CUTTING OF CIRCULAR AND RECTANGULARHOLLOW SECTIONS FOR LATTICE STRUCTURES

LE A D IN G C OMPA N Y FOR P R OD U C TION A N D E R E C TION O F S TE E L S TR U C TU R E S IN TH E S O U TH -E A ST E U R O P E

TECHNICAL INFORMATION 02 May 21, 2008

CUTTING OF CIRCULAR AND RECTANGULAR HOLLOW SECTIONS FOR LATTICE STRUCTURES Prof. d-r Tihomir Nikolovski, Development adviser

1. Introduction The mathematical relations shown below can be used for preparation of templates for cutting ends of hollow circular and rectangular sections for bracing members (diagonal and vertical elements) of lattice structures, as well as for other purposes. FAKOM AD possesses a modern 3D computer aided machine for cutting of circular sections having a diameter up to 600 mm. Consequently, these relations could be evaluated as considerably anachronistic and unnecessary. However, they may be very useful in some cases, especially when cutting circular pipes for structures and pipelines with very large diameters. In this moment, such need appears in FAKOM – Production unit.

2. Cutting of circular hollow sections 2.1 Cutting of diagonal member "d" in full contact with chord member "D" This case of contact of diagonal member to chord member is preferred in connections of lattice structures made of hollow circular sections (See Item 4: Provisions of prEN 1090-2).

2.11 Cutting length "m"

e

1

d/ 2y

β

y1

α o

α e

e

d x

α D

a

m

D/2-y2

a

d

ϕ

y2

Input parameters:

e=

d / 2 − y1 ; tgα

e0 =

d ; 2tgα

a=

D / 2 − y2 ; sin α

y1 =

d cos β ; 2

y2 =

D cos ϕ 2

It follows from the above drawing:

m =a−e=

D / 2 − y 2 d / 2 − y1 − sin α tgα

FAKOM AD - Skopje

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Technical information 02: Cutting of circular hollow sections

Substituting y1 and y2 we obtain:

m=

D ⎧ d ⎫ ⎨1 − cos ϕ − cos α (1 − cos β ) ⎬ 2 sin α ⎩ D ⎭

The equation should be independent of the angle ϕ. Substituting “x”:

x=

D d sin ϕ = sin β 2 2

→

sin ϕ =

d d sin β → cos ϕ = 1 − ( ) 2 sin 2 β D D

Which leads to the final relation:

m=

D 2 sin α

⎧⎪ ⎫⎪ d 2 2 d ⎨1 − 1 − ( ) sin β − cos α (1 − cos β ) ⎬ D D ⎩⎪ ⎭⎪

When the diameters of both diagonal and chord member are equal (D = d):

m=

D D (1 − cos β )(1 − cos α ) {1 − cos β − cos α (1 − cos β )} = 2 sin α 2 sin α

And when α = 90o (T-connection):

m=

D ⎧⎪ d 2 2 ⎨1 − 1 − ( ) sin β 2 ⎪⎩ D

⎫⎪ ⎬ ⎪⎭

(Definition: m = distance from the external edge of circular hollow section having a diameter "d" (for angle β = 0) to the contact to chord circular hollow section "D" for an arbitrary angle β ( 0 ≤ β ≤ 2π )

2.12 Geometrical length "L" of circular hollow section (Definition: mmax = maximum distance (continuation) of the circular hollow section "d" for an angle β = βmax. βmax

α

L = geometrical length of diagonal

e

1

D/2-vd

member "d"

α

l = system length (distance

e

2

d

between lattice joints) of diagonal member "d" D/2

e

D/2

K

d

1

e

2

vd

L

m

α

l

m

ax

d/ 2

L = l − Kd − K g

Кd, Kg = “rests” of bottom and top end of diagonal member "d"

chord axis

For the bottom end:

K d = e1 − e2 − m max

lattice joint

Input parameters:

e1 =

D / 2 − vd ; sin α

e2 =

d ; 2tgα

e1 − e2 =

D 2 sin α

2v d ⎧ ⎫ ⎨1 − ( ) − ( ) cos α ⎬ D D ⎩ ⎭

Using the equation for "m", for β = βmax we obtain: FAKOM AD - Skopje

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Technical information 02: Cutting of circular hollow sections

