Formulary for Composite MaterialsInstitute for Carbon Composites - LCC

16. Juni 2014Prof. Dr.-Ing. Klaus Drechsler

Lehrstuhl fr Carbon Composites

1

1 MicromechanicsFor isotropic materials

Eiso = 2(1 + )Giso (1)

1.1 Rule of MixturesE1 = Ef1 f + Em (1 f ), (2)

1E2

= fEf2

+ (1 f )Em

. (3)

12 = f12 f + m (1 f ), (4)

1G12

= fGf12

+ (1 f )Gm

. (5)

1.2 Square-Cylinder Model

E2 = Em (

f

1f (1 Em/Ef2) + (1f )

). (6)

1.3 Chamis Model

E2 =Em Ef2

Emf + (1f ) Ef2 . (7)

1.4 Circular-Cylinder-Assemblage (Hashin)

E22 =4G23K23

K23 +G23 + 4212G23K23E11

(8)

with

G(+)23 = Gm

(1 + (1 + cm)f

c3 f (1 + 3c2m(1f )21+c23f

)

), (9)

G()23 = Gm +

f1

Gf,23Gm +Km,23+2Gm

2Gm(Km,23+Gm)(1 f ), (10)

K23 = Km,23 +f

1Kf,23Km,23 +

1fKm,23+Gm

, (11)

Km,23 = Km +Gm3 , Kf,23 =

Ef,222(1 f,32 2f,12f,21) , Km =

Em3(1 2m) , (12)

and

2

cf =Kf,23

Kf,23 + 2Gf,23, cm =

Km,23Km,23 + 2Gm

, c1 =Gf,23Gm

, c2 =cm c1cf1 + c1cf

, c3 =c1 + cmc1 1 .

1.5 Halpin-TsaiM

Mm= 1 + f1 f , with =

(Mf/Mm) 1(Mf/Mm) +

, (13)

E2Em

= 1 + f1 f ,with =(Ef2/Em) 1(Ef2/Em) +

, (14)

2 MesomechanicsWaviness (crimp) of yarn

Fig. 1: Crimp interval for circu-lar yarns.

c = lyarnlfabric

1. (15)

Knockdown factor =

{1 + 2

(pid

)2[ ExGxy

2(1 + xy)]}1

, (16)

3

3 Macromechanics3.1 Lamina PropertiesCompliance matrix

= S (17)1

2

12

=S11 S12 0S12 S22 00 0 S66

1

2

12

(18)with

S11 =1E1

S22 =1E2

(19)

S12 = 12E1

= 21E2

= S21 S66 =1G12

(20)

Stiffness matrix = Q (21)

1

2

12

=Q11 Q12 0Q12 Q22 0

0 0 Q66

1

2

12

(22)with

Q11 =E1

1 1221 Q22 =E2

1 1221 (23)

Q12 =21E1

1 1221 =12E2

1 1221 = Q21 Q66 = G12 (24)

3.2 Transformation of Stress and Strainxy = T112 (25)

xy = RT1R112 = T T 12 (26)

Reuter matrix R:

R =

1 0 00 1 00 0 2

(27)

4

Tranformation matrix with s = sin, c = cos:

T =

c2 s2 2scs2 c2 2scsc sc c2 s2

T1 =c2 s2 2scs2 c2 2scsc sc c2 s2

(28)Transformation of Stiffness and Compliance Matrix

Q = T1 Q TT (29)S = T T S T (30)

with

S11 = S11cos4 + (2S12 + S66)sin2cos2 + S22sin4S12 = S12(sin4 + cos4) + (S11 + S22 S66)sin2cos2S22 = S11sin4 + (2S12 + S66)sin2cos2 + S22cos4S16 = (2S11 2S12 S66)sincos3 (2S22 2S12 S66)sin3cosS26 = (2S11 2S12 S66)sin3cos (2S22 2S12 S66)sincos3S66 = 2(2S11 + 2S22 4S12 S66)sin2cos2 + S66(sin4 + cos4)

and

Q11 = Q11cos4 + 2(Q12 + 2Q66)sin2cos2 +Q22sin4Q12 = Q12(sin4 + cos4) + (Q11 +Q22 4Q66)sin2cos2Q22 = Q11sin4 + 2(Q12 + 2Q66)sin2cos2 +Q22cos4Q16 = (Q11 Q12 2Q66)sincos3 + (Q22 +Q12 + 2Q66)sin3cosQ26 = (Q11 Q12 2Q66)sin3cos + (Q22 +Q12 + 2Q66)sincos3Q66 = (Q11 +Q22 2Q12 2Q66)sin2cos2 +Q66(sin4 + cos4)

