1. GENERAL

1.1 ScopeThis teaching document shows the calculation of sign board frame installation on a given offshore structure. The case study is that the sign board is required to be installed in a cellar deck of a platform.The scope of work is to design sign board frame and structure that can withstand wind load of 35.8 m/s. The structure should also withstand ground motion of UBC 1997 region 3.

1.2 Executive SummaryThe sign board frame is designed as a steel structure. Structure computer modelling is using SAP 2000 is shown in Figure 1.

Figure 1. Sign Board Frame Structure Front View

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Figure 2. Sign Board Frame Structure Back View

The frame to support the sign board will be installed on the edge of cellar deck, bolted on its leg, and clamped to the tubular column. As shown in Figure 2, there are four legs that had been modelled as fix restraint and 2 clamps in the middle of the frame board that had been modelled vertically as hinge restraint.

Figure 3. Key Plan

1.3 ReferencesThe structural design calculation and analysis was performed in compliance with the following references at minimum.

1.3.1 Design Codes and StandardsAISC ANSI 360-05 Specification for Structural Steel BuildingsAISC LRFD Manual of Steel Construction – Load & Resistance Factor

Design

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SNI 03-1726-2002 Indonesian Standard for Seismic LoadUBC 1997 Uniform building CodeACI 318 – 05 Building Code Requirements for Structural Concrete

1.4 UnitsInternational system of units (S.I.) was used for all dimensions and weights in the design and details.

1.5 LanguageThe report was written in English.

2. COMPUTER MODELING

The entire structure was modelled as a 3D space frame.

3. LOADING

The following loads were considered for the computer analysis:

3.1 Dead Load & Super Imposed Dead Load (SIDL)Dead load comprised the structural self-weight automatically generated by the SAP computer program. Super imposed dead load comprised the weight of the board (without frame). The board is aluminium with density of 2700 kg/m3 and thickness of 2 mm.

SIDL = 2700 kg/m3 x 0.002 mSIDL = 5.4 kg/m2 = 5.4 x 9.8 kg.f/m2

= 0.52 kN/m2

F = 0.52 kN/m2 x (sign board area) = 0.52 x (3.05 x 1.5) = 2.379 kNF, per 4 length = 2.379 /4 = 0.59475 kNF, per length = 0.59475 / 3.05 = 0.195 kN/m = 1.95 x 10-4 kN/mm

Figure 4. SIDL Assigned on Frame (Y-Z View)

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3.2 Live LoadLive load is assumed working on top frame during installation and maintenance. This is based on assumption that only one worker will stand on top of the frame hen assembling the structure.

Point load P = 100 kg = 980 kg.f = 9.61 kN

Figure 5. Live Load Assigned on Frame (Y-Z View)

3.3 Rain LoadRefer to ASCE, rain load is designed 10 kg/m2. This load is distributed on top of the frame.

Lw = 10 kg/m2 = 0.96 kN/m2

Prain = 0.96 kN/m2

Upper frame area = 3050 x 75 = 228750 mm2 = 0.22875 m2

Frain = 0.96 kN/m2 x 0.22875 m2 = 0.2196 kN = 7.08 x 10-5 kN/mm

Figure 6. Rain Load Assigned on Frame (Y-Z View)

3.4 Wind loadDesign wind pressures for buildings and structures and elements therein shall be determined for any height in accordance with the following formula:

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P = Ce Cq qs Iw WhereP = Design wind pressure (kN/m2)Ce = Combined height, exposure and gust factor coefficient (Ref. UBC 1997 Chapter

16 Table 16-G). The exposure is categorized in Exposure D due to the fact that the sign board is installed in the offshore area.

Cq = Pressure coefficient for the structure or portion of structure (Ref. UBC1997 Chapter 16 Table 16-H), see Figure 4.

qs = Wind stagnation pressure at the standard height of 10 m.= 0.78556 kN/m2 for Basic Wind Speed : 35.8 m/s (= 129 km/h) (Ref. UBC1997

Chapter 16 Table 16-F). The Basic Wind Speed refer to the Exposure D definition. (Ref. UBC1997 Chapter 16 Section 1616).

Iw = Importance factor (Ref. UBC1997 Chapter 16 Table 16-K and S-900-1220-001 “Project Design Data” section 2.1)

Figure 7. Wind pressure Coefficient Cq

Wind CalculationBased on the above equation, wind pressure was calculated on for every side of board. The board is assumed as a wall with small segment of z axis as the thickness. This value of thickness is considered too small to has an impact from z direction wind (Cq = 0). Due to this adjustment, the leeward wind is also assumed has the same number of windward Cq

and not reduced to 0.5. Figure 5 shows wall naming and Table 1 shows wind pressures acting on each side of wall.

