general construction notes and guidance

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Tel: 01287 348404 Mob: 07716569093 E-mail: [email protected] Website: www.pmce.co.uk

PM Consulting Engineers LTD 79 - 81 Church Street Guisborough TS14 6HG

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________________________________________________________________________________________________________________ Project: 18125 - 76 Runswick Avenue

General Construction Notes and Guidance:

1. Calculations are not to be used for the purpose of ordering materials and should only be used for Building Regulations submissions. All dimensions should be checked by the contractor on site.

2. All steelwork to be mechanically wire brushed and painted two coats of red oxide. Steelwork located in the cavity or below DPC to be suitably protected with 2 coats of bituminous paint.

3. All timber to be graded C24 (SC4) unless stated otherwise. Preservative treated to Architect´s details 4. To be read in conjunction with architect’s drawings, any inconsistencies should be reported. 5. For details of fire protection to steelwork, see Architects drawings. 6. The contractor is to ensure that all existing construction is adequately supported, using needles and props as required. Where a new beam supports the existing construction adequate pre-load is to be applied and suitable packs such as driven dry-state introduced, then pointed up with mortat. 7. All blockwork to be 7.3 N/mm2 in class III mortar below DPC in accordance with BS5628:Part3:2005 or suitable 7.0 N/mm2 foundation quality blocks in class II mortar in accordance with the manufacturer’s instructions. All brickwork below DPC to be engineering bricks DPC in accordance with BS 5628: Part 3: 2005. 8. The builder is to take into consideration the placement of the structural elements, ensuring that the method of lifting and placement is safely carried out. Responsibility for this element lies with the Contractor. As the existing walls need to be propped in order to introduce some of the lintels, this should also be considered in relationship to the risk assessment of the Contractor. Safe working procedures must be adopted. Responsibility for this element lies with the Contractor. Splice details for long-span beams can often be accommodated if required.

Party Wall Act 1996 If part of the work is adjacent to the boundary, the adjacent neighbours right to support could be affected; the issues associated with Party Wall Act may need to be considered. This may include providing information to the adjoining owner, giving sufficient notice of works in compliance with the Act. If the following list applies to this project then the Party Wall Act will apply. 1. Installing a new beam into the shared wall between properties 2. Demolishing, building or under-pinning an existing shared wall

3. Building a new wall at or on the boundary or junction of two properties 4. Damp-proofing all the way through a party wall 5. Digging foundations that are within 3m of a Party Wall, where the new foundations are deeper than the existing ones 6. Where the new foundations are within 6m and lower than a 45° line from the bottom of the existing foundations.

Codes

BS EN 1990+A1:2006/NA: 2005-06 Basis of structural design BS EN 1991-1-1 Part 1-1: General actions - Self-weight, imposed loads for buildings BS EN 1991-1-3/NA: 2005-12 Part 1-3: General actions - Snow loads BS EN 1995-1-1+A1:2008/NA: 2006 Part 1-1: General - Common rules and rules for buildings BS EN 1995-1-2 Part 1-2: General - Structural fire design BS EN 1991-1-4/NA: 2006 Part 1-4: General actions - Wind loads BS EN 14080:2013-08 Timber structures - Glued laminated timber and solid timber - Requirements

BS EN 338:2010-03 Structural timber - Strength classes

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Loading

Dead Live

Roof1c

Tiles g1= 0.65 kN/m2

Rafters,felt,insulation etc g2= 0.30 kN/m2

Plasterboard g3= 0.25 kN/m2

g0= 1.20 kN/m2

Roof pitch a= 35 °

gk= g0/COS(a)+g4 = 1.46kN/m2

Attic g4= 0.25 kN/m2

Roof snow loading qk= 0,60*((60-a)/30) = 0.75 kN/m2

Floor

Joist & boarding,finishes g1= 0.25 kN/m2

Plasterboard g2= 0.25 kN/m2

gk= g1+g2 = 0.50kN/m2

Imposed qk= = 1.50 kN/m2

Walls

W1 External cavity blockwork g1= 2,7*(2,1+1,4)= 9.45 kN/m

Plasterboard g2= 2,4*0,25 0.60 kN/m

gk= g1+g2 = 10.05 kN/m

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Plans:

