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 free-body 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:

which means that

Or that

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)]:

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

Now, for static equilibrium in the y' direction, it is necessary that

Or

We now write an expression for ' as

F



x'= 0 P' P cos  = 0 P' = P cos 



'   = P' A' = Pcos A cos = P Acos2   = cos2 F_y'



= 0 =V' + Psin V' = Psin   = V' A'

which is just Equation 6.4b. = Psin A cos = P Asin cos =  sin cos

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., cos2 _{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 modulus for Cu is given in Table 6.1 as 110 GPa (or 110  109 N/m2). Combining Equations 6.1 and 6.5 and solving for the strain yields

 = E = F A₀ E  F A₀E = 44,500 N

(

2.90104m2

)(

110109N/m2

)

= 1.39  10 -3

6.4 A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa (30 × 106 _{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

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

l₀ = l  = l  E E l E F A₀ = l E d0 2     2 F = l Ed₀2 4F

Substitution into this expression, values of d₀, l, E, and F given in the problem statement yields the following

= 0.475 m = 475 mm (18.7 in.) A₀ =  d0 2     2 l₀ =

(

0.25103m

)

(

207109N/m2

)

()

(

10.2103m

)

2 (4)(8900 N)

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 lbf) 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 = , 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

Substitution into this expression, values of l₀, b₀l, and F given in the problem statement yields the following

b₀2 E =   = F A₀ l l₀ = Fl0 b₀2l E = (66,700 N)

(

125103m

)

(

16.5103m

)

2

(

0.43103m

)

 71.2  109 N/m2  71.2 GPa (10.4  106 psi)

6.6 Consider a cylindrical nickel wire 2.0 mm (0.08 in.) in diameter and 3 × 104 _{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

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

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 = l₀ = l₀ E= l₀F EA₀= l₀F E d0 2     2  4l₀F Ed₀2 A₀ =  d0 2     2 l = (4)(30 m)(300 N)

(

207109N/m2

)

()

(

2103m

)

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 mm2

(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

= 44,850 N (10,000 lb_f)

(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

Incorporating values of l₀, , and E provided in the problem statement leads to the computation of l_i as follows: F_y = _yA₀ =

(

345  106 N/m2

)(

133 mm2) 1 m 103 mm     2 l_i



l₀l₀l₀(



 1) l_i = l₀  E  1     l_i= (76 mm) 345 MPa 103  103MPa  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 lbf). 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:

And upon simplification

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

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

= 9.5  103 m = 9.5 mm (0.376 in.)  = F A₀= F  d02 4       = E= El l₀ F  d02 4       El l₀ d₀ = 4l0F El d₀ = (4)

(

500103m

)

(11,100 N) ()

(

207109N/m2

)(

0.38103m

)

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

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 ₂₁

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

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.

E = 2  1 2  1  492.4 MPa - 0 MPa 0.0049 - 0  100,500 MPa  100.5 GPa E = 21 2 1  916.7 MPa  0 MPa 0.0045  0  203, 700 MPa  203.7 GPa

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

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. E = 2  1

2  1

 193.6 MPa  0 MPa

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

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

E = 2  1

2  1

 160.6 MPa  0 MPa

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 lbf) 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  d0 2     2= 65,250 N  8.5103m 2       2 = 1.15  10 9 _N/m2_{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:

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

0.0006  0 = 86,700 MPa = 86.7 GPa (12.6  10 6 _psi)

(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 0.0006  0 = 100,000 MPa = 100 GPa (14.5  10 6 _psi)

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

where and 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 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

Using the values of and from Table 3.4 for Al (63.7 GPa and 76.1 GPa, respectively) 1 E_uvw 1 E₁₀₀ 3 1 E₁₀₀  1 E₁₁₁          2_2_2_2_2_2





  E₁₀₀   E₁₁₁   E_110 1 E_110 = 1 E_100 3 1 E_100 1 E_111      _ (0.707)2(0.707)2 (0.707)2(0)2(0)2(0.707)2_ = 1 E_100 (0.75) 1 E_100 1 E_111       E_100 E_111 1 E_110 = 1 63.7 GPa  (0.75) 1 63.7 GPa  1 76.1 GPa      

Which leads to, = 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,

Thus, = 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 E100 and E111,

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

E_110 1 E_110= 1 66.7 GPa  (0.75) 1 66.7 GPa  1 191.1 GPa    _ E_110 1 E_110 = 1 125.0 GPa  (0.75) 1 125.0 GPa  1 272.7 GPa       E_110

6.13 In Section 2.6, it was noted that the net bonding energy EN between two isolated positive and negative

ions is a function of interionic distance r as follows:

(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,

Derive an expression for the dependence of the modulus of elasticity on these A, B, and n parameters (for the two-ion 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 r0 corresponds to the value of r at the

minimum of the EN-versus-r curve (Figure 2.10b), take the derivative dEN/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 dE_N/dr in order to obtain an expression for the force F; this is accomplished as follows:

