base plate design - fixed
TRANSCRIPT
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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