Proposed Extension

Profile Padstone

B01 203x203 UC 46 +

300x10 Bottom Plate +

Stiffener Plates @ 700 c/c

215 long x 300 wide 2 Course Eng. Bwk

Padstone

B02 203x203 UC 46 +

300x10 Bottom Plate +

Stiffener Plates @ 700 c/c

215 long x 300 wide 2 Course Eng. Bwk

Padstone

P01 152x152 UC 23 [Grade 43]

Pad Footing For P01

Use 1.2x1.2m Size Pad Footing,

0.45m Deep, Concrete Min C20/25

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Pos: B01

Span length l= 2.30 m

Loading: Dead Live

Roof 1c g= 4.00 / 2 * 1.46 = 2.92 kN/m

Floor g= 4.00 / 2 * 0.50 = 1.00 kN/m

Wall Ext g= = 10.05 kN/m

13.97 kN/m

Roof 1c q= 4.00 / 2 * 0.75 = 1.50 kN/m

Floor q= 4.00 / 2 * 1.50 = 3.00 kN/m

4.50 kN/m

AXIAL WITH MOMENTS (MEMBER)

B01

Member 1 (N.3-N.4) @ Level 1 in Load Case 1

Member Loading and Member Forces Loading Combination : 1 UT + 1.4 D1 + 1.6 L1

D1 UDLY -013.970 ( kN/m )

L1 UDLY -004.500 ( kN/m )

Member Forces in Load Case 1 and Maximum Deflection from Load Case 2

Mem

ber

No.

Node

End1

End2

Axial

Force

(kN)

Torque

Moment

(kN.m)

Shear Force

(kN)

Bending Moment

(kN.m)

Maximum Moment

(kN.m @ m)

Maximum

Deflection

(mm @ m) x-x y-y x-x y-y x-x y-y

1 3 0.08T 0.00 30.77 0.00 0.00 0.00 17.69 0.47

4 0.08T 0.00 -30.77 0.00 0.00 0.00 @ 1.150 @ 1.150

Classification and Properties (BS 5950: 2000) Section (68.48 kg/m) 203x203 UC 46 + 285x10 B Plate 68.48 [Grade 43] Class = Fn(b/T,d/t,py,F,Mx,My) 9.25, 22.33, 275, 0, 17.69, 0 (Axial: Non-Slender) Compact Auto Design Load Cases 1

Local Capacity Check Fvx/Pvx 0.003 / 241.402 = 0 Low Shear Mcx = py.Sxx1.2 py.Zxx 275 x 599.21.2 x 275 x 494.8 = 163.284 kN.m

Ae = Fn(Ag,A.net,py,Us) 87.23,87.23,275,410 87.23 cm² Pz = Ae.py 87.23x275 2398.825 kN n = F/Pz -0.079 / 2398.825 = 0.000 OK Srx = Fn(Sxx, n) 599.2, 0 599.2 cm³ Mrx = Srx.py 599.2 x 275 163.284 kN.m (Mx/Mrx)Z1+(My/Mry)Z2 (17.692/163.284)²+(0)1= 0.012 OK

Equivalent Uniform Moment Factors mLT, mx, my and myx mLT=0.2+(.15M2+.5M3+.15M4)/Mmax 0.2+(.15x13+.5x18+.15x13)/18 0.44 0.925 Table 18

my=0.2+(.1M2+.6M3+.1M4)/Mmax 0.2+(.1x0+.6x0+.1x0)/0 .8x0/0 1 Table 26

mx=0.2+(.1M2+.6M3+.1M4)/Mmax 0.2+(.1x13+.6x18+.1x13)/18 .8x18/18 0.95 Table 26 myx=0.2+(.1M2+.6M3+.1M4)/Mmax 0.2+(.1x0+.6x0+.1x0)/0 .8x0/0 1 Table 26

Lateral Buckling Check Mb Le = 1.2L+2D 1.2 x 2.3 + 2 x 0.203 = 3.166 m λ = Le/ryy 3.166 / 6.31 50.18 OK

