HW #4

Problem 6.8

a. To Find:

The diameter of the test cylinder,d0.

b. Given:

Load, F = 6660 N (1500 lb)

Modulus of elasticity, E = 110 GPa (16 *106

psi)

Yield strength, y = 240 MPa (35,000 psi)

Initial length of rod,l0 = 380 mm (15.0 in)

Elongation, l = 0.50 mm (0.020 in)

c. Assumptions:

(i) The applied stress is in the elastic regime, i.e, the deformation is elastic. If this assumption is

valid, F/( * d02/4) < y

(ii) The applied force is uniaxial and tensile/compressive and in a direction parallel to the length

of the rod.

(iii) The material is isotropic and hence, the value of E is independent of the direction in which

elongation occurs.

(iv) The experiment is conducted at room temperature. This is relevant because we do not wish

to consider time-dependent plastic deformation (creep) that occurs at y and comes into the

picture at higher temperatures.

d. Solution:

A0 = * d02/4

4

== 2

00

d

F

A

F …(1)

0

= l

lE

…(2)

From (1) and (2),

4

2

0d

F =

0

l

lE

lE

Fld

00

4 =

d0= 7.65 10-3

m = 7.65 mm (0.30 in.)

Problem 6.22

a. To Find:

(a) Specimen elongation

(b) Reduction in specimen diameter

b. Given:

(a) Stress-strain behavior of the alloy is as shown in Fig. 6.12

(b) Diameter of cylinder, d0 = 6 mm (0.24 in.)

(c) Length of specimen,l0 = 50 mm (2 in.)

(d) Tensile force, F=5000N (1125 lb)

(e) Poisson’s ratio = 0.3

c. Assumptions:

(i) Fig.6.12 is an accurate representation of the stress vs. strain behavior for the given

alloy.

(ii) The applied force is uniaxial and in a direction parallel to the length of the cylinder.

(iii) The material is isotropic.

(iv) The experiment is conducted at room temperature. This is relevant because we do not

wish to consider time-dependent plastic deformation that comes into the picture at

higher temperatures.

d. Solution:

(a) Applied stress,

=F

A0

=F

d0

2

2=

5000 N

6 103 m

2

2= 177  106 N/m2 =177 MPa (25,000 psi)

From the stress-strain curve in Figure 6.12, at this stress, strain ≈ 2.0 10-3.

= l/l

in.) 10 (4 mm 0.10 = mm) (5010 2.0== -3-3

0 )( ll

(b)

Fig.1. The cylinder before and after (red color) elongation.

In part (a), the elongation/strain is in a direction parallel to the direction of the applied

force/stress. In a lateral direction, from equation 6.8,

xz …(1)

Also, by definition,

x = ∆d/d0 …(2)

From (1) and (2),

d = d0x = d0z = (6 mm)(0.30)(2.0 10-3)

∆d = –3.6 10-3 mm (–1.4 10-4 in.)

al = 0.1 mm (4 x 10-3

in.)

bd = –3.6 10-3 mm (–1.4 10-4 in.)

Problem 6.29

a. To Find:

We are required to plot and interpret the engineering stress vs. engineering strain curve for

aluminum.

b. Given:

Diameter of test specimen, d0 = 12.8 mm

Length of test specimen, l0 = 50.8 mm

Load vs. elongation data for the test specimen.

c. Assumptions:

(i) Given data is accurate.

(ii) The applied force is uniaxial and in a direction parallel to the length of the cylinder.

(iii) The material is isotropic.

(iv) The experiment is conducted at room temperature. This is relevant because we do not

wish to consider time-dependent plastic deformation that comes into the picture at

higher temperatures.

d. Solution:

(Engineering) Stress = Load/A0

A0 = π* d02/4

A0 = 128.61 mm2

Stress = Load (N) / 128.61 mm2

= Load / 128.61 MPa …(1)

(Engineering) Strain = [Length (mm) – l0 (mm)] / l0 (mm)

Strain = [Length – 50.8 ] / 50.8 …(2)

From (1) and (2), the stress and strain values corresponding to the given (load, length) data

points are calculated and tabulated below.

