Materials Science and Engineering An Introduction (9th EDITION) by William D. Callister, Jr & David G. Rethwisch

CHAPTER 6 MECHANICAL PROPERTIES OF METALS PROBLEM SOLUTIONS

Concepts of Stress and Strain 6.1 Using mechanics-of-materials principles (i.e., equations of mechanical equilibrium applied to a freebody diagram), derive Equations 6.4a and 6.4b. Solution This problem asks that we derive Equations 6.4a and 6.4b, using mechanics of materials principles. In Figure (a) below is shown a block element of material of cross-sectional area A that is subjected to a tensile force P. Also represented is a plane that is oriented at an angle  referenced to the plane perpendicular to the tensile axis; the area of this plane is A' = A/cos . In addition, the forces normal and parallel to this plane are labeled as P' and V', respectively. Furthermore, on the left-hand side of this block element are shown force components that are tangential and perpendicular to the inclined plane. In Figure (b) are shown the orientations of the applied stress , the normal stress to this plane ', as well as the shear stress ' taken parallel to this inclined plane. In addition, two coordinate axis systems in represented in Figure (c): the primed x and y axes are referenced to the inclined plane, whereas the unprimed x axis is taken parallel to the applied stress.

Normal and shear stresses are defined by Equations 6.1 and 6.3, respectively. However, we now chose to express these stresses in terms (i.e., general terms) of normal and shear forces (P and V) as

 =

P A

=

V A

For static equilibrium in the x' direction the following condition must be met:

 F x' = 0 which means that

P'  P cos  = 0 Or that

P' = P cos  Now it is possible to write an expression for the stress

'

in terms of P' and A' using the above expression and the

relationship between A and A' [Figure (a)]:

P' A'

 =

=

Pcos P = cos 2 A A cos

However, it is the case that P/A = ; and, after making this substitution into the above expression, we have Equation 6.4a--that is

  =  cos 2 Now, for static equilibrium in the y' direction, it is necessary that

F

y'

=0

=  V' + Psin  Or

V' = Psin  We now write an expression for ' as

 =

V' A'

=

=

P sin  A cos

P sin  cos A

=  sin  cos which is just Equation 6.4b.

6.2 (a) Equations 6.4a and 6.4b are expressions for normal (σ′) and shear (τ′) stresses, respectively, as a function of the applied tensile stress (σ) and the inclination angle of the plane on which these stresses are taken (θ of Figure 6.4). Make a plot showing the orientation parameters of these expressions (i.e., cos 2 θ and sin θ cos θ) versus θ. (b) From this plot, at what angle of inclination is the normal stress a maximum? (c) At what inclination angle is the shear stress a maximum? Solution (a) Below are plotted curves of cos2 (for

'

) and sin cos  (for ') versus .



(b) The maximum normal stress occurs at an inclination angle of 0. (c) The maximum shear stress occurs at an inclination angle of 45.

Stress–Strain Behavior 6.3 A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm (0.60 in. × 0.75 in.) is pulled in tension with 44,500 N (10,000 lbf) force, producing only elastic deformation. Calculate the resulting strain. Solution This problem calls for us to calculate the elastic strain that results for a copper specimen stressed in tension. The cross-sectional area is just (15.2 mm)  (19.1 mm) = 290 mm2 (= 2.90  10-4 m2 = 0.45 in.2); also, the elastic 2

modulus for Cu is given in Table 6.1 as 110 GPa (or 110  109 N/m ). Combining Equations 6.1 and 6.5 and solving for the strain yields

F A  F = = 0 E E A0 E

=

44,500 N

(2.90  10

4

m 2 )(110  109 N/m 2 )

= 1.39  10-3

6.4 A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa (30 × 10 6 psi) and an original diameter of 10.2 mm (0.40 in.) experiences only elastic deformation when a tensile load of 8900 N (2000 lbf) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.25 mm (0.010 in.). Solution We are asked to compute the maximum length of a cylindrical nickel specimen (before deformation) that is deformed elastically in tension. For a cylindrical specimen the original cross-sectional area A0 is equal to 2

 d  A0 =   0   2

where d0 is the original diameter. Combining Equations 6.1, 6.2, and 6.5 and solving for l0 leads to 2

 d  l E  0  l E d02  2 l l l E l0 = = E = =  F  F 4F E A0 Substitution into this expression, values of d0, l, E, and F given in the problem statement yields the following

l0 =

(0.25  103 m )(207  109 N/m 2 ) () (10.2  103 m )2 (4)(8900 N)

= 0.475 m = 475 mm (18.7 in.)

6.5 An aluminum bar 125 mm (5.0 in.) long and having a square cross section 16.5 mm (0.65 in.) on an edge is pulled in tension with a load of 66,700 N (15,000 lb f) and experiences an elongation of 0.43 mm (1.7 × 10 –2 in.). Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum. Solution This problem asks us to compute the elastic modulus of aluminum. For a square cross-section, A = 0

,

b02

where b0 is the edge length. Combining Equations 6.1, 6.2, and 6.5 and solving for the modulus of elasticity, E, leads to

F A Fl0  E = = 0= l b2  l  0 l0 Substitution into this expression, values of l0, b0 l, and F given in the problem statement yields the following

E=

(66,700 N) (125  103 m )

(16.5  103 m )2 (0.43  103 m )

 71.2  109 N/m 2  71.2 GPa (10.4  106 psi)

6.6 Consider a cylindrical nickel wire 2.0 mm (0.08 in.) in diameter and 3 × 10 4 mm (1200 in.) long. Calculate its elongation when a load of 300 N (67 lbf) is applied. Assume that the deformation is totally elastic. Solution In order to compute the elongation of the Ni wire when the 300 N load is applied we must employ Equations 6.2, 6.5, and 6.1. Combining these equations and solving for ∆l leads to the following

 l F l = l0 = l0 = 0 = E EA0

l0 F 2

 d  E  0   2



4l0 F E d02

since the cross-sectional area A0 of a cylinder of diameter d0 is equal to 2

 d  A0 =   0   2

Incorporating into this expression values for l0, F, and d0 in the problem statement, and realizing that for Ni, E = 207 GPa (30  106 psi) (Table 6.1), and that 3  104 mm = 30m, the wire elongation is

l =

(4)(30 m)(300 N)

( 207  109 N/m 2 )( )(2  103 m )2

= 0.0138 m = 13.8 mm (0.53 in.)

6.7 For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa (15.0 × 106 psi). (a) What is the maximum load that can be applied to a specimen with a cross-sectional area of 130 mm 2 (0.2 in.2) without plastic deformation? (b) If the original specimen length is 76 mm (3.0 in.), what is the maximum length to which it can be stretched without causing plastic deformation? Solution (a) This portion of the problem calls for a determination of the maximum load that can be applied without plastic deformation (Fy). Taking the yield strength to be 345 MPa, and employment of Equation 6.1 leads to



Fy =  y A0 = (345  106 N/m 2 )(133 mm 2 ) 

1m 

2

 103 mm  

= 44,850 N (10,000 lbf) (b) The maximum length to which the sample may be deformed without plastic deformation is determined by combining Equations 6.2 and 6.5 as

li   l0  l0  l0 (  1)     1  E 

li = l0 

Incorporating values of l0, , and E provided in the problem statement leads to the computation of li as follows:



345 MPa

li = (76 mm) 

 103 

103 MPa

  1 = 76.25 mm (3.01 in.) 

6.8 A cylindrical rod of steel (E = 207 GPa, 30 × 106 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a load of 11,100 N (2500 lb f). If the length of the rod is 500 mm (20.0 in.), what must be the diameter to allow an elongation of 0.38 mm (0.015 in.)? Solution This problem asks us to compute the diameter of a cylindrical specimen of steel in order to allow an elongation of 0.38 mm. Employing Equations 6.1, 6.2, and 6.5, assuming that deformation is entirely elastic, we may write the following expression:

F = A0

 =

F  d2   0   4 

= E = E

l l0

And upon simplification

F 

E

d 02 





l l0

 4 

Solving for the original cross-sectional area, d0, leads to

d0 =

4 l0 F  El

Incorporation of values for l0, F, E, and l provided in the problem statement yields the following:

d0 =

(4) (500  103 m ) (11,100 N) () ( 207  109 N/m 2 )(0.38  10 3 m ) 3

= 9.5  10

m = 9.5 mm (0.376 in.)

6.9 Compute the elastic moduli for the following metal alloys, whose stress-strain behaviors may be observed in the Tensile Tests module of Virtual Materials Science and Engineering (VMSE): (a) titanium, (b) tempered steel, (c) aluminum, and (d) carbon steel. How do these values compare with those presented in Table 6.1 for the same metals? Solution The elastic modulus is the slope in the linear elastic region (Equation 6.10) as

E=

  2  1 =   2  1

Since stress-strain curves for all of the metals/alloys pass through the origin, we make take 1 = 0 and 1 = 0. Determinations of 2 and 2 are possible by moving the cursor to some arbitrary point in the linear region of the curve and then reading corresponding values in the “Stress” and “Strain” windows that are located below the plot. (a) A screenshot for the titanium alloy in the elastic region is shown below.

Here the cursor point resides in the elastic region at a stress of 492.4 MPa (which is the value of 2) at a strain of 0.0049 (which is the value of 2). Thus, the elastic modulus is equal to

E=

 2  1 492.4 MPa - 0 MPa   100,500 MPa  100.5 GPa  2  1 0.0049 - 0

The elastic modulus for titanium given in Table 6.1 is 107 GPa, which is in reasonably good agreement with this value. (b) A screenshot for the tempered steel alloy in the elastic region is shown below.

Here the cursor point resides in the elastic region at a stress of 916.7 MPa (which is the value of 2) at a strain of 0.0045 (which is the value of 2). Thus, the elastic modulus is equal to

E=

 2  1 916.7 MPa  0 MPa   203, 700 MPa  203.7 GPa  2  1 0.0045  0

The elastic modulus for steel given in Table 6.1 is 207 GPa, which is in good agreement with this value. (c) A screenshot for the aluminum alloy in the elastic region is shown below.

Here the cursor point resides in the elastic region at a stress of 193.6 MPa (which is the value of 2) at a strain of 0.0028 (which is the value of 2). Thus, the elastic modulus is equal to

E=

 2  1 193.6 MPa  0 MPa   69,100 MPa  69.1 GPa  2  1 0.0028  0

The elastic modulus for aluminum given in Table 6.1 is 69 GPa, which is in excellent agreement with this value. (d) A screenshot for the carbon steel alloy in the elastic region is shown below.

Here the cursor point resides in the elastic region at a stress of 160.6 MPa (which is the value of 2) at a strain of 0.0008 (which is the value of 2). Thus, the elastic modulus is equal to

E=

 2  1 160.6 MPa  0 MPa   200,800 MPa  200.8 GPa  2  1 0.0008  0

The elastic modulus for steel given in Table 6.1 is 207 GPa, which is in reasonable agreement with this value.

6.10 Consider a cylindrical specimen of a steel alloy (Figure 6.22) 8.5 mm (0.33 in.) in diameter and 80 mm (3.15 in.) long that is pulled in tension. Determine its elongation when a load of 65,250 N (14,500 lb f) is applied. Solution This problem asks that we calculate the elongation ∆l of a specimen of steel the stress-strain behavior of which is shown in Figure 6.22. First it becomes necessary to compute the stress when a load of 65,250 N is applied using Equation 6.1 as

=

F = A0

F 2

 d0   2 



65,250 N

=

2

 8.5  103 m   2  

= 1.15  109 N/m 2  1150 MPa (170,000 psi)



Referring to Figure 6.22, at this stress level we are in the elastic region (shown in the inset) on the stress-strain curve, which corresponds to a strain of 0.0054. Now, utilization of Equation 6.2 to compute the value of l:

 l =  l0 = (0.0054)(80 mm) = 0.43 mm (0.017 in.)

6.11 Figure 6.23 shows the tensile engineering stress–strain curve in the elastic region for a gray cast iron. Determine (a) the tangent modulus at 25 MPa (3625 psi) and (b) the secant modulus taken to 35 MPa (5000 psi). Solution (a) This portion of the problem asks that the tangent modulus be determined for the gray cast iron, the stress-strain behavior of which is shown in Figure 6.23. In the figure below is shown a tangent draw on the curve at a stress of 25 MPa.