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v = Fn (x,Le,ryy,λ) 18.763, 3.166, 6.31, 50.18 1.267 Table 19 λLT= u.v.λ.√βW 0.692 x 1.267 x 50.18 √ 1 44.01 pb = Fn (py,λLT) 275, 44.01 252.39 N/mm² Table 16 Mb = Sxx.pb Mc 599.2 x 252.39 163.284 = 151.233 kN.m

Combined Axial Compression and Bending to Annex I rb=mLT.MLT/Mb 0.925x17.7/151.2 0.108 rc=Fc/Pcy 0/2129.7 0.000 λr=(rbλLT+rcλy)/(rb+rc) (0.108•44+0•36.5)/(0.108+0) 44.014 λro=17.15 ε (2rb+rc)/(rb+rc) 17.15•1(2•0.108+0)/(0.108+0) 34.300 Mob= Mb(1-Fc/Pcy) 151.233(1-0/2129.7) 151.233 Mxy= Mcx(1-Fc/ Pcy)

½ 163.284(1-0/2129.7)½ 163.284 Mox= Mcx(1-Fc/Pcx)/(1+0.5Fc/Pcx) 163.284(1-0/2318.7)/(1+0.5•0/2318.7) 163.284 Moy= Mcy(1-Fc/Pcy)/(1+ky(Fc/Pcy)) 80.520(1-0/2129.7)/(1+1.0(0/2129.7)) 80.520 Mab=fn( λr, λro, ε, Mxy, Mob) 44.014, 34.300, 1.000, 163.284, 151.233 161.011 Max=fn( λx, ε, Mrx, Mox) 26.136, 1.000, 163.284, 163.284 163.284 May=fn( λy, ε, Mry, Moy) 36.450, 1.000, 80.520, 80.520 80.520 mx.Mx/Max 0.95x17.7/163.3 0.103 OK mLT.MLT/Mab 0.925x17.7/161 0.102 OK mx.Mx/Max 0.95x17.7/163.3 0.103 OK Compare with Simplied to 4.8.3.3 0.124, 0.108, 0.108 0.124 Compare with MoreExact to 4.8.3.3 0.103, 0.108, 0.1 0.108

Deflection Check - Load Case 2 δ Span/360 0.47 2300 / 360 0.47 mm OK

Section (68.48 kg/m) 203x203 UC 46 + 285x10 B Plate 68.48 [Grade 43]

Consider Bearings:

Max Load Rd= 30.80 kN

Assuminggm

=3.5 fk = 3.5

Bearing b = 200 mm

= 102.67 mmRequired Bearing length Rd*E3*gm/( fk*1,5)/b=

Provide 215 long x 300 wide 2 Course Eng. Bwk Padstone

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Pos: B02

Span length l= 3.15 m

Loading: Dead Live

Roof 1c g= 4.00 / 2 * 1.46 = 2.92 kN/m

Floor g= 4.00 / 2 * 0.50 = 1.00 kN/m

Wall Ext g= = 10.05 kN/m

13.97 kN/m

Roof 1c q= 4.00 / 2 * 0.75 = 1.50 kN/m

Floor q= 4.00 / 2 * 1.50 = 3.00 kN/m

4.50 kN/m


B02



D1 UDLY -013.970 ( kN/m )

L1 UDLY -004.500 ( kN/m )

Member Forces in Load Case 1 and Maximum Deflection from Load Case 2

Mem

ber

No.

Node

End1

End2

Axial

Force

(kN)

Torque

Moment

(kN.m)

Shear Force

(kN)

Bending Moment

(kN.m)

Maximum Moment

(kN.m @ m)

Maximum

Deflection


2 4 0.00C 0.00 42.14 0.00 0.00 0.00 33.19 0.00 1.67

5 0.00C 0.00 -42.14 0.00 0.00 0.00 @ 1.575 @ 0.000 @ 1.575

Classification and Properties (BS 5950: 2000) Section (68.48 kg/m) 203x203 UC 46 + 285x10 B Plate 68.48 [Grade 43] Class = Fn(b/T,d/t,py,F,Mx,My) 9.25, 22.33, 275, 0, 33.19, 0 (Axial: Non-Slender) Compact Auto Design Load Cases 1

