composite design and analysis

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Example # 01 The purpose of this problem is to illustrate how laminate modulii are determined and to compare the modulii of some composite laminates with those of some common metals. Material A: 90% 0 o , 10 % 90 o ‘s Graphite-Epoxy Type III Material B: 50% 0 o , 50% 45 o ‘s Class-I (Use Fig II-2, II- 4) Material C: 100% 45 o ‘s Answer Material E x (Msi) E xy (Msi) Spec. Wt (lb/in 3 ) A 15.5 0.9 0.055 B 10 2.65 0.055 C 3 4.35 0.055 Titanium 16 5 0.165 Aluminum 10.3 3.8 0.100 Example # 02 Determine the Lay up of following laminate in terms of Number of Plies Ply Orientations Stacking Sequence Answer: For axial loading, the design guideline require 60%/30%/10% for 0 o / 45 o / 90 o plies.

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Page 1: Composite Design and Analysis

Example # 01

The purpose of this problem is to illustrate how laminate modulii are determined and to compare the modulii of some composite laminates with those of some common metals.

Material A: 90% 0o, 10 % 90o ‘s Graphite-Epoxy Type III

Material B: 50% 0o, 50% 45o ‘s Class-I (Use Fig II-2, II-4)

Material C: 100% 45o ‘s

Answer

Material Ex (Msi) Exy (Msi) Spec. Wt (lb/in3)

A 15.5 0.9 0.055B 10 2.65 0.055C 3 4.35 0.055Titanium 16 5 0.165Aluminum 10.3 3.8 0.100

Example # 02

Determine the Lay up of following laminate in terms of

Number of Plies

Ply Orientations

Stacking Sequence

Answer:

For axial loading, the design guideline require 60%/30%/10% for 0o / 45o / 90o plies.

Use 60%/30%/10% 0o ,45%, 90o ---------- refer to laminate guidelines GR-EP, Type Class-1

Ex= 12.3X106 psi from Fig II-1

t = .0035 from Table II-3

t Ex

Page 2: Composite Design and Analysis

6 ) = 43,050 PSI

= t Ex : N= ( t Ex) T = 43050T

N= 43,050 T lb/in

T = Laminate thickness

43,050T=6267

T= .146”

t ply=.0056 from Table II-1

T=n/t : n= T/t ply

= No. of plies= =26

For Axial loading

Use 0o / 45o / 90o 16/8/2 61%/31%/8%

+ 9 – 0000 + 0000 - - 0000 + 0000 – 9 + [45, 90,-45, 04, 45, 04, - 45] s

Note: the 45o plies and 90o plies are used to resist shear and transverse normal stresses that may exist. However, it is assumed that the normal stress in the 0-degree direction is predominant, and if we design for it while using 60/30/10 configuration the panel will be “safe” relative to all stresses.

Analysis:

Only 0o s, 45o s and 90o s have been used to reduce complexity.

The 90o s are used to carry any axial load in the transverse direction not accounted for.

The 45o s are used to carry shear.

The layup is symmetric to eliminate coupling between in-plane strains are curvatures/twists. [B]=0

It is balanced to eliminate coupling between normal strains and shear strains. A16 = A26 = 0

The O’S and 90’ s are separated by 45’s to reduce poission’s mismatch (and possible delamination).

No more than 4 plies in any one orientation are stacked together (to prevent possible delamination between the similarly oriented ply stack and the adjacent plies).

We have avoided placing a similarly oriented ply stack at the outer surface, as such a stuck is more susceptible to damage.

Page 3: Composite Design and Analysis

We have used 45/90/-45 sequence at the outer surface because this sequence is more damage tolerant.

Major load carrying plies (O’S) are protected from possible damage by avoiding placing them at the outer surface.

It may be necessary to use fabric for the outer ply to provide abrasion resistance (however, this was not done here).

EXAMPLE 3

Determine a layup that will carry a load of Nx = 4300 lb/in

List all the properties of the panel.

