Scantling Calculation
Hull Geometry: Length between perpendicular L = 125 m B/L = 0.1333 H/L = .00533 CB = 0.600 Cpr = 0.614 Cx =0.977 Cs = 0.710 LWL/L = 1.0167
Particulars: Length between Perpendicular LWL = 127.1 m Breadth of ship Draft Height Block coefficient Volume of Displacement ∇ = 8530 tonnes Frame spacing a=0.5m Web frame spacing Material factor Yield strength ReH = 355 N/mm2
(H=.66D+.9)
(Ref.Sec-2, A.2)
Bottom Shell Plating: Length co-efficient, CL = 1 for L ≥ 90 m Service coefficient CRW = 1 for unlimited service range Distribution factor, CF x CD = 1 for midship Wave coefficient, Co = Wave Coefficient 1..5 300 L = 10 .75 CRW 100
(Ref.Sec-4, Table-4.1) (Ref.Sec-4, A 3)
1.5 300 125 = 10 .75 x1 100
= 8.43 nf = 0.83 for longitudinal framing (Ref.Sec-6, A.2) Probability factor, f = 1 for girders and girder systems of the outer hull (web frames, stringers, grillage systems) (Ref.Sec-4, A.3) CL = Length coefficient =1 Po = Basic external dynamic load (Ref.Sec-4, A 3) = 2.1(CB+0.7) Co × CL× f = 2.1(0.6+0.7) × 8.43 × 1 × 1 = 23.014 KN/m2 PB = Load at bottom = 10H + Po × CF = 10 × 6.66 + 23.014 × 1 = 89.614 KN/m2
(Ref.Sec-4, B-5)
Now,
perm =230/ k KN/m2
LB
pl
(Ref.Sec-6, A.2)
=230/0.72 =319.44 KN/m2 = 120/k (for L≥90m ) =166.77 KN/m2 =
2 perm
3 L
2
-0.89
Corrosion addition tk = 2.5 mm,
LB =
319.44
2
3.(0) 2 -0.89 ×166.67=171.01 N/mm2
(Ref.Sec-3, G)
tB1= 18.3x nf x a x tB2 =1.21 x a x
PB
pl
+ tk = 18.3 x 0.83 x .5 x
89 .89 + 1.5 =7.0 mm 171 .01
(Ref.Sec-6, B-1.2)
( PB .K ) + tK = 1.21×0.5× (89.614x 0.72) +2.5=6 mm
Minimum plate thickness = C1
L.K + tk = 1 x 125 0.72 + 0.25=9.73 mm
(Ref.Sec-6, B-2)
So, we accept the bottom plate thickness = 10 mm
Side Plate thickness:
Ps = 10(T-z) + Po .CF (1+z/T) (Ref.Sec-4, B-2) Where, z = vertical distance of the structure's load centre above base line [m] = D/2 = 8.73/2 = 4.365 Ps = 10(6.66-4.365) + 23.014 x 1 x (1+4.365/6.66) = 61.05 kN/m2 Side plating thickness, ts = 1.9 nf. a √(Ps .k) + tk (Ref.Sec-6, C-1) = 1.9 x 0.83 X .5 X √(61.05 X 0.72) + 2.5 =7.73 mm As plates can be thinner,so, we take the side plate thickness = 8 mm
Bilge thicknes: We have taken bilge thikness same as the bottom plate thikness,TB ie, bilge thikness = 10 mm
Flat Keel Plate: Width Of Keel Plate =800+5L = 1325 mm =1.325 m The thickness of flat plate keel should not be less than tFK =tB+2.0 =10+2.0=12mm
(Ref.Sec-6, B 5.1)
So we take the thickness of our flat plate keel as tFK=12 mm
Deck Plate: P = PD = Pressure on ship’s deck 20 T = Po × ×CD (10 Z T ) H =25.29 × (20×6.46) ×1/(10+4.925-6.46) ×9.85 =39.19 KN/m2
(Ref. Sec4, B-1.1)
The thickness of strength deck plating tE for 0.1L from the ends & between hatchways is not to be less than tE1 = 1.21*a*(Pd*k)0.5+tk = 4.71 mm tE min= (5.5+0.02xL)√k = 8.96 mm (Ref. Sec7,A-7.1) we take deck plate thickness = 9 mm Where, Z = vertical distance of the structure's load centre above base line [m] = D/2 = 9.85/2 =4.925 m CD = distribution factor = 1 H=depth of ship
Bulkhead f=
=√
p
39.24
cp √ Vertical stiffener spacing=0.712 m, Corossion addition tk = 1.5 mm
Thickness of bulkhead t = √ =0.81 √ So,we take the thickness = 5 mm
(Ref. Sec11,Table-11.1 & B-2.1)
Web frame and Side Stringers׃ P=Ps=Load on ship side = 10(T-Z) + Po ×CF (1+
(Ref. Sec4,B-2.1.1) Z ) T
= 10(6.46-4.925) +25.29×1× (1+
4.925 ) 6.46 .
