875-E399-01 R1 ABS Structural Calculations.pdf

REV ZONE DRAWN CHKD

1 - TEL

GENERAL NOTES

RESERVATIONS

NO.

SIZE REV

A 1

DATE: SCALE: SHEET 1 OF 15

DRAWN BY: CHECKED BY:

BARGE MODIFICATION

TEL

NTS

GWB

DATE: BHGI PROJ #:

MARINE DESIGN CENTER

APP'D BY: PROJECT NO. DWG. NO.

3219 875-E399-01

ABS STRUCTURAL CALCULATIONS14JUN21 21709

DESCRIPTION DWG NO. REV.

REFERENCES

1. Vessel is designed to ABS Rules for Building and Classing Steel Barges.

REVISION HISTORY

DESCRIPTION DATE

20OCT211. Removed crane mat load reduction

2. Changed deck beam to 16"x4"

3. Changed longest XVRS span to 12.5'

4. Added Appendix B

5. Added plate thickness mitigation

JTH /

TEL

LOADS

GBP = 17928 psf Lift 4, Critical Angle p = 17928 psf For use in later calculations

Structure is designed around the loads imparted by crane through its treads onto the deck. At direction of the client, 3D Lift Plan was used to calculate these ground bearing pressures for the

Manitowoc 16000 with 6 load load cases at 2 degrees of list. These are the unburdened crane, the four client defined lifts, and an additional worst case lift that is the heaviest lift at the maximum radius the crane is rated to lift it to. Before these loads are transmitted to the barge structure, it goes through crane mats. However, as these crane mats are longitudinally aligned, there is no reduction in loads applied to the barge deck. All outputs are hard ground bearing pressure, and the maximum, from lift 4 at the critical angle, is 17928 psf, is used for design.

Member h h b t l s In tank? seffective ratio NA INA SMPLATE SMSTIFFENER SMMIN hreq'd treq'd SMreq'd ABS Barge Rules JAN 2020 SF in in in in in in (yes/no) in in in in in in in in

SM or t

ABS RULES AS WRITTEN

Deck Plate PL 1 1/8 1.15 3-2-3/1.5, 3-2-3/1.1.4 0.98 FAIL *

Deck Beam FP 16 4 1 30 25 no 10.00 16.0 6.68 1156.1 173.0 110.7 110.7 2.5 12.4 3-2-4/3.9, 5-1-1/3.3 8.90 PASS

Deck Longitudinal Girder FP 22 6 1 75 30 no 25.00 22.0 7.18 3853.1 536.8 241.6 241.6 6.3 97.3 3-2-4/3.9, 5-1-1/7 2.48 PASS

Deck Transverse Girder Typical Span FP 34 10 1 150 75 no 50.00 34.0 9.67 15983.1 1653.3 627.8 627.8 12.5 972.7 3-2-4/3.9, 5-1-1/7 0.65 FAIL **

Deck Transverse Girder Longest Span FP 34 10 1 150 75 no 50.00 34.0 9.67 15983.1 1653.3 627.8 627.8 12.5 972.7 3-2-4/3.9, 5-1-1/7 0.65 FAIL **

DISTRIBUTING LOAD ON PLATE

Deck Plate PL 1 1/8 1.11 3-2-3/1.5, 3-2-3/1.1.4 1.01 PASS

DISTRIBUTING LOAD ON TRANSVERSES

Deck Transverse Girder Typical Span FP 34 10 1 150 75 no 50.00 34.0 9.67 15983.1 1653.3 627.8 627.8 12.5 311.3 3-2-4/3.9, 5-1-1/7 2.02 PASS

Deck Transverse Girder Longest Span FP 34 10 1 150 75 no 50.00 34.0 9.67 15983.1 1653.3 627.8 627.8 12.5 311.3 3-2-4/3.9, 5-1-1/7 2.02 PASS

FIRST PRINCIPLES OF TRANSVERSES

Deck Transverse Girder Typical Span FP 34 10 1 150 75 no 50.00 34.0 9.67 15983.1 1653.3 627.8 627.8 12.5 487.6 - 1.29 PASS

Deck Transverse Girder Longest Span FP 34 10 1 150 75 no 50.00 34.0 9.67 15983.1 1653.3 627.8 627.8 12.5 487.6 - 1.29 PASS

** This assumes the full preasure load of 17928 PSF over the full area of the deck supported by the deck transverse, which is grossly conservative and not within the design scope of the barge. Two less conservative methods were used to show adequacy of the structure, one distributing the load imparted by the crane over the entire span of the member, and the other going by first principles. Both methods show the members pass with margin (See DISTRIBUTING LOAD ON TRANSVERSES and FIRST PRINCIPLES OF TRNASVERSES).

* This assumes the full preasure load of 17928 PSF over the full area of the deck supported by the deck transverse, which is grossly conservative and not within the design scope of the barge. A less conservative method was used to show the adequacy of the structure by taking the greatest average load imparted on a single panel. This method shows the plate passes (See DISTRIBUTING LOAD ON PLATE).

Barge Rules January 2021

Part 3: Hull Construction and Equipment

Chapter 2: Hull Structures and Arrangements

Section 1: Longitudinal Strength (3-2-1)

1. General

3. Longitudinal Hull Girder Strength

3.1 Strength Standard

The required hull girder section modulus SMR amidships, to the deck and bottom is to be obtained from the following equation:

SM R = KSM b in where

K = 0 . 629 + M s /(f p SM b ) but is not to be taken less than 1.0

M s = maximum still-water bending moment in the governing loaded or ballasted condition in Ltf-ft

When still-water bending moment calculations are not submitted, K will be taken as 1.0 f p = 11.33 Ltf/in

SM b = C 1 C 2 L 2 B(C b + 0 . 7) in

-ft

C 1 = 4.11 100 ≤ L < 150 ft

16 . 33(L/328)

− 15 . 47(L/328) + 7 . 77 150 ≤L < 310 ft

10 . 75 − ((984 − L/328))

1.5 310 ≤L≤ 984 ft

10.75 984 < L < 1148 ft

10 . 75 − ((984-L)/328) 1.5

C 2 = 1.44 ×10 -4

L = length, in feet, as defined in 3-1-1/3

B = breadth, in feet, as defined in 3-1-1/5

C b = block coefficient, as defined in 3-1-1/31

K = 1

M s = 12148 Lt-ft f p = 11.33 Ltf/in

SM b = 4316.909 in

-ft

Barges intended to be classed for unrestricted ocean service are to have longitudinal strength in accordance with the requirements of this Section. A breadth to depth ratio up to approximately 4 is acceptable in association with longitudinal bulkheads and trusses arranged at suitable intervals. A breadth to depth ratio greater than 4 can be specially considered provided the transverse strength of the barge is adequate.

