Plastics Engineering 3E Episode 3 potx

When using this pseudo-elastic design approach it should be remem- bered that the creep curves used to derive modulus values have normally been obtained on test pieces which are essentia

54 Mechanical Behaviour of Plastics

examples When using this pseudo-elastic design approach it should be remem-

bered that the creep curves used to derive modulus values have normally been

obtained on test pieces which are essentially isotropic In practice the manu-

facture of the end-product by injection moulding or extrusion, etc will have

resulted in some degree of anisotropy This may make the predictions inaccu-

rate because the creep data for the material is no longer appropriate for the

structural morphology introduced by the moulding method Similar comments

could, of course, also be made about metals in that the test data may have

been obtained on specimens of the material which do not accurately reflect the

nature of the material in the end-product Therefore, pseudo-elastic design is a

valid analytical procedure but one should always be cautious about the way in

which the manufacturing method has affected the behaviour of the material

Example 2.1 A ball-point pen made from polypropylene has the clip design

shown in Fig 2.1 1 When the pen is inserted into a pocket, the clip is subjected

to a deflection of 2 mm at point A If the limiting strain in the material is to

be 0.5% calculate (i) a suitable thickness, d, for the clip (ii) the initial stress in

the clip when it is first inserted into the pocket and (iii) the stress in the clip

when it has been in the pocket for 1 week The creep curves in Fig 2.5 may

be used and the short-term modulus of polypropylene is 1.6 GN/m2

? +I , width 6 mm

Fig 2.1 1 Ball-point pen clip design

Solution Strain, E , is given by the ratio of stress, u, to modulus, E In the

case of the pen clip, it is effectively a cantilever of length 40 mm

(i) Hence

MY

stress = -

I

where M = bending moment (WL)

y = half beam depth (d/2)

WLd

strain, E = -

Mechanical Behaviour of Plastics 55

Also, the classical elastic equation for the end deflection of a cantilever is:

stress = EE = 1.6 x lo9 x 0.005 = 8 MN/m2

(iii) After 1 week (6.1 x lo5 seconds), the isometric curves (Fig 2.8) derived from the creep curves show that at a strain of 0.5% the stress would have decayed to about 3.3 MN/m2

Example 2.2 A polypropylene beam is 100 m m long, simply supported at each end and is subjected to a load W at its mid-span If the maximum permis- sible strain in the material is to be 1.5%, calculate the largest load which may

be applied so that the deflection of the beam does not exceed 5 mm in a service life of 1 year For the beam I = 28 mm4 and the creep curves in Fig 2.5 should

56 Mechanical Behaviour of Plastics

0

Strain ("74

Fig 2.13 Isochronous curve for polypropylene (1 year)

The only unknown on the right hand side is a value for modulus E For the plastic this is time-dependent but a suitable value may be obtained by reference

to the creep curves in Fig 2.5 A section across these curves at the service life

of 1 year gives the isochronous graph shown in Fig 2.13 The maximum strain

is recommended as 1.5% so a secant modulus may be taken at this value and

is found to be 347 MN/m2 This is then used in the above equation

service conditions at some other temperature, then allowance would need to be made for the thermal strains set up in the beam These could be obtained from

a knowledge of the coefficient of thermal expansion of the beam material This

type of situation is illustrated later

Mechanical Behaviour of Plastics 57 For some plastics, particularly nylon, the moisture content can have a signifi- cant effect on the creep behaviour For such plastics, creep curves are normally available in the wet and dry states and a knowledge of the service conditions enables the appropriate data to be used

For convenience so far we have referred generally to creep curves in the above examples It has been assumed that one will be using the correct curves for the particular loading configuration In practice, creep curves obtained under tensile and flexural loading conditions are quite widely available Obviously it is important to use the creep curves which are appropriate to the particular loading situation Occasionally it is possible to obtain creep curves for compressive or

shear loading but these are less common

If only one type of data is available (e.g tensile creep curves) then it is possible to make conversions to the other test modes It should always be remembered, however, that these may not always be absolutely accurate for plastics under all situations

Generally there is a stiffening effect in compression compared to tension As

a first approximation one could assume that tension and compression behaviour are the same Thomas has shown that typically for PVC, the compression modulus is about 10% greater than the tensile modulus However, one needs to

be careful when comparing the experimental data because normally no account

is taken of the changes in cross-sectional area during testing In tension, the area will decrease so that the true stress will increase whereas in compression the opposite effect will occur

