Tutorial Questions and Answers - 2011

7/28/2019 Tutorial Questions and Answers - 2011

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TUTORIAL QUESTIONS

Dr. S. Guo

Nov. 2009

Question 04-13.1

Determine the transverse modulus E2 of a carbon/epoxy composite ply using the

micromechanics and the Halpin-Tsai approach withx1=1 respectively. The material

properties are given as follows:

E2f=14.8 GPa; Em =3.45 GPa; vm =0.36; Vf=0.65

Answers: E2 =7.56 GPa (or 6.88 if ignoring vm); E2 =8.13 GPa

Question 04-13.2

Determine the in-plane shear modulus G12 of a glass/epoxy composite ply using the

micromechanics and the Halpin-Tsai approach withx2 =1 respectively. The material

properties are given as follows:

Gf=28.3 GPa; Gm =1270 MPa; Vf=0.55

Answers: G12 =2.68 GPa; G12 =3.84 GPa

Question 04-13.3

Plot a curve for the ratio E2 / Em against Vf=0 ~ 1 for one of the following composite

ply using the Halpin-Tsai approach withx1=1.

material Boron

/epoxy

E-glass

/epoxy

Boron

/aliminum

SiC/aluminium Carbon/epoxy Kevlar

/epoxy

E2f / Em 120 20 6 5 4.3 2

Question 04-13.4

Plot a curve for the ratio G12 / Gm against Vf=0 ~ 1 for one of the following composite

ply using the Halpin-Tsai approach withx2 =1.

material Boron

/epoxy

E-glass

/epoxy

Boron

/aliminum

SiC/aluminium Carbon/epoxy Kevlar

/epoxy

G12f/ Gm 130 23 6.4 5.5 22 16


2/10

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Question 04-14.1

Determine the longitudinal modulus E1 and tensile strength Xt of a unidirectional solicon

cardide/ceramic composite with the following properties:

Vf=0.4 Ef=172 GPa; Em =97 GPa; fft =1930 MPa; fmt =138 MPa

Answers: E1=127 GPa; Xt =1425.1 MPa -- Eq(14.1); Xt =180.7 MPa -- Eq(14.2)

Question 04-14.2

Determine the longitudinal modulus E1 and tensile strength Xt for a unidirectional E-

glass/ epoxy composite with the following properties:

Vf=0.65 Ef=69 GPa; Em =3.45 GPa; fft =3450 MPa; fmt =104 MPa

Answers: E1= 46.06 GPa; Xt = 2303 MPa -- Eq(14.1); Xt = 1388 MPa -- Eq(14.2)

Question 04-16.1

For a graphite/epoxy unidirectional lamina of properties E1=181 GPa, E2=10.3 GPa,

v12=0.28, G12=7.17 GPa, determine

1. the compliance matrix;

2. minor Poissons ratio v21;

3. reduced stiffness matrix;

4. strains in the 1-2 coordinate system if the applied

stresses aref1=2 MPa, f2= -3 MPa,f12 = 4 MPa

(hint: 1. Apply Eq(16.3); 2. use Eq(16.4);

3. use Eq(16.1); 4. use Eq(16.6))

Question 04-17.1

For a 60o angle graphite/epoxy unidirectional lamina of properties E1=181 GPa, E2=10.3

GPa, v12=0.28, G12=7.17 GPa, determine

1. the transformed compliance matrix;2. transformed reduced stiffness matrix;

If the applied stress isfx=2 MPa, fy= -3 MPa,

fxy = 4 MPa,

also find

3. Global strains (in x-y-axis);

4. Local strains and stress (in 1-2-axis);

(Answers: 3. ex=0.5534 x 10-4, ey= -0.3078 x 10

-3, exy = 0.5328 x 10-3,

4. e1=0.1367 x 10-4, e2= -0.2662 x 10

-3, e12 = -0.5809 x 10-3

5. f1=1.714 MPa, f2= -2.714 MPa,f12 = -4.165 MPa

f1=2 MPa

F2 = -3 MPa

f12=4 MPa

f1=2 MPa

F2 = -3 MPa

f12=4 MPa

f1=2 MPa

F2 = -3 MPa

f12=4 MPa

f1=2 MPa

F2 = -3 MPa

f12=4 MPa


3/10

3

Question 04-18.1

Find the Eng constants for a 60o angle graphite/epoxy unidirectional lamina of properties

E1=181 GPa, E2=10.3 GPa, v12=0.28, G12=7.17 GPa

(Answers:

Ex= 12.42 GPa, vxy = 0.09783, mx = 5.854;Ey= 28.78 GPa, my = 8.5, Gxy = 8.761 GPa)

Question 04-19.1

Determine the maximum value ofF>0 when a stressfx= 2F,fy= -3Fandfxy = 4FMPa is

applied to a 60o angle graphite/epoxy unidirectional lamina. Use the maximum stress,

maximum strain and Tsai-Hill failure theory respectively. The properties of the lamina

are given below:

