Sixth GE SI Edition MECHANICS OF MATERIALS - · PDF file · 2016-02-29Stress Concentration: Hole Stress Concentration: Fillet ... • Considering structures as deformable allows determination

1

MECHANICS OF

MATERIALS

Sixth GE SI Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

John T. DeWolf

David F. Mazurek

Lecture Notes:

J. Walt Oler

Texas Tech University

Modified by:

Prof. Keun Park

SeoulTech, Korea

CHAPTER

© 2012 The McGraw-Hill Companies, Inc. All rights reserved.

2 Stress and Strain

– Axial Loading


MECHANICS OF MATERIALS

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Beer • Johnston • DeWolf • Mazurek

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Contents

Stress & Strain: Axial Loading

Normal Strain

Stress-Strain Test

Stress-Strain Diagram: Ductile Materials

Stress-Strain Diagram: Brittle Materials

Hooke’s Law: Modulus of Elasticity

Deformations Under Axial Loading

Example 2.01

Sample Problem 2.1

Static Indeterminacy

Example 2.04

Thermal Strains

Thermal Stresses

Sample Problem 2.4

Poisson’s Ratio

Generalized Hooke’s Law

Dilatation: Bulk Modulus

Shearing Strain

Example 2.10

Relation Among E, n, and G

Sample Problem 2.5

Saint-Venant’s Principle

Stress Concentration: Hole

Stress Concentration: Fillet

Example 2.12

2_2_axial_loading.ppt



















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Stress & Strain: Axial Loading

• Suitability of a structure or machine may depend on the deformations in

the structure as well as the stresses induced under loading. Statics

analyses alone are not sufficient.

• Considering structures as deformable allows determination of member

forces and reactions which are statically indeterminate.

• Determination of the stress distribution within a member also requires

consideration of deformations in the member.

• Chapter 2 is concerned with deformation of a structural member under axial loading. Later chapters will deal with torsional and pure bending loads.



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Normal Strain

Fig. 2.1

Fig. 2.3

Fig. 2.4







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Stress-Strain Test : Tensile Test

Experimental setup for the tensile test

Tensile specimen

(dog-bone shape)

• P: Centric tensile load

• L0: gage length before

loading

• L : distance b/w two

gage marks

• : elongation (= L - L0)

▶strain () calculated

• A0: initial diameter

▶ stress () calculated

• Stress-strain values can

be recorded at every

loading step



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Stress-Strain Diagram: Ductile Materials (연성재료)

• Ductile materials: characterized by their ability to yield at normal temperature (typical

structural steel and many metal alloys)

• Under increasing axial load, its length first increases stiffly and linearly (elastic region)

• After a critical value of Y (Yield stress), the specimen undergoes a large deformation with

a relatively small and nonlinear increase (plastic region)

• After a critical value of U (Ultimate stress or tensile strength) has been reached, the

diameter of a certain region begins to decrease (necking)







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Stress-Strain Diagram: Brittle Materials (취성재료)

• Brittle materials: characterized by the fact that rupture (파열) occurs without any

noticeable prior change in the range of elongation (glass, stone, cement, etc.)

• No difference between the ultimate strength and the breaking strength (no necking)

• The strain at the rupture is much smaller than ductile materials



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Hooke’s Law: Modulus of Elasticity

• Below the yield stress

• Strength is affected by alloying,

heat treating, and manufacturing

process but stiffness (Modulus of

Elasticity) is not.

Fig 2.16 Stress-strain diagrams for iron and different grades of steel.







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Elastic vs. Plastic Behavior

• If the strain disappears when the

stress is removed, the material is

said to behave elastically.

• When the strain does not return

to zero after the stress is

removed, the material is said to

behave plastically.

• The largest stress for which this

occurs is called the elastic limit.

Fig. 2.18



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Deformations Under Axial Loading

• From Hooke’s Law:

• From the definition of strain:

• Equating and solving for the deformation,

• With variations in loading, cross-section or

material properties,

Fig. 2.22







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Example 2.01

Determine the deformation of

the steel rod shown under the

given loads.

SOLUTION:

• Divide the rod into components at

the load application points.

• Apply a free-body analysis on each

component to determine the

internal force

• Evaluate the total of the component

deflections.



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SOLUTION:

• Divide the rod into three

components:

• Apply free-body analysis to each

component to determine internal forces,

• Evaluate total deflection,







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Sample Problem 2.1

The rigid bar BDE is supported by two

links AB and CD.

