Base Plate - fixed

DESIGN FOR BASE PLATE (FIXED) (Shed columns on grid 23)

Design Data Assumed section Depth of section Width of section Thickness of flange Maximum Vertical load

=

D

Case ( i )

= =

T F

= UC 356 x 171 x 67 = 363.4 mm = 173.2 mm = 15.7 mm = 90 kN P-42 Att II

Maximum Moment

Case ( ii )

=

M1

=

189

Vertical load

Case ( ii )

=

F1

=

90

kN

Minimum vertical load

Case ( iii )

=

F2

=

20

kN P-40 Att II

Moment

Case ( iii )

=

M2

=

189

kN.m

= = = =

Fs L B t

= = = =

112 700 500 30

kN mm mm mm

s

=

12.0

mm

Resultant horizontal shear Length of plate Width of plate Thickness of plate

Assumed thickness of weld for base = plate & column connection

kN.m P-42 Att II

Design strength of the plate

=

pyp

=

265

N/mm2

Grade of concrete Modular ratio

= =

f cu

= =

40 15

N/mm2

Diameter of bolts Grade of bolt Number of bolts

=

db

32 A36 4

mm

N

= = =

Tensile area of bolt Tensile capacity of the single bolt Shear capacity of the single bolt

=

At

=

644 177.47 72.68

mm2

Shear strength of bolt

=

160

N/mm2

m

ps

=

kN kN

"SES B55-E01" "SES B55-E01"

Table 30

pt N/mm2 Tensile strength of bolt = = 240 Bond coefficient = 0.28 (BS 8110-P1, 65) THEREFORE TENSION OCCURS IN THE HOLDING DOWN BOLTS L = 700 700

4 - 32dia bolts

B = 500 500 UC 356 x 171 x 67 30mm Thick plate

THEREFORE TENSION DOE'S NOT OCCUR IN THE HOLDING DOWN BOLTS Edge distance provided = n = 60 mm mm2 Tension bolt area = As = 1288 Lever arm

=

d

=

640

H:\71146 Sabtank JPE\11\Deliverable\Calc\PIPERACK\EW piperack1\Connection\39738609.xls

mm

refer 4.13.2.4

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Bearing pressure Case ( i ) Loading

>

Base pressure

=

w

=

0.26

Permissible bearing pressure

=

wp

=

0.6 f cu

N/mm2 < OKAY

= 24.0 N/mm2 NOT OKAY > Actu Actual al base base pres pressu sure re OKAY Case ( ii ) Loading The eccentricity

=

e'

= =

L/6 116.7

mm >

M1/F1 = = 2100.0 > L/6 THEREFORE TENSION OCCURS IN THE HOLDING DOWN BOLTS

Actual eccentricity

=

e

d1

< mm

= 0.5( 0.5(d d - n) + M1 M1 / F1 2390 mm

A1

-5250

= 6 m d1 As / B mm2

= 5.5E+05 y is the solution of y3 - 3(d - d1)y2 + A1y - A1d By solving above equation Pressure

= =

Permissible pressure

=

y w wp

= 0.0 = 209.1 mm = 6 d1 F / L y (3d - y) > =

5.15

=

0.6 f cu

N/mm2 < OKAY

= 24.00 N/mm2 NOT OKAY > Actu Actual al bas base e pres pressu sure re (w) (w) OKAY Case ( iii ) Loading The eccentricity

=

e'

= =

L/6 116.7

mm >

M2/F2 = = 9450.0 > L/6 THEREFORE TENSION OCCURS IN THE HOLDING DOWN BOLTS

Actual eccentricity

=

e

d1

< mm

= 0.5(d - n) + M2 / F2 9740 9740 mm

A1

= 6 m d1 As / B

-27300

=

y

mm2

2.3E+06

0.0E+00

y is the solution of y3 - 3(d - d1)y2 + A1y - A1d By solving above equation

=

y

= =

Pressure

=

w

= 6 d1 F / L y (3d - y)

Permissible pressure

=

wp

0.0 191.9

=

5.04

=

0.6 f cu

mm >

N/mm2 <


OKAY 2 of 4

= 24.00 N/mm2 NOT OKAY > Actu Actual al bas base e pres pressu sure re (w) (w) OKAY Bolt capacity under tension due to moment Case ( ii ) Loading Bolt stress

=

Fbt

Tensile Force in bolt due to bending Tensile capacity of single bolt

f t

=

= m w [(d / y) - 1] =

176.4

N/mm2 >

=

113.6

kN <

Pnom = 177.47 > Actu Actual al forc force e on on bol boltt

OKAY

Case ( iii ) Loading Bolt stress

=

f t

= m w [(d / y) - 1] =

176.4

N/mm2

Tensile Force in bolt due to bending

Fbt

=

113.6

kN

Force in bolt due to direct tension

Fdt

= F2 / N = 0.0

kN

= Fbt + Fdt = 113.6

kN

Ft

Total tension in the bolt Tensile capacity of single bolt

=

Pnom = 177.47 > Actu Actual al forc force e on on bol boltt

OKAY OKAY

Check for horizontal shear Shear capacity of one bolt

=

Actual horizontal shear / bolt Safety index

Ps

=

72.68

Fs

= =

28.0 2.6

1.4

=

kN NOT OKAY 1 OKAY > <

Check for combined shear and tension Fs Ps

+

Ft Pnom

<

OKAY

1.03

NOT OKAY <

1.4

OKAY

Check for thickness of base plate N/mm2

Maximum base pressure By assuming constant pressure

=

w

=

5.15

Maximum B.M. Assumed thickness of the plate

= =

M1 t

= =

36.5 30

kN.m mm 1

Plate modulus

=

Z

=

7.5E+04

mm3 >

Moment capacity

=

Mc

= 1.5 pyp Z = 29.8 = 0.82

Safety index

< kN.m OKAY NOT OKAY NOT OKA

Column / base plate weld


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Maximum tension in flange

= = = = = =

Length of weld for tension flange Weld shear (Moment) Weld shear (Horizontal force) Resultant weld shear

Max of (case 1 , case 2) 543.6 kN 346.4 mm 1. 1.57 kN/mm 0.32 kN/mm > 1. 1 .60 kN/mm <

= 0.7 s Pw OKAY = 1.85 kN/mm NOT OKAY > Resu Result ltan antt weld weld shea shear r OKAY

Weld capacity

Check for edge distance Minimum edge distance

= max. of 4d & 115 mm = 128 mm

Minimum edge distance required as per = Non-ductile design (in inch)

m

Edge distance provided

= D * SQRT{f ut / [73*SQRT(f'c)]} =

107.6

mm

=

130

mm

OKAY

Check for anchorage length under tension Anchorage required as per BS 8110 recommendations. f bu = Beeta(f cu)1/2 Design ultimate anchorage bond stress = = 1.77 Anchorage required

=

LA

=

N/mm2

(Ft)s

>

f bu PI d

< mm OKAY

= 638.3 Anchorage required as per "Sabic Engineering" recommendations. Minimum required anchorage = 12 * bolt dia. = 384 Minimum embedment required as per La = SQRT(a2+b2) = Non-ductile design 184 Lap = Anchorage length provided as per Sabic 800

mm

mm mm

OKAY

Engineering Engineering standard B50-F01-13


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Base Plate - fixed

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