CHAPTER 5: MECHANICAL PROPERTIES

5.1

(i) Constant stressing rate

Therefore

Differentiating

Therefore

dO" = R = constant dt

I dO" Maxwell model' Equation (5.44) (page 324): de dt

- --

de dt

R

e Rt

+ -11

since 0" = J R dt

de Voigt model, Equation (5.52) (page327): dt

+

d2e 1 dO" E de ----dt 11 dt 11 dt

Ede 11 dt

R = 0 11

. R dO" SInce =-dt

Edt

0"

0" + -

Ee

11

11

This is a differential equation which can be solved in one of several ways. One is the

'convolution integral' method outlined by J.G. Williams in Stress Analysis of Polymers. The

simplest method is to 'guess' the result as follows:

Therefore

where T = !J.. o E

R. J. Young et al., Introduction to Polymers© Springer-Verlag US 1992

52

and

The equation then becomes

- ~ 1 + E R [1 -exp [ - ~ 1 ] ~ ~ E ~

Cancelling throughout the relation between e and t becomes

R - - = ° T]

Hence it has been shown that the solution that was guessed is correct.

(ii) Constant strain rate

de = S = constant dt

Maxwell model, Equation (5.44) (page 324):

Therefore

or d<1 (S~ - (1)

Integrating gives

In (S~ - (1) = Et + A ~

where A is the constant of integration.

Since when t = 0, <1 = ° then A = In (ST])

de dt

= Edt ~

1 <1 +

Edt

R

~

Hence

Therefore

53

Et

11

where To = !!.. E

Rearranging gives

Therefore

5.2

a 10-+-----,

Voigt model: de dt

(J Ee = S 11

(J = Ee + S11

(J = ESt + S11 . S e SInce =-t

o;-------~------------------------------e -----

o~------~--------~---------o 100 200 tis

(a) t=100s Thus 1(100 sj = 10 [1 - exp [ - ~~ ~]J

Therefore J(100 s) = 2(1 - e-O.5) = 0.787 m2 GN-1

But

e J=­(J

therefore e = J (J

Therefore e(100 s) = 0.01 GN m-2 x 0.787 m2 GN-1 = 0.00787

54

(b) If stress was not removed the strain after 200 s would have been

e(200 s) = 0.01 GN m-2 X 2 m2 GN-1 X (1 - e-1) = 0.01264

The removal of stress is equivalent to the addition of a negative strain of

0.00787 after (2oo-1oo)s.

Therefore residual strain = 0.01264 - 0.00787

Strain after 200 s = 0.00477

5.3

The Maxwell model for stress relaxation is given by Equation (5.49) (page 326) as

Therefore

Therefore

5.4

Therefore

In one cycle

CTO = strain X short-term modulus

CTO = 0.2 x 3 MN m-2 = 0.6 MN m-2

Leaking pressure = 0.3 MNm-2 = 0.6 MNm-2 exp [ - t 1 300 days

t = 208 days

e = eo sin wt

CT = CTo sin (wt + 0)

CT = CTo sin wt cos 0 + CTo cos wt sin 0

flU = J CT de

21rlw

flU = f CT de dt t dt

55

But de dt = eow cos wt

Therefore

2",/",

!::..U = ! eow coswt (0"0 sinwt coso + 0"0 coswt sino) dt

2",/",

= eowO"o ! (coso sinwt coswt + sino cos2wt) dt

Therefore

5.5

co co co

Evaluate f f f -00-00-00 (1)

Given that

co

f exp( - (32X2) x 2dx = 11"112/2{33 -co

(2)

and

co f exp( - (32X2) dx = 11"112/ {3 (3)

-co

Multiplying out the first term within the integral (1) gives

56

00 00 00

J J J ~5 [O .. ~ - 1)x2 exp( _/32X2) exp( -/32y2) exp( _/32Z2) + ...••..•..•••••• ] dxdydz -0)-00-00

Integrating w.r.t. x using (2) gives for the term inside the square brackets

[ (Ai - 1) 1r1l2 exp( -/32y2) exp( _/32Z2) + ••••.....••...•...•••..••.• J 2/33

Integrating w.r.t. y using (3) gives

[ 2 1r1l2 1r1l2 2 2 J (Al - 1) 2/33 73 exp( -/3 z) + .••••••••••••••••••.•••••••••

Integrating w.r.t. z using (3) gives

[0,; - 1) 2~ ";" + ....................•.. J

Performing the same exercise for all terms leads to the integral becoming

5.6

w = II2G(AI2 + A22 + A/ - 3)

For two dimensional straining, volume is constant

Therefore Al A2A3 = 1

and so Al = A2 = A, A3 = lIA2

therefore w = jG [2),,' + :. - 3]

Hence

a'

57

The forces per unit area in the unstrained state are It, h and h in each direction.

