Base Plate - Fixed
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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 )
= =
Maximum Moment
Case ( ii )
=
T F M1
Vertical load
Case ( ii )
=
Minimum vertical load
Case ( iii )
Moment
Case ( iii )
= UC 356 x 171 x 67 = 363.4 mm = 173.2 mm = 15.7 mm = 90 kN P-42 Att II =
189
F1
=
90
kN
=
F2
=
20
kN P-40 Att II
=
M2
=
189
kN.m
Resultant horizontal shear = Length of plate = Width of plate = Thickness of plate = Assumed thickness of weld for base = plate & column connection
Fs L B t
= = = =
112 700 500 30
kN mm mm mm
s
=
12.0
mm
Design strength of the plate
=
pyp
=
265
N/mm2
Grade of concrete Modular ratio
= =
fcu
40 15
N/mm2
Diameter of bolts Grade of bolt Number of bolts
=
m db
= =
32 A36 4
mm
Tensile area of bolt Tensile capacity of the single bolt Shear capacity of the single bolt
=
N At
= = = =
644 177.47 72.68
Shear strength of bolt
=
mm2 kN kN N/mm2
ps
=
160
kN.m P-42 Att II
"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
4 - 32dia bolts
B = 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 1 of 4
Bearing pressure Case ( i ) Loading Base pressure
=
Permissible bearing pressure
Case ( ii ) Loading The eccentricity Actual eccentricity
=
=
w wp
=
e'
= =
=
0.26 0.6 fcu
> N/mm < 2
= OKAY = 24.0 N/mm2 NOT OKAY > Actual base pressure OKAY
e
L/6 116.7 M1/F1
mm >
= < = 2100.0 mm > L/6 THEREFORE TENSION OCCURS IN THE HOLDING DOWN BOLTS d1 = 0.5(d - n) + M1 / F1 2390 A1
-5250
mm
= 6 m d1 As / B = 5.5E+05
y is the solution of y3 - 3(d - d1)y2 + A1y - A1d By solving above equation Pressure
= =
y w
mm2
= 0.0 = 209.1 mm = 6 d1 F / L y (3d - y) > =
5.15 0.6 fcu
N/mm2 <
Permissible pressure
=
wp
= OKAY 2 = 24.00 N/mm NOT OKAY > Actual base pressure (w) OKAY
Case ( iii ) Loading The eccentricity
=
e'
= =
Actual eccentricity
=
mm >
= < = 9450.0 mm > L/6 THEREFORE TENSION OCCURS IN THE HOLDING DOWN BOLTS d1 = 0.5(d - n) + M2 / F2 A1
e
L/6 116.7 M2/F2
9740 mm = 6 m d1 As / B
-27300
=
y y is the solution of y - 3(d - d1)y2 + A1y - A1d By solving above equation
=
y
Pressure
=
w
= =
0.0 191.9 mm = 6 d1 F / L y (3d - y) > =
=
mm2
0.0E+00
3
Permissible pressure
2.3E+06
wp
=
5.04 0.6 fcu
N/mm2 <
H:\71146 Sabtank JPE\11\Deliverable\Calc\PIPERACK\EW piperack1\Connection\39738609.xls
OKAY 2 of 4
= 24.00 N/mm2 NOT OKAY > Actual base pressure (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
ft
= m w [(d / y) - 1] =
176.4
N/mm2 >
=
113.6
kN <
= Pnom = 177.47 > Actual force on bolt
OKAY
Case ( iii ) Loading Bolt stress
=
ft
= m w [(d / y) - 1] =
Tensile Force in bolt due to bending
Fbt
Force in bolt due to direct tension
Fdt Ft
Total tension in the bolt Tensile capacity of single bolt
176.4
N/mm2
113.6 = F2 / N = 0.0 = Fbt + Fdt
kN
=
kN
=
kN
113.6
= Pnom = 177.47 OKAY > Actual force on bolt
OKAY
Check for horizontal shear Shear capacity of one bolt
=
