dimenzioniranje čelika

CALCULATIONS

PROJECT PROJECT No.

PART OF STRUCTURE SHEET No. REVISION

BY DATE CHECKED BY

8-Apr-23

ELEMENT Glavni lučni nosač 1

Raspon cca 10.5 m

Slučaj opterećenja "NADOLJE"tj. Maksimalno vertikalno opterećenje

PODACI O Čelik Č0361 240 N/mm²PRESJEKU t≤40mm 92.9

Duljine izvijanja Lix = 10.00 mLiy = 2.50 m

Pretpostavljeni profil = IPB400Površina A = 198 cm²

Moment otpora Sx = 1620Moment otpora Wx = 2880 Wy = 721Moment inercije Ix = 57680 Iy = 10820

357

Radijus inercije ix = 17.1 cm iy = 7.4 cm

rezultat statičke Max. moment Mx = 6.3 kNmanalize My = 1.9 kNm

Max. uzdužno N = 32.5 kNMax. poprečna Vx = 8.0 kN

Vy = 4.0 kN

Koeficijenti oblika momentnog dijagrama

1.00

1.00

JUS U.E7.081 DOZVOLJENI NAPONI( 1986. g. )

= 240 = 180 N/mm²

n 1.33

= = 104 N/mm²

II slučaj opterećenja - osnovno+dopunsko

JUS U.E7.096 PROVJERA OSNOVNOG MATERIJALA:( 1986. g. )

s = N + Mx + MyA Wx Wy

= 32.5 + 6.3 + 1.9198.0 2880.0 2880.0

= 1.6 + 2.2 + 0.6

Čelik Č.0361 sv = lv =

cm3

cm3 cm3

cm4 cm4

Torzijski moment inercije ID = cm4

bx =

by =

s dop = sv

t dop = sv

n * Ö3

= 4.5 N/mm² < 180 N/mm² OK

JUS U.E7.081 PROVJERA IZVIJANJA( 1986. g. )

krivulja izvijanja presjeka : A

a = = 0.206

`s = = 0.009

= 058 / 093 = 0.629

= 015 / 093 = 0.157

= 1.485

= 0.880

JUS U.E7.096 = 1.089

( 1986. g. ) = 0.991

= 1.136

= 1.004

= 1.000

= 1.000

s = + +

= 1.8 + 2.2 + 0.6

= 4.6 N/mm² < 180 N/mm² OK

PROVJERA PROGIBA

Prema rezultatima statičkog proračuna

f max = 251 mmL = 10000 mm

f / L = 1 / 39.8 NE VALJA

g sN* / sV

`l x = lx / lV

`l y = ly / lV

b = 1+a (`l-0.2 ) +`l2

K = 2 / ( b + Ö (b2 - 4 `l2 ))

kN = 1+a (`lx-0.2 ) / (1-`lx2 `s)

1+a (`ly-0.2 ) / (1-`ly2 `s)

1/K

kMx = bx / (1-`lx2 `s)

kMy = by / (1-`ly2 `s)

J = sV / sD

kN s N J kMx sMx kMy sMy


Raspon cca 10.5 m




Pretpostavljeni profil = 108 diam x 4 debljinaPovršina A = 13.1 cm²

Moment otpora W = 32.8Moment inercije I = 177Radijus inercije i = 3.68 cm

rezultat statičke Max. uzdužno N = 23 kNanalize

JUS U.E7.081 DOZVOLJENI NAPONI( 1986. g. ) = 240 = 160 N/mm²

n 1.50

= = 92 N/mm²

I slučaj opterećenja - osnovno


s = N = 23.0 = 17.6 N/mm²A 13.1

= 17.6 N/mm² < 160 N/mm² OK


krivulja izvijanja presjeka : Da = = 0.756

= 1.609

= 1.609

= 4.653

= 0.250

s = = 17.6 =0.250

= 70.4 N/mm² < 160 N/mm² OK

sv = lv =

cm3

cm4

s dop = sv

t dop = sv

n * Ö3

`l x = lx / lV

`l y = ly / lV

b = 1+a ( `lx-0.2 ) + `lx2

K = 2 / ( b + Ö ( b2 - 4 `l2 ))