D 2 sin α

Kd =

⎧⎪ d 2 2 d 2 v ⎫⎪ ⎨ 1 − ( ) sin β max + cos α cos β max − ( ) ⎬ D D D ⎪⎭ ⎪⎩

2.13 Determination of "βmax" Definition: βmax = angle for which the cutting length is maximum: m = mmax.

Starting from the equation:

m=

D ⎧⎪ d 2 2 d ⎪⎫ ⎨1 − 1 − ( ) sin β − cos α (1 − cos β ) ⎬ 2 sin α ⎩⎪ D D ⎭⎪

dm =0 → dβ

β = β max

We obtain:

⎧ d 2 ⎫ 2( ) sin β cos β ⎪ ⎪⎪ dm D ⎪ D d = − cos α sin β ) ⎬ = 0 ⎨ d β 2 sin α ⎪ D d ⎪ 2 1 − ( ) 2 sin 2 β D ⎩⎪ ⎭⎪

sin β [

c

d d cos β − cos α 1 − ( ) 2 sin 2 β ] = 0 D D

sin β = 0 →

mmin =

d

β min = 0

D d d 2 cos α = 2 sin α D tgα

d d cos β − cos α 1 − ( ) 2 sin 2 β = 0 D D (

d 2 d d ) cos 2 β = cos 2 α [1 − ( ) 2 + ( ) 2 cos 2 β ] D D D

cos β max =

d 1 1 − ( )2 d D ( )tgα D

sin β max =

→

1 d ( ) sin α D

(

d 2 ) − cos 2 α D

CONDITION: mmax may exist only if:

(

d 2 ) − cos 2 α ≥ 0 D

→

cos α ≤

d D

→ α ≥ arccos(

d ) D

Substituting into the equation for "m":

mmax =

•

D 2 sin α

1− (

⎧⎪ ⎫⎪ d 2 2 d ⎨1 − 1 − ( ) sin β max − cos α (1 − cos β max ) ⎬ D D ⎩⎪ ⎭⎪

d 2 2 d 1 d 1 d ) sin β max = 1 − ( ) 2 × [( ) 2 − cos 2 α ] = 1 − ( )2 d D D sin α D ( ) 2 sin 2 α D D

• d cos α (1 − cos β max ) = d cos α [1 − D

FAKOM AD - Skopje

D

1 (

d )tgα D

1− (

d 2 ) ] D

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Technical information 02: Cutting of circular hollow sections

Finally:

mmax =

=

D 2 sin α

1 d 2 d cos 2 α d ⎪⎫ ⎪⎧ 1 1 ( ) cos 1 − ( )2 ⎬ = − − − + α ⎨ sin α D D D ⎭⎪ ⎩⎪ sin α

D ⎧⎪ d 2 d ⎪⎫ ⎨1 − sin α 1 − ( ) − cos α ⎬ 2 sin α ⎩⎪ D D ⎭⎪

Substituting "mmax" in the equation for Kd:

K d = e1 − e2 − mmax = Kd =

D 2

D ⎧⎪ 2v d d 2 d ⎪⎫ ⎨1 − ( ) − ( ) cos α − 1 + sin α 1 − ( ) + cos α ⎬ 2 sin α ⎪⎩ D D D D ⎭⎪

⎧⎪ d 2 1 2 v ⎫⎪ ( )⎬ ⎨ 1− ( ) − D sin α D ⎪⎭ ⎪⎩

The total length of the member "L" is obtained from:

L = l – Kd - Kg where, in order to obtain Kg the same expression may be used, but after substituting the corresponding values for D, α and v for the top (other) chord.

2.2 Cutting of circular hollow section "d" with transverse joint plate This case of contact is recommended if overlapping of two adjacent diagonal members in connection to the chord members is to be avoided (See Item 4: Provisions of prEN 1090-2).