= 59.92 KN/m
2
Where, Z = vertical distance of the structure's load centre above base line [m] =4.925 m T = Draft =6.46 m CF = distribution factor=1
Co= Wave Coefficient = 8.03 Po = Basic external dynamic load = 25.29 KN/m2 Web frame spacing, e = 2 m l = Length of unsupported span = same as web frame spacing =2m Section modulus, w = 0.55 × e × l2 × P × nc × K
(Ref.Sec-9, A- 5.3.1)
= 0.55 × 2 × 2²× 59.92× 1 × 0.72 = 189.83 ≈ 200cm3 Where, nc = Reduction coefficient, K = Material factor Dimension ׃T- 189x50x5
(Ref. Zulfikar.M.M. (2008).The Inland Shipping Laws & Rules, P-170)
Bottom structure (Keelson) ׃ Depth, h = 350 + 45×B
(Ref.Sec-8, B- 2.2.1)
= 350 + (45×16.15) = 1076.75 mm ha= depth of certer girder as buildt Again, double bottom depth=B/15 = 16.15/15= 1.08 m=1080 mm So, we take ha= 1.08 m h h Thickness, t = ( + 1) × K for h≤1200 (Ref.Sec-8, B- 2.2.2) ha 100 1076.75 1076.75 ( + 1) × 1080 100 = 9.95 ≈ 10 mm
=
0.72
Face plate width ≈ 10×10=100 mm Dimension ׃T- 1000× 100× 10
(Ref. Zulfikar.M.M. (2008).The Inland Shipping Laws & Rules, P-170)
Deck Web and Deck Girder׃ Section Modulus, W = c × e × l2 × P × K
(Ref.Sec-10, B 4.1)
= 0.75 × 2 × 22 × 39.19 × 0.72 = 169.3 cm3≈180 cm3 Where, P = PD = Pressure on ship’s deck =39.39 KN/m2 l = Length of unsupported span = 2m c = 0.75 for beams, girders and transverses which are simply supported on one or both ends
Dimension ׃T- 178x50x5 Bottom Longitudinal׃ Po = Basic external dynamic load = 25.29 KN/m2 a = frame spacing = 0.5 m
(Ref. Zulfikar.M.M. (2008).The Inland Shipping Laws & Rules, P-170)
f = Probability factor = 0.75 for secondary stiffening members P = PB = Load at bottom = 89.89 KN/m2 0 .5 a a 0 .5 2 ma = 0.204 × [4– ( )2] = 0.204 × [4–( ) ] = 0.191 2 l l 2 mk = 1 – Σ
l KU 0 .5 × sin2α = 1- 4 × ×(sin45)2 = 0.524 l 2
m = mk2 – ma2 = 0.238
σpr =
230 = 319.44 k
Sectional modulus, W =
83 .3
pr
× m × a × l2 × P = 11.16 cm3≈ 16cm3
Dimension ׃L – 60×40 ×5
(Ref. Zulfikar.M.M. (2008).The Inland Shipping Laws & Rules, P-164)
Side Longitudinal׃ Ps = Load on side = 10(T-Z) + Po × CF (1 +
Z ) T
(Ref.Sec-4, B.2)
= 10(6.66-4.925) + 23.143 × 1 × (1 + = 57.61 KN/m2
4.925 ) 6.66
83 .3 83.3 × m × a × l 2 × Ps = × 0.238×.5×22 ×57.61 pr 319.44 3 = 7.15 cm3 ≈ 16 cm
Section Modulus, W =
Dimension ׃L – 60 × 40 × 5
(Ref. Zulfikar.M.M. (2008).The Inland Shipping Laws & Rules, P-164)
Deck Longitudinal׃ P = PD = Pressure on ship’s deck 20 T = Po × ×CD = 46.69 KN/m2 (Ref.Sec4-, B.1) (10 Z T ) H 83 .3 83.3 W= × m × a × l2 × P D = × 0.238 × 22 ×0.5 × 46.69 = 5.8 ≈ 16cm3 pr 319.44
Dimension =L- 60×40×5
(Ref. Zulfikar.M.M. (2008).The Inland Shipping Laws & Rules, P-164)
Deck Beam׃ P = PD = Pressure on ship’s deck 20 T = Po × ×CD = 46.69 KN/m2 (Ref.Sec4-, B.1) (10 Z T ) H c = 0.75 for beam & girders Sectional modulus, Wd = c × a × l2 × k × P = 50.42≈ 43cm3
Dimension ׃L- 90× 60× 6
(Ref. Zulfikar.M.M. (2008).The Inland Shipping Laws & Rules, P-165)
Dimension of Brackets Here for bottom frames Unsupported span The section modulus of bottom and inner bottom frames should not be less than
For flanged bracket The thickness of the bracket should not be less than √
√ we take the thickness of the brackets as Now
√ √
(Ref.Sec3-,B.3.5.2.3)
The arm length of brackets should not be less than √ √
