BRIDGE STRUDLMANUAL November 1~73

APPEIDIX C

HEffi ER PROPER~IES

CONTENTS

Section Page

cl Torsional Constants

I Formulas for Standard Sections C-2

II Formulas for Built Up Sections C-9

III Multi-Celled Sections C-10

a Solution by Method of Sirnultashyneous Equations

b Solution by Method of Successive Corrections

c Solution by Approximate Hethod

d Solution by Finite Element Alalysi s

c 2 Shear Constants - Standard Shapes C-29

c-1

AP PED IX C

gtlE1BER PROPERTIES

Cl Torsional Constants

I Formulas for Standa~d Sections

For the proper interaction between torsional and bending moments in a STRUDL analysis the torsional propshyerties of the ma~ers must be specified The torsional rigidities IX for many standard shapes of members have been documented in many texts and will be included in this appendix for your convenience

TORSIONAL CONSTANTS IX

MxLSection Formulas for IX in 8 =

IXG

General IX = E bH 3I I

IX

IX =

c-2

I

b t3IX =

3

b

I I

b+d)t3r=+I

d IX = 3

H T d t 32 b tj 0

IX =

i IT i

Id

~ I

C-3

3

IX 2btf+ht~ 3

Solid circular

IX = 7T r4 -2-

So I i d e II i pt i c a I

IX

IX = 01406 a 4

Solid square

IX = o rt 3

3

Open circular tube C-4

I IX = ab 3 [ 16 -3 36Q (1- b44 )]3 middot a 12o____----[Jzb

j 2 a bull 1

Solid rectangle

= a43IX 80

Equilateral triangle

c-s

-middot Standard Closed

e =

IX=

Ts =

f =

lI

Ci I t

Section TL

EsiX

4 [AJ2

~~ ts

T 2 [A] t s

T 2A

Formulas

A = area enclosed within mean dimensions

ds = length of particular segment of section

ts = ave ragamp thickness of segment at point ( s)

T5 =shear stress at point ( s)

IX = torsional resistance in 4

E5 = modulus of elasticity in shear

(steel = 12000000

e = angular twist (radians)

L = length of member (inches)

f = unit shear force

f ~ =

stress at i of b

T 2 [A] tb 2

T bd tb

IX middot= 4 b2 d2

b 2d b fb + td + T

IX

c-6

c

IX=

IX =

b

IX 2 t bz d 2 b + d

IX 4 bz dz

b + 2d b+shyt tl

b

X c 4

2o b -t- + Tb

C-7

4rlt21f+2c) IX

2ot-7rr +~ t tr

4 4 77(rz-r 1 )

IX 2

Hollow circle

2 a

IX =

ah = bw ere middot= k = -lshyOz b2 I

Hollow ellipse

IX = L

where Am= approximate area Thin walled tube of arbitary shape within median

L =length of median

C-8

IX

Thin wolfed elliptical tube

2t t (o -t ) 2 (b-t )2I 2 2 IIX =

Rectongulor tube

II Fo~ulas for Built Uo Sections

The problem of finding torsional rigidities for some Bridge Department standard cress-sectional sha~es ~ecomes difficult when standarc fc~ulas de no~ applbullmiddot n the following discussion methods are developed to obtain torsional rigidities for shapes related to Bridge Design

The following assumptions are made in developing the e~uations used

(1) Plane sections =~~ain plane

(2) The material is homogeneous isotropic and linearly elastic

(3) Saint Venants principle applies

~orsional Ricidities

The torsional igidity of an ope1 thin wallec section may be calculated from the following e~uation

E (ef Bl)

i=l

T~~s e~ua~~on is for t~e ge~e=al case a= a section w~th el a-ne ts

C-9

Examoles

bz bt

f tz I 14 bullI ttl I r-~ r tb2

bt ~ ~ b3 ~W1

1-1

~~ b4 bull

The torsional constant for a single thin walled closed section may be obtained from

IX = (Ref Bl)

where n is the total area enclosed by the center line of the walls of the closed section The integral term represents the sum of the various lengths of wall section divided b7 their respective thicknesses

ExamOle

-------- I ~ I I I 1-- I I I ~--------J

I I

4(b1 bz) 2 2 ( b bz) 2 IX =

2bt 2bz bt + bz-+-shyt 1 tz t I t 2

C-10

For a hybrid section of a closed section plus outshystard~ng ins ~~e formula for IX becomes

1 3 (Ref Bl)-+ E middot bi ti3 i=l

Exarnoles

4h-r2)2 4IX shy= bt tl21TL 3

t

IX = 27rr 3 t ~ bt t~3

I Multi Celled Sec~ions

Torsion of two or more cells cornected a= the valls is a statically indete~i~ate probla~ T~e general me=hcd to find the torsional rigidity X (Re Bl) is as follows Ass~~e an n celled closed thin walled section

1------- ---I ( I II

----- ----- II

II I flt II

II [22 II

11 I I I i) i ______

------- ------shybt

_______

c-11

The equation of equilibrium for n cells is

n (1) Mt = 2 t q in i (1

i= I

where qi is the shear flow in cell i and n i is the area inclosec by the center line of the walls inclosing the cell and Mt is the twisting moment applied to the cell

The equations of consistent deformation are

(2) (2)

ds _l dswhere Sji - ~ ss s jk

J I t G ss j k t

ds Ls j j 1G fsjj

G is the shear moduls o= elasticity

dst is the sum of the length of cell wall com--ao~ to cells j and i divided by its thickness

dst is the sum of the length of cell wall common to cells j and k divided by its thicknesses

dst is the sum of the lengths of cell walls common5Smiddot middot J J to cell j divided by their respective thicknesses

is the angle of twist in radianse

Equation (2) will yield n equations for n unknown shear flows and can be solved for the shear flows qi in terms of G and the angle of twist e Knowing qi and Ili the torshysional constant IX may be calcul~ted from

( 3) 2 ( 3)IX = Ge

C-12

The following examples are attached to show four methods of solution that may be used for a box girder section The example section chosen was a standard three celled box girder with sloping erterior girders

a Solution bv Method of Simultaneous Ecuations

The torsional constant IX for a three celled box gi~der may be calculated by the method of simultaneous equashytions which is based upon the following facts

I

1) The summation of external torsional moments and the internal resisting shear flow system must be equal to zero

(2) The angle of twist must be the same for each cell

These facts are ~sed to write one equation for each cell in te~s of the shear flow q for that particular cell The resulting shear flows are then used to calculate the Torsional Constant IX The method used in the following example uses three simultaneous equations to solve for the unknown shear flOIS q

EXlZbullPLE CALCwJJtTICN FOR TORSIONAL CONSTiT IX

Assume Box Girder

6 I I I

i I j 36 92u a- 8~ 9middot 2 ~6 i t-1---=-------~~-----____i

Idealize Box Girder 892 867 892

~~_=fL= ( 619 + 8921 ( 548) = 41 4o sq ftJ l 2 Top slob thickness = 56 ft

112 548)(667) = 4 7 51 sq ft Bottom slob thickness = 48ft

C-13

AP PED IX C

gtlE1BER PROPERTIES

Cl Torsional Constants

I Formulas for Standa~d Sections

For the proper interaction between torsional and bending moments in a STRUDL analysis the torsional propshyerties of the ma~ers must be specified The torsional rigidities IX for many standard shapes of members have been documented in many texts and will be included in this appendix for your convenience

TORSIONAL CONSTANTS IX

MxLSection Formulas for IX in 8 =

IXG

General IX = E bH 3I I

IX

IX =

c-2

I

b t3IX =

3

b

I I

b+d)t3r=+I

d IX = 3

H T d t 32 b tj 0

IX =

i IT i

Id

~ I

C-3

3

IX 2btf+ht~ 3

Solid circular

IX = 7T r4 -2-

So I i d e II i pt i c a I

IX

IX = 01406 a 4

Solid square

IX = o rt 3

3

Open circular tube C-4

I IX = ab 3 [ 16 -3 36Q (1- b44 )]3 middot a 12o____----[Jzb

j 2 a bull 1

Solid rectangle

= a43IX 80

Equilateral triangle

c-s

-middot Standard Closed

e =

IX=

Ts =

f =

lI

Ci I t

Section TL

EsiX

4 [AJ2

~~ ts

T 2 [A] t s

T 2A

Formulas

A = area enclosed within mean dimensions

ds = length of particular segment of section

ts = ave ragamp thickness of segment at point ( s)

T5 =shear stress at point ( s)

IX = torsional resistance in 4

E5 = modulus of elasticity in shear

(steel = 12000000

e = angular twist (radians)

L = length of member (inches)

f = unit shear force

f ~ =

stress at i of b

T 2 [A] tb 2

T bd tb

IX middot= 4 b2 d2

b 2d b fb + td + T

IX

c-6

c

IX=

IX =

b

IX 2 t bz d 2 b + d

IX 4 bz dz

b + 2d b+shyt tl

b

X c 4

2o b -t- + Tb

C-7

4rlt21f+2c) IX

2ot-7rr +~ t tr

4 4 77(rz-r 1 )

IX 2

Hollow circle

2 a

IX =

ah = bw ere middot= k = -lshyOz b2 I

Hollow ellipse

IX = L

where Am= approximate area Thin walled tube of arbitary shape within median

L =length of median

C-8

IX

Thin wolfed elliptical tube

2t t (o -t ) 2 (b-t )2I 2 2 IIX =

Rectongulor tube

II Fo~ulas for Built Uo Sections

The problem of finding torsional rigidities for some Bridge Department standard cress-sectional sha~es ~ecomes difficult when standarc fc~ulas de no~ applbullmiddot n the following discussion methods are developed to obtain torsional rigidities for shapes related to Bridge Design

The following assumptions are made in developing the e~uations used

(1) Plane sections =~~ain plane

(2) The material is homogeneous isotropic and linearly elastic

(3) Saint Venants principle applies

~orsional Ricidities

The torsional igidity of an ope1 thin wallec section may be calculated from the following e~uation

E (ef Bl)

i=l

T~~s e~ua~~on is for t~e ge~e=al case a= a section w~th el a-ne ts

C-9

Examoles

bz bt

f tz I 14 bullI ttl I r-~ r tb2

bt ~ ~ b3 ~W1

1-1

~~ b4 bull

The torsional constant for a single thin walled closed section may be obtained from

IX = (Ref Bl)

where n is the total area enclosed by the center line of the walls of the closed section The integral term represents the sum of the various lengths of wall section divided b7 their respective thicknesses

ExamOle

-------- I ~ I I I 1-- I I I ~--------J

I I

4(b1 bz) 2 2 ( b bz) 2 IX =

2bt 2bz bt + bz-+-shyt 1 tz t I t 2

C-10

For a hybrid section of a closed section plus outshystard~ng ins ~~e formula for IX becomes

1 3 (Ref Bl)-+ E middot bi ti3 i=l

Exarnoles

4h-r2)2 4IX shy= bt tl21TL 3

t

IX = 27rr 3 t ~ bt t~3

I Multi Celled Sec~ions

Torsion of two or more cells cornected a= the valls is a statically indete~i~ate probla~ T~e general me=hcd to find the torsional rigidity X (Re Bl) is as follows Ass~~e an n celled closed thin walled section

1------- ---I ( I II

----- ----- II

II I flt II

II [22 II

11 I I I i) i ______

------- ------shybt

_______

c-11

The equation of equilibrium for n cells is

n (1) Mt = 2 t q in i (1

i= I

where qi is the shear flow in cell i and n i is the area inclosec by the center line of the walls inclosing the cell and Mt is the twisting moment applied to the cell

The equations of consistent deformation are

(2) (2)

ds _l dswhere Sji - ~ ss s jk

J I t G ss j k t

ds Ls j j 1G fsjj

G is the shear moduls o= elasticity

dst is the sum of the length of cell wall com--ao~ to cells j and i divided by its thickness

dst is the sum of the length of cell wall common to cells j and k divided by its thicknesses

dst is the sum of the lengths of cell walls common5Smiddot middot J J to cell j divided by their respective thicknesses

is the angle of twist in radianse

Equation (2) will yield n equations for n unknown shear flows and can be solved for the shear flows qi in terms of G and the angle of twist e Knowing qi and Ili the torshysional constant IX may be calcul~ted from

( 3) 2 ( 3)IX = Ge

C-12

The following examples are attached to show four methods of solution that may be used for a box girder section The example section chosen was a standard three celled box girder with sloping erterior girders

a Solution bv Method of Simultaneous Ecuations

The torsional constant IX for a three celled box gi~der may be calculated by the method of simultaneous equashytions which is based upon the following facts

I

1) The summation of external torsional moments and the internal resisting shear flow system must be equal to zero

(2) The angle of twist must be the same for each cell

These facts are ~sed to write one equation for each cell in te~s of the shear flow q for that particular cell The resulting shear flows are then used to calculate the Torsional Constant IX The method used in the following example uses three simultaneous equations to solve for the unknown shear flOIS q

EXlZbullPLE CALCwJJtTICN FOR TORSIONAL CONSTiT IX

Assume Box Girder

6 I I I

i I j 36 92u a- 8~ 9middot 2 ~6 i t-1---=-------~~-----____i

Idealize Box Girder 892 867 892

~~_=fL= ( 619 + 8921 ( 548) = 41 4o sq ftJ l 2 Top slob thickness = 56 ft

112 548)(667) = 4 7 51 sq ft Bottom slob thickness = 48ft

C-13

I

b t3IX =

3

b

I I

b+d)t3r=+I

d IX = 3

H T d t 32 b tj 0

IX =

i IT i

Id

~ I

C-3

3

IX 2btf+ht~ 3

Solid circular

IX = 7T r4 -2-

So I i d e II i pt i c a I

IX

IX = 01406 a 4

Solid square

IX = o rt 3

3

Open circular tube C-4

I IX = ab 3 [ 16 -3 36Q (1- b44 )]3 middot a 12o____----[Jzb

j 2 a bull 1

Solid rectangle

= a43IX 80

Equilateral triangle

c-s

-middot Standard Closed

e =

IX=

Ts =

f =

lI

Ci I t

Section TL

EsiX

4 [AJ2

~~ ts

T 2 [A] t s

T 2A

Formulas

A = area enclosed within mean dimensions

ds = length of particular segment of section

ts = ave ragamp thickness of segment at point ( s)

T5 =shear stress at point ( s)

IX = torsional resistance in 4

E5 = modulus of elasticity in shear

(steel = 12000000

e = angular twist (radians)

L = length of member (inches)

f = unit shear force

f ~ =

stress at i of b

T 2 [A] tb 2

T bd tb

IX middot= 4 b2 d2

b 2d b fb + td + T

IX

c-6

c

IX=

IX =

b

IX 2 t bz d 2 b + d

IX 4 bz dz

b + 2d b+shyt tl

b

X c 4

2o b -t- + Tb

C-7

4rlt21f+2c) IX

2ot-7rr +~ t tr

4 4 77(rz-r 1 )

IX 2

Hollow circle

2 a

IX =

ah = bw ere middot= k = -lshyOz b2 I

Hollow ellipse

IX = L

where Am= approximate area Thin walled tube of arbitary shape within median

L =length of median

C-8

IX

Thin wolfed elliptical tube

2t t (o -t ) 2 (b-t )2I 2 2 IIX =

Rectongulor tube

II Fo~ulas for Built Uo Sections

The problem of finding torsional rigidities for some Bridge Department standard cress-sectional sha~es ~ecomes difficult when standarc fc~ulas de no~ applbullmiddot n the following discussion methods are developed to obtain torsional rigidities for shapes related to Bridge Design

