PowerPoint slides of chapter 3

MEEN2313 - Mechanics of Solids

Chapter 3: Mechanical Properties of Materials

 The strength of a material depends on its ability to sustain a load.

 This property of materials is determined by a tension or

compression test.

 A universal testing machine (UTM) is used to stretch the

specimen at a constant rate until it fails.

 The elongation δ = L - Lo measured by an extensometer or an

electrical resistance strain gauge

 At frequent interval during loading, the applied load and the

elongation of the specimen are recorded

Strain gauge

Strain gauge mounted specimen

UTM

The Stress–Strain Diagram

The Stress–Strain Diagram

The Stress–Strain Diagram

The Stress–Strain Diagram

Stress–Strain Behavior of Ductile and Brittle Materials

Hooke’s Law

Hooke’s Law

Strain Energy

Strain Energy

The stress–strain diagram for an aluminum alloy that is used for

making aircraft parts is shown in Figure. 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.

Problem 1

Solutions

• 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,

450 75.0 GPa

0.006 E  

From triangle CBD,

BD E

CD 

This strain represents the amount of

recovered elastic strain.

The permanent strain is

Computing the modulus of resilience,

0.023 0.008OC  

 

 

1

2

1

2

r initial pl pl

r final pl pl

u

u

 

 





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

   

6

9 600 10

75.0 10

0.008 mm/mm

CD

CD

 

 

0.0150 mm/ (mm Ans)

  

  

3

3

1 450 0.006 (Ans)

2

1 600 0.0

1.35 MJ/m

2.40 MJ/m08 (Ans) 2

 

 

Determine the elongation of the square hollow bar when it is

subjected to the axial force P = 100 kN. If this axial force is increased to P = 360 kN and released, find the permanent elongation of the bar.

The bar is made of a metal alloy having a stress–strain diagram

which can be approximated as shown.

Problem 2

Solutions

Normal Stress and Strain:

The cross-sectional area of

the hollow bar is

A = 0.052 - 0.042

= 0.9(10-3)m2

When, P = 100 kN

1

P σ =

A

From the stress–strain diagram shown in Fig, the slope

of the straight line OA which represents the modulus of

elasticity of the metal alloy is 6250(10 )

E= 0.00125

Since σ1 < 250 MPa, Hooke’s Law can be applied. Thus

1 1σ = Eε

Thus, the elongation of the bar is 1 1δ = ε L

When, P = 360 kN 1

P σ =

A

3

-3

100(10 ) = 1 =

0 11.1

.9(10 ) 1 MPa

200 = GPa

6 9

1

1

-3

111.11(10 ) = 200(10 )ε

0.5556(10 ) m mε = m/m

-3= 0.5556(10 )(6 0.300) = 33 mm

3

-3

360(10 ) = = 4

0.9(10 ) 00 MPa

From the geometry of the stress–strain diagram

2 0.0305

0. 5

/

02 - 0.00125 -0.00125 = 400 - 250 500 - 250

mm mm



 

When P = 360 kN is removed, the strain recovers linearly along line

BC, Figure, parallel to OA. Thus, the elastic recovery of strain is

given by

2 rE 

The permanent set is

2p r   

Thus, the permanent elongation of the bar is

p pL 

   6 9400 10 200 10 0.002 /

r

r mm mm







 

0.0305 0.002 0.0285 /mm mm  

 0.0285 60 17 10 . mm 

Poisson’s Ratio (υ)

 When a deformable body is subjected to an axial tensile loading, it

elongates and contracts laterally

 Similarly, under compressive loading, its length shortens and expands

laterally

'

,long latLongitudinal strain and Lateral strain L r

    

 With in elastic limit the ratio of these strains is a constant called the

Poisson’s ratio (named after French Scientist S.D Poisson)

 Poisson’s ratio is a dimensionless quantity

 For ideal material, having no lateral deformation when it is stretched

or compressed, the Poisson’s ratio is “zero”

 The maximum possible value of “υ” is 0.5

Shear Stress-Strain Diagram

 When a material is subjected to

pure shear, shear stress will be

developed

 The resulting shear strain γxy measures the angular distortion of

each small element relative to the

original sides

𝑷𝒐𝒊𝒔𝒔𝒐𝒏′𝒔 𝒓𝒂𝒕𝒊𝒐, 𝝂 = − 𝜺𝒍𝒂𝒕 𝜺𝒍𝒐𝒏𝒈

 For most engineering materials, the elastic behaviour is linear, so

Hooks law for shear can be written as

 G is called the shear modulus of elasticity or modulus of rigidity

 Its value represents the slope of τ-γ diagram

 The three material constants E, υ and G are related by the equation

pl

pl

or G and G 

    

  

 2 1 E

G 

 

Problem 3 A specimen of titanium alloy is tested in torsion and the shear stress- strain diagram is shown in Figure. Determine the shear modulus G, the proportional limit, and the ultimate shear stress. Also, determine the maximum distance d that the top of a block of this material, shown in Figure, could be displaced horizontally if the material behaves elastically when acted upon by a shear force V. What is the magnitude of V necessary to cause this displacement?

 By inspection, the graph ceases to be linear at point A. Thus, the proportional limit is

 The maximum shear stress is at point B. Thus the ultimate stress is

 Since the angle is small, the top of the specimen will be displaced horizontally by

 The shear force V needed to cause the

displacement is

Solutions

504 MPau 

360 MPa pl 

 tan 0.008 rad 0.008 0.4 mm 50 mm

d d   

   ; 360 MPa

75 100 2700 kN avg V

V V

A     

Problem 4 A bar made of A-36 steel has the dimensions shown in Figure. 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. The value of E from tables is obtained as 200 Gpa, and ν as 0.32

Solutions

 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

z

P

A   

z z

stE

   

zz zL  

    

  3

6 80 10

16.0 10 Pa 0.1 0.05



   

  6

6

6

16.0 10 80 10 m/m

200 10



  680 10 1.5 120 m (Ans)   

 The contraction strains in both the x and y directions are

 The changes in the dimensions of the cross section are

x y st zv     

x x xL  

 60.32 80 10 25.6 m/m    

  

  

6

6

25.6 10 0.1 2.56 m (Ans)

25.6 10 0. 05 1.28 m (Ans)









     

      y y yL  

Creep and Fatigue  Creep

 When a material has to support a load for a very long period of

time, it may continue to deform until a sudden fracture occurs

 Normally creep is considered when materials are used for

structural members that are subjected to high temperatures

 An important mechanical property that is used in this regard is

called the creep strength

 Fatigue

 When a material is subjected to repeated cycles of stress or strain,

it causes its structure to breakdown, ultimately leading to failure

 Example of fatigue failure is the kind of failures in connecting

rods, crankshafts of engines and rolling bearings

 Fatigue strength of a materials determines its ability to support

stress cycles with out failures