Chapter 3 Strain And Materials Properties Fi.docx

STRAIN AND MATERIAL PROPERTIES

Chapter 3

3.1 Deformation 3.2 Normal Strain 3.3 Shear Strain 3.4 Thermal Stress 3.5 Components of Strain 3.6 Stress–Strain Diagrams 3.7 Hooke’s Law 3.8 Poisson’s Ratio 3.9 Generalized Hooke’s Law 3.10 Strain Energy

3.1 Deformation: Consider a body subjected to external forces that cause it to take up the positions shown

by

dashed

3.1.Observed (straining)

can

that, be

lines

in

the

deformation

described

Figure

by

the

following two types: 1. Rigid body displacement • Translation • Rotation • Combination of these 2. Elastic deformation • Occurs due to a change in shape when acted up on by a load (strain) in the axial, bending or torsion (angle of twist) directions. Deformation versus strain (ε or γ): – Change in the relative position of two points within a body • Axial strain, ε (elongation (+) or contraction (-)) • Bending strain, ε (can be zero, positive or negative) • Torsional strain, γ (can be positive or negative) • Combination of the above Principle of superposition: is the small displacement assumption and linear behaviour of materials, which is valid whenevere the following conditions are satisfied: – The quantity (displacement or stress) to be determined is directly proportional to the loads that produce it (linear elasticity; be in the linear elastic range (Hooke‘s law) of the stress strain diagram). – The loading does not significantly change the size and shape of the member.

1 6

3.2 Normal Strain: The concept of normal strain is illustrated by considering the deformation of a prismatic bar as shown in Figure 3.2: (a) original bar; with original length, L. (b) loaded bar, the length increases an amount δ. Defining the normal strain ε, as the unit change in length. Strain in the axial direction is called normal strain; and is calculated as:

𝜀=

𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ,𝛿 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ,𝐿

,𝜀 =

𝛿 𝐿

… … … … … … … . (3.1)

Normal strain is positive(+) when there is elongation, otherwise (contraction) negative (-).

Example 3.1: Given: A stepped round bar as shown is subjected to a load P that produces axial deformation in each portion of the bar as shown in Figure 3.3.

Find: the strain in portions AB and BC and the total strain in the bar. Assumptions: the bar weight is neglected. SOLUTION: 0.0045 m = 0.009 0.5 m 0.0135 m = = 0.0087 1.5 m

εAB = εBC

Using the superposition principle: total Strain, 𝜀𝑡𝑜𝑡𝑎𝑙 = εAB + εBC = 0.009 + 0.0087 = 0.0177 On the otherhand, if we use the definition of strain (change in length / original length) as applied to the entire bar, 𝜀𝐴𝐶 =

(0.0045 m + 0.013 m) = 0.00875 (1.5 m + 0.5 m)

Why is this result different than the previous one? Which is correct? How do we explain this?

Comments:

Example 3.2:

1 7

Given: The ends of bar shown in Figure 3.4, are subjected to an axial load P. The total eleongation of the bar betoween joints A and C is 0.15 in. in segment (2), The normal strain is measured as 1,300 µin/in. Find: 1. The elongation of segment (2). 2. The normal strain in segment (1). Assumptions: The bar weight is neglected. SOLUTION:

3.3 Shear Strain: 

The shearing or shear strain (γ) is the tangent of the total change in

angle occuring between two originally perpendicular lines (n and t) in a body during deformation as shown in Figure 3.5. 

Shearing strain is positive (+) if the right angle between the reference

lines decrease, otherwise, the shearing strain is negative(-). 𝛾𝑛𝑡 = 

𝜋 2

– 𝜃 ′ … … … … . . (3.2)

Both normal and shear strains are indicated as dimensionless quantities.

1 8

Example 3.3: Given: A thin, triangular plate ABC is uniformly deformed into a shape ABC′, as shown by the dashed lines. (Figure 3.6) Find: (a) The normal strain along the centerline OC. (b) The normal strain along the edge AC. (c) The shear strain between the edges AC and BC. Assumptions: the edge AB is rigidly attached to the frame. The deformed edges 𝐴𝐶́ =́ 𝐵𝐶́ are straight lines. SOLUTION:

1 9

3.4 Components of Strains: 

For one-dimensional strain, the concidering deformation of a bar under axial loading as shown in Figure 3.2, is represented by the end points of the lines which experience displacement u and u+ Δu. Thus the normal strain at a point is: ∆𝑢 𝜕𝑢 = … … … … . (3.3) ∆𝑥→0 ∆𝑥 𝜕𝑥

𝜀𝑥 = lim 

In the case of two-dimentional strain as shown in Figure 3.7, the normal and shear strains at a point are: 𝜀𝑥 =



𝜕𝑢 𝜕𝑥

,

𝜀𝑦 =

𝜕𝑣 𝜕𝑦

, 𝛾𝑥𝑦 =

𝜕𝑢 𝜕𝑣 + … … (3.4) 𝜕𝑥 𝜕𝑦

Average Strains: the values of normal and shear strains are the same as the strain at a point. The average stains are,

𝜀𝑥 =

∆𝑢 ∆𝑥

,

𝜀𝑦 =

∆𝑣 ∆𝑦

, 𝛾𝑥𝑦 =

∆𝑢 ∆𝑥

+

∆𝑣 ∆𝑦

… … (3.5), and u and v

resprctively, are the x- and y-directed displasmens of a point.