E1=181 GPa,E2=10.3 GPa, v12=0.28, G12=7.17 GPaXt=1500 MPa,Xc=1500 MPa, Yt = 40 MPa, Yc= 246 Mpa, S= 68 Mpa

Use ext = Xt / E1; exc = Xc / E1; eyt = Yt / E2 ; eyc = Yc / E2; es = S/ G12 ;

(Answers: 0< F


4/10

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Question 04-20.2

For the three-ply [-45/30/0] carbon /epoxy laminate, the material strength of the ply is

given as Xt = 1370 MPa, Xc = 1000 MPa, Yt = 42 MPa, Yc=200 MPa, S=60 MPa.

Based on the [A], [B], [D] for the laminate obtained in Q20.1,

a) determine the compliance matrices [a], [b], [d]

b) determine the strain {e} of the laminate (mid-plane plus curvature)c) given a load Nx = 500 kN/m, Ny=500 kN/m, Nxy=50 kN/m, Mx=My=Mxy=0,

determine the strain e1, e2, e12 and stress f1, f2, f12 in the 0o ply

d) determine the F.I. for the 0o ply

5 mm

0o

30o

-45o

5 mm

5 mm

5 mm

0o

30o

-45o

5 mm

5 mm

Answers:************** THE COMPLIANCE MATRIX [abd] ***************

0.1297E-08 -0.9843E-09 -0.6039E-09 0.1777E-06 -0.1098E-06 0.1627E-06

-0.9843E-09 0.4476E-08 -0.1559E-09 -0.1480E-06 -0.2187E-06 0.2474E-06

-0.6039E-09 -0.1559E-09 0.3712E-08 -0.5488E-07 0.3119E-06 -0.3934E-06

0.1777E-06 -0.1480E-06 -0.5488E-07 0.5967E-04 -0.3000E-04 0.3011E-04

-0.1098E-06 -0.2187E-06 0.3119E-06 -0.3000E-04 0.2471E-03 0.7388E-04

0.1627E-06 0.2474E-06 -0.3934E-06 0.3011E-04 0.7388E-04 0.3130E-03

Strain {e}={eo}-z{k} in X, Y & XY in each layer (z-coordinate is measured from mid-plane of the layer)

0.1865E-03 0.9948E-03 0.7327E-030.1260E-03 0.1738E-02 -0.1943E-03

0.6539E-04 0.2481E-02 -0.1121E-02

********** THE STRESS IN EACH LAYER ************

PLY-NO S1 S2 t12

1 0.4355E+08 0.1055E+08 -0.5795E+07

2 0.8499E+08 0.1597E+08 0.9314E+07

3 0.1908E+08 0.2586E+08 -0.8040E+07

Max FI from Max Stress Criterion in 1, 2 & 1-2: 0.0620 0.6158 0.1552 in the layers: 2 3 2


5/10

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Question 04-21.1

Determine the in-plane (membrane) and flexural (bending) Eng elastic constants for a

three-ply [0/90]s graphite/epoxy laminate of properties E1=181 GPa, E2=10.3 GPa,

v12=0.28, G12=7.17 GPa. Each ply thickness is 5 mm.

(Answers:

in-plane: Ex=124.5 GPa, Ey=67.43 GPa, Gxy=7.17 GPa, vxy=0.04292, vyx =0.02323;

bending: Ex=175.0 GPa, Ey=16.65 GPa, Gxy=7.17 GPa, vxy=0.1735, vyx =0.01651 )

Question 04-22.1

A three-ply [-45/30/0] graphite/epoxy laminate of ply thickness 5 mm is subjected to a

load of Nx = Ny =1000 N/m. The properties of the laminate are E1=181 GPa, E2=10.3

GPa, v12=0.28, G12=7.17 GPa and strength Xt=1500 MPa, Xc=1500 MPa, Yt = 40 MPa,

Yc = 246 MPa, S=68 MPa.

Determine

1. Mid-pane strains and curvature;

2. Global and local stresses on upper surface of 30o ply;

3. Percentage of load Nx taken by each ply;

4. F.I. of each ply of the laminate (choose any two of the criteria to compare)

5 mm

0o

30o

-45o

5 mm

5 mm

5 mm

0o

30o

-45o

5 mm

5 mm

Answers:

1. exo =3.123 x 10

-7 (m/m); eyo =3.492 x 10

-6 (m/m); exyo = -7.598 x 10

-7 (m/m);

kx=2.971 x 10-5 (1/m); ky = -3.285 x 10

-4 (1/m); kxy = 4.101 x 10-4 (1/m);