3.3 ABD-Matrix{nm

}=

[A BB D

]{0

}(31){

0

}=

[A BB D

]1{nm

}=[a bb d

]{nm

}(32)

5

Nx

Ny

Nxy

Mx

My

Mxy

=

A11 A12 A16 B11 B12 B16

A12 A22 A26 B12 B22 B26

A16 A26 A66 B16 B26 B66

B11 B12 B16 D11 D12 D16

B12 B22 B26 D12 D22 D26

B16 B26 B66 D16 D26 D66

0x

0y

0xy

x

y

xy

(33)

A =nk=1

Qk (zk zk1) =nk=1

Qk tk (34)

B =12

nk=1

Qk (z2k z2k1) =nk=1

Qk tk (zk tk2

)(35)

D =13

nk=1

Qk (z3k z3k1) =nk=1

Qk (t3k12 + tk

(zk tk2

)2)(36)

CLT-Engineering constants (E, G, ) for symmetrical and balanced laminates

Ex =1

(A1)11 t , Ey =1

(A1)22 t , (37)

Gxy =1

(A1)66 t , xy =(A1)12(A1)22

, yx =(A1)12(A1)11

. (38)

Calculation of ply deformations

{}k ={0}

+ {} zk (39)

6

4 Composite Failure4.1 Basic UD Lamina StrengthXt = R+|| Fiber tensile strengthXc = R|| Fiber compressive strengthYt = R+ Matrix tensile strengthYc = R Matrix compressive strengthSL = R|| In-plane shear strengthST = R Transverse shear strength

4.2 Max Stress Criterion (2D)

|1|{< Xt for 1 > 0< Xc for 1 < 0

|2|{< Yt for 2 > 0< Yc for 2 < 0

(40)

Stress exposure fE :

fE = max(|1|Xt/c

,|2|Yt/c

,|12|SL

). (41)

4.3 Max Strain Criterion (2D)

|1|{< fr1t for 1 > 0< fr1c for 1 < 0

|2|{< fr2t for 2 > 0< fr2c for 2 < 0

|12| < fr12 (42)

Stress exposure fE :

fE = max

1fr1t/c

,2

fr2t/c,12

fr12

. (43)4.4 Tsai-Wu Criterion (2D)Failure criterion: ( 1

Xt 1Xc

)1 +

( 1Yt 1Yc

)2 +

21Xt Xc

+ 22

Yt Yc 1 2

Xt Xc Yt Yc+

212S2L

= 1.(44)

Stress exposure fE for Tsai-Wu Criterion:

1fE

= 1a

(b

b2 + a

)(45)

7

with

a = 21

|Xt Xc| 1 2

Xt Xc Yt Yc+

22

|Yt Yc| +(12SL

)2, (46)

b = 12

[1

( 1Xt 1Xc

)+ 2

( 1Yt 1Yc

)](47)

4.5 Tsai-Wu Criterion (3D)Failure criterion:( 1Xt 1Xc

)1 +

( 1Yt 1Yc

)(2 + 3) +

21Xt Xc

+ 22 + 23Yt Yc

+ 2( 1Yt Yc

12Yt Yc Zt Zc

)223

+212 + 213S2L

1 2 + 1 3Xt Xc Yt Yc

+ 2 3Yt Yc Zt Zc

= 1.

(48)

Stress exposure fE for Tsai-Wu Criterion:

1fE

= 1a

(b

b2 + a

)(49)

with

a = 21

|XtXc| 1 2 + 1 3Xt Xc Yt Yc

2 3Yt Yc Zt Zc

+ 22

|Yt Yc| + 2( 1Yt Yc

12Yt Yc Zt Zc

)223 +

212 + 213S2L

,

(50)

b = 12

[1

( 1Xt 1Xc

)+ (2 + 3)

( 1Yt 1Yc

)](51)

4.6 Hashin Criterion (2D)

FF tension(1Xt

)2+(12SL

)2= 1 1 0

FF compression (1Xc

)= 1 1 < 0

IFF tension(2Yt

)2+(12SL

)2= 1 2 0

IFF compression(2

2ST

)2+[(

Yc2ST

)2 1

]2Yc

+(12SL

)2= 1 2 < 0.