Example 3.4: Given: A 0.4-m by 0.4-m square ABCD is drawn on a thin panel or flat plate of an aircraft prior to loading. Subsequent to loading, the square has the dimentions shown by dashed lines in Figure 3.9. Find: The average values of strain components at A? Assumption: Deformation of the panel is uniform. SOLUTION:

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3.5 Thermal Stress: Most engineering materials expand when heated and contract when cooled. The strain due to a temperature change of ΔT is called thermal strain and is obtained by: 𝜀𝑇 = 𝛼∆𝑇,

𝑎𝑛𝑑 𝛿𝑇 = 𝛼∆𝑇𝐿 … … … … . (3.6)

Where 𝜀𝑇 : 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛, 𝛼: 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑒𝑥𝑝𝑎𝑛𝑡𝑖𝑜𝑛, 𝐿: 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ, 𝑎𝑛𝑑 𝛿𝑇 : 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛. Total Strains: strains caused by temperature changes and strains caused by applied loads are

essentially independent.the total normal strain in a body acted on both temperature changes and applied load is given by : 𝜀𝑡𝑜𝑡𝑎𝑙 = 𝜀𝜎 + 𝜀𝑇 … … … … … (3.7)

Example 3.5: Given: At a temperature of 40°F, a 0.08-in. gap exists between the ends of the two bars shown in Fig. P2.18. Bar (1) is an aluminum alloy [α = 12.5 × 10−6/°F] and bar ( 2) i s s t a i n l e s s s t e e l [ α = 9 . 6 × 10−6/°F]. The supports at A and C are rigid. Find: The lowest temperature at which the two bars contact each other? SOLUTION:

2 1

3.6 Mechanical Properties of Materials: Mechanical properties of engineering materials loaded in tension and compression are determined as per: 1. ASTM : American Society for Testing of Materials 2. JIS : Japanese Industrial Standards 3. CNS : Chinese National Standards,..... etc.  Ductile materials can undergo large inelastic strains prior to fracture. For example, structural steel and many alloys of other metals, and nylon, are characterized by their ability to yield at normal temperatures. Percentage elongation is 5 or more.  Brittle materials (for example, cast iron or concrete) exhibits little deformation before rupture and, as a result, fails suddenly without visible warning. Percentage elongation is less than around 5.  A homogeneous solid displays identical properties throughout. For example, if a bar made of the same material is divided in to several small pieces (of any shape and size) and the density (weight or mass / volume) of each piece is determined to be same then the material is homegeneous.Exceptions?  If the properties of a material are identical in all directions at a point, the material is said to be isotropic (Ex. mild steel).  A nonisotropic, or anisotropic, material displays direction-dependent properties (Ex. Composite materials).  An orthotropic material is a special case of anisotropic material in which the material properties differ in two mutually perpendicular directions (Ex. graphite, wood).  By definition, an orthotropic material has at least 2 orthogonal

planes

of

symmetry,

where

material

properties are independent of direction within each plane.  In contrast, a material without any planes of symmetry is fully anisotropic.  A material with an infinite number of symmetry planes (i.e. every plane is a plane of symmetry) is isotropic.

2 2

Conventional stress strain diagram: 

The tensile test: The most common test is tension test for metals, to obtain the stress-strain diagram of materials.



Yield strength



Offset Yield strength (Proof strength): is the stress determined by the intrsection of the line (drawn through a strain of 0.002 or 0.2%, parallel to the initial slope of the curve) -ε curve (point B- Figure 3.16).



Ultimate and fracture stresses



Geometry change of specimen: The ductility of material in tension can be characterized by its:

i.

𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 =

Lf− Lo Lo

× 100

𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑎𝑟𝑒𝑎 =

ii. 