The following assumptions are made in developing the e~uations used

(1) Plane sections =~~ain plane

(2) The material is homogeneous isotropic and linearly elastic

(3) Saint Venants principle applies

~orsional Ricidities

The torsional igidity of an ope1 thin wallec section may be calculated from the following e~uation

E (ef Bl)

i=l

T~~s e~ua~~on is for t~e ge~e=al case a= a section w~th el a-ne ts

C-9

Examoles

bz bt

f tz I 14 bullI ttl I r-~ r tb2

bt ~ ~ b3 ~W1

1-1

~~ b4 bull

The torsional constant for a single thin walled closed section may be obtained from

IX = (Ref Bl)

where n is the total area enclosed by the center line of the walls of the closed section The integral term represents the sum of the various lengths of wall section divided b7 their respective thicknesses

ExamOle

-------- I ~ I I I 1-- I I I ~--------J

I I

4(b1 bz) 2 2 ( b bz) 2 IX =

2bt 2bz bt + bz-+-shyt 1 tz t I t 2

C-10

For a hybrid section of a closed section plus outshystard~ng ins ~~e formula for IX becomes

1 3 (Ref Bl)-+ E middot bi ti3 i=l

Exarnoles

4h-r2)2 4IX shy= bt tl21TL 3

t

IX = 27rr 3 t ~ bt t~3

I Multi Celled Sec~ions

Torsion of two or more cells cornected a= the valls is a statically indete~i~ate probla~ T~e general me=hcd to find the torsional rigidity X (Re Bl) is as follows Ass~~e an n celled closed thin walled section

1------- ---I ( I II

----- ----- II

II I flt II

II [22 II

11 I I I i) i ______

------- ------shybt

_______

c-11

The equation of equilibrium for n cells is

n (1) Mt = 2 t q in i (1

i= I

where qi is the shear flow in cell i and n i is the area inclosec by the center line of the walls inclosing the cell and Mt is the twisting moment applied to the cell

The equations of consistent deformation are

(2) (2)

ds _l dswhere Sji - ~ ss s jk

J I t G ss j k t

ds Ls j j 1G fsjj

G is the shear moduls o= elasticity

dst is the sum of the length of cell wall com--ao~ to cells j and i divided by its thickness

dst is the sum of the length of cell wall common to cells j and k divided by its thicknesses

dst is the sum of the lengths of cell walls common5Smiddot middot J J to cell j divided by their respective thicknesses

is the angle of twist in radianse

Equation (2) will yield n equations for n unknown shear flows and can be solved for the shear flows qi in terms of G and the angle of twist e Knowing qi and Ili the torshysional constant IX may be calcul~ted from

( 3) 2 ( 3)IX = Ge

C-12

The following examples are attached to show four methods of solution that may be used for a box girder section The example section chosen was a standard three celled box girder with sloping erterior girders

a Solution bv Method of Simultaneous Ecuations

The torsional constant IX for a three celled box gi~der may be calculated by the method of simultaneous equashytions which is based upon the following facts

I

1) The summation of external torsional moments and the internal resisting shear flow system must be equal to zero

(2) The angle of twist must be the same for each cell

These facts are ~sed to write one equation for each cell in te~s of the shear flow q for that particular cell The resulting shear flows are then used to calculate the Torsional Constant IX The method used in the following example uses three simultaneous equations to solve for the unknown shear flOIS q

EXlZbullPLE CALCwJJtTICN FOR TORSIONAL CONSTiT IX

Assume Box Girder

6 I I I

i I j 36 92u a- 8~ 9middot 2 ~6 i t-1---=-------~~-----____i

Idealize Box Girder 892 867 892

~~_=fL= ( 619 + 8921 ( 548) = 41 4o sq ftJ l 2 Top slob thickness = 56 ft

112 548)(667) = 4 7 51 sq ft Bottom slob thickness = 48ft

C-13

IX 2btf+ht~ 3

Solid circular

IX = 7T r4 -2-

So I i d e II i pt i c a I

IX

IX = 01406 a 4

Solid square

IX = o rt 3

3

Open circular tube C-4

I IX = ab 3 [ 16 -3 36Q (1- b44 )]3 middot a 12o____----[Jzb

j 2 a bull 1

Solid rectangle

= a43IX 80

Equilateral triangle

c-s

-middot Standard Closed

e =

IX=

Ts =

f =

lI

Ci I t

Section TL

EsiX

4 [AJ2

~~ ts

T 2 [A] t s

T 2A

Formulas

A = area enclosed within mean dimensions

ds = length of particular segment of section

ts = ave ragamp thickness of segment at point ( s)

T5 =shear stress at point ( s)

IX = torsional resistance in 4

E5 = modulus of elasticity in shear

(steel = 12000000

e = angular twist (radians)

L = length of member (inches)

f = unit shear force

f ~ =

stress at i of b

T 2 [A] tb 2

T bd tb

IX middot= 4 b2 d2

b 2d b fb + td + T

IX

c-6

c

IX=

IX =

b

IX 2 t bz d 2 b + d

IX 4 bz dz

b + 2d b+shyt tl

b

X c 4

2o b -t- + Tb

C-7

4rlt21f+2c) IX

2ot-7rr +~ t tr

4 4 77(rz-r 1 )

IX 2

Hollow circle

2 a

IX =

ah = bw ere middot= k = -lshyOz b2 I

Hollow ellipse

IX = L

where Am= approximate area Thin walled tube of arbitary shape within median

L =length of median

C-8

IX

Thin wolfed elliptical tube

2t t (o -t ) 2 (b-t )2I 2 2 IIX =

Rectongulor tube

II Fo~ulas for Built Uo Sections

The problem of finding torsional rigidities for some Bridge Department standard cress-sectional sha~es ~ecomes difficult when standarc fc~ulas de no~ applbullmiddot n the following discussion methods are developed to obtain torsional rigidities for shapes related to Bridge Design

The following assumptions are made in developing the e~uations used

(1) Plane sections =~~ain plane

(2) The material is homogeneous isotropic and linearly elastic

(3) Saint Venants principle applies

~orsional Ricidities

The torsional igidity of an ope1 thin wallec section may be calculated from the following e~uation

E (ef Bl)

i=l

T~~s e~ua~~on is for t~e ge~e=al case a= a section w~th el a-ne ts

C-9

Examoles

bz bt

f tz I 14 bullI ttl I r-~ r tb2

bt ~ ~ b3 ~W1

1-1

~~ b4 bull

The torsional constant for a single thin walled closed section may be obtained from

IX = (Ref Bl)

where n is the total area enclosed by the center line of the walls of the closed section The integral term represents the sum of the various lengths of wall section divided b7 their respective thicknesses

ExamOle

-------- I ~ I I I 1-- I I I ~--------J

I I

4(b1 bz) 2 2 ( b bz) 2 IX =

2bt 2bz bt + bz-+-shyt 1 tz t I t 2

C-10

For a hybrid section of a closed section plus outshystard~ng ins ~~e formula for IX becomes

1 3 (Ref Bl)-+ E middot bi ti3 i=l

Exarnoles

4h-r2)2 4IX shy= bt tl21TL 3

t

IX = 27rr 3 t ~ bt t~3

I Multi Celled Sec~ions

Torsion of two or more cells cornected a= the valls is a statically indete~i~ate probla~ T~e general me=hcd to find the torsional rigidity X (Re Bl) is as follows Ass~~e an n celled closed thin walled section

1------- ---I ( I II

----- ----- II

II I flt II

II [22 II

11 I I I i) i ______

------- ------shybt

_______

c-11

The equation of equilibrium for n cells is

n (1) Mt = 2 t q in i (1

i= I

where qi is the shear flow in cell i and n i is the area inclosec by the center line of the walls inclosing the cell and Mt is the twisting moment applied to the cell

The equations of consistent deformation are

(2) (2)

ds _l dswhere Sji - ~ ss s jk

J I t G ss j k t

ds Ls j j 1G fsjj

G is the shear moduls o= elasticity

dst is the sum of the length of cell wall com--ao~ to cells j and i divided by its thickness

dst is the sum of the length of cell wall common to cells j and k divided by its thicknesses

dst is the sum of the lengths of cell walls common5Smiddot middot J J to cell j divided by their respective thicknesses

is the angle of twist in radianse

Equation (2) will yield n equations for n unknown shear flows and can be solved for the shear flows qi in terms of G and the angle of twist e Knowing qi and Ili the torshysional constant IX may be calcul~ted from

( 3) 2 ( 3)IX = Ge

C-12

The following examples are attached to show four methods of solution that may be used for a box girder section The example section chosen was a standard three celled box girder with sloping erterior girders

a Solution bv Method of Simultaneous Ecuations

The torsional constant IX for a three celled box gi~der may be calculated by the method of simultaneous equashytions which is based upon the following facts

I

1) The summation of external torsional moments and the internal resisting shear flow system must be equal to zero

(2) The angle of twist must be the same for each cell

These facts are ~sed to write one equation for each cell in te~s of the shear flow q for that particular cell The resulting shear flows are then used to calculate the Torsional Constant IX The method used in the following example uses three simultaneous equations to solve for the unknown shear flOIS q

EXlZbullPLE CALCwJJtTICN FOR TORSIONAL CONSTiT IX

Assume Box Girder

6 I I I

i I j 36 92u a- 8~ 9middot 2 ~6 i t-1---=-------~~-----____i

Idealize Box Girder 892 867 892

~~_=fL= ( 619 + 8921 ( 548) = 41 4o sq ftJ l 2 Top slob thickness = 56 ft

112 548)(667) = 4 7 51 sq ft Bottom slob thickness = 48ft

C-13

I IX = ab 3 [ 16 -3 36Q (1- b44 )]3 middot a 12o____----[Jzb

j 2 a bull 1

Solid rectangle

= a43IX 80

Equilateral triangle

c-s

-middot Standard Closed

e =

IX=

Ts =

f =

lI

Ci I t

Section TL

EsiX

4 [AJ2

~~ ts

T 2 [A] t s

T 2A

Formulas

A = area enclosed within mean dimensions

ds = length of particular segment of section

ts = ave ragamp thickness of segment at point ( s)

T5 =shear stress at point ( s)

IX = torsional resistance in 4

E5 = modulus of elasticity in shear

(steel = 12000000

e = angular twist (radians)

L = length of member (inches)

f = unit shear force

f ~ =

stress at i of b

T 2 [A] tb 2

T bd tb

IX middot= 4 b2 d2

b 2d b fb + td + T

IX

c-6

c

IX=

IX =

b

IX 2 t bz d 2 b + d

IX 4 bz dz

b + 2d b+shyt tl

b

X c 4

2o b -t- + Tb

C-7

4rlt21f+2c) IX

2ot-7rr +~ t tr

4 4 77(rz-r 1 )

IX 2

Hollow circle

2 a

IX =

ah = bw ere middot= k = -lshyOz b2 I

Hollow ellipse

IX = L

where Am= approximate area Thin walled tube of arbitary shape within median

L =length of median

C-8

IX

Thin wolfed elliptical tube

2t t (o -t ) 2 (b-t )2I 2 2 IIX =

Rectongulor tube

II Fo~ulas for Built Uo Sections

The problem of finding torsional rigidities for some Bridge Department standard cress-sectional sha~es ~ecomes difficult when standarc fc~ulas de no~ applbullmiddot n the following discussion methods are developed to obtain torsional rigidities for shapes related to Bridge Design

The following assumptions are made in developing the e~uations used

(1) Plane sections =~~ain plane

(2) The material is homogeneous isotropic and linearly elastic

(3) Saint Venants principle applies

~orsional Ricidities

The torsional igidity of an ope1 thin wallec section may be calculated from the following e~uation

E (ef Bl)

i=l

T~~s e~ua~~on is for t~e ge~e=al case a= a section w~th el a-ne ts

C-9

Examoles

bz bt

f tz I 14 bullI ttl I r-~ r tb2

bt ~ ~ b3 ~W1

1-1

~~ b4 bull

The torsional constant for a single thin walled closed section may be obtained from

IX = (Ref Bl)

where n is the total area enclosed by the center line of the walls of the closed section The integral term represents the sum of the various lengths of wall section divided b7 their respective thicknesses

ExamOle

-------- I ~ I I I 1-- I I I ~--------J

I I

4(b1 bz) 2 2 ( b bz) 2 IX =

2bt 2bz bt + bz-+-shyt 1 tz t I t 2

C-10

For a hybrid section of a closed section plus outshystard~ng ins ~~e formula for IX becomes

1 3 (Ref Bl)-+ E middot bi ti3 i=l

Exarnoles

4h-r2)2 4IX shy= bt tl21TL 3

t

IX = 27rr 3 t ~ bt t~3

I Multi Celled Sec~ions

Torsion of two or more cells cornected a= the valls is a statically indete~i~ate probla~ T~e general me=hcd to find the torsional rigidity X (Re Bl) is as follows Ass~~e an n celled closed thin walled section

1------- ---I ( I II

----- ----- II

II I flt II

II [22 II

11 I I I i) i ______

------- ------shybt

_______

c-11

The equation of equilibrium for n cells is

n (1) Mt = 2 t q in i (1

i= I

where qi is the shear flow in cell i and n i is the area inclosec by the center line of the walls inclosing the cell and Mt is the twisting moment applied to the cell

The equations of consistent deformation are

(2) (2)

ds _l dswhere Sji - ~ ss s jk

J I t G ss j k t

ds Ls j j 1G fsjj

G is the shear moduls o= elasticity

dst is the sum of the length of cell wall com--ao~ to cells j and i divided by its thickness

dst is the sum of the length of cell wall common to cells j and k divided by its thicknesses

dst is the sum of the lengths of cell walls common5Smiddot middot J J to cell j divided by their respective thicknesses

is the angle of twist in radianse

Equation (2) will yield n equations for n unknown shear flows and can be solved for the shear flows qi in terms of G and the angle of twist e Knowing qi and Ili the torshysional constant IX may be calcul~ted from

( 3) 2 ( 3)IX = Ge

C-12

The following examples are attached to show four methods of solution that may be used for a box girder section The example section chosen was a standard three celled box girder with sloping erterior girders

a Solution bv Method of Simultaneous Ecuations

The torsional constant IX for a three celled box gi~der may be calculated by the method of simultaneous equashytions which is based upon the following facts

I

1) The summation of external torsional moments and the internal resisting shear flow system must be equal to zero

(2) The angle of twist must be the same for each cell

These facts are ~sed to write one equation for each cell in te~s of the shear flow q for that particular cell The resulting shear flows are then used to calculate the Torsional Constant IX The method used in the following example uses three simultaneous equations to solve for the unknown shear flOIS q

EXlZbullPLE CALCwJJtTICN FOR TORSIONAL CONSTiT IX

Assume Box Girder

6 I I I

i I j 36 92u a- 8~ 9middot 2 ~6 i t-1---=-------~~-----____i

Idealize Box Girder 892 867 892

~~_=fL= ( 619 + 8921 ( 548) = 41 4o sq ftJ l 2 Top slob thickness = 56 ft

112 548)(667) = 4 7 51 sq ft Bottom slob thickness = 48ft

C-13

-middot Standard Closed

e =

IX=

Ts =

f =

lI

Ci I t

Section TL

EsiX

4 [AJ2

~~ ts

T 2 [A] t s

T 2A

Formulas

A = area enclosed within mean dimensions

ds = length of particular segment of section

ts = ave ragamp thickness of segment at point ( s)