s N / K


Raspon cca 10.5 m




Pretpostavljeni profil = 139.7 dia x 4 deb.stij.Površina A = 17.1 cm²

Moment otpora W = 56.2Moment inercije I = 392.9Radijus inercije i = 4.79 cm

rezultat statičke Max. moment Mx = 6.3 kNmanalize My = 1.9 kNm

Max. uzdužno N = 32.5 kNMax. poprečna Vx = 8.0 kN

Vy = 4.0 kN


1.00

1.00


= 240 = 180 N/mm²

n 1.33

= = 104 N/mm²

II slučaj opterećenja - osnovno+dopunsko


s = N + Mx + MyA Wx Wy

= 32.5 + 6.3 + 1.917.1 56.2 56.2

= 19.0 + 112.4 + 32.9

= 164.3 N/mm² < 180 N/mm² OK


cm3

cm4

bx =

by =

s dop = sv

t dop = sv

n * Ö3



a = = 0.206

`s = = 0.106

= 209 / 093 = 2.247

= 052 / 093 = 0.562

= 6.472

= 0.180

JUS U.E7.096 = 1.904

( 1986. g. ) = 1.077

= 5.564

= 2.142

= 1.034

= 1.000

s = + +

= 36.2 + 240.7 + 34.0

= 310.9 N/mm² > 180 N/mm² NE VALJA

PROVJERA PROGIBA


f max = 251 mmL = 10000 mm

f / L = 1 / 39.8 NE VALJA

g sN* / sV

`l x = lx / lV

`l y = ly / lV

b = 1+a (`l-0.2 ) +`l2

K = 2 / ( b + Ö (b2 - 4 `l2 ))

kN = 1+a (`lx-0.2 ) / (1-`lx2 `s)

1+a (`ly-0.2 ) / (1-`ly2 `s)

1/K

kMx = bx / (1-`lx2 `s)

kMy = by / (1-`ly2 `s)

J = sV / sD

kN s N J kMx sMx kMy sMy


Raspon cca 10.5 m




Pretpostavljeni profil = 139.7 dia x 4 deb.stij.Površina A = 17.1 cm²

Moment otpora W = 56.2Moment inercije I = 392.9Radijus inercije i = 4.79 cm

rezultat statičke Max. moment Mx = 6.3 kNmanalize Max. uzdužno N = 32.5 kN

Max. poprečna Vx = 8.0 kNVy = 4.0 kN


1.00


= 240 = 200 N/mm²

n 1.20

= = 115 N/mm²

III slučaj opterećenja - izuzetno


s = N + MxA Wx

= 32.5 + 6.317.1 56.2

= 19.0 + 112.4

= 131.4 N/mm² < 200 N/mm² OK


cm3

cm4

bx =

s dop = sv

t dop = sv

n * Ö3



a = = 0.206

`s = = 0.095

= 209 / 093 = 2.247

= 052 / 093 = 0.562

= 6.472

= 0.180

JUS U.E7.096 = 1.811

( 1986. g. ) = 1.077

= 5.564

= 1.923

= 1.000

s = +

= 34.4 + 216.0

= 250.4 N/mm² > 200 N/mm² NE VALJA

PROVJERA PROGIBA


f max = 251 mmL = 10000 mm

f / L = 1 / 39.8 NE VALJA

g sN* / sV

`l x = lx / lV

`l y = ly / lV

b = 1+a (`l-0.2 ) +`l2

K = 2 / ( b + Ö (b2 - 4 `l2 ))

kN = 1+a (`lx-0.2 ) / (1-`lx2 `s)

1+a (`ly-0.2 ) / (1-`ly2 `s)

1/K

kMx = bx / (1-`lx2 `s)

J = sV / sD

kN s N J kMx sMx

REFERENCES - RAYTHEON DRAWING No.

- STAAD III STRUCTURAL ANALYSIS

Bowstring truss, restraint wires at 5m vertical centresfor latereral stability.

Main Column (Worst Case Moment)

DESIGN Steel Grade fyk = 240 N/mm²PARAMETERS Effective length Sk = 5 m

Assumed Section Size = 160 x 80 x 5.0 RHSArea A = 22.9 cm²

Elastic modulus zz = 62.9 (BS Notation Zy)Elastic modulus zx = 94.1

Second moment of area Izz = 251 (BS Notation Iyy)Second moment of area Ixx = 753

Radius of gyration rzz = 3.31 cm (Bs Notation ryy)Radius of gyration rxx = 5.74 cm

ref staad III Max. MomentMz max = 12.5 kNM load case 8 (Stad Notation My)Max. MomentMx max = 12.5 KNM load case 8 (Stad Notation Mz)

analysis results Max. Axial N max = 4 kN load case 8Max. Shear Vz max = 12.5 kN load case 8 (Stad Notation Vy)Max. Shear Vx max = 12.5 kN load case 8 (Stad Notation Vz)

