Design example: connection between steelwork and a ... · Design example: connection between...

60 cm

30 cm

40 cm

30 cm

MMEdEd

NNEdEd VVEdEd

ZEd

DEd

Vij,Ed

Vij,Ed

Vfr,Ed

VEd, NEd, MEd

dh c

e 1

1

z =

0.9

· d

© 2014 HALFEN · TI-HUC 13-E · www.halfen.com

fig.: standard connection→ page 10 Technical Product Information HALFEN HUC Universal Connection

Design example: connection between steelwork and a concrete column using the HALFEN HSC-B Concrete Steel Connection References:

These static calculations are according to DIN EN 1992-1-1, DIN EN 1993-1-1 and DIN EN 1993-1-8. References to formulae refer to the Technical Product Information: “HALFEN HUC Universal connection” and “HALFEN HSC Stud Connector” as well as to the national technical approvals Z-21.8-1974 HSC-B Concrete Steel connection and Z-21.8-1973 HALFEN HSC Stud Connector.

Concrete C25/30B500Bbcol = 40 cm, hcol = 60 cmMain reinforcement: 12 Ø 25 mmStirrup reinforcement: dsw = 12 mm, s = 8 cmEnvironmental classification: XC1Concrete cover: cv = 2.5 cmFace plate dimensions: bc = 30 cm, hc = 30 cm

Forces: predominantly staticMEd = 160 kNmNEd = 120 kNVEd = 700 kN(Internal static forces as present between positioning and face plate – design deter-mining cross-section 1-1)Positioning plate is installed flush with the formwork

All publications referred to in this example are available for download in PDF format at www.halfen.com

Specifications

Fig.: positioning plate flush with the formwork → page 6 Technical Product Information HALFEN HUC Universal Connection

1 design cross-section 1-1

Face plate

3 mm positioning plate

Gravity axis


Design Example

HALFEN HUC UNIVERSAL CONNECTION

2

friction coefficients μinf μsup

frictional component considered 0.1

0.2frictional component not considered 0

for verification in ULS

Table: page 10 Technical Product Information HALFEN HUC Universal Connection (Z-21.8-1974, Tab. 2)

Tensile force

ZEd = z

MEd + NEd ⋅ (0.5 · hc - 0.1 · d)

z

Assumption: edge distance of the upper bolt row e1 = 50 mmd = hc – e1 = 30.0 cm – 5.0 cm = 25.0 cmz = 0.9 ⋅ d = 0.9 ⋅ 25.0 cm = 22.5 cm

ZEd = 16000 kNcm

22.5 cm + 120 kN ·

(0.5 ⋅ 30 cm – 0.5 ⋅ 25 cm)22.5 cm

= 777.8 kN

Compressive force

DEd = z

MEd - NEd ⋅ z

(d - 0.5 ⋅ hc)

= 22.5 cm

16000 kNcm - 120 kN ⋅

22.5 cm(25 cm - 0.5 ⋅ 30 cm)

= 657.8 kN

Frictional component

Vfr,Ed,inf = DEd ⋅ μinf

Assumption:Frictional component will be considered

Vfr,Ed,inf = 657.8 kN ⋅ 0.1 = 65.8 kN(for verification of bolts, sleeves, local concrete failure)

Vfr,Ed,sup = DEd ⋅ μsup = 657.8 kN ⋅ 0.2 = 131.6 kN(for verification of concrete edge failure)

→ page 10 Technical Product Information HALFEN HUC Universal Connection (Z-21.8-1974, eq. 4)

→ page 10 Technical Product Information HALFEN HUC Universal Connection (Z-21.8-1974, eq. 3b)

→ page 10 Technical Product Information HALFEN HUC Universal Connection (Z-21.8-1974, appendix 3, pg. 3)

0

0

20

20

40

60

80

100

120

140

160

180

200

220

240

40 60 80 100 120 140 160 180 200 220 240 260 280 300 320

M 27 10.9

M 27 5.6

M 27 4.6

HSC-B 25

Nij,

Rd [

kN]