T5 =shear stress at point ( s)

IX = torsional resistance in 4

E5 = modulus of elasticity in shear

(steel = 12000000

e = angular twist (radians)

L = length of member (inches)

f = unit shear force

f ~ =

stress at i of b

T 2 [A] tb 2

T bd tb

IX middot= 4 b2 d2

b 2d b fb + td + T

IX

c-6

c

IX=

IX =

b

IX 2 t bz d 2 b + d

IX 4 bz dz

b + 2d b+shyt tl

b

X c 4

2o b -t- + Tb

C-7

4rlt21f+2c) IX

2ot-7rr +~ t tr

4 4 77(rz-r 1 )

IX 2

Hollow circle

2 a

IX =

ah = bw ere middot= k = -lshyOz b2 I

Hollow ellipse

IX = L

where Am= approximate area Thin walled tube of arbitary shape within median

L =length of median

C-8

IX

Thin wolfed elliptical tube

2t t (o -t ) 2 (b-t )2I 2 2 IIX =

Rectongulor tube

II Fo~ulas for Built Uo Sections

The problem of finding torsional rigidities for some Bridge Department standard cress-sectional sha~es ~ecomes difficult when standarc fc~ulas de no~ applbullmiddot n the following discussion methods are developed to obtain torsional rigidities for shapes related to Bridge Design

The following assumptions are made in developing the e~uations used

(1) Plane sections =~~ain plane

(2) The material is homogeneous isotropic and linearly elastic

(3) Saint Venants principle applies

~orsional Ricidities

The torsional igidity of an ope1 thin wallec section may be calculated from the following e~uation

E (ef Bl)

i=l

T~~s e~ua~~on is for t~e ge~e=al case a= a section w~th el a-ne ts

C-9

Examoles

bz bt

f tz I 14 bullI ttl I r-~ r tb2

bt ~ ~ b3 ~W1

1-1

~~ b4 bull

The torsional constant for a single thin walled closed section may be obtained from

IX = (Ref Bl)

where n is the total area enclosed by the center line of the walls of the closed section The integral term represents the sum of the various lengths of wall section divided b7 their respective thicknesses

ExamOle

-------- I ~ I I I 1-- I I I ~--------J

I I

4(b1 bz) 2 2 ( b bz) 2 IX =

2bt 2bz bt + bz-+-shyt 1 tz t I t 2

C-10

For a hybrid section of a closed section plus outshystard~ng ins ~~e formula for IX becomes

1 3 (Ref Bl)-+ E middot bi ti3 i=l

Exarnoles

4h-r2)2 4IX shy= bt tl21TL 3

t

IX = 27rr 3 t ~ bt t~3

I Multi Celled Sec~ions

Torsion of two or more cells cornected a= the valls is a statically indete~i~ate probla~ T~e general me=hcd to find the torsional rigidity X (Re Bl) is as follows Ass~~e an n celled closed thin walled section

1------- ---I ( I II

----- ----- II

II I flt II

II [22 II

11 I I I i) i ______

------- ------shybt

_______

c-11

The equation of equilibrium for n cells is

n (1) Mt = 2 t q in i (1

i= I

where qi is the shear flow in cell i and n i is the area inclosec by the center line of the walls inclosing the cell and Mt is the twisting moment applied to the cell

The equations of consistent deformation are

(2) (2)

ds _l dswhere Sji - ~ ss s jk

J I t G ss j k t

ds Ls j j 1G fsjj

G is the shear moduls o= elasticity

dst is the sum of the length of cell wall com--ao~ to cells j and i divided by its thickness

dst is the sum of the length of cell wall common to cells j and k divided by its thicknesses

dst is the sum of the lengths of cell walls common5Smiddot middot J J to cell j divided by their respective thicknesses

is the angle of twist in radianse

Equation (2) will yield n equations for n unknown shear flows and can be solved for the shear flows qi in terms of G and the angle of twist e Knowing qi and Ili the torshysional constant IX may be calcul~ted from

( 3) 2 ( 3)IX = Ge

C-12

The following examples are attached to show four methods of solution that may be used for a box girder section The example section chosen was a standard three celled box girder with sloping erterior girders

a Solution bv Method of Simultaneous Ecuations

The torsional constant IX for a three celled box gi~der may be calculated by the method of simultaneous equashytions which is based upon the following facts

I

1) The summation of external torsional moments and the internal resisting shear flow system must be equal to zero

(2) The angle of twist must be the same for each cell

These facts are ~sed to write one equation for each cell in te~s of the shear flow q for that particular cell The resulting shear flows are then used to calculate the Torsional Constant IX The method used in the following example uses three simultaneous equations to solve for the unknown shear flOIS q

EXlZbullPLE CALCwJJtTICN FOR TORSIONAL CONSTiT IX

Assume Box Girder

6 I I I

i I j 36 92u a- 8~ 9middot 2 ~6 i t-1---=-------~~-----____i

Idealize Box Girder 892 867 892

~~_=fL= ( 619 + 8921 ( 548) = 41 4o sq ftJ l 2 Top slob thickness = 56 ft

112 548)(667) = 4 7 51 sq ft Bottom slob thickness = 48ft

C-13

c

IX=

IX =

b

IX 2 t bz d 2 b + d

IX 4 bz dz

b + 2d b+shyt tl

b

X c 4

2o b -t- + Tb

C-7

4rlt21f+2c) IX

2ot-7rr +~ t tr

4 4 77(rz-r 1 )

IX 2

Hollow circle

2 a

IX =

ah = bw ere middot= k = -lshyOz b2 I

Hollow ellipse

IX = L

where Am= approximate area Thin walled tube of arbitary shape within median

L =length of median

C-8

IX

Thin wolfed elliptical tube

2t t (o -t ) 2 (b-t )2I 2 2 IIX =

Rectongulor tube

II Fo~ulas for Built Uo Sections

The problem of finding torsional rigidities for some Bridge Department standard cress-sectional sha~es ~ecomes difficult when standarc fc~ulas de no~ applbullmiddot n the following discussion methods are developed to obtain torsional rigidities for shapes related to Bridge Design

The following assumptions are made in developing the e~uations used

(1) Plane sections =~~ain plane

(2) The material is homogeneous isotropic and linearly elastic

(3) Saint Venants principle applies

~orsional Ricidities

The torsional igidity of an ope1 thin wallec section may be calculated from the following e~uation

E (ef Bl)

i=l

T~~s e~ua~~on is for t~e ge~e=al case a= a section w~th el a-ne ts

C-9

Examoles

bz bt

f tz I 14 bullI ttl I r-~ r tb2

bt ~ ~ b3 ~W1

1-1

~~ b4 bull

The torsional constant for a single thin walled closed section may be obtained from

IX = (Ref Bl)

where n is the total area enclosed by the center line of the walls of the closed section The integral term represents the sum of the various lengths of wall section divided b7 their respective thicknesses

ExamOle

-------- I ~ I I I 1-- I I I ~--------J

I I

4(b1 bz) 2 2 ( b bz) 2 IX =

2bt 2bz bt + bz-+-shyt 1 tz t I t 2

C-10

For a hybrid section of a closed section plus outshystard~ng ins ~~e formula for IX becomes

1 3 (Ref Bl)-+ E middot bi ti3 i=l

Exarnoles

4h-r2)2 4IX shy= bt tl21TL 3

t

IX = 27rr 3 t ~ bt t~3

I Multi Celled Sec~ions

Torsion of two or more cells cornected a= the valls is a statically indete~i~ate probla~ T~e general me=hcd to find the torsional rigidity X (Re Bl) is as follows Ass~~e an n celled closed thin walled section

1------- ---I ( I II

----- ----- II

II I flt II

II [22 II

11 I I I i) i ______

------- ------shybt

_______

c-11

The equation of equilibrium for n cells is

n (1) Mt = 2 t q in i (1

i= I

where qi is the shear flow in cell i and n i is the area inclosec by the center line of the walls inclosing the cell and Mt is the twisting moment applied to the cell

The equations of consistent deformation are

(2) (2)

ds _l dswhere Sji - ~ ss s jk

J I t G ss j k t

ds Ls j j 1G fsjj

G is the shear moduls o= elasticity

dst is the sum of the length of cell wall com--ao~ to cells j and i divided by its thickness

dst is the sum of the length of cell wall common to cells j and k divided by its thicknesses

dst is the sum of the lengths of cell walls common5Smiddot middot J J to cell j divided by their respective thicknesses

is the angle of twist in radianse

Equation (2) will yield n equations for n unknown shear flows and can be solved for the shear flows qi in terms of G and the angle of twist e Knowing qi and Ili the torshysional constant IX may be calcul~ted from

( 3) 2 ( 3)IX = Ge

C-12

The following examples are attached to show four methods of solution that may be used for a box girder section The example section chosen was a standard three celled box girder with sloping erterior girders

a Solution bv Method of Simultaneous Ecuations

The torsional constant IX for a three celled box gi~der may be calculated by the method of simultaneous equashytions which is based upon the following facts

I

1) The summation of external torsional moments and the internal resisting shear flow system must be equal to zero

(2) The angle of twist must be the same for each cell

These facts are ~sed to write one equation for each cell in te~s of the shear flow q for that particular cell The resulting shear flows are then used to calculate the Torsional Constant IX The method used in the following example uses three simultaneous equations to solve for the unknown shear flOIS q

EXlZbullPLE CALCwJJtTICN FOR TORSIONAL CONSTiT IX

Assume Box Girder

6 I I I

i I j 36 92u a- 8~ 9middot 2 ~6 i t-1---=-------~~-----____i

Idealize Box Girder 892 867 892

~~_=fL= ( 619 + 8921 ( 548) = 41 4o sq ftJ l 2 Top slob thickness = 56 ft

112 548)(667) = 4 7 51 sq ft Bottom slob thickness = 48ft

C-13

4rlt21f+2c) IX

2ot-7rr +~ t tr

4 4 77(rz-r 1 )

IX 2

Hollow circle

2 a

IX =

ah = bw ere middot= k = -lshyOz b2 I

Hollow ellipse

IX = L

where Am= approximate area Thin walled tube of arbitary shape within median

L =length of median

C-8

IX

Thin wolfed elliptical tube

2t t (o -t ) 2 (b-t )2I 2 2 IIX =

Rectongulor tube

II Fo~ulas for Built Uo Sections

The problem of finding torsional rigidities for some Bridge Department standard cress-sectional sha~es ~ecomes difficult when standarc fc~ulas de no~ applbullmiddot n the following discussion methods are developed to obtain torsional rigidities for shapes related to Bridge Design

The following assumptions are made in developing the e~uations used

(1) Plane sections =~~ain plane

(2) The material is homogeneous isotropic and linearly elastic

(3) Saint Venants principle applies

~orsional Ricidities

The torsional igidity of an ope1 thin wallec section may be calculated from the following e~uation

E (ef Bl)

i=l

T~~s e~ua~~on is for t~e ge~e=al case a= a section w~th el a-ne ts

C-9

Examoles

bz bt

f tz I 14 bullI ttl I r-~ r tb2

bt ~ ~ b3 ~W1

1-1

~~ b4 bull

The torsional constant for a single thin walled closed section may be obtained from

IX = (Ref Bl)

where n is the total area enclosed by the center line of the walls of the closed section The integral term represents the sum of the various lengths of wall section divided b7 their respective thicknesses

ExamOle

-------- I ~ I I I 1-- I I I ~--------J

I I

4(b1 bz) 2 2 ( b bz) 2 IX =

2bt 2bz bt + bz-+-shyt 1 tz t I t 2

C-10

For a hybrid section of a closed section plus outshystard~ng ins ~~e formula for IX becomes

1 3 (Ref Bl)-+ E middot bi ti3 i=l

Exarnoles

4h-r2)2 4IX shy= bt tl21TL 3

t

IX = 27rr 3 t ~ bt t~3

I Multi Celled Sec~ions

Torsion of two or more cells cornected a= the valls is a statically indete~i~ate probla~ T~e general me=hcd to find the torsional rigidity X (Re Bl) is as follows Ass~~e an n celled closed thin walled section

1------- ---I ( I II

----- ----- II

II I flt II

II [22 II

11 I I I i) i ______

------- ------shybt

_______

c-11

The equation of equilibrium for n cells is

n (1) Mt = 2 t q in i (1

i= I

where qi is the shear flow in cell i and n i is the area inclosec by the center line of the walls inclosing the cell and Mt is the twisting moment applied to the cell

The equations of consistent deformation are

(2) (2)

ds _l dswhere Sji - ~ ss s jk

J I t G ss j k t

ds Ls j j 1G fsjj

G is the shear moduls o= elasticity

dst is the sum of the length of cell wall com--ao~ to cells j and i divided by its thickness

dst is the sum of the length of cell wall common to cells j and k divided by its thicknesses

dst is the sum of the lengths of cell walls common5Smiddot middot J J to cell j divided by their respective thicknesses

is the angle of twist in radianse

Equation (2) will yield n equations for n unknown shear flows and can be solved for the shear flows qi in terms of G and the angle of twist e Knowing qi and Ili the torshysional constant IX may be calcul~ted from

( 3) 2 ( 3)IX = Ge

C-12

The following examples are attached to show four methods of solution that may be used for a box girder section The example section chosen was a standard three celled box girder with sloping erterior girders

a Solution bv Method of Simultaneous Ecuations

The torsional constant IX for a three celled box gi~der may be calculated by the method of simultaneous equashytions which is based upon the following facts

I

1) The summation of external torsional moments and the internal resisting shear flow system must be equal to zero

(2) The angle of twist must be the same for each cell

These facts are ~sed to write one equation for each cell in te~s of the shear flow q for that particular cell The resulting shear flows are then used to calculate the Torsional Constant IX The method used in the following example uses three simultaneous equations to solve for the unknown shear flOIS q

EXlZbullPLE CALCwJJtTICN FOR TORSIONAL CONSTiT IX

Assume Box Girder

6 I I I

i I j 36 92u a- 8~ 9middot 2 ~6 i t-1---=-------~~-----____i

Idealize Box Girder 892 867 892

~~_=fL= ( 619 + 8921 ( 548) = 41 4o sq ftJ l 2 Top slob thickness = 56 ft

112 548)(667) = 4 7 51 sq ft Bottom slob thickness = 48ft

C-13

IX

Thin wolfed elliptical tube

2t t (o -t ) 2 (b-t )2I 2 2 IIX =

Rectongulor tube

II Fo~ulas for Built Uo Sections

The problem of finding torsional rigidities for some Bridge Department standard cress-sectional sha~es ~ecomes difficult when standarc fc~ulas de no~ applbullmiddot n the following discussion methods are developed to obtain torsional rigidities for shapes related to Bridge Design

The following assumptions are made in developing the e~uations used

(1) Plane sections =~~ain plane

(2) The material is homogeneous isotropic and linearly elastic

(3) Saint Venants principle applies

~orsional Ricidities

The torsional igidity of an ope1 thin wallec section may be calculated from the following e~uation

E (ef Bl)

i=l

T~~s e~ua~~on is for t~e ge~e=al case a= a section w~th el a-ne ts

C-9

Examoles

bz bt

f tz I 14 bullI ttl I r-~ r tb2

bt ~ ~ b3 ~W1

1-1

~~ b4 bull

The torsional constant for a single thin walled closed section may be obtained from