Unity check based on elastic / elastic method

DIN 18800 Pt1 LIMIT STRESS[746]

= 240 = 218 N/mm²1.1

= = 126 N/mm²

DIN 18800 Pt1 CHECKS:[747]

Mz + NZz A

= 1E+07 + 4000 = 20062900 2290

Mx + NZx A

= 1E+07 + 4000 = 13594100 2290

= 200 = 0.92 < 1 OK218

= 135 = 0.62 < 1 OK= 218

= Vz = 10.90.5 A

= 10.92 = 0.09 < 1 OK126

= Vx = 10.90.5 A

= 10.92 = 0.09 < 1 OK1.26E+02

cm3

cm3

cm4

cm4

s R,D = fy,d = fyk

gm

tR,D fy,d

Ö3

sz =

N/mm2

sx =

N/mm2

sz s R,D

sx s R,D

txy N/mm2

txytR,D

tyz N/mm2

tyztR,D

DIN 18800 Pt1 FAILURE CRITERIA[748]

((200² +135²) -(200*135)+(357.6 = 179 N/mm²

= 179 = 0.820 < 1 OK218

DIN 18800 Pt2 OUT OF PLANE STABILITY CHECK[304] Stability check for design axial compression

Nki =Sk²

x 2.1E5 x 7.530E+06 = 624 KN5000²

Npl = A x fyk

= 2290 x 240 = 550 KN

=

(550 / 624) = 0.939

> 0.2

0.21 hot rolled RHS

k =

DIN 18800 Pt2 = 0.5 [ 1 + (0.939- 0.2) + 0.939²] = 1.018Table 4 & 5

x = 1

x = 1( 1.018² - 0.939² ) + 1.018 = 0.709

N = 4.0

0.709 x 2290 x 240/1.1 = 0.009 < 1 OK

160 x 80 x 5.0 RHSis aqequate

s v = Ö (sx²+sy²+sz²-sxsy-sxsz-sysz+3txy2+3txz2+3tyz2)

= Ö

s v

s R,D

p².E.I

= p²

lk Ö ( Npl / Nki )

= Ö

lk

where a =

0.5 [ 1 + a ( lk - 0.2 ) + lk² ]

lk (lk + a)

x. Npl/gm

CALCULATIONS

PROJECT PROJECT No.

PLIVA RESEARCH INSTITUTE , ZAGREB 393300PART OF STRUCTURE SHEET No. REVISION

Headblock Wall 3 of 8BY DATE CHECKED BY

RH 8-Apr-23



Bowstring truss, restraint wires at 5m vertical centresfor latereral stability.

Main Column (Worst Case Axial Force)

DESIGN Steel Grade fyk = 240 N/mm²PARAMETERS Effective length Sk = 5 m

Assumed Section Size = 160 x 80 x 5.0 RHSArea A = 22.9 cm²

Elastic modulus zz = 62.9 (BS Notation Zy)Elastic modulus zx = 94.1

Second moment of area Izz = 251 (BS Notation Iyy)Second moment of area Ixx = 753

Radius of gyration rzz = 3.31 cm (Bs Notation ryy)Radius of gyration rxx = 5.74 cm

ref staad III Max. MomentMz max = 1 kNM load case 8 (Stad Notation My)Max. MomentMx max = 1 KNM load case 8 (Stad Notation Mz)

analysis results Max. Axial N max = 110 kN load case 8Max. Shear Vz max = 5 kN load case 8 (Stad Notation Vy)Max. Shear Vx max = 5 kN load case 8 (Stad Notation Vz)



= 240 = 218 N/mm²1.1

= = 126 N/mm²


Mz + NZz A

= 1E+06 + 110000 = 6462900 2290

Mx + NZx A

= 1E+06 + 110000 = 5994100 2290

= 64 = 0.29 < 1 OK218

= 59 = 0.27 < 1 OK= 218

= Vz = 4.40.5 A

= 4.37 = 0.03 < 1 OK126

= Vx = 4.40.5 A

= 4.37 = 0.03 < 1 OK1.26E+02

cm3

cm3

cm4

cm4

s R,D = fy,d = fyk

gm

tR,D fy,d

Ö3

sz =

N/mm2

sx =

N/mm2

sz s R,D

sx s R,D

txy N/mm2

txytR,D

tyz N/mm2

tyztR,D

CALCULATIONS

PROJECT PROJECT No.


