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.