IX = (Ref Bl)

where n is the total area enclosed by the center line of the walls of the closed section The integral term represents the sum of the various lengths of wall section divided b7 their respective thicknesses

ExamOle

-------- I ~ I I I 1-- I I I ~--------J

I I

4(b1 bz) 2 2 ( b bz) 2 IX =

2bt 2bz bt + bz-+-shyt 1 tz t I t 2

C-10

For a hybrid section of a closed section plus outshystard~ng ins ~~e formula for IX becomes

1 3 (Ref Bl)-+ E middot bi ti3 i=l

Exarnoles

4h-r2)2 4IX shy= bt tl21TL 3

t

IX = 27rr 3 t ~ bt t~3

I Multi Celled Sec~ions

Torsion of two or more cells cornected a= the valls is a statically indete~i~ate probla~ T~e general me=hcd to find the torsional rigidity X (Re Bl) is as follows Ass~~e an n celled closed thin walled section

1------- ---I ( I II

----- ----- II

II I flt II

II [22 II

11 I I I i) i ______

------- ------shybt

_______

c-11

The equation of equilibrium for n cells is

n (1) Mt = 2 t q in i (1

i= I

where qi is the shear flow in cell i and n i is the area inclosec by the center line of the walls inclosing the cell and Mt is the twisting moment applied to the cell

The equations of consistent deformation are

(2) (2)

ds _l dswhere Sji - ~ ss s jk

J I t G ss j k t

ds Ls j j 1G fsjj

G is the shear moduls o= elasticity

dst is the sum of the length of cell wall com--ao~ to cells j and i divided by its thickness

dst is the sum of the length of cell wall common to cells j and k divided by its thicknesses

dst is the sum of the lengths of cell walls common5Smiddot middot J J to cell j divided by their respective thicknesses

is the angle of twist in radianse

Equation (2) will yield n equations for n unknown shear flows and can be solved for the shear flows qi in terms of G and the angle of twist e Knowing qi and Ili the torshysional constant IX may be calcul~ted from

( 3) 2 ( 3)IX = Ge

C-12

The following examples are attached to show four methods of solution that may be used for a box girder section The example section chosen was a standard three celled box girder with sloping erterior girders

a Solution bv Method of Simultaneous Ecuations

The torsional constant IX for a three celled box gi~der may be calculated by the method of simultaneous equashytions which is based upon the following facts

I

1) The summation of external torsional moments and the internal resisting shear flow system must be equal to zero

(2) The angle of twist must be the same for each cell

These facts are ~sed to write one equation for each cell in te~s of the shear flow q for that particular cell The resulting shear flows are then used to calculate the Torsional Constant IX The method used in the following example uses three simultaneous equations to solve for the unknown shear flOIS q

EXlZbullPLE CALCwJJtTICN FOR TORSIONAL CONSTiT IX

Assume Box Girder

6 I I I

i I j 36 92u a- 8~ 9middot 2 ~6 i t-1---=-------~~-----____i

Idealize Box Girder 892 867 892

~~_=fL= ( 619 + 8921 ( 548) = 41 4o sq ftJ l 2 Top slob thickness = 56 ft

112 548)(667) = 4 7 51 sq ft Bottom slob thickness = 48ft

C-13

Examoles

bz bt

f tz I 14 bullI ttl I r-~ r tb2

bt ~ ~ b3 ~W1

1-1

~~ b4 bull

The torsional constant for a single thin walled closed section may be obtained from

IX = (Ref Bl)

where n is the total area enclosed by the center line of the walls of the closed section The integral term represents the sum of the various lengths of wall section divided b7 their respective thicknesses

ExamOle

-------- I ~ I I I 1-- I I I ~--------J

I I

4(b1 bz) 2 2 ( b bz) 2 IX =

2bt 2bz bt + bz-+-shyt 1 tz t I t 2

C-10

For a hybrid section of a closed section plus outshystard~ng ins ~~e formula for IX becomes

1 3 (Ref Bl)-+ E middot bi ti3 i=l

Exarnoles

4h-r2)2 4IX shy= bt tl21TL 3

t

IX = 27rr 3 t ~ bt t~3

I Multi Celled Sec~ions

Torsion of two or more cells cornected a= the valls is a statically indete~i~ate probla~ T~e general me=hcd to find the torsional rigidity X (Re Bl) is as follows Ass~~e an n celled closed thin walled section

1------- ---I ( I II

----- ----- II

II I flt II

II [22 II

11 I I I i) i ______

------- ------shybt

_______

c-11

The equation of equilibrium for n cells is

n (1) Mt = 2 t q in i (1

i= I

where qi is the shear flow in cell i and n i is the area inclosec by the center line of the walls inclosing the cell and Mt is the twisting moment applied to the cell

The equations of consistent deformation are

(2) (2)

ds _l dswhere Sji - ~ ss s jk

J I t G ss j k t

ds Ls j j 1G fsjj

G is the shear moduls o= elasticity

dst is the sum of the length of cell wall com--ao~ to cells j and i divided by its thickness

dst is the sum of the length of cell wall common to cells j and k divided by its thicknesses

dst is the sum of the lengths of cell walls common5Smiddot middot J J to cell j divided by their respective thicknesses

is the angle of twist in radianse

Equation (2) will yield n equations for n unknown shear flows and can be solved for the shear flows qi in terms of G and the angle of twist e Knowing qi and Ili the torshysional constant IX may be calcul~ted from

( 3) 2 ( 3)IX = Ge

C-12

The following examples are attached to show four methods of solution that may be used for a box girder section The example section chosen was a standard three celled box girder with sloping erterior girders

a Solution bv Method of Simultaneous Ecuations

The torsional constant IX for a three celled box gi~der may be calculated by the method of simultaneous equashytions which is based upon the following facts

I

1) The summation of external torsional moments and the internal resisting shear flow system must be equal to zero

(2) The angle of twist must be the same for each cell

These facts are ~sed to write one equation for each cell in te~s of the shear flow q for that particular cell The resulting shear flows are then used to calculate the Torsional Constant IX The method used in the following example uses three simultaneous equations to solve for the unknown shear flOIS q

EXlZbullPLE CALCwJJtTICN FOR TORSIONAL CONSTiT IX

Assume Box Girder

6 I I I

i I j 36 92u a- 8~ 9middot 2 ~6 i t-1---=-------~~-----____i

Idealize Box Girder 892 867 892

~~_=fL= ( 619 + 8921 ( 548) = 41 4o sq ftJ l 2 Top slob thickness = 56 ft

112 548)(667) = 4 7 51 sq ft Bottom slob thickness = 48ft

C-13

For a hybrid section of a closed section plus outshystard~ng ins ~~e formula for IX becomes

1 3 (Ref Bl)-+ E middot bi ti3 i=l

Exarnoles

4h-r2)2 4IX shy= bt tl21TL 3

t

IX = 27rr 3 t ~ bt t~3

I Multi Celled Sec~ions

Torsion of two or more cells cornected a= the valls is a statically indete~i~ate probla~ T~e general me=hcd to find the torsional rigidity X (Re Bl) is as follows Ass~~e an n celled closed thin walled section

1------- ---I ( I II

----- ----- II

II I flt II

II [22 II

11 I I I i) i ______

------- ------shybt

_______

c-11

The equation of equilibrium for n cells is

n (1) Mt = 2 t q in i (1

i= I

where qi is the shear flow in cell i and n i is the area inclosec by the center line of the walls inclosing the cell and Mt is the twisting moment applied to the cell

The equations of consistent deformation are

(2) (2)

ds _l dswhere Sji - ~ ss s jk

J I t G ss j k t

ds Ls j j 1G fsjj

G is the shear moduls o= elasticity

dst is the sum of the length of cell wall com--ao~ to cells j and i divided by its thickness

dst is the sum of the length of cell wall common to cells j and k divided by its thicknesses

dst is the sum of the lengths of cell walls common5Smiddot middot J J to cell j divided by their respective thicknesses

is the angle of twist in radianse

Equation (2) will yield n equations for n unknown shear flows and can be solved for the shear flows qi in terms of G and the angle of twist e Knowing qi and Ili the torshysional constant IX may be calcul~ted from

( 3) 2 ( 3)IX = Ge

C-12

The following examples are attached to show four methods of solution that may be used for a box girder section The example section chosen was a standard three celled box girder with sloping erterior girders

a Solution bv Method of Simultaneous Ecuations

The torsional constant IX for a three celled box gi~der may be calculated by the method of simultaneous equashytions which is based upon the following facts

I

1) The summation of external torsional moments and the internal resisting shear flow system must be equal to zero

(2) The angle of twist must be the same for each cell

These facts are ~sed to write one equation for each cell in te~s of the shear flow q for that particular cell The resulting shear flows are then used to calculate the Torsional Constant IX The method used in the following example uses three simultaneous equations to solve for the unknown shear flOIS q

EXlZbullPLE CALCwJJtTICN FOR TORSIONAL CONSTiT IX

Assume Box Girder

6 I I I

i I j 36 92u a- 8~ 9middot 2 ~6 i t-1---=-------~~-----____i

Idealize Box Girder 892 867 892

~~_=fL= ( 619 + 8921 ( 548) = 41 4o sq ftJ l 2 Top slob thickness = 56 ft

112 548)(667) = 4 7 51 sq ft Bottom slob thickness = 48ft

C-13

The equation of equilibrium for n cells is

n (1) Mt = 2 t q in i (1

i= I

where qi is the shear flow in cell i and n i is the area inclosec by the center line of the walls inclosing the cell and Mt is the twisting moment applied to the cell

The equations of consistent deformation are

(2) (2)

ds _l dswhere Sji - ~ ss s jk

J I t G ss j k t

ds Ls j j 1G fsjj

G is the shear moduls o= elasticity

dst is the sum of the length of cell wall com--ao~ to cells j and i divided by its thickness

dst is the sum of the length of cell wall common to cells j and k divided by its thicknesses

dst is the sum of the lengths of cell walls common5Smiddot middot J J to cell j divided by their respective thicknesses

is the angle of twist in radianse

Equation (2) will yield n equations for n unknown shear flows and can be solved for the shear flows qi in terms of G and the angle of twist e Knowing qi and Ili the torshysional constant IX may be calcul~ted from

( 3) 2 ( 3)IX = Ge

C-12

The following examples are attached to show four methods of solution that may be used for a box girder section The example section chosen was a standard three celled box girder with sloping erterior girders

a Solution bv Method of Simultaneous Ecuations

The torsional constant IX for a three celled box gi~der may be calculated by the method of simultaneous equashytions which is based upon the following facts

I

1) The summation of external torsional moments and the internal resisting shear flow system must be equal to zero

(2) The angle of twist must be the same for each cell

These facts are ~sed to write one equation for each cell in te~s of the shear flow q for that particular cell The resulting shear flows are then used to calculate the Torsional Constant IX The method used in the following example uses three simultaneous equations to solve for the unknown shear flOIS q

EXlZbullPLE CALCwJJtTICN FOR TORSIONAL CONSTiT IX

Assume Box Girder

6 I I I

i I j 36 92u a- 8~ 9middot 2 ~6 i t-1---=-------~~-----____i

Idealize Box Girder 892 867 892

~~_=fL= ( 619 + 8921 ( 548) = 41 4o sq ftJ l 2 Top slob thickness = 56 ft

112 548)(667) = 4 7 51 sq ft Bottom slob thickness = 48ft

C-13

The following examples are attached to show four methods of solution that may be used for a box girder section The example section chosen was a standard three celled box girder with sloping erterior girders

a Solution bv Method of Simultaneous Ecuations

The torsional constant IX for a three celled box gi~der may be calculated by the method of simultaneous equashytions which is based upon the following facts

I

1) The summation of external torsional moments and the internal resisting shear flow system must be equal to zero

(2) The angle of twist must be the same for each cell

These facts are ~sed to write one equation for each cell in te~s of the shear flow q for that particular cell The resulting shear flows are then used to calculate the Torsional Constant IX The method used in the following example uses three simultaneous equations to solve for the unknown shear flOIS q

EXlZbullPLE CALCwJJtTICN FOR TORSIONAL CONSTiT IX

Assume Box Girder

6 I I I

i I j 36 92u a- 8~ 9middot 2 ~6 i t-1---=-------~~-----____i

Idealize Box Girder 892 867 892

~~_=fL= ( 619 + 8921 ( 548) = 41 4o sq ftJ l 2 Top slob thickness = 56 ft

112 548)(667) = 4 7 51 sq ft Bottom slob thickness = 48ft

C-13

Using equations for multi-celled sections bullle may obtain the following (See Ref 1 for additional details)

-548 G

( 548 + 548 + 867 + 867) ~ I I 56 48 4450 ~22 GG

From Equation 2 we obtain

4044 or- 548 q 2 + o ct 3= 828 Ge

-548 ~~ + 4450q2 - 548q 3 = sso2 Ge

ct1 = 242 Ge

q2 = 2 73 G e

q 3 = 242 G9

And from Equafion 3 we obtain

IX = _pound(242 bull 4140 + 273middot 4751 + 242middot 4140) Ge

IX = 6601 ft 4

b Solution bv Method of S~ccessive Corrections

The Torsional Constant IX for a multiple cell box gi=der may be calculated by the method of successive correcshytions This method is similar to the moment distri~ution method used in frw~e analysis The method is based on the following facts

C-14

---~lf

~~~middot(1 The s~Tmation of the external torsional moment and the internal resisting shear flow force system must equal 0

(2) The angle of twist must be the sa~e for each cell

These facts are used to write one equation for each cell in terms of the shear flow q The resulting equatiors are tee~ solved by the method of successive corrections

The relation between shear flow and twist per urit length is given by

(3) q 2AGe (Ref = 2)EJ

T

~middot1here

G = Modulus of rigidity

L = Length of any cell wall of constant thickness

t = Thickness

9 = ~ist per unit length

A = Area of cell interior

Assu~~ng G9 = 1 equation (3) can be written

This equation then solves for q for each cell ~ndependently T~e resulting q for eac~ cell is the ass~~ed q tha~ is adjus~ed by the successile approximations method

carry over factors are determined for each cell from the follOIing equations

(1-) web ( 1-2) (~)web (2- 3)COFt2middot1l = CO Fc3middot 21 = (E ~)cell ( I ) (t ~)cell ( 2)

(t) web (2-1) (~)web (J I)COF c1- 21 = COF(i -J l = (E i) cell (2) ( LEr) cell J )

~ltpound~-(1) web (3middot2) bull -~=--

COFltZmiddot3l = -~~

(t~)cetl (3

C-15

The carry over operation is performed until the desired precision is reached The final q for each cell is equal to the initial q plus all the carry overs from adjacent cells It will be noted in the example problem that the carry over from cell 2 in the third step is computed from the ~ of car~J overs to cell 2 in the previous step The torshysional constant IX is then computed from the following equation

4) IX = 2 E Qj Aj (4) j I

Exatole Problem

Given Box girder section with cell areas and wall thickshynesses taken from previous example problal

Required Compute the torsional constant IX by the method of successive corrections and compare with results obtained from solving the simultaneous equations

reg 0 8 I At 4140 sq ft poundh = 4044T

A2 = 4751 E-=-T = 4450

A3 =4140 tbT 4044

Assuming Ge = I equation 3 may be written

2 4140~ q3= (I - 2047 X 10-2E L 4044

T

2A2 2 4751 qz= = = 2135 X 10-z

E L 4450 T

548COFz = = 1234044

548COFz-1 = coz-3 = = 1364450

548 co F3middot2 = 1234044

C-16

CELL I CELL 2 CELL 3

COF 123 136 123

2047 x2135 ~047

co 290 ~252 252 290

co 36 ~6868

co 10 10

Total 2121 x 1omiddotz 2415 bull 1omiddot2

From equation (4) we may obtain

IX = 2 ( 2415 4140 + 2727 X 4751 + 2415 bull 4140)

IX = 659044 tt 4

c Solution bv Aaoroxirnate Method

k~ aaoroxi~ate method to fi~d the torsional co~stant o= a multiple box girder would be to assume that the interior web members were not effective in torsion The torsional constant could then be calculated from standard fornulas pubshylished in Engi~eering Handbooks A good reference for torsional constants is Design of Welded Structures by Blodgett The following example uses the di~ensions as stated in the other methods but neglects the ef=ects of interior web rna~ers for torsional considerations

Approximate Me~~od Example

Box Idealized

(No webs)

from Design of Welded Structures Page 210-4

2651 I

----- -------I

R 1548 Ill

--- ___ _

2105 _I I I

C-17

= ( 2651 2105)A 548 = 13031

A2 = 1698070

4 A2 = 67922 80

2651 + 1226 + 2 05Jd = = 1034556 I 48

R = 6792280 = 656 6 ft4 10345

c Solution by Finite Element Analysis

The solution by finite elements requires the use of a torsional analysis program available in Bridge Computer Services A complete discussion of the theort and use of the program is also available in Bridge Computer Services The follmmiddotling example shows the steps needed to obtain a torshysional constant by the finite element method

Examole Problem

Given Box girder section with cell areas and wall thickshyness taken from previous example problem (See sketch)

Required Compute the torsional constant IX by the finite Y element method

1 5 9 13 17 26 35 39 43 47 51 55 57

2 6 10 14 18 ~--~-------------

27 36 40 44 48 52 56 58

19 28

59

60

67

68

69

20 29

21 30

22 31

3

4

7

8

It

12

15

16

23 24

25

32 33

34

37

38

41

42

45

46

49

50

53

54 74 X

Fl NITE ELEMENT MODEL

C-18

---~lf