The slope of this line (i.e., ∆/∆), the tangent modulus, is computed as follows:

 57 MPa  5 MPa = = 86,700 MPa = 86.7 GPa (12.6  106 psi)  0.0006  0 (b) The secant modulus taken from the origin is calculated by taking the slope of a secant drawn from the origin through the stress-strain curve at 35 MPa (5,000 psi). This secant modulus is drawn on the curve shown below:

The slope of this line (i.e., ∆/∆), the secant modulus, is computed as follows:

 60 MPa  0 MPa = = 100,000 MPa = 100 GPa (14.5  106 psi)  0.0006  0

6.12 As noted in Section 3.15, for single crystals of some substances, the physical properties are anisotropic—that is, they depend on crystallographic direction. One such property is the modulus of elasticity. For cubic single crystals, the modulus of elasticity in a general [uvw] direction, Euvw, is described by the relationship

1 Euvw

where

E 100

and

E 111





1 E 100

1

 3

 E 100





1



  2  2   2 2   2 2 E 111  



are the moduli of elasticity in the [100] and [111] directions, respectively; α, β, and γ are

the cosines of the angles between [uvw] and the respective [100], [010], and [001] directions. Verify that the

E110

  values for aluminum, copper, and iron in Table 3.4 are correct. 

Solution

We are asked, using the equation given in the problem statement, to verify that the modulus of elasticity values along [110] directions given in Table 3.4 for aluminum, copper, and iron are correct. The , , and  parameters in the equation correspond, respectively, to the cosines of the angles between the [110] direction and [100], [010] and [001] directions. Since these angles are 45, 45, and 90, the values of , , and  are 0.707, 0.707, and 0, respectively. Thus, the given equation takes the form

1 E110 =

1 E100



1

 3

 E100



2 2 2 2 2 2   (0.707) (0.707)  (0.707) (0)  (0) (0.707)  E111  

=

Using the values of

E100

and

E111



1

1 E100



1

 (0.75) 

 E100





1



E111 

from Table 3.4 for Al (63.7 GPa and 76.1 GPa, respectively)

  1 1 1 1 =  (0.75)    E110 63.7 GPa 63.7 GPa 76.1 GPa  

Which leads to,

E110

= 72.6 GPa, the value cited in the table.

For Cu, having values of 66.7 GPa and 191.1 GPa, respectively, for

E100

and

E111

,

  1 1 1 1 =  (0.75)    E110 66.7 GPa 66.7 GPa 191.1 GPa   Thus,

E110

= 130.3 GPa, which is also the value cited in the table.

Similarly, for Fe, having values of 125.0 GPa and 272.7 GPa, respectively, for

1 E110 And

E110

=

  1 1 1  (0.75)   125.0 GPa 272.7 GPa   125.0 GPa

= 210.5 GPa, which is also the value given in the table.

E100

and

E111

,

6.13 In Section 2.6, it was noted that the net bonding energy E N between two isolated positive and negative ions is a function of interionic distance r as follows:

A B EN    r rn

(6.31)

where A, B, and n are constants for the particular ion pair. Equation 6.31 is also valid for the bonding energy between adjacent ions in solid materials. The modulus of elasticity E is proportional to the slope of the interionic force–separation curve at the equilibrium interionic separation; that is,

 dF   dr   r

E

o

Derive an expression for the dependence of the modulus of elasticity on these A, B, and n parameters (for the twoion system), using the following procedure: 1. Establish a relationship for the force F as a function of r, realizing that

F

dEN dr

2. Now take the derivative dF/dr. 3. Develop an expression for r0, the equilibrium separation. Because r 0 corresponds to the value of r at the minimum of the EN-versus-r curve (Figure 2.10b), take the derivative dE N/dr, set it equal to zero, and solve for r, which corresponds to r0. 4. Finally, substitute this expression for r0 into the relationship obtained by taking dF/dr. Solution This problem asks that we derive an expression for the dependence of the modulus of elasticity, E, on the parameters A, B, and n in Equation 6.31. It is first necessary to take dEN/dr in order to obtain an expression for the force F; this is accomplished as follows:

 A  B d   d dE  r  rn  F N  dr dr dr

=

A r

(1 + 1)



nB r

(n + 1)

The second step is to set this dE /dr expression equal to zero and then solve for r (= r0). The algebra for this N procedure is carried out in Problem 2.18, with the result that 1/(1  n)

 A r0 =   nB  

Next it becomes necessary to take the derivative of the force (dF/dr), which is accomplished as follows:

 A

d

 r2 

 dF  dr dr 

2A r

3



 nB  d  n  r  1   dr

(n)(n  1)B r (n + 2)

Now, substitution of the above expression for r0 into this equation yields

 dF  2A (n)(n  1)B    dr  3/(1  n) (n  2)/(1  n)  r  A  A 0  nB   nB    which is the expression to which the modulus of elasticity is proportional.

Elastic Properties of Materials 6.14 Using the solution to Problem 6.13, rank the magnitudes of the moduli of elasticity for the following hypothetical X, Y, and Z materials from the greatest to the least. The appropriate A, B, and n parameters (Equation 6.31) for these three materials are tabulated below; they yield EN in units of electron volts and r in nanometers: Material

A

B

n

X

1.5

7.0 × 10–6

8

Y

2.0

1.0 × 10

–5

9

Z

3.5

4.0 × 10–6

7

Solution This problem asks that we rank the magnitudes of the moduli of elasticity of the three hypothetical metals X, Y, and Z. From Problem 6.13, it was shown for materials in which the bonding energy is dependent on the interatomic distance r according to Equation 6.31, that the modulus of elasticity E is proportional to

E

2A 3/(1  n)

 A  nB  



(n)(n  1)B (n  2)/(1  n)

 A  nB  

For metal X, A = 1.5, B = 7  10-6, and n = 8. Therefore,

(2)(1.5)

E  

1.5 

(8) (7  106 ) 

3/(1  8)

  



(8)(8  1) (7  106 )

+





(8  2)/(1  8)





1.5 

 (8) (7  106 )  = 830

For metal Y, A = 2.0, B = 1  10-5, and n = 9. Hence

(2)(2.0)

E   

(9) (1  105 )  





3/(1  9)

2.0 



(9)(9  1) (1  105 )

+ 

(9  2)/(1  9)





2.0 

 (9) (1  105 ) 

= 683 And, for metal Z, A = 3.5, B = 4  10-6, and n = 7. Thus

(2)(3.5)

E   

(7) ( 4  106 )  







3/(1  7)

3.5 



(7)(7  1) (4  106 )

+ 

(7  2)/(1  7)



3.5 



 (7) ( 4  106 )  = 7425

Therefore, metal Z has the highest modulus of elasticity.

6.15 A cylindrical specimen of steel having a diameter of 15.2 mm (0.60 in.) and length of 250 mm (10.0 in.) is deformed elastically in tension with a force of 48,900 N (11,000 lb f). Using the data contained in Table 6.1, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Will the diameter increase or decrease? Solution (a) We are asked, in this portion of the problem, to determine the elongation of a cylindrical specimen of steel. To solve this part of the problem requires that we use Equations 6.1, 6.2 and 6.5. Equation 6.5 reads as follows:

 = E Substitution the expression for  from Equation 6.1 and the expression for  from Equation 6.2 leads to

F 

=E

d02 





l l0

 4 

In this equation d0 is the original cross-sectional diameter. Now, solving for l yields

l =

4 F l0  d02 E

And incorporating values of F, l0, and d0, and realizing that E = 207 GPa (Table 6.1), leads to

l =

(4)(48,900 N) (250  10 3 m ) 3

() (15.2  10 m ) (207  10 N/m 2

9

2

)

 3.25  104 m = 0.325 mm (0.013 in.)

(b) We are now called upon to determine the change in diameter, d. Using Equation 6.8 (the definition of Poisson's ratio)

= 

x d/d0 =  z  l/l0

From Table 6.1, for steel, the value of Poisson's ratio,  is 0.30. Now, solving the above expression for ∆d yields

d = 

 l d0 (0.30)(0.325 mm)(15.2 mm) =  l0 250 mm 3

= –5.9  10 The diameter will decrease since d is negative.

4

mm (–2.3  10

in.)

6.16 A cylindrical bar of aluminum 19 mm (0.75 in.) in diameter is to be deformed elastically by application of a force along the bar axis. Using the data in Table 6.1, determine the force that produces an elastic reduction of 2.5 × 10–3 mm (1.0 × 10–4 in.) in the diameter. Solution This problem asks that we calculate the force necessary to produce a reduction in diameter of 2.5  10-3 mm for a cylindrical bar of aluminum. For a cylindrical specimen, the cross-sectional area is equal to

A0 =

 d02 4

Now, combining Equations 6.1 and 6.5 leads to

 =

F F   E z A0  d2 0 4

And, since from Equation 6.8

d  d d z   x   0      d0 Substitution of this equation into the previous one gives

 d   E     d0   d02 4 F

And, solving for F leads to

F = 

d0 d E 4

Incorporating into this equation values of d0 and d from the problem statement and realizing that, for aluminum,  = 0.33 and E = 69 GPa (Table 6.1) allows us to determine the value of the force F as follows:

F = 

(19  103 m )( 2.5  106 m ) () (69  109 N/m 2 ) (4)(0.33)

= 7,800 N (1,785 lbf)

6.17 A cylindrical specimen of a metal alloy 10 mm (0.4 in.) in diameter is stressed elastically in tension. A force of 15,000 N (3,370 lbf) produces a reduction in specimen diameter of 7 × 10–3 mm (2.8 × 10–4 in.). Compute Poisson’s ratio for this material if its elastic modulus is 100 GPa (14.5 × 106 psi). Solution This problem asks that we compute Poisson's ratio for the metal alloy. From Equations 6.5 and 6.1

z =

 F F 4F = = = 2 E A0 E  d02 E  d   0  E  2

Since the transverse strain x is equal to

x =

d d0

and Poisson's ratio is defined by Equation 6.8, then

= 

x d / d0 d d E =  =  0  4F  z 4F   2   d0 E 

Now, incorporating values of d0, d, E and F from the problem statement yields the following value for Poisson's ratio

=

(10  103 m)(7  106 m ) () (100  109 N/m 2 ) (4)(15,000 N)

= 0.367

6.18 A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 and 30.04 mm, respectively, and its final length is 105.20 mm, compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 65.5 and 25.4 GPa, respectively. Solution This problem asks that we compute the original length of a cylindrical specimen that is stressed in compression. It is first convenient to compute the lateral strain x using the Poisson's ratio expression for this strain:

x =

d 30.04 mm  30.00 mm = = 1.33  103 d0 30.00 mm

In order to determine the longitudinal strain z we need Poisson's ratio, which may be computed using Equation 6.9; using values for E and G in the problem statement and solving this equation for  yields

=

E 65.5  103 MPa 1 =  1 = 0.289 2G (2) (25.4  103 MPa )

Now z may be computed from Equation 6.8 as

z = 

x 1.33  103 =  =  4.60  10-3  0.289

Using this value of z we solve for l using Equation 6.2 and the value of li given in the problem statement: 0

l0 =

=

li 1  z

105.20 mm 1 ( 4.60  103 )

= 105.69 mm

6.19 Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 10.0 mm (0.39 in.). A tensile force of 1500 N (340 lbf) produces an elastic reduction in diameter of 6.7 × 10–4 mm (2.64 × 10–5 in.). Compute the elastic modulus of this alloy, given that Poisson’s ratio is 0.35. Solution This problem asks that we calculate the modulus of elasticity of a metal that is stressed in tension. Combining Equations 6.5 and 6.1 leads to

F A  F E= = 0  =  z  z A0 z

F 2

 d   z  0   2

=

4F

 z d02

From the definition of Poisson's ratio, (Equation 6.8) and realizing that for the transverse strain,

z = 

x 

d d0

leads to

x d/d0 d =    d0

Therefore, substitution of this expression for z into the above equation for E yields

E=

4F

 z d02

=

4F 4F =    d0 d d 2   d    d0  0  

Incorporation of values for F, , d0, and d given in the problem statement (and realizing that d is negative) allows us to calculate the modulus of elasticity as follows:

E= 

(4)(1500 N)(0.35) 3

7

 (10  10 m )( 6.7  10 m )

=1011 Pa = 100 GPa (14.7  106 psi)

6.20 A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elastic modulus of 110 GPa (16.0 × 10 6 psi). A cylindrical specimen of this alloy 15.2 mm (0.60 in.) in diameter and 380 mm (15.0 in.) long is stressed in tension and found to elongate 1.9 mm (0.075 in.). On the basis of the information given, is it possible to compute the magnitude of the load necessary to produce this change in length? If so, calculate the load; if not, explain why. Solution We are asked to ascertain whether or not it is possible to compute, for brass, the magnitude of the load necessary to produce an elongation of 1.9 mm (0.075 in.). It is first necessary to compute the strain at yielding from the yield strength and the elastic modulus, and then the strain experienced by the test specimen. Then, if

(test) < (yield) deformation is elastic, and the load may be computed using Equations 6.1 and 6.5. However, if

(test) > (yield) computation of the load is not possible inasmuch as deformation is plastic and we have neither a stress-strain plot nor a mathematical expression relating plastic stress and strain. The value of (test) is determined using Equation 6.2 as follows:

 (test) =

l 1.9 mm = = 0.005 l0 380 mm

and for (yield), Equation 6.5 is used—i.e.,

 (yield) =

y E

=

240 MPa 110  103 MPa

= 0.0022

Therefore, computation of the load is not possible since (test) > (yield).

