Plastics Engineering 3E Episode 14 ppsx

This may be converted to an equivalent shear stress by the relation... From the 1 day isochronous at this stress, From the 0.417% isometric curve, the stress after 1 year is 1.45 MN/m2 e

From 1 year isochronous at 0.5% strain, E = 370 MN/m2

16 x 370 x 2S3 x 4 3(1 - u)(5 + v ) P - 3 x 0.6 x 5.4 x 324

M y W L 150x200

= 5.1 MN/m2

c = - = - -

I 16d2 - 16dz

W = {0.48(?) (;)“”>

(2.16) This is a stress relaxation problem but the isometric curves may be used

From 2% isometric, after 10 seconds, E = - 16*75 - 837.5 MN/m2

0.02

= 3 N

837.5 x 34 x 10 128(1 Ed4S + w)R3N - - 128(1.4)53 x 10

(2.17) From the 1 day isochronous curve, the maximum stress at which the material is

linear is 4 MN/m2 This may be converted to an equivalent shear stress by the relation

Equivalent tensile stress = u = 2.15 x 2 x 1.4 = 6 MN/m2

From the 1 day isochronous at this stress,

From the 0.417% isometric curve, the stress after 1 year is 1.45 MN/m2

exceeded the stress in the pipe wall after 1 year

The pipe would leak if the hoop stress caused by atmospheric pressure (0.1 MN/m2)

(2.20) This is a stress relaxation problem, but the question states that creep data may

be used

Then from a 1.6% isometric taken from the creep curves it may be determined that the stress after 10 seconds is 15.1 MN/m2 and after 1 year it is 5.4 MN/m2

(a) Initial pressure at interface = p = (ha)/R = 3.02 MN/m2

Thus normal force at interface, F = p x 2xRL = 3.02 x 2x x 5 x 15

= 1.423 kN

So axial force, W = p F = 0.3 x 1.423 kN = 0.427 kN

(b) Similarly, after 1 year for u = 5.4 MN/m2 the axial force W = 0.153 kN

(2.21) (a) As illustrated in Example 2.7, the extraction force, F, is given by

Hence, the hoop strain, &e, is given by

Hence the nylon bush would need to be cooled by 143°C to achieve the necessary

contraction to have easy assembly This suggests a cooled tempem- of -123°C

(c) At the bore of the bush, Benham et al shows the stresses to be

2 k 2 p a@ =

Therefore the internal diameter of the bush will be 35 - 2(0.3) = 34.4 mm

(d) If the long-term modulus of the nylon is 1 GN/m2 then for the same interference conditions, the interface pressure would be reduced to half its initial value ie 2 MN/m2

This means that the separation force would be half the design value ie 600 N

(2.22) If the acetal ring is considered as a thick wall cylinder, then at the inner surface there will be hoop stresses and radial stresses if it is constrained in a uniform manner:

hoop stress, ae = p { z}

radial stress, a, = - p where p is the effective internal pressure

k is the ratio of the outer to inner radii

Hence, the hoop strain, E @ , at the inner surface is given by

A r ae a,

&e = - = - - v-

Weight of solid beam = 14.27’ x loe6 x 909 = 185.1 g

Weight of foamed beam =

384EI - 384 x 466.7 x 3126.3 (2.27) Weight of solid beam = 909 x 12 x 8 x 300 x

Weight of composite = (909 x 2 x 2 x 12 x 300 x 10-6)+(500 x h12 x 300 x

h = 7.27 nun, so composite beam depth = 11.27 mm

The ratio of stifhesses will be equal to the ratio of second moment of area

Consider a flexural loading situation as above,

s L3 cx E I

_ -

So ratio foam: solid: composite = 8:12:15.9

(2.29) (a) Sandwich Beam of Minimum Weight for a given stiffness

KWL3

KI

where K is a factor depending on loading and supports

For E s ~ , , >> E,, EI Ebdh2/2

2KWL3

S =

E&,,bdh2

Total mass

m = pCoEbhL + p*,,bdL

From deflection equation (1)

d = - 2KWL3 Esbnbh26

(b) Sandwich Beam of Minimum Weight for a Given Strength

Assuming the core does not fail in shear, failure occurs in bending when the stress in the skin reaches its yield strength cy, ie

(c) Geometry of Deformation

E = ~1 + and ~2 = 8 3 From above

The solution to this differential equation will have the form

where T = q / h and a0/6l is the initial strain when the stress is applied

The unrelaxed creep modulus is obtained by putting t = 0

Hence, unrelaxed modulus E" = - -

The relaxed modulus is obtained by putting t = 00

Finally from the expression for the 4-element model

BT*+'

A(n + 1) This will be a non-linear response for all non-zero values of n

(2.37) For the Maxwell Model

Note that in this question an alternative solution may be carried out using the creep

modulus but this causes slight inaccuracies

Now for Maxwell Model E ( t ) = -

So for case in question, the strain hstory is

From a plot of this data

Initial tangent modulus = - lo - 17.54 GN/m2

0.00057

10.98 0.001 0.1% secant modulus = - - 10.98 GN/mz

(2.43) From the information provided

After 10 cycles in the given sequence (t' = 1500 seconds)