C 1 = 4.408608

C 2 = 0.000144

L = 200 ft

B = 100 ft

C b = 1

SM R = 4316.909 in

-ft

Section 3: Decks (3-2-3)

1. Deck Plating

1.1 Deck Plating Thicknesses

1.1.1

1.1.1 (a) t = 0.01s + 0.09 in. for s <= 30 in.

t = 0.0066s + 0.192 in. for s > 30 in.

1.1.1 (b) t = s(L + 150) / (25L + 19950) in.

1.1.2

1.1.2 (a) t = 0.009s + 0.095 in. for s <=30 in.

t = 0.006s + 0.185 in. for s > 30 in.

1.1.2 (b) t = s(L + 160) / (26L + 28482) in. for L <= 600 ft t = 80s / (5300 - 1.1L) in. for L > 600 ft

The strength deck plating is to be of the thickness necessary to obtain the required hull girder section modulus. The thickness outside the longitudinal line of the openings, or completely across the barge where there are no centerline openings, is not to be less than obtained from the equations in 3-2-3/1.1.1 or 3-2-3/1.1.2 below, except within steel deckhouses, where the plating may be 1 mm (0.04 in.) less. The requirements of equations 3-2-3/1.1.1(b) and 3-2-3/1.1.2(b) are to extend over 0.4L amidships, beyond which they may be gradually reduced in the manner permitted for strength deck area in 3-

2-1/3.3. The thickness requirements of 3-2-3/1.1.1(a) and 3-2-3/1.1.2(a) are to extend over the 0.8L amidships.

Thickness of Strength Decks with Transverse Beams in Barges Over 300 ft up to 600 ft in Length

Thickness of strength decks with transverse beams in barges over 300 ft up to

600 ft in length is to be not less than 3-2-3/1.1.1(a) and 3-2-3/1.1.1(b) below:

Thickness of Strength Decks with Longitudinal Beams, Strength Decks with

Transverse Beams in Barges 300 ft or Less and Forecastle Decks in Barges over

400 ft

Thickness of strength decks with longitudinal beams is to be not less than 3-2-

3/1.1.2(a) and 3-2-3/1.1.2(b), below. Thickness of strength decks with transverse beams in barges 300 ft or less is to be not less than 3-2-3/1.1.2(a) and 3-2-3/1.1.1(b). Thickness of forecastle decks in barges over 400 ft is to be not less than 3-2-3/1.1.2(a), below.

1.1.3 t = 0.01s + 0.035 in. for s <= 30 in.

t = 0.0067s + 0.134 in. for s > 30 in.

1.1.4 Thickness of Platform Decks in Enclosed Cargo Spaces

t = 0.00218s √h + 0.06 in. but not less than 0.20 in.

where p = uniformly distributed deck loading, in lbf/ft

1.1.5 Thickness of Enclosed Platform Decks not Intended for Cargo t = 0.0058s + 0.04 in. but not less than 0.18 in.

where

L = length of the barge as defined in 3-1-1/3, in ft s = spacing of the beams, in in.

1.3 Plating within Line of Openings

1.5 Plating In Way of Deck Cargo

s = 25 in p = 17928 psf h = 398.40 ft t = 1.148 in

By inspection, the crane load will be driving the design of the deck thickness, therefore equation 3-2-3/1.1.4 is used. Plate not in way of the crane is of the same thickness or is original to the barge and therefore is assumed to be sufficient.

Thickness of Exposed Strength Decks within Line of Openings, Forecastle Decks in Barges of 400 ft in Length or Less h = tween deck height in feet. When a design load is specified, h is to be taken as p/45 ft.

Within the longitudinal line of openings, the thickness of exposed strength deck plating is to be not less than obtained from the equation in 3-2-3/1.1.3 amidships. At the forward and after ends it is to be as required by 3-2-3/1.7. Within steel deckhouses, the plating may be of the thickness obtained from the equation in 3-2-3/1.1.4.

Where cargo is to be carried on deck, the deck thickness is not to be less than obtained from the appropriate equations in 3-2-3/1.1.1, 3-2-3/1.1.2 or 3-2-3/1.1.3, nor is the deck thickness to be less than is required for platform decks in enclosed cargo spaces per 3-2-

3/1.1.4.

1.7 Deck Plating at Ends

t = 0.00036L + 0.0036s + 0.11 in.

where

L = length of the barge as defined in 3-1-1/3, in ft s = spacing of the beams, in in.

The thickness of the deck plating forward or aft of 0.1L from the ends is not to be less than 0.01 in. for each inch of frame spacing. In way of deck cargo, the requirements of 3-

2-3/1.5 also apply. Exposed freeboard deck plating forward of 0.05L from the forward perpendicular is also not to be less in thickness than given by the following equation:

Part 5: Specific Barge Types

Chapter 2: Dry Cargo Barges

Section 1: Deck Cargo Barges (5-1-1)

1. General

3. Deck Beams

3.1

SM = 0.0041c h s l where k = SM R Y / I A

Y = distance, in ft, from the neutral axis to the deck at side

I A = hull girder moment of inertia of the barge amidships s = spacing of beams, in ft

L = length of the barge as defined in 3-1-1/3

SM R = hull girder section modulus amidships required by 3-2-1/3.1, in in

-ft

The values of I A and Y are to be those obtained using the area of the longitudinal beams given by the above equation.

l = unsupported span in meters (feet) as shown in 5-1-1/15.11 FIGURE 1, 5-1-

1/15.11 FIGURE 2 and 5-1-1/15.11 FIGURE 3. Transverse beams fitted with end brackets having a slope of approximately 45° and a thickness given in 5-1-

1/3.1 TABLE 1, the length l may be measured to a point on the bracket equal to 25% of the length of the bracket.

h = 0.02L + 2.5 feet

= 9.5 feet where L < 350 feet where L >= 350 feet

The requirements of this Section are generally intended to apply to single deck cargo barges of the type illustrated by 5-1-1/15.11 FIGURE 1,5-1-1/15.11 FIGURE 2 and 5-1-

1/15.11 FIGURE 3.