6.21 A cylindrical metal specimen 15.0 mm (0.59 in.) in diameter and 150 mm (5.9 in.) long is to be subjected to a tensile stress of 50 MPa (7250 psi); at this stress level, the resulting deformation will be totally elastic. (a) If the elongation must be less than 0.072 mm (2.83 × 10–3 in.), which of the metals in Table 6.1 are suitable candidates? Why? (b) If, in addition, the maximum permissible diameter decrease is 2.3 × 10 –3 mm (9.1 × 10–5 in.) when the tensile stress of 50 MPa is applied, which of the metals that satisfy the criterion in part (a) are suitable candidates? Why? Solution (a) This part of the problem asks that we ascertain which of the metals in Table 6.1 experience an elongation of less than 0.072 mm when subjected to a tensile stress of 50 MPa. The maximum strain that may be sustained, (using Equation 6.2) is just

=

l 0.072 mm = = 4.8  10-4 l0 150 mm

Since the stress level is given (50 MPa), using Equation 6.5 it is possible to compute the minimum modulus of elasticity that is required to yield this minimum strain. Hence

E=

 50 MPa = = 104.2 GPa  4.8  104

Which means that those metals with moduli of elasticity greater than this value are acceptable candidates--namely, Cu, Ni, steel, Ti and W. (b) This portion of the problem further stipulates that the maximum permissible diameter decrease is 2.3  -3

10

mm when the tensile stress of 50 MPa is applied. This translates into a maximum lateral strain x(max) as

 x(max) =

d 2.3  103 mm = =  1.53  10-4 d0 15.0 mm

But, since the specimen contracts in this lateral direction, and we are concerned that this strain be less than 1.53 

 10-4, then the criterion for this part of the problem may be stipulated as

d  1.53  10-4 . d0

Now, Poisson’s ratio is defined by Equation 6.8 as



x z

For each of the metal alloys let us consider a possible lateral strain,

x 

d d0

. Furthermore, since the deformation is

elastic, then, from Equation 6.5, the longitudinal strain, z is equal to

z 

 E

Substituting these expressions for x and z into the definition of Poisson’s ratio we have

d x d     0  z E which leads to the following:



d    d0 E

Using values for  and E found in Table 6.1 for the six metal alloys that satisfy the criterion for part (a), and for  =

 50 MPa, we are able to compute a





d d0

for each of Cu, Ni, steel, Ti and W as follows:

d (0.34)(50  106 N/m 2 ) (copper)   1.55  10 4 d0 110  109 N/m 2

d (0.34)(50  106 N/m 2 ) (titanium)   1.59  104 9 2 d0 107  10 N/m



d (0.31)(50  106 N/m 2 ) (nickel)   7.49  10 5 d0 207  109 N/m 2





d (0.30)(50  106 N/m 2 ) (steel)   7.25  105 d0 207  109 N/m 2

d (0.28)(50  106 N/m 2 ) (tungsten)   3.44  105 9 2 d0 407  10 N/m

Thus, copper and titanium alloys will experience a negative transverse strain greater than 1.53  104. This means that the following alloys satisfy the criteria for both parts (a) and (b) of this problem: nickel, steel, and tungsten.

6.22 A cylindrical metal specimen 10.7000 mm in diameter and 95.000 mm long is to be subjected to a tensile force of 6300 N; at this force level, the resulting deformation will be totally elastic. (a) If the final length must be less than 95.040 mm, which of the metals in Table 6.1 are suitable candidates? Why? (b) If, in addition, the diameter must be no greater than 10.698 mm while the tensile force of 6300 N is applied, which of the metals that satisfy the criterion in part (a) are suitable candidates? Why? Solution (a) This part of the problem asks that we ascertain which of the metals in Table 6.1 experience an increase in length from 95.000 mm (l0) to 95.040 mm (li) when subjected to a tensile force of 6300 N. We may make this determination by computing the minimum modulus of elasticity required using data given in the problem statement; those metals having values greater than this minimum satisfy the criterion. Modulus of elasticity is determined using Equation 6.5 ( = E), which requires that we first compute values for the stress and strain. The maximum strain that may be sustained, (using Equation 6.2) is just

=

li  l0 95.040 mm  95.000 mm = = 4.21  104 l0 95.000 mm

The maximum stress level is computed using Equation 6.1; original cross-sectional area (A0) may be determined from original specimen diameter (d0 = 10.7000 mm). Thus, the stress is equal to



F  A0

F 2

 d   0  2



6300 N 2

 10.7000  103 m    2  

70.1  106 N/m 2  70.1 MPa We now calculate the minimum modulus of elasticity incorporating these stress and strain values into Equation 6.5. Hence

E=

 70.1 MPa = = 167  103 MPa  167 GPa  4.21  104

Which means that those metals with moduli of elasticity greater than this value are acceptable candidates--namely, Ni, steel, and W.

(b) For this part of the problem it is necessary to ascertain which of nickel, steel, and tungsten will have diameters of no greater than 10.698 while the 6300 N force is being applied. This determination can be made by computing the maximum allowable Poisson's ratio (Equation 6.8) from the ratio of transverse and longitudinal strains (x and z, respectively). The strain calculated in part (a) is a longitudinal strain—i.e.,

 z = 4.21  104 Transverse strain (x)is computed using the transverse analog of Equation 6.2—i.e.,

x 

di  d0 d0

In this equation di is the diameter while the stress is being applied—i.e., 10.698 mm. Thus

x 

10.698 mm  10.7000 mm  0.002 mm   1.87  104 10.700 mm 10.700 mm

We now compute Poisson's ratio using Equation 6.8 as follows:

 

x  1.87  10 4   0.444 z 4.21  104

This result means that for deformed diameters to be less than 10.698 mm, the value of Poisson's ratio must be less than 0.444—therefore, all of nickel ( = 0.31), steel ( = 0.30), and tungsten ( = 0.28) are candidates.

6.23 Consider the brass alloy for which the stress–strain behavior is shown in Figure 6.12. A cylindrical specimen of this material 10.0 mm (0.39 in.) in diameter and 101.6 mm (4.0 in.) long is pulled in tension with a force of 10,000 N (2250 lbf). If it is known that this alloy has a value for Poisson’s ratio of 0.35, compute (a) the specimen elongation and (b) the reduction in specimen diameter. Solution (a) This portion of the problem asks that we compute the elongation of the brass specimen. The first calculation necessary is that of the applied stress using Equation 6.1, as

 =

F = A0

F 2

 d   0   2

=

10,000 N 2

 10  103 m    2  

= 127 MPa (17,900 psi)

From the stress-strain plot in Figure 6.12, this stress corresponds to a strain of about 1.5  103. From the definition of strain, Equation 6.2

l =  l0 = (1.5  10-3 ) (101.6 mm) = 0.15 mm (6.0  103 in.) (b) In order to determine the reduction in diameter ∆d, it is necessary to use Equation 6.8 and the definition of lateral strain (i.e., x = ∆d/d0) as follows

d = d0 x =  d0  z = (10 mm)(0.35) (1.5  10 3 ) = –5.25  103 mm (–2.05  104 in.)

6.24 A cylindrical rod 120 mm long and having a diameter of 15.0 mm is to be deformed using a tensile load of 35,000 N. It must not experience either plastic deformation or a diameter reduction of more than 1.2 × 10 –2 mm. Of the following materials listed, which are possible candidates? Justify your choice(s). Material Aluminum alloy Titanium alloy Steel alloy Magnesium alloy

Modulus of Elasticity (GPa) 70 105 205 45

Yield Strength (MPa) 250 850 550 170

Poisson’s Ratio 0.33 0.36 0.27 0.35

Solution This problem asks that we assess the four alloys relative to the two criteria presented. The first criterion is that the material not experience plastic deformation when the tensile load of 35,000 N is applied; this means that the stress corresponding to this load not exceed the yield strength of the material. Upon computing the stress

 =

F = A0

F 2

 d0   2 



35,000 N

=

2

 15  103 m   2  

= 200  106 N/m 2 = 200 MPa



Of the alloys listed, the Al, Ti and steel alloys have yield strengths greater than 200 MPa. Relative to the second criterion (i.e., that d be less than 1.2  102 mm), it is necessary to calculate the change in diameter ∆d for these three alloys. It is first necessary to generate an expression for Poisson's ratio from Equation 6.8, where x = d/d0, and, from Equation 6.5, y = E. Thus

d x d E d =  =  0   z  d0 E Now, solving for ∆d from this expression,

d = 

  d0 E

From the data presented in the table in the problem statement, we now compute the value of ∆ d for each of the four alloys. For the aluminum alloy (E = 70 GPa,  = 0.33, and taking the stress  to be 200 MPa)

d = 

(0.33)(200 MPa)(15 mm) 3

70  10 MPa

=  1.41  103 mm

Therefore, the Al alloy is not a candidate because its ∆d is greater than 1.2  102 mm. For the steel alloy (E = 205 GPa,  = 0.27)

d = 

(0.27)(200 MPa)(15 mm) 205  103 MPa

=  0.40  10 2 mm

Therefore, the steel is a candidate. For the Ti alloy (E = 105 GPa,  = 0.36)

d = 

(0.36)(200 MPa)(15 mm) 105  103

Hence, the titanium alloy is also a candidate.

MPa

=  1.0  102 mm

6.25 A cylindrical rod 500 mm (20.0 in.) long and having a diameter of 12.7 mm (0.50 in.) is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than 1.3 mm (0.05 in.) when the applied load is 29,000 N (6500 lb f), which of the four metals or alloys listed in the following table are possible candidates? Justify your choice(s).

Material Aluminum alloy Brass alloy Copper Steel alloy

Modulus of Elasticity (GPa) 70 100 110 207

Yield Strength (MPa) 255 345 210 450

Tensile Strength (MPa) 420 420 275 550

Solution This problem asks that we ascertain which of four metal alloys will not (1) experience plastic deformation, and (2) elongate more than 1.3 mm when a tensile load of 29,000 N is applied. It is first necessary to compute the stress using Equation 6.1; a material to be used for this application must necessarily have a yield strength greater than this value. Thus,

 =

F = A0

F 2

 d   0  2



29,000 N 2

 12.7  103 m    2  

= 230 MPa

Of the metal alloys listed, aluminum, brass and steel have yield strengths greater than this stress. Next, we must compute the elongation produced in each of aluminum, brass, and steel by combining Equations 6.2 and 6.5 in order to determine whether or not this elongation is less than 1.3 mm. For aluminum (E = 70 GPa = 70  103 MPa)

l =  l0 =

 l0 (230 MPa)(500 mm) = = 1.64 mm E 70  103 MPa

Thus, aluminum is not a candidate because its l is greater than 1.3 mm. For brass (E = 100 GPa = 100  103 MPa)

l =

 l0 (230 MPa)(500 mm) = = 1.15 mm E 100  103 MPa

Thus, brass is a candidate. And, for steel (E = 207 GPa = 207  103 MPa)

l =

 l0 (230 MPa)(500 mm) = = 0.56 mm E 207  103 MPa

Therefore, of these four alloys, only brass and steel satisfy the stipulated criteria.

Tensile Properties 6.26 Figure 6.22 shows the tensile engineering stress–strain behavior for a steel alloy. (a) What is the modulus of elasticity? (b) What is the proportional limit? (c) What is the yield strength at a strain offset of 0.002? (d) What is the tensile strength? Solution (a) Shown below is the inset of Figure 6.22.

The elastic modulus is just the slope of the initial linear portion of the curve; or, from Equation 6.10

E=

 2  1  2  

Inasmuch as the linear segment passes through the origin, let us take both  and  to be zero. If we arbitrarily take

 = 0.005, as noted in the above plot,  = 1050 MPa. Using these stress and strain values we calculate the elastic modulus as follows:

E=

 2   1 1050 MPa  0 MPa =  210  103 MPa = 210 GPa  2   0.005  0

The value given in Table 6.1 is 207 GPa. (b) The proportional limit is the stress level at which linearity of the stress-strain curve ends, which is approximately 1370 MPa (200,000 psi). (c) As noted in the plot below, the 0.002 strain offset line intersects the stress-strain curve at approximately 1600 MPa (232,000 psi).

(d) The tensile strength (the maximum on the curve) is approximately 1950 MPa (283,000 psi), as noted in the following plot.