Offset yield strength



True stress and true strain:

Ao− Af Ao

× 100

Engineering strain (ε) and True strain (𝜺𝒕 ): 𝑬𝒏𝒈𝒊𝒏𝒆𝒆𝒓𝒊𝒏𝒈 𝒔𝒕𝒓𝒂𝒊𝒏 𝜀 =

𝐿𝑓− 𝐿𝑜 𝐿𝑜

… … . . (3.12)

where Lf: final length, and Lo: original length. True strain (𝜺𝒕 ): the increment of this strain is defined as, 𝑑𝜀𝑡 =

𝑑𝐿 𝐿

… … … … . (3.13)

and from equation (3.13) we get, 𝐿 = 1 + 𝜀 … … … … . . (3.14) 𝐿𝑜 By integration of equation (3.13) and using equation (3.14), the true strain is: 𝐿

𝑑𝐿 𝐿 = 𝑙𝑛 = 𝑙𝑛(1 + 𝜀) … … … (3.15) 𝐿𝑜 𝐿𝑜 𝐿𝑜

𝜀𝑡 = ∫

Where Lo : original length, L: instantaneous length, ε: is the engineering strain.

2 3

Engineering stress()and True stress(t): 𝑬𝒏𝒈𝒊𝒏𝒆𝒆𝒓𝒊𝒏𝒈 𝒔𝒕𝒓𝒆𝒔𝒔, 𝜎 =

𝑃 … … . . (3.16). 𝐴𝑜

Assumed material is incompressible and the volume constan. As a result, 𝐴𝑜 . 𝐿𝑜 = 𝐴𝐿 … … … … . . (3.17) , where the left and right sides of this equations represent the original and the instantaeous volume, respectively. If P is the current load, then, 𝜎𝑡 =

𝑃

= 𝐴

𝑃𝐿 𝐴𝑜 𝐿𝑜

𝐿

= 𝜎 𝐿 = 𝜎(𝜀 + 1) … … . . (3.18) 𝑜

Where Ao : original area, A: instantaneous area, and : is the engineering stress. Figure 3.21 illustrated the the stress-strain diagrams for a mild steel: 

The curve, OABCF, represent the true stress-true strain relationship.



The curve, OABCDF, represent the engineering stress-engineering strain relationship.



The differences between the curves are magnified in the strain hardening rate. In particular, there is a large divergence within the necking region. Why?

Elastic vs. Plastic Behavior: • If the strain disappears when the stress is removed, the material is said to behave elastically. • The largest stress for which this occurs is called the elastic limit. When the strain does not return to zero after the stress is removed, the material is said to behave plastically.

3.7 Hooke’s Law: 

Is the linear relationship between stress and strain for linear elastic homogeneous and isotropic materials



For Below the yield stress:

  E , E  Youngs Modulus or Modulus of Elasticity 

The slope of -ε diagram beyond the yield stress is called tangent modulus,

2 4

Et : 𝐸𝑡 = 𝑑𝜎⁄𝑑𝜀. Wherease, the ratio of stress to strain at any point on the curve above the yield stress is called secant modulus, Es: 𝐸𝑠 = 𝜎⁄𝜀 , as shown in Figure 3.24. •

For the shear stress-strain diagrams,   G ,G  shear modulus of elasticity or modulus of rigidity .

• Strength is affected by alloying, heat treating, and manufacturing process but stiffness (Modulus of Elasticity) is not.

3.8 POISSON’S RATIO: • For a slender bar subjected to axial loading: • x 

x E

,& y   z  0

• The elongation in the x-direction is accompanied by a contraction in the other directions. Assuming that the material is isotropic (no directional dependence),  y   z  0 • Poisson’s ratio is defined as,  

y lateral strain    z axial strain x x

• Typical values of  for steel ranges from about 0.25 to 0.3, while for aluminum materials it is of the order of 0.33. Maximum value of  for linear elastic materials is 0.5. VOLUME CHANGE – dilatation: for the axial loading condition represented in Figure 3.27, we have 𝜎𝑦 =

𝜎𝑧 = 0 , and 𝜎𝑥 is the axial stress. The transverse strains are releated to the axial strain as,

 y   z   x

 

x E

.............( a ) .

The initial volume of the element, 𝑉𝑜 = 𝑑𝑥𝑑𝑦𝑑𝑧 The final volume, 𝑉𝑓 = (1 + 𝜀𝑥 )𝑑𝑥 . (1 + 𝜀𝑦 )𝑑𝑦 . (1 + 𝜀𝑧 )𝑑𝑧 Introducing eq.(a), we have: 𝑉𝑓 = [1 + (εx + εy + εz )]𝑑𝑥𝑑𝑦𝑑𝑧 = Vo (1 + εx − 2𝜈 εx ) ∴ 𝛥𝑉 = 𝑉𝑓 − 𝑉𝑜 = 𝑉𝑜 εx (1 − 2ν) … … … . . (𝑏) The unit volume change, or dilatation is defined as, 𝑒=

