REFERENCE
OUTPUT
CALCULATIONS TOWER BASE CONNECTION Check base plate and bolts supplied by Tower manufacturer considering typical pedestal with maximum reaction loads
700x700 base plate 900x900x300 column plinth pile cap Anchor bolt
ELEVATION
c
b
a PLAN
DESIGN DATA EN 1993‐1‐5:2005
Grade of Base plate
Table 3.1
Design strength (thickness ≤ 40 mm), py =
275 N/mm2
Table 3.1
Bearing capacity of base plate, Pbs =
460 N/mm2
Grade of Bolt =
8.8
EN 1993‐1‐8:2005
Shear strength of bolts, Ps =
640 N/mm2
Table 3.1
Tension strength of bolt, Fub =
800 N/mm2
REFERENCE
S275
CALCULATIONS CASE 1: COMPRESSION LOADING
EN 1993‐1‐8:2005
BASE PLATE DESIGN
Table ‐A1
At Node 1, maximum reaction forces
OUTPUT
Axial, N =
574 kN
Lateral, Fh =
38.00 kN
Moment due to lateral load, M = Fh*La
43.32 kNm
At ULS with Factor of Safety, FS = Axial Load, Nult = N*FS
774.9 kN
Lateral Load, Fh =
51.3 kN
Moment, Mult = M*FS EN 1993‐1‐1:2005
1.35
69.26 kNm
Bearing Pressures Check base plate dimensions: 700 mm x 700 mm Length, L =
700 mm
Width, B =
700 mm
Bearing Pressures 1.58 N/mm2
Pb = Nult/L*B = Steel Designers' Manual, 6th Edition (2003)
Bearing Strength of concrete (allowable), fp = 0.6 x fcK= Pb (actual) < fp (allowable)
15 N/mm2 Bearing Pressure OK
Plate Thickness Nominal gusset plates shall be provided for fixing;
700
700
3 0 m m a n c h o r b o lt 1 5 0 x 1 5 0 a n g le
FIG. 6: BASE PLATE PLAN Consider 1 mm strip of base plate as shown in Fig. 6 Lever arm, ℓ =
275 mm
Bearing pressure, Pb = w =
1.58 N/mm2
Maximum moment on base plate, M = wℓ2/2 = Bearing Strength of concrete, fp = 0.6 x fck= REFERENCE
CALCULATIONS
59797.77 Nmm 15.00 N/mm2 OUTPUT
Using the effective area method, Area of plate required Areq= Nult/fp = 774.9x1000/0.6xf ck Compute actual bearing area as a function of outstand, c Refer to figure A5 above, (2c+15)(150‐15) + (2c+15)(2c+150) = 57400
51660 mm
2
4c2 + 600c ‐ 53125 = 0 solving the equation gives c = 62.5
62.5 mm
Plate thickness Steel Designers' Manual, 6th Edition (2003)
Thickness of base required, tp = c(3w/py)0.5 were w=fp Provide 40 mm thick base plate BOLTS SELECTION Bolt Design for Shear; Check 8No. 30 mm Ø Anchor bolts Shear force to be resisted, Fh =
EN 1993‐1‐1:2005
Shear capacity of bolts, P = PsAb
Bolt Data Steel Designers' Manual, 6th Edition (2003)
Ps = shear strength per bolt
25.28 mm 40 mm
30 mm 51.3 kN
640 N/mm2
For Grade 8.8 bolts, Ps = n = no. of bolts = Effective Diameter (Excluding 3mm for shank) = Db = Ab = Effective Area of bolts = πDb²/4 = Total shear capacity of bolts, n*P =
8 No. 27 mm 572.6 mm2 2931.7 kN Provide 8No. 30 mm bolts
Bolt Design for Tension; Minimum tensile strength, fyk =
EN 1993‐1‐8:2005
Ft = Ft per bolt = Effective Area of bolts, As = πDb²/4 =
800 N/mm2 774.9 kN 96.9 572.6 mm2
Actual bolt tensile load, T = Ab*ft =
55.46 kN
Allowable tensile load, Ft,Rd = K2*Fub AS / ym2 572.6 mm2 800 N/mm2
Table 3.4
Effective Area of bolts, As = πDb²/4 =
Table 3.1
Fub= Bolt tensile strenght =
Table 3.4
K2 =
Table 2.1
Ym2 = partial factor of safety =
1.25
Allowable bolt tensile stress, Ft,Rd=
330 kN
0.9
Steel Design Guide, John Minimum embedment length, Lm = 17d> 100mm = T Dewolf. (2003)
Adopt Bolt length, L =
510 mm 900.00 mm
Provide a Total Bolt Length of 900 mm CASE 2: TENSION LOADING Consider tension load on base due to load reversal Max. Tension = Max. Compression Tensile force, Ft = Nult = Table 34
Tension carried by each bolt = Ft/8 =
EN 1993‐1‐1:2005
Tension capacity of bolts, Pt = 0.8FubAt =
774.90 kN 96.86 kN 366.46 kN
Bolt Data
Total tensile capacity = n*Pt =
2932 kN
Check combined shear and tension; EN 1993‐1‐1:2005
Combined shear and tension; Fh/Ps + Ft/Pt ≤ 1.4 where; Ps = shear strenght of bolt =
2931.70 N/mm2
Pt = Tensile capacity of bolt =
2931.71 kN
Ft = Tensile Force force to be resisted = Fh= shear force to be resisted= Fh/Ps + Ft/Pt ≤ 1.4
774.90 kN 51.30 kN 0.28
OK