~~~middot(1 The s~Tmation of the external torsional moment and the internal resisting shear flow force system must equal 0

(2) The angle of twist must be the sa~e for each cell

These facts are used to write one equation for each cell in terms of the shear flow q The resulting equatiors are tee~ solved by the method of successive corrections

The relation between shear flow and twist per urit length is given by

(3) q 2AGe (Ref = 2)EJ

T

~middot1here

G = Modulus of rigidity

L = Length of any cell wall of constant thickness

t = Thickness

9 = ~ist per unit length

A = Area of cell interior

Assu~~ng G9 = 1 equation (3) can be written

This equation then solves for q for each cell ~ndependently T~e resulting q for eac~ cell is the ass~~ed q tha~ is adjus~ed by the successile approximations method

carry over factors are determined for each cell from the follOIing equations

(1-) web ( 1-2) (~)web (2- 3)COFt2middot1l = CO Fc3middot 21 = (E ~)cell ( I ) (t ~)cell ( 2)

(t) web (2-1) (~)web (J I)COF c1- 21 = COF(i -J l = (E i) cell (2) ( LEr) cell J )

~ltpound~-(1) web (3middot2) bull -~=--

COFltZmiddot3l = -~~

(t~)cetl (3

C-15

The carry over operation is performed until the desired precision is reached The final q for each cell is equal to the initial q plus all the carry overs from adjacent cells It will be noted in the example problem that the carry over from cell 2 in the third step is computed from the ~ of car~J overs to cell 2 in the previous step The torshysional constant IX is then computed from the following equation

4) IX = 2 E Qj Aj (4) j I

Exatole Problem

Given Box girder section with cell areas and wall thickshynesses taken from previous example problal

Required Compute the torsional constant IX by the method of successive corrections and compare with results obtained from solving the simultaneous equations

reg 0 8 I At 4140 sq ft poundh = 4044T

A2 = 4751 E-=-T = 4450

A3 =4140 tbT 4044

Assuming Ge = I equation 3 may be written

2 4140~ q3= (I - 2047 X 10-2E L 4044

T

2A2 2 4751 qz= = = 2135 X 10-z

E L 4450 T

548COFz = = 1234044

548COFz-1 = coz-3 = = 1364450

548 co F3middot2 = 1234044

C-16

CELL I CELL 2 CELL 3

COF 123 136 123

2047 x2135 ~047

co 290 ~252 252 290

co 36 ~6868

co 10 10

Total 2121 x 1omiddotz 2415 bull 1omiddot2

From equation (4) we may obtain

IX = 2 ( 2415 4140 + 2727 X 4751 + 2415 bull 4140)

IX = 659044 tt 4

c Solution bv Aaoroxirnate Method

k~ aaoroxi~ate method to fi~d the torsional co~stant o= a multiple box girder would be to assume that the interior web members were not effective in torsion The torsional constant could then be calculated from standard fornulas pubshylished in Engi~eering Handbooks A good reference for torsional constants is Design of Welded Structures by Blodgett The following example uses the di~ensions as stated in the other methods but neglects the ef=ects of interior web rna~ers for torsional considerations

Approximate Me~~od Example

Box Idealized

(No webs)

from Design of Welded Structures Page 210-4

2651 I

----- -------I

R 1548 Ill

--- ___ _

2105 _I I I

C-17

= ( 2651 2105)A 548 = 13031

A2 = 1698070

4 A2 = 67922 80

2651 + 1226 + 2 05Jd = = 1034556 I 48

R = 6792280 = 656 6 ft4 10345

c Solution by Finite Element Analysis

The solution by finite elements requires the use of a torsional analysis program available in Bridge Computer Services A complete discussion of the theort and use of the program is also available in Bridge Computer Services The follmmiddotling example shows the steps needed to obtain a torshysional constant by the finite element method

Examole Problem

Given Box girder section with cell areas and wall thickshyness taken from previous example problem (See sketch)

Required Compute the torsional constant IX by the finite Y element method

1 5 9 13 17 26 35 39 43 47 51 55 57

2 6 10 14 18 ~--~-------------

27 36 40 44 48 52 56 58

19 28

59

60

67

68

69

20 29

21 30

22 31

3

4

7

8

It

12

15

16

23 24

25

32 33

34

37

38

41

42

45

46

49

50

53

54 74 X

Fl NITE ELEMENT MODEL

C-18

The carry over operation is performed until the desired precision is reached The final q for each cell is equal to the initial q plus all the carry overs from adjacent cells It will be noted in the example problem that the carry over from cell 2 in the third step is computed from the ~ of car~J overs to cell 2 in the previous step The torshysional constant IX is then computed from the following equation

4) IX = 2 E Qj Aj (4) j I

Exatole Problem

Given Box girder section with cell areas and wall thickshynesses taken from previous example problal

Required Compute the torsional constant IX by the method of successive corrections and compare with results obtained from solving the simultaneous equations

reg 0 8 I At 4140 sq ft poundh = 4044T

A2 = 4751 E-=-T = 4450

A3 =4140 tbT 4044

Assuming Ge = I equation 3 may be written

2 4140~ q3= (I - 2047 X 10-2E L 4044

T

2A2 2 4751 qz= = = 2135 X 10-z

E L 4450 T

548COFz = = 1234044

548COFz-1 = coz-3 = = 1364450

548 co F3middot2 = 1234044

C-16

CELL I CELL 2 CELL 3

COF 123 136 123

2047 x2135 ~047

co 290 ~252 252 290

co 36 ~6868

co 10 10

Total 2121 x 1omiddotz 2415 bull 1omiddot2

From equation (4) we may obtain

IX = 2 ( 2415 4140 + 2727 X 4751 + 2415 bull 4140)

IX = 659044 tt 4

c Solution bv Aaoroxirnate Method

k~ aaoroxi~ate method to fi~d the torsional co~stant o= a multiple box girder would be to assume that the interior web members were not effective in torsion The torsional constant could then be calculated from standard fornulas pubshylished in Engi~eering Handbooks A good reference for torsional constants is Design of Welded Structures by Blodgett The following example uses the di~ensions as stated in the other methods but neglects the ef=ects of interior web rna~ers for torsional considerations

Approximate Me~~od Example

Box Idealized

(No webs)

from Design of Welded Structures Page 210-4

2651 I

----- -------I

R 1548 Ill

--- ___ _

2105 _I I I

C-17

= ( 2651 2105)A 548 = 13031

A2 = 1698070

4 A2 = 67922 80

2651 + 1226 + 2 05Jd = = 1034556 I 48

R = 6792280 = 656 6 ft4 10345

c Solution by Finite Element Analysis

The solution by finite elements requires the use of a torsional analysis program available in Bridge Computer Services A complete discussion of the theort and use of the program is also available in Bridge Computer Services The follmmiddotling example shows the steps needed to obtain a torshysional constant by the finite element method

Examole Problem

Given Box girder section with cell areas and wall thickshyness taken from previous example problem (See sketch)

Required Compute the torsional constant IX by the finite Y element method

1 5 9 13 17 26 35 39 43 47 51 55 57

2 6 10 14 18 ~--~-------------

27 36 40 44 48 52 56 58

19 28

59

60

67

68

69

20 29

21 30

22 31

3

4

7

8

It

12

15

16

23 24

25

32 33

34

37

38

41

42

45

46

49

50

53

54 74 X

Fl NITE ELEMENT MODEL

C-18

CELL I CELL 2 CELL 3

COF 123 136 123

2047 x2135 ~047

co 290 ~252 252 290

co 36 ~6868

co 10 10

Total 2121 x 1omiddotz 2415 bull 1omiddot2

From equation (4) we may obtain

IX = 2 ( 2415 4140 + 2727 X 4751 + 2415 bull 4140)

IX = 659044 tt 4

c Solution bv Aaoroxirnate Method

k~ aaoroxi~ate method to fi~d the torsional co~stant o= a multiple box girder would be to assume that the interior web members were not effective in torsion The torsional constant could then be calculated from standard fornulas pubshylished in Engi~eering Handbooks A good reference for torsional constants is Design of Welded Structures by Blodgett The following example uses the di~ensions as stated in the other methods but neglects the ef=ects of interior web rna~ers for torsional considerations

Approximate Me~~od Example

Box Idealized

(No webs)

from Design of Welded Structures Page 210-4

2651 I

----- -------I

R 1548 Ill

--- ___ _

2105 _I I I

C-17

= ( 2651 2105)A 548 = 13031

A2 = 1698070

4 A2 = 67922 80

2651 + 1226 + 2 05Jd = = 1034556 I 48

R = 6792280 = 656 6 ft4 10345

c Solution by Finite Element Analysis

The solution by finite elements requires the use of a torsional analysis program available in Bridge Computer Services A complete discussion of the theort and use of the program is also available in Bridge Computer Services The follmmiddotling example shows the steps needed to obtain a torshysional constant by the finite element method

Examole Problem

Given Box girder section with cell areas and wall thickshyness taken from previous example problem (See sketch)

Required Compute the torsional constant IX by the finite Y element method

1 5 9 13 17 26 35 39 43 47 51 55 57

2 6 10 14 18 ~--~-------------

27 36 40 44 48 52 56 58

19 28

59

60

67

68

69

20 29

21 30

22 31

3

4

7

8

It

12

15

16

23 24

25

32 33

34

37

38

41

42

45

46

49

50

53

54 74 X

Fl NITE ELEMENT MODEL

C-18

= ( 2651 2105)A 548 = 13031

A2 = 1698070

4 A2 = 67922 80

2651 + 1226 + 2 05Jd = = 1034556 I 48

R = 6792280 = 656 6 ft4 10345

c Solution by Finite Element Analysis

The solution by finite elements requires the use of a torsional analysis program available in Bridge Computer Services A complete discussion of the theort and use of the program is also available in Bridge Computer Services The follmmiddotling example shows the steps needed to obtain a torshysional constant by the finite element method

Examole Problem

Given Box girder section with cell areas and wall thickshyness taken from previous example problem (See sketch)

Required Compute the torsional constant IX by the finite Y element method

1 5 9 13 17 26 35 39 43 47 51 55 57

2 6 10 14 18 ~--~-------------

27 36 40 44 48 52 56 58

19 28

59

60

67

68

69

20 29

21 30

22 31

3

4

7

8

It

12

15

16

23 24

25

32 33

34

37

38

41

42

45

46

49

50

53

54 74 X

Fl NITE ELEMENT MODEL

C-18

The following coa~ng describes the elements that make up the cross section to be analyzed The first four cards give the title of the problem the loads to be applied the material properties and the number of node and element cards to follow in the input The subsequent cards describe the position of each node in XY coordinates and the last group of cards describes hew the el~~ents are connected at the nodes

The attached output listing shows the results that may be expected from the input data shown For additional information on input and output for this program contact Bridge Computer Services

C-19

-_ middot bull ~ lt bull T tCLlfl rtbull~middot TltIampL tlI~TA bullbullamp-lll

middot bull

1~-UtE lt140U ~nbullr Tl4~ XbulltbullfrTtnNbullbullbull tonoooo

t C1poundamp1t bullnbullrF TNII v-bullHbull bullrT fill( bull bullbull 10011000

11-CbullIIAF ~ S F bullbullbull no T-CIIAT bullobulle~~= bullbullbull nn

TVt~TfttR __Nbull bullbullbull tnooooo

ILAltTI( UIO~OTIF~ TWF bullATFOIAL

lrtbullJS 11lAS1ClTYbullbullbull 1nnonoo

~SOlt UTIbullbullbull OU(I

HCIbull ~middotllbullrlt bullbullbull ~~

I ll bull bullbull011111 z bull middotooll bullbullbullbull011 I oo 41101 J nll bull nanO middotniiiCI ll X bull oomiddotmiddot~no P II

bull nlfl bull ~11111 ll II bull 11~1()0 l)n II bull ~ bullonoo II bull Imiddot middot440ft l1 bull ll 04~00 l2 I bull middot a II1- I ll bull 1Jtl1 bullbull 1100 11101 14 II bull loll1111 14400_

15 I ~ 04111)11 I~ ll lo8110 oo I r bull llonnmiddot0 I~ bull I 140~ 1441111 ll bull 1111nO nnoo

1MI ll bull ~no bullono11Z Zt I 10000middot0 - WIIIF I 1111111 zooon ll bull 1(t to11nn

24 I bull middot1111111 04800 11 middot~ 2 oo ~ ll bull bull bullonoo

-cnnC 11 bull 4ACtfl ~44iW l 711 I onM nbull 70 ll bull middotmiddot411~nn middot00 I (I bullbull 41t1ifr) 1nnn -~ I II bull bull~on 2111110 - bull bull 41 tOCIOn 1 ll bull nn11 nbullllno 1

w bull middot middot ~middot _bull middot IIAftft 04~00 -middot bull bull middotmiddot X bull middotmiddotmiddot middotbull bull 6001111

ronne w ---~ bullbullbullnamiddot11no t 41 ll - ~ lltoaiiO 42 t bull bullbull On

-~ X bull T-~00 middotoooo _ I bull P~oo ~4400 ll bull obullbullunnmiddot~ middotmiddotf I TJIInO on ~--0 bullbullbull ~no middot11460tl 40 ll bull ~ 10 bull 04110