FdEN dr  d A r     dr  d B rn     dr = A r(1 + 1)  nB r(n + 1) E_N A r  B rn E dF dr    _r o

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

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

dF dr  d A r2     dr  d  nB rn  1     dr 2A r3  (n)(n  1)B r(n + 2)

Now, substitution of the above expression for r₀ into this equation yields dF dr    _r 0   2A A nB     3/(1 n) (n)(n  1)B A nB     (n  2)/(1 n)

which is the expression to which the modulus of elasticity is proportional. r₀ = A nB     1/(1n)

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 A nB     3/(1 n) (n)(n  1)B A nB     (n  2)/(1 n)

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

= 830

For metal Y, A = 2.0, B = 1  10-5, and n = 9. Hence E   (2)(1.5) 1.5 (8)

(

7106

)

          3/(18) + (8)(81)

(

7106

)

1.5 (8)

(

7106

)

        (82)/(18) E   (2)(2.0) 2.0 (9)

(

1105

)

          3/(19) + (9)(91)

(

1105

)

2.0 (9)

(

1105

)

        (92)/(19)

= 683

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

= 7425

Therefore, metal Z has the highest modulus of elasticity. E   (2)(3.5) 3.5 (7)

(

4106

)

          3/(17) + (7)(71)

(

4106

)

3.5 (7)

(

4 106

)

        (72)/(17)

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 lbf). 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:

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

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

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

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

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



= E



F  d02 4       =E l l₀ l = 4F l0 d₀2E l = (4)(48,900 N)

(

250103m

)

()

(

15.2103m

)

2

(

207109N/m2

)

3.25  10 4 _{m = 0.325 mm (0.013 in.)}  = x _z =  d/d₀ l/l₀

= –5.9  103 mm (–2.3  104 in.)

The diameter will decrease since d is negative.

d =  l d0

l₀ = 

(0.30)(0.325 mm)(15.2 mm) 250 mm

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

Now, combining Equations 6.1 and 6.5 leads to

And, since from Equation 6.8

Substitution of this equation into the previous one gives

And, solving for F leads to

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: A₀ =d0 2 4  = F A₀  F d₀2 4  E_z _z x    d d₀     d d₀ F d₀2 4 E  d d₀     F = d0dE 4 F = 

(

19103m

)(

2.5106m

)

()

(

69109N/m2

)

(4)(0.33)

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

Since the transverse strain _x is equal to

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

 = x _z =  d / d₀ 4F d₀2E       =  d0dE 4F

Now, incorporating values of d₀, d, E and F from the problem statement yields the following value for Poisson's ratio _z =  E= F A₀E= F  d0 2     2 E = 4F d₀2E _x=d d₀  = 

(

10103m

)(

7106m

)

()

(

100109N/m2

)

(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 d₀=

30.04 mm  30.00 mm

30.00 mm = 1.33  10

3

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 2G1 =

65.5103 MPa

(2)

(

25.4103 MPa

)

1 = 0.289

Now _z may be computed from Equation 6.8 as

_z = x

 = 

1.33  103

0.289 = 4.60  10 -3

Using this value of _z we solve for l₀ using Equation 6.2 and the value of l_i given in the problem statement:

l₀ = li 1 _z

= 105.20 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

E =  _z = F A₀ _z  F A₀_z = F _z d0 2     2 = 4F _zd₀2

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

_x d d_{0 leads to} _z = x  = d/d₀    d d₀

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

E = 4F _zd₀2= 4F  d d₀    d0 2 =  4F d₀d

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)



(

10103m

)(

6.7107m

)

=10

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 × 106 _{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 l₀ =

1.9 mm

380 mm= 0.005

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

(yield) =y E =

240 MPa

110103MPa = 0.0022

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 l₀=

0.072 mm

150 mm = 4.8  10 -4

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

4.8104 = 104.2 GPa

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 

10-3 mm when the tensile stress of 50 MPa is applied. This translates into a maximum lateral strain _x(max) as

_x(max) = d d₀ =

2.3103 mm

15.0 mm =1.53  10 -4

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

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

d_{0. 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

  x _z   d d₀  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

d_{0 for each of Cu, Ni, steel, Ti and W as follows:}

 d d₀(copper) (0.34)(50106N/m2) 110109N/m2 1.5510 4  d d₀(titanium) (0.34)(50106 N/m2) 107109N/m2  1.5910 4  d d₀(nickel) (0.31)(50106 N/m2) 207109N/m2  7.4910 5

 d d₀ (steel) (0.30)(50106 N/m2) 207109 N/m2  7.2510 5  d d₀(tungsten) (0.28)(50106 N/m2) 407109 N/m2  3.4410 5

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 (l_i) when subjected to a tensileforce 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 l₀ =