Vij,Rd [kN]

M 27 8.8

Design Example


3© 2014 HALFEN · TI-HUC 13-E · www.halfen.com

Tension flange bolts:

Maximum tension loadmax ZEd = 213.4 kN

ntie,req = Nij,Rd

ZEd = 213.4 kN777.8 kN

= 3.6

Selected: ntie = 4

Stresses to tension flange bolts:

Nij,Ed = ntie

ZEd = 4

777.8 kN = 194.4 kN

Remaining shear force resistance:

Vij,Rd ≈ 120 kN

Total number of bolts:

ntot,req = Vij,Rd

VEd − Vfr,Ed,inf = 120 kN

700 kN − 65.8 kN = 5.3

Selected: ntot = 6

Shear stresses to bolts:

Vij,Ed = ntot

VEd − Vfr,Ed,inf = 6

700 kN − 65.8 kN = 105.7 kN

Strut: nstrut = ntot − ntie = 6 − 4 = 2

→ page 12 Technical Product Informa- tion HALFEN HUC Universal Con- nection “Pre-selection”

→ page 11 Technical Product Informa- tion HALFEN HUC Universal Con- nection (Z-21.8-1974, eq. 5 and 10)

→ page 11 Technical Product Informa- tion HALFEN HUC Universal Con- nection (Z-21.8-1974, eq. 6 and 11)

= yield force; reinforcement steel Fyd,25, see graph

(Taken from graph)

Determination of required number of sleeves and boltsGiven: HSC-B 25, bolts M 27, strength class 8.8

Fyd, 25 = 213.4 kN

90 (> 86)

70 (> 66) 80 (> 66)

220 (> 66)

70 (> 66)40

40

50 5040

50u

200

u = 50 mm (> min u = 36 mm)

40

[mm]

dL

u

aij

aij

aij

aij,edge

aij,edge


Design Example


4

Dimensions, geometric regulations

HSC-B Bolt minimal distances hole diam.

ds[mm] [-]

aij,edge[mm]

aij[mm]

u[mm]

max dL[mm]

12 M12 40 30 21 13

16 M16 50 38 21 17

20 M20 63 48 27 21

25 M27 86 66 36 28.5

Fig.: page 9 Technical Product Informa-tion HALFEN HUC Universal Con-nection “Design of the hole pattern, minimum distances”

Hole pattern design

Except for the lower protrusion “u” the minimum and maximum distances of the bolts at the steelwork member have to be checked separately.

Table: page 9 Technical Product Infor-mation HALFEN HUC Universal Connec-tion (Z-21.8-1974, appendix 2)

face plate

Design Example



Stress areas of the bolts

bolt size Asp [mm²]

M 12 0.84

M 16 1.57

M 20 2.45

M 27 4.59

bolt resistances

strength fy,b,k [kN/cm²]

fu,b,k[kN/cm²]

αv

4.6 24 40 0.60

5.6 30 50 0.60

8.8 64 80 0.60

10.9 90 100 0.50

Verification for the bolts

Nij,Ed = 194.4 kN (per bolt)

Vij,Ed = 105.7 kN (per bolt)

Nij,Rd = k2 ⋅ m2

fub ⋅ ASp = 0.9 ⋅ 1.25

80 kN/cm² ⋅ 4.59 cm² = 264.4 kN

Vij,Rd = v ⋅ m2

fub ⋅ ASp = 0.60 ⋅ 1.25

80 kN/cm² ⋅ 4.59 cm² = 176.3 kN

→ page 11 Technical Product Information HALFEN HUC Universal connection (Z-21.8-1974, eq. 8)