Where structural arrangements are not provided for in these Rules, the ABS Rules for

Building and Classing Marine Vessels (Marine Vessel Rules) are to apply.

c = 0.585 for transverse beams, for longitudinal beams of platform decks and between hatches at all decks

= 1 / (1.709 - 0.651k) for longitudinal beams of strength decks and the effective lower decks

Each deck beam in association with the plating to which it is attached, is to have a section modulus SM not less than that obtained from the following equation:

3.3

SM = 0.0041c h s l where c = 0.585 h = p/45 ft p = uniformly distributed deck loading in lbf/ft p = 64 x d x C b lbf/ft d = draft as defined in 3-1-1/11 in ft

Cb = block coefficient as defined in 3-1-1/31 s = as defined in 5-1-1/3.1, in ft l = as defined in 5-1-1/3.1, in ft c = 0.585 p = 17928 psf h = 398.40 ft s = 2.08 ft l = 2.5 ft

SM = 12.44 in

3.5

3.7

Where cargo is to be carried on deck, the uniformly distributed deck loading is to be indicated on the drawings and the transverse or longitudinal deck beams, in addition to complying with 5-1-1/3.1, are also to comply with the following:

Where the uniformly distributed deck loading is not indicated on the drawings of flat top single deck cargo barges the following loading is to be assumed:

Where the space below the deck is a ballast or deep tank, the longitudinal or transverse deck beams, in addition to complying with the requirements of 5-1-1/3.1 and 5-1-1/3.3, are also to satisfy the requirements of 5-2-1/7, except that the head, h, may be taken to a point located at two-thirds of the distance from the top of the tank to the top of the overflow or 3.0 feet, whichever is greater.

The welded connections of the deck stiffeners to their deep supporting frame members are to be checked with contributory area loading attributed to uniform or isolated deck cargo loads.

By inspection, the crane load will be driving the design of the deck beams, therefore equation 5-1-1/3.3 is used with a known uniformly distributed pressure.

5. Container Loading

SM = M / f where f = permissible maximum bending stress as given in 5-1-1/5 TABLE 2

Ltf/in

A = F / q in where

F = shearing force at the point under consideration, in Ltf q = allowable average shear in the web, not to exceed 6.7 Ltf/in

Where it is intended to carry containers, the exact locations of the container pads and the maximum total static load on the pads are to be indicated on the plans. Where the pads are not in line with the supporting structures, headers are to be provided to transmit the loads to these members.

Each deck member intended to support containers is to have a section modulus SM, in cm3 (in3 ), not less than obtained from the following equation:

M = maximum bending moment due to maximum static container loading, in

Ltf-in.

In determining the maximum bending moment, members may be considered fixed-ended provided the member is continuous over the adjacent spans or is effectively attached to a bulkhead stiffener or frame. Where this is not the case, the member is to be considered simply-supported. Where weather deck containers are supported by pedestals, a section modulus is to be determined for the deck beam in accordance with 5-1-1/3.1, except that the head, h, is to be the distance between the deck and the underside of the container, but need not be greater than 50% of the appropriate h value in 5-1-1/3.1. This section modulus is to be added to the above required section modulus.

The net sectional area, A , of the web of the member, including effective brackets where applicable, is to be not less than obtained from the following equation:

Effective longitudinal members

Transverse members and longitudinal members inside the line of openings

TABLE 2

Values of f (Ordinary-strength steel)

7. Deck Girders, Chords and Transverses

SM = 0.0025 c h s l where c = 1.0 (1.1 for channel construction. See 5-1-1/15.11 FIGURE 1.)

s = mean breadth of the area of deck supported, in feet c = 1 h = 398.40 ft s = 2.50 ft Longitudinal Girder

6.25 ft Transverse Girder, typical

6.25 ft Transverse Girder, longest span l = 6.25 ft Longitudinal Girder

12.50 ft Transverse Girder, typical

12.50 ft Transverse Girder, longest span

SM = 97.27 in 3 Longitudinal Girder

972.66 in 3 Transverse Girder, typical

972.66 in 3 Transverse Girder, longest span l = unsupported span, in feet, as shown in 5-1-1/15.11 FIGURE 1, 5-1-1/15.11

FIGURE 2 and 5-1-1/15.11 FIGURE 3. Deck transverses fitted with end brackets having a slope of approximately 45° and a thickness given in 5-1-1/3.1 TABLE

1, the length l may be measured to a point on the bracket equal to 25% of the length of the bracket.

h = head as required by 5-1-1/3 or 5-1-1/3.3, whichever is greater, for the deck beams supported, in feet

Each deck girder, chord and transverse, in association with any plating to which it is attached, is to have a section modulus SM not less than that obtained from the following equation:

VESSEL SECTION MODULUS

Member Number b h y Area y*Area ILOCAL INA

Bottom Shell 1 1188 0.5 0.25 594 148.5 12.375 1852033

Side Shell 2 0.5 132 72 66 4752 95832 16710.31

Long'l BHD 37.5' OCL 2 0.375 144 72 54 3888 93312 13672.08

Long'l BHD 25' OCL 2 0.375 144 72 54 3888 93312 13672.08

Long'l BHD 12.5' OCL 2 0.375 134.375 67.1875 50.39063 3385.62 75823.78 6207.888

CL Long'l BHD 1 0.375 134.375 67.1875 50.39063 3385.62 75823.78 6207.888

Deck Plate 2 294.375 0.5 144.25 147.1875 21231.8 3.066406 1144016

SUM 1387.547 77824.95 792401.9 4246798

Area 1387.547 in

Neutral Axis 56.08816 in

I 5039200 in

34994 in -ft

SMTOP 57321 in

4777 in -ft PASS

SMBOTTOM 89844 in

7487 in -ft PASS

Required SM 4316.909 in -ft

SM calculated only using continuous decks and bulkheads. Notable omissions include the lowered crane deck with its corresponding stiffeners as these do not remain consistent for the entire length of the vessel. This creates a very conservative estimate.