2. fx =6.930 x 104 (Pa);fy =7.391 x 10

4 (Pa);fxy = 3.381 x 104 (Pa);

f1 =9.973 x 104 (Pa);f2 =4.348 x 10

4 (Pa);f12 = 1.890 x 104 (Pa);

3. Percentage of Nx taken by 0o ply =22.32 %, by 30 o ply = 53.15 %,

by -45 o ply =24.52%

5 mm

5 mm

5 mm

0o

0o90o

5 mm

5 mm

5 mm

0o

0o90o


6/10

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Question 04-29.1

Determine the axial forces carried in the elements of a C-section subjected to a B.M. of

Mx =1 x 106 Nmm. The extensional and bending compliance matrices for the flanges are

The extensional and bending compliance matrices for the web are

Answers:

Flange Ez = 1 / ta11 = 1/ 1x0.0185 = 54.05 x 103 N/mm2

Ixx = A y2 = 1 x 100 x 100

2 = 106 mm4 = 10-6 m4

Web Ez = 1 / ta11 = 1/ 0.5x0.1127 = 17.75 x 103 N/mm2

Ixx = b h3 /12 = 0.5 x 200

3 / 12 = 0.333 x 106 mm4 = 0.333 x 10

-6 m4

Ixx = ( EzIxx ) = 2 x54.05x109x10

-6 + 17.75x109x 0.333x10

-6 mm4 = 114.02x103 Nm2

Since Ixy = 0,My = 0 andMx =Mx = 103 Nm

Flange: fz =Ez (Mx Y /Ixx) = 54.05 x109

x (103x 0.1 / 114.02 x 10

3) = 47.4 MPa

Force Pz =fz (10-3

x 0.1) = 4.74 kN

Web: fz =Ez (Mx Y /Ixx) = 17.75 x109

x (103Y/ 114.02 x10

3) = 155.67 Y MPa

The web force intensity varies linearly fromNz =fzx0.1 x 0.5 x 10-3 = 7.78 kN/m at the

top and to zero at the middle and then7.78 kN/m at the bottom of the web. The force

at any point of the web isPz =NzdY.

(Double Check Internal

1.0

0

2363 105.01067.15522.01074.4 dYYMx

= 0.948 x10

3

+ 2 x 77.83 x10

3

x 1/3(0.1)

3

= 0.948 x10

3 + 51.9 x103x10

-3= 999.9 Nm 1000 Nm )

kNmmAa /

0483.000

00185.00058.0

00058.00185.0

][][ 1

mmkNDd

/1

7376.01598.00199.0

1598.05396.00759.0

0199.00759.00336.0

][][1

kNmmA /

0549.000

01127.00873.0

00873.01127.01

mmkND

/1

59.443.143.1

43.187.575.3

43.175.387.51

100 mm

200 x

z

y

Web lay-up

[45/-45]s

Flange lay-up

[0/45/-45/90]s

Flange t =1 mm

Web t =0.5 mm

Mx

100 mm

200 x

z

y

Web lay-up

[45/-45]s

Flange lay-up

[0/45/-45/90]s

Flange t =1 mm

Web t =0.5 mm

Mx


7/10

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Question 04-30.1

A symmetric rectangular box section is

subjected to a shear force Sy= 10 kN

acting through the shear centre of thesection. The lay-up of the flanges is

[0/45/-45/90]s and the properties are:

equivalent membraneEx =54.1 kN/mm2,

t=1 mm and the lay-up and properties

for the webs are [45/-45]s and

Ex =17.7 kN/mm2, t=0.5 mm.

Determine the shear flow distribution

in the wall of the section.

Answers:

Ixx = 1.37 x 104 Nm2,

basic shear flows: q12 = -0.987 x 106S1 (S1=0, q12 =0; S1=0.1, q12 = -9.87 x 10

4 N/m);

q23 = -16.15 x 104S2 + 3.23 x 10

6S22 - 9.87 x 10

4

(S2=0: q23 =-9.87 x 104; S2=2.5 x 10

-2 m: q23 = -10.07 x 104 N/m;

S2=5 x 10-2 m: q23 = -9.87 x 10

4 N/m);

q34 = (-54.1 x 109x 10

4 / 1.37 x 104 )(-2.5 x 10

-2) x 10-3S3 - 9.87 x 10

4

(S3=0: q34 = -9.87 x 104 N/m; S3=0.1 m: q34 = 0)

The rest of walls have symmetrical distribution but positive sign.

Since the Sy is acting at the centre of the symmetrical section where cut has been made

and there is no external torque, qo = 0. Hence the final shear flow remains the same asthe basic shear flow.

(check: internal horizontal force Px=0; Transverse Py = 2 x10 x 104

x 50 x 10-3 =10 kN)

12

4 3

Sy=10 kN

200 mm

50 mmX

Y

Z

xy

x

y1

2

4 3

Sy=10 kN

200 mm

50 mmX

Y

Z

xy

x

y

-9.87x10

-9.87x104

-10.07x10

9.87x104

9.87x10

10.07x10

+

+

+

1 2

345

6


8/10

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Question 04-31.1

Determine the maximum shear stress and

twist angle of a 1 m long laminated C-section

beam subject to a torque T=1 kNmm.The property of the laminate is Gxy=20.9 GPa.

