Lecture Strength of Materials I: Chapter 3 - PhD. Tran Minh Tu

Chapter 3 - Axially loaded members. The following will be discussed in this chapter: Normal stress and normal strain, tension and compression test, poisson’s ratio, shearing strain, allowable stress – factor of safety, statically indeterminate problem.

STRENGTH OF MATERIALS

TRAN MINH TU -University of Civil Engineering,

CHAPTER

Axially loaded

members

3.1 Introduction3.2.Normal Stress and Normal Strain3.3 Tension and Compression Test3.4 Poisson’s ratio

3.5 Shearing Strain3.6 Allowable Stress – Factor of Safety3.7 Statically Indeterminate Problem

3.1 Introduction

• Suitability of a structure or machine may depend on thedeformations in the structure as well as the stresses inducedunder loading Statics analyses alone are not sufficient

• Considering structures as deformable ones allows us todeterminate the member forces and reactions which are staticallyindeterminate.

• Determination of the stress distribution within a member alsorequires the consideration of deformations in the member

• Chapter 3 is concerned with the stress and deformation of astructural member under axial loading Later chapters will dealwith torsional and pure bending loads

cross-• Axial force: Load directed along the axis of the member

• Axial force can be tensile or compressive

3.1 Introduction

 Kinematic assumptions

3.2 Normal stress and normal strain

 Normal stress

z z

N A

z const

 – normal stress at any point on

the cross-sectional area

Nz – internal resultant normal force

A – cross-sectional area

1 The axis of the member remains straight

2 Cross sections which are plane and are perpendicular to the axis beforedeformation, remain plane and remain perpendicular to the axis afterdeformation And the cross sections do not rotate about the axis

3.2 Normal stress and normal strain

 Elongation of the bar:

Consider the bar, which has a cross-sectional area that gradually variesalong its length L The bar is subjected to concentrated loads at its endsand variable external load distributed along its length

0

( ) ( )

L

N z dz L

EA z



3.2 Normal stress and normal strain

 Elongation of the bar – constant load

and cross-sectional area:

z

N L L

EA



  

zN

y x

• Strength of a material can only be determined by experiment

• One test used by engineers is the tension or compression test

• This test is used primarily to determine the relationship between theaverage normal stress and the average normal strain in commonengineering materials, such as metals, ceramics, polymers andcomposites

3.3 Tension and Compression Test

Performing the tension or compression test

• Specimen of material is made into “standard” shape and size

• Before testing, 2 small punch marks are identified along the specimen’slength

• Measurements are taken for both the specimen’s initial x-sectional area

A0 and the gauge-length distance L0; between the two marks

Performing the tension or compression test

• Seat the specimen into a testing machine as shown below

- The machine will stretch the specimen

at a slow constant rate until the breaking

point

- At frequent intervals during test, the

data is recorded of the applied load P

3.3 Tension and Compresion Test

• The Elongation δ = L − L0 is measured

by using either a caliper or an

extensometer

• δ is used to calculate the normal strain

in the specimen

• Sometimes, the strain can also be read

directly by using an electrical-resistance

strain gauge

3.3 Tension and Compresion Test

• A stress-strain diagram is obtained by plotting the various values of

the stress and corresponding strain in the specimen

Conventional stress-strain diagram

• Using recorded data, we can determine the nominal or engineeringstress by

3.3 Tension and Compression Test

Conventional stress-strain diagram

• This Figure shows the characteristic stress-strain diagram for steel, acommonly used material for structural members and mechanicalelements

3.3 Tension and Compresion Test

Conventional stress-strain diagram

Elastic behavior.

• the straight line

• The stress is proportional to the

strain, i.e., linearly elastic

• Upper stress limit, or proportional

limit; σpl

• If the load is removed upon

reaching the elastic limit , the

specimen will return to its original

shape

Yielding.

• The material deforms permanently;

yielding; plastic deformation

• Yield stress, σY

• Once the yield point is reached, the specimen continues to elongate

(strain) without any increase in load

3.3 Tension and Compresion Test

Conventional stress-strain diagram

Figure 3-4

Strain hardening.

• Ultimate stress, σu

• While the specimen is elongating,

its z-sectional area will decrease

• Decrease in area is fairly uniform

over entire gauge length

Necking.