Moment Capacity Check Mc Fv/Pv 0.003 / 241.402 = 0 Low Shear Mc = py.Sxx1.2 py.Zxx 275 x 599.21.2 x 275 x 494.8 = 163.284 kN.m

MA/Mc 33.187 / 163.284 = 0.203 OK

Equivalent Uniform Moment Factor mLT mLT=0.2+(.15M2+.5M3+.15M4)/Mmax 0.2+(.15x25+.5x33+.15x25)/33 0.44 0.925 Table 18

Lateral Buckling Check Mb Le = 1.2L+2D 1.2 x 3.15 + 2 x 0.203 = 4.186 m λ = Le/ryy 4.186 / 6.31 66.35 OK v = Fn (x,Le,ryy,λ) 18.763, 4.186, 6.31, 66.35 1.168 Table 19 λLT= u.v.λ.√βW 0.692 x 1.168 x 66.35 √ 1 53.63 pb = Fn (py,λLT) 275, 53.63 229.08 N/mm² Table 16 Mb = Sxx.pb Mc 599.2 x 229.08 163.284 = 137.262 kN.m MA/(Mb/mLT) 33.187 / (137.262 / 0.925 ) 0.224 OK

Deflection Check - Load Case 2 δ Span/360 1.67 3150 / 360 1.67 mm OK

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Section (68.48 kg/m) 203x203 UC 46 + 285x10 B Plate 68.48 [Grade 43]

Consider Bearings:

Max Load Rd= 42.00 kN

Assuminggm

=3.5 fk = 3.5

Bearing b = 200 mm

= 140 mmRequired Bearing length Rd*E3*gm/( fk*1,5)/b=

Provide 215 long x 300 wide 2 Course Eng. Bwk Padstone


P01



Member Forces in Load Case 1

Mem

ber

No.

Node

End1

End2

Axial

Force

(kN)

Torque

Moment

(kN.m)

Shear Force

(kN)

Bending Moment

(kN.m)

Maximum Moment

(kN.m @ m)

Maximum

Deflection


4 2 72.92C 0.00 0.23 0.00 -0.70 0.00 0.00

4 72.92C 0.00 0.23 0.00 0.00 0.00 @ 1.575

Additional Nominal Moments MxUp, MyUp 5.422 kN.m, 4.337 kN.m

Classification and Properties (BS 5950: 2000) Section (22.95 kg/m) 152x152 UC 23 [Grade 43] Class = Fn(b/T,d/t,py,F,Mx,My) 11.19, 21.31, 275, 72.92, 5.42, 4.34 (Axial: Non-Slender) SemiComp Effective Properties Area=29.24 cm², Sxx=176.29(182) cm³, Syy=71.41(80.2) cm³ Auto Design Load Cases 1

Local Capacity Check Fvx/Pvx 0.232 / 145.847 = 0.002 Low Shear Mcx = py.Sxx1.2 py.Zxx 275 x 176.291.2 x 275 x 164.13 = 48.481 kN.m

Fvy/Pvy 0 / 307.383 = 0 Low Shear Mcy = py.Syy1.2 py.Zyy 275 x 71.411.2 x 275 x 52.67 = 17.381 kN.m

Pz = Ag.py 29.24 x 275 = 804.1 kN n = F/Pz 72.916 / 804.1 = 0.091 OK F/Ag.py+Mx/Mcx+My/Mcy 72.916 / 804.1 + 5.42 / 48.481 + 4.337 / 17.381 = 0.452 OK

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Compression Resistance Pc λx = Lex/rxx 100x1x3/6.54 = 45.9 OK Pcx = Area.pcx 29.24x242.421/10 = 708.839 kN Table 24 b λy = Ley/ryy 100x1x3/3.7 = 81.1 OK Pcy = Area.pcy 29.24x159.14/10 = 465.319 kN Table 24 c