The design guidelines requires %age of ;

Use 60% O’s, 30% 45’s 10% o’s

T= Thickness of Lamination = Fx/Ex t

Ex= 11.6 X 106 psi from Fig II-2

Page 4: Composite Design and Analysis

= .0035 from Table II-3

x= t Ex

x= .0035 (11.6 x 106 ) = 40,600psi limit stress

x :: Nx = xt

4300 = 40,600t

T=.106 t ply = .0059 from Table II-1 GR-EP Type-III, Class-1, Grade 145

No. of plies = .106/.0059 = 17.95 or 18 (refer to note in Ex-2)

Use [12/4/2] [0000 + 9 – 0000 - 9 + 0000] [O4/+/9/-/O2]

The properties of this panel [67% / 22% / 11%] are:

Ex =12.4 x 106 from fig II-2

Ey = 3.9 x 106 from fig II-2

Gxy = 1.65 x 106 from fig II-4

x = .31 from Fig II – 7 x = .7 x 10-6 from fig II-9

y = Ey/Ex x x = .0975 y = 4.5 x 10-6 from fig II-9

EXAMPLE 4

Material E (psi) Area (in2 ) A(equir)A 8 X 106 .25 .25 (8/12) = .167B 12 X 106 .25 .25 (12/12) = .25

A Sandwich structure is subjected to

2000 lb/in load in the laminate causes extension (no bending) as above

Determine the tensile force in each laminate.

Page 5: Composite Design and Analysis

Determine the average tensile stress in each laminate.

Determine the location of the tensile load resultant.

A = B

FA FB

AA EA AB EB

FA FB

AA AB

FA + FB = F = 2000 lbs

FA + AB FA = F

AA

FA = AA x F

AA + AB

FB = AB x F

AA +AB

FA = (2000X2.5) = 2000 lb

FA = (2000X2.5) = 3000 lb

A= = 8000 psi

B= = 12000 psi

= AAA + BAB

AA+AB

= 0.659.167)+0.05(0.25)

Page 6: Composite Design and Analysis

0.167+0.25

= 0.29”

EXAMPLE 5

Determine the valve of the ultimate tensile load for this panel based on the strain allowable (Table II-3) Assume that bending stresses are negligible.

Skin + Cover: Gr-Ep, Type IV, Class 2, style 3k-70-pw

Inner + outer chords : Gr-Ep, Type III, Class 1, Grade 145

Skin: [( 45) / (0,90) / ( 45)3 /(0,90) / ( 45)]s

Cover: [( 45)2 / (0,90)2 / ( 45)2 /(0,90)2 / ( 45)]s Fabric

Inner Chord: [04/+45/902/-45/02]s

Inner Chord: [0s/+45/-45/90/+45/-45/02]s

Thickness (Table II-1)

Skin: 2/10/2 14(.0083) =0 .116”

Cover: 18(.0083) = 0.149”

Inner Chord: 20(.0059) = 0.118”

Outer Chord: 24(.0059) = 0.142”

Modulus &Poisson’s Ratio (Fig II-2,3 + Fig II – 7 + 8)

Skin: 4/20/4 14.3%/71.4%/14.3% Ex = 4.65 x 106x= .44

Cover: 8/20/8 22%/56%/22% Ex = 5.7 x 106x= .34

Inner Chord: 12/4/4 60%/20%/20% Ex = 11.4 x 106x= .21

Outer Chord: 14/8/2 58%/33%/8% Ex = 11.3 x 106x= .38

Page 7: Composite Design and Analysis

Note:

- = .44 - .34 = 0.1 there will be no excessive

- = .44 - .38 = .06 trouble from Poisson’s mismatch

- = .34 - .21 = .13

Areas

Skin: 0.116(7) = .812

Cover: 0.149 [7+2(1.15-.116-.149)] = 1.307

Inner Chord 0 .118[1.45-2(.149)] = .136

Outer Chord 0 .142[1.45-2(.149)] = .163

Loads

P=E A

Skin: P= 4.65 x (.0035) (.812) = 13215 lbs.

Cover: P= 5.7 x (.0035) (1.307) = 26075 lbs.

Inner chord: P= 11.4 x (.0035) (.136) = 5426 lbs.

Outer chord: P= 11.3 x (.0035) (.163) = 6447 lbs.

51163 lbs.

