Column Base Plate (Fixed Type)

(1)

:

Job No.

JAC0701

Project Title

Jurong Aromatics Complex Project

CALCULATION REPORT

FOR

COLUMN BASE PLATE

Location

Jurong Island, in Singapore

(2)

Jurong Aromatics Complex Project

CALCULATION

COLUMN BASE PLATE

1. SKETCH OF COLUMN BASE PLATE

2. CALCULATION OF BOLT TENSION AND CONCRETE COMPRESSION

2-1. Moment plus maximum axial force

2-2. Moment plus minimum axial force

3. BASE PLATE THICKNESS

3-1. Compression side bending

3-2. Tension side bending

4. HOLDING DOWN BOLTS AND ANCHORAGE

4-1. Holding down bolts

4-2. Anchorage to concrete

5. SHEAR TRANSFER TO CONCRETE

6. WELDING CHECK

6-1. Tension flange weld

6-2. Compression flange weld

6-3. Web weld

(3)

1. SKETCH OF COLUMN BASE PLATE

Design loads

Axial force (Fc)

Mininum, Fc,min = 219.2 kN Maximum, Fc,max = 882.7 kN

Shear force (Fv) = 65.7 kN 60% of shear capa.

Moment (M) = 67.9 kNm Design data Column size Depth Dc 200.0 mm Width Bc 100.0 mm Flange thickness Tc 8.0 mm Web thickness tc 5.5 mm Base plate Depth Dp 400.0 mm Width Bp 300.0 mm Thickness Tp 35.0 mm Flange weld sw 8.0 mm Web weld s' 8.0 mm Anchor bolt Bolt size 20 mm No. of bolt 4 ea Bolt to bolt distance Gh 300.0 mm

Bolt to edge distance 50.0 mm

Length 450.0 mm

Materials

Base plate SM490A refer to GIS G3106

Min. tensile strength Fu 490.0 MPa

Min. yielding strength Fy 325.0 MPa t < 16

Fy 315.0 MPa 16 < t < 40

Anchor bolt refer to BS 4190

◆ DESIGN RESULT SUMMARY Shear strength ps 160.0 N/mm2 Tensile capacity 78.4 kN

1. Plate thickness ∴ O.K!

2. Anchor bolt ∴ O.K! Bedding Concrete fcu 40.0 N/mm2 refer to JES-43A1

3. Shear transfer ∴ O.K!

4. Welds ∴ O.K! Welding

Welding capacity Pw 215.0 N/mm2

H200X100X5.5X8

Grade 4.6

(4)

:

CALCULATION

: 07261D

FOR COLUMN BASE PLATE (H-SHAPE)

:

TITLE: Base Plate for H - Shape Steel Column (Fixed Type A) REFERENCE

Doc No

Page Rev Job No

2. CALCULATION OF BOLT TENSION AND CONCRETE COMPRESSION

STEP 1

An interface compression force is coupled with a tensile force in the bolts to balance the applied axial compression and bending moment. The moment may act in either direction and symmetrical details are chosen.

The distribution of forces gives the equations which must be satisfied simultaneously for a simple base with one row of bolts on each side.

2-1. Moment plus maximum axial force

To determine the eccentricity leads to an indication of the necessary base size if no bolt tension was available.

Check whether there is tension in the bolts;

eccentricity b = M / N = 67925 kNmm / 882.7 kN

= 77.0 mm

distance to edge of compressive stress block X / 2 = Dp / 2 - b = 400 / 2 - 77

= 123 mm X = 246.0 mm Compression = (area of compressive stress block) x (cuve strength of concrete) ≥ (axial force)

= (2 x 123 x 300) x (0.6 x 40 N/mm² / 10³)

= 1771 kN ≥ 882.7 kN ☞ No tension in the bolts!

Projecting portion of base as a cantilever (see Figure 1):

required design stress

= (Axial force) / (area of compressive stress block) = 882.7 x 10³ / (300 x 123 x 2)

= 11.96 N/mm2

e = L1 - 0.8·sw = 100 - (0.8 x 8)

= 93.6 mm

Figure 1: Uniform pressure on cantilever Moment per mm width applied to plate from stress block,

e2 11.96 x 93.6²

2 2 = 52393.6 Nmm (per mm width)

2-2. Moment plus minimum axial force

Substituting defined values into the equilibrium equation M = Ta + Cb becomes:

X hp

2 2

Substituting values results in the quadratic equation: (-3600) X² + (2520000) X + (-100798120) = 0

Solving for X gives: X = 42.6 or 657.4

∴ X = 42.6 mm Figure 2: Uniform pressure on part of cantilever )

) N (h -(required design stress) x = mc =

M = 0.6·fcu·bp·X·(h -

(5)

Substituting into the equations for C and T gives:

C = 0.6·fcu·Bp·X = 0.6 x 40 x 300 x 42.6 / 10³ = 306.7 kN

T = C - N = 306.7 - 219.2 = 87.6 kN

Moment per mm width applied to plate from stress block, mc = 0.6·fcu·X (e - X/2)

= 0.6 x 40 x 42.6 x (93.6 - 42.6/2) = 73919.5 Nmm (per mm width)

3. BASE PLATE THICKNESS

STEP 2

Plate bending on either the tension side or the compression side may govern.