Equivalent Uniform Moment Factors mLT, mx, my and myx mLT=0.2+(.15M2+.5M3+.15M4)/Mmax 0.2+(.15x1+.5x2+.15x4)/5 0.44 0.549 Table 18 my=0.2+(.1M2+.6M3+.1M4)/Mmax 0.2+(.1x1+.6x2+.1x3)/4 .8x3/4 0.6 Table 26

mx=0.2+(.1M2+.6M3+.1M4)/Mmax 0.2+(.1x1+.6x2+.1x4)/5 .8x4/5 0.574 Table 26

myx=0.2+(.1M2+.6M3+.1M4)/Mmax 0.2+(.1x1+.6x2+.1x3)/4 .8x3/4 0.6 Table 26

Lateral Buckling Check Mb Le = 1.00 L 1 x 3 = 3 m λ = Le/ryy 3 / 3.7 81.08 OK v = Fn (x,Le,ryy,λ) 20.292, 3, 3.7, 81.08 0.864 Table 19 λLT= u.v.λ.√βW 0.84 x 0.864 x 81.08 √ 0.969 57.9 pb = Fn (py,λLT) 275, 57.9 218.47 N/mm² Table 16 Mb = Sxx.pb Mc 176.29 x 218.47 48.481 = 38.515 kN.m

Simplified Approach py.Zx 275x164.13 45.136 kN.m py.Zy 275x52.67 14.484 kN.m F/Pc+mx.Mx/py.Zx+my.My/py.Zy 72.916/465.319+0.574x5.4/45.1+0.6x4.3/14.5 0.405 OK F/Pcy+mLT.MLT/Mb+my.My/py.Zy 72.916/465.319+0.549x5.4/38.5+0.6x4.3/14.5 0.414 OK

More Exact Approach Max=Mcx/(1+.5F/Pcx) 48.5/(1+.5x72.9/708.8) 46.109 kN.m May=Mcy/(1+F/Pcy) 17.4/(1+72.9/465.3) 15.026 kN.m F/Pcx+mx.Mx/Max+.5myx.My/Mcy 72.9/708.8+0.574x5.4/46.1+.5x0.6x4.3/17.4 0.245 OK F/Pcy+mLT.MLT/Mb+my.My/May 72.9/465.3+0.549x5.4/38.5+0.6x4.3/15 0.407 OK Max=Mcx(1-F/Pcx)/(1+.5F/Pcx) 48.5(1-72.9/708.8)/(1+.5x72.9/708.8) 41.366 kN.m May=Mcy(1-F/Pcy)/(1+F/Pcy) 17.4(1-72.9/465.3)/(1+72.9/465.3) 12.672 kN.m

m.Mx/Max+m.My/May 0.549x5.42/41.366+0.6x4.337/12.672 0.277 OK

Section (22.95 kg/m) 152x152 UC 23 [Grade 43]

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Baseplate for Post P01

Base-Plate Connection to BS 5950

LOADING CASE 001 : DEAD PLUS LIVE (ULTIMATE)

Basic Data Applied Forces at Interface Resultant Forces M, Fv, F Moment +0.3 kNm, Shear +0.1 kN, Axial +72.9 kN Forces taken from Support Reaction (Left side in tension, Axial Compression)

Basic Dimensions Column: 152x152UC23 [43] D=152.4, B=152.2, T=6.8, t=5.8, r=7.6, py=275 Bolts 16 mm Ø in 17 mm holes Grade 8.8 Bolts Plates S 275, Welds E 35 Data grout, Fcu, Fcv, py, slope 15 N/mm², 25 N/mm², 0.35 N/mm², 275 N/mm², 30 deg to vertical Design to BS 5950-1: 2000 and the SCI Green Book: Joints in Steel Construction : Moment Connections: SCI-P-207/95 Tk limited by Axial only elastic stress 4.13.2.3 Column Capacities Mc, Fvc, Fc 50.1 kN.m, 145.8 kN, 804.1 kN Fc = 804.1 kN OK

Summary of Results (Unity Ratios) Concrete Pressure (under Axial only) 0.17 OK Base-Plate thickness in Compression (under Axial only) 0.43 OK Concrete Pressure 0.17 OK Base-Plate thickness in Compression 0.35 OK Horizontal Shear 0.00 OK Flange & Web Welds 0.00 0.00 OK