EXAMPLE 6

If the total load on the panel of problem 5 is 40,000lb, determine the loads in each element. Also determine the strain and the margin of safety. (Based on ultimate load)

Equivalent Areas (Transform all element areas to an equivalent amount of skin area)

Skin: (.812) = .812

Page 8: Composite Design and Analysis

Cover: (1.307) = 1.602

Inner Chord: (.136) = .333

Outer Chord: (.163) =

= (F) = (F)

= x (F)

Loads Stresses

= 10,334/.812 = 12,727 psi

= 20,388/1.307 = 15,600 psi

= 4,238/.136 = 31,162 psi

= 5,040/.163 = 30,920 psi

40,000 lbs

Strain Margin of Safety

MS= - 1

Note: This may also be determining with the loads from an individual element. e.g

Page 9: Composite Design and Analysis

From an individual element. e.g

EXAMPLE 7

The aluminum skin and stiffeners is to be replaced by the composite. The basic aerodynamic cross section is to remain unchanged. Also, the “Stiffness’s” EA and GA are to remain unchanged. (Note: since the basic geometric shape is to remain unchanged, we can work with EA and GA instead of EI and GI. Since the tensional shear is only reacted by the skin, calculate GA for the skin only.) Use Figures II-2 and II-4 in the solution.

Aluminum Panel stiffnesses

.08

Since the composite stiffener spacing is 6”:

EA = 10.3 x 106 (6 x .20) = 12.36 x 106

The torsional stiffness is determined from the skin only:

GA = 3.9 x 106 (6 x .12) = 2.81 x 106

About the Solution

The determination of the number of plies and the layup configuration for each of the elements of the elements of the composite replacement is and iterative procedure. This example shows only the final calculations.

E and G for elements 2,3 and 4

No. of plies Percentage

Element

2 2 14 2 18 .1062 11 78 11 4.9x106 3.6 x 106

3 16 4 2 22 .1298 73 18 9 13.3 x 106 1.5 x 106

4 20 4 2 26 .1534 77 15 8 13.9 x 106 1.4 x 106

Page 10: Composite Design and Analysis

Possible layup for elements 2,3 and 4

2: 45 / 90 / 45 / 0/ 45 / 45]s

3: [04 / +45 / 04/ -45 / 90]s

4: [05 / +45/ 05/ -45/ 90]s

AE for element 1

12.36 x 106 = + 1.7 + 1.7)(.1062)4.9 x 106 +(1.2-.1062-.1062)(.1298)13.3 x 106

+ (1.2-.1062-.1062)(.1532)13.9 x 106

12.36 x 106 = (AE)1 + 4.98 x 106 + 1.71 x 106 + 2.11 x 106

(AE)1 = 3.65 x 106

AG for element 1

2.81 x 106 = + [6-(1.2-.1062-.1062)](.1062)(3.6 x 106)

+ (1.2-.1062-.1062)(.1298)(1.5 x 106)

2.81 x 106 = + 1.92x 106 + .19x106

= .70x 106

Note: The shear is reacted by the “skin” only.

E and G for element 1

Assume that the sin is II plies; t = II (.0059) = .0049”

A = .0049 (6) = .3894

= 3.65 x 106 from before = .70 x 106 from before

= .3894 = .3894

Layup for element 1

Page 11: Composite Design and Analysis

For

For a balanced layup we use 4 plies of 45 degrees, which corresponds to 4/11 = 36%

For 36% 45’s and E = 9.37 x 106 , the % 0’s is 44%, which yields 5 plies.

Thus, we have 5/4/2 => 45.5% / 36.4% / 18.1%

This gives E = 9.6 x and G = 2.15 x

A possible layup is [0/+45/90/-45/04]s

Final check for matching EA and GA

Note: The solution used here is an exercise based on stiffness. Strength, durability, damage tolerance, buckling, etc are also important criteria.

EXAMPLE# 8

A fiberglass skin is bundle to a Gr-EP chord. Determine the axial stress in both members when going from a ground temp of 11F to a cruise temp of -65F. Ignore bending.

