag = As + Ac
400 x 400 = As + Ac
δls = δlc
σs / Es = σc / Ec ------------ σ = P /A

Q.4 RCC column is 300 x 300 mm in section. It
is provided with 8 bars of 20 mm diameter.
Determine the stresses induced in concrete
and steel bars, if it carries a load of 180 kN.
Take Es = 210 Gpa and Ec = 14 Gpa.
 Given :
P = 180 kN = 180 x 103 N
Es = 210 Gpa
= 210 x Mpa
Ec = 14 Gpa
= 14 x 103 Mpa

1 ) Total Load
P = Ps + Pc
P = σ s As + σc Ac
As= n x л/4 x d2
= 8x л/4 x 202
Ag= 300 x300
Ag = As + Ac
300 x 300 = As + Ac
δls = δlc
σs / Es = σc / Ec ------------ σ = P /A

 Q. 5 A square R.C.C. column of 300mm X
300 mm in section with 8 steel bars of 20
mm diameter carries a load of 360 kN. Find
the stresses induced in steel and concrete.
Take modular ratio = 15.
 Given
 A=300×300 mm2 , d =20 mm
 No. of steel bar = 8,
 P =360kN,
 m =15
 Find: σc , σs ,

 Q.6 A RCC Column 450 mm in diameter is
reinforced with 6 bar of 16 mm diameter.
Find safe load that column carry. If
permissible stresses in concrete and steel are
5 N/mm2 and 125 N/mm2 respectively.
Take Ec = 0.14 x 105 N / mm2
Es = 2.1 x 105 N / mm2

Q.7 A steel tube with 40 mm inside diameter
and 4 mm thickness is filled with concrete.
Determine load shared by each material due
to axial thrust of 60 kN.
Take E steel = 210 N/mm2
E concrete = 14 x 103 N/mm2

Q. 8 A brass bar of 250 mm length and 20 mm
diameter is fixed inside a steel tube of 40 mm
external and 20 mm internal diameter and of
same length. The composite bar is subjected to
an axial force pull of 140 kN. Find the stress in
each metal. Take Es = 200 Gpa and Eb =
110Gpa.

Q.9 A composite bar of length 500 mm consist
of a mild steel circular rod of 20 mm diameter
enclosed in a brass tube of 30 mm external
diameter and 22 mm internal diameter. The
composite bar is subjected to an axial pull of
60 kN. Find Stresses in mild steel rod and
brass tube. Es = 210 Gpa and Ebr = 100 Gpa

Temperature stresses and
strain
 Changes in temperature produce expansion
or contraction of materials and result in
thermal strains and thermal stresses
 For most structural materials, thermal strain
εT is proportional to the temperature change
ΔT :
εT = α (ΔT)
α : Coefficient of thermal expansion.

 Temperature Stress :
σ = E α (ΔT)
Temperature Strain :
e = α t
Free deformation or expansion
δl = L α T

 Q.1 Square Rod 10 mm x10mm in cross
section and 100 mm long is fixed at both
ends. Determine end reaction due to rise in
temperature of 50ºc. Take E = 2 x 105 N /
mm2 and α = 12 x106 /ºC

 Q.2 A rod of 10 m long at 10ºc is heated to
70ºc. If the free expansion is prevented, find
the magnitude and nature of stress induced.
Take E = 2.1 x 105 N /mm2. and
α = 12 x106 /ºC

Q.3 A rod is 2 m long at 10ºc. Find the
expansion of the rod when temperature is
raised to 80ºc. If this expansion is prevented
the stress in the material. Take E = 2 x 105
N /mm2 and α = 0.000012 Per ºc.

 Q.4 A steel rod 15m long is at
temperature of 15ºc. Find the free
expansion of the length when the
temperature is raised to 65ºc. Find the
temperature stresses when the expansion
of the rod is fully prevent.
Take E = 2 x 105 N /mm2. and
α = 12 x106 /ºC

 Q.5 An aluminum rod of 22 mm diameter
is fixed at both end at the temperature of
150ºC. Find the stress and force induced
along with nature in rod when temperature
fall to 100ºC and 30ºC.
 Take Ea = 70 Gpa and α = 23 x10-6 /ºC

 Q.6 A metal rod 16 mm diameter, 1500 mm
long is loosely held. If support slip by 0.5
mm due to rise of temperature of 60ºC, Find
stress developed in the rod and its nature.
Take E = 110 Gpa and Coefficient of
thermal expansion α = 10 x10-6 /ºC.

Strain Energy
 When a body is subjected to gradual, sudden
or impact load, the body deforms and work is
done upon it. If the elastic limit is not exceed,
this work is stored in the body. This work
done or energy stored in the body is called
strain energy
 Energy is stored in the body during
deformation process and this energy is called
“Strain Energy”.
 Strain energy = Work done

Resilience
 Total strain energy stored in a body is called
resilience.
 ∴ 𝐮 = σ² / 2E x V
Where, σ = stress V = volume of the body
Proof Resilience :
Maximum strain energy which can be stored
in a body is called proof resilience.
𝐮p = (𝛔𝐄)² x V/ 𝟐𝐄

Modulus of Resilience :
 Maximum strain energy which can be stored
in a body per unit volume, at elastic limit is
called modulus of resilience.
𝐮m = (𝛔𝐄)² / 𝟐

Application of load
1.Gradually applied Load
It is the type of loading in which load is
applied on the bar from zero and slowly
increasing uniformly is called as gradually
applied load.
∴ σ = P / A
2. Suddenly applied Load
3. Impact Load

Unit 2 simple stresses and strain

More Related Content

What's hot (20)

Similar to Unit 2 simple stresses and strain (20)

Recently uploaded (20)

Unit 2 simple stresses and strain