6.27 A cylindrical specimen of a brass alloy having a length of 100 mm (4 in.) must elongate only 5 mm (0.2 in.) when a tensile load of 100,000 N (22,500 lb f) is applied. Under these circumstances, what must be the radius of the specimen? Consider this brass alloy to have the stress–strain behavior shown in Figure 6.12. Solution We are asked to calculate the radius of a cylindrical brass specimen in order to produce an elongation of 5 mm when a load of 100,000 N is applied. It first becomes necessary to compute the strain corresponding to this elongation using Equation 6.2 as

=

l 5 mm = = 5  102 l0 100 mm

From Figure 6.12, a stress of 335 MPa (49,000 psi) corresponds to this strain. Since for a cylindrical specimen, stress, force, and initial radius r0 are related (Equation 6.1) as

 =

F F  A0  r 2 0

then taking F = 10,000 N and  = 335 MPa = 335  106 N/m2, r0 is calculated as follows

r0 =

F = 

100,000 N  (335  106 N/m 2 )

= 0.0097 m = 9.7 mm (0.38 in.)

6.28 A load of 140,000 N (31,500 lbf) is applied to a cylindrical specimen of a steel alloy (displaying the stress–strain behavior shown in Figure 6.22) that has a cross-sectional diameter of 10 mm (0.40 in.). (a) Will the specimen experience elastic and/or plastic deformation? Why? (b) If the original specimen length is 500 mm (20 in.), how much will it increase in length when this load is applied? Solution This problem asks us to determine the deformation characteristics of a steel specimen, the stress-strain behavior for which is shown in Figure 6.22. (a) In order to ascertain whether the deformation is elastic or plastic, we must first compute the stress, then locate it on the stress-strain curve, and, finally, note whether this point is on the elastic or elastic + plastic region. Thus, from Equation 6.1

 =

F F 140,000 N =  = 1,782 MPa (250,000 psi) 2 2 A0  d0   10  10 3 m      2  2  

The 1782 MPa point is beyond the linear portion of the curve, and, therefore, the deformation will be both elastic and plastic. (b) This portion of the problem asks us to compute the increase in specimen length. From the stress-strain curve, the strain at 1782 MPa is approximately 0.017. Thus, from Equation 6.2

l =  l0 = (0.017)(500 mm) = 8.5 mm (0.34 in.)

6.29 A bar of a steel alloy that exhibits the stress–strain behavior shown in Figure 6.22 is subjected to a tensile load; the specimen is 375 mm (14.8 in.) long and has a square cross section 5.5 mm (0.22 in.) on a side. (a) Compute the magnitude of the load necessary to produce an elongation of 2.25 mm (0.088 in.). (b) What will be the deformation after the load has been released? Solution (a) We are asked to compute the magnitude of the load necessary to produce an elongation of 2.25 mm for the steel displaying the stress-strain behavior shown in Figure 6.22. First, calculate the strain, and then the corresponding stress from the plot. Using Equation 6.2, the strain  is equal to

=

l 2.25 mm = = 0.006 l0 375 mm

This is within the elastic region; from the inset of Figure 6.22, this corresponds to a stress of about 1250 MPa (180,000 psi). Now, from Equation 6.1

F =  A0 =  b2 in which b is the cross-section side length. Thus,

F = (1250  106 N/m 2 ) (5.5  103 m )2 = 37,800 N (8500 lbf ) (b) After the load is released there will be no net deformation since the material was strained only elastically when the load was applied.

6.30 A cylindrical specimen of stainless steel having a diameter of 12.8 mm (0.505 in.) and a gauge length of 50.800 mm (2.000 in.) is pulled in tension. Use the load–elongation characteristics shown in the following table to complete parts (a) through (f). Load N 0 12,700 25,400 38,100 50,800 76,200 89,100 92,700 102,500 107,800 119,400 128,300 149,700 159,000 160,400 159,500 151,500 124,700

Length lbf 0 2,850 5,710 8,560 11,400 17,100 20,000 20,800 23,000 24,200 26,800 28,800 33,650 35,750 36,000 35,850 34,050 28,000

mm 50.800 50.825 50.851 50.876 50.902 50.952 51.003 51.054 51.181 51.308 51.562 51.816 52.832 53.848 54.356 54.864 55.880 56.642

in. 2.000 2.001 2.002 2.003 2.004 2.006 2.008 2.010 2.015 2.020 2.030 2.040 2.080 2.120 2.140 2.160 2.200 2.230

Fracture (a) Plot the data as engineering stress versus engineering strain. (b) Compute the modulus of elasticity. (c) Determine the yield strength at a strain offset of 0.002. (d) Determine the tensile strength of this alloy. (e) What is the approximate ductility, in percent elongation? (f) Compute the modulus of resilience. Solution This problem calls for us to make a stress-strain plot for stainless steel, given its tensile load-length data, and then to determine some of its mechanical characteristics. (a) The data are plotted below on two plots: the first corresponds to the entire stress-strain curve, while for the second, the curve extends to just beyond the elastic region of deformation.

(b) The elastic modulus is the slope in the linear elastic region (Equation 6.10)—i.e.,

E=

 2  1  2  1

Because the stress-strain curve passes through the origin, to simplify the computation let us take both  and  to be zero. If we select  = 400 MPa, its corresponding strain on the plot  is about 0.002. Thus the elastic modulus is equal to

E=

400 MPa  0 MPa = 200  103 MPa = 200 GPa (29  106 psi) 0.002  0

(c) For the yield strength, the 0.002 strain offset line is drawn dashed in the lower plot. It intersects the stress-strain curve at approximately 750 MPa (112,000 psi ). (d) The tensile strength is approximately 1250 MPa (180,000 psi), corresponding to the maximum stress on the complete stress-strain plot. (e) Ductility, in percent elongation, is just the plastic strain at fracture, multiplied by one-hundred. The total fracture strain at fracture is 0.115; subtracting out the elastic strain (which is about 0.003) leaves a plastic strain of 0.112. Thus, the ductility is about 11.2%EL. (f) From Equation 6.14, the modulus of resilience is just

Ur =

 2y 2E

which, using values of y and E computed above, (750 MPa = 750  106 N/m2 and 200 GPa = 200  109 N/m2, respectively) give a modulus of resilience of

Ur =

(750  106 N/m 2 )2 (2) (200  109 N/m 2 )

= 1.40  106 N/m 2 = 1.40  106 J/m 3

(210 in.-lbf /in.3 )

6.31 A specimen of magnesium having a rectangular cross section of dimensions 3.2 mm × 19.1 mm ( in. ×

3 4

1 8

in.) is deformed in tension. Using the load–elongation data shown in the following table, complete parts

(a) through (f). Load lbf 0 310 625 1265 1670 1830 2220 2890 3170 3225 3110 2810

Length N 0 1380 2780 5630 7430 8140 9870 12,850 14,100 14,340 13,830 12,500

in. 2.500 2.501 2.502 2.505 2.508 2.510 2.525 2.575 2.625 2.675 2.725 2.775

mm 63.50 63.53 63.56 63.62 63.70 63.75 64.14 65.41 66.68 67.95 69.22 70.49

Fracture (a) Plot the data as engineering stress versus engineering strain. (b) Compute the modulus of elasticity. (c) Determine the yield strength at a strain offset of 0.002. (d) Determine the tensile strength of this alloy. (e) Compute the modulus of resilience. (f) What is the ductility, in percent elongation? Solution This problem calls for us to make a stress-strain plot for a magnesium, given its tensile load-length data, and then to determine some of its mechanical characteristics. (a) The data are plotted below on two plots: the first corresponds to the entire stress-strain curve, while for the second, the curve extends just beyond the elastic region of deformation.

(b) The elastic modulus is the slope in the linear elastic region (Equation 6.10)—i.e.,

E=

 2  1  2  1

Because the stress-strain curve passes through the origin, to simplify the computation let us take both  and  to be zero. If we select  = 50 MPa, its corresponding strain on the plot  is about 0.001. Thus the elastic modulus is equal to

E=

 2  1 50 MPa  0 MPa =  50  103 MPa = 50 GPa (7.3  106 psi)  2  1 0.001  0

(c) For the yield strength, the 0.002 strain offset line is drawn dashed. It intersects the stress-strain curve at approximately 140 MPa (20,300 psi). (d) The tensile strength is approximately 230 MPa (33,350 psi), corresponding to the maximum stress on the complete stress-strain plot. (e) From Equation 6.14, the modulus of resilience is just

Ur =

 2y 2E

which, using data computed above, yields a value of

Ur =

(140  106 N/m 2 )2 (2) (50  10 N/m 9

2

)

= 1.96  105 N/m 2 = 1.96  105 J/m3

(28.4 in.-lbf /in.3 )

(f) The ductility, in percent elongation, is just the plastic strain at fracture, multiplied by one-hundred. The total fracture strain at fracture is 0.110; subtracting out the elastic strain (which is about 0.003) leaves a plastic strain of 0.107. Thus, the ductility is about 10.7%EL.

6.32 A cylindrical metal specimen 15.00 mm in diameter and 120 mm long is to be subjected to a tensile force of 15,000 N. (a) If this metal must not experience any plastic deformation, which of aluminum, copper, brass, nickel, steel, and titanium (Table 6.2) are suitable candidates? Why? (b) If, in addition, the specimen must elongate no more than 0.070 mm, which of the metals that satisfy the criterion in part (a) are suitable candidates? Why? Base your choices on data found in Table 6.1. Solution (a) In order to determine which of the metals in Table 6.2 do not experience plastic deformation we need to compute the applied engineering stress; any metal in the table that has a yield strength greater than this value will not plastically deform. The applied engineering stress is computed using Equation 6.1 as follows:



F A0

which, for a cylindrical specimen of diameter d0 takes the form:

F



2

 d0   2 



For F = 15,000 N and d0 = 15.00 mm (15  103 m) the stress is equal to

15,000 N



2

 15  103 m   2  

 84.9  106 N/m 2  84.9 MPa



Of those 6 alloys in Table 6.2, only nickel, steel, and titanium have yield strengths greater than 84.9 MPa. (b) Because none of these three metals has experienced plastic deformation it is possible to compute the elongation l by combining Equations 6.2 and 6.5. From Equation 6.5 it is the case that



 E

in which E is the modulus of elasticity. If we incorporate the definition of the strain (Equation 6.2) into this expression, the following results:

l   l0 E And solving for the elongation l leads to

l 

l0 E

In order to calculate the elongation for each of these metals we incorporate the original length 120 mm (120  103 m), the stress computed above (84.9 MPa = 84.9  106 N/m2), and, for each of the three metals, its modulus of elasticity found in Table 6.1. For Ni (E = 207 GPa = 207  109 N/m2 ), therefore:

l 

(120  103 m)(84.9  106 N/m 2 ) 207  109 N/m 2

 4.9  10 5 m = 0.049 mm

Thus, Ni satisfies this criterion since its elongation is less than 0.070 mm. For steel (E = 207 GPa = 207  109 N/m2) the elongation will also be 0.049 mm since its elastic modulus is the same as Ni—i.e., steel meets the criterion). For titanium (E = 107 GPa = 107  109 N/m2)

l 

(120  103 m)(84.9  106 N/m 2 ) 107 

109

N/m 2

 9.5  10 5 m = 0.095 mm

And, titanium does not satisfy the criterion because its elongation (0.095 mm) is greater than 0.070 mm.

6.33 For the titanium alloy whose stress–strain behavior can be observed in the Tensile Tests module of Virtual Materials Science and Engineering (VMSE), determine the following: (a) the approximate yield strength (0.002 strain offset) (b) the tensile strength (c) the approximate ductility, in percent elongation How do these values compare with those for the two Ti-6Al-4V alloys presented in Table B.4 of Appendix B? Solution (a) A screenshot of the entire stress-strain curve for the Ti alloy is shown below.

The intersection of the straight line drawn parallel to the initial linear region of the curve and offset at a strain of 0.002 with this curve is at approximately 720 MPa. (b) The maximum stress on the stress-strain curve is 1000 MPa, the value of the tensile strength. (c) The approximate percent elongation corresponds to the strain at fracture multiplied by 100 (i.e., 12%) minus the maximum elastic strain (i.e., value of strain at which the linearity of the curve ends multiplied by 100—in this case about 0.5%); this gives a value of about 11.5%EL.

From Table B.4 in Appendix B, yield strength, tensile strength, and percent elongation values for the annealed Ti-6Al-4V are 830 MPa, 900 MPa, and 14%EL, while for the solution heat treated and aged alloy, the corresponding values are 1103 MPa, 1172 MPa, and 10%EL. Thus, tensile strength and percent elongation values for the VMSE alloy are slightly lower than for the annealed material in Table B.4 (720 vs 830 MPa, and 11.5 vs. 14 %EL), whereas the tensile strength is slightly higher (1000 vs. 900 MPa).