The classical relationship between moduli in tension, compression and flexure is

(2.14) where

MR = Ec/ET

It may be seen that if E, = ET then Eflex = ET However, if E, = 1 1 E ~ The classical relationship between the shear modulus G, and the tensile (for example), then Eflex = 1.05E~

modulus, E, for an isotropic material is

(2.15) where v = Poissons ratio

58 Mechanical Behaviour of Plastics

The bulk modulus is appropriate for situations where the material is subjected

to hydrostatic stresses The proof of equations (2.15) and (2.16) is given by

Benham et al

Example 23 A cylindrical polypropylene tank with a mean diameter of 1 m

is to be subjected to an internal pressure of 0.2 MN/m2 If the maximum strain

in the tank is not to exceed 2% in a period of 1 year, estimate a suitable value

for its wall thickness What is the ratio of the hoop strain to the axial strain in

Solution The maximum strain in a cylinder which is subjected to an internal pressure, p, is the hoop strain and the classical elastic equation for this is

The modulus term in this equation can be obtained in the same way as

in the previous example However, the difference in this case is the term

u For elastic materials this is called Poissons Ratio and is the ratio of the transverse strain to the axial strain (See Appendix C) For any particular

metal this is a constant, generally in the range 0.28 to 0.35 For plastics

u is not a constant It is dependent on time, temperature, stress, etc and

so it is often given the alternative names of Creep Contraction Ratio or Lateral Strain Ratio There is very little published information on the creep

contraction ratio for plastics but generally it varies from about 0.33 for hard

plastics (such as acrylic) to almost 0.5 for elastomers Some typical values are given in Table 2.1 but do remember that these may change in specific loading situations

Using the value of 0.4 for polypropylene,

h = - ( 2 - ~ ) PR

~ E E

6.5 0.02

Mechanical Behaviour of Plastics 59

Table 2.1 Qpical tensile and shear moduli for a range of polymers

Material

Tensile Shear modulus modulus Poisson’s Density (E)? ( G ) ratio ( u )

(kg/m3) (GN/m2) (GN/mZ) Polystyrene (PS)

Polymethyl Methacrylate (PMMA)

Polyvinyl Chloride (PVC)

(Unplasticised)

Nylon 66 (at 65% RH)

Acetal Homopolymer (POM)

Acetal Copolymer (POM)

Polyethylene - High Density (HDPE)

Polyethylene - Low Density (LDPE)

0.99 1.16 1.13 0.34 1.15 0.93 0.39 0.1 1

0.55

0.40 0.98

0.33 0.33 0.39 0.44 0.41 0.39 0.34 0.45 0.36 0.40 0.41

-

tl00 second modulus at 20°C for small strains (t0.2%)

Note that the ratio of the ratio of the hoop stress ( p R / h ) to the axial stress

(pR/2h) is only 2 From the data in this question the hoop stress will be 8.12 MN/m2 A plastic cylinder or pipe is an interesting situation in that it is

an example of creep under biaxial stresses The material is being stretched in the hoop direction by a stress of 8.12 MN/mz but the strain in this direction is restricted by the perpendicular axial stress of OS(8.12) MN/m2 Reference to any solid mechanics text will show that this situation is normally dealt with by

calculating an equivalent stress, a, For a cylinder under pressure a, is given

by OSa& where a0 is the hoop stress This would permit the above question

to be solved using the method outlined earlier

Example 2.4 A glass bottle of sparkling water has an acetal cap as shown

in Fig 2.14 If the carbonation pressure is 375 kN/m2, estimate the deflection

at the centre of the cap after 1 month The value of Poissons ratio for acetal may be taken as 0.33

Solution The top of the bottle cap is effectively a plate with clamped edges The central deflection in such a situation is given by Benham et al as

60 Mechanical Behaviour of Plastics

640 - 64 x io00 x (1.213

S = l m m

Example 2.5 In a small polypropylene pump the flange on the cover plate

is 2 mm thick When the rigid clamping screws are tightened, the flange is

reduced in thickness by 0.03 mm Estimate the initial stress in the plastic and

the stress after 1 week

Solution The strain in the material is given by

curve shown in Fig 2.8 it may be seen that the initial stress is 16 MN/m2 and

the stress after 1 week is 7 MN/m2

Mechanical Behaviour of Plastics 61

Fig 2.15 Creep curves for acetal (20°C)