Then the work done by these forces will be in general:

dw = ItdAl + hdA2 + hdA3

But It = h = f and h = 0

Therefore

and so

These forces are related to the true stresses i.e. the forces on the strained area O'h 0'2 and 0'3 by

Therefore

where O't is the two-dimensional true stress.

Therefore

or

5.7

0' or f = .-!.

A

The stress-strain curve for the polymer is given by

0' = aeb I (1)

58

where (Jt = true stress, e = nominal strain and a and b are constants.

The Considere construction for a neck to form is given by Equation (5.153) (page 359) as

---(1+e)

Hence, from (1)

d(Jt b 1 = abe -

de

Also from (1)

(1 +e) (1 +e)

Hence, for necking

ae b ---

(1+e)

Cancelling gives

b 1 e (1 +e)

or e = b/(1 - b)

Yield will occur when, putting (5) into (1)

5.8

(2)

(3)

(4)

(5)

The pressure-dependent von Mises yield criterion is given by Equation (5.161) (page 364)

Let the tensile yield stress = (Jyt and the compressive yield stress = Uyc

Therefore the criterion can be written as

59

ACTy, + 2BCT;, = 1 (tension)

-ACTyc + 2BCT~ = 1 (compression)

Therefore

Hence the criterion becomes

Let the critical hydrostatic pressure to cause yielding be Pc, then Equation (5.2)(page 312) gives

Pc = %(CT1 + CT2 + CT3)

and CTl = CT2 = CT3

The criterion becomes

Therefore

5.9

Plane stress deformation CT3 = 0

Yield criterion, Equation (5.161) (page 364): A[CTI + CT2] + B[(CTI - CT2)2 + CT22 + CT12] = 1

Craze criterion, Equation (5.170) (page 370): CT1 - UCT2 = X + YI(CTI + CT2)

From 5.8

But

Therefore

and

and B = 1I(2CTycCTy,)

(J = 92 MN m-2 and (J = 73 MN m-2 yc yt

A = 2.829 X 10-3 m2 MN-1

B = 7.445 X 10-5 m4 MN-2

60

Crazing: 0"1 = 47 MN m-2, 0"2 = 0

and 0"1 = 45 MN m-2, 0"2 = 45 MN m-2

Therefore

47 MN m-2 = X + Y/47 MN m-2

45 MN m-2 - (0.33 x 45 MN m-2) = X + Y/(45 +45) MN m-2

Therefore

x = 11.73 MN m-2

Y = 1657.6 MW m-4

Therefore the two equations become

Shear yielding:

Crazing:

Crazing and shear yielding will take place simultaneously at values of 0"1 and 0"2 which satisfy

both of the above equations. They can be solved by plotting the two functions graphically.

Crazing \

-100 -60

, , , ,

100

-60

-100

, , , ,

-2 m

61

This gives U1 = 59 MN m-2 and U2 = - 30 MN m-2

and U1 = - 30 MN m-2 and U2 = 59 MN m-2

The curve for crazing has not been plotted on the opposite side of the U1 = -U2 line, since

crazing can only take place when the overall hydrostatic stress is tensile.

5.10

The chain direction repeat in polyethylene is 2.54 A

Therefore

Therefore

u/MN m-2

0

40

80

120

100

200

dOO2 = 1.270 A (un strained)

The modulus E = stress/strain = u/e

Strain, e = tJ.d/d but A = 2d sin ()

d = A/(2 sin ()

()/degrees d/A

37.483 1.26700

37.477 1.26717

37.471 1.26734

37.466 1.26749

37.460 1.26766

37.454 1.26783

tJ.d/A tJ.d/d

0 0

0.00017 0.000134

0.00034 0.000268

0.00049 0.000387

0.00066 0.000521

0.00083 0.000655

Therefore the modulus - 300 GN m-2

5.11

This problem is taken from:

Bevis, M. (1978), Coll. Polym. Sci. , 256, 234 and

Young, R.I. et ai, (1979) J.Poiym.Sci: Poiym. Phys. Ed. , 17, 1325.