6.34 For the tempered steel alloy whose stress-strain behavior can be observed in the Tensile Tests module of Virtual Materials Science and Engineering (VMSE), determine the following: (a) the approximate yield strength (0.002 strain offset) (b) the tensile strength (c) the approximate ductility, in percent elongation How do these values compare with those for the oil-quenched and tempered 4140 and 4340 steel alloys presented in Table B.4 of Appendix B? Solution (a) A screenshot of the entire stress-strain curve for the tempered steel alloy is shown below.

The intersection of the straight line drawn parallel to the initial linear region of the curve and offset at a strain of 0.002 with this curve is at approximately 1450 MPa. (b) As noted on this stress-strain curve, the maximum stress is about 1650 MPa, the value of the tensile strength.

(c) The approximate percent elongation corresponds to the strain at fracture multiplied by 100 (i.e., 14.8%) minus the maximum elastic strain (i.e., value of strain at which the linearity of the curve ends multiplied by 100—in this case about 0.8%); this gives a value of about 14.0%EL. For the oil-quenched and tempered 4140 and 4340 steel alloys, yield strength values presented in Table B.4 of Appendix B are 1570 MPa and 1620 MPa, respectively; these values are somewhat larger than the 1450 MPa for the tempered steel alloy of VMSE. Tensile strength values for these 4140 and 4340 alloys are, respectively 1720 MPa and 1760 MPa (compared to 1650 MPa for the VMSE steel). And, finally, the respective ductilities for the 4140 and 4340 alloys are 11.5%EL and 12%EL, which are slightly lower than the 14%EL value for the VMSE steel.

6.35 For the aluminum alloy whose stress strain behavior can be observed in the Tensile Tests module of Virtual Materials Science and Engineering (VMSE), determine the following: (a) the approximate yield strength (0.002 strain offset) (b) the tensile strength (c) the approximate ductility, in percent elongation How do these values compare with those for the 2024 aluminum alloy (T351 temper) presented in Table B.4 of Appendix B? Solution (a) A screenshot of the entire stress-strain curve for the aluminum alloy is shown below.

The intersection of a straight line drawn parallel to the initial linear region of the curve and offset at a strain of 0.002 with this curve is at approximately 300 MPa. (b) As noted on this stress-strain curve, the maximum stress is about 480 MPa, the value of the tensile strength.

(c) The approximate percent elongation corresponds to the strain at fracture multiplied by 100 (i.e., 22.4%) minus the maximum elastic strain (i.e., value of strain at which the linearity of the curve ends multiplied by 100—in this case about 0.5%); this gives a value of about 21.9%EL. For the 2024 aluminum alloy (T351 temper), the yield strength value presented in Table B.4 of Appendix B is 325, which is slightly larger than the 300 MPa for the aluminum alloy of VMSE. The tensile strength value for the 2024-T351 is 470 MPa (compared to 480 MPa for the VMSE alloy). And, finally, the ductility for 2024-T351 is 20%EL, which is about the same as for the VMSE aluminum (21.9%EL).

6.36 For the (plain) carbon steel alloy whose stress-strain behavior can be observed in the Tensile Tests module of Virtual Materials Science and Engineering (VMSE), determine the following: (a) the approximate yield strength (b) the tensile strength (c) the approximate ductility, in percent elongation Solution (a) A screenshot of the entire stress-strain curve for the plain carbon steel alloy is shown below.

Inasmuch as the stress-strain curve displays the yield point phenomenon, we take the yield strength as the lower yield point, which, for this steel, is about 225 MPa. (b) As noted on this stress-strain curve, the maximum stress is about 275 MPa, the value of the tensile strength. (c) The approximate percent elongation corresponds to the strain at fracture multiplied by 100 (i.e., 43.0%) minus the maximum elastic strain (i.e., value of strain at which the linearity of the curve ends multiplied by 100—in this case about 0.6%); this gives a value of about 42.4%EL.

6.37 A cylindrical metal specimen having an original diameter of 12.8 mm (0.505 in.) and gauge length of 50.80 mm (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 8.13 mm (0.320 in.), and the fractured gauge length is 74.17 mm (2.920 in.). Calculate the ductility in terms of percent reduction in area and percent elongation. Solution This problem calls for the computation of ductility in both percent reduction in area and percent elongation. Percent reduction in area, for a cylindrical specimen, is computed using Equation 6.12 as 2

 A0  Af 

%RA = 

A0 

2  df   d0       2  2 

   100 

2



 d   0   2

 100

in which d0 and df are, respectively, the original and fracture cross-sectional areas. Thus, 2

%RA =

2

 12.8 mm   8.13 mm      2 2    



2

 12.8 mm    2  

 100 = 60%

While, for percent elongation, we use Equation 6.11 as

 l f  l0 

%EL =  

=

l0

  100 

74.17 mm  50.80 mm  100 = 46% 50.80 mm

6.38 Calculate the moduli of resilience for the materials having the stress–strain behaviors shown in Figures 6.12 and 6.22. Solution This problem asks us to calculate the moduli of resilience for the materials having the stress-strain behaviors shown in Figures 6.12 and 6.22. According to Equation 6.14, the modulus of resilience Ur is a function of the yield strength and the modulus of elasticity as

Ur =

 2y 2E

The values for  and E for the brass in Figure 6.12 are determined in Example Problem 6.3 as 250 MPa (36,000 y

psi) and 93.8 GPa (13.6  106 psi), respectively. Thus

Ur =

(250 MPa)2 (250  106 N/m 2 )2  (2) (93.8 GPa ) (2)(93.8  109 N/m 2 )

= 3.32  105 N/m 2 = 3.32  105 J/m 3

(48.2 in.-lbf /in.3)

Values of the corresponding parameters for the steel alloy (Figure 6.22) are determined in Problem 6.26 as 1600 MPa (232,000 psi) and 210 GPa (30.5  106 psi), respectively, and therefore

Ur =

(1600 MPa)

2

(2) ( 210 GPa )

=

(1600  106 N/m 2 )2 (2)(210  109 N/m 2 )

 6.10  106 N/m 2 = 6.10  106 J/m 3 (882 in.-lbf /in.3 )

6.39 Determine the modulus of resilience for each of the following alloys: Yield Strength Material Steel alloy Brass alloy Aluminum alloy Titanium alloy

MPa 830 380 275 690

psi 120,000 55,000 40,000 100,000

Use the modulus of elasticity values in Table 6.1. Solution The moduli of resilience of the alloys listed in the table may be determined using Equation 6.14, that is

Ur =

 2y 2E

Yield strength values are provided in this table, whereas the elastic moduli are tabulated in Table 6.1. For steel (E = 207 GPa)

Ur =

(830 MPa)2 (830  106 N/m 2 )2  (2)(207 GPa) (2) ( 207  109 N/m 2 )

 1.66  106 N/m 2  1.66  106 J/m 3

(240

in.-lbf /in.3 )

For the brass (E = 97 GPa)

Ur =

(380 MPa) (380  106 N/m 2 )2  (2)(97 GPa) (2) (97  109 N/m 2 )

= 7.44  105 N/m 2 = 7.44  105 J/m 3

(108

in.-lbf /in.3 )

For the aluminum alloy (E = 69 GPa)

Ur =

(275 MPa)2 ( 275  106 N/m 2 )2  (2)(69 GPa) (2) (69  109 N/m 2 )

= 5.48  105 N/m 2 = 5.48  105 J/m 3 And, for the titanium alloy (E = 107 GPa)

(80.0

in.-lbf /in.3 )

Ur =

(690 MPa)2 (690  106 N/m 2 )2  (2)(107 GPa) (2) (107  109 N/m 2 )

= 2.22  106 N/m 2 = 2.22  106 J/m 3

(323

in.-lbf /in.3 )

6.40 A steel alloy to be used for a spring application must have a modulus of resilience of at least 2.07 MPa (300 psi). What must be its minimum yield strength? Solution The modulus of resilience, yield strength, and elastic modulus of elasticity are related to one another through Equation 6.14, that is

Ur 

 2y 2E

Solving for y from this expression yields

y=

2U r E

The value of E for steel given in Table 6.1 is 207 GPa. Using this elastic modulus value leads to a yield strength of

y=

2U r E =

(2)(2.07 MPa) ( 207  103 MPa )

= 926 MPa (134,000 psi)

6.41 Using data found in Appendix B, estimate the modulus of resilience (in MPa) of cold-rolled 17-7PH stainless steel. Solution To solve this problem it is necessary to use Equation 6.14—viz.

Ur 

 2y 2E

From Table B.2 (Appendix B) the modulus of elasticity for 17-7PH stainless steel is 204 GPa (204  103 MPa). Furthermore, the yield strength for this cold-rolled stainless steel is 1210 MPa (Table B.4). Insertion of these two values into Equation 6.14 leads to the following:

Ur 

(1210 MPa)2 (2)(204  103 MPa)

 3.59 MPa

True Stress and Strain 6.42 Show that Equations 6.18a and 6.18b are valid when there is no volume change during deformation. Solution To show that Equation 6.18a is valid, we must first rearrange Equation 6.17 as

Ai =

A0 l0 li

Substituting this expression into Equation 6.15 yields

T =

 l  F F F  li   = = i   Ai A0l0 A0  l0   l0  li

From Equation 6.2

=

li  l0 li   1 l0 l0

Or

li = 1+ l0 Substitution of this expression into the equation above leads to

 li   =  (1 +  )  l0 

T =   which is Equation 6.18a.

For Equation 6.18b, true strain is defined in Equation 6.16 as

 li    l0 

 T = ln 

Substitution the of the li-l0 ratio of Equation 6.2b leads to Equation 6.18b—viz.

 T = ln (1 +  )

(6.2b)

6.43 Demonstrate that Equation 6.16, the expression defining true strain, may also be represented by

∈T = ln

 A0   A  i

when the specimen volume remains constant during deformation. Which of these two expressions is more valid during necking? Why? Solution This problem asks us to demonstrate that true strain may also be represented by

 A0    Ai 

 T = ln  Rearrangement of Equation 6.17 yields

li A0 = l0 Ai Substitution of A0/Ai for li/l0 in the definition of true strain, Equation 6.16 leads to

 l     T = ln   i  = ln   l0   

The expression

 A0    Ai 

 T = ln 

A    0    Ai  

  valid during necking because A is taken as the area of the neck. is more i

6.44 Using the data in Problem 6.30 and Equations 6.15, 6.16, and 6.18a, generate a true stress–true strain plot for stainless steel. Equation 6.18a becomes invalid past the point at which necking begins; therefore, measured diameters are given in the following table for the last three data points, which should be used in true stress computations. Load N 159,500 151,500 124,700

Length lbf 35,850 34,050 28,000

mm 54.864 55.880 56.642

Diameter in. 2.160 2.200 2.230

mm 12.22 11.80 10.65

Solution Both true and engineering stress-strain data for this stainless steel are plotted below.

in. 0.481 0.464 0.419

6.45 A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.16 is produced when a true stress of 500 MPa (72,500 psi) is applied; for the same metal, the value of K in Equation 6.19 is 825 MPa (120,000 psi). Calculate the true strain that results from the application of a true stress of 600 MPa (87,000 psi). Solution We are asked to compute the true strain that results from the application of a true stress of 600 MPa (87,000 psi); other true stress-strain data are also given. It first becomes necessary to solve for n in Equation 6.19. Taking logarithms of this expression leads to

log  T  log K  nlog  T Next we rearrange this equation such that n is the dependent variable:

n=

log  T  log K log  T

We now solve for n using the following data given in the problem statement: T = 500 MPa

T = 0.16 K = 825 MPa Thus

n=

log (500 MPa)  log (825 MPa) = 0.273 log (0.16)

We now rearrange Equation 6.19 such that T is the dependent variable; we first divide both sides of the Equation 6.19 by K, which leads to the following expression:

 Tn 

T K

 becomes the dependent variable by taking the 1/n root of both sides of this expression, as 1/n

   T =  T   K 

Finally, using values of K and n, we solve for the true strain at a true stress of 600 MPa:

1/n

   T =  T   K

1/0.273

 600 MPa   825 MPa  

=

= 0.311

6.46 For some metal alloy, a true stress of 345 MPa (50,000 psi) produces a plastic true strain of 0.02. How much does a specimen of this material elongate when a true stress of 415 MPa (60,000 psi) is applied if the original length is 500 mm (20 in.)? Assume a value of 0.22 for the strain-hardening exponent, n. Solution We are asked to compute how much elongation a metal specimen will experience when a true stress of 415 MPa is applied, given the value of n and that a given true stress produces a specific true strain. Solution of this problem requires that we utilize Equation 6.19. It is first necessary to solve for K from the given true stress and strain. Rearrangement of this Equation 6.19 so that K is the dependent parameter yields

K=

T ( T ) n

The value of K is determined by substitution of the first set of data in the problem statement, as follows:

K=

345 MPa = 816 MPa (118,000 psi) (0.02 )0.22

Next we must solve for the true strain produced when a true stress of 415 MPa is applied, also using Equation 6.19. Algebraic manipulation of Equation 6.19 such that true strain is the dependent variable leads to 1/n

   T =  T   K

Incorporation into this expression the value of K and the applied true stress (415 MPa) results in the following value for true strain: 1/0.22

 415 MPa  T =   816 MPa  

= 0.0463

And from the definition of true strain, Equation 6.16, we have

 l   T = ln  i   0.0463  l0  Another way to write this expression is as follows:

e0.0463 

li l0

Now, solving for li gives

li = l0e0.0463 = (500 mm)e0.0463 = 523.7 mm (20.948 in.) And, since l0 = 500 mm, we solve for ∆l as follows:

l = li  l0 = 523.7 mm  500 mm = 23.7 mm (0.948 in.)