Step 1: Base-Plate Pressure (under Axial only) Allowable Pressure=0.40•Fcu 0.40•15 6.0 N/mm² Base ecc=M/F 0.0/72.9 0.0 mm Pressure Configuration Compression Only Proj, Xcc 61.3, 137.5 L Zones X1, X2, X3, X4, X5 0.0, 129.4, 16.2, 129.4, 0.0 W Zones Wstiff, Wflange, Wweb 0.0, 274.8, 128.4 Ac=x2•wf+x3•ww+x4•wf 129.40•274.80 + 16.20•128.40 + 129.40•274.80 732.0 cm² Conc Cap C=0.40•Fcu•Ac 0.40•15•73198.3 439.2 kN OK Pressure=P•1000/Ac 72.9•1000/73198 1.00 N/mm² OK note: Clause 4.13.2.3 requires you to also check base under Axial only to Cl 4.13.2.5

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Step 2a: Plate Compression Bending (under Axial only) e=L1 61.3 61.3 mm Mapp=p•e²/2 1.0•61.3²/2 1871 Nmm/mm tp=√(6•Mapp/py) √(6•1871/275) 6.4 mm OK Note: Axial Load Axial Using Elastic Modulus Zp (4.13.2.2)

Step 1: Base-Plate Pressure Allowable Pressure=0.40•Fcu 0.40•15 6.0 N/mm² Base ecc=M/F 0.3/72.9 4.1 mm Pressure Configuration Compression Only Proj, Xcc 61.3, 133.4 L Zones X1, X2, X3, X4, X5 0.0, 129.4, 16.2, 121.4, 0.0 W Zones Wstiff, Wflange, Wweb 0.0, 274.8, 128.4 Ac=x2•wf+x3•ww+x4•wf 129.40•274.80 + 16.20•128.40 + 121.44•274.80 710.1 cm² Conc Cap C=0.40•Fcu•Ac 0.40•15•71009.7 426.1 kN OK Pressure=P•1000/Ac 72.9•1000/71010 1.03 N/mm² OK

Step 2a: Plate Compression Bending e=L1 61.3 61.3 mm Mapp=p•e²/2 1.0•61.3²/2 1929 Nmm/mm tp=√(4•Mapp/py) √(4•1929/275) 5.3 mm OK

Step 4: Shear Base Friction Friction Fr=0.30•Fc 0.30•+72.9 kN 21.9 kN

Bolt Bearing Bolt Shear Bs=Fn(ShrCap, tg) 58.9, 35 58.9 kN Concrete Bearing Cb=3•Ø²•0.4•Fcu 3•16²•0.4•15 (No Shear Reinf.) 4.6 kN Plate Bearing Pb=Fn(pb,e,Ø,t,kbs) 460, 35, 16, 15, 1.00 110.4 kN Bolt Bearing Bb=pb•Ø•tk 1000•16•15 240.0 kN Pss=Min(Bs, Cb, Pb, Bb)•nbs Min(58.9, 4.6, 110.4, 240.0) = 4.6•2 9.2 kN Pts=Min(Bsten, Cb, Pb, Bb)•nbt Min(58.9, 4.6, 110.4, 240.0) = 4.6•2, (no tension) 9.2 kN

Total Shear Capacity Total Cap=Fr+Pss+Pts 21.9 + 9.2 + 9.2 40.3 kN OK

Step 5: Flange & Web Welds Load dispersal Flanges resist Moment and Axial, Web resists Axial and Shear. Direct Bearing therefore design for tensile forces only. Areas A, Af, Aw 29.2, 2 x 10.3, 8.1 cm²

Flange Welds Fapp=M/la-F•Af/A 0.3/(152.4 - 6.8) - 72.9•10.3/29.2 0.0 kN No Resultant Tensile Force

Web Welds Web weld load=Fv/(D-2(fw+T)) 0.1/(152.4 - 2(6 +6.8)) 0.00 kN/mm Fcap w=2•0.7•leg•Py 2•0.7•6•220 1.85 kN/mm OK

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Pad Footing For Post P01

Use 1.2x1.2m Size Pad Footing, 0.45m Deep, Concrete Min C20/25

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B01, B02 and Post P01 Connection Detail

View from Below:

general construction notes and guidance

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