∆𝑉 … … … … … … . . (3.19) 𝑉𝑜

Substitution of eqs. (b) and (a) into (3.19) yields 𝑒 = (1 − 2𝜈)𝜀𝑥 =

1 − 2𝜈 𝜎𝑥 … … … … . (3.20) 𝐸

2 5

•

For element subjected to uniform hydrostatic pressure,

31  2  p  E k E k  bulk modulus 31  2 

e  p

•

Subjected to uniform pressure, dilatation must be negative, therefore 0   

1 2

3.9 Generalized Hooke’s law: 

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:

 x  y

E  x  y  E E 1  2   E  y  x   y  y  x , y  E E 1  2  xy  G xy ..................( 3.22 )

x 





, x 

Relation Among E, , and G:

E  G............( 3.23 ) 21   

2 6

3.10 Strain energy: Uniaxial Stress: 𝜎2

𝑆𝑡𝑟𝑎𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦, 𝑈 = ∫ 2𝐸𝑥 𝑑𝑣 𝜎2

𝑆𝑡𝑟𝑎𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝑈𝑜 , 𝑑𝑈⁄𝑑𝑉 , 𝑖𝑠: 𝑈 = 2𝐸𝑥 =

𝐸𝜀𝑥2 2

Strain energy (shear stress): 2 𝜏𝑥𝑦

𝑆𝑡𝑟𝑎𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦, 𝑈 = ∫ 2𝐺 𝑑𝑣 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝑈𝑜 , 𝑑𝑈⁄𝑑𝑉 , 𝑖𝑠: 𝑈=

2 𝜏𝑥𝑦

2𝐺

1

= 2 𝜏𝑥𝑦 𝛾𝑥𝑦

Example 3.10 PROBLEM (3.31): Given:

Figure P3.31 shows a steel block subjected to an axial compression load of

400 kN. After loading, if dimensions b and L are changed to 40.02 and 199.7 mm, respectively, and a = 60 mm,

b = 40 mm,

L = 200 mm.

y

Find:

b=40 mm x a=60 mm

400 kN

(a) Poisson's ratio. (b) The modulus of elasticity.

z

(c) The final value of the dimension a. (d) The shear modulus of elasticity. SLUTION:

2 7

L=200 mm

Example 3.11 PROBLEM (3.33) The following data are obtained from a tensile test of a 12.7-mm-diameter aluminum specimen having a gage length of 50 mm. After the specimen ruptures, the minimum (neck) diameter is found to be 8.8 mm. --------------------------------------------------------------------------------------------------Stress, MPa Strain Stress, MPa S train ---------------------------------------------------------------------------------------------------35 0.0005 284 0.0062 70 0.0010 305 0.02 104 0.0014 319 0.05 139 0.0017 326 0.08 172 0.0024 312 0.12 207 0.0030 291 0.15 242 0.0035 256 0.20 259 0.0039 (Fracture) --------------------------------------------------------------------------------------------------Plot the engineering stress-strain diagram and determine: (a) The modulus of elasticity. (b) The proportional limit. (c) The yield strength at 0.2 %. (d) The ultimate strength. (e) The percent elongation in 50 mm. (f) The percent reduction in area. (g) The true ultimate stress. (h) The tangent and secant modules at a stress level of 310 MPa.

2 8

SOLUTION:

 (MPa ) Et 326

310 300

275

242

200

Es

100

0

0.002

0.004

0.006

0.05

0.1

0.15

(a)

242(106 ) E  69 GPa 0.0035

(b)

 p  242 MPa

(c)

 y  275 MPa

(d)

 u  326 MPa

(e)

50  0.2(50)  50 (100)  20 % 50

 (f)

4

(12.7)2 

 4



(8.8) 2

4

(12.7)

( u )t  326 4

 4

(h)

N  M

(100)  52 %

2

 (g)

0.2

(12.7) 2 (8.8)

 679 MPa

2

(330  290)106  0.8 GPa 0.05 310 106 Es   10 GPa 0.031 Et 

Example 3.12 PROBLEM (3.54)

The aluminum rod, 50 mm in diameter and 1.2 m in length, of a hydraulic

ram is subjected to the maximum axial loads of ±200 kN. What are the largest diameter and the 0.05 m

largest volume of the rod during service?

200 kN

Given: E = 70 GPa,  = 0.3. SOLUTION:

1.2 m

2 9

Vo 

 4

(0.052 )(1.2)  2355(106 ) m3

x 

200(103 )  101.9 MPa  (0.052 ) 4

x 

101.9(106 )  1456  70(109 )

 t  0.3(1456)(106 )  437  d   t d  437(106 )(0.05)  0.022 mm dmax  50.022 mm

Also,

e  (1  2 ) x  0.4(1456)(106 )  582(106 ) V  eVo  582(106 )(2355)(106 )  1.371106 m3

Vmax  Vo  V  (2355  1.371)106  2356.371(103 ) mm3

3 0