DISTRIBUTING LOAD ON PLATE

p = design pressure pl = lower pressure pe = effective pressure l = length of pressure along the treads s = unsupported length of plate panels he = effective head t = required plate thickness pl = p (l - s) / l pe = (p + pl) / 2 he = pe / 45 p = 17928 psf l = 15.26 ft from lift 4, critical case (see Appendix A) s = 25 in pl = 15480 psf

Pe = 16704 psf he = 371.20 ft

ABS STEEL BARGES 3-2-3/1

s = 25 in h = 371.20 ft t = 1.11 in

This method finds the greatest average pressure and therefore the effective head as if the load from the treads on the panel was evenly distributed. Then, ABS rules are used using this reduced pressure.

The panel is 25 inches long, and assumes the peak is at one end.

DISTRIBUTING LOAD ON TRANSVERSES

pe = effective pressure he = effective head x = distance along transverse pressure is acting l = span of the member pe = p x / l he = pe / 45 p = 17928 psf x = 4.00 ft l = 12.50 ft

12.50 ft pe = 5736.96 psf

5736.96 psf he = 127.49 ft

127.49 ft

ABS STEEL BARGES 5-1-1/7

c = 1 h = 127.49 ft Transverse Girder, typical

127.49 ft Transverse Girder, longest span s = 6.25 ft Transverse Girder, typical

6.25 ft Transverse Girder, longest span l = 12.50 ft Transverse Girder, typical

12.50 ft Transverse Girder, longest span

SM = 311.25 in 3 Transverse Girder, typical

311.25 in 3 Transverse Girder, longest span

This method finds the effective pressure and therefore the effective head as if the load from the treads on the members was evenly distributed along the entire span of the member, rather than concentrated at the ends. Then, ABS rules are used using those reduced heads.

FIRST PRINCIPLES OF TRANSVERSES

s = deep transverse girder spacing x = distance along transverse pressure is acting

W = force acting on transverse l = span of the member

Sy = Yield strength,ssumed to be 34 ksi

Sallowable = allowable stress, 60% of yield a = x b = l - x p = 17928 psf s = 6.25 ft x = 4 ft

W = 448200 lb

ROARK, CHAPTER 8, TABLE 3, CASE 32

Both ends fixed, Intermediate load a = 48 in l = 150 in Transverse Girder, typical

150 in Transverse Girder, longest span b = 102 in Transverse Girder, typical

102 in Transverse Girder, longest span

M = 9947889 in-lb Transverse Girder, typical

9947889 in-lb Transverse Girder, longest span

Sy = 34000 psi

Sallowable = 20400 psi

SMREQ'D = 487.64 in

Transverse Girder, typical

487.64 in 3 Transverse Girder, longest span

This method finds the total force of the crane acting on a deep transverse girder, for conservatism places it as a point load at the far end of crane dunnage tread, and back solves for a required section modulus to not exceed 60% of yield.

Appendix A

875-E399-01

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A-7

Appendix B

875-E399-01

Crane Stop and Tie-Down Calculations

B-1

Crane Acceleration Calculations - In Transit

Principal Particulars d = 3.1 ft Principal particulars drawn fron Rhino 3D hull model and Orca run.

GM = 275.6 ft

B = 100.0 ft

D = 12.0 ft

LOA = 200.0 ft

LCF = 105.8 ft

ABS Guide for Certification of

Container Securing Systems

Section 6: Securing System Design Principles

3 Design Loads

3.1 General

3.3 Wind Loads

3.3.1 Wind Load

Wind forces are to be applied to exposed containers. The wind pressure, PW, shall be taken as:

P w = 0.0101 LT/ft

Pw = 0.0101 LT/ft

L = 36.75 ft Length of crane (side facing wind) h = 10.00 ft Crane Body Height

A= 367.50 ft

Fw = 3.71 LT

Although the barge is an inland vessel, loads were needed for tie down and end stop calculations. Therefore

ABS Guide for Certification of Container Securing Systems 6-3 is used as a general guide. These are dynamic transit loads, therefore no static lift load is included.

The wind load is assumed equally distributed over the side of the container. The vertical center of pressure should be taken at the mid height of the container, and the longitudinal center of pressure should be taken at the mid length of the container.

The basic loads to be taken into account in container securing calculations include gravitational forces, dynamic forces associated with ship motions, wind forces, and lashing or other securing forces. Sea loads and green water impact are not explicitly considered in the securing system design criteria. Adequate protection from green water impact shall be provided.

B-2

3.5 Design Ship Motions

3.5.1 Ship Conditions

GM = transverse metacentric height for the actual load condition, in ft. Where calculations are carried out for representative conditions for presentation in the Cargo Securing

Manual, GM values should be evaluated over the expected operating range.

d = draft to the summer load line, in ft d = 3.14 ft

GM = 275.65 ft

3.5.2 Ship Motions

3.5.2(a) Roll Motion. The natural roll period (full cycle) is to be obtained from the following equation:

T R = sec where k 1 = 1.104 k r = roll radius of gyration, in ft, and may be taken as 0.40B

GM = transverse metacentric height, in ft

B = 100.00 ft k1 = 1.10 kr = 40.00

TR = 2.66 s

For service without restrictions, the accelerations and loads on containers are to be determined from the ship motions. The formulas for ship motions and accelerations assume that parametric rolling is avoided, either through The formulas for ship motions and accelerations assume that parametric rolling is avoided, either through design or through vessel operations. For more information on parametric rolling, see the ABS Guide for the Assessment of Parametric Roll Resonance in the Design of Container

Carriers. In particular, these formulas do not account for the extreme roll motions and simultaneous occurrence of extreme roll and pitch induced accelerations that may occur in head sea parametric rolling.