------------------------------------------------- -------------- ----------------------------------------

Skin (GL-Ep)

Page 12: Composite Design and Analysis

(in/in)

Chord (Graphite-Ep, Type III, Class 1, Grade 145

Direction No. of plies %

2 20

6 60

2 20

Axial Load

(

Stress

EXAMPLE 9

Determine the margin of safety for the panel with the elliptic hole (ref fig II – 10, 11)

______________________ ______ ______________________________

Panel Properties

No, of plies = 16 Gr-Ep, Type IV, Class II, 3K-70PW

Ply thickness = .0083” (0, 90)4 / (+- 45)2 / (0, 90)2 s

Page 13: Composite Design and Analysis

Panel thickness = 16(.0083) = .1328”

%O’S = 12/32 = 37.5

%45’S = 8/32 = 25

% 90’S = 12/32 = 37.5

E= 7.25 x

Allowable load

P = f(W D) t

= E (W D) t would be the load if we did not consider stress concentration.

Is the load (taking into account stress concentration)

Margin of Safety

EXAMPLE # 10

The addition of holes, whether for fasteners or for cutouts, decrees the load that the laminate can take. It is important to take into account this reduced capability. For the joint shown, determine the pad thickness required at the fasteners to sustain the load.

Page 14: Composite Design and Analysis

Modes of Failure

Bearing:

Net Tension:

Basic Panel Properties

No. of plies = 10

Ply thickness = .015”

Basic panel thickness = 10(.015)=.150”

%O’S=30,%45’s=40,%90’S=30

Bearing

Assume each row takes 50% of the load.

Net Tension

Figure II-12,

=

(4800(1.5)=(1.5-.25)t(25000)

3600 = .25t (60,000) 7200=1.25t(25000)

t = .24” t=.23”

Pad Configuration

t=.24

t ply=.015

Rivet Shear

Allowable load/fastener=4660 lb

Page 15: Composite Design and Analysis

Actual load / fastener=3600 lb

Rivet shear is safe.

Basic Panel Tension New Net Tension (for pad)Use = .005 in/in or allowable strainallowable load = E A

Basic panel tension is safe

= 5250 lb/in

%45’s=50,

= 4800 lb/inNote: D/t = .25/.24 = 1.04 Ok, fig II-12

= 1.5/.25 = 6

Example 11

Design a blade – stiffened tension panel from Gr-EP, Type II, Class 1 and Grade 145

The skin is to contain integral built-up planks for damage arrestment.

Stiffener spacing is 6”

Running load is 12, 700 lb/in

Strain allowable is .0035 in/in (Table II -3)

The skin-plank is to carry 70% of the total section load.

Blade

Unknowns:

1. b --------------------try b=2

2. c --------------------try no. of plies = 25

3. d --------------------use 60/30/10 percentages try b=2

4. e --------------------A=P/E

1. b=2

2. t=25(.0056)=.14

3. 15/8/2 60%/32%/8% E=12.23x from Fig II-1

Page 16: Composite Design and Analysis

4. A= P/ E

Bt + 2th = P/(12.23 x )).0035)

P=.30(12700)6=22860lbs

H=.9

[+9-000+000-000-000+000-9+]

Skin and Plank

Unknowns:

1. ts --------------------------- try 50 plies

2. tp --------------------------- try 70 plies, the added plies are 0’s

3. d --------------------------- Ply drop off = 6 plies per .2”

4. Es --------------------------- 10/80/10 percentage

5. Ep --------------------------- The skin is to contain the integral built-up plank of

interleaved 0-degree plies for crack

6. w ---------------------------- A=P/E

1. Ts = 50(.0056) = .28”

2. tp = 70 (.0056) = .392”

3. 6 plies per .2”

d=.67”

P=.70(12700)6=53340 lbs

4. 10/80/10 4/40/6 8/80/12 5.1

5. The added plies are O’s because the plank functions in the same manner as the stiffener flange, and the 0-degree plies act as a damage arrester for the skin.

4+20/40/6 = 24/40/6 34/57/9 E=8.6

Page 17: Composite Design and Analysis

6. AE = P/ P/

EXAMPLE # 12

Determine the allowable flexural load on the beam. The material is Gr –Ep, Type –III, Class –I grade 145. The flange has 14 plies (4/8/2) and the web has 8 plies (4/4/0).

