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DESIGN FOR BASE PLATE (FIXED)(Shed columns on grid 23)Design DataAssumed section = UC 356 x 171 x 67Depth of section = D = 363.4 mmWidth of section = 173.2 mmThickness of flange = T = 15.7 mmMaximum Vertical load Case ( i ) = F = 90 kN P-42 Att II

Maximum Moment Case ( ii ) = = 189 kN.m P-42 Att II

Vertical load Case ( ii ) = = 90 kN

Minimum vertical load Case ( iii ) = = 20 kN P-40 Att II

Moment Case ( iii ) = = 189 kN.m

Resultant horizontal shear = = 112 kNLength of plate = L = 700 mmWidth of plate = B = 500 mmThickness of plate = t = 30 mm

= s = 12.0 mm

Design strength of the plate = = 265

Grade of concrete = = 40Modular ratio = m = 15

Diameter of bolts = = 32 mmGrade of bolt = A36Number of bolts N = 4

Tensile area of bolt = = 644Tensile capacity of the single bolt 177.47 kN "SES B55-E01"

Shear capacity of the single bolt 72.68 kN "SES B55-E01"

Shear strength of bolt = = 160 Table 30

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

L = 700

4 - 32dia bolts

B = 500UC 356 x 171 x 67

30mm Thick plate

THEREFORE TENSION DOE'S NOT OCCUR IN THE HOLDING DOWN BOLTSEdge distance provided = n = 60 mm

Tension bolt area = = 1288Lever arm = d = 640 mm refer 4.13.2.4

M1

F1

F2

M2

Fs

Assumed thickness of weld for base plate & column connection

pyp N/mm2

fcu N/mm2

db

At mm2

ps N/mm2

pt N/mm2

As mm2

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Bearing pressure

Case ( i ) Loading >Base pressure = w = 0.26 <

Permissible bearing pressure = = OKAY= 24.0 NOT OKAY> Actual base pressure OKAY

Case ( ii ) LoadingThe eccentricity = e' = L/6

= 116.7 mm >

Actual eccentricity = e = <= 2100.0 mm> L/6

THEREFORE TENSION OCCURS IN THE HOLDING DOWN BOLTS

= 0.5(d - n) + M1 / F12390 mm

-5250 = 6 m d1 As / B= 5.5E+05

= 0.0By solving above equation = y = 209.1 mmPressure = w = 6 d1 F / L y (3d - y) >

= 5.15 <

Permissible pressure = = OKAY= 24.00 NOT OKAY> Actual base pressure (w) OKAY

Case ( iii ) LoadingThe eccentricity = e' = L/6

= 116.7 mm >

Actual eccentricity = e = <= 9450.0 mm> L/6

THEREFORE TENSION OCCURS IN THE HOLDING DOWN BOLTS

=9740 mm

-27300 == 2.3E+06

y 0.0E+00

= 0.0By solving above equation = y = 191.9 mm

Pressure = w = >= 5.04 <

Permissible pressure = = OKAY

N/mm2

wp 0.6 fcu

N/mm2

M1/F1

d1

A1

mm2

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

N/mm2

wp 0.6 fcu

N/mm2

M2/F2

d1 0.5(d - n) + M2 / F2

A1 6 m d1 As / Bmm2

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

6 d1 F / L y (3d - y)N/mm2

wp 0.6 fcu

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= 24.00 NOT OKAY> Actual base pressure (w) OKAY

Bolt capacity under tension due to momentCase ( ii ) LoadingBolt stress = = m w [(d / y) - 1]

= 176.4 >

Tensile Force in bolt due to bending = 113.6 kN <

Tensile capacity of single bolt = = 177.47> Actual force on bolt OKAY

Case ( iii ) LoadingBolt stress = = m w [(d / y) - 1]

= 176.4

Tensile Force in bolt due to bending = 113.6 kN

Force in bolt due to direct tension == 0.0 kN

Total tension in the bolt == 113.6 kN

Tensile capacity of single bolt = = 177.47 OKAY> Actual force on bolt OKAY

Check for horizontal shearShear capacity of one bolt = = 72.68

Actual horizontal shear / bolt = 28.0 kN NOT OKAYSafety index = 2.6 1 OKAY

>Check for combined shear and tension <

+ < 1.4 = 1.03OKAY

NOT OKAY< 1.4 OKAY

Check for thickness of base plateMaximum base pressure = w = 5.15By assuming constant pressure

Maximum B.M. = = 36.5 kN.mAssumed thickness of the plate = t = 30 mm 1Plate modulus = Z = 7.5E+04 >

Moment capacity = = <= 29.8 kN.m OKAY

Safety index = 0.82 NOT OKAY NOT OKAY

Column / base plate weld

N/mm2

ft

N/mm2

Fbt

Pnom

ft

N/mm2

Fbt

Fdt F2 / N

Ft Fbt + Fdt

Pnom

Ps

Fs

Fs Ft

Ps Pnom

N/mm2

M1

mm3

Mc 1.5 pyp Z

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Maximum tension in flange = Max of (case 1 , case 2)= 543.6 kN

Length of weld for tension flange = 346.4 mmWeld shear (Moment) = 1.57 kN/mmWeld shear (Horizontal force) = 0.32 kN/mm >Resultant weld shear = 1.60 kN/mm <

Weld capacity = OKAY= 1.85 kN/mm NOT OKAY> Resultant weld shear OKAY

Check for edge distanceMinimum edge distance = max. of 4d & 115 mm

= 128 mm

= m =

= 107.6 mmEdge distance provided = 130 mm OKAY

Check for anchorage length under tension Anchorage required as per BS 8110 recommendations.

Design ultimate anchorage bond stress = == 1.77

Anchorage required = =>

<= 638.3 mm OKAY

Anchorage required as per "Sabic Engineering" recommendations.Minimum required anchorage = 12 * bolt dia.

= 384 mm

= =

184 mmAnchorage length provided as per Sabic = 800 mm OKAYEngineering standard B50-F01-13

0.7 s Pw

Minimum edge distance required as per Non-ductile design (in inch) D * SQRT{fut / [73*SQRT(f'c)]}

fbu Beeta(fcu)1/2

N/mm2

LA(Ft)s

fbu PI d

Minimum embedment required as per Non-ductile design La SQRT(a2+b2)

Lap