Load Length Stress Strain

N mm MPa

0 50.8 0 0

7,330 50.851 56.9921 0.001

15,100 50.902 117.405 0.00201

23,100 50.952 179.607 0.00299

30,400 51.003 236.365 0.004

34,400 51.054 267.466 0.005

38,400 51.308 298.567 0.01

41,300 51.816 321.115 0.02

44,800 52.832 348.328 0.04

46,200 53.848 359.213 0.06

47,300 54.864 367.766 0.08

47,500 55.88 369.321 0.1

46,100 56.896 358.436 0.12

44,800 57.658 348.328 0.135

42,600 58.42 331.223 0.15

36,400 59.182 283.017 0.165

Table 1. (Engineering) Stress and (Engineering) Strain values calculated from the load and

length values.

(a)

Fig.1. Plot of engineering stress vs. engineering strain (from Table 1).

0

50

100

150

200

250

300

350

400

0 0.05 0.1 0.15 0.2

Engi

ne

eri

ng

Stre

ss (

MP

a)

Engineering Strain

(b)

Fig.2. Plot of engineering stress () vs. engineering strain () in and slightly beyond the linear

elastic region (from Table 1).

The modulus of elasticity is the slope in the linear elastic region.

E = MPa – 0 MPa) / (0.00299 – 0) ≈ 6 x 104 MPa ≈ 60 GPa

(c)

The 0.002 strain offset line (red colored line) intersects the stress-strain curve at a stress value of

approximately 280 MPa. Thus, the yield strength, y ≈ 280 MPa.

(d)

The tensile strength (also called Ultimate Tensile Strength or UTS ) is approximately 370 MPa

corresponding to the maximum stress on the stress-strain plot.

0

50

100

150

200

250

300

350

0 0.002 0.004 0.006 0.008 0.01 0.012

Engi

ne

eri

ng

Stre

ss (

MP

a)

Engineering Strain

(e)

The ductility, in percent elongation, equals the plastic strain at fracture, multiplied by one-

hundred.

The plastic strain may be estimated as follows:

Fig,3. Elastic, plastic and total strain at fracture.

In the above figure (Fig.3), the point at which fracture occurs is indicated by

The perpendicular from this point to the strain-axis (red colored line), strikes this axis at strain =

0.14. Thus, the total strain at fracture = 0.166

A line parallel to the linear elastic part of the plot passing through this point (blue line) intersects

the strain-axis at strain = 0.161. Thus, the plastic strain at fracture = 0.161.

( Elastic strain = Total strain – plastic strain = 0.005)

Thus, ductility = 0.161 * 100 = 16.1 % elongation.

(f)

From Equation 6.14, the modulus of resilience,

Ur = y

2

2E

From the values calculated in parts (b) and (c)

Ur = (280 MPa) 2/ (2) * ( 6 x 10

4 MPa) = 0.65 MPa = 0.65 x 10

6 N/m

2 = 6.5 x 10

5 J/m

3

(a) Engineering stress vs. engineering strain curve: Fig.1

(b) E ≈ 6 x 104 MPa ≈ 60 GPa

(c) ys = 280 MPa

(d) UTS = 370 MPa

(e) Ductility ≈ 16.1 % elongation

(f) Ur = 6.5 x 105 J/m

3

Problem 6.41

a. To Find:

We are required to plot the true stress vs. true strain curve for aluminum.

b. Given:

Diameter of test specimen, d0 = 12.8 mm

Length of test specimen, l0 = 50.8 mm

Load vs. elongation data for the test specimen.

t = F/Ai , where t is the trues stress, F is the load and Ai is the instantaneous cross-sectional

area (Equation 6.15)

t = ln (li/l0), where t is the true strain and li is the instantaneous length of the test specimen

(Equation 6.16)

t = (1+) , where is the engineering stress and is the engineering strain (Equation 6.18a)

Equation 6.18a is invalid once necking begins.Hence, the values of the measured diameter of the

specimen given in the question are to be used to calculate true stress, for the last four data points.

c. Assumptions:

(i) Given data is accurate.

(ii) The applied force is uniaxial and in a direction parallel to the length of the cylinder.

(iii) The material is isotropic.

(iv) The experiment is conducted at room temperature. This is relevant because we do not

wish to consider time-dependent plastic deformation that comes into the picture at

higher temperatures.

d. Solution:

Except for the last four data points,

t : Using the values of and Table 1/Problem 6.29) and Equation 6.18a, calculate t = (1+)

For the last four data points,

Using the given value of the instantaneous diameter di, calculate Ai = π *di2

/ 4

From Equation 6.15, calculate t = F/Ai

For all data points,

Using the given values of li, l0 and Equation 6.16, calculate t = ln (li/l0)

Note –Equation 6.18b may be used to calculate t for all but the last four data points. However,

the question explicitly asks us to use Equations 6.15, 6.16 and 6.18a. Hence, it would not be

appropriate to use 6.18b here.