(Answer: GJ=3.5 x 105 Nmm2; = 2.9 rad,fxy = 30 MPa)

Question 04-31.2

A symmetric rectangular section box beam

fixed at one end and free at the other is

subjected to a torque = 10 Nm. The layup

of the top and bottom walls is [0/45/-45/90]s

and the properties are: equivalent membrane

Ex =54.1 kN/mm2, Gxy =20.7 kN/mm

2 t=1 mm.

The layup and properties for the webs are

[45/-45]s andEx =17.7 kN/mm2, t=0.5 mm

and Gxy =36.4 kN/mm2.

Determine the shear flow distribution in the

wall of the section and the maximum twist angle.

Answers:

Shear flow q = T/ 2Am = 10000 / 2x50x200 = 0.5 N/mm

Top and bottom walls: a33 = 1/t Gxy = 1/ 20.7x103 = 0.483 x 10

-4 mm/N

Webs: a33 = 1/t Gxy = 1/ 0.5x36.4x103 = 0.55 x 10

-4 mm/N

Sectional )1055.0210483.02/(44200

0

50

042 dsdsAJG m

=4x(50 x 200)2/(2 x 0.483 x 10

-4x 200 + 2 x 0.55 x 10

-4x 50)

=4 x 108 / (0.0193 + 0.0055)= 161.3 x 10

8 = 16.13 x 109 Nmm2

69 1062.01013.16/10000// JGTdzd

(deg)0355.0)(1062.01062.0 36 radL

100 mm

200 x

z

y

flangeand web

t =0.5 mm

T

100 mm

200 x

z

y

flangeand web

t =0.5 mm

T

12

4 3

T =10 Nm

200 mm

50 mmX

Y

Z

xy

x

y1

2

4 3

T =10 Nm

200 mm

50 mmX

Y

Z

xy

x

y


9/10

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Question 04-31.3

A thin-walled box beam of length L=1 m

with the cross section shown is made of

graphite epoxy with lay-up [452/012/452].The material properties are E1=148 GPa,

E2 = 9.65 GPa, G12 = 4.55 GPa, v12 = 0.3,

ply thickness t =0.1 mm. Gxy = 18.0 GPa,

The beam, built in at one end and free at

another, is subjected to a torque T = 500 Nm

acting at the free end. Determine the

maximum twist angle and the q.

Answers:

Shear flow q = T/ 2Am = 500x103 / 2x50x70 = 71.23 N/mm

For all walls: a33 = 1/t Gxy = 1/ 2 x 18.0x103 = 27.77 x 10

-6 mm/N

Sectional )1077.2721077.272/(4650

0

70

062 dsdsAJG m

=4x(50 x 70)2/(2 x 27.77 x 10

-6x 50 + 2 x 27.77 x 10

-6x 70)

=4 x 1.225x107 / (2.777 + 3.888) x 10

-3 = 7.352 x 10

9Nmm2

)/(1068.010352.7/10500// 493 mradJGTdzd

(deg)95.1)(1068.01068.0 14 radL

x

y

50 mm

70 mm

2 mm

T

x

y

50 mm

70 mm

2 mm

T


10/10

10

Question 04-25.1

Determine the buckling loadNbx of a laminated plate simply-supported along all edges

when subjected to the following load conditions. The plate dimensions are a=273.6 mm,

b=100 mm and its bending stiffness matrix is

(a) a uniaxial load

(b) a biaxial load with a tensile load

Ny =100 N/m ;

(c) a shear load Nxy in addition to

the uniaxial load with a of ratio 0.7:1

Answers:

(a)From ESDU 80023 Table 3.1, use Fig.1

414.1517.0736.2)468.1

1048.0

(100

6.273

)(

4/14/1

11

22

D

D

b

a

From Fig.1 curve C=2, Ko=24.5

mNC

Nbx /1229)68.261.9(1001.0

)0152.020315.0(

01.0

)1048.0468.1(5.24 222/1

(b)Use Fig.2 according to Table 3.154.9

1048.0

101002

22

2

D

bNby

From Fig.2, Ko=6 mNNby /0.503)68.235.2(102

Question 04-25.2

Determine the buckling loadNbx of a laminated plate simply-supported along all edges

when subjected to biaxial loadings withNy = - 410 N/m. The plate dimensions are a=225

mm, b=173 mm and its bending stiffness matrix is

NmD

0521.000

01048.00315.0

00315.0468.1

NmD

42.100

0645.029.1

029.164.7

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Tutorial Questions and Answers - 2011