B-3

The roll angle (single amplitude) is to be obtained from the following equation:

θ = deg where

B = molded breadth of the vessel, in ft

For vessels with bilge keels

C = 0.75 if T R ≥ 18 sec

C = 0.75 + 0.10 (18 – T R ) if T R < 18 sec, but need not be taken greater than 0.90

For vessels without bilge keels

C = 1.00

B = 100.00 ft

C = 1.00 θABS = 29.88 ° θImmersion = θImmersion = 10.05 °

For vessels with active stabilizing systems, C may be specially considered.

The roll center, R CTR , is to be taken at the vertical center of gravity of the vessel, measured in ft above baseline. When the calculated vertical center of gravity of the vessel is not submitted, R CTR may be estimated from the following formula:

R CTR = ft above baseline where

D = molded depth at side, ft d = draft as defined in 6/3.5.1 d = 3.14 ft

D = 12.00 ft

RCTR = 4.57 ft above BL

Note: the ABS roll angle is calculated for seafaring vessels. To find a more realistic value, the barge's angle of deck immersion was found. All following references to θ will use this deck immersion angle. As period is independent of amplitude, the roll period will remain the same.

10340

� + 246 tan��

�/2

B-4

3.5.2(b) Pitch Motion. The natural pitch period (full cycle) is to be obtained from the following equations:

T P = 7 + 0.00375 × (L – 600.4) sec for L in ft

The single pitch amplitude to be taken as:

φ = 7 deg where L ≤ 394 ft φ = 6 deg where 394 ft < L < 902 ft φ = 5 deg where L ≥ 902 ft where

L = length between perpendiculars, in ft

LOA = 200.00 ft

TP = 5.50 s φABS = 7.0 ° φImmersion = φImmersion = 5.06 °

The pitch center of the vessel P CTR is to be taken at the longitudinal center of flotation. When the calculated longitudinal center of floatation is not submitted, P CTR may be estimated as 0.45L forward of the aft perpendicular.

PCTR = LOA - LCF ft

PCTR = 94.24 ft fwd of AP

Note: the ABS pitch angle is calculated for seafaring vessels. To find a more realistic value, the barge's angle of deck immersion was found. All following references to φ will use this deck immersion angle. As period is independent of amplitude, the pitch period will remain the same.

tan��

�/2

B-5

3.7 Accelerations

Containers and their securing systems shall be capable of withstanding the forces generated by the following load combinations:

Condition A: The maximum roll condition generating maximum across-the-deck accelerations, expected in quartering stern or beam seas.

Condition B: The maximum pitch condition generating maximum normal-to-deck accelerations, expected in head or near head seas.

The following definitions apply to both the Condition A and the Condition B load combinations:

x c = longitudinal distance to the center of gravity of the container, in ft, forward of the aft perpendicular abs(x c -P CTR ) = absolute value of the longitudinal distance from the vessel’s pitch center to the center of gravity of the container, in ft y C = transverse distance to the center of gravity of the container, in ft, from the vessel’s centerline abs(y c ) = absolute value of the transverse distance from the vessel’s centerline to the center of gravity of the container, in ft z C = vertical distance to the center of gravity of the container, in ft, from the vessel’s baseline abs(z C -R CTR ) = absolute value of the vertical distance from the vessel’s roll center to the center of gravity of the container, in ft a 0 = common acceleration parameter, in g’s

= 0.2012 k C = 0.0214 for x C , y C , and z C in ft k 3 = force factor accounting for longitudinal position of container stack, where

= for x C < 0.2L

= 0 for 0.2L≤x C ≤0.7L

= for x c >0.7L

The designer is to ensure that the stowage system satisfies all of the strength criteria for both Condition

A and Condition B accelerations. For conventional lashed systems on the deck with containers having properties as listed in Section 3, Table 3, Condition A governs and Condition B need not be evaluated.

For containers stowed in cell guides, Condition B governs and Condition A need not be evaluated. For other configurations, including block stowage in the holds of bulk carriers, Condition A and Condition B should be evaluated.

0.5 × ( 0.2� − ��

0.2�

0.7 × ( �� − 0.7�

0.3�

B-6

Lift Position 1 2 3 4 xc = 82.875 20.125 30.58 51.5 ft abs(xc-PCTR) = 11.36 74.11 63.66 42.74 ft yC = 0 0 0 0 ft abs(yc) = 0.00 0.00 0.00 0.00 ft zC = 14.72 14.72 14.72 14.72 ft abs(zC-RCTR) = 10.15 10.15 10.15 10.15 ft a0 = 0.20 g's kC = 0.02

1 2 3 4 k3 = -0.54 0.25 0.12 -0.14 for xC<0.2L

0.00 0.00 0.00 0.00 for 0.2L<=xC<=0.7L

-0.67 -1.40 -1.28 -1.03 for xC>=0.7L

Highlighted cells indicate which k3 will be used, based on location of crane lift position.

As the crane is evaluated at four different locations, all four must be considered here. Calculations are similar for all four locations.

B-7

3.7.1 Condition A – Roll and Heave

A T = a GT + k C a RT + (1 + k 3 )a 0 sin θ in g’s

The maximum vertical acceleration is obtained from the following equation:

A VMAX = a GRV + k C a RV + (1 + k 3 )a 0 cos θ in g’s

The minimum vertical acceleration is obtained from the following equation:

A VMIN = a GRV – k C a RV + (1 – k 3 )a 0 cos θ in g’s

A VMIN is not to be taken greater than 1.0 where a GT = transverse static gravitational acceleration component, in g’s

= sin θ a RT = transverse roll acceleration component, in g’s a GRV = vertical static gravitational acceleration component, in g’s

= cos θ a RV = vertical roll acceleration component, in g’s

Lift Position 1 2 3 4 aGT = 0.17 0.17 0.17 0.17 g's aRT = 14.42 14.42 14.42 14.42 g's aGRV = 0.98 0.98 0.98 0.98 g's aRV = 0.00 0.00 0.00 0.00 g's

AT = 0.52 0.47 0.52 0.52 g's

AVMIN = 1.00 1.00 1.00 1.00 g's

The transverse and vertical accelerations at any point are to be obtained from the following formulas.