Accurately performed relaxation tests in which the strain in the material was maintained constant and the decaying stress monitored, would give slightly lower values than those values obtained from the isometric data

It should also be noted that in this case the material was loaded in compre- sion whereas the tensile creep curves were used The vast majority of creep data which is available is for tensile loading mainly because this is the simplest and most convenient test method However, it should not be forgotten that the material will behave differently under other modes of deformation In compres- sion the material deforms less than in tension although the effect is small for

strains up to 0.5% If no compression data is available then the use of tensile data is permissible because the lower modulus in the latter case will provide a conservative design

2.6 Thermal Stresses and Strains

It is quite common in modem engineering designs, for plastics to be used in conjunction with other materials, particularly metals In such cases it is wise to consider the possibility of thermal stresses being set up due to the differences

in the thermal expansion (or contraction) in each material

The change in shape of a material when it is subjected to a change in

temperature is determined by the coefficient of thermal expansion, a ~ Normally for isotropic materials the value of CYT will be the same in all directions For convenience this is often taken to be the case in plastics but one always needs

62 Mechanical Behaviour of Plastics

to bear in mind that the manufacturing method may have introduced anisotropy which will result in different thermal responses in different directions in the material

The coefficient of thermal expansion, UT, is given by

(2.17)

where SL is the change in length in the material

L is the original length

AT is the change in temperature

There are standard procedures for determining U T (e.g ASTM 696) and typical values for plastics are given in Table 1.2 It may be observed that the

coefficients of thermal expansion for plastics are higher than those for metals

Thus if 50 mm lengths of polypropylene and stainless steel are each heated up

by 60°C the changes in length would be

stress = modulus x strain

If the modulus of the material is 1.2 GN/m2 at the final temperature, then the stress in the material would be given by

stress = 1.2 x io9 (;E) - = -7.2 MN/m2 Note that the stress is compressive because the material is effectively compressed by 0.3 mm

Example 2.6 The bobbin shown in Fig 2.16 has been manufactured by sliding the acetal ring on to the steel inner and then placing the end-plate in position At 20°C there are no stresses in the acetal and the distance between the metal end-plates is equal to the length of the acetal ring If the whole assembly is heated to lOO"C, calculate the axial stress in the acetal It may be assumed that there is no friction between the acetal and the steel The coeffi-

cients of thermal expansion for the acetal and the steel are 80 x 10-6"C-' and

1 1 x 10-60C-1 respectively The modulus of the acetal at 100°C is 1.5 GN/m*

Mechanical Behaviour of Plastics 63 Steel end-plate rigidly

/

LG-4

Fig 2.16 Metal bobbin with plastic sleeve

Solution Under free conditions, the acetal would expand more than the steel but in the configuration shown they will both expand to the same extent Hence, the acetal will effectively be put in compression by an amount given

Thus there will be a compressive stress of 8.3 MN/m2 in the acetal It should

be noted that the above analysis ignores the effect of the constraining effect which the acetal has on the thermal expansion of the steel However, as the modulus of the steel is over 100 times greater than the acetal, this constraining

64 Mechanical Behaviour of Plastics effect will be very small Thus the above analysis is perfectly acceptable for engineering design purposes

Example 2.7 A nylon ring with a nominal inside diameter of 30 mm, an outer diameter of 50 mm and a width of 5 m m is to be made an interference fit

on a metal shaft of 30 mm diameter as shown in Fig 2.17 The design condition

is that the initial separation force is to be 1 kN Calculate (a) the interference

on radius needed between the ring and the shaft and (b) the temperature to which the nylon must be heated to facilitate easy assembly What will be the maximum stress in the nylon when it is in position on the shaft? The coefficient

of friction between nylon and steel is 0.25 The short-term modulus of the nylon

is 1 GN/m2, its Poisson's ratio is 0.4 and its coefficient of thermal expansion

Mechanicd Behaviour of Plastics 65

internal pressure (see Appendix D) At the inner surface of the ring there will

Hoop stress, cr,, = p { ;}

Radial stress, or = -p

where k = ratio of outer to inner radius

S O

E

This is the interference needed on radius to give the desired separation force

(b) To facilitate easy assembly, the nylon should be heated so that the inner

-

0.32

100 x x 15

( c ) The maximum stress in the nylon ring when i t is on the shaft will be the

0 0 = p

66 Mechanical Behaviour of Plastics

As a practical point, it should be noted that the separation force will change with time This is because the modulus, E, will decrease with time Suppose that in this Example, the assembly is to be maintained in position for 1 year

and that during this time the modulus decreases to half its initial value (the 1

year modulus would be obtained from the creep curves in the normal way) The above analysis shows that the interface pressure would then be half its

initial value (because 6r is fixed) and this in turn means that the separation force would become 500 N instead of 1 kN