Both sides must be investigated and the required plate thickness is the larger value resulting from these checks.

3-1. Compression side bending

Projecting portion of base as a cantilever (see Figure 1):

moment per mm width applied to plate from stress block,

mc = maximum value of 2-1. and 2-2. above = 73919.5 Nmm

4·mc 4 x 73919.5

Fy 315 = 30.6 mm

3-2. Tension side bending

Plate thickness to resist bolt tension is based on a calculation for a pure cantilever, with no prying assumed.

Plate bending across the corners may only be avoided by ensuring bolts are positioned within lines 45˚ from the corner of the column flange. (see Figure 3)

mr = T x m = 87.6 x 43.6 x 10³ Figure 3: Plate bending on tension side

= 3817871.8 Nmm m = L1 - k - 0.8 sw = 100 - 50 - 0.8 x 8

= 43.6 mm

4·mr 4 x 3817871.8

Fy·bp 315 x 300 = 12.7 mm

Larger plate thickness tp = 30.6 mm ≤ 35 mm ∴ O.K! ☞ Plate thickness is sufficient! required base plate thickness:

required base plate thickness: required base plate thickness:

tp = =

= tp =

(6)

:

CALCULATION

: 07261D

FOR COLUMN BASE PLATE (H-SHAPE)

:

TITLE: Base Plate for H - Shape Steel Column (Fixed Type A) REFERENCE

Doc No

Page Rev Job No

4. HOLDING DOWN BOLTS AND ANCHORAGE

STEP 3

4-1. Holding down bolts

Force T is assumed to be shared equally between all the bolts in the tension row:

T 87.6

(number of bolts in tension) 2

= 43.8 kN ≤ 78.4 kN ∴ O.K!

4-2. Anchorage to concrete

Assume an effective depth of the holding down bolts, L = 450 - 50 (cover to reinforcement) = 400 mm

Using anchor plate and check the concrete base for punching shear in accordance with BS 8110.

Anchor plate size (for Grade 4.6 bolts) = 5d x 5d x 0.6d thk. = 100 x 100 x 12 mm

Perimeter for punching shear

P = (12 x L) + (total perimeter of anchor plate)

= (12 x 400) + (50 x 2 + 300 x 1) x 2 + (100 x 2) = 5800.0 mm Basic requirement, fv ≤ vc

Average shear stress

T 87.6 x 10³

P x L 5800 x 400 = 0.038 N/mm2

Obtaining design concrete shear stress, vc in accordance with table 3.8 of BS 8110,

an area of tension reinforcement is assumed to its minimum value of 0.15% at the effective depth of 400 mm.

vc = 0.34 N/mm2 > 0.038 N/mm2 ∴ O.K! ☞ Anchor bolts are sufficient!

5. SHEAR TRANSFER TO CONCRETE

STEP 4

Most moment connections are able to rely on friction. However, if high shear is combined with low moment and low axial compression, or if there is axial tension, it is the safest to provide a direct shear connection, either by setting the base plate in a shallow pocket which is filled with concrete or by providing a shear key welded to the underside of the plate.

Check if the horizontal shear is transferred by friction, assuming available resistance to 0.3 of axial compression where axial tension is not applied.

available shear resistance = 0.3 x Fc,min = 65.7 kN ≥ Fv = 65.7 kN ∴ O.K! ☞ The horizontal shear is transferred by friction! force per bolt =

fv = =

=

(7)

6. WELDING CHECK

STEP 5

6-1. Tension flange weld

For most small and medium sized columns, the tension flange welds will be symmetrical, full strength fillet welds. Once the leg length of the required fillet weld exceeds 12mm then a partial penetration butt welds with

superimposed fillet welds, or full penetration butt welds will probably be a more economical solution.

tension capacity of the flange = Bc x Tc x Fy

= 100 x 8 x 325 / 10³ = 260 kN

force in the tension flange M Af

Dc - Tc Ac

67.9 x 10³ 800

200 - 8 2716

= 289.2 kN where, Af : area of the column flange = Bc x Tc ,mm

2

Ac : column cross-sectional area, mm 2

Therefore, weld force per mm = 289.2 / (2 x 100 - 5.5) 1.487 kN/mm weld throat required at 215 N/mm² = 1.487 x 10³ / 215 6.9 mm flange weld thickness = 8.0 mm ≥ 6.9 mm ∴ O.K!

☞ 8mm full strength fillet weld is required!

6-2. Compression flange weld

Assuming bearing contact, nominal welds only are required. However, since the moment is reversible, the tension weld must be made to both flanges.