6.47 The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi) 60,000 70,000

True Strain 0.15 0.25

What true stress is necessary to produce a true plastic strain of 0.21? Solution For this problem, we are given two values of T and T, from which we are asked to calculate the true stress that produces a true plastic strain of 0.21. From Equation 6.19, we want to set up two simultaneous equations with two unknowns (the unknowns being K and n). Taking logarithms of both sides of Equation 6.19 leads to

log  T  log K  nlog  T The two simultaneous equations using data provided in the problem statement are as follows:

log (60,000 psi) = log K + n log (0.15) log (70,000 psi) = log K + n log (0.25) Solving for n from these two expressions yields

n=

log (60,000)  log (70,000) = 0.302 log (0.15)  log (0.25)

We solve for K by substitution of this value of n into the first simultaneous equation as



log K  log  T  nlog  T  log(60,000)  (0.302)[log(0.15)]  5.027

Thus, the value of K is equal to

K  105.027  106,400 psi

The true stress required to produce a true strain of 0.21 is determined using Equation 6.19 as follows:

 T = K ( T )n = (106,400 psi)(0.21)0.302 = 66,400 psi (460 MPa)

6.48 For a brass alloy, the following engineering stresses produce the corresponding plastic engineering strains prior to necking: Engineering Stress (MPa) 315 340

Engineering Strain 0.105 0.220

On the basis of this information, compute the engineering stress necessary to produce an engineering strain of 0.28. Solution For this problem we first need to convert engineering stresses and strains to true stresses and strains so that the constants K and n in Equation 6.19 may be determined. Since

 T   (1   )

(Equation 6.18a), we convert the

two values of engineering stress into true stresses as follows:

 T1 = (315 MPa)(1 + 0.105) = 348 MPa

 T 2 = (340 MPa)(1 + 0.220) = 415 MPa Similarly for strains—we convert engineering strains to true strains using Equation 6.18b [i.e.,

 T  ln(1   )

] as

follows:

 T1 = ln (1 + 0.105) = 0.09985

 T 2 = ln (1 + 0.220) = 0.19885 Taking logarithms of both sides of Equation 6.19 leads to

log  T  log K  nlog  T which allows us to set up two simultaneous equations for the above pairs of true stresses and true strains, with K and n as unknowns. Thus

log (348) = log K + n log (0.09985) log (415) = log K + n log (0.19885) Solving for n from these two expressions yields

n=

log (348)  log (415) = 0.256 log (0.09985)  log (0.19885)

We solve for K by substitution of this value of n into the first simultaneous equation as

log K  log  T  nlog  T  log(348)  (0.256)[log(0.09985)]  2.7977 Thus, the value of K is equal to

K  102.7977  628 MPa We now, converting  = 0.28 to true strain using Equation 6.18b as follows:

 T = ln (1 + 0.28) = 0.247 The corresponding T to give this value of T (using Equation 6.19) is just

 T = K  Tn = (628 MPa)(0.247)0.256 = 439 MPa Now converting this value of T to an engineering stress using Equation 6.18a gives

 =

T 439 MPa = = 343 MPa 1   1  0.28

6.49 Find the toughness (or energy to cause fracture) for a metal that experiences both elastic and plastic deformation. Assume Equation 6.5 for elastic deformation, that the modulus of elasticity is 103 GPa (15 × 10 6 psi), and that elastic deformation terminates at a strain of 0.007. For plastic deformation, assume that the relationship between stress and strain is described by Equation 6.19, in which the values for K and n are 1520 MPa (221,000 psi) and 0.15, respectively. Furthermore, plastic deformation occurs between strain values of 0.007 and 0.60, at which point fracture occurs. Solution This problem calls for us to compute the toughness (or energy to cause fracture). This toughness is equal to the area beneath the entire stress-strain curve. The easiest way to make this computation is to perform integrations in both elastic and plastic regions using data given in the problem statement, and then add these values together. Thus, we may define toughness using the following equation:

Toughness =   d Toughness in the elastic region is determined by integration of Equation 6.5, integrating between strains of 0 and 0.007 (per the problem statement). Thus 0.007

Toughness (elastic) =

E 2 = 2

=

0

E d

0.007

0

103  109 N/m 2  2  (0.007 ) 2



(0)2 

= 2.52  106 N/m 2 = 2.52  106 J/m 3 Now for the plastic region of the stress-strain curve, we integrate Equation 6.19 between the strain limits of 0.007 and 0.60, as follows: 0.60

Toughness (plastic) =



0.007

K  n d

K =  ( n1) (n  1)

=

0.60

0.007

1520  106 N/ m 2  (0.60)1.15  (0.007)1.15  (1.0  0.15) 

 730  106 N/m 2  7.30  106 J/m 3 And, finally, the total toughness is the sum of elastic and plastic values, which is equal to

Toughness  Toughness(elastic) + Toughness(plastic)  2.52  106 J/m 3 + 730  106 J/m 3

 733  106 J/m 3  7.33  108 J/m 3 (1.07  105 in.-lbf /in.3 )

6.50 For a tensile test, it can be demonstrated that necking begins when

d T  T d T

(6.32)

Using Equation 6.19, determine an expression for the value of the true strain at this onset of necking. Solution This problem asks that we determine the value of T for the onset of necking assuming that necking begins when

d T = T d T Let us take the derivative of Equation 6.19, set it equal to T, and then solve for T from the resulting expression. Differentiating Equation 6.19 leads to the following:

d  K ( T ) n  d T

= Kn( T )(n1) =  T

n

However, from Equation 6.19, T = K( ) , which, when substituted into the above expression, yields T

Kn( T )(n  1) = K( T ) n Now solving for T from this equation leads to  =n T

as the value of the true strain at the onset of necking.

6.51 Taking the logarithm of both sides of Equation 6.19 yields log σT = log K + n log ∈T

(6.33)

Thus, a plot of log σT versus log ∈T in the plastic region to the point of necking should yield a straight line having a slope of n and an intercept (at log σT = 0) of log K. Using the appropriate data tabulated in Problem 6.30, make a plot of log σ T versus log ∈T and determine the values of n and K. It will be necessary to convert engineering stresses and strains to true stresses and strains using Equations 6.18a and 6.18b. Solution This problem calls for us to utilize the appropriate data from Problem 6.30 in order to determine the values of n and K for this material. From Equation 6.33 the slope of a log T versus log T plot will yield n. However, Equation 6.19 is only valid in the region of plastic deformation to the point of necking; thus, only the 8th, 9th, 10th, 11th, 12th, and 13th data points may be utilized. These data are tabulated below:

 (MPa)

(MPa)

log 





log 

797

803

2.905

0.0075

7.47  103

2.127

838

846

2.927

0.010

9.95  103

2.002

0.015

1.488 

102

1.827

102

1.703

928

942

2.974

997

1017

3.007

0.020

1.980 

1163

1210

3.083

0.040

3.92  102

1.407

1236

1310

3.117

0.060

5.83  102

1.234

The log-log plot with these data points is given below.

From Equation 6.33, the value of n is equal to the slope of the line that has been drawn through the data points; that is

n

 log  T  log T

From this line, values of log T corresponding to log T (1) = 2.20 and log T (2) = 1.20 are, respectively, log T (1) = 2.886 and log T (2) = 3.131. Therefore, we determine the value of n as follows:

n

 log  T log  T (1)  log  T (2)   log  T log  T (1)  log T (2)



2.886  3.131  0.245 2.20  (1.20)

In order to determine the value of K we rearrange Equation 6.33 to read as follows:

log K  log  T  nlog  T It is possible to compute log K by inserting into this equation the value for a specific log  and its corresponding log . (Of course we have already calculated the value of n.) If we use log  (1) and log T (1) values from above, then K is determined as follows:

log K  log  T (1)  nlog  T (1)  2.886  (0.245)(2.20)  3.425

Which means that

K  103.425  2660 MPa

Elastic Recovery after Plastic Deformation 6.52 A cylindrical specimen of a brass alloy 10.0 mm (0.39 in.) in diameter and 120.0 mm (4.72 in.) long is pulled in tension with a force of 11,750 N (2640 lbf); the force is subsequently released. (a) Compute the final length of the specimen at this time. The tensile stress–strain behavior for this alloy is shown in Figure 6.12. (b) Compute the final specimen length when the load is increased to 23,500 N (5280 lbf) and then released. Solution (a) In order to determine the final length of the brass specimen when the load is released, it first becomes necessary to compute the applied stress using Equation 6.1; thus

 =

F F 11,750 N = = =150  106 N/m 2  150 MPa (22,000 psi) 2 2 A0 3  d  10  10 m   0    2 2    

Upon locating this point on the stress-strain curve (Figure 6.12), we note that it is in the linear, elastic region; therefore, when the load is released the specimen will return to its original length of 120 mm (4.72 in.). (b) In this portion of the problem we are asked to calculate the final length, after load release, when the load is increased to 23,500 N (5280 lbf). Again, computing the stress

 =

23,500 N 2

 10  103 m    2  

= 300  106 N/m 2  300 MPa (44,200 psi)

The point on the stress-strain curve corresponding to this stress is in the plastic region. We are able to estimate the amount of permanent strain by drawing a straight line parallel to the linear elastic region; this line intersects the strain axis at a strain of about 0.012 which is the amount of plastic strain. The final specimen length li may be determined from a rearranged form of Equation 6.2 as li = l0(1 + ) = (120 mm)(1 + 0.012) = 121.44 mm (4.78 in.)

6.53 A steel alloy specimen having a rectangular cross section of dimensions 19 mm × 3.2 mm (

3 4

in. ×

1 8

in.) has the stress–strain behavior shown in Figure 6.22. This specimen is subjected to a tensile force of 110,000 N (25,000 lbf). (a) Determine the elastic and plastic strain values. (b) If its original length is 610 mm (24.0 in.), what will be its final length after the load in part (a) is applied and then released? Solution (a) We are asked to determine both the elastic and plastic strain values when a tensile force of 110,000 N (25,000 lbf) is applied to the steel specimen and then released. First it becomes necessary to determine the applied stress using Equation 6.1; thus

 =

F F = A0 b0 d0

where b0 and d0 are cross-sectional width and depth [19 mm (19  103 m) and 3.2 mm (3.2  103 m), respectively). Thus

 =

110,000 N = 1.810  109 N/m 2  1810 MPa (265,000 psi) (19  10 m )(3.2  103 m ) 3

From Figure 6.22, this point is in the plastic region so the specimen will be both elastic and plastic strains. The total strain at this point, t, is about 0.020. From this point on the stress-strain curve we are able to estimate the amount of strain recovery e from Hooke's law, Equation 6.5 as

e =

 E

And, since E = 207 GPa for steel (Table 6.1)

e =

1810 MPa = 0.0087 207  103 MPa

The value of the plastic strain, p is just the difference between the total and elastic strains; that is

p = t – e = 0.020 – 0.0087 = 0.0113

(b) If the initial length is 610 mm (24.0 in.) then the final specimen length li may be determined from a rearranged form of Equation 6.2 using the plastic strain value as li = l0(1 + p) = (610 mm)(1 + 0.0113) = 616.9 mm (24.26 in.)