Link AB is made of aluminum (E = 70

GPa) and has a cross-sectional area of 500

mm2. Link CD is made of steel (E = 200

GPa) and has a cross-sectional area of (600

mm2).

For the 30-kN force shown, determine the

deflection a) of B, b) of D, and c) of E.

SOLUTION:

• Apply a free-body analysis to the bar

BDE to find the forces exerted by

links AB and DC.

• Evaluate the deformation of links AB

and DC or the displacements of B

and D.

• Work out the geometry to find the

deflection at E given the deflections

at B and D.



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Sample Problem 2.1

Free body: Bar BDE

SOLUTION: Displacement of B:

Displacement of D:

mm 514.0B

mm 300.0D







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Displacement of D:

Sample Problem 2.1

mm 928.1E



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Static Indeterminacy (부정정 문제)

• Structures for which internal forces and reactions

cannot be determined from statics alone are said

to be statically indeterminate.

• Deformations due to actual loads and redundant

reactions are determined separately and then added

or superposed.

• Redundant reactions are replaced with

unknown loads which along with the other

loads must produce compatible deformations.

• A structure will be statically indeterminate

whenever it is held by more supports than are

required to maintain its equilibrium.







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Static Indeterminacy (부정정 문제)



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Example 2.04

Determine the reactions at A and B for the steel

bar and loading shown, assuming a close fit at

both supports before the loads are applied.

• Solve for the reaction at A due to applied loads

and the reaction found at B.

• Require that the displacements due to the loads

and due to the redundant reaction be compatible,

i.e., require that their sum be zero.

• Solve for the displacement at B due to the

redundant reaction at B.

SOLUTION:

• Consider the reaction at B as redundant, release

the bar from that support, and solve for the

displacement at B due to the applied loads.







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Example 2.04 SOLUTION:

• Solve for the displacement at B due to the applied

loads with the redundant constraint released,

• Solve for the displacement at B due to the redundant

constraint,



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• Require that the displacements due to the loads and due to

the redundant reaction be compatible,

Example 2.04

• Find the reaction at A due to the loads and the reaction at B

kN577

kN323

B

A

R

R







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Thermal Strains

• A temperature change results in a change in length or thermal strain.

• There is no stress associated with the thermal strain unless the

elongation is restrained by the supports.

Thermal deformation Thermal strain

: coefficient of thermal expansion



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Thermal Stresses

• If the elongation is restrained by the supports,

the part undergoes thermal stress.

• Treat the additional support as redundant and apply

the principle of superposition.

• The thermal deformation and the deformation from

the redundant support must be compatible.







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Thermal Stresses



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Thermal Stresses







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Sample Problem 2.4

The rigid bar CDE is attached to a pin

support at E and BD, at 20oC.

Rod AC is made of steel (E = 200 GPa,

d = 22 mm, = 11.7 10-6/oC)

Cylinder BD is made of brass (E = 105

GPa, d = 30 mm, = 20.9 10-6/oC)

When the brass cylinder is heated to

50oC, determine stress in the cylinder

(neglect heat transfer effect)

SOLUTION:

• Apply a free-body analysis to the bar

CDE to find the relation of reactions

between RA and RB.

• Use method of superposition to

calculate deflection of each member

according to thermal deformation of

the cylinder BD.

• Calculate thermal deformation of

the cylinder BD

• Find the reaction force at B and the

resulting stress in the cylinder BD



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Sample Problem 2.4

Method of superposition for each deflection:







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Sample Problem 2.4

Thermal deformation:



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Poisson’s Ratio

• For a slender bar subjected to axial loading:

• The elongation in the x-direction is

accompanied by a contraction in the other

directions. Assuming that the material is

isotropic (no directional dependence),

• Poisson’s ratio is defined as







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Poisson’s Ratio



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• For an element subjected to multi-axial loading,

the normal strain components resulting from the

stress components may be determined from the

principle of superposition. This requires:

1) strain is linearly related to stress

2) deformations are small

• With these restrictions:







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Dilatation: Bulk Modulus

• Relative to the unstressed state, the change in volume is

• For element subjected to uniform hydrostatic pressure,

• Subjected to uniform pressure, dilatation must be

negative, therefore







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Shearing Strain

• A cubic element subjected to a shear stress will

deform into a rhomboid. The corresponding shear

strain is quantified in terms of the change in angle

between the sides,

• A plot of shear stress vs. shear strain is similar to the

previous plots of normal stress vs. normal strain

except that the strength values are approximately

half. For small strains,

where G is the modulus of rigidity or shear modulus.