-f bull bull bullbull111110 1111 ll o~ ampoonn ~ bullbull bull~~tJO middotbullbullonlOnE ll 1141111~middot bullbull _ ( oo ( tnlllln amponnof ~ w bull tnbull~nn ~ bullon bull 118 6nono 1111~nn oo bullbull 1middot011 ~nnn 1~1 bullmiddotbullbullno A I bull oonn 12-ll11bullbullnc- bull bull 411111)1 Nltft~ middot~ f bull 11141)1) lfIF a 2011111 X 11104110 y bull middot f w 10141111 10000 1 bull Uln-nn a H

X bull nn amplnoo M f bull _11 ~ ao Nf 0 1 bull 11 111101) yenF TO I middot011 bulloono

11 U41o01) 01)00 bullll u

y bull 000 f 1t non Mill _ bull bull 011middoto~n

c-20

bullbull

bullbull

bullbull bullbull

lltrlll _tNT bull bull bull 74

bullbull1bull11 bullt l

bull 1

IT

41 41

~1

u

bullbull ~

49

n II tbull 20 nzbull tbull

n I

2

111 4 11

2T

40

44

S2

~ 40 61 ~- 64

bullbullbullbull shy34 zo Z1 Z2

n

2 16 lZ

I

10 14

u ZT 40 44 2 0 n Tt

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bullbull 771001 21UYI-06 _middotmiddot1- II -z bull 20-01middot- 011 Zo1f4mbull06 1bullbullbull0~bullbull n ~~~ middotbullbull 12111111

110 middot-11n 16011111-04 -zbullmiddot~-obull 7 bullbull 100 z~t~obull middot~-2bullbull _0 114104611bull0 1111rlllll II 4Pt0ttt)r no 11114Qeii-G6 bull obull 10211bull01 111)1111 1)1 zzu11n-ltmiddot 1141gt111-116 11111614-oampmiddotIll t71bullotl l 1 _06 l bull bullbulllt~bullCJI- II I middotlmiddotbullbullbullbullbullo-oz

onll~ ttbull4Z nt l 11421Abull06 bullbull bullbullbull t-112 -t44Uo-oz

IUOIIrn t bullbullbull- bull z1 I 110bullbull 0011 1161111-06 bull1Uooampll)oo03 bull2642271)bull02middot Ill 0111111 no middot 06111-114 11111 ttoAl1102tmiddotbull IIll 2 ftMG 10 J 207110bull1)6 ll61_11Z 01)IOG-02 61 ~onnn 110 46114bull117 G-1)2middot tl ll 1)1 6211111-01 1 HlbullOlbull06 0~1)1-ll bull~_ztllt ~ Ill 741111111)-11 1oZ1114411-o6 2 u_1 I 01011-02 32bull 11middotl-01 bullZUZII)-01011111-1 1011~411-06 ~ z 40110 til-Ill z I6411-G6 1641111bull11 Jtbullllo--0bull I Z4110111-l Zo42ZIZII-GII bull 71_1 -bull zbull21 to-bull ~ on 741111111_01 2ZttUII-G6 bullunnzt~-nt 11~7110G-111 zbullllnnt-(11 tU7lbull06 101~1ft-l) -lnbullbullo-oz~-bullmiddotnO on 5111011 lll 4112510bull06 middot_ _02

~0 161116111-06 z bullbull 0bull112 bullJ oeozbull n 00middotmiddot middot bullon IIlii Otl 1middotmiddot06 Zollll- bulllo411bull2TIl-GZ 74IIOIIM ()0 ~ middot 6711Pioll-116 ~llllltbullCI -bull-bull215_01 Imiddot nn 1o6ZP71f1bull06 2zbulln2ft-llgt middot252gt-01bull bullbull000111-(lt tVIIIbull06 tbullUH-111 t bullbullbullu_z z bull111111 _ t tA16111bull06 ZTIIII_I)t ~_1middotmiddot 01 74000111bull111 1 1 bullbull 04611111 z bullz~bullzmiddotlit 10

2 M 1 bull rooootn-o1 middotZ_ 41 ~11-11 0-0l middot110 4111101-1 middot H ubulln-oT ZOZ1411bull111 4711RIel)oo04 bullbull 0- 0011-ltll 1 64C711-ltI 4o7Un111-ll __ ltlo-01

C-23

~ 4 ~

6 ~

10 11 Z 13 4 I~

16 7 1~

tbull 10 21 2 n ~ 11gt 2 za

2 n

~middot ~ 36 17

~ 40 42 43 46 47 48 4lto ~0

11 ~2

~4 ~

17

110 61 u 63 64 6S lt6 67 64 69 TO 11 2 1l

WampIIOtt JltrfiTN

142~fimiddotO~ t 74-namp

-1 14A44n-o6 -1 ~3~~11-11

116-116t _Illgt

bulllT1~06 -I 1221411bull06 t 161middot06 I 40 5-0

-151gt42bull011 bull1oOO~Zrl-04

1017bull06 1 )441 7rl-06

bull 1Hft7lbull06 -1 ~middot06 SIH~tl-07 7 6636_0T 61lllbullbull_nT 241010~0

bull4A64CI0-07 bull10700)bull07 -8~0610-07 bull1115~11-)6

-1101411-06 801716~07

6zn~07 middot 211 011)-07241 2411bull07

-15011-0l _ 421 17-01

-middotmiddot -6107bull10ll64Q-06 -041171gt)-06 middot~middot07 13tltQ

-lllo46806 -114115Q-06 11080~11-06 1131 01110bull06

-11150211-06 -11161~06

11514_06 11degHI_06

bull11 USII0-06 bulllltMlbull06

lol15fta-06 1177bull-o-()4gt

n1nn-oT bull12UD-07 t OSUGo-06 108bull4306 bull_0

bull702~110-0

~ 61 17lbullOl 172501)- Ol 61 014607 bullbull 122061)-0 414028)-07 431281-07 3 I ~61-0 7 middot3111bull08

-1610160-07 bull3 011-0T 61)bull07 -611610111bull0 lbull521D-07 4_07 47Zltt1~07 z n~1lton-o 1 2ltgt1133_ ns

-2 ~~ 01161-1)7 1-Q3211bull07 bull 5 564 Hrgt- 01

C-24

1

n enbullbull

~l~f0-01 lo110IIII 00 ~u110~ 11n 1l1110IIft 00 bullbull 11110011 00 bullbull ~ 00-zbullbullll Ol zbull- tlfl 10 bullbull ~ 00 tll110t 0 1111001 01 t7tbullbulln 111

middot- 01 t~~~O~ 0 121gt10111 Ill t2ofll) 01 t61111 Ol tlUIIIUI 01 tOT1-111 01 toPbullon 01 ~ 00 21 (

middot- 00 middotmiddot- 00middot 1111

middoton _ 00 3111 00

~41DC)C)I

middot no )100ft onmiddotmiddotctt

1110ft 011 z 01 t(l on bullbull ~011011-0 t

(wtl bullbull tft

77111111 Ill 11710011 00 bull 11001 00 721110 middot T7fl011 00 middot T7fl0flf 1111middot ~~~~ bull OIIIll 00

1111middot 771111111 72( middot T71lllltl n-1 bullbull 11110111nrvbull 10 bullbull P7001ft 122n no 45)11 00 middot1100 011 71111111011 00 1111111 110 middotmiddotll-01 24000111~1 ~1100bull0l onotn-o 1 zoono1_1 24flfl(llll-lll middotbullOilllllbulll ~7ZOtllfl 00 1100011 nn

middot~ nomiddot~OOftl 0 t~nlom If middot01_1 2~nontn t Z4nOntCI-bull1 I 7 40fl01ft-l I 2o4000l_l 7bull1lOCII-IJI

vbull

tno-bull I 26 11 1 noean-o6bullt_m OTbull01 a-middot~middotU-07 t 4-04 middot 1017-06 lo11111_06 t 117~_06 bull4tleiIbull07 bull _07 bullbull 411middot~-07 ~10111bull)7

-0 zbullbull2_07 _011

bullZIIZ811-17

-middot 2681-0P bullbull~1-07 middotbull~bull4t7f)oOT bullll~fOII~P

bulllOOitZ11-G4 bull11~-06 -t71~1111-06 bulltoe4bull2bulll6 207 bullbull Jl213-0P 7~160bull011

_07 -t~~~~~~middotOP -bull~ontO-o bulll06120ft-G6 bulllo2HI4D-06 -1bullbullbullbull71)-16 bull1 6~1011bull06 -1 71bull tan-oo

Ttl

-bull n-ot -I ~_11 bullbull~211-11 bullbull~-1 -~1062117

7o1ZlZ-111 lo27M4n-lltmiddot ~_oz

-~1214411-112 middotImiddotbullIZbulln-nl middot2middot0206n-lll -671-bull1 -middotUilll-111-middotmiddot middot~ _middot 4f7llll-112 middot-middotIll Zo0lt114llii-OI zoou411-at loiObullII-01 2772~1 14041~1 1oPIJ6111bull0 z~bullt tot-middot111 ~bull _112

-1 1aon-bulln bullI bull-bull11411-02 _ zzbull 07111~-1111

middotmiddotbullu2_

~~ozt 771 11111bull07 1middot~ bullz_l 2noebullll bullbullbulln-ltmiddotzbullzn-

~

middotZbull~uo-n~ o11l~IWbullOJ

bull1llii-O] 1 bull131 OZ 17275611-01 middot ZbullUfl~o-oz

bullbullbull 6146CI-01 lbull o-01bullbull 6_bull 70_(~ tl)ePQIJI)-01

-1 ~ bull _ 73_02 t2amp0411-0 D-Ill bullbullnn-ot 211111)-11 4oU07o-1Jl 1614111-ll 2~611-01 I l~IIZifl-01

bulllo1112~4

middot0_loOU6D-01

bullbullbull 0-01

middot 41-IJ2 bullmiddot~~oo-o 1 nbullbull34_0l middot 4110lt941)-l 1 2-uo-o1 ~ M06-- l bullbull lle(ll)70_t t 011bull~1) I 1bullbull~nebulln-oz

bulllt1middot227411-31 1 24 u l_fl

C-25

Nf1C IIA qiTNt FUNrT nl

1 oa 1 oo 3 oo 4 no

bull24~4010 Ill

Igt bull140l1T no z 2~111 011

8 zMbull 00 11~ n

10 -4)15171 00 bull

11 40441) on 12 middot-~middot0 00 1 -13042~ 00 14 -middotmiddot7271 01 H 166610 00 16 bullbull 0612111 00 17 -1103 01 u -lbullttm 10 I Oil-e 201 21 -11100 Oil n middot 7021-01 u 5 1211 no 13 )813111 onzbull 105~11 01 25 1211121 01 16 -12603 01 21 -bull 63~7 on 7~ -Tnn11l oo 2 -3232HgtI 00 0 lnbull~zl-o1 II 1 ~Ao no 12 bull 101610 00 H 1 1) 60 01 14 t T2440 01 35 -141111 Cll -11651)1) 01 1111~61) 01 1A 1 117040 0

_ n ~1

~-41) -13061110 rll 41 tI1H~n 01 41 lA4tbulllO 01 4 - 01 44 -1 5381 01 41 1 bull 1 n1 102780 01 -zztbull151l 01 -17240 01 1o140TIIl 01 0 z 364141) 01 II -2462240 01

52 -111 01

z 151650 01

14 z6U70 01 -z 70516 o1

16 -2008801 01 -z 8760 ot bullZmiddotl 01 -1100l 01

~0 ot-z 1 61 -20615 01 u -l141 01 6 -t448ltn oo 64 bullmiddotz 1111 61 1 705111) 01 66 ZZOflllfiiO 01 -96lt9000 01 6A -zn~o Ill -174112 I) I 71 bull411177~1) 00 71 -4371130-01 71 ~142Tn 00 71 1amp757~1 01 2 IHtD Ill

C-26

J TAL AlltO 1lfTIG bullbull

TH 11

51)0_01 1lOOC no ~ HOOO DO non oo 3~0Qn 00 11111lt00 00 ~ 00 lll-0 00 a1z-n 00 lbullbull 00 1011000 Ot t I HOl 01 tztZil Ot tnbull1ZI 01 1ol0tJ 01 t2to501lr) Ot tZ11ntl0 01 t 16lt1001) 0 tl5liO Ot 10~11011 01 tot~~ 01 bullJlO 00 bullbull32 00 112 00 ll- Ill 1BlOOn ll 4]10000 00 bullbull nntlOfl

middot 0)OOOI on bullbull 31001)0 1)11 bullHnOotl 00 bullbull HOOOf Otl ])1000 00 2])1)00 0 IJ111101 ~~~

4 HOOO_Ol

middotnonan II nooo ~0 ~ 70000 00 middot nooan 10 lt~noooo no ~ 1000 00 middot 701171 00 ~ 10000 00 1120000 00 middot nooon 00 cnoooo 00 510001 00 middot 72001 00 middot 10000 00 1220000 00 bullbull 5001100 00 ]COOOOO 00 z 500()1 00 1 5001)1 00 bullbull notltn-ot Z 4000 IIbull 1 Z 4000 I0-lt11 zbull(loo1n-ot z bullnoot0-lt11 2410011-01 z bull on1ro-ltt ~2200011 on bullmiddotbullonooo oo )IIOnOO 00 7500001 00 t OOOCJI ll 741100tl-lt11 z 4000 11-lt11 zbullntlliO-ot zbulloootn-ot zbullOOOIbullOl zbullooa1o-nt

bullt~n 00 bull311Hil 00 bullbullbull 1amp00(1 00 -~ ttbullbull4o 00 -tonno 01 -1 21611 Ot -1414111 01 -16450 tH bull11~1 0 -z Ot bullll 01 -z~bullbullTD 01-zmiddotbullbulln 01 -zmiddotzbullo ot -2140510 01 -1 bull 400 01 bull1 631161) 00 1~17011 00 1H4UO bullH ~~nm ~1 zbullbull~ll 01 znt uo 01 zoHnn 01 o 01 tnzn ot ]211~Tl 01

-middotmiddot54 00 bullbullnHn oo -t 511400 00

zbullbullbull 00 bullbull UIIUn 0 middot200 00 1toYZI 01~ 00 bullmiddotubull11o 00 ]11172 0 I ZZIII 00

-z ~ubullzl)oo-z ~_ bullZTUCIbullll-z bull n-ot -1 bullbullst40bull11 -2 bullbullbull_1 -z 4011 -z bullnuo-ot bullZ 4e1Uil-t11 -zbullbullTut-ot -zbullbulln_o1 bull7 4 Tlo-O t -zbullbull~ T15t1-02

middot 04bullmiddotbulltn-nz bullmiddotzbulluoz bullbull 2411_07 middot 511bullso-n7 116201bull01 J 06~11Y_OI tbull36-01 zbullotlo4lbullt11 zbullu_o1 z bullnbull~_1 zbullnbull1_01

bull171CY~D-IIZ

-1 bull ~~~zn-111 bulltZZHffbulllt1t bullmiddot12~-o1 _ middotbulltooto-nz z~obullbullm-tlt 11~lIbullbullH 1zz_ 2~- zzY6o1

-bull702~~ - bull bull iltlc--04 bullI 5365o-Ol

tbull-no-oz 4 lampft51o-Ol

-tuuot 1tUt-)1

bull1216050bull04 10-lt15

bullbullbull 5462)-(6 1 034_~

bull4 _04 1 3C11trgt-OZ bullbullbullbullbull_02 t zbullbullbullo-ot tZ4bullTI)n-01 1zbullbullbullbullo-o1 1 zbullubull-1 12bullbullbullbull~gt-~1 1ouuo-ot bull ll213o-oz

bullZ17104o-)3 31120n-04

-zbull1aa1n-44 1 Y37_1]