The calculated values are tabulated below.

Load Length Diameter Engineering Stress

Engineering

Strain

True

Strain

True

Stress

N mm mm MPa MPa

0 50.8 0.000 0.000 0.0000 0.000

7,330 50.851 56.992 0.001 0.0010 57.049

15,100 50.902 117.405 0.002 0.0020 117.641

23,100 50.952 179.607 0.003 0.0030 180.144

30,400 51.003 236.365 0.004 0.0040 237.310

34,400 51.054 267.466 0.005 0.0050 268.803

38,400 51.308 298.567 0.010 0.0100 301.553

41,300 51.816 321.115 0.020 0.0198 327.537

44,800 52.832 348.328 0.040 0.0392 362.261

46,200 53.848 359.213 0.060 0.0583 380.766

47,300 54.864 367.766 0.080 0.0770 397.187

47,500 55.88 369.321 0.100 0.0953 406.253

46,100 56.896 11.71 0.1133 428.270

44,800 57.658 11.26 0.1266 450.123

42,600 58.42 10.62 0.1398 481.162

36,400 59.182 9.4 0.1527 524.778

Table 1. True stress, true strain: input data and calculated values

Fig.1. True stress vs. true strain curve (From Table 1)

0.000

100.000

200.000

300.000

400.000

500.000

600.000

0.0000 0.0200 0.0400 0.0600 0.0800 0.1000 0.1200 0.1400 0.1600

Tru

e S

tre

ss (

MP

a)

True Strain

Problem 6.D2 (parts a,b,c)

a. To Find:

(a) Minimum thickness of the wall of the cylindrical tube, x.

(b) Circumferential stress on the wall of the pressurized tube, .

(c) Suitability of the cylinder of wall thickness ‘x’ at 300 0C.

b. Given:

Fig. 1. Thin-walled cylindrical tube: elevation view.

Pressure of hydrogen inside the tube, p1 = 1.013 MPa

Pressure of hydrogen outside the tube, p2 = 0.01013 MPa

Temperature = 300 0C

Radius of cylinder, r = 0.1 m

Maximum diffusion flux = 1 x 10-7

Concentration of hydrogen,

CH 30.8 pH2exp

12.3 kJ/mol

RT

…(1)

Diffusion coefficient,

DH 4.76 107 exp 39.56 kJ/mol

RT

…(2)

Circumferential stress, x

pr

=

y of Ni (at room temperature) = 100 MPa

y decreases by 5MPa for every 50 0C increase in temperature.

c. Assumptions:

(i) Linear concentration profile, i.e., steady state diffusion condition.

(ii) One-dimensional diffusion, i.e. diffusion occurs only along the axis parallel to the

thickness of the plate.

(iii) The relationship between temperature and yield strength is described accurately.

(iv) Material is isotropic.

d. Solution:

(a)

Using equation 5.3:

J = - D* CH / x

J = - D * (CH1-CH2) / (x1-x2) or - D * (CH2-CH1) / (x2-x1) …(4)

x2 – x1 = x …(5)

From, (1), (2), (4) and (5):

J = [ - 4.76*10-7

* exp (-39560 (J/mol)/RT) * 30.8 *exp (-12300 (J/mol) / RT) *

( (p1)0.5

- (p2)0.5

) ] / -x

x smol/m101

1 =

27

(4.76 10-7) exp 39,560 J /mol

(8.31 J/mol- K)(300 273 K)

01013.0 1013.1)K 273300)(K-J/mol 31.8(

J/mol 300,12exp(30.8) MPaMPa

x = 0.0025 m = 2.5 mm

(b)

x

pr

= =

) 0025.0(

) 01013.0 013.1)( 10.0(=

m

MPaMPam = 40.11 MPa

[Note – If

=r p

4x is used, =10 MPa]

(c)

y = 100 MPa 5 MPa

50CT Tr

Where, y is the yield strength at temperature T and Tr = room temperature .

At 300 0C,

y = 100 MPa (0.1 MPa/C) (300C 20C) = 72 MPa

The circumferential stress (40.11 MPa) is less than the yield strength (72 MPa). Hence, this

thickness is suitable.

(a)x = 0.0025 m = 2.5 mm

(b) = 40.11 MPa

(c) Suitable since <y