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 :



CHAPTER 1




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.




CHAPTER 1




External Forces

External Forces

Surface Forces

caused by direct contact of

other body’s surface

Bod Forcesother body exerts a force


w ou con ac



CHAPTER 1




Reactions

Surface forces developed at the supports/points of contact between bodies.




CHAPTER 1




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.


ese wo con ons can e expresse ma ema ca y as:

( )atanypoint

0

0

0

x

y

F

F

M

=

=

=

∑∑

∑



CHAPTER 1




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


torque, T

Bending moment, M



CHAPTER 1




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


Magnitude of the resultant of the

distributed load,

( )( ) N540618021 == F

which acts from C ( ) m2631

=

6 9



CHAPTER 1




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

=

− =

=

+ ↑ =

− =

∑


540N [ans]

0;

0 [ans]

x

V

F

=

← + =

=

∑



CHAPTER 1




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.


F M E



CHAPTER 1




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


0lim

y

zy A

τ

→

∆ →

∆=

∆

C F M E



CHAPTER 1




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.


CHAPTER 1 FACULTY OFMECHANICALENGINEERING



CHAPTER 1




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


( )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




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





CHAPTER 1




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


a) Single Shear b) Double Shear




CHAPTER 1




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


( )

( ) N240003000 ;0

N180003000 ;0

54

53

=⇒=−=↑+

=⇒=−=→+

∑∑

BC BC y

AB AB x

F F F

F F F




CHAPTER 1




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


( )( )(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




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

≈'


+ε line elongate

-ε line contracts





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

→= −


θ <90 + shear strain

θ>90 - shear strain

a ong t →




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


relative to the x and y axes.





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


( )

( )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






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

− = = −


CHAPTER 1






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 =

=

.

.


allow

fail

allow

S F τ

τ =.

CHAPTER 1







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.


CHAPTER 1







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


Area of glue

( ) 23m1052.11120.0096.0 −×=== lw A

CHAPTER 1







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.


CHAPTER 1







Stress-strain Relation

Gτ γ =

LinearElastic

Material

σ ε = yε ν

ε =


E = modulus of elasticity , G = modulus of rigidity or shear modulus, and v = Poisson’s ratio

CHAPTER 1







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


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







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.


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







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.


.

displacements indicated. Determine the shear

strain along the edges of the plate at A and B.

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1-Concept of Stress and Strain.pdf