The longitudinal accelerations are taken as zero for this condition. The transverse acceleration is obtained from the following equation:

θ θ

B-8

3.7.2 Condition B – Pitch and Heave

The longitudinal and vertical accelerations at any point are to be obtained from the following formulas.

The transverse accelerations are taken as zero for this condition.

The longitudinal acceleration is obtained from the following equation:

A L = a GL + k C a PL + a 0 sin φ in g’s

The maximum vertical acceleration is obtained from the following equation:

A VMAX = a GPV + k C a PV + a 0 cos φ in g’s

The minimum vertical acceleration is obtained from the following equation:

A VMIN = a GPV – k C a PV + a 0 cos φ in g’s

A VMIN is not to be taken greater than 1.0 where a GL = longitudinal static gravitational acceleration component, in g’s

= sin φ a PL = longitudinal pitch acceleration component, in g’s a GPV = vertical static gravitational acceleration component, in g’s

= cos φ a PV = vertical pitch acceleration component, in g’s

Lift Position 1 2 3 4 aGL = 0.09 0.09 0.09 0.09 g's aPL = 1.70 1.70 1.70 1.70 g's aGPV = 1.00 1.00 1.00 1.00 g's aPV = 1.90 12.42 10.66 7.16 g's

AVMIN = 1.00 0.93 0.97 1.00 g's

AL = 0.14 0.14 0.14 0.14 g's

B-9

Crane Acceleration Calculations - Lifting Conditions

ABS Guide for Certification of

Container Securing Systems

Section 6: Securing System Design Principles

3 Design Loads

3.1 General

3.3 Wind Loads

3.3.1 Wind Load

Wind forces are to be applied to exposed containers. The wind pressure, PW, shall be taken as:

P w = 0.0101 LT/ft

Pw = 0.0101 LT/ft

L = 36.75 ft Length of crane (side facing wind) h = 10.00 ft Crane Body Height

A= 367.50 ft

Fw = 3.71 LT

To calculate accelerations experienced by the crane in lifting conditions, the same process is repeated replacing both θ and φ with the maximum crane operating angle of 2°. This is conservative, as these calculations will find a dynamic load, and the crane operating angle is intended to be static. As period is independent of amplitude, the pitch and roll periods will remain the same.

The basic loads to be taken into account in container securing calculations include gravitational forces, dynamic forces associated with ship motions, wind forces, and lashing or other securing forces. Sea loads and green water impact are not explicitly considered in the securing system design criteria. Adequate protection from green water impact shall be provided.

The wind load is assumed equally distributed over the side of the container. The vertical center of pressure should be taken at the mid height of the container, and the longitudinal center of pressure should be taken at the mid length of the container.

B-10

3.5 Design Ship Motions

3.5.1 Ship Conditions

GM = transverse metacentric height for the actual load condition, in ft. Where calculations are carried out for representative conditions for presentation in the Cargo Securing

Manual, GM values should be evaluated over the expected operating range.

d = draft to the summer load line, in ft d = 3.14 ft

GM = 275.65 ft

3.5.2 Ship Motions

3.5.2(a) Roll Motion. The natural roll period (full cycle) is to be obtained from the following equation:

T R = sec where k 1 = 1.104 k r = roll radius of gyration, in ft, and may be taken as 0.40B

GM = transverse metacentric height, in ft

B = 100.00 ft k1 = 1.10 kr = 40.00

TR = 2.66 s θOperating = 2.00 °

For service without restrictions, the accelerations and loads on containers are to be determined from the ship motions. The formulas for ship motions and accelerations assume that parametric rolling is avoided, either through The formulas for ship motions and accelerations assume that parametric rolling is avoided, either through design or through vessel operations. For more information on parametric rolling, see the ABS Guide for the Assessment of Parametric Roll Resonance in the Design of Container

Carriers. In particular, these formulas do not account for the extreme roll motions and simultaneous occurrence of extreme roll and pitch induced accelerations that may occur in head sea parametric rolling.

B-11

For vessels with active stabilizing systems, C may be specially considered.

The roll center, R CTR , is to be taken at the vertical center of gravity of the vessel, measured in ft above baseline. When the calculated vertical center of gravity of the vessel is not submitted, R CTR may be estimated from the following formula:

R CTR = ft above baseline where

D = molded depth at side, ft d = draft as defined in 6/3.5.1 d = 3.14 ft

D = 12.00 ft

RCTR = 4.57 ft above BL

3.5.2(b) Pitch Motion. The natural pitch period (full cycle) is to be obtained from the following equations:

TP = 7 + 0.00375 × (L – 600.4) sec for L in ft where

L = length between perpendiculars, in ft

LOA = 200.00 ft

TP = 5.50 s φOperating = 2.00 °

The pitch center of the vessel P CTR is to be taken at the longitudinal center of flotation. When the calculated longitudinal center of floatation is not submitted, P CTR may be estimated as 0.45L forward of the aft perpendicular.

PCTR = LOA - LCF ft

PCTR = 94.24 ft fwd of AP

B-12

3.7 Accelerations

Containers and their securing systems shall be capable of withstanding the forces generated by the following load combinations:

Condition A: The maximum roll condition generating maximum across-the-deck accelerations, expected in quartering stern or beam seas.

Condition B: The maximum pitch condition generating maximum normal-to-deck accelerations, expected in head or near head seas.

