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7/27/2019 1-Concept of Stress and Strain.pdf
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
V{tÑàxÜ D
Concept of
Stress & Strain
MEC411 – MECHANICS OF MATERIALS Ch 1 - 1
1. Ferdinand P. Beer, E. Russell Johnston,Jr, John T. Dewolf, David F. Mazurek “ Mechanics of Materials”
5th Edition in SI units
2. R.C.Hibbeler “ Mechanics of Materials “ Seventh Edition
Materials for this chapter are taken from :
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
Introduction
1. Mechanics of materials is a study of the relationship between the external loads on a
body and the intensity of the internal loads within the body.
2. This subject also involves the deformations and stability of a body when subjected to
external forces.
MEC411 – MECHANICS OF MATERIALS Ch 1 - 2
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
External Forces
External Forces
Surface Forces
caused by direct contact of
other body’s surface
Bod Forcesother body exerts a force
MEC411 – MECHANICS OF MATERIALS Ch 1 - 3
w ou con ac
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
Reactions
Surface forces developed at the supports/points of contact between bodies.
MEC411 – MECHANICS OF MATERIALS Ch 1 - 4
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
Equation of Equilibrium
The condition of statics are:
1. the algebraic sum of all forces (or components of forces) in any directionmust equal to zero or ∑ F = 0
2. the algebraic sum of the moments of the forces about any axis or point must
equal to zero or ∑ M = 0.
MEC411 – MECHANICS OF MATERIALS Ch 1 - 5
ese wo con ons can e expresse ma ema ca y as:
( )atanypoint
0
0
0
x
y
F
F
M
=
=
=
∑∑
∑
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
Objective of FBD is to determine the resultant force and moment acting within a
body.
Equilibrium of a Deformable Body
In general, there are 4 different types
of resultant loadings:
Normal force, N
Shear force, V
MEC411 – MECHANICS OF MATERIALS Ch 1 - 6
torque, T
Bending moment, M
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
Example 1.1
Determine the resultant internal loadings
acting on the cross section at C of the
Solution
.
Distributed loading at C is found by
proportion,
270w
MEC411 – MECHANICS OF MATERIALS Ch 1 - 7
Magnitude of the resultant of the
distributed load,
( )( ) N540618021 == F
which acts from C ( ) m2631
=
6 9
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
Example 1.1
Free Body Diagram Applying the equations of
equilibrium we have
( )540 2 0
1080 Nm [ans]
0;
540 0
c
c
c
y
M
M
F
V
=
− =
=
+ ↑ =
− =
∑
MEC411 – MECHANICS OF MATERIALS Ch 1 - 8
540N [ans]
0;
0 [ans]
x
V
F
=
← + =
=
∑
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
What is Stress ?
Distribution of internal loading is important in mechanics of materials.
We will consider the material to be continuous.
This intensity of internal force at a point is called stress.
MEC411 – MECHANICS OF MATERIALS Ch 1 - 9
F M E
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
Stress (Cont.)
Normal Stress , σ
• Force per unit area acting normal to ∆A
Shear Stress ,
• Force per unit area acting tangent to ∆A
0lim z
z A
σ ∆ →
=∆
lim zxτ ∆
=
τ
MEC411 – MECHANICS OF MATERIALS Ch 1 - 10
0lim
y
zy A
τ
→
∆ →
∆=
∆
C F M E
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
Example 1.2
Each of the four vertical links has an 8
x 36 mm uniform rectangular cross
section and each of the four pins has a
16 mm diameter. Determine the
maximum value of the average normal
stress in the links connecting (a)
points B and D, (b) points C and E.
MEC411 – MECHANICS OF MATERIALS Ch 1 - 11
CHAPTER 1 FACULTY OFMECHANICALENGINEERING
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
Example 1.2
i. Use bar ABC as a free body
ii. Solve for F BD and F CE
( )( ) 010*2004.0025.004.0;0 3 =+−=∑ BDC F M
MEC411 – MECHANICS OF MATERIALS Ch 1 - 12
( )on][Compressi N105.12
010*20025.004.0;0
[Tension] N105.32
3
3
3
×−=
=−−=
×=
∑CE
CE B
BD
F
F M
F
CHAPTER 1 FACULTY OFMECHANICALENGINEERING
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
Example 1.2
iii. Calculation of net area
( )016.0036.0008.0)(link onefor −=tension
iv. Calculation of stress
1056.10110*5.32 6
3
×===−
F BD BDσ
( )
26
26
26
26
m10576)(linkstwofor
m10288
036.0008.0)(link onefor
m10320)(linkstwofor
m10160
−
−
−
−
×=
×=
=
×=×=
ncompressio
ncompressio
tension
[ans]MPa7.21
107.21
10*576
10*5.12
[ans]MPa6.101
6
6
3
−=
×−=−
==
=
−
−
A
F CE CE
net
σ
MEC411 – MECHANICS OF MATERIALS Ch 1 - 13
CHAPTER 1 FACULTY OFMECHANICALENGINEERING
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
Average Shear Stress
The average shear stress distributed over each sectioned area that develops a
shear force.
