mom chapter 3-mechanical properties of materials

7/29/2019 MOM Chapter 3-Mechanical Properties of Materials

1/22

Chapter Objectives Understand how to measure the stress and strain

through experiments

Correlate the behavior of some engineering materials

to the stress-strain diagram.

Copyright 2011 Pearson Education South Asia Pte Ltd


2/22

1. Check homework, if any

2. Reading Quiz

3. Applications

4. Stress-Strain diagram5. Strength parameters

6. Poissons ratio

7. Shear Stress-strain diagram

8. Concept Quiz

In-class Activities



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TENSION AND COMPRESSION TEST



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READING QUIZ

1) The modulus of elasticity E is a measure ofthe linear relationship between stress and

strain. The common unit is:

a) kN/mm2

b) MPa

c) GPa

d) All of them



5/22

READING QUIZ (cont)

2) The Poissons ratio, vof commonengineering materials lies in the range:

a) 0 v 1

b) 0 v 0.5

c) -1 v 1

d) -0.5 v 0.5



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APPLICATIONS



7/22

APPLICATIONS (cont)



8/22

STRESS STRAIN DIAGRAM


Note the critical status for strength specification

proportional limit

elastic limit

yield stress

ultimate stress

fracture stress


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STRENGTH PARAMETERS


Modulus of elasticity (Hookes Law)

Modulus of Resistance

Modulus of Toughness

It measures the enter areaunder the stress-strain diagram

E

Eu plplplr

2

21

21


10/22

EXAMPLE 1


The stressstrain diagram for an aluminum alloy that is used

for making aircraft parts is shown in Fig. 319. If a specimen

of this material is stressed to 600 MPa, determine the

permanent strain that remains in the specimen when the load

is released. Also, find the modulus of resilience both before

and after the load application.


11/22

EXAMPLE 1 (cont)


When the specimen is subjected to the load, the strain is approximately

0.023 mm/mm.

The slope of line OA is the modulus of elasticity,

From triangle CBD,

Solutions

mm/mm008.0

100.7510600 9

6

CDCDCD

BDE

GPa0.75

006.0

450E


12/22

EXAMPLE 1 (cont)


This strain represents the amount ofrecovered elastic strain.

The permanent strain is

Computing the modulus of resilience,

Note that the SI system of units is measured in joules, where 1 J = 1 N

m.

Solutions

(Ans)MJ/m40.2008.060021

2

1

(Ans)MJ/m35.1006.04502

1

2

1

3

3

plplfinalr

plplinitialr

u

u

(Ans)mm/mm0150.0008.0023.0 OC


13/22

POISSONs RATIO


long

latv


14/22

EXAMPLE 2


A bar made of A-36 steel has the dimensions shown in Fig.

322. If an axial force of P = 80kN is applied to the bar,

determine the change in its length and the change in the

dimensions of its cross section after applying the load. The

material behaves elastically.


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EXAMPLE 2 (cont)


The normal stress in the bar is

From the table for A-36 steel, Est= 200 GPa

The axial elongation of the bar is therefore

Solutions

mm/mm108010200

100.16 66

6

st

zz

E

Pa100.1605.01.0

1080 63

A

Pz

(Ans)m1205.11080 6z zz L


16/22

EXAMPLE 2 (cont)


The contraction strains in both thexand ydirections are

The changes in the dimensions of the cross section are

Solutions

m/m6.25108032.0 6 zstyx v

(Ans)m28.105.0106.25

(Ans)m56.21.0106.25

6

6

yyy

xxx

L

L


17/22

SHEAR STRESS-STRAIN DIAGRAM


Strength parameter G Shear modulus of elasticity or the

modules of rigidity

G is related to the modulus of elasticity E and Poissonsratio v.

G

vE

G

12


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EXAMPLE 3


A specimen of titanium alloy is tested in torsion and the

shear stress strain diagram is shown in Fig. 325a.

Determine the shear modulus G, the proportional limit, and

the ultimate shear stress. Also, determine the maximum

distance dthat the top of a block of this material, shown in

Fig. 325b, could be displaced horizontally if the material

behaves elastically when acted upon by a shear force V.What is the magnitude ofV necessary to cause this

displacement?


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EXAMPLE 3 (cont)


By inspection, the graph ceases to be linear at pointA. Thus, the

proportional limit is

This value represents the maximum shear stress, point B. Thus the

ultimate stress is

Since the angle is small, the top of

the will be displaced horizontally by

Solutions

(Ans)MPa504u

(Ans)MPa360pl

mm4.0mm50

008.0rad008.0tan dd


20/22

EXAMPLE 3 (cont)


The shear force Vneeded to cause the displacement is

Solutions

(Ans)kN2700

10075MPa360; V

V

A

Vavg


21/22

CONCEPT QUIZ

1) The head H is connected to the cylinder of acompressor using six steel bolts. If the

clamping force in each bolt is 4000N,

determine the normal strain in the bolts.

Each bolt has a diameter of 5 mm. Ify= 280

MPa and Est = 210GPa, what is the strain ineach bolt when the nut is unscrewed so that

the clamping force is released?



22/22

CONCEPT QUIZ (cont)

a) 0.970

b) 0.203

c) 0.970(10-3)

d) Insufficient information to

determine because the

stress is beyond yield point


mom chapter 3-mechanical properties of materials

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