Actual horizontal shear / bolt Safety index Check for combined shear and tension Fs Ft + < Ps Pnom
Ps
=
72.68
Fs
= =
28.0 2.6
1.4
=
1.03
<
1.4
Check for thickness of base plate Maximum base pressure By assuming constant pressure
=
w
=
Maximum B.M. Assumed thickness of the plate Plate modulus
= = =
M1
= = =
Moment capacity
=
Safety index
t Z Mc
5.15
kN NOT OKAY 1 OKAY > < OKAY NOT OKAY OKAY
N/mm2
36.5 30 7.5E+04 = 1.5 pyp Z = 29.8 = 0.82
kN.m mm 1 mm3 > < kN.m OKAY NOT OKAY NOT OKAY
Column / base plate weld H:\71146 Sabtank JPE\11\Deliverable\Calc\PIPERACK\EW piperack1\Connection\39738609.xls
3 of 4
Maximum tension in flange
= = = = = =
Max of (case 1 , case 2) 543.6 kN 346.4 mm 1.57 kN/mm 0.32 kN/mm > 1.60 kN/mm < = 0.7 s Pw OKAY = 1.85 kN/mm NOT OKAY > Resultant weld shear OKAY
Length of weld for tension flange Weld shear (Moment) Weld shear (Horizontal force) Resultant weld shear 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{fut / [73*SQRT(f'c)]} =
107.6
mm
=
130
mm
OKAY
Check for anchorage length under tension Anchorage required as per BS 8110 recommendations. 1/2 Design ultimate anchorage bond stress = fbu = Beeta(fcu)
Anchorage required
=
LA
= 1.77 N/mm2 (Ft)s = fbu PI d
= 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 Engineering standard B50-F01-13
H:\71146 Sabtank JPE\11\Deliverable\Calc\PIPERACK\EW piperack1\Connection\39738609.xls
> < mm OKAY
mm
mm mm
OKAY
4 of 4
Design Data Assumed section Depth of section Width of section Thickness of flange Maximum Vertical load
=
D
Case ( i )
= =
Maximum Moment
Case ( ii )
=
T F M1
Vertical load
Case ( ii )
=
Minimum vertical load
Case ( iii )
Moment
Case ( iii )
= UC 356 x 171 x 67 = 363.4 mm = 173.2 mm = 15.7 mm = 90 kN P-42 Att II =
189
F1
=
90
kN
=
F2
=
20
kN P-40 Att II
=
M2
=
189
kN.m
Resultant horizontal shear = Length of plate = Width of plate = Thickness of plate = Assumed thickness of weld for base = plate & column connection
Fs L B t
= = = =
112 700 500 30
kN mm mm mm
s
=
12.0
mm
Design strength of the plate
=
pyp
=
265
N/mm2
Grade of concrete Modular ratio
= =
fcu
40 15
N/mm2
Diameter of bolts Grade of bolt Number of bolts
=
m db
= =
32 A36 4
mm
Tensile area of bolt Tensile capacity of the single bolt Shear capacity of the single bolt
=
N At
= = = =
644 177.47 72.68
Shear strength of bolt
=
mm2 kN kN N/mm2
ps
=
160
kN.m P-42 Att II
"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
4 - 32dia bolts
B = 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 1 of 4
Bearing pressure Case ( i ) Loading Base pressure
=
Permissible bearing pressure
Case ( ii ) Loading The eccentricity Actual eccentricity
=
=
w wp
=
e'
= =
=
0.26 0.6 fcu
> N/mm < 2
= OKAY = 24.0 N/mm2 NOT OKAY > Actual base pressure OKAY
e
L/6 116.7 M1/F1
mm >
= < = 2100.0 mm > L/6 THEREFORE TENSION OCCURS IN THE HOLDING DOWN BOLTS d1 = 0.5(d - n) + M1 / F1 2390 A1
-5250