95.040 mm95.000 mm

95.000 mm = 4.21  10 4

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  d0 2     2  6300 N  10.7000  103 m 2       2 70.1  106 N/m2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 4.21104 = 167  10 3 _{MPa167 GPa}

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 d_i is the diameter while the stress is being applied—i.e., 10.698 mm. Thus

_x10.698 mm  10.7000 mm

10.700 mm  

0.002 mm

10.700 mm  1.87  10 4

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

  x _z  

1.87  104

4.21  104 0.444

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  d0 2     2 = 10,000 N  10103m 2       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

)

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 Modulus of Elasticity (GPa) Yield Strength (MPa) Poisson’s Ratio

Aluminum alloy 70 250 0.33

Titanium alloy 105 850 0.36

Steel alloy 205 550 0.27

Magnesium alloy 45 170 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  d0 2     2 = 35,000 N  15103m 2       2 = 200  10 6 _N/m2 _{= 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

 = x _z =  d d₀  E  Ed d₀ 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)

70103MPa = 1.41  10

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103MPa = 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103MPa = 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 Modulus of Elasticity (GPa) Yield Strength (MPa) Tensile Strength (MPa)

Aluminum alloy 70 255 420

Brass alloy 100 345 420

Copper 110 210 275

Steel alloy 207 450 550

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  d0 2     2  29,000 N  12.7  103m 2       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 E =

(230 MPa)(500 mm)

70  103MPa = 1.64 mm Thus, aluminum is not a candidate because its l is greater than1.3 mm.

For brass(E = 100 GPa = 100  103 MPa)

l = l0

E =

(230 MPa)(500 mm)

100  103MPa = 1.15 mm

l =l0

E =

(230 MPa)(500 mm)

207  103MPa = 0.56 mm 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 ₂_ = 1050 MPa0 MPa 0.005  0 210  10 3 _{MPa = 210 GPa}

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 lbf) 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 l₀ =

5 mm

100 mm= 510

2

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 A₀ 

F

r₀2

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

r₀ = F

 =

100,000 N

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

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

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 l₀ =

2.25 mm

375 mm = 0.006

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/m2

)

(

5.5103 m

)

2= 37,800 N (8500 lb_f)

(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 Length N lbf mm in. 0 0 50.800 2.000 12,700 2,850 50.825 2.001 25,400 5,710 50.851 2.002 38,100 8,560 50.876 2.003 50,800 11,400 50.902 2.004 76,200 17,100 50.952 2.006 89,100 20,000 51.003 2.008 92,700 20,800 51.054 2.010 102,500 23,000 51.181 2.015 107,800 24,200 51.308 2.020 119,400 26,800 51.562 2.030 128,300 28,800 51.816 2.040 149,700 33,650 52.832 2.080 159,000 35,750 53.848 2.120 160,400 36,000 54.356 2.140 159,500 35,850 54.864 2.160 151,500 34,050 55.880 2.200 124,700 28,000 56.642 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

0.002  0 = 200  10

3 _{MPa = 200 GPa (29  10}6 _psi)

(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

U_r =y 2 2E

which, using values of _y and E computed above, (750 MPa = 750  106 _N/m2 _{and 200 GPa = 200  10}9 _N/m2_, respectively) give a modulus of resilience of

U_r = (750  106 N/m2)2

(2)

(

200109N/m2

)

= 1.40  10

6 _N/m2_{= 1.40  106 J/m}3

₍

_{210 in.-lb} f/in.3

)

6.31 A specimen of magnesium having a rectangular cross section of dimensions 3.2 mm × 19.1 mm (18

in. × 34 _{in.) is deformed in tension. Using the load–elongation data shown in the following table, complete parts}

(a) through (f). Load Length lbf N in. mm 0 0 2.500 63.50 310 1380 2.501 63.53 625 2780 2.502 63.56 1265 5630 2.505 63.62 1670 7430 2.508 63.70 1830 8140 2.510 63.75 2220 9870 2.525 64.14 2890 12,850 2.575 65.41 3170 14,100 2.625 66.68 3225 14,340 2.675 67.95 3110 13,830 2.725 69.22 2810 12,500 2.775 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 ₂  ₁ =

50 MPa0 MPa

0.0010  50  10

3 _{MPa = 50 GPa (7.3  10}6 _psi)

(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

U_r = y 2 2E

which, using data computed above, yields a value of

U_r =

(

140106N/m2

)

2

(2)

(

50109N/m2

)

= 1.96  10

5 _N/m2_{= 1.96  10}5 _J/m3

₍

_{28.4 in.-lb} f/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  d0 2     2

For F = 15,000 N and

d

₀ = 15.00 mm (15  103 m) the stress is equal to

  15,000 N  15  103 m 2       2  84.9  10 6 _N/m2_{ 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

  

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/m2)

207  109 N/m2 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/m2)

107  109 N/m2 9.5  10

5 _{m = 0.095 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.