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Figure 8. Wall Designation

Table 1. Wind Pressure

Side Cq Ceqs

(kN/m2) IwP

(kN/m2) Direction

A Wind 0o

Board

X1 0.8 1.73 0.78556 1 1.087 inward

X2 0.8 1.73 0.78556 1 1.087 outwardB Wind 180o

Board

X1 0.8 1.73 0.78556 1 1.087 outward

X2 0.8 1.73 0.78556 1 1.087 inward

P = 1.73 x 0.8 x 0.78556 x 1 = 1.087 kN/m2

F = P x A = 1.087 x (3.05 x 1.5) = 4.974 kNWind load is assigned in kN per meter. The load is distributed uniformly in all frame section.Length of frame in total = 22.7 mP, per frame length = 4.974 / 22.7 = 0.22 kN/m = 2.2 X 10-4 kN/mmW0 = Px1 + Px2 = 2.2 X 10-4 + 2.2 X 10-4 = 4.4 X 10-4 kN/mmW180 = Px1 + Px2 = 2.2 X 10-4 + 2.2 X 10-4 = 4.4 X 10-4 kN/mm

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Figure 9. Wind 0o Load Assigned on Frame (Y-Z View)

Figure 10. Wind 180o Load Assigned on Frame (Y-Z View)

3.5 Seismic LoadSeismic load is applied by using response spectrum analysis according to UBC ‘97. The following criteria are used as input to the SAP2000 software.

- Response Spectrum CurveSeismic region = 3

Soil condition = SD

Ca = 0.36Cv = 0.54

Damping ratio = 0.05- Importance Factor (I) = 1- Reduction factor (R) = 4.5 (Ordinary Moment Frame)- Load configuration

Seismic load configuration is given in two major directions as shown on Figure 6.

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Figure 11. Seismic Load Configuration

3.6 Load CombinationLoad combination is assigned based on AISC LRFD manual of steel construction. The following load combinations are used:1 1.4 D + 1.4 SIDL2 1.2 D + 1.2 SIDL + 1.6L + 0.5 R3 1.2 D + 1.2 SIDL + 1.6L + 0.5 R + 0.8 W04 1.2 D + 1.2 SIDL + 1.6L + 0.5 R + 0.8 W1805 1.2 D + 1.2 SIDL + 1.3W0 + 0.5L + 0.5 R 6 1.2 D + 1.2 SIDL + 1.3W180 + 0.5L + 0.5 R 7 1.2 D + 1.2 SIDL + 1EX + 0.5L 8 1.2 D + 1.2 SIDL + 1EY + 0.5L 9 0.9 D + 0.9 SIDL + 1.3W010 0.9 D + 0.9 SIDL + 1.3W180

4. MEMBER DESIGN AND STRESS RATIO

The maximum stress ratio occurred are shown in the table below:

Table 2. Required Member Profile

No. Description Stress Ratio Remark

1 Base frame L 75x75x8 0.27105 Pass

2 Main frame L 75x75x8 0.35125 Pass

3 Inner frame L 30x30x3 0.144577 Pass

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Y axis

X axis

30% Eq X

100% Eq Y

Y axis

X axis

100% Eq X

30% Eq Y

Figure 12. Deformed Shape

5. DEFLECTION AND SWAY

Maximum deflection in X and Y direction is 5.461 mm and 0.002 mm, less than max deflection of beam L/200 = 7.5 mm (ok). In Z direction, the maximum deflection is 0.234 mm, less than max deflection of beam L/200 = 7.5 mm. (ok).The maximum sway of the upper structure is happened on the X direction with the value of 5.461 mm, less than max sway of the structure h/200 = 7.5 mm. (ok).The particular frames with the maximum deflection are show in Table 3.

Table 3. Joint DisplacementsLoad combination: Envelope

Frame Joint DirectionDiplacement

(mm)L/200 (mm) Remark

12 U1 5.4607

U2 0.0016

U3 0.2342

23 U1 0

U2 0

U3 0

23 U1 0

U2 0

U3 0

8 U1 5.4607

U2 0.0016

U3 0.2342

20,21,22

23,24,25

OK

OK

OK

OK

OK

OK

7.5

7.5

7.5

7.5

7.5

7.5

X 5.461

Y 0.002

Y 0.002

Z 0.234

Deflection (mm) Deflection (mm)

X 5.461

Z 0.234

U3

U2

U1

0.234

0.002

5.461

U1 -5.461

U2 -0.002

U3 -0.234

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Figure 13. Deformed Shape and Displacement

6. JOINT REACTION

The summary of joint reaction calculation result is shown in the table below.