Answer

Plies % 0’s

% 45’s

E t A Aeq y Aeq y

Flange

14 29 57 7.6 X 106

0.0826 1.15(.0826) .095(7.6/7.6)=0.95 1.3413 .1274

Web 8 50 50 10 X 106 .0472 1.3(.0472) .0614(10/7.6)=.0808

.65 .0527

Maximum Load based on allowable based strain

The web is transformed into an equivalent amount of flange material by multiplying the thickness by the ratio of E’s.

+Aeqd2 )f + +Aeqd2 )w

I = negligible +.0952(.3183)2 + )

+0.811(3.73)2

Page 18: Composite Design and Analysis

=0+.0096+.0114+.0113

=.0323 in4

Ff =

Ef f = =

M=1843 lb-in

1843 = (24)

P = 154 lbs

Maximum load based allowable web strain

we know change the flange into an equivalent amount of web material.

I =0.323 = .0246 in4

Ew w=

10 x 106 (0.0035) =

M=842

842 = (24)

P = 70 lbs therefore we limit the load to 70 lbs

1-Determine the layup: 4630 lb/in ultimate

No. of plies

Ply orientations

Stacking Sequence GR-EP,

Class-1,

Grade145

Page 19: Composite Design and Analysis

TypeII

2- use the figure (skin chord) shown in example 8. If the axial

Tensile load is 8000 lbs ultimate, determine the stress in the chord

Assume the axial strain is uniform over the entire cross section.

Determine the M.S in the chord, omit temperature effects.

Element 1 Element 2Material TypeIII, Class I, Grade 190 Type II, ClassI, Grade

145Layup (+9-00+ - 00 + -0)s (+99-0000+0000-00)

The total axial tensile load acting on the cross section shown in P lb/in determine the axial force and stress acting on each solid laminate (assume axial strains are identical)

Determine the margin of safety if P=9000 lb/in

4- Determine the ultimate tensile load on this panel, based on the strain allowable from table II-3. Determine the ultimate load and average ultimate strength for each of the three elements.

1- (+ 9 – 9 + 9 -+ -+ -+ -+ -+ -+ -+ 0 – 0) 50 plies

2- (+ 9 – 9 + 9 -+ -+ -+ -+ -+ - 0+ 00-00+00-00+00-0) 70 plies

3- (+ 9 – 000 + 000 – 000 – 000+000 – 9+) 25 plies

5 – Repeat the problem 4, but change the skin and plank to type II material. The blade will remain type III

6 – If the load acting on the section of example 5 is 8000 lb/in, determine the axial load and stress on each element.

7- Determine the allowable load, W lb/in for the mechanically fasten joint. Neglect the fastener shear

Page 20: Composite Design and Analysis

8- Determine the ultimate load P for the panel with the 3” D Cutout. Use the layup for the problem 7 ( use

t = .0064 in / in)

9-

Skin1 Skin 2Material

Type III, Class I, Grade 95 Type III, Class I, Grade 95

Layup (+9, 00)s (+ 9, 000 )

The sandwich panel is subjected to an axial load of 2000 lb/in as shown. Determine the axial force carried by each skin ) assume uniform strain over the cross section) determine the stress resultant in skin I. include in your description a flow chart of the analysis.

Use the same material and layup information as given in problem -3.

Determine the average compressive stress in element- 2 from the moment M in-lbs

Determine the maximum tensile stress in element 1 from the bending moment M

Determine the allowable value for M based on the strains given in table II-3

Page 21: Composite Design and Analysis

Answers:

1- No of plies =40 (10% 0’s, 80% -45’s, 10 % -90,s)

2- Skin= 5390 psi, MS= 0.82

3- Chord= 11750 psi, MS= 0.82

4- 1 = 20900 psi, 2 = 28400psi, MS = 054

5- P skin = 13940 lbs, P plank = 34700 lbs, P blade = 23950 lbs

6- P = 75370 lbs total

7- Skin= 10470 psi, Plank = 18950 psi , blade = 25850psi, MS= .57

8- W = 1125 lb/in (bearing critical)

9- = 825, tB = 3.2, tu = 20360 lbs

10- Skin = 820 lb/in

11- M = 5365 in-lb (based on element 2)