2.21 Cutting length "m" βk

β

b

α

m m

ck

D/2-vd

d

b

mk

α

h L

l

D

vd e

chord axis

Kd

lattice jo int

Input parameters:

m = btgα ;

b=

d (1 − cos β ) 2

We obtain:

m=

d tgα (1 − cos β ) 2

m=

D 2 sin α

for

0 ≤ β ≤ β k and 360 0 − β k ≤ β ≤ 360 0

⎧⎪ ⎫⎪ d 2 2 d h ⎨1 − 1 − ( ) sin β − cos α (1 − cos β ) ⎬ + D D ⎪⎩ ⎪⎭ sin α

FAKOM AD - Skopje

for β k ≤ β ≤ 360 0 − β k

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2.22 Determination of "βk" Definition: βk = angle for which the diagonal member "d" abandons the transverse joint plate and enters into a contact with chord member "D".

2 d/

mk

m

h

α

ck

α

D/2-vd

d” r“ e b e m a xi s

e

lattice joint

( D/2-vd) / tgα

Input parameters:

mk = ( ck + h ) sin α

D d (1 − 1 − ( ) 2 sin 2 β ) 2 D d D 2v h= − (1 − d ) + etgα D 2 cos α 2 Equalising the expressions for "mk" d ( ck + h ) sin α = tgα (1 − cos β k ) 2 ck =

By substituting "ck" и "h" in the above expression, a quadratic equation is obtained:

(

d 2 2 d 2e 2 vd 2e ) sin α ⋅ cos 2 β k + 2( ) sin α [( ) + ( )] ⋅ cos β k + [( ) + D D D Dtgα D d 2 vd 2 2 +( )] sin α − [1 − ( ) 2 ]cos 2 α = 0 Dtgα D

and finally ("е" is positive with an orientation as in the drawing):

cos β k = −

⎧⎪ 2 e 2 vd d 2e 2 vd 2 ⎫⎪ [( ) + ( )] − cos α 1 − ( ) 2 + [( ) + ( )] ⎬ ⎨ d α α D Dtg D D Dtg ⎪ ⎪⎭ ( ) sin α ⎩ D 1

2.23 Determination of "mmax" and "mk" Definitions: "mmax" = maximum value of distance "m". "mk" = distance corresponding to angle βk for which the diagonal member "d" abandons the transverse joint plate and enters into a contact with chord member "D".

2.231 If βmax < βk, that is cosβmax > cosβk then: mmax = mk (за β = βmax = βk) mmax = mk =

d tgα (1 − cos β k ) 2

2.232 If βmax > βk, that is cosβmax < cosβk then: mmax = m (за β = βmax) mmax =

D ⎧⎪ d d 2 1 2e 2 vd ⎪⎫ [( ) + ( )] ⎬ ⎨ ( )tgα − 1 − ( ) − 2 ⎩⎪ D D cos α D Dtgα ⎭⎪

2.24 Geometrical length "L" of circular hollow section The total length "L" is obtained from:

L = l – Kd - Kg FAKOM AD - Skopje

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Technical information 02: Cutting of circular hollow sections

K d = e1 − ( e2 −

h ) − mk (see the drawing) sin α

)

ax

m (m

e2 d

mk

α

h L

D/2-vd e1

vd e

l c hord axis

D

Kd

lattice joint

D / 2 − vd D 2v d D d = [1 − ( d )]; e2 = = ( ) cos α ; sin α 2 sin α D 2 tgα 2 sin α D h d D 2v 1 = − (1 − d ) + e sin α 2 sin α cos α 2 sin α D cos α

e1 =

and finally ("е" is positive with an orientation as in the drawing):

Kd =

D⎧ d 1 2e ⎫ ( ) ⎬ − m max ⎨ ( )tgα + 2 ⎩ D cos α D ⎭

To obtain Kg the same expression may be used, but after substituting the corresponding values for D, α, v and mmax for the top (other) chord.

3. Cutting of rectangular hollow sections

S

2

Due to simplicity of mathematical operations, the derivation of formulae is not shown. It goes without saying that for this type of cutting chord members are rectangular (or square) hollow sections. The bracing members (diagonals) may also be circular sections. In that case, strict attention of their orientation in relation to the lattice plane should be paid.