8

Stress exposure fE :

FF tensionfE =

(1Xt

)2+(12SL

)2(52)

FF compressionfE = 1

Xc(53)

IFF tensionfE =

(2Yt

)2+(12SL

)2(54)

IFF compression1fE

= 1a

(bb2 + a), (55)

with

a =(2

2ST

)2+(12SL

)2, (56)

b = 12

[(Yc

2ST

)2 1

]2Yc. (57)

4.7 Pucks Criterion (2D)4.7.1 Stress exposure fE Fiber tension

fE =1Rt

for 1 > 0 (58)

Fiber compressionfE = 1

Rcfor 1 < 0 (59)

Inter Fiber Failure Mode A (2 0)

fE =

[( 1RtptR

)2

]2+(12R

)2+ptR

2. (60)

9

Inter Fiber Failure Mode B (2 < 0 and 0 212

RA12,c)

fE =

( 12R

)2+( pcR

2

)2+pcR

2. (61)

Inter Fiber Failure Mode C (2 < 0 and 0 212 >

RA12,c)

fE =[(

122(1 + pc)R

)2+(2Rc

)2] Rc2 . (62)

where: 12,c = R

1 + 2pc, RA =Rc

2(1+pc), p

cRA

=pcR

.

Inclination Parameters (Recommendations)

pt pc p

t p

c

GFRP/Epoxy 0.3 0.25 0.2-0.25 0.2-0.25CFRP/Epoxy 0.35 0.3 0.25-0.3 0.25-0.3

4.7.2 Fracture plane angle

cos fp =

12(1 + pc)

[(RA12R2

)2+ 1

]. (63)

4.7.3 Weakening due to 1

fE1 =fE0w1

(64)

w1 =c (a c2 (a2 s2) + 1 + s

(c a)2 + 1 c =fE0

f(FF )E

a = 1 s1m2

s: Section of ellipse start; m: Minimum of weakening factor.

Indices:

w: weakening 0: unweakend 1: weakening due to 1

10

4.8 Puck 3D: Master Fracture Body4.8.1 Stress Transformation

n()nt()n1()

= c2 s2 2cs 0 0sc sc (c2 s2) 0 0

0 0 0 s c

23233121

(65)

c = cos s = sin

n =2nt + 2n1

= arctan n1nt : abitrary angle fp : angle of the fracture plane

4.8.2 General Inter Fiber Failure (IFF)

For n 0 the general fracture condition is defined as:

fe() =

[( 1Rt p

tRA

)n()

]2+(nt()RA

)2+(n1()R

)2+ptRA

n()

For n < 0 the general fracture condition is defined as:

fe() =

(nt()RA

)2+(n1()R

)2+( pcRA

n())2

+pcRA

n()

withpt,cRA

= pt,cRA

cos2 +pt,cR

sin2

cos2 = 2nt

2nt + 2n1and sin2 =

2n1

2nt + 2n1

11

5 Damage and Material Degradation5.1 Continuum MechanicsUndamaged compliance matrix H0 and stiffness matrix C0

H0 =

1E1

21E2 012E1 1E2 0

0 0 1G12

; C0 =

E111221

21E111221 0

21E111221

E211221 0

0 0 G12

; 12E1 = 21E2 (66)

5.2 Continuum Damage MechanicsDamage compliance matrix

H =

1

(1df)E1 21E2

0

12E1 1(1dm)E2 00 0 1(1ds)G12

(67)

Damage stiffness matrix

C = 1D

(1 df )E1 (1 df ) (1 dm) 21E1 0

(1 df ) (1 dm) 12E2 (1 dm)E2 00 0 D (1 ds)G12

(68)

where D = 1 (1 df ) (1 dm) 1221

and ds = 1 (1 dft) (1 dfc) (1 dmt) (1 dmc)

5.3 Energy-based Degradation Laws

G = 211

2 (1 d1)E1 +222

2 (1 d2)E2 12E1

1122 +212

2 (1 d6)G12 (69)

gM =GMcl

=0

G

dMd dt =

1

G

dM dMfME

dfME (70)

Critical element length

l 2GMc EMR2M

= lcrit (71)

12

6 Interlaminar Shear Stress

Transverse shear

= Q S(z)Iy b(z) (72)

Neutral axis

zg =B11A11

(73)

First moment of area

Sy(z) =e1z

b()d (74)

Shear Flux

t(z) = (z) b(z) (75)

with the following variables:

Q shear forceIy second moment of areab(z) width of beam distance from neutral axise1 height

7 Notch Analysis - Stress ConcentrationsThe stress concentration factor

KT = ny (R,0)

ny(76)

with:ny (R,0) stress occuring close to the holeny far field stress

Calculation of KT by Lekhnitskii

13

for an orthotropic material

KT = 1 +

2(E1E2 12

)+ E1G12

(77)

for a laminate

KT = 1 +

2A11

(A11A22 A12 + A11A22 (A12)

2

2A66

)(78)

Hole-Size effect - Correction factor by Peterson

KTKT =

2 +(1 dw

)33(1 dw

) (79)with:d diameter of the holew width of the plate

Strength reduction factor SR:

SR = nN

n0(80)

with:nN Strength of the notched laminate

n0 Strength of the unnotched laminate

8 Stability8.1 Column Buckling

Fcritpi2

4 EIL2 pi2 EIL2 pi2

(0.7)2 EIL2 4pi2 EIL2

14

General equation for critical load

Fcrit =C pi2 EI

L2(81)

with:C := end fixity coefficientEI := bending stiffness

8.2 Plate Buckling8.2.1 All edges supported

Support Conditions Buckling Load Ncr

pi2

b2

[D11 b2 (ma )2 + 2 (D12 + 2D66) +D22 ( 1b )2 ( am )2

]

pi2

b2

[D11 b2 (ma )2 + 2.33 (D12 + 2D66) + 2.441D22 ( 1b )2 ( am )2

]

pi2

b2

[D11 b2 (ma )2 + 2.33 (D12 + 2D66) + 5.139 D22 ( 1b )2 ( am )2

]

8.2.2 One edge free

Support Conditions Buckling Load Ncr

pi2D11a2 +

12D66b2

pi2D11(0.5a)2 +

12D66b2

1.25a2D22m2a2 +

pi2m2D11a2 +

12D66b2

15

8.2.3 Long plate approximation

Support Conditions Requirement Buckling Load Ncr

b > 3 a 4

D22D11

pi2D11a2

a > 3 b 4

D11D22

pi2

b2

[2 D11D22 + 2 (D12 + 2D66)

]

8.3 Plate buckling - "Girlandenkurve"

Ncr = K pi2

b2D11 D22 + C pi

2

b2 (D12 + 2 D66) (82)

8.4 Plate buckling - shear loading

Ns,cr = ks pi2

b2 4D11 D322 (83)

16

9 Sandwich

Chamfer edgecorner,peak =

32nixrc

(84)

17

9.1 Analytical Approach Shear Stress

The constant shear stress in the core can be calculated and compared to the maximumbearable core shear stress:

core =Q

b d core,cr (85)whith the following abbreviation:

Q transverse force d sandwich thickness b width of sandwich panel core,cr Maximum bearable core shear stress core current core shear stress

18

10 Joining

10.1 Mechanical Joints10.1.1 Huths Formula for single lap joint

Cf =(t1 + t2

2db

)a b(

1t1E1

+ 1t2E2

+ 12t1Ef+ 12t2Ef

)(86)

10.1.2 Huths Formula for double lap joint

Cf =(ts + tp

2db

)a b2

(1

tpEp+ 1tsEs

+ 12tpEf+ 12tsEf

)(87)

C : Fastener Flexibility

E : Youngs Modulus

db : fastener diameter

f : fastener

p : center plate

s : side plate

t : lap thickness

a : empirical constant depending on the material of the parts

b : empirical constant depending on the fastener type For bolted metallic parts: a = 23 and b = 3 For riveted metallic parts: a = 25 and b = 2.2 For bolted carbon FRP parts: a = 23 and b = 4.2

ti : thickness of part i

10.1.3 Design guidlines for fastener placemente

d> 3; p

d> 4; w

d> 5; (88)

e : edge distance

p : pitch

w : width

d : fastener diameter

10.1.4 Bearing evaluationFfdb t bear,allow; bear,allow 350MPa; (89)

Ff : fastener load

db : fastener diameter

t : lap thickness

19

10.1.5 Fastener failure - shear

Shear =F

d24 i

Allowable (90)

10.1.6 Fastener failure - bending

Bending = 16 F bd3 Bending,Allowable (91)

For double lap joints:bending moment arm b = t12 +

t24 + , gap distance and diameter d.