The following definitions apply to both the Condition A and the Condition B load combinations:

x c = longitudinal distance to the center of gravity of the container, in ft, forward of the aft perpendicular abs(x c -P CTR ) = absolute value of the longitudinal distance from the vessel’s pitch center to the center of gravity of the container, in ft y C = transverse distance to the center of gravity of the container, in ft, from the vessel’s centerline abs(y c ) = absolute value of the transverse distance from the vessel’s centerline to the center of gravity of the container, in ft z C = vertical distance to the center of gravity of the container, in ft, from the vessel’s baseline abs(z C -R CTR ) = absolute value of the vertical distance from the vessel’s roll center to the center of gravity of the container, in ft a 0 = common acceleration parameter, in g’s

= 0.2012 k C = 0.0214 for x C , y C , and z C in ft k 3 = force factor accounting for longitudinal position of container stack, where

= for x C < 0.2L

= 0 for 0.2L≤x C ≤0.7L

= for x c >0.7L

The designer is to ensure that the stowage system satisfies all of the strength criteria for both Condition

A and Condition B accelerations. For conventional lashed systems on the deck with containers having properties as listed in Section 3, Table 3, Condition A governs and Condition B need not be evaluated.

For containers stowed in cell guides, Condition B governs and Condition A need not be evaluated. For other configurations, including block stowage in the holds of bulk carriers, Condition A and Condition B should be evaluated.

0.5 × ( 0.2� − ��

0.2�

0.7 × ( �� − 0.7�

0.3�

B-13

Lift Position 1 2 3 4 xc = 82.88 20.13 30.58 51.50 ft abs(xc-PCTR) = 11.36 74.11 63.66 42.74 ft yC = 0.00 0.00 0.00 0.00 ft abs(yc) = 0.00 0.00 0.00 0.00 ft zC = 14.72 14.72 14.72 14.72 ft abs(zC-RCTR) = 10.15 10.15 10.15 10.15 ft a0 = 0.20 g's kC = 0.02

1 2 3 4 k3 = -0.54 0.25 0.12 -0.14 for xC<0.2L

0.00 0.00 0.00 0.00 for 0.2L<=xC<=0.7L

-0.67 -1.40 -1.28 -1.03 for xC>=0.7L

Highlighted cells indicate which k3 will be used, based on location of crane lift position.

As the crane is evaluated at four different locations, all four must be considered here. Calculations are similar for all four locations.

B-14

3.7.1 Condition A – Roll and Heave

A T = a GT + k C a RT + (1 + k 3 )a 0 sin θ in g’s

The maximum vertical acceleration is obtained from the following equation:

A VMAX = a GRV + k C a RV + (1 + k 3 )a 0 cos θ in g’s

The minimum vertical acceleration is obtained from the following equation:

A VMIN = a GRV – k C a RV + (1 – k 3 )a 0 cos θ in g’s

A VMIN is not to be taken greater than 1.0 where a GT = transverse static gravitational acceleration component, in g’s

= sin θ a RT = transverse roll acceleration component, in g’s a GRV = vertical static gravitational acceleration component, in g’s

= cos θ a RV = vertical roll acceleration component, in g’s

Lift Position 1 2 3 4 aGT = 0.03 0.03 0.03 0.03 g's aRT = 2.87 2.87 2.87 2.87 g's aGRV = 1.00 1.00 1.00 1.00 g's aRV = 0.00 0.00 0.00 0.00 g's

AT = 0.10 0.09 0.10 0.10 g's

AVMIN = 1.00 1.00 1.00 1.00 g's

The transverse and vertical accelerations at any point are to be obtained from the following formulas.

The longitudinal accelerations are taken as zero for this condition. The transverse acceleration is obtained from the following equation:

θ θ θ

B-15

3.7.2 Condition B – Pitch and Heave

The longitudinal and vertical accelerations at any point are to be obtained from the following formulas.

The transverse accelerations are taken as zero for this condition.

The longitudinal acceleration is obtained from the following equation:

A L = a GL + k C a PL + a 0 sin φ in g’s

The maximum vertical acceleration is obtained from the following equation:

A VMAX = a GPV + k C a PV + a 0 cos φ in g’s

The minimum vertical acceleration is obtained from the following equation:

A VMIN = a GPV – k C a PV + a 0 cos φ in g’s

A VMIN is not to be taken greater than 1.0 where a GL = longitudinal static gravitational acceleration component, in g’s

= sin φ a PL = longitudinal pitch acceleration component, in g’s a GPV = vertical static gravitational acceleration component, in g’s

= cos φ a PV = vertical pitch acceleration component, in g’s

Lift Position 1 2 3 4 aGL = 0.03 0.03 0.03 0.03 g's aPL = 0.67 0.67 0.67 0.67 g's aGPV = 1.00 1.00 1.00 1.00 g's aPV = 0.75 4.90 4.21 2.83 g's