The following two cases of connection of the diagonal members are given:

β

tgα =

h1 ⋅ l0 + b l02 − b 2 + H 12 l02 − b 2

tg β =

H1 − H 0 l0

α

S

1

Ho

b

α b/sinα

FAKOM AD - Skopje

H1

L

α−β

H1 b + sin α tgα b b ; S1 = S2 = tgα tg (α − β ) L=

lo

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S

2

Technical information 02: Cutting of circular hollow sections

β

α

L

α−β

h1 l0

tg β =

H1 − H 0 l0

H1

Ho

b

tgα =

1

L=

l0 H1 = cos α sin α

S

S1 = b ⋅ tgα ;

S2 =

α

b tg (α − β )

lo

The following rules should be respected when using the above expressions: 1. The distance l0 is a "pure" (net) distance, but not the distance between the chord joints of the lattice.

2. The heights H0 and H1 are "pure" (net) distances between the chords of the lattice.

3. The height H0 is always from the side of the origin of the diagonal member; 4. The height H0 may be larger than H1, the angle β may be negative.

4. Provisions of prEN 1090-2:2007-08 (Е) and EN ISO 9692-1 The provisions dealing with details and preparation of ends as well as welding of hollow sections are given in Annex E: Welded connections of hollow sections of the European prEN 1090-2: Execution of steel structures and aluminium structures, Part 2: Technical requirements for steel structures, issued August 2007.

It is recommended for the construction of joints: 1. Case A with separated diagonals and not overlapping welds.

2. If the diagonals are overlapped (Case B), it should be specified which diagonal is to be cut in order to wrap the other diagonal. The last diagonal may, but not necessarily should be welded to the chord, which also should be specified.

3. Welded joints with separated diagonals but overlapped welds (Case C) should be avoided.

g

(а) Recomm ended

(b) Acceptable

(c) To be avoided

Welding of joints may be carried out using but welds or fillet welds, which is to be taken into account for the preparation of ends. When using but welds, the permitted tolerances are significantly severe. At the other side, it should be taken into consideration that for maximum fillet weld аmax = t (wall depth of the diagonal member) the bearing capacity of the fillet welds amounts to 75% of the bearing capacity if the diagonal member.

FAKOM AD - Skopje

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Technical information 02: Cutting of circular hollow sections

The drawing bellow shows the permitted tolerances for but welds. In order to achieve better fitting of the diagonal member to the chord member, it is obvious that the cut the diagonal member using a diameter "d – 2t" (t = wall depth of the diagonal member) instead of its nominal diameter "d".

d

C

1-2mm D

A

B

C

A

D

1-2mm 1-2mm 2-4mm

Detail А, В d<D

2mm

Detail А, В d=D

2-4mm Detail С

β

2-4mm

Fillet weld for β < 60

o

Detail D

To remember only, bellow are given the correct start and stop positions of the welding sequences of bracing members (circular or rectangular hollow section) to the chord member.

3 2 1 4

3

1

2

4

5. Recommendation for cutting 1. For connections wit overlapping diagonal members (Case B – acceptable detail) combined expressions for cutting (a) diagonal to chord member, and (b) Diagonal to diagonal member may be used. However, first of all the contact angle of the diagonal members should be obtained.

2. For the correct fitting in the connections of circular hollow sections, the bracing members (diagonals and verticals) should be taken into mathematical expression given in Item 2. with their internal diameter "d – 2t" (t = wall depth of the diagonal member). Before cutting, diameters of circular sections "D" and "d", or better circumferences "dπ" and depth "t" should be checked.

3. For the area of acute angle, for angles β < 600 (see drawing above), it could be better to take the external diameter "d" into the expression, and to perform fillet weld instead of but weld with partial penetration.

4. Small memory capacity is necessary for the calculation of expressions. Beside EXCEL program which originally gives the results in tabular form, for this purpose an ordinary programmable calculator may be sufficient. Appropriate are steps of 15-300 (24 to 12 points).

References [1] Т.Nikolovski, G.Todoroski: Proposal on construction of computer aided machine for cutting of circular hollow sections for lattice structures. Principles of work and mathematical bases. FCE Skopje, 1997 (in Macedonian).

[2] ЕN 1993-1-8:2005 EUROCODE 3 Design of steel structures, Part 1-8 Design of joints (translation in Serbian, 2006, Yugoslav Association of Structural Engineers (YuASE) and Faculty of Civil Engineering, Belgrade) [3] prEN 1090-2:2007-08 (E): Execution of steel structures and aluminium structures, Part 2: Technical requirements for steel structures, Annex E: Welded joints in hollow sections, CEN TC 135 N 154 rev, 2007 [4] EN ISO 9692-1 (ЕN 29692-1): Welding and allied processes, Recommendation for joint preparation, Part 1: Manual metal-arc welding, gas shielded metal-arc welding, gas welding, TIG welding and beam welding of steels.