So, after lo00 seconds, ~ ( t ) = 0.523%

x=lO

~ ( 1 5 x 104) = (0.523) +OS23 [ ( 1 5 ~ ) ~ " ~ - ( 1 5 ~ - l)0."4] = 0.691%

x= 1

Then if a straight line is drawn from the point 0.523, lo00 to 0.691, 1O,o00 on Fig 2.4

then this may be extrapolated to 1% strain which occurs at approximately t = 9 x

16 seconds This is the total creep time (ignoring recovery) and so the number of cycles for this time is

Total creep time

(2.45) For the PP creep curves in Fig 2.4, n = 0.114 at Q = 7 MN/m2, ~ ~ ( 2 1 6 x

104) = 0.99% after 11 cycles of creep (10 cycles of load removal), t = 66 hrs = 2.37 x 1 6 s (using equation in text or from computer p g r a m ) ~ ~ ( 2 3 7 x 1 6 ) = 1.105%

A line joining 0.99, 2.16 x 10'' to 1.105, 2.375 x lo6 on Fig 2.4 allows strain at

365 days (ie 365 days @ 6 hrs per day = 7.88 x lo6 seconds) to be extrapolated to 1.29% The computer program predicts

As shown in the previous question, the governing equation for this type of Standard

Linear Solid is given by:

52 tj

& + - E = - + - -

rl 51

Multiplying top and bottom by the conjugate of the denominator,

comparing this with t = A-Bu

t = toeu0IRT and B = y/RT

From Fig 3.10 creep rupture strength = 8 MN/m2

Using a safety factor of 1.5 the design stress = 8/1.5 = 5.33 MN/m2 for the pipe, hoop stress = p R / h

Table 3.1 gives KI, for acrylic as 0.9 - 1.6 MN/m-"")

Now at 1 x lo7 cycles of = 16 MN/mz

also 1 x lo7 cycles at 5 H f represents 2 x I d seconds

(i) loss of energy due to friction, etc = (44.145)(0.3 - 0.29) = 0.44 J

(ii) energy absorbed due to specimen fracture = (44.145)(0.29 - 0.2) = 3.973 J

= 165.5 kJ/m2 3.973

(12 x 2)104 impact strength =

(iii) Initial pendulum energy = 0.25 x 44.145 = 11.036 J

Loss of energy due to friction + specimen fracture = 0.44 + 3.947 = 4.414 J

Remaining energy = 11.036 - 4.414 = 6.6218 J

= 0.15 m 6.6218

44.145 (2.59)

from the graph - da = 2 x

Hence the following table may be drawn up

From this table it may be seen that carbon fibre has low energy absorbing capability compared with the other fibres However, the other fibres are not as stiff Hence it is

quite common to use hybrid fibre composites, eg glass and carbon fibres in order to

get a better combination of properties

(3.2) It is necessary to calculate the volume fraction for each of the fibres Firstly for the carbon

Hence for 1 kg epoxy, the weight of carbon = 2.52 kg

(3.4) Using rule of mixtures

(3.9) The local compliance matrix is

(3.10) The properties of the ply in the fibre and transverse directions are

El = 200 GN/m2 E2 = 11 GN/m2 G12 = 8 GN/m2

a22 h' x y - a66 h all

It may be Seen that these agree with the previous values

For the applied Force N,

[;;,I = a [;;,I ( = a [;I ")

(3.1 1)

(i) Symmetric

(ii) Non-symmetric (there is an uneven number of plies)

(iii) Symmetric (could also be written as [0/90/45/ - 4531 - 453/45/90/0],

(3.12) This is treated as a 4 layer situation with

= -0.7, hi = -0.2, h2 = 0, h3 = 0.2, h4 = 0.7 mm

For Material A Material B

Compliance Matrix Stamss Matrix

S =

1

-G 2

Overall SrifSness Matrix Overall Compliance Matrix

-

The Extension Stiffness Matrix is then obtained from

[AI = 0.@(-60) + 0.4Q(-30) + 0.4Q(O) + 0.4Q(30) + 0.4Q(60) + 0.4Q(90) which gives

The Extension Stiffness matrix is then given by

and the strains may be obtained as

1

1

8.77 x 1O4 2.98 x 1O4 -4.31 x 1O4 2.98 x 1O4 2.22 x 1O4 -1.35 x 1O4 -4.31 x 104 -1.35 x 104 2.96 x 104

8.77 x 1O4 2.98 x 1O4 -4.31 x 1O4

2.98 x 1O4 2.22 x 1O4 -1.35 x 1O4 -4.31 x 104 -1.35 x 104 2.% x 104 8.77 x 104 2.98 x 104 4.31 x 104 4.31 x 104 1.35 x 104 2.96 x 104

2.98 x 104 2.22 x 104 1.35 x 104 The A, B , D matrices are