base plate design
DESCRIPTION
fTRANSCRIPT
BASE PLATE DESIGN (BS CODE)Design for Axial, Shear Load and Moments.
Column Section 305x305x118
dc bfColumn Data 305x305x118 314.50 307.40
Tw Tf12.00 18.70
5748.38Base Plate Size
Length 600 mm
Width 600 mm
Concrete Pedestal Data
Length L' 2000 mmWidth W' 2000 mm
Area of Steel in tension zone As 1500
Anchor Bolt Data
Dia of Bolt d 24 mmAnchor Bolt Grade 8.8Number of Bolts n 8 nos
Anchor Bolt Embedded Length without Cover 400 mm
Edge Distance k 75 mm
Edge Distance h 525 mm
Center to Center distance 75 mm
150 mm
150 mm
150 mm
75 mm
Dist from flange to egde of plate 142.75 mm
Weld Data
Weld Size 10Material Stress
Yield Stress of Plate 275
Bearing strength of Plate 460
Bearing Stress of Anchor Bolt 1000
Tensile strength of Anchor bolt 560
Shear Stress of anchor bolt 375
Concrete cube strength 40
Yield stress of column Py 275
Weld Strength 215Load Data
Axial Force Fy (N) 300 KNMoment about - X Mx 400 KN-MMoment about - Z Mz KN-MShear along - X Fx 75 KNShear along - Z Fz KNResultant Shear Fs 75 KN
Step - 1Check for Tension in Bolts
b = M/N 1333.33 mmDistance to the edge of compressive block
X/2 = L/2 - b -1033.33 mm
Compression -29760C < N
Tension in Anchor Bolts, Ignore Step-1 goto Step -2
lp /(hp)
bp
mm2
Le
l1l2l3l4l5L1
Sw
Pyp N/mm2
Pbp N/mm2
Pbb N/mm2
Pt N/mm2
Ps N/mm2
fcu N/mm2
N/mm2
Pyw N/mm2
C = 0.6 fcu bp X
Step-2
Eq-1
T = C - N Eq-2
Eq-3
From Eq-1 & 2 into 3
7200 h(sq) - 7560000 h + 467500000 = 0
1 h(sq) - 1050 h 64930.56 = 0
X = 984.0165.99
X = 65.99 mmSubstituting the Value in Eq-1 and Eq-2
C 950.19 KNT 650.19 KN
Step - 3
(a) Compression Side Bending
134.75
161147.83
217890.75
217890.75 Nmm Per mm width
56.30 mm
Step -4
Tension Side Bending
59.75 mm
38848799.33 Nmm
30.69 mm
Adopted Thickness of Plate = 56.30 mm
Step - 5
Anchorage
Number of Bolts in Tension 4.00 Nos
Tension force per bolt 162.55 KN
Anchor plate Lap 1380.00 mm
Check for free edges 600.00 mm
Available 700.00 mm
Edgedistancesatified
P 6180.00 mm
0.26
0.29
0.39 use 0.39
1.00 use 1.00
fv < Vc, OK
Anchor Bolt bond along the embedded length
From BS 8110 cl. 3.12.8.3 & 3.12.8.4 the basic requirement is
5.39
C = 0.6 fcu bp X
M = T (h-hp/2) + C (hp-X/2)
M = 0.6fcu bp X (h - X/2) - N(h-hp/2)
e = L1 - 0.8sw
mc = 0.6fcuX (e-X/2)
mc = 0.6fcu e2/2
mc =
tp = required base plate thickness = (4mc /Pyp)0.5
m = L1 -k - 0.8sw
mt =T x m
tp = required base plate thickness tp = (4mt/Pypbp)0.5
nt = n/2
T/nt
Req (1.5 Le )
fv = average shear stress over effective depth = T/(P x Le) N/mm2
Vc = design concrete shear stress
Vc = 0.79/1.25 x [100As/P Le]1/3 x [400/Le]
1/4 x [fcu/25]1/3 N/mm2
If [100As/P Le]1/3 is < 0.15, use 0.15
If [400/Le]1/4 is < 1, use 1.
fb = anchorage bond stress = T / (n d L)
5.60
fb < fbu, OK
fbu = design ultimate anchorage bond stress = 0.28fcu0.5
Step - 6
Tension in Bolts
Allowable Tension per Bolt 253.21 KN
Actual Tension per Bolt 162.55 KN
Ok
Step - 7
Shear
Check if Shear is transferred through friction 0.3 N 90.00 OK
Design horizontal shear force
169.56 KN
621.52 KN
138.24 KN
138.24 KN
0.4 ps As for bolt shear or 150.00 KN
621.52 KN
138.24 KN
138.24 KN
1105.92 KN
#NAME?
Step - 8
Shear Lugs (AISC Approach)
Cylinderical Strength of Concrete fc' 32.00
Shear lug area required Al =Fs/(0.8fc') 2930.60
Assume Width of Shear lug < Base plate width 300.00 mm
Embeddment Depth 9.77 mm
Using Cantilever model of lug thickness
Thickness of Grout G 25.00
Ml 88.24 KN-mm
Section Modulus
Thickness of Shear Lug required 10.99 mm
Concrete failure width a 2000.00 mm
Edge distance of shear lug 500.00 mm
b 509.77 mm
Projected area of failure Av 1016606.71
Shear Capacity of Concrete in front of lug 1432.29 KN
Ok
Step - 9
Tension Weld Design
Tension Capacity of Flange bf Tf Py 1580.80 KN
Forces in tension flange 1235.92 KN
OK
Weld force per mm 2.05 KN/mm
Weld throat required 9.54 mm
Ftall = Pt*d2*3.14/4
Ftacc =T/nt
H = ns pss + nt pts
psAs for bolt shear or
dtp Pb for bolt bearing on base plate or
6d2fcu for bolt bearing on concretePss
dtpPb for bolt bearing on base plate
6d2fcu for bolt bearing on concrete
Pts
H = ns pss + nt pts
N/mm2
mm2
blug
dlug
Ml = V (G + dlug/2)
Ml= Pyp Z
Z = blugtlug2/4
tlug
elug
mm2
Vu = 4fc0.5 Av
Tfla = M/(dc-tf) - N x Af/Ac
wf = Tfla /(2bf-tw)
wf/pyw
BASE PLATE DESIGN (BS CODE)Design for Axial, Shear Load and Moments.