Hardness 6.54 (a) A 10-mm-diameter Brinell hardness indenter produced an indentation 2.50 mm in diameter in a steel alloy when a load of 1000 kg was used. Compute the HB of this material. (b) What will be the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used? Solution (a) We are asked to compute the Brinell hardness for the given indentation. It is necessary to use the equation in Table 6.5 for HB, where P = 1000 kg, d = 2.50 mm, and D = 10 mm. Thus, the Brinell hardness is computed as

HB =

=

2P  D  D  

D2  d2  

(2)(1000 kg) = 200.5  ()(10 mm) 10 mm  (10 mm)2  (2.50 mm)2   

(b) This part of the problem calls for us to determine the indentation diameter d that will yield a 300 HB when P = 500 kg. Solving for d from the equation in Table 6.5 gives

d=

=



2

2

(2)(500 kg)  (10 mm)   10 mm  = 1.45 mm (300)()(10 mm)   2



2P  D   D (HB) D   2

6.55 (a) Calculate the Knoop hardness when a 500-g load yields an indentation diagonal length of 100 m. (b) The measured HK of some material is 200. Compute the applied load if the indentation diagonal length is 0.25 mm. Solution (a) Knoop microhardness is related to the applied load (P in kg) and indentation length (l in mm) according to the equation given in Table 6.5—viz.

HK 

14.2P l2

The applied load is 500 g = 0.500 kg, whereas the indentation length is 100 m = 0.100 mm. Incorporating these values into the equation above leads to

HK 

(14.2)(0.500 kg)  710 (0.100 mm)2

(b) This portion of the problem asks for the applied load for HK = 200 and l = 0.25 mm. Assigning P in the above equation to be the dependent variable yields the following expression:

P

(HK)l 2 14.2

When we insert the above values for HK and l into this expression, the value of the P is computed as follows:

P

(200)(0.25 mm)2  0.880 kg 14.2

6.56 (a) What is the indentation diagonal length when a load of 0.60 kg produces a Vickers HV of 400? (b) Calculate the Vickers hardness when a 700-g load yields an indentation diagonal length of 0.050 mm. Solution (a) The equation given in Table 6.5 for Vickers microhardness is as follows:

HV 

1.854P d12

Here P is the applied load (in kg) and d1 is the diagonal length (in mm). This portion of the problem asks that we compute the indentation diagonal length. Rearranging the above equation to make d1 the dependent variable gives

d1 

1.854 P HV

Using values for P and HV given in the problem statement yields the following value for d1:

d1 

(1.854)(0.60 kg)  0.0527 mm 400

(b) For P = 700 g (0.700 kg) and d1 = 0.050 mm, the Vickers hardness value is

HV 



1.854P d12

(1.854)(0.700 kg)  519 (0.050 mm) 2

6.57 Estimate the Brinell and Rockwell hardnesses for the following: (a) The naval brass for which the stress–strain behavior is shown in Figure 6.12. (b) The steel alloy for which the stress–strain behavior is shown in Figure 6.22. Solution This problem calls for estimations of Brinell and Rockwell hardnesses. (a) For the brass specimen, the stress-strain behavior for which is shown in Figure 6.12, the tensile strength is 450 MPa (65,000 psi). From Figure 6.19, the hardness for brass corresponding to this tensile strength is about 125 HB or 85 HRB. (b) The steel alloy (Figure 6.22) has a tensile strength of about 1950 MPa (283,000 psi) [Problem 6.26(d)]. This corresponds to a hardness of about 560 HB or ~55 HRC from the line (extended) for steels in Figure 6.19.

6.58 Using the data represented in Figure 6.19, specify equations relating tensile strength and Brinell hardness for brass and nodular cast iron, similar to Equations 6.20a and 6.20b for steels. Solution This problem calls for us to specify expressions similar to Equations 6.20a and 6.20b for nodular cast iron and brass. These equations, for a straight line, are of the form TS = C + (E)(HB) where TS is the tensile strength, HB is the Brinell hardness, and C and E are constants, which need to be determined. One way to solve for C and E is analytically--establishing two equations using TS and HB data points on the plot, as (TS)1 = C + (E)(HB)1 (TS)2 = C + (E)(HB)2 Solving for E from these two expressions yields

E=

(TS)1  (TS)2 (HB) 2  (HB)1

For nodular cast iron, if we make the arbitrary choice of (HB) 1 and (HB)2 as 200 and 300, respectively, then, from Figure 6.19, (TS)1 and (TS)2 take on values of 600 MPa (87,000 psi) and 1100 MPa (160,000 psi), respectively. Substituting these values into the above expression and solving for E gives

E=

600 MPa  1100 MPa = 5.0 MPa/HB (730 psi/HB) 200 HB  300 HB

We may determine the value for the constant C by inserting this value of E into either of the (TS)1 or (TS)2 equations cited above. Using the (TS)1 expression, we solve for C as follows: C = (TS)1 – (E)(BH)1 = 600 MPa - (5.0 MPa/HB)(200 HB) = – 400 MPa (– 59,000 psi) Thus, for nodular cast iron, these two equations take the form TS(MPa) = – 400 + 5.0  HB TS(psi) = – 59,000 + 730  HB

Now for brass, we take (HB) 1 and (HB)2 as 100 and 200, respectively, then, from Figure 6.19, (TS)1 and (TS)2 take on values of 370 MPa (54,000 psi) and 660 MPa (95,000 psi), respectively. Substituting these values into the above expression for E gives

E=

370 MPa  660 MPa = 2.9 MPa/HB (410 psi/HB) 100 HB  200 HB

We determine the value of C in using the same procedure outlined above: C = (TS)1 – (E)(BH)1 = 370 MPa – (2.9 MPa/HB)(100 HB) = 80 MPa (13,000 psi) Thus, for brass these two equations that relate tensile strength and Brinell hardness take the form TS(MPa) = 80 + 2.9  HB TS(psi) = 13,000 + 410  HB

Variability of Material Properties 6.59 Cite five factors that lead to scatter in measured material properties. Solution The five factors that lead to scatter in measured material properties are the following: (1) test method; (2) variation in specimen fabrication procedure; (3) operator bias; (4) apparatus calibration; and (5) material inhomogeneities and/or compositional differences.

6.60 The following table gives a number of Rockwell G hardness values that were measured on a single steel specimen. Compute average and standard deviation hardness values. 47.3 52.1 45.6 49.9 47.6 50.4

48.7 50.0 46.2 48.3 51.1 46.7

47.1 50.4 45.9 46.4 48.5 49.7

Solution The average of the given hardness values is calculated using Equation 6.21 as 18

HRG =

=

 HRG i i1

18

47.3  52.1  45.6 . . . .  49.7 = 48.4 18

And we compute the standard deviation using Equation 6.22 as follows:

  HRG i  HRG  18

s=

i1

2

18  1

 ( 47.3  48.4)2  (52.1  48.4)2  . . . .  (49.7  48.4)2  =  17  

=

64.95 = 1.95 17

1/2

6.61 The following table gives a number of yield strength values (in MPa) that were measured on the same aluminum alloy. Compute average and standard deviation yield strength values. 274.3

277.1

263.8 267.5

258.6

271.2 255.4

266.9

257.6 270.8

260.1

264.3 261.7

279.4

260.5 Solution The average of the given yield strength values is calculated using Equation 6.21 as 15

y=



 ( y)i i1

15

274.3 + 267.5 + 255.4 + . . . . + 260.5  265.9 MPa 15

And we compute the standard deviation using Equation 6.22 as follows:

  ( y)i   y  15

s=

i1

2

15  1

 (274.3  265.9)2  (267.5  265.9)2  . . . .  (260.5  265.9)2  =  14  

=

753.75 = 7.34 MPa 14

1/2

Design/Safety Factors 6.62 Upon what three criteria are factors of safety based? Solution The criteria upon which factors of safety are based are (1) consequences of failure, (2) previous experience, (3) accuracy of measurement of mechanical forces and/or material properties, and (4) economics.

6.63 Determine working stresses for the two alloys that have the stress–strain behaviors shown in Figures 6.12 and 6.22. Solution The working stresses for the two alloys the stress-strain behaviors of which are shown in Figures 6.12 and 6.22 are calculated by dividing the yield strength by a factor of safety, which we will take to be 2. For the brass alloy (Figure 6.12), since y = 250 MPa (36,000 psi), the working stress is 125 MPa (18,000 psi), whereas for the steel alloy (Figure 6.22), y = 1600 MPa (232,000 psi) (from the solution to Problem 6.26), and, therefore, w = 800 MPa (116,000 psi).

DESIGN PROBLEMS 6.D1 A large tower is to be supported by a series of steel wires; it is estimated that the load on each wire will be 13,300 N (3000 lbf). Determine the minimum required wire diameter, assuming a factor of safety of 2.0 and a yield strength of 860 MPa (125,000 psi) for the steel. Solution For this problem the working stress is computed using Equation 6.24 with N = 2, as

w =

y 2

=

860 MPa = 430 MPa (62,500 psi ) 2

Since the force is given, the area may be determined from Equation 6.1, and subsequently the original diameter d0 may be calculated as 2

A0 =

 d  F =  0  w  2

And

d0 =

4F =  w

(4)(13,300 N)  ( 430  106 N/m 2 )

 6.3  10 3 m  6.3 mm (0.25 in.)

6.D2 (a) Consider a thin-walled cylindrical tube having a radius of 65 mm is to be used to transport pressurized gas. If inside and outside tube pressures are 100 and 2.0 atm (10.13 and 0.2026 MPa), respectively, compute the minimum required thickness for each of the following metal alloys. Assume a factor of safety of 3.5. (b) A tube constructed of which of the alloys will cost the least amount?

Yield Strength, y

Density, 

(MPa)

(g/cm3)

($US/kg)

Steel (plain)

375

7.8

1.65

Steel (alloy)

1000

7.8

4.00

Cast iron

225

7.1

2.50

Aluminum

275

2.7

7.50

Magnesium

175

1.80

15.00

Alloy

Unit Mass Cost,

c

Solution (a) To solve this problem we use a procedure similar to that used for Design Example 6.2. Alloy yield strength, tube radius and thickness, and the inside-outside pressure difference are related according to Equation 6.26; furthermore, labels for the tube dimensions are represented in the sketch of Figure 6.21. For this problem all parameters in this equation are provided except for the tube thickness, t. Rearranging Equation 6.26 such that t is the dependent variable leads to

t

Nrp y

(6.26b)

Tube wall thickness will vary from alloy to alloy since each alloy has a different yield strength, whereas N, r, and p are constant. Their values are provided in the problem statement as follows: N = 3.5 r = 65 mm = 65  103 m p = 10.13 MPa  0.2026 MPa = 9.927 MPa Using Equation 6.26b, we compute the tube wall thickness for these five alloys as follows:

t (plain steel) 

(3.5)(65  103 m)(9.927 MPa)  0.00602 m = 6.02 mm 375 MPa

t (alloy steel) 

(3.5)(65  103 m)(9.927 MPa)  0.00226 m = 2.26 mm 1000 MPa

t (cast iron) 

(3.5)(65  10 3 m)(9.927 MPa)  0.0100 m = 10.0 mm 225 MPa

t (aluminum) 

(3.5)(65  103 m)(9.927 MPa)  0.00821 m = 8.21 mm 275 MPa

t (magnesium) 

(3.5)(65  10 3 m)(9.927 MPa)  0.0129 m = 12.9 mm 175 MPa

(b) This portion of the problem asks that we determine at tube constructed of which of the alloys will cost the least amount. We begin by considering Equation 6.28, which gives tube volume as a function of t. Furthermore, when the expression for t, Equation 6.26b, is incorporated into Equation 6.28, the following equation for tube volume V results:

V   (2rt + t2 )L



2  Nrp  Nrp      2r      L     y   y    

Now tube cost per unit length is dependent on volume, alloy density, and cost per unit mass values according to Equation 6.29. Substitution of the above expression for V into Equation 6.29 yields

 V  (c)  1000  

Cost   

 2      Nrp Nrp     L  c   2r         y     y     1000 

 





 



Using this equation, let us now determine the cost per unit length for the plain steel.

Cost (plain steel) 

 



2

 (3.5)(65  10 3 m)(9.927 MPa)   (3.5)(65 10 3 m)(9.927 MPa)     375 MPa 375 MPa   

3    (2)(65  10 m)    

   (106 cm 3 /m 3 )(1 m)(7.8 g/cm 3 )  (1.65 $US/kg)   

1000 g/kg

= $33.10 We will not present cost computations for the other alloys. However, the following table lists costs for all five alloys (along with wall thickness values calculated above.

Alloy

t (mm)

Cost ($US)

Steel (plain)

6.02

33.10

Steel (alloy)

2.26

29.30

Cast iron

10.0

78.40

Aluminum

8.21

72.20

Magnesium

12.9

156.50

Hence, the alloy steel is the least expensive, the plain steel is next; the most expensive is the magnesium alloy.