Fig. 2-46

Fig. 2-47



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Example 2.10

A rectangular block of material with

modulus of rigidity G = 630 MPa is

bonded to two rigid horizontal plates.

The lower plate is fixed, while the

upper plate is subjected to a horizontal

force P. Knowing that the upper plate

moves through 1.0 mm. under the

action of the force, determine a) the

average shearing strain in the material,

and b) the force P exerted on the plate.

SOLUTION:

• Determine the average angular

deformation or shearing strain of

the block.

• Use the definition of shearing stress to

find the force P.

• Apply Hooke’s law for shearing stress

and strain to find the corresponding

shearing stress.







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• Determine the average angular deformation

or shearing strain of the block.

• Apply Hooke’s law for shearing stress and

strain to find the corresponding shearing

stress.

• Use the definition of shearing stress to find

the force P.

kN2.156P



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Relation Among E, n, and G

• An axially loaded slender bar will

elongate in the axial direction and

contract in the transverse directions.

• Components of normal and shear strain are

related,

• If the cubic element is oriented as in the

bottom figure, it will deform into a

rhombus. Axial load also results in a shear

strain.

• An initially cubic element oriented as in

top figure will deform into a rectangular

parallelepiped. The axial load produces a

normal strain.







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Sample Problem 2.5

A circle of diameter d = 225 mm is scribed on

an unstressed aluminum plate of thickness t =

18 mm. Forces acting in the plane of the plate

later cause normal stresses x = 84 MPa and

z = 140 MPa.

For E = 70 GPa and n = 1/3, determine the

change in:

a) the length of diameter AB,

b) the length of diameter CD,

c) the thickness of the plate, and

d) the volume of the plate.



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SOLUTION:

• Apply the generalized Hooke’s Law to

find the three components of normal

strain.

• Evaluate the deformation components.

mm12.0AB

mm36.0DC

mm0192.0t

• Find the change in volume

3mm2733V

mm/mm10600.1

mm/mm10067.1

mm/mm10533.0

MPa1403

10MPa84

MPa1070

1

3

3

3

3

EEE

EEE

EEE

zyxz

zyxy

zyxx

nn

nn

nn







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Saint-Venant’s Principle

• Loads transmitted through rigid

plates result in uniform distribution

of stress and strain.

• Saint-Venant’s Principle:

Stress distribution may be assumed

independent of the mode of load

application except in the immediate

vicinity of load application points.

• Stress and strain distributions

become uniform at a relatively short

distance from the load application

points.

• Concentrated loads result in large

stresses in the vicinity of the load

application point.



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Stress Concentration: Hole

(a) Flat bars with holes

Discontinuities of cross section may result in

high localized or concentrated stresses.







http://imagesearch.naver.com/search.naver?sm=ext&viewloc=1&where=idetail&rev=17&query=%EB%9D%BC%EB%A9%B4%EC%8A%A4%ED%94%84&section=image&sort=0&res_fr=0&res_to=0&start=89&ie=utf8&img_id=cafe25445736%7C211%7C174335_3&face=0&color=0&ccl=0&viewtype=2&aq=0&spq=0&nx_search_query=%EB%9D%BC%EB%A9%B4%EC%8A%A4%ED%94%84&nx_and_query=&nx_sub_query=&nx_search_hlquery=&nx_search_fasquery=

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Stress Concentration: Fillet

(b) Flat bars with fillets



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Example 2.12

Determine the largest axial load P

that can be safely supported by a

flat steel bar consisting of two

portions, both 10 mm thick, and

respectively 40 and 60 mm wide,

connected by fillets of radius r = 8

mm. Assume an allowable normal

stress of 165 MPa.

SOLUTION:

• Determine the geometric ratios and

find the stress concentration factor

from Fig. 2.64b.

• Apply the definition of normal stress to

find the allowable load.

• Find the allowable average normal

stress using the material allowable

normal stress and the stress

concentration factor.







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• Determine the geometric ratios and

find the stress concentration factor

from Fig. 2.64b.

• Find the allowable average normal

stress using the material allowable

normal stress and the stress

concentration factor.

• Apply the definition of normal stress

to find the allowable load.

kN3.36P

(b) Flat bars with fillets



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Sixth GE SI Edition MECHANICS OF MATERIALS - · PDF file · 2016-02-29Stress Concentration: Hole Stress Concentration: Fillet ... • Considering structures as deformable allows determination