Headblock Wall 4 of 8 1of 8BY DATE CHECKED BY

RH 8-Apr-23


((64² +59²) -(64*59)+(57.3+57.3) = 63 N/mm²

= 63 = 0.287 < 1 OK218


Nki =Sk²

x 2.1E5 x 7.530E+06 = 624 KN5000²

Npl = A x fyk

= 2290 x 240 = 550 KN

=

(550 / 624) = 0.939

> 0.2

0.21 hot rolled RHS

k =

DIN 18800 Pt2 = 0.5 [ 1 + (0.939- 0.2) + 0.939²] = 1.018Table 4 & 5

x = 1

x = 1( 1.018² - 0.939² ) + 1.018 = 0.709

N = 110.0

0.709 x 2290 x 240/1.1 = 0.257 < 1 OK

160 x 80 x 5.0 RHSis aqequate

s v = Ö (sx²+sy²+sz²-sxsy-sxsz-sysz+3txy2+3txz2+3tyz2)

= Ö

s v

s R,D

p².E.I

= p²

lk Ö ( Npl / Nki )

= Ö

lk

where a =

0.5 [ 1 + a ( lk - 0.2 ) + lk² ]

lk (lk + a)

x. Npl/gm

CALCULATIONS

PROJECT PROJECT No.



RH 8-Apr-23


=

#REF! #REF! N/mm²

= #REF! = #REF! #REF! ###180


Nki = x 2.1E5 x 3.929E+06Sk² 10000²

= 81 kN

Npl = A x fyk = 1710 x 240

= 410 kN

= (410 / 81)

= 2.250

k = 0.21 hot rolled CHS

= 0.5 [ 1 + 0.21(2.25- 0.2) + 2.25²]

= 3.247 DIN 18800 Pt2Table 4 & 5

> 0.2

=> x = 1 = 0.179

Ö ( 3.247² - 2.25² ) + 3.247

N = 1.9

0.179 x 410/1.33333333333333

= 0.03 < 1 OK

=> 139.7x4 CHS###

s v Ö s² + 3 x t

= Ö

s v

s R,D

p².E.I = p²

lk Ö ( Npl / Nki ) = Ö

0.5 [ 1 + a ( lk - 0.2 ) + lk² ] where a =

lk

x. Npl/gm

CALCULATIONS

PROJECT PROJECT No.



RH 8-Apr-23



Bowstring truss, restraint at 2.5m vertical centres for lateral stability

Bowstring Strut (Worst possible Case)

DESIGN Steel Grade fyk = 240 N/mm²PARAMETERS Effective length Sk = 1.6 m

Assumed plate Size = 12.5 cm x 2 cm thkArea A = 25 cm²

Elastic modulus z = 52.08Second moment of area I = 325.52 cm4

Radius of gyration i = 3.61 cm

ref staad III Max. Moment M max = 2 kN load case 11analysis results Max. Axial N max = 13 kN load case 11

Max. Shear V max = 4 kN load case 11



= 240 = 218 N/mm²

1.1

= = 126 N/mm²


s = M + NZ A

= 2E+06 + 1300052080 2500

= 44 N/mm²

s = 44 = 0.20 < 1 OK218

t = V = 4000 = 3 N/mm²0.5 x A 0.5 x2500

t = 3.00 = 0.02 < 1 OK

126

cm3

s R,D = fy,d = fyk

gm

tR,D fy,d

Ö3

s R,D

tR,D

CALCULATIONS

PROJECT PROJECT No.



RH 8-Apr-23


=

(44² + 3 x 3²) = 44 N/mm²

= 44 = 0.202 < 1 OK218


Nki = x 2.1E5 x 3.255E+06Sk² 1600²

= 2635 kN

Npl = A x fyk = 2500 x 240

= 600 kN

= (600 / 2635)

= 0.477

k = 0.49 solid sections

= 0.5 [ 1 + 0.49(0.477- 0.2) + 0.477²]

= 0.682 DIN 18800 Pt2Table 4 & 5

> 0.2

=> x = 1 = 0.855

Ö ( 0.682² - 0.477² ) + 0.682

N = 13.0

0.855 x 600/1.1

= 0.03 < 1 OK

=> 12.5x2 cm plateis aqequate

s v Ö s² + 3 x t

= Ö

s v

s R,D

p².E.I = p²

lk Ö ( Npl / Nki ) = Ö

0.5 [ 1 + a ( lk - 0.2 ) + lk² ] where a =

lk

x. Npl/gm

dimenzioniranje čelika

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