-tbulluo-oz t~o4Dto-oz

t 40SD-OZ 151l410n-QZ 1 M4too-oz 1 5lll 1_02 1 bulln40ImiddotI)Z 42~10-03

t bullobullObullbullH -z t Z41D-03 -t 3D-Q4 -~ 30~_

C-27

ampbullbullamp SEC TTCfo bullbull 4o7U~obull 01

1-C~-OtfampT 0 CPNT-nto O~

v-r~bull~tfampT ~- tfNrbullntn middot-~711P no

J~fT OF tNbullTtamp l3~Z~6F 01

v-~~ cbull ffbullTtA -17bull11f ~1

ftfMirT llltfITt amp bull bullbull O 0

ampNrLf rn tNttbullu arn bullbullmiddotbull oo

r-rnnbullOINT OF S~AR rbullNTPbullbullbullbull

T-C~DNampT 0bull ~MampR C~T~bullbullbull

SMampbull cnebulltrtffT axxbullbullbull 11112~~ 00

z bullbull etn oo

HOAbull tnFFIC(NT ampXT OoO

TrITAL TWST IlliG IOfjiiiNT bull bullbull 111l14~ 01

Ttll $1101 ampL ClltSTANT bull bull 7 obullbullzl OZ

~Mll CDFOICT~T YYoobull

C-28

middot~~-middot

c2 Shear Constants - Standard Shaoes

The following tables are provided to dete~ine the shear constants that may be needed for a STRUDL analysis

SHEAR SHAPE FACTORS f

dJfVdx GAx

wherebull

v shearing force G shearing modulus A crossmiddot sectional area f shear shope foetor

Seaion fI D

12solid rectoncuiar

0 111solid

circular

r II f~ Af

I I I

W bending where A total area about minor Af tlance a reo OlliS I -T- i

I f ~

AwI - IW bendinc I wherebull A rotol area

I iabout major Aw weo area

oais

C-29

REFERENCES

1 Hechanics of Elastic Structures by J T Oden 1967

2 Analysis and Desiqn of Airplane Structures by E F Bruhn

3 Desicrn of Welded Str1ctures by 0 i-T Blodgett

~ middot

middot __

C-30

1

n enbullbull

~l~f0-01 lo110IIII 00 ~u110~ 11n 1l1110IIft 00 bullbull 11110011 00 bullbull ~ 00-zbullbullll Ol zbull- tlfl 10 bullbull ~ 00 tll110t 0 1111001 01 t7tbullbulln 111

middot- 01 t~~~O~ 0 121gt10111 Ill t2ofll) 01 t61111 Ol tlUIIIUI 01 tOT1-111 01 toPbullon 01 ~ 00 21 (

middot- 00 middotmiddot- 00middot 1111

middoton _ 00 3111 00

~41DC)C)I

middot no )100ft onmiddotmiddotctt

1110ft 011 z 01 t(l on bullbull ~011011-0 t

(wtl bullbull tft

77111111 Ill 11710011 00 bull 11001 00 721110 middot T7fl011 00 middot T7fl0flf 1111middot ~~~~ bull OIIIll 00

1111middot 771111111 72( middot T71lllltl n-1 bullbull 11110111nrvbull 10 bullbull P7001ft 122n no 45)11 00 middot1100 011 71111111011 00 1111111 110 middotmiddotll-01 24000111~1 ~1100bull0l onotn-o 1 zoono1_1 24flfl(llll-lll middotbullOilllllbulll ~7ZOtllfl 00 1100011 nn

middot~ nomiddot~OOftl 0 t~nlom If middot01_1 2~nontn t Z4nOntCI-bull1 I 7 40fl01ft-l I 2o4000l_l 7bull1lOCII-IJI

vbull

tno-bull I 26 11 1 noean-o6bullt_m OTbull01 a-middot~middotU-07 t 4-04 middot 1017-06 lo11111_06 t 117~_06 bull4tleiIbull07 bull _07 bullbull 411middot~-07 ~10111bull)7

-0 zbullbull2_07 _011

bullZIIZ811-17

-middot 2681-0P bullbull~1-07 middotbull~bull4t7f)oOT bullll~fOII~P

bulllOOitZ11-G4 bull11~-06 -t71~1111-06 bulltoe4bull2bulll6 207 bullbull Jl213-0P 7~160bull011

_07 -t~~~~~~middotOP -bull~ontO-o bulll06120ft-G6 bulllo2HI4D-06 -1bullbullbullbull71)-16 bull1 6~1011bull06 -1 71bull tan-oo

Ttl

-bull n-ot -I ~_11 bullbull~211-11 bullbull~-1 -~1062117

7o1ZlZ-111 lo27M4n-lltmiddot ~_oz

-~1214411-112 middotImiddotbullIZbulln-nl middot2middot0206n-lll -671-bull1 -middotUilll-111-middotmiddot middot~ _middot 4f7llll-112 middot-middotIll Zo0lt114llii-OI zoou411-at loiObullII-01 2772~1 14041~1 1oPIJ6111bull0 z~bullt tot-middot111 ~bull _112

-1 1aon-bulln bullI bull-bull11411-02 _ zzbull 07111~-1111

middotmiddotbullu2_

~~ozt 771 11111bull07 1middot~ bullz_l 2noebullll bullbullbulln-ltmiddotzbullzn-

~

middotZbull~uo-n~ o11l~IWbullOJ

bull1llii-O] 1 bull131 OZ 17275611-01 middot ZbullUfl~o-oz

bullbullbull 6146CI-01 lbull o-01bullbull 6_bull 70_(~ tl)ePQIJI)-01

-1 ~ bull _ 73_02 t2amp0411-0 D-Ill bullbullnn-ot 211111)-11 4oU07o-1Jl 1614111-ll 2~611-01 I l~IIZifl-01

bulllo1112~4

middot0_loOU6D-01

bullbullbull 0-01

middot 41-IJ2 bullmiddot~~oo-o 1 nbullbull34_0l middot 4110lt941)-l 1 2-uo-o1 ~ M06-- l bullbull lle(ll)70_t t 011bull~1) I 1bullbull~nebulln-oz

bulllt1middot227411-31 1 24 u l_fl

C-25

Nf1C IIA qiTNt FUNrT nl

1 oa 1 oo 3 oo 4 no

bull24~4010 Ill

Igt bull140l1T no z 2~111 011

8 zMbull 00 11~ n

10 -4)15171 00 bull

11 40441) on 12 middot-~middot0 00 1 -13042~ 00 14 -middotmiddot7271 01 H 166610 00 16 bullbull 0612111 00 17 -1103 01 u -lbullttm 10 I Oil-e 201 21 -11100 Oil n middot 7021-01 u 5 1211 no 13 )813111 onzbull 105~11 01 25 1211121 01 16 -12603 01 21 -bull 63~7 on 7~ -Tnn11l oo 2 -3232HgtI 00 0 lnbull~zl-o1 II 1 ~Ao no 12 bull 101610 00 H 1 1) 60 01 14 t T2440 01 35 -141111 Cll -11651)1) 01 1111~61) 01 1A 1 117040 0

_ n ~1

~-41) -13061110 rll 41 tI1H~n 01 41 lA4tbulllO 01 4 - 01 44 -1 5381 01 41 1 bull 1 n1 102780 01 -zztbull151l 01 -17240 01 1o140TIIl 01 0 z 364141) 01 II -2462240 01

52 -111 01

z 151650 01

14 z6U70 01 -z 70516 o1

16 -2008801 01 -z 8760 ot bullZmiddotl 01 -1100l 01

~0 ot-z 1 61 -20615 01 u -l141 01 6 -t448ltn oo 64 bullmiddotz 1111 61 1 705111) 01 66 ZZOflllfiiO 01 -96lt9000 01 6A -zn~o Ill -174112 I) I 71 bull411177~1) 00 71 -4371130-01 71 ~142Tn 00 71 1amp757~1 01 2 IHtD Ill

C-26

J TAL AlltO 1lfTIG bullbull

TH 11

51)0_01 1lOOC no ~ HOOO DO non oo 3~0Qn 00 11111lt00 00 ~ 00 lll-0 00 a1z-n 00 lbullbull 00 1011000 Ot t I HOl 01 tztZil Ot tnbull1ZI 01 1ol0tJ 01 t2to501lr) Ot tZ11ntl0 01 t 16lt1001) 0 tl5liO Ot 10~11011 01 tot~~ 01 bullJlO 00 bullbull32 00 112 00 ll- Ill 1BlOOn ll 4]10000 00 bullbull nntlOfl

middot 0)OOOI on bullbull 31001)0 1)11 bullHnOotl 00 bullbull HOOOf Otl ])1000 00 2])1)00 0 IJ111101 ~~~

4 HOOO_Ol

middotnonan II nooo ~0 ~ 70000 00 middot nooan 10 lt~noooo no ~ 1000 00 middot 701171 00 ~ 10000 00 1120000 00 middot nooon 00 cnoooo 00 510001 00 middot 72001 00 middot 10000 00 1220000 00 bullbull 5001100 00 ]COOOOO 00 z 500()1 00 1 5001)1 00 bullbull notltn-ot Z 4000 IIbull 1 Z 4000 I0-lt11 zbull(loo1n-ot z bullnoot0-lt11 2410011-01 z bull on1ro-ltt ~2200011 on bullmiddotbullonooo oo )IIOnOO 00 7500001 00 t OOOCJI ll 741100tl-lt11 z 4000 11-lt11 zbullntlliO-ot zbulloootn-ot zbullOOOIbullOl zbullooa1o-nt

bullt~n 00 bull311Hil 00 bullbullbull 1amp00(1 00 -~ ttbullbull4o 00 -tonno 01 -1 21611 Ot -1414111 01 -16450 tH bull11~1 0 -z Ot bullll 01 -z~bullbullTD 01-zmiddotbullbulln 01 -zmiddotzbullo ot -2140510 01 -1 bull 400 01 bull1 631161) 00 1~17011 00 1H4UO bullH ~~nm ~1 zbullbull~ll 01 znt uo 01 zoHnn 01 o 01 tnzn ot ]211~Tl 01

-middotmiddot54 00 bullbullnHn oo -t 511400 00

zbullbullbull 00 bullbull UIIUn 0 middot200 00 1toYZI 01~ 00 bullmiddotubull11o 00 ]11172 0 I ZZIII 00

-z ~ubullzl)oo-z ~_ bullZTUCIbullll-z bull n-ot -1 bullbullst40bull11 -2 bullbullbull_1 -z 4011 -z bullnuo-ot bullZ 4e1Uil-t11 -zbullbullTut-ot -zbullbulln_o1 bull7 4 Tlo-O t -zbullbull~ T15t1-02

middot 04bullmiddotbulltn-nz bullmiddotzbulluoz bullbull 2411_07 middot 511bullso-n7 116201bull01 J 06~11Y_OI tbull36-01 zbullotlo4lbullt11 zbullu_o1 z bullnbull~_1 zbullnbull1_01

bull171CY~D-IIZ

-1 bull ~~~zn-111 bulltZZHffbulllt1t bullmiddot12~-o1 _ middotbulltooto-nz z~obullbullm-tlt 11~lIbullbullH 1zz_ 2~- zzY6o1

-bull702~~ - bull bull iltlc--04 bullI 5365o-Ol

tbull-no-oz 4 lampft51o-Ol

-tuuot 1tUt-)1

bull1216050bull04 10-lt15

bullbullbull 5462)-(6 1 034_~

bull4 _04 1 3C11trgt-OZ bullbullbullbullbull_02 t zbullbullbullo-ot tZ4bullTI)n-01 1zbullbullbullbullo-o1 1 zbullubull-1 12bullbullbullbull~gt-~1 1ouuo-ot bull ll213o-oz

bullZ17104o-)3 31120n-04

-zbull1aa1n-44 1 Y37_1]

-tbulluo-oz t~o4Dto-oz

t 40SD-OZ 151l410n-QZ 1 M4too-oz 1 5lll 1_02 1 bulln40ImiddotI)Z 42~10-03

t bullobullObullbullH -z t Z41D-03 -t 3D-Q4 -~ 30~_

C-27

ampbullbullamp SEC TTCfo bullbull 4o7U~obull 01

1-C~-OtfampT 0 CPNT-nto O~

v-r~bull~tfampT ~- tfNrbullntn middot-~711P no

J~fT OF tNbullTtamp l3~Z~6F 01

v-~~ cbull ffbullTtA -17bull11f ~1

ftfMirT llltfITt amp bull bullbull O 0

ampNrLf rn tNttbullu arn bullbullmiddotbull oo

r-rnnbullOINT OF S~AR rbullNTPbullbullbullbull

T-C~DNampT 0bull ~MampR C~T~bullbullbull

SMampbull cnebulltrtffT axxbullbullbull 11112~~ 00

z bullbull etn oo

HOAbull tnFFIC(NT ampXT OoO

TrITAL TWST IlliG IOfjiiiNT bull bullbull 111l14~ 01

Ttll $1101 ampL ClltSTANT bull bull 7 obullbullzl OZ

~Mll CDFOICT~T YYoobull

C-28

middot~~-middot

c2 Shear Constants - Standard Shaoes

The following tables are provided to dete~ine the shear constants that may be needed for a STRUDL analysis

SHEAR SHAPE FACTORS f

dJfVdx GAx

wherebull

v shearing force G shearing modulus A crossmiddot sectional area f shear shope foetor

Seaion fI D

12solid rectoncuiar

0 111solid

circular

r II f~ Af

I I I

W bending where A total area about minor Af tlance a reo OlliS I -T- i

I f ~

AwI - IW bendinc I wherebull A rotol area

I iabout major Aw weo area

oais

C-29

REFERENCES

1 Hechanics of Elastic Structures by J T Oden 1967

2 Analysis and Desiqn of Airplane Structures by E F Bruhn

3 Desicrn of Welded Str1ctures by 0 i-T Blodgett

~ middot

middot __

C-30

Nf1C IIA qiTNt FUNrT nl

1 oa 1 oo 3 oo 4 no

bull24~4010 Ill

Igt bull140l1T no z 2~111 011

8 zMbull 00 11~ n

10 -4)15171 00 bull

11 40441) on 12 middot-~middot0 00 1 -13042~ 00 14 -middotmiddot7271 01 H 166610 00 16 bullbull 0612111 00 17 -1103 01 u -lbullttm 10 I Oil-e 201 21 -11100 Oil n middot 7021-01 u 5 1211 no 13 )813111 onzbull 105~11 01 25 1211121 01 16 -12603 01 21 -bull 63~7 on 7~ -Tnn11l oo 2 -3232HgtI 00 0 lnbull~zl-o1 II 1 ~Ao no 12 bull 101610 00 H 1 1) 60 01 14 t T2440 01 35 -141111 Cll -11651)1) 01 1111~61) 01 1A 1 117040 0

_ n ~1

~-41) -13061110 rll 41 tI1H~n 01 41 lA4tbulllO 01 4 - 01 44 -1 5381 01 41 1 bull 1 n1 102780 01 -zztbull151l 01 -17240 01 1o140TIIl 01 0 z 364141) 01 II -2462240 01