• At ultimate stress, x-sectional

area begins to decrease in a

localized region

• As a result, a constriction or

“neck” tends to form in this region

as the specimen elongates further

• The Specimen finally breaks at fracture stress, σf

3.3 Tension and Compresion Test

Stress – Strain Diagram: Ductile materials

• Defined as any material that can be subjected to large strains before itrupture, e.g., mild steel

• Such materials are used because of its capacity of absorbing shock orenergy so that it, will exhibit a large deformation before failing

• Ductility of material is to report its percent elongation or percentreduction in area at time of fracture

Ductile materials

• Percent elongation is the specimen’s fracture strain expressed as a

percent

• Percent reduction in area is defined within necking region as

• Most metals do not exhibit constant yielding behavior beyond the

elastic range, e.g aluminum

• It does not have a well-defined yield point, thus it is standard practice

to define its yield strength using a graphical procedure called the offset

Stress – Strain Diagram: Brittle Materials

• Materials that exhibit little or no yielding before failure are referred to asbrittle materials, e.g., gray cast iron

• Brittle materials do not have a well-defined tensile fracture stress, sincethe appearance of an initial cracks in a specimen is quite random

3.3 Tension and Compression Test

3.3 Tension and Compresion Test

• Most engineering materials

exhibit a linear relationship

between the stress and strainwith the elastic region

• Discovered by Robert Hooke in

1676 using springs, known as

Hooke’s law

σ = E 

• E represents the constant of

proportionality, also called the

modulus of elasticity or Young’s modulus

• E has units of stress, i.e., pascals,

MPa or GPa

Hooke’s Law: Modulus of Elasticity

3.3 Tension and Compresion Test

• If the strain disappears whenthe stress is removed, thematerial is said to behave

elastically.

• When the strain does notreturn to zero after thestress is removed, thematerial is said to behave

3.3 Tension and Compresion Test

3.5 Allowable Stress – Factor of Safety

• When designing a structural member or mechanical element, the

stress in it must be restricted to safe level

• Choose an allowable load that is less than the load the member can

fully support

• One method used is the factor of safety (F.S.)

F.S = F Ffail

allow

• If the load applied is linearly related to the stress developed within

the member, then F.S can also be expressed as:

F.S = σfail

σallow F.S = τfail

τallow

• In all the equations, F.S is chosen to be greater than 1, to avoid

potential for failure

• Specific values will depend on the type of material used and its

intended purpose

3.5 Allowable Stress – Factor of Safety

• Fail Stress: 0 – stress coresponding lossing ability to sustain a load

  0

.

fail allow

• Three main problems:

1 Determine if the condition of strength is satisfied or not

2 Determine the required cross-sectional area

3 Determine the maximum applied load

3.6 Statically Indeterminate Problem

Sample Problem 3.1

The A-36 steel bar shown in the Figure is made from two segments having the cross-sectional areas of AAB=1cm 2 ; ABD=2cm 2 Draw the axial force diagram and determine the maximum normal stress and the vertical displacement of end A

1 Use the method of section

N

kN cm A

BC

N

kN cm A

CD

N

kN cm A

     

2 max 15( kN cm / )

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 thebar 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 findthe deflection at E given thedeflections at B and D

Sample Problem 3.2

Displacement of B:

   

m 10 514

Pa 10 70 m

10 500

m 3 0 N 10 60

6

9 2

6 - 3

Pa 10 200 m

10 600

m 4 0 N 10 90

6

9 2

6 - 3

F

tension F

F M

AB

AB

CD

CD B

kN 60

m 2 0 m

4 0 kN 30 0

0 M

kN 90

m 2 0 m

6 0 kN 30 0

Displacement of D:

mm 7 73

mm 200 mm

0.300

mm 514 0

BH D

D

B B

mm 7 73

mm 7 73 400 mm

300 0

D

E E





Sample Problem 3.2

b aB

A2

A1

Consider the bar made from two segments

having the cross-sectional areas of A1 and A2

1 Draw the axial force diagram

2 Determine the max normal stress

3 Determine the displacement of end D

with F 1 =10kN; F 2 =25kN; A 1 =5cm 2 ; A 2 =8cm 2

a=b=1m; E=2.10 4 kN/cm 2

SOLUTION

1 Using method of section, the internal axial

force in each segments are:

z1

F1D

Axial force diagram:

2 Determine the maximum normal stress

2 1

10

5

CD CD

N

kN cm A

N

kN cm A

1 15.10 10.10

0, 0625.10 ( ) 2.10 8 5

The steel bars CD an CE with Young’s modulus E,

each have a cross-sectional area of A, are joined

at C with a pin Determine the axial forces in each

bars and the displacement of point C cause by

load P.

SOLUTION:

1 Determine axial forces:

Using method of join FBD of join C

Sample Problem 3.4

2 Displacement of join C:

D

N L L

Bài 2.4: Three steel rods, each have a

cross-sectional area of A=5cm 2 , jointly support the

load P= 50kN Determine the axial forces in

these rods and the displacement of join C.

PH

(3)

30o 30o

CA

Sample Problem 3.6

Sample Problem 3.6

Sample Problem 3.7

Sample Problem 3.7

3

2 9

0,00020,01

Sample Problem 3.7

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