6.D3 (a) Gaseous hydrogen at a constant pressure of 0.658 MPa (5 atm) is to flow within the inside of a thin-walled cylindrical tube of nickel that has a radius of 0.125 m. The temperature of the tube is to be 350°C and the pressure of hydrogen outside of the tube will be maintained at 0.0127 MPa (0.125 atm). Calculate the minimum wall thickness if the diffusion flux is to be no greater than 1.25 × 10 –7 mol/m2.s. The concentration of hydrogen in the nickel, CH (in moles hydrogen per cubic meter of Ni), is a function of hydrogen pressure, P H2 (in MPa), and absolute temperature T according to

 12,300 J/mol  CH  30.8 pH exp    2 RT  

(6.34)

Furthermore, the diffusion coefficient for the diffusion of H in Ni depends on temperature as

 39,560 J/mol  DH (m 2 /s)  (4.76  107 )exp    RT  

(6.35)

(b) For thin-walled cylindrical tubes that are pressurized, the circumferential stress is a function of the pressure difference across the wall (Δp), cylinder radius (r), and tube thickness ( x) according to Equation 6.25— that is



rp x

(6.25a)

Compute the circumferential stress to which the walls of this pressurized cylinder are exposed. (Note: the symbol t is used for cylinder wall thickness in Equation 6.25 found in Design Example 6.2; in this version of Equation 6.25 (i.e., 6.25a) we denote wall thickness by x.) (c) The room-temperature yield strength of Ni is 100 MPa (15,000 psi), and σ y diminishes about 5 MPa for every 50°C rise in temperature. Would you expect the wall thickness computed in part (b) to be suitable for this Ni cylinder at 350°C? Why or why not? (d) If this thickness is found to be suitable, compute the minimum thickness that could be used without any deformation of the tube walls. How much would the diffusion flux increase with this reduction in thickness? However, if the thickness determined in part (c) is found to be unsuitable, then specify a minimum thickness that you would use. In this case, how much of a decrease in diffusion flux would result? Solution (a) This portion of the problem asks for us to compute the wall thickness of a thin-walled cylindrical Ni tube at 350C through which hydrogen gas diffuses. The inside and outside pressures are, respectively, 0.658 and 0.0127 MPa, and the diffusion flux is to be no greater than 1.25  107 mol/m2-s. This is a steady-state diffusion problem, which necessitates that we employ Equation 5.2. The concentrations at the inside and outside wall faces

may be determined using Equation 6.34, and, furthermore, the diffusion coefficient is computed using Equation 6.35. Solving for x (using Equation 5.2)

x = 

= 

(4.76

DC J

1 1.25  10

7

mol/m 2 -s



  39,560 J/mol  10-7 ) exp     (8.31 J/mol-K)(350  273 K)  

  12,300 J/mol (30.8)exp    (8.31 J/mol-K)(350  273 K)  



0.0127 MPa  0.658 MPa



= 0.00366 m = 3.66 mm (b) Now we are asked to determine the circumferential stress:

 =

=

r p x

(0.125 m)(0.658 MPa  0.0127 MPa) 0.00366 m = 22.0 MPa

(c) Now we are to compare this value of stress to the yield strength of Ni at 350 C, from which it is possible to determine whether or not the 3.66 mm wall thickness is suitable. From the information given in the problem, we may write an equation for the dependence of yield strength (y) on temperature (T) as follows:

 y = 100 MPa 

5 MPa T  Tr 50C





where Tr is room temperature and for temperature in degrees Celsius. Thus, at 350C

 5 MPa   y = 100 MPa   (350C  20C) = 67 MPa  50C  

Inasmuch as the circumferential stress (22.0 MPa) is much less than the yield strength (67 MPa), this thickness is entirely suitable. (d) And, finally, this part of the problem asks that we specify how much this thickness may be reduced and still retain a safe design. Let us use a working stress by dividing the yield stress by a factor of safety, according to Equation 6.24. On the basis of our experience, let us use a value of 2.0 for N. Thus

w =

y N

=

67 MPa = 33.5 MPa 2

Using this value for w and Equation 6.25a, we now compute the tube thickness as

x =



r p w

(0.125 m)(0.658 MPa  0.0127 MPa) (33.5 MPa)

= 0.00241 m = 2.41 mm Substitution of this value into Fick's first law we calculate the diffusion flux as follows:

J =D

C x

  39,560 J/mol =  ( 4.76  107 ) exp      (8.31 J/mol-K)(350  273 K)    12,300 J/mol (30.8)exp     (8.31 J/mol-K)(350  273 K)  0.00241 m



0.0127 MPa 

0.658 MPa



 1.90  107 mol/m 2 -s   Thus, the flux increases by approximately a factor of 1.5, from 1.25  10 7 to 1.90  10 7 mol/m2-s with this

reduction in thickness.

6.D4 Consider the steady-state diffusion of hydrogen through the walls of a cylindrical nickel tube as described in Problem 6.D3. One design calls for a diffusion flux of 2.5 × 10 –8 mol/m2.s, a tube radius of 0.100 m, and inside and outside pressures of 1.015 MPa (10 atm) and 0.01015 MPa (0.1 atm), respectively; the maximum allowable temperature is 300°C. Specify a suitable temperature and wall thickness to give this diffusion flux and yet ensure that the tube walls will not experience any permanent deformation. Solution This problem calls for the specification of a temperature and cylindrical tube wall thickness that will give a diffusion flux of 2.5  108 mol/m2-s for the diffusion of hydrogen in nickel; the tube radius is 0.100 m and the inside and outside pressures are 1.015 and 0.01015 MPa, respectively.

There are probably several different

approaches that may be used; and there is not one unique solution. Let us employ the following procedure to solve this problem: (1) assume some wall thickness, and, then, using Fick's first law for diffusion (which also employs Equations 5.2 and 6.35), compute the temperature at which the diffusion flux is that required; (2) compute the yield strength of the nickel at this temperature using the dependence of yield strength on temperature as stated in Problem 6.D3; (3) calculate the circumferential stress on the tube walls using Equation 6.25a; and (4) compare the yield strength and circumferential stress values--the yield strength should probably be at least twice the stress in order to make certain that no permanent deformation occurs. If this condition is not met then another iteration of the procedure should be conducted with a more educated choice of wall thickness.  As a starting point, let us arbitrarily choose a wall thickness of 2 mm (2  10 3 m). The steady-state

diffusion equation, Equation 5.2, takes the form

J =D

C x

 2.5  108 mol/m 2 -s  39,560 J/mol  =  (4.76  107 ) exp     (8.31 J/mol-K)(T) 





 12,300 J/mol  (30.8)exp   0.01015 MPa  1.015 MPa   (8.31 J/mol-K)(T)  0.002 m



Solving this expression for the temperature T gives T = 500 K = 227C; this value is satisfactory inasmuch as it is less than the maximum allowable value (300C). The next step is to compute the stress on the wall using Equation 6.25a; thus

 =

=

r p x

(0.100 m)(1.015 MPa  0.01015 MPa)

(2

 103 m )

50.2 MPa Now, the yield strength (y) of Ni at this temperature may be computed using the expression

 y = 100 MPa 

5 MPa T  Tr 50C





where Tr is room temperature. Thus, y = 100 MPa – 0.1 MPa/C (227C – 20C) = 79.3 MPa Inasmuch as this yield strength is about 50% greater than the circumferential stress, a wall thickness of 2 mm is a satisfactory design parameter; furthermore, 227C is a satisfactory operating temperature.

FUNDAMENTALS OF ENGINEERING QUESTIONS AND PROBLEMS 6.1FE A steel rod is pulled in tension with a stress that is less than the yield strength. The modulus of elasticity may be calculated as (A) Axial stress divided by axial strain (B) Axial stress divided by change in length (C) Axial stress times axial strain (D) Axial load divided by change in length Solution The modulus of elasticity of a steel rod placed under a tensile stress that is less than the yield strength (i.e., when deformation is totally elastic) may be calculated using Equation 6.5 as

E

 

That is, the modulus is equal to the axial stress divided by axial strain. Therefore, the correct answer is A.

6.2FE A cylindrical specimen of brass that has a diameter of 20 mm, a tensile modulus of 110 GPa, and a Poisson’s ratio of 0.35 is pulled in tension with force of 40,000 N. If the deformation is totally elastic, what is the strain experienced by the specimen? (A) 0.00116

(C) 0.00463

(B) 0.00029

(D) 0.01350

Solution This problem calls for us to calculate the elastic strain that results for a brass specimen stressed in tension. The cross-sectional area A0 is equal to 2

2  20  10 3 m   d0  4 2     3.14  10 m  2 2  

A0   

Here d0 is the original cross-sectional diameter. Also, the elastic modulus for brass is given in the problem statement as 110 GPa (or 110  109 N/m2). Combining Equations 6.1 and 6.5 and solving for the strain yields

=

 F = E A0 E

Now incorporating values for stress, cross-sectional area, and modulus of elasticity into this expression leads to the following value of strain:

=

which is answer A.

40,000 N =1.16  10 3  0.00116 (3.14  10 m 2 )(110  10 9 N/m 2 ) 4

6.3FE The following figure shows the tensile stress-strain curve for a plain-carbon steel.

(a) What is this alloy’s tensile strength? (A) 650 MPa

(C) 570 MPa

(B) 300 MPa

(D) 3,000 MPa

(b) What is its modulus of elasticity? (A) 320 GPa

(C) 500 GPa

(B) 400 GPa

(D) 215 GPa

(c) What is the yield strength? (A) 550 MPa

(C) 600 MPa

(B) 420 MPa

(D) 1000 MPa

Solution (a) The tensile strength (the maximum on the curve) is approximately 570 MPa. Therefore, C is the correct answer. (b) The elastic modulus is just the slope of the initial linear portion of the curve, which may be determined using Equation 6.10:

E

 2  1  2  1

If we take  and  to be zero and  = 300 MPa, then its corresponding strain  (from the inset) is about 0.0014. Thus, we calculate the modulus of elasticity as follows:

E

300 MPa  0 MPa  215  103 MPa = 215 GPa 0.0014  0

Thus, D is the correct answer. (c) The 0.002 strain offset line intersects the stress-strain curve at approximately 550 MPa, which means that the correct answer is A.

6.4FE A specimen of steel has a rectangular cross section 20 mm wide and 40 mm thick, an elastic modulus of 207 GPa, and a Poisson’s ratio of 0.30. If this specimen is pulled in tension with a force of 60,000 N, what is the change in width if deformation is totally elastic? (A) Increase in width of 3.62  106 m (B) Decrease in width of 7.24  106 m (C) Increase in width of 7.24  106 m (D) Decrease in width of 2.18  106 m Solution This problem calls for us to calculate the change in width that results for a steel specimen stressed in tension. The first step is to calculate the tensile strain. The cross-sectional area is (20 mm)  (40 mm) = 800 mm2 = 8  104 m2; also, the elastic modulus for steel is given as 207 GPa (or 207  109 N/m2). Combining Equations 6.1 and 6.5 leads to the following:

F  A0 F z = = = E E A0 E Incorporation of values for force, cross-sectional area, and the modulus of elasticity yields the following for the axial strain:

z =

(8.0  10

4

60,000 N = 3.62  10 4 2 9 2 m )(207  10 N/m )

The lateral strain is calculated by rearranging Equation 6.8 to give

 x =   z =  (0.30)(3.62  10 4 ) =  1.09  10 4 Finally, the change in width, w, is calculated using a rearranged and modified form of Equation 6.2 as

w w0 x Here w0 is the original specimen width (20 mm). Therefore, using the above equation, w is calculated as

w (20 mm)(  1.09  10 4 ) =  2.18  10 3 mm =  2.18  10 6 m Therefore, the width decreases (a minus w) by 2.18  106 m, which means that the correct answer is D.

6.5FE A cylindrical specimen of undeformed brass that has a radius of 300 mm is elastically deformed to a tensile strain of 0.001. If Poisson’s ratio for this brass is 0.35, what is the change in specimen diameter? (A) Increase by 0.028 mm (B) Decrease by 1.05  104 m (C) Decrease by 3.00  104 m (D) Increase by 1.05  104 m Solution This problem calls for us to calculate the change in diameter that results for a brass specimen stressed in tension. The lateral strain (x) is calculated using a rearranged form of Equation 6.8 as

 x =   z =  (0.35)(1.00  10 3 ) =  3.50  10 4 The change in diameter, d, is calculated using a modified and rearranged form of Equation 6.2:

d  d0 x where d0 is the original diameter (300 mm). Now we determine the change in diameter using the above equation as follows:

d  (300 mm)(  3.5  10 4 ) =  0.105 mm =  1.05  10 4 m Therefore, the diameter decreases (a minus d) by 1.05  10-4 m, which means that the correct answer is B.