FAKOM AD - Skopje

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Technical information 02: Cutting of circular hollow sections

ANNEX А А.1 Cutting of circular hollow section "d" having an eccentricity "с" in relation tо circular hollow section "D" This case may appear very rarely in lattice structures, but it can be applied in various types of installations and pipelines with larger diameter. Obviously, it is a general case of cutting described in Item 2.1.

β β

d/ 2-

e

1

y

x

А.11 Cutting length "m"

y1

α

a

d m

e

o

α

D/2-y2

a

δ

α D

d x

y2

ϕ c

d 2 D [1 - 1 - (- ) ] 2Sinα D

δ =-

The input parameters and general expression for “m” are the same. It follows:

m =a−e=

D 2 sin α

d ⎧ ⎫ ⎨1 − cos ϕ − cos α (1 − cos β ) ⎬ D ⎩ ⎭

Similarly, the expression should be independent of the angle ϕ. By substituting “x”:

x=

D d 2c d 2c d sin ϕ − c = sin β → sin ϕ = + sin β → cos ϕ = 1 − ( + sin β ) 2 2 2 D D D D

Finally:

m=

D 2 sin α

⎧⎪ ⎫⎪ 2c d d 2 ⎨1 − 1 − ( + sin β ) − cos α (1 − cos β ) ⎬ D D D ⎪⎩ ⎪⎭

But only if: c +

d D ≤ → 2c ≤ D − d 2 2

When с = 0 (no eccentricity), the equation given in Item 2.1 is obtained.

References [5] Johnson: Welding Design, Part 5.10 – Connections for tubular construction.

(Remark: There is an error in original expression – equation (1) on page 5.10-11: the algebraic sign in parenthesis under the square root should be "+". Also, the developed view of cutting given in Figure 16 should be replaced with that one given in Figure 19 and vice versa. The equation (2) given in this reference and the equation given in Item 2.1 of this TI use different symbols, however they are completely identical.)

FAKOM AD - Skopje

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Technical information 02: Cutting of circular hollow sections

ANNEX B B.1 Developed view of penetration of circular hollow section "d" having an eccentricity "с", through circular hollow section "D" Mathematical expressions given bellow may be used for preparation of templates for marking and cutting of circular hollow section "D" through which a circular hollow section "d" penetrates. In general, the expressions originate from the case given in Annex A. In the special case (eccentricity c = 0), the expressions originate from the case given in Item 2.1. For simplicity, the expressions are given in parametric coordinates.

Definitions: "n" = ordinate of penetration of the hollow section "d" through the hollow section "D". "S" = abscissa of penetration – developed length of arc (x + c) for angle "ϕ".

x

В.11 Parametric equations for "n" and "S" β

1

y

β

α n2

y1

n1 n n1

d

α α

D/2-y2

n2

S D

y2

d x

ϕ c

S = x+c = Dπ-

ϕ 360

o

Input parameters are the same as in Annex A. It follows:

n1 =

y1 d cos β = ⋅ sin α 2 sin α

n = n1 + n2 =

n2 =

D 1 − cos ϕ ⋅ 2 tan α

D ⎧ d cos β ⎫ ⎨1 − cos ϕ + ⋅ ⎬ 2 tan α ⎩ D cos α ⎭

The expression should be independent from ϕ. Using the same procedure as in Annex А:

cos ϕ = 1 − (

2c d + sin β ) 2 D D

Finally:

n=

D 2 tan α

2c d d cos β ⎫⎪ ⎪⎧ 2 ⎨1 − 1 − ( + sin β ) + ⋅ ⎬ D D D cos α ⎪⎭ ⎩⎪

Similarly:

x=

D d D 2c d sin ϕ − c = sin β → x + c = sin ϕ → sin ϕ = + sin β 2 2 2 D D

ϕo Dπ 2c d S = xq arcsin( + sin β ) + c = D ⋅π = o 360 360 D D FAKOM AD - Skopje

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