6-3. Web weld

The capacity of the column web welds for horizontal shear forces should be taken as:

Psw = 2 x 0.7 x s' x Pw x Lws

= 2 x 0.7 x 8 x 215 x (200 - 8 x 2) / 10³

= 443.1 kN ≥ 65.7 kN ∴ O.K!

where, Lws : length of web welds between fillets, mm

☞ 8mm fillet weld is required! =

- N x =

(8)

AVAILABLE SECTION LIST & PROPERTY for JAC

H

Bf

Tw

Tf

r

A

Ix

Iy

(mm)

(cm

2

)

(cm

4

)

(cm

4

)

1 H194X150X6X9

194

150

6

9

13

39.01 2630

507 2 H200X100X5.5X8

200

100

5.5

8

11

27.16 1810

134 3 H244X175X7X11

244

175

7

11

16

56.24 6040

984 4 H250X125X6X9

250

125

6

9

12

37.66 3960

294 5 H294X200X8X12

294

200

8

12

18

72.38 11100

1600

6 H300X150X6.5X9

300

150

6.5

9

13

46.78 7210

508 7 H340X250X9X14

340

250

9

14

20

101.5 21200

3650

8 H350X175X7X11

350

175

7

11

14

63.14 13500

984 9 H390X300X10X16

390

300

10

16

22

135 37900

7200

10 H400X200X8X13

400

200

8

13

16

84.12 23500

1740

11 H440X300X11X18

440

300

11

18

24

157.4 54700

8110

12 H450X200X9X14

450

200

9

14

18

96.76 32900

1870

13 H488X300X11X18

488

300

11

18

26

163.5 68900

8110

14 H500X200X10X16

500

200

10

16

20

114.2 46800

2140

15 H588X300X12X20

588

300

12

20

28

192.5 114000

9010

16 H600X200X11X17

600

200

11

17

22

134.4 75600

2270

17 H700X300X13X24

700

300

13

24

28

235.5 197000

10800

18 H800X300X14X26

800

300

14

26

28

267.4 286000

11700

19 H100X100X6X8

100

6

8

10

21.9

378

134 20 H125X125X6.5X9

125

6.5

9

10

30.31

839

293 21 H150X150X7X10

150

7

10

11

40.14 1620

563 22 H200X200X8X12

200

8

12

13

63.53 4720

1600

23 H250X250X9X14

250

9

14

16

92.18 10700

3650

24 H300X300X10X15

300

10

15

18

119.8 20200

6750

25 H350X350X12X19

350

12

19

20

173.9 39800

13600

26 H400X400X13X21

400

13

21

22

218.7 66600

22400

Where,

Zx, Zy

Modulus of Section

Sx, Sy

Plastic Modulus of Section

Sv

Plastic Modulus of Web, means Plastic Modulus of Shear Area Av

For Calculation of Section Capacity, Refer BS5950-1:2000, Section 4

SM490A

t < 16

t < 40

t > 40

py

325

315

295 Mpa

(9)

181

26.7

209

41.9

46.6

1

214.5

67.9

495

112

558

173

86.2

1

333.1

181.4

317

47

366

73.1

80.7

1

292.5

119.0

756

160

859

247

145.8

2

458.6

279.2

481

67.7

542

105

129.2

3

380.3

156.3 1250

292 1410

447

219.0

2

596.7

458.3

771

112

868

174

188.3

3

477.8

250.6 1940

480 2190

733

320.4

3

760.5

630.5 1170

174 1330

268

279.8

3

624.0

380.3 2490

540 2820

828

448.8

1

943.8

916.5 1460

187 1680

291

400.7

3

789.8

474.5 2820

540 3230

830

561.8

1 1046.8

1049.8

1870

214 2180

335

547.6

3

975.0

607.8 3890

601 4490

928

900.9

3 1375.9

1264.3

2520

227 2980

361

881.0

3 1287.0

819.0 5640

721 6460

1120

1381.6

3 1774.5

1833.0

7160

781 8240

1220

1958.3

₃

_2184.0

_2327.0

75.6

26.7

87.6

41.2

10.6

1

117.0

28.5

134

46.9

154

71.9

18.6

1

158.4

50.1

216

75.1

246

115

29.6

1

204.8

80.0

472

160

525

244

62.0

2

312.0

170.6

860

292

960

444

110.9

2

438.8

312.0 1350

450 1500

684

182.3

3

585.0

438.8 2280

776 2550

1180

292.0

2

819.0

828.8

(10)

High Shear

85.3

52.8

153.3

92.7

231.8

128.3

387.1

209.8

543.8

319.6

746.9

387.7

840.5

474.1 1036.2

608.8 1486.5

1844.2

25.0

44.0

70.3

150.5

276.0

399.3

711.3

961.5

※assume Fv/Pv=1, ρ=[2(Fv/Pv)-1] 2

= 1, which is about 20% consevative than Fv/Pv=0.6

(11)

2 H200X100X5.5X8 3 H244X175X7X11 4 H250X125X6X9 5 H294X200X8X12 6 H300X150X6.5X9 7 H340X250X9X14 8 H350X175X7X11 9 H390X300X10X16 10 H400X200X8X13 11 H440X300X11X18 12 H450X200X9X14 13 H488X300X11X18 14 H500X200X10X16 15 H588X300X12X20 16 H600X200X11X17 17 H700X300X13X24 18 H800X300X14X26 19 H100X100X6X8 20 H125X125X6.5X9 21 H150X150X7X10 22 H200X200X8X12 23 H250X250X9X14 24 H300X300X10X15 25 H350X350X12X19 26 H400X400X13X21