MECA-H-406 Composite structures - Exercises 2: solutions

Exercise 1

T

L

sT

eL

(a) The load case is represented in the diagram. The most generalapproach is based on the compliance matrix, [S], in the natural(orthotropy) axes {L, T} :

εL

εL

γLT

=

1/EL −νLT /EL 0−νLT /EL 1/ET 0

0 0 1/GLT

.

σL

σT

τLT

= [S] .

σL

σT

τLT

(1)

For the load case considered, we have σL=0, τLT =0, σT =90MPa, εL=-2.1e-4 (and εT 6=0).We can inject these values in eq.(1) ; the first equation gives

εL = −νLT

EL.σT ⇔ νLT = −εL.

EL

σT= 0.117 [/] , (2)

where we have used the rule of mixtures for computing EL = 50.05GPa. Note : εL is negativeas the ply is compressed in the longitudinal direction.

Other approach : we first focus on the stress-strain relation in the transverse direction :

εT =σT

ET= 0.0161 [/] , (3)

where we have used Halpin-Tsaı model to approximate ET =5.59GPa (see session 1). Then,we combine this result with the definition of the minor Poisson’s ratio (we have to use thisone because the load is in the transverse direction)

εL = −νTL.εT ⇔ νTL = − εL

εT= +1.3e− 2 [/] . (4)

Finally, we use the equation that relates the major and the minor Poisson’s ratios :

νLT

EL=

νTL

ET⇔ νLT = νTL

EL

ET= 0.117 [/] . (5)

Explain why this second approach is less general than the first one.

1

(b) When a ply is subjected to compression in the longitudinal direction, in most practicalcases, the failure comes from the matrix. The composite breaking strain is then given by (slide11 of part 2, eq.(3.43) of the reference book)

εTu = εmu.(1− V

1/3f

)=

σmu

Em.(1− V

1/3f

)= 1.537e− 3 [/] . (6)

We need to determine the maximum compressive load that can be applied to the compositebefore it breaks, using the hypotheses E

′L = EL and E

′T = ET and the load case σL < 0 and

σT = τLT = 0. Therefore, we have

σL = EL.εL = −EL.εT

νLT⇒ σ

′Lu = −E

′L.

εTu

νLT= −657.63 [MPa] , (7)

where we have used the definition of the major Poisson’s ratio εT = −νLT .εL [/] (explain whywe did not use that of the minor Poisson’s ratio).

2

Exercise 2

T

L

z

z [m]

-h/2

h/2

t [°C]

t0

Dt

(a) We make the hypothesis that the distribution of temperatures in the ply follows thatdepicted in the diagram above : t0 is the average temperature variation with respect to thereference temperature of the ply tref (that at which it has been manufactured), and ∆tis a (spatial) temperature gradient that varies linearly with the normal coordinate z. Thecorresponding thermal strain is given by

ε = ε(z) = α. (t− tref ) = α.(t0 + ∆t .

z

h

)[/] . (8)

With respect to the theory of Euler-Bernoulli, the strain at any point of a beam can be written

ε = ε(z) = ε0 − k.z [/] , (9)

where ε0 is the average strain and k is the local curvature of the beam, defined by

k =∂2w

∂x2= w

′′[1/m] , (10)

where w is the transverse displacement of the beam along the z axis. Combining eq.(8) to(11), one obtains

k = −α∆t

h[1/m] . (11)

Approach 1 : going back to the theory of Euler-Bernoulli, the transverse displacements of abeam subjected to pure bending by a torque M(x) follows

w′′

= −M

EI[1/m] ⇔ M = −EIk ⇔ M = EI.

α∆t

h[Nm] . (12)

Following the same approach, it is left as an exercise to show that t0 induces non-zero normalforces in the beam, and that the resulting displacements are in general small with respect tothose induced by the bending torque.

Approach 2 : we use the definition of the resultant bending moment over a cross-section ofthe ply :

M =∫∫

A

z.σdA [Nm] . (13)

which leads to the same result (this is left as a good exercise).

3

(b) The expression of the displacement field is obtained by integrating twice the curvature ofeq.(11) over the length of the beam and applying the boundary conditions (simply supported)to determine the value of the constants. One obtains

w(x) =kx

2(x− L) [m] , (14)

which is the equation of a parabola with its maximum a mid-span given by

wmax = w(x = L/2) =∣∣∣∣−

kL2

8

∣∣∣∣ =α∆tL2

8h[m] . (15)

It is interesting to notice that the displacements are a function of α and not of E. Give aphysical explanation.

Using the numerical values given, one finally obtains the results below. Notice how the (tai-lored) properties of the composites allow changing notably the response of the beam to thesame load.

Material wmax [m]

Aluminium 0.074

Kevlar-epoxy -0.013

Graphite-epoxy 0.0067

4

Exercise 3

The data allows us to write easily the compliance matrix in the natural axes {L, T} usingeq.(1). However, the load case is given in the structural axes {x, y} and we want to determinethe component εy of the strain tensor. This requires the use of the transformation matrix[T ] from the {x, y} axes to the {L, T} axes (slide 14 of part 3, Section 5.3.4 of the referencebook). The steps are described below.

We start by expressing the stresses in the {L, T} frame :

σL

σT

τLT

= [T ] .

10e600

=

7.52.54.33

[MPa] , (16)

Then we use eq.(1) :

εL

εL

γLT

= [S] .

σL

σT

τLT

=

67.5e− 6227.5e− 6721.7e− 6

[/] , (17)

Finally, we apply the inverse transformation, using the property [T (θ)]−1 = [T (−θ)] :

εx

εy12γxy

= [T (θ)]−1 .

εL

εL12γLT

=

420e− 6−125e− 6499.4e− 6

[/] , (18)

Hence, we have εy = 125e− 6.

5

Exercise 4[Q

]is the stiffness matrix expressed in the structural axes {x, y} (slides 13-16 of part 3,

Section 5.3.4 of the reference book) :

σx

σy

τxy

=

[Q

].

εx

εy

γxy

[MPa] . (19)

The components Qij as a function of θ are detailed in the course notes. They have beenimplemented in Matlab to plot the curves in the figure below. Identify and comment physicallyon the particular values of θ.

0

20

40

60

80

100

120

10 20 30 40 50 60 70 80 90

q [°]

Q11

Q22

Q66

Q12

Q16

Q26

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