Table 4. Maximum Support Reaction

F1 F2 F3 M1 M2 M3

KN KN KN KN-mm KN-mm KN-mm3.154 -0.224 15.376 0.000 0.000 0.000

Compression Shear Shear Moment Moment Moment4.516 -0.163 0.842 0.000 0.000 0.000

Compression Shear Shear Moment Moment Moment1.599 0.211 0.646 0.358 633.265 16.763Shear Shear Tension Moment Moment Moment1.058 0.077 0.403 1.032 149.741 5.922Shear Shear Tension Moment Moment Moment1.058 0.077 0.403 1.032 149.741 5.922Shear Shear Tension Moment Moment Moment1.599 0.211 0.646 0.358 633.265 16.763Shear Shear Tension Moment Moment Moment

FixRestraint 4

HingeRestraint 44

HingeRestraint 23

FixRestraint 43

FixRestrain 41

FixRestraint 6

Maximum Joint Forces

Joint Type

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Figure 14. Joint Reaction

7. CONNECTION

7.1 Welded ConnectionBecause of the restriction to do welding activity in platform, the board and the frame are assembled together in onshore area. The welded connection is used to enhance the board with the main and inner frame. A relatively small depth of material will become molten, and upon cooling, the structural steel and the weld material will act as one continuous part where they are joined. In this computer modelling, the welded connections are assigned as fix connection.

Figure 15. Frame Front View

7.2 Bolted ConnectionType of bolt that is being used in the connection design is the High Strength Bolt A325. This bolt relies on initial force called proof load during the early fastening. The proof load

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will provide friction so that the bolt connection performs serviceability limit state before it goes to strength limit state under bigger load. For sign board designing, high strength bolts are sufficient because it is best used to resist alternating force.

Figure 16. Frame Deck Connection

Table 5. Bolt Profile

Type Quality db (mm) Proof Stress (MPa) Ultimate Strength, fu

(MPa)

A325

High strength bolt 10 mm 70% fu = 585 825

Figure 17. Sample Type of Tubular Clamp

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Figure 18. Bolt Connection Measurement

Vu = 1.599 x 103 NMu = 6.33265 x 105 N.mmTu = 646 NProof stress = 585 MPan = 2Without screw on the sliding plane

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Shear and tensile combination check against Strength Limit State : Proof load = 0.75 x Ab x Proof Stress

= 0.75 x ¼ x π x 102 x 585 = 3.44 ton = 33735.048 NVn = 1.13 x μ x Proof load x m = 1.13 x 0.35 x 3.44 x 1 = 1.36 ton = 13337.112 Nφ = 1 , as in standard boltφ Vn = 1 x 13337.112 = 13337.112 NVu/n = 1599 / 2 = 799.5 N

φVn(1− Tun

1.13 Proofload )=13337.112(1− 6462

1.13×33735.048 ) = 13224.11 N

799.5 N < 13224.11 N -------- OK

Shear check against Serviceability Limit State :

f uv=Vu

n× Ab= 1599

2×π4×102

=10.2MPa

0.5 x φ x fub x m = 0.5 x 0.75 x 825 * 1 = 309.375 MPa

f uv<302.375MPa ---OK

Moment check:ft = 807 - 1.5 fuv ≤ 621 MPa = 807 - 1.5(10.2) ≤ 621 MPa = 791.5 MPa ≤ 621 MPaft = 621 MPaa x fy x b = nl x (ni x Ab x ft)a = nl x (ni x Ab x ft) / (fy x b) = 2 x (1 x 78.5 x 621) / (240 x 75) = 5.42 mmMn = (ni x Ab x ft) (41+130) – (a x fy x b)a/2 = 48748.5 (41+130) – 2643875.6 = 5692117.9 = 0.57 t.m = 5589819 N.mmMd = φ x Mn = 0.75 x 5589819 = 4192364.25 N.mm Mn > Mu (=6.33265 x 105 N.mm) -------- OK

Bolt connection in the design can safely perform against maximum shear, tensile, and moment at the joint that connecting frame and platform deck.

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