10.2 Bonding

Volkersen-Equationbb

= 2

[cosh(x

lb)

sinh(2) (1)sinh(

xlb

)(1 + )cosh(2)

](92)

Fig. 2: Sinus-, Cosinus- and Cotangenshy-perbolicus

with

2 = (1 + ) Gb l2b

E1 t1 tb (93)

= E1 t1E2 t2 (94)

b =F

lb w (95)

and

cosh = ex + ex

2 sinh =ex ex

2 (96)

For a critical joining length lb the following has to apply for the bonding index: 5.

11 Fatigue

11.1 Palmgren-Miner-Rule

Linear accumulation of fatigue damage D with number of load cycles. Failure is assumedto occur, when the critical value of damage (usually 1) is reached:

D =ki=1

niNi

20

with

ni: number of load cycles with constant periodic stress loading

Ni: to ni corresponding total number of load cycles which generate pure fatigue failure

where each ith sequence of ni cycles causes a partial damage of Di. The total damagecan be calculated by adding up the partial damages Di.

12 Liquid Composite Moulding

Fig. 3: Reference geometries for developing rectilinear and radial flow equations.

12.1 Basic equations for 1D Flow

Relationship between Darcy and Flow front velocity:udx = vx = (1 Vf ) vx (97)

Darcys law and Darcys velocity:

udx = Kxx dpdx, udx =

Q

A(98)

Flow front velocity:vx =

dxfdt (99)

21

12.2 Rectlinear filling

Constant pressure injection:

p(x) = pinj(

1 xxf

)(100)

Constant flowrate injection:

p(x) = QAK

(xf x) (101)

Pressure vs. Flowrate equation

pinj =QxfKA

(102)

Constant flow rate:t(x) = A(1 Vf )xf

Q(103)

tfill =A(1 Vf )L

Q(104)

pinj(t) =Q2

K(1 Vf )(A)2 t (105)

Constant pressure:t(x) =

(1 Vf )x2f2pinjK

(106)

tfill =(1 Vf )L2

2pinjK(107)

Q(t) = ApinjK(1 Vf )

2t (108)

12.3 Radial filling

Pressure vs. Flowrate equation

p(r) = Q2pihK ln(rfr

)(109)

Constant flow rate:t(r) =

pi(r2f r2i )h(1 Vf )Q

(110)

22

tfill =pi(r20 r2i )h(1 Vf )

Q(111)

pinj(t) =Q

4pihK

(ln ( Qt(1 Vf )pihr2i

) + 1)

(112)

Constant pressure:

t(rf ) =(1 Vf )pinjK

(r2f2 ln

(rfri

) r

2f r2i

4

)(113)

tfill =(1 Vf )pinjK

(r202 ln

(r0ri

) r

20 r2i

4

)(114)

13 Process Induced Deformations

13.1 Resin Kinetics

Change in degree of conversion:

dX

dt= k Xm (1X)n

with m,n approximation parameters,k kinetics rate constant, 1

k= 1

kA+ 1

kD, where

kA = a exp( ERTC

)Arrhenius dependency,

a const,E Activation Energy,TC Cure Temperature,R Gas const

kD diffusion dependet kinetics rate.

13.2 Spring-In

Analytical calculation of Spring-In due to resin shrinkage and thermal effects:

= thermal + CS = 0 ((xC zC)T

1 + xCT

)+ 0

(xC zC1 + xC

)(115)

23

13.3 Lamina Homogenization

Thermal expansion coefficient for a lamina in fiber direction:

1 =1FE1FF + MEM(1 F )

E1FF + EM(1 F ) (116)

Thermal expansion coefficient for a lamina transversal to the fiber direction:

2 = F 2F + (1 F )M (117)Cure Shrinkage Coefficient in fiber direction:

CSC1 =EM(1 f )

E11ff + EM(1 f ) (118)

Cure Shrinkage Coefficient transversal to fiber direction:

CSC2 = (1 F ) (119)Poisson ratio:

23 = F23,F + (1 F )M (1 + M 21 EME1 )

(1 2M + M21 EME1 )

(120)Poisson ratio (linear):

23 = F 23,F + (1 F ) M (121)

13.4 Laminate Homogenization13.4.1 Thermal expansion coefficients

In-Plane thermal expansion coefficients of the laminate:

C =0

T (122)

with[A]1 {N} = {0} (123)

equivalent loads:

{N} =Nn=1

hn2

hn2

[Q]nnTdz (124)