Nij,Rd

Nij,Ed = 264.4 kN194.4 kN

= 0.74

Vij,Rd

Vij,Ed = 176.3 kN105.7 kN

= 0.60

Nij,Rd

Nij,Ed

2

+ Vij,Rd

Vij,Ed

2

= 0.742 + 0.602 = 0.91

Table: page 11 Technical Product Information HALFEN HUC Universal Connection


→ page 11 Technical Product Informa- tion HALFEN HUC Universal Con- nection (Z-21.8-1974, eq. 7)


0 20 40 60 80 100 120 140 160 180 200 220 240

0

20

40

60

80

100

120

140

160

180

200

220

240

260 280 300 320

Nij,

Rd [

kN]

Vij,Rd [kN]

HSC-B 25

HSC-B 20

HSC-B 16HSC-B 12


Design Example


6

Sleeve verification

Nij,Ed = 194.4 kN (per sleeve)Vij,Ed = 105.7 kN (per sleeve)

Verification: values pair (Vij,Ed, Nij,Ed) is between diameter-specific curve limit (given: HSC-B 25) and coordinate axes, see chart

Verification of the reinforcement steel

Nij,Ed = 194.4 kN (force per bar)

Fyd,25 = γs

fyk ⋅ As = 1.15

50 kN/cm² ⋅

4π

⋅ (2.5 cm)2

= 213.4 kN > 194.4 kN = Nij,Ed

Please note: anchorage has to be verified separately → see page 10

→ page 11 Technical Product Informa- tion HALFEN HUC Universal Con- nection (Z-21.8-1974, appendix 3, page 5, 6 )

→ Z-21.8-1974, diagram 1(105.7 kN, 194.4 kN)

xb

Design Example



Dimensions HSC-B

LM

Sw

ds

ds [mm] Sw [mm] LM [mm]

12 19 36

16 24 48

20 30 60

25 41 75

Localised concrete failure

Vij,Ed = 105.7 kN (per sleeve)

Resistance per sleeve Vij,loc,Rd = c

1.44 ⋅ Sw

2 ⋅ (fc ⋅ Rp,0.2)0.5

Sw = 41 mm (see table)

fck = 25 N/mm² ≤ 50 N/mm²

Rp,0,2 = 440 N/mm²

Vij,c,loc,Rd = 1.5

1.44 ⋅ 412 ⋅ (25 ⋅ 440)0,5 = 169253 N = 169.3 kN

Vij,loc,Rd

Vij,Ed = 169.3105.7

= 0.62 < 1.0

Concrete edge failure

Actions

Vconc,Ed = 2

VEd + Vfr,Ed,sup = 2

700 kN + 131.6 kN = 415.8 kN

Resistances

Vconc,Rd = 15 ⋅ cα

⋅ bc ⋅ LM ⋅ fck0.25

bc = 300 mm (Face plate width)

LM = 75 mm (see table page 4)

fck = 25 N/mm² ≤ 50 N/mm²

Vconc,Rd = 15 ⋅ 1.50.85

⋅ 300 ⋅ 75 ⋅ 250.25 = 427648 N = 427.6 kN

Vconc,Rd

Vconc,Ed = 427.6415.8

= 0.97 < 1.0


→ page 13 Technical Product Informa- tion HALFEN HUC Universal Con- nection (Z-21.8-1974, eq. 14, 15)

Fig.: page 13 Technical Product Information HALFEN HUC Universal Connection

Fig.: page 13 Technical Product Information HALFEN HUC Universal Connection

Table: pg. 13 Technical Product Informa-tion HALFEN HUC Universal Connection

55

1010

Asl,bet

Asl,edge

aaijij

bbcolcol


Design Example


8

Primary reinforcement preventing tensile splitting

Asw,1 = 0.25 ⋅ ntot

VEd ⋅ 1 − aij,max

Sw ⋅ fyk

1.15

VEd = 700.0 kN ntot = 6 (total number of sleeves) Sw = 41 mm (wrench size see table page 7) aij,max = maximum distance between two sleeves in one rowTie: aij,max = 80 mm (upper position)Strut: aij,max = 220 mm (lower position)