2.7 Multi-layer Mouldings

It is becoming common practice to have the cross-section of a plastic moulding made up of several different materials This may be done to provide a perme- ation barrier whilst retaining attractive economics by having a less expensive material making up the bulk of the cross-section To perform stress analysis in such cases, it is often convenient to convert the cross-section into an equivalent section consisting of only one material This new section will behave in exactly the same way as the multi-layer material when the loads are applied A very common example of this type of situation is where a solid skin and a foamed core are moulded to provide a very efficient stiffnesdweight ratio This type

of situation may be analysed as follows:

Example 2.8 A polypropylene sandwich moulding is 12 mm thick and

consists of a foamed core sandwiched between solid skin layers 2 mm thick A beam 12 mm wide is cut from the moulding and is subjected to a point load, W,

at mid-span when it is simply supported over a length of 200 mm Estimate the depth of a solid beam of the same width which would have the same stiffness when loaded in the same way Calculate also the weight saving by using the foam moulding The density of the solid polypropylene is 909 kg/m3 and the density of the foamed core is 600 kg/m3

Solution The first step in analysing the foamed sandwich type structure is

to calculate the second moment of area of the cross-section This is done by converting the cross section to an equivalent section of solid plastic This is shown in Fig 2.18

The equivalent width of the flange in the I section is given by

(2.18)

where E, and E, refer to the modulus values for the core ( c ) and solid (s)

material In most cases there is very little information available on the modulus

of foamed plastics but fortunately an empirical relationship has been found to

Mechanical Behaviour of Plastics 67

Fig 2.18 Equivalent section for sandwich moulding

exist between density (p) and modulus (Moore et ul.)

W , = 12 x 11.3 x x 909 x lo3 = 123 g The weight per unit length of the foamed beam is

W f = (909 x 2 x 12 x 2 x + (600 x 12 x 8 x

= 101.2 g

Hence the weight saving is 17.7%

Once the foamed plastic moulding has been converted to an equivalent section of solid plastic then the long term design procedures illustrated in the

68 Mechanical Behaviour of Plastics previous questions can be used For example, to determine the 1 year deflec- tion of the beam in this question, the appropriate 1 year modulus from the polypropylene creep curves could be used

Throne has reported that the relationship between foam modulus and density

can be generalised to other properties such as tensile strength, fatigue strength,

creep properties as well as shear and compression modulus Thus if X is the general material property then

X, (2.20)

Example 2.9 A solid polyethylene beam is 10 mm thick and 15 mm wide

If it is to be replaced with a sandwich section with solid polyethylene in the two outer skins and polyethylene foam (density = 200 kg/m3) in the centre,

calculate the dimensions of the sandwich beam if it is to have optimum stiffness

at the same weight as the solid beam If the foam material costs 20% more than

the solid material, calculate the increase or decrease in cost of the sandwich beam

Solution The weight per unit length, W , for the solid beam is

the section to an equivalent ‘I’ beam as in the previous example

The flexural stiffness of the beam will be proportional to EZ So converting

- - + ( ; ) 2 $ ] 12 h)3 bh3 E,I = E ,

Substituting for h from above, the stiffness may be optimised for constant W ,

by differentiating this expression and equating to zero This gives the optimum beam depth as

Mechanical Behaviour of Plastics 69

Equivalent solid section

Using the information given

70 Mechanical Behaviour of Plastics

per unit length of 140 g/m (equivalent to a solid thickness of 10 mm) If a

different solid thickness is of interest then the optimum skin thickness would

need to be scaled accordingly For example, for a 5 mm thick solid beam which

is to be converted to a foam sandwich beam of the same weight per unit length,

the optimum skin thickness would be half of the value shown in Fig 2.20 The

optimum skidcore thickness ratios are shown in Fig 2.21

Densny ratk (core:-)

Fig 2.20 Variation of optimum skin t h i c h with core: skin density ratio