General sketch of twins in plane of shear:

u/e MN m-2

298507

298507

310077

307101

305343

(hk2»

62

The (hkO) planes bend by an angle of 24>

across the twin boundary.

(a) If the planes of molecules that bend or kink are of the type (hkO) the twinning plane is

(hk2) and hence:

Plane of molecules

(200)

(020)

(110)

Twinning Plane

(202)

(022)

(112)

(b) The angle is 24> and from the sketch it can be seen that:

tan 4> = cl2dhkO (c = 2.537 A)

The values of dhkO can be calculated from the crystal geometry as shown in Question 4.1

(hkO) d,dA tan 4> 24> (hk2)

200 3.700 0.343 37.8° 202

020 2.465 0.515 54.5° 022

110 4.103 0.309 34.4° 112

(c) The direction of shear lies in the twinning plane, (hk2) and in the plane of shear.

(hkO) (hk2) Plane of Shear Direction of Shear

200 202 (010) [202]

020 022 (010) [022]

110 112 Irrational -[112]

The plane of shear and direction of shear are irrational for the (112) twin.

63

(d) The magnitude of the shear, s, is given by

s = 2 tancf>

Therefore the twin with the lowest shear is the (112) twin.

5.12

(1f = 85 MN m-2

Kc = 1.25 MN m-3/2

~ 25mm

a

The sheet contains 'inherent flaws' equivalent to small cracks of length, ll{). The equation for

Kc in Table 5.5 (page 404) for a single-edge notched specimen can therefore be used.

i.e. L 112

K = _a_ [1.99 - 0.41(alb) + 18.7(alb? - 38.48(alb)3 + 53.85(alb)4] c bt

As the flaws are very small a ~ b

Therefore

112 Kc - Lj ao x (1.99Ibt)

where Lj is the fracture load

Therefore

or

112 -1 ao = (1j x (Kj1.99)

Hence

64

a~12 = 1.25 MN m-3/2/(85 MN m-2 x 1.99)

or

ao = 55 x 10-6 m = 55 J.'m

5.13

The data show Uf decreases with increasing crack length, a.

It should be assumed that the width of the polystyrene specimen is very much greater than the

crack length, hence equations (5.193) and (5. 195)(page 401) can be used

[ ]

112

EGc (J = -f 7f'a

(1)

(2)

Hence it is necessary to plot uf versus a- Ifz to first determine Gc

(JclMPa almm a-'hlm-1f,

17.2 10 10.00

17.5 9 10.54

17.9 8 11.18

18.4 7 11.95

19.0 6 12.91

20.0 5 14.14

22.0 4 15.81

25.0 3 18.25

31.0 2 22.36

40.0 1 31.12

41.0 0.5 44.72

42.0 0.1 100.0

50

40

co 30 0.. ~

'" -b 20

10

65

-- -o---------------~

o 10 20 30 40 50 60 70 80 90 100

a-1/2/m-1I2

It is possible to fit a straight line to the initial data points. The final 3 points fallon a different

(dashed) line.

The slope of the initial line is 1.117 ± 0.03 MPa mlh

a) Hence from (1)

[ E~, 1 In • slope. 1.117 MPa mIn

Therefore with E = 3 GPa

G = 7r X 1.1172 MPa2 m c 3 x 1()3 MPa

Gc = 1.307 X 10-3 MPa m

G = 1 307 J m-2 c

66

b) From (2)

Kc = 1.98 MN m-312

c) The inherent flaw size, ao can be determined from the change in slope of the lines on the

graph (ie at a-'h = 33 m-1h)

5.14

For the elastomer sheet

Hence ao = 1 mm.

S = KaW

and da = qsn = q(Ka wy dt

If a sheet is held at constant extension

1 dt = ---

x da

The crack grows from ao to a where a approaches the sheet width in time t.t,

then

Therefore

But n is generally large and a ~ ao

67

Hence, the equation for t.t approximates to

t - 1 f - 1

qKlIWll(n - 1)a;-