24

Transformation of thermal coefficients:xyxy

= cos

2 sin2 sin cos sin2 cos2 cos sin

2 sin cos 2 sin cos cos2

1212

(125)Thermal expansion coefficient in thickness direction for a lamina:

3r = T 131E1T

232E2T

(126)

Stresses in each layer:

1 =E1T

1 1221 (1C 1) +21E1T1 1221 (2C 2) (127)

2 =21E1T1 1221 (1C 1) +

E2T1 1221 (2C 2) (128)

Average across laminate thickness:

3C =1h

n

hn [3r]n (129)

13.4.2 Cure Skrinkage coefficients

In-Plane Cure Shrinkage Coefficients of the laminate:

CSCij =0sr

(130)

with sr resin strain due to cure shrinkage and

[A]1 {N} = {0}equivalent loads:

{N} =Nn=1

hn2

hn2

[Q]nCSCn

srdz (131)

(CSC transformation to global coordinate system equivalent to transformation of thermalextension coefficient)Cure shrinkage coefficient in thickness direction for a lamina:

CSC3r = CSC2 131E1sr

232E2sr

(132)

Stresses in each layer:

25

1 =E1

sr

1 1221 (CSC1C CSC1) +21E1

sr

1 1221 (CSC2C CSC2) (133)

2 =21E1

sr

1 1221 (CSC1C CSC1) +E2

sr

1 1221 (CSC2C CSC2) (134)

Average across laminate thickness:

CSC3C =1h

n

hn [CSC3r]n (135)

Linear Cure Shrinkage Strain:

sr = (V sr + 1)13 1 (136)

Effective Cure Shrinkage Strain:

CSiC = 0.2srCSCiC (137)

14 Draping and Braiding

14.1 Kinematic Draping Simulation - Mesh Computation

(xi,j xi1,j)2 + (yi,j yi1,j)2 + (zi,j zi1,j)2 = a2i ,(xi,j xi,j1)2 + (yi,j yi,j1)2 + (zi,j zi,j1)2 = b2i , (138)

F (xi,j,yi,j,zi,j) = 0.

26

14.2 Picture Frame Test

Material shear angle

= pi2 2 cos1( 1

2+ d2LR

) material shear angled head displacementLR side length of the frame

Fig. 4: Picture Frame Test

14.3 Bias Extension Test

Material shear angle = pi2 2 cos

1(D + d

2D

)

Fig. 5: Bias Extension Test

14.4 Braiding

nHG Rotational speed of the horn gearNHG Number of horn gearsv Velocity of the mandrelU Circumference of the mandreld Diameter

27

b Braiding yarn width Braiding angle

Angular velocity of the horn gear:

HG = 2pi nHG. (139) Angular velocity of the carrier:

C = HG 2NHG

. (140)

Cycle duration of the carrier:

TC =2piC

. (141)

Length:

l = v TC . (142) Braiding angle:

= arctan Ul. (143)

Number of yarns:

n = 2pid cosb

. (144)

15 Winding

15.1 Definition of geodesic line15.1.1 Cylinder

15.1.2 Truncated cone

15.2 Compaction pressure15.2.1 Cylinder

28

(z) 0 = Ccr20 C2c

z z0r0

, (145)

whereCc = r sin. (146)

(z) 0 = 1sin 0(

arccos Ccr(z) arccos

Ccr0

). (147)

15.2.2 Truncated cone

15.2.3 Convex toroid section

For r0 = 0:qN(z) =

R

r(z) > 0. (154)

15.2.4 Concave toroid section

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Normal pressure at a cylinder segment:

qN (z) =R

r0sin2 (z) 0, (148)

with a meridional section

r(z) = r0 for < z

Normal pressure at a convex toroid section:

qN (z) =R

r(z)

[1 r0

r(z) sin2 (z)

]> 0, (152)

with a meridional section

r(z) = r0 +r2c (rc z + z0)2 (153)

forz0 z z0 + 2rc.

Normal pressure at a concave toroid section:

qN (z) =R

rc

[r0 + rcr(z) sin

2 (z) 1], (155)

with a meridional section

r(z) = r0 + rc r2c (z z0)2 (156)

forz0 rc z z0 + rc.

Line of constant sliding resistance on a cylinder[r0 CRC0

r0(z z0)

]sin = C0. (162)

(z) 0 = 1CR

{arccosh r0

C0 arccosh

[r0C0 CR(z z0)

r0

]}(163)

withC0 = r0 sin0, CR 6= 0 and C0 6= 0.

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