52 -111 01

z 151650 01

14 z6U70 01 -z 70516 o1

16 -2008801 01 -z 8760 ot bullZmiddotl 01 -1100l 01

~0 ot-z 1 61 -20615 01 u -l141 01 6 -t448ltn oo 64 bullmiddotz 1111 61 1 705111) 01 66 ZZOflllfiiO 01 -96lt9000 01 6A -zn~o Ill -174112 I) I 71 bull411177~1) 00 71 -4371130-01 71 ~142Tn 00 71 1amp757~1 01 2 IHtD Ill

C-26

J TAL AlltO 1lfTIG bullbull

TH 11

51)0_01 1lOOC no ~ HOOO DO non oo 3~0Qn 00 11111lt00 00 ~ 00 lll-0 00 a1z-n 00 lbullbull 00 1011000 Ot t I HOl 01 tztZil Ot tnbull1ZI 01 1ol0tJ 01 t2to501lr) Ot tZ11ntl0 01 t 16lt1001) 0 tl5liO Ot 10~11011 01 tot~~ 01 bullJlO 00 bullbull32 00 112 00 ll- Ill 1BlOOn ll 4]10000 00 bullbull nntlOfl

middot 0)OOOI on bullbull 31001)0 1)11 bullHnOotl 00 bullbull HOOOf Otl ])1000 00 2])1)00 0 IJ111101 ~~~

4 HOOO_Ol

middotnonan II nooo ~0 ~ 70000 00 middot nooan 10 lt~noooo no ~ 1000 00 middot 701171 00 ~ 10000 00 1120000 00 middot nooon 00 cnoooo 00 510001 00 middot 72001 00 middot 10000 00 1220000 00 bullbull 5001100 00 ]COOOOO 00 z 500()1 00 1 5001)1 00 bullbull notltn-ot Z 4000 IIbull 1 Z 4000 I0-lt11 zbull(loo1n-ot z bullnoot0-lt11 2410011-01 z bull on1ro-ltt ~2200011 on bullmiddotbullonooo oo )IIOnOO 00 7500001 00 t OOOCJI ll 741100tl-lt11 z 4000 11-lt11 zbullntlliO-ot zbulloootn-ot zbullOOOIbullOl zbullooa1o-nt

bullt~n 00 bull311Hil 00 bullbullbull 1amp00(1 00 -~ ttbullbull4o 00 -tonno 01 -1 21611 Ot -1414111 01 -16450 tH bull11~1 0 -z Ot bullll 01 -z~bullbullTD 01-zmiddotbullbulln 01 -zmiddotzbullo ot -2140510 01 -1 bull 400 01 bull1 631161) 00 1~17011 00 1H4UO bullH ~~nm ~1 zbullbull~ll 01 znt uo 01 zoHnn 01 o 01 tnzn ot ]211~Tl 01

-middotmiddot54 00 bullbullnHn oo -t 511400 00

zbullbullbull 00 bullbull UIIUn 0 middot200 00 1toYZI 01~ 00 bullmiddotubull11o 00 ]11172 0 I ZZIII 00

-z ~ubullzl)oo-z ~_ bullZTUCIbullll-z bull n-ot -1 bullbullst40bull11 -2 bullbullbull_1 -z 4011 -z bullnuo-ot bullZ 4e1Uil-t11 -zbullbullTut-ot -zbullbulln_o1 bull7 4 Tlo-O t -zbullbull~ T15t1-02

middot 04bullmiddotbulltn-nz bullmiddotzbulluoz bullbull 2411_07 middot 511bullso-n7 116201bull01 J 06~11Y_OI tbull36-01 zbullotlo4lbullt11 zbullu_o1 z bullnbull~_1 zbullnbull1_01

bull171CY~D-IIZ

-1 bull ~~~zn-111 bulltZZHffbulllt1t bullmiddot12~-o1 _ middotbulltooto-nz z~obullbullm-tlt 11~lIbullbullH 1zz_ 2~- zzY6o1

-bull702~~ - bull bull iltlc--04 bullI 5365o-Ol

tbull-no-oz 4 lampft51o-Ol

-tuuot 1tUt-)1

bull1216050bull04 10-lt15

bullbullbull 5462)-(6 1 034_~

bull4 _04 1 3C11trgt-OZ bullbullbullbullbull_02 t zbullbullbullo-ot tZ4bullTI)n-01 1zbullbullbullbullo-o1 1 zbullubull-1 12bullbullbullbull~gt-~1 1ouuo-ot bull ll213o-oz

bullZ17104o-)3 31120n-04

-zbull1aa1n-44 1 Y37_1]

-tbulluo-oz t~o4Dto-oz

t 40SD-OZ 151l410n-QZ 1 M4too-oz 1 5lll 1_02 1 bulln40ImiddotI)Z 42~10-03

t bullobullObullbullH -z t Z41D-03 -t 3D-Q4 -~ 30~_

C-27

ampbullbullamp SEC TTCfo bullbull 4o7U~obull 01

1-C~-OtfampT 0 CPNT-nto O~

v-r~bull~tfampT ~- tfNrbullntn middot-~711P no

J~fT OF tNbullTtamp l3~Z~6F 01

v-~~ cbull ffbullTtA -17bull11f ~1

ftfMirT llltfITt amp bull bullbull O 0

ampNrLf rn tNttbullu arn bullbullmiddotbull oo

r-rnnbullOINT OF S~AR rbullNTPbullbullbullbull

T-C~DNampT 0bull ~MampR C~T~bullbullbull

SMampbull cnebulltrtffT axxbullbullbull 11112~~ 00

z bullbull etn oo

HOAbull tnFFIC(NT ampXT OoO

TrITAL TWST IlliG IOfjiiiNT bull bullbull 111l14~ 01

Ttll $1101 ampL ClltSTANT bull bull 7 obullbullzl OZ

~Mll CDFOICT~T YYoobull

C-28

middot~~-middot

c2 Shear Constants - Standard Shaoes

The following tables are provided to dete~ine the shear constants that may be needed for a STRUDL analysis

SHEAR SHAPE FACTORS f

dJfVdx GAx

wherebull

v shearing force G shearing modulus A crossmiddot sectional area f shear shope foetor

Seaion fI D

12solid rectoncuiar

0 111solid

circular

r II f~ Af

I I I

W bending where A total area about minor Af tlance a reo OlliS I -T- i

I f ~

AwI - IW bendinc I wherebull A rotol area

I iabout major Aw weo area

oais

C-29

REFERENCES

1 Hechanics of Elastic Structures by J T Oden 1967

2 Analysis and Desiqn of Airplane Structures by E F Bruhn

3 Desicrn of Welded Str1ctures by 0 i-T Blodgett

~ middot

middot __

C-30

J TAL AlltO 1lfTIG bullbull

TH 11

51)0_01 1lOOC no ~ HOOO DO non oo 3~0Qn 00 11111lt00 00 ~ 00 lll-0 00 a1z-n 00 lbullbull 00 1011000 Ot t I HOl 01 tztZil Ot tnbull1ZI 01 1ol0tJ 01 t2to501lr) Ot tZ11ntl0 01 t 16lt1001) 0 tl5liO Ot 10~11011 01 tot~~ 01 bullJlO 00 bullbull32 00 112 00 ll- Ill 1BlOOn ll 4]10000 00 bullbull nntlOfl

middot 0)OOOI on bullbull 31001)0 1)11 bullHnOotl 00 bullbull HOOOf Otl ])1000 00 2])1)00 0 IJ111101 ~~~

4 HOOO_Ol

middotnonan II nooo ~0 ~ 70000 00 middot nooan 10 lt~noooo no ~ 1000 00 middot 701171 00 ~ 10000 00 1120000 00 middot nooon 00 cnoooo 00 510001 00 middot 72001 00 middot 10000 00 1220000 00 bullbull 5001100 00 ]COOOOO 00 z 500()1 00 1 5001)1 00 bullbull notltn-ot Z 4000 IIbull 1 Z 4000 I0-lt11 zbull(loo1n-ot z bullnoot0-lt11 2410011-01 z bull on1ro-ltt ~2200011 on bullmiddotbullonooo oo )IIOnOO 00 7500001 00 t OOOCJI ll 741100tl-lt11 z 4000 11-lt11 zbullntlliO-ot zbulloootn-ot zbullOOOIbullOl zbullooa1o-nt

bullt~n 00 bull311Hil 00 bullbullbull 1amp00(1 00 -~ ttbullbull4o 00 -tonno 01 -1 21611 Ot -1414111 01 -16450 tH bull11~1 0 -z Ot bullll 01 -z~bullbullTD 01-zmiddotbullbulln 01 -zmiddotzbullo ot -2140510 01 -1 bull 400 01 bull1 631161) 00 1~17011 00 1H4UO bullH ~~nm ~1 zbullbull~ll 01 znt uo 01 zoHnn 01 o 01 tnzn ot ]211~Tl 01

-middotmiddot54 00 bullbullnHn oo -t 511400 00

zbullbullbull 00 bullbull UIIUn 0 middot200 00 1toYZI 01~ 00 bullmiddotubull11o 00 ]11172 0 I ZZIII 00

-z ~ubullzl)oo-z ~_ bullZTUCIbullll-z bull n-ot -1 bullbullst40bull11 -2 bullbullbull_1 -z 4011 -z bullnuo-ot bullZ 4e1Uil-t11 -zbullbullTut-ot -zbullbulln_o1 bull7 4 Tlo-O t -zbullbull~ T15t1-02

middot 04bullmiddotbulltn-nz bullmiddotzbulluoz bullbull 2411_07 middot 511bullso-n7 116201bull01 J 06~11Y_OI tbull36-01 zbullotlo4lbullt11 zbullu_o1 z bullnbull~_1 zbullnbull1_01

bull171CY~D-IIZ

-1 bull ~~~zn-111 bulltZZHffbulllt1t bullmiddot12~-o1 _ middotbulltooto-nz z~obullbullm-tlt 11~lIbullbullH 1zz_ 2~- zzY6o1

-bull702~~ - bull bull iltlc--04 bullI 5365o-Ol

tbull-no-oz 4 lampft51o-Ol

-tuuot 1tUt-)1

bull1216050bull04 10-lt15

bullbullbull 5462)-(6 1 034_~

bull4 _04 1 3C11trgt-OZ bullbullbullbullbull_02 t zbullbullbullo-ot tZ4bullTI)n-01 1zbullbullbullbullo-o1 1 zbullubull-1 12bullbullbullbull~gt-~1 1ouuo-ot bull ll213o-oz

bullZ17104o-)3 31120n-04

-zbull1aa1n-44 1 Y37_1]

-tbulluo-oz t~o4Dto-oz

t 40SD-OZ 151l410n-QZ 1 M4too-oz 1 5lll 1_02 1 bulln40ImiddotI)Z 42~10-03

t bullobullObullbullH -z t Z41D-03 -t 3D-Q4 -~ 30~_

C-27

ampbullbullamp SEC TTCfo bullbull 4o7U~obull 01

1-C~-OtfampT 0 CPNT-nto O~

v-r~bull~tfampT ~- tfNrbullntn middot-~711P no

J~fT OF tNbullTtamp l3~Z~6F 01

v-~~ cbull ffbullTtA -17bull11f ~1

ftfMirT llltfITt amp bull bullbull O 0

ampNrLf rn tNttbullu arn bullbullmiddotbull oo

r-rnnbullOINT OF S~AR rbullNTPbullbullbullbull

T-C~DNampT 0bull ~MampR C~T~bullbullbull

SMampbull cnebulltrtffT axxbullbullbull 11112~~ 00

z bullbull etn oo

HOAbull tnFFIC(NT ampXT OoO

TrITAL TWST IlliG IOfjiiiNT bull bullbull 111l14~ 01

Ttll $1101 ampL ClltSTANT bull bull 7 obullbullzl OZ

~Mll CDFOICT~T YYoobull

C-28

middot~~-middot

c2 Shear Constants - Standard Shaoes

The following tables are provided to dete~ine the shear constants that may be needed for a STRUDL analysis

SHEAR SHAPE FACTORS f

dJfVdx GAx

wherebull

v shearing force G shearing modulus A crossmiddot sectional area f shear shope foetor

Seaion fI D

12solid rectoncuiar

0 111solid

circular

r II f~ Af

I I I

W bending where A total area about minor Af tlance a reo OlliS I -T- i

I f ~

AwI - IW bendinc I wherebull A rotol area

I iabout major Aw weo area

oais

C-29

REFERENCES

1 Hechanics of Elastic Structures by J T Oden 1967

2 Analysis and Desiqn of Airplane Structures by E F Bruhn

3 Desicrn of Welded Str1ctures by 0 i-T Blodgett

~ middot

middot __

C-30

ampbullbullamp SEC TTCfo bullbull 4o7U~obull 01

1-C~-OtfampT 0 CPNT-nto O~

v-r~bull~tfampT ~- tfNrbullntn middot-~711P no

J~fT OF tNbullTtamp l3~Z~6F 01

v-~~ cbull ffbullTtA -17bull11f ~1

ftfMirT llltfITt amp bull bullbull O 0

ampNrLf rn tNttbullu arn bullbullmiddotbull oo

r-rnnbullOINT OF S~AR rbullNTPbullbullbullbull

T-C~DNampT 0bull ~MampR C~T~bullbullbull

SMampbull cnebulltrtffT axxbullbullbull 11112~~ 00

z bullbull etn oo

HOAbull tnFFIC(NT ampXT OoO

TrITAL TWST IlliG IOfjiiiNT bull bullbull 111l14~ 01

Ttll $1101 ampL ClltSTANT bull bull 7 obullbullzl OZ

~Mll CDFOICT~T YYoobull

C-28

middot~~-middot

c2 Shear Constants - Standard Shaoes

The following tables are provided to dete~ine the shear constants that may be needed for a STRUDL analysis

SHEAR SHAPE FACTORS f

dJfVdx GAx

wherebull

v shearing force G shearing modulus A crossmiddot sectional area f shear shope foetor

Seaion fI D

12solid rectoncuiar

0 111solid

circular

r II f~ Af

I I I

W bending where A total area about minor Af tlance a reo OlliS I -T- i

I f ~

AwI - IW bendinc I wherebull A rotol area

I iabout major Aw weo area

oais

C-29

REFERENCES

1 Hechanics of Elastic Structures by J T Oden 1967

2 Analysis and Desiqn of Airplane Structures by E F Bruhn

3 Desicrn of Welded Str1ctures by 0 i-T Blodgett

~ middot

middot __

C-30

middot~~-middot

c2 Shear Constants - Standard Shaoes

The following tables are provided to dete~ine the shear constants that may be needed for a STRUDL analysis

SHEAR SHAPE FACTORS f

dJfVdx GAx

wherebull

v shearing force G shearing modulus A crossmiddot sectional area f shear shope foetor

Seaion fI D

12solid rectoncuiar

0 111solid

circular

r II f~ Af

I I I

W bending where A total area about minor Af tlance a reo OlliS I -T- i

I f ~

AwI - IW bendinc I wherebull A rotol area

I iabout major Aw weo area

oais

C-29

REFERENCES

1 Hechanics of Elastic Structures by J T Oden 1967

2 Analysis and Desiqn of Airplane Structures by E F Bruhn

3 Desicrn of Welded Str1ctures by 0 i-T Blodgett

~ middot

middot __

C-30

REFERENCES

1 Hechanics of Elastic Structures by J T Oden 1967

2 Analysis and Desiqn of Airplane Structures by E F Bruhn

3 Desicrn of Welded Str1ctures by 0 i-T Blodgett

~ middot

middot __

C-30