Tie:

Asw,1,tie = 0.25 ⋅ 6

700.0 kN ⋅ 1 −

80 mm41 mm

⋅ 50.0 kN/cm²

1.15 = 0.33 cm²

Strut:

Asw,1,strut = 0.25 ⋅ 6

700.0 kN ⋅ 1 −

220 mm41 mm

⋅ 50.0 kN/cm²

1.15 = 0.54 cm²

The primary reinforcement is selected together with the secondary reinforcement for prevention of tensile splitting.

Secondary reinforcement preventing tensile splitting

Tie:

Asw,2,tie = 0.25 ⋅ 2

VEd ⋅ 1 − bcol

∑ aij ⋅ fyk

1.15

VEd = 700.0 kN ∑aij = 2 ⋅ 70 mm + 80 mm = 220 mm bcol = 400 mm fyk = 50 kN/cm²

Asw,2,tie = 0.25 ⋅ 2

700.0 kN ⋅ 1 −

400 mm220 mm

⋅ 50 kN/cm²

1.15 = 0.91 cm²

Strut:

Asw,2,strut = 0.25 ⋅ Vconc,Ed ⋅ 1 − bcol

∑ aij ⋅ fyk

1.15

Vconc,Ed = 415.8 kN ∑aij = 220 mm

Asw,2,strut = 0.25 ⋅ 415.8 kN ⋅ 1 − 400 mm220 mm

⋅ 50 kN/cm²

1.15 = 1.08 cm²

Min

mum

con

stru

ctio

nal d

ista

nces

[cm

]


Fig.: page 14 Technical Product Information HALFEN HUC Universal Connection “minimum constructional distances”



Design Example



Min. diameter, stirrup and main reinforcement

HSC-B [mm]

dsw,2 [mm]

dsl,edge[mm]

dsl,bet[mm]

12 6 12 10

16 6 12 10

20 8 12 12

25 12 20 20

Selection of the tensile splitting reinforcementMinimum diameter dsw,2 = 12 mm for HSC-B 25

Tie:

Asw,tie,tot = Asw,1,tie + Asw,2,tie = 0.33 cm² + 0.91 cm² = 1.24 cm²

Selected: 2 stirrups with dsw = 12 mm where: Asw,prov = 2.26 cm² > 1.24 cm² = Asw,req

Strut:

Asw,strut,tot = Asw,1,strut + Asw,2,strut = 0.55 cm² + 1.08 cm² = 1.63 cm²

Selected: 2 Stirrups with dsw = 12 mm where: Asw,prov = 2.26 cm² > 1.63 cm² = Asw,req

Layout: one stirrup is always placed directly below the sleeves, the second stirrup at a distance of 8 cm.The stirrups enclose the longitudinal column reinforcement.

Longitudinal reinforcement

The minimum diameters according to approval are dsl,edge = 20 mm and dsl,bet = 20 mm, see table.

The reinforcement has to be anchored above and below the steelwork element.

Existing: 5 ∅ 25 mm, continuous longitudinal reinforcement

→ see fig.: complete overview page 11


bcolhcol

hbe

am


Design Example


10

Minimum dimensionsanchor diam. Column

concreteclassdHSC

[mm]bcol,min [mm]

hcol,min [mm]

ds,col,min [mm]

12 240 240 12 C20/25 - C70/85

16 240 240 12 C20/25 - C70/85

20300 300

16C20/25 - C35/45

240 240 C40/50 - C70/85

25

300 400

20

C20/25

300 350 C25/30 - C30/37

300 300 C35/45 - C70/85

Anchorage of the tensile reinforcement

Degree of utilization of the reinforcement steel (see verification reinforcement steel on page 6 also)

213.4 kN194.4 kN

= 0.91

Selected: full anchorage of the reinforcement with HALFEN HSC Anchor heads according to approval Z-21.8-1973 (type HSC-B SH)→ Simplified anchorage verification by keeping to construction regulations