2 different types of shear:
avg =τ
= average shear stress
V = internal resultant shear force A = area at that section
τ
MEC411 – MECHANICS OF MATERIALS Ch 1 - 14
a) Single Shear b) Double Shear
CHAPTER 1 FACULTY OFMECHANICALENGINEERING
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
Example 1.3
The inclined member is subjected to a
compressive force of 3000 N. Determine the
average compressive stress along the smooth
areas of contact defined by AB and BC , and the
average shear stress along the horizontal plane
defined by EDB.
i. The compressive forces acting on the areas of contact are
MEC411 – MECHANICS OF MATERIALS Ch 1 - 15
( )
( ) N240003000 ;0
N180003000 ;0
54
53
=⇒=−=↑+
=⇒=−=→+
∑∑
BC BC y
AB AB x
F F F
F F F
CHAPTER 1 FACULTY OFMECHANICALENGINEERING
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CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
Example 1.3
ii. The shear force acting on the sectioned horizontal plane EDB is
N1800 ;0 ==→+ ∑ V F x
iii. Average compressive stresses along the AB and BC planes are
( )( )
( )( )(Ans) N/mm20.1
4050
2400
(Ans) N/mm80.1
4025
1800
2
2
==
==
BC
AB
σ
σ
MEC411 – MECHANICS OF MATERIALS Ch 1 - 16
( )( )(Ans) N/mm60.0
4075
1800 2==avg τ
iv. Average shear stress acting on the BD plane is
CHAPTER 1 F ACULTY OFMECHANICALE NGINEERING
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CONCEPT OF STRESS & STRAIN
FDIVISION OF ENGINEERING MECHANICS
What is Strain ?
Normal Strain
• The elongation / contraction of a line segment per unit of length is referred to as
norma s ra n.• Average normal strain is defined as;
• If the normal strain is known, then the approximate final length is:
s
s savg
∆
∆−∆=
'ε
≈'
MEC411 – MECHANICS OF MATERIALS Ch 1 - 17
+ε line elongate
-ε line contracts
CHAPTER 1 F ACULTY OFMECHANICALE NGINEERING
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CONCEPT OF STRESS & STRAIN
FDIVISION OF ENGINEERING MECHANICS
Units
Normal strain is a dimensionless quantity since it is a ratio of two lengths.
Strain (Cont.)
Change in angle between 2 line segments that were perpendicular to one another
refers to shear strain.
lim '2 along
nt B A n
π γ θ
→= −
MEC411 – MECHANICS OF MATERIALS Ch 1 - 18
θ <90 + shear strain
θ>90 - shear strain
a ong t →
CHAPTER 1 F ACULTY OFMECHANICALE NGINEERING
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CONCEPT OF STRESS & STRAIN DIVISION OF ENGINEERING MECHANICS
Example 1.4
The plate is deformed into the dashed
shape. If, in this deformed shape,horizontal lines on the plate remain
horizontal and do not change their
length, determine (a) the average
normal strain along the side AB, and (b)
the average shear strain in the plate
2 mm
3 mm
300 mm
3 0 0
m m
x
B
C
D
A
MEC411 – MECHANICS OF MATERIALS Ch 1 - 19
relative to the x and y axes.
CHAPTER 1 F ACULTY OFMECHANICALE NGINEERING
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CONCEPT OF STRESS & STRAIN DIVISION OF ENGINEERING MECHANICS
Example 1.4
the average normal strain along the side AB
Line AB, coincident with the y axis,
,
the length of this line is
( )2 2' 250 2 3 248.018 mm AB = − + =
The average normal strain for AB is
therefore
MEC411 – MECHANICS OF MATERIALS Ch 1 - 20
( )
( )3
' 248.018 250250
7.93 10
AB avg AB AB
ABε
−
− −= =
= − mm/mm (Ans)
The Negative Sign
Indicates The Strain Causes A Contraction
Of AB.
CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
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CONCEPT OF STRESS & STRAIN DIVISION OF ENGINEERING MECHANICS
Example 1.4
the average shear strain in the plate
relative to the x and y axes.
As noted, the once 90°angle BAC between
the sides of the plate, referenced from the x, y
axes, changes to θ ’ due to the displacement of
B to B’.
1 3tan 0.121
250 2rad (Ans) xyγ
− = = −
MEC411 – MECHANICS OF MATERIALS Ch 1 - 21
CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
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CONCEPT OF STRESS & STRAIN DIVISION OF ENGINEERING MECHANICS
Factor of Safety
Many unknown factors that influence the actual stress in a member.
A factor of safety is needed to obtained allowable load.
The factor of safety (F.S.) is a ratio of the failure load divided by the
allowable load
fail
allow
fail
S F
F
F S F
σ =
=
.
.
MEC411 – MECHANICS OF MATERIALS Ch 1 - 22
allow
fail
allow
S F τ
τ =.
CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
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CONCEPT OF STRESS & STRAIN DIVISION OF ENGINEERING MECHANICS
Example 1.5
The two wooden members shown, which
su ort a 20 kN load are oined b l wood
splices fully glued on the surfaces in
contact. The ultimate shearing stress in the
glue is 2.8 MPa and the clearance between
the members is 8 mm. Determine the factor
of safety, knowing that the length of each
s lice is L = 200 mm.
MEC411 – MECHANICS OF MATERIALS Ch 1 - 23
CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
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CONCEPT OF STRESS & STRAIN DIVISION OF ENGINEERING MECHANICS
Example 1.5
There are 4 separate areas of glue.
Each glue area must transmit 10 kN of
shear load.
Ultimate load
( )1052.11108.2 36 ××== − A P U U τ
N10103
×= P
Length of splice
l = length of glue and c = clearance.
( ) ( ) m096.0008.02.021
21
2
=−=−=
+=
c Ll
cl L
N1032.2563
×=
Factor of safety
][ans23.31010
10256.32.3
3
=×
×== P
P S F U
MEC411 – MECHANICS OF MATERIALS Ch 1 - 24
Area of glue
( ) 23m1052.11120.0096.0 −×=== lw A
CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
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CONCEPT OF STRESS & STRAIN DIVISION OF ENGINEERING MECHANICS
Deformation due to Stress & Strain
When a force is applied to a body, it will change the body’s shape and size.
These changes are deformation.
Note the before
and after positions
of 3 line segments where the material
is subjected to
tension.
MEC411 – MECHANICS OF MATERIALS Ch 1 - 25
CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
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CONCEPT OF STRESS & STRAIN DIVISION OF ENGINEERING MECHANICS
Stress-strain Relation
Gτ γ =
LinearElastic
Material
σ ε = yε ν
ε =
MEC411 – MECHANICS OF MATERIALS Ch 1 - 26
E = modulus of elasticity , G = modulus of rigidity or shear modulus, and v = Poisson’s ratio
CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
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DIVISION OF ENGINEERING MECHANICS
Supplementary Problem 1
1. Determine the resultant internal loadings
acting on the cross section through point D
of member AB.
2. A force of 80 N is supported by the bracket asA
MEC411 – MECHANICS OF MATERIALS Ch 1 - 27
s own. e erm ne e resu an n erna oa ngs
acting on the section through point A.
80 N
45o
30o
0.1 m
CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
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Supplementary Problem 1
3. The lever is held to the fixed shaft using a
tapered pin AB, which has a mean diameter of
4. Part of a control linkage for an airplane consists
of a rigid member CBD and a flexible cable AB.
6 mm. If a couple is applied to the lever,
determine the average shear stress in the pin
between the pin and lever.
MEC411 – MECHANICS OF MATERIALS Ch 1 - 28
If a force is applied to the end D of the member
and causes it to rotate by θ = 0.3°, determine
the normal strain in the cable. Originally the
cable is unstretched.
CHAPTER 1
CONCEPT OF STRESS & STRAIN
F ACULTY OFMECHANICALE NGINEERING
DIVISION OF ENGINEERING MECHANICS
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Supplementary Problem 1
5. The square deforms into the position shown
by the dashed lines. Determine the shear
strain at each of its corners, A, B, C, and D.Side D'B‘ remains horizontal.
MEC411 – MECHANICS OF MATERIALS Ch 1 - 29
.
displacements indicated. Determine the shear
strain along the edges of the plate at A and B.