mm
= 6 m d1 As / B = 5.5E+05
y is the solution of y3 - 3(d - d1)y2 + A1y - A1d By solving above equation Pressure
= =
y w
mm2
= 0.0 = 209.1 mm = 6 d1 F / L y (3d - y) > =
5.15 0.6 fcu
N/mm2 <
Permissible pressure
=
wp
= OKAY 2 = 24.00 N/mm NOT OKAY > Actual base pressure (w) OKAY
Case ( iii ) Loading The eccentricity
=
e'
= =
Actual eccentricity
=
mm >
= < = 9450.0 mm > L/6 THEREFORE TENSION OCCURS IN THE HOLDING DOWN BOLTS d1 = 0.5(d - n) + M2 / F2 A1
e
L/6 116.7 M2/F2
9740 mm = 6 m d1 As / B
-27300
=
y y is the solution of y - 3(d - d1)y2 + A1y - A1d By solving above equation
=
y
Pressure
=
w
= =
0.0 191.9 mm = 6 d1 F / L y (3d - y) > =
=
mm2
0.0E+00
3
Permissible pressure
2.3E+06
wp
=
5.04 0.6 fcu
N/mm2 <
H:\71146 Sabtank JPE\11\Deliverable\Calc\PIPERACK\EW piperack1\Connection\39738609.xls
OKAY 2 of 4
= 24.00 N/mm2 NOT OKAY > Actual base pressure (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
ft
= m w [(d / y) - 1] =
176.4
N/mm2 >
=
113.6
kN <
= Pnom = 177.47 > Actual force on bolt
OKAY
Case ( iii ) Loading Bolt stress
=
ft
= m w [(d / y) - 1] =
Tensile Force in bolt due to bending
Fbt
Force in bolt due to direct tension
Fdt Ft
Total tension in the bolt Tensile capacity of single bolt
176.4
N/mm2
113.6 = F2 / N = 0.0 = Fbt + Fdt
kN
=
kN
=
kN
113.6
= Pnom = 177.47 OKAY > Actual force on bolt
OKAY
Check for horizontal shear Shear capacity of one bolt
=
Actual horizontal shear / bolt Safety index Check for combined shear and tension Fs Ft + < Ps Pnom
Ps
=
72.68
Fs
= =
28.0 2.6
1.4
=
1.03
<
1.4
Check for thickness of base plate Maximum base pressure By assuming constant pressure
=
w
=
Maximum B.M. Assumed thickness of the plate Plate modulus
= = =
M1
= = =
Moment capacity
=
Safety index
t Z Mc
5.15
kN NOT OKAY 1 OKAY > < OKAY NOT OKAY OKAY
N/mm2
36.5 30 7.5E+04 = 1.5 pyp Z = 29.8 = 0.82
kN.m mm 1 mm3 > < kN.m OKAY NOT OKAY NOT OKAY
Column / base plate weld H:\71146 Sabtank JPE\11\Deliverable\Calc\PIPERACK\EW piperack1\Connection\39738609.xls
3 of 4
Maximum tension in flange
= = = = = =
Max of (case 1 , case 2) 543.6 kN 346.4 mm 1.57 kN/mm 0.32 kN/mm > 1.60 kN/mm < = 0.7 s Pw OKAY = 1.85 kN/mm NOT OKAY > Resultant weld shear OKAY
Length of weld for tension flange Weld shear (Moment) Weld shear (Horizontal force) Resultant weld shear 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{fut / [73*SQRT(f'c)]} =
107.6
mm
=
130
mm
OKAY
Check for anchorage length under tension Anchorage required as per BS 8110 recommendations. 1/2 Design ultimate anchorage bond stress = fbu = Beeta(fcu)
Anchorage required
=
LA
= 1.77 N/mm2 (Ft)s = fbu PI d
= 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 Engineering standard B50-F01-13
H:\71146 Sabtank JPE\11\Deliverable\Calc\PIPERACK\EW piperack1\Connection\39738609.xls
> < mm OKAY
mm
mm mm
OKAY
4 of 4
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