Requirements:Minimum dimensions of the column, see table:

bcol,min = 300 mm < bcol,prov = 400 mm hcol,min = 350 mm < hcol,prov = 600 mm

Minimum longitudinal reinforcement ratio of the column

ρcol,min = bcol ⋅ hcol

As1,col = bcol ⋅ hcol

As2,col ≥ 0.5 %

ρcol,1 = ρcol,2 = 40 cm ⋅ 60 cm

5 ⋅ π/4 ⋅ (2.5 cm)2

= 1.02 % ≥ 0.5 %

Minimum diameter of the longitudinal reinforcement from table: ds,col,min = 20 mm < ds,col,prov = 25 mm

Minimum stirrup reinforcement in the joint area, see figure :dsw = 8 mm, s = 100 mmExisting: dsw = 12 mm, s = 80 mm

The anchor heads are placed behind the longitudinal rein-forcement. See figure above

Load transfer

Load transfer has to be verified by the design engineer.This is also required if an alternative anchorage for the tension reinforcement e.g. bent reinforcement bars has been selected.

Fig.: page 6 Technical Product Informa-tion HALFEN HSC Stud Connector “Geometry, Minimum stirrup reinforce-ment” (Z-21.8-1973, appendix 3)

→ Page 6 Technical Product Informa- tion HALFEN HSC Stud Connector (Z-21.8-1973, eq. 1)

→ Page 6 Technical Product Informa- tion HALFEN HSC Stud Connector (Z-21.8-1973, appendix 7, page 2)

Table page 7 Technical Product Information HALFEN HSC Stud Con-nector (Z-21.8-1973, tab. 2)

Stirr

ups

diam

. 8 s

= 1

00

at h

col

Stirr

ups di

am. 8

s = 10

0

at h

beam

Stirr

ups

diam

. 8 s

= 10

0

at h

col

Stirrup

s diam

. 8 s = 100

at hbeam

L

ds

Design Example



Overview of the selected reinforcement

Tie reinforcement:

4 x Type HSC-B SH, ds = 25 mmL = hcol − tp − cv = 600 mm − 3 mm − 25 mm = 572 mmSleeves zinc electroplated

Strut reinforcement:

2 x Type HSC-B SH, ds = 25 mmL = 572 mmSleeves zinc electroplated

Fig.: overview of the selected reinforcement (except for the tensile splitting reinforcement no stirrups are shown)

Fig.: page 7 Technical Product Informa-tion HALFEN HUC Universal Connec-tion “product overview” „HSC-B female bar with anchor head”

310

300

310

300

tp = 3 mm

45

4540

40

70

8070

u = 50 mm

u50

5555

Liebigstr. 14 ⋅ 40764 Langenfeld ⋅ GERMANYTel.: +49 - (0)2173 / 970-9031Fax: +49 - (0)2173 / 970-420E-Mail: [email protected]

HALFEN GmbH

The Quality Management System of Halfen GmbH is certified for the locations in Germany, France, the Netherlands, Austria, Poland, Switzerland and the Czech Republic according to DIN EN ISO 9001:2008, Certificate No. QS-281 HH.

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Design Example


Face plate HSC-B FP:Measurements: thickness tc = … (determined by design engineer), Other measurements see figureSteel S 460 mill finish(determined by design engineer)Bolts M27

Positioning plate HSC-B P:Dimensions see figure(to facilitate installation the positioning plate is 5 mm wider on all sides than the face plate)Corrosion protection: zinc electroplatedAssembly flush with the formworkBolts M27Nailing holes: d = 3 mm

Not scope of delivery:Longitudinal and stirrup reinforcement6 pcs. bolts M 27 8.8 with washers

Fig.: face plate and positioning plate [all dimensions are in mm]

tc

Explanatory notes on order

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