AVMIN = 1.00 1.00 1.00 1.00 g's

AL = 0.06 0.06 0.06 0.06 g's

B-16

Forces Exerted Due to Crane Accelerations

Counterweight 148.2 LT

Carbody 53.6 LT

Largest Lift Load 64.5 LT

Total Weight (Transit) 201.8 LT

Total Weight (Lifting) 266.3 LT

Transit Condition

Condition A - Roll and Heave

Lift Position 1 2 3 4

Transverse Force FH = 104.6 94.7 104.6 104.6 LT

Vertical Force FV = 201.8 201.8 201.8 201.8 LT

Wind Force (Xvrs) FW = 3.7 3.7 3.7 3.7 LT

Max Long'l Force A (No Long'l Force) FX = 0.0 LT

Max Xvrs Force A (FH + FW) FY = 108.3 LT

Min. Vertical Force A (FV) FZ = 201.8 LT

Condition B - Pitch and Heave

Lift Position 1 2 3 4

Longitudinal Force FH = 28.7 28.7 28.7 28.7 LT

Vertical Force FV = 201.8 187.8 195.4 201.8 LT

Wind Force (Xvrs) FW = 3.7 3.7 3.7 3.7 LT

Max Long'l Force B (FH) FX = 28.7 LT

Max Xvrs Force B (FW) FY = 3.7 LT

Min. Vertical Force B (FV) FZ = 187.8 LT

Maximum Forces - Transit

FX,MAX = 28.7 LT

FY,MAX = 108.3 LT

B-17

Lifting Condition

Condition A - Roll and Heave

Lift Position 1 2 3 4

Transverse Force FH = 27.5 24.9 27.5 27.5 LT

Vertical Force FV = 266.3 266.3 266.3 266.3 LT

Wind Force (Xvrs) FW = 3.7 3.7 3.7 3.7 LT

Max Long'l Force A (No Long'l Force) FX = 0.0 LT

Max Xvrs Force A (FH + FW) FY = 31.2 LT

Min. Vertical Force A (FV) FZ = 266.3 LT

Condition B - Pitch and Heave

Lift Position 1 2 3 4

Longitudinal Force FH = 15.0 15.0 15.0 15.0 LT

Vertical Force FV = 266.3 266.3 266.3 266.3 LT

Wind Force (Xvrs) FW = 3.7 3.7 3.7 3.7 LT

Max Long'l Force B (FH) FX = 15.0 LT

Max Xvrs Force B (FW) FY = 3.7 LT

Min. Vertical Force B (FV) FZ = 266.3 LT

Maximum Forces - Lifting

FX,MAX = 15.0 LT

FY,MAX = 31.2 LT

Maximum Forces - Both Cases

FX,MAX = 28.7 LT

FY,MAX = 108.3 LT

Minimum Vertical Force - Both Cases

FZ,MIN = 187.8 LT

As the transit condition provides greater forces, it will be used in the following strength calculations.

B-18

Crane Tie-Down Calculations Tie-down strength evaluated using ship motions in transit, omitting end blocks.

Tie-Down Angles

Found using geometry of crane tie-down lines in Rhino model α = 43.0 ° between line and X-axis β = 66.1 ° between line and Y-axis γ = 56.7 ° between line and Z-axis

Tension Force FT = 15.6 LT Rated working load

Peck&Hale 70MTC/818A

FT,X =

FT,X = 11.4 LT

FT,Y =

FT,Y = 6.3 LT

FT,Z =

FT,Z = 8.6 LT

Coeff. of Friction µ = 0.7 Steel on rubber coefficient of static friction

Max. Friction Force FF = -µ x (F Z,MIN +F T,Z ) LT

FF,X = LT

FF,Y = LT

FX + FF,X ≤ FT,X and FY + FF,Y ≤ FT,Y where

FX = FX,MAX

FY = FY,MAX

From the previous crane force calculations

Friction force found using the minimum vertical downward force for most conservative estimate.

The tie down rated strength and max. static frictional force are sufficient to overcome crane force when the forces exerted by the crane, shown with a positive magnitude; and the maximum possible force of friction, shown with a negative magnitude; sum to not more than the rated safe working load of the tie-down.

,- sin ,0

,1

,- cos ,0

,1

,4 cos5

,4 cos6

,4 cos 7

B-19

Condition A - Roll and Heave

Min. Vertical Force FZ,MIN = 187.8 LT

Max. Friction Force FF = -147.3 LT

FF,X = 0.0 LT

FF,Y = -147.3 LT

Force Comparison

Crane Force (LT) Max. Friction (LT) Rated Load (LT)

F FF FT

Long'l 0.0 0.0 11.4 PASS

Xvrs 108.3 -147.3 6.3 PASS

Condition B - Pitch and Heave

Min. Vertical Force FZ,MIN = 187.8 LT

Max. Friction Force FF = -137.5 LT

FF,X = -132.7 LT

FF,Y = -36.1 LT

Force Comparison

Crane Force (LT) Max. Friction (LT) Rated Load (LT)

F FF FT

Long'l 28.7 -132.7 11.4 PASS

Xvrs (Wind) 3.7 -36.1 6.3 PASS

Conclusion: In the hypothetical absence of any end block stops, the selected tie-downs can secure the crane in transit up to the angle of deck immersion in both roll and pitch, and lifting conditions up to the 2° rating of the crane. The maximum possible static friction is never exceeded, so the tie downs would not need to exert force. For the purposes of FEA analysis, the safe working load of the tie downs was applied to the deck.

B-20

Crane End Stop Calculations

ASTM A36 Steel

Yield Stress σyield = 36000 psi

Safety Factor 0.5 σyield,adj = 18000 psi

Shear Force Fshear,x = 14.4 LT 1/2 of Max Long'l Force

Coeff. of Friction µ = 0.70 Steel on Steel

Fshear,y = =Fshear,x x (1-µ)

Fshear,y = 4.3 LT

Plate Calculations - ABS STEEL BARGES 3-2-3/1 p = design pressure pl = p (l - s) / l pl = lower pressure pe = (p + pl) / 2 pe = effective pressure he = pe / 45 l = length of pressure along the treads p = Fx/A s = unsupported length of plate panels he = effective head t = required plate thickness

Permanent End Block Stop

Shear Area A = 98.5 in

Shear stress τx = 326.81 psi PASS τy = 98.04 psi PASS

Narrowest cross-sectional area A = 39.9 in

Compressive stress σ = 0.4 psi PASS

Plate Calculations A = 5 ft p = 45 psf l = 5 ft s = 30 in pl = 22 psf

Pe = 34 psf he = 0.75 ft trequired = 0.12 in

Thickness Used t = 0.5 in PASS

Friction applied using FX as the normal force, see FEA report for full rationale.

As shown in tie-down calculations, the force of friction between the tracks and deck is sufficient to counteract the force exerted by the crane in pitch and roll. For purposes of this calculation, all longitudinal force is treated as if it were acting upon the end stops.

B-21

Portable End Block Stop

Shear Area A = 76.0 in

Shear stress τx = 423.57 psi PASS τy = 127.07 psi PASS

Narrowest cross-sectional area A = 21.3 in

Compressive stress σ = 0.68 psi PASS

Plate Calculations A = 5.0 ft p = 45 psf l = 5.0 ft s = 14 in pl = 34 psf

Pe = 39 psf he = 0.9 ft trequired = 0.1 in

Thickness Used t = 0.5 in PASS

Conclusion: Even without the assistance of the frictional force or the tie-downs, the crane stops are sufficient to hold a fully loaded crane in place at the angle of deck immersion in pitch. FEA was not ran on the portable crane stops as they rest on Axion Struxure timbers, and are not structurally connected to the vessel.

B-22