· /// A──┘ ooo BS5268-2:2002. ├─────────── L ───────────┤ The following assumptions are made in these calculations: that the

sheet 1

Pitched Roof: Open Vaulted weight No kN/m2 Timber Floor weight No kN/m2 natural slate 20*10 0.40 1 0.40 softwood boards 20mm 0.10 -

plain clay-concrete tile 0.77 - chipboard 22mm 0.15 - Interlocking concrete tile 0.51 - tiles & adhesive total 15mm 0.30 -

battens 0.04 1 0.04 sound insulated flooring 0.10 - TEK 172 SIP panel at 600 c/c 0.39 - UFH 0.10 -

PV panels 0.20 - insulation 0.06 - plywood 12mm 0.126 - plasterboard and skim 0.120 -

tyvek 0.02 1 0.02 lath and plaster 0.25 - 125mm celetex 0.06 1.2 0.07 joists 100*50 at 400 c/c 0.08 -

plasterboard and skim 0.120 1 0.12 joists 125*50 at 400 c/c 0.09 - softwood boards 20mm 0.10 - joists 150*50 at 400 c/c 0.11 -

Rafters 100*50 at 400 c/c 0.08 - joists 175*50 at 400 c/c 0.13 - Rafters 125*50 at 400 c/c 0.09 - Joists 200*50 at 400 c/c 0.15 - Rafters 150*50 at 400 c/c 0.11 1 0.11 Joists 220*50 at 400 c/c 0.17 -

Rafters 80*80 oak at 500 c/c 0.10 - Joists 220*75 at 400 c/c 0.25 - 0.76 DL= -

pitch angle 25 deg. 0.91 1 0.84 LL= 1.50 pitch angle 30 deg. 0.87 - pitch angle 35 deg. 0.82 - Block and Beam Floor - domestic - kN/m2 pitch angle 40 deg. 0.77 - pitch angle 45 deg. 0.71 - 4.60

DL= 0.84 4.70 pitch 30 LL= 0.6 4.90

LL= 1.50 Pitched Roof: Conventional weight No kN/m2

natural slate 20*10 0.40 1 0.40 Masonry walls: 100mm leafplain clay-concrete tile 0.77 - Brick/conc. block + plasterboard 2.2 -

Interlocking concrete tile 0.51 - LW Block + plasterboard 1.2 - battens 0.04 1 0.04 tyvek 0.02 1 0.02 Timber Wall: External weight No kN/m2

Rafters 100*50 at 400 c/c 0.08 - plywood 12mm 0.084 1 0.08 Rafters 125*50 at 400 c/c 0.09 - plywood 9mm 0.06 - Rafters 150*50 at 400 c/c 0.11 1 0.11 insulation 0.10 1 0.10

0.57 TEK 172 SIP 0.30 - pitch angle 25 deg. 0.91 1 0.63 plasterboard and skim 0.120 1 0.12 pitch angle 30 deg. 0.87 - lath and plaster 0.25 - pitch angle 35 deg. 0.82 - studs 100*50 at 400 c/c 0.08 - pitch angle 40 deg. 0.77 - studs 125*50 at 400 c/c 0.09 - pitch angle 45 deg. 0.71 - studs 150*50 at 400 c/c 0.11 1 0.11

DL= 0.63 noggins 1 0.03 pitch 30 LL= 0.6 battens 0.04 1 0.04

eml and render 0.66 - Loft Ceiling weight No kN/m2 timber cladding 0.19 1 0.19

insulation 0.06 1 0.06 natural slate 0.30 - chipboard 22mm 0.15 1 0.15 DL= 0.67

plasterboard and skim 0.120 1 0.12 lath and plaster 0.25 - Timber Wall: Internal weight No kN/m2

joists 100*50 at 400 c/c 0.08 - plywood 12mm 0.084 - joists 125*50 at 400 c/c 0.09 - plywood 9mm 0.06 - joists 150*50 at 400 c/c 0.11 1 0.11 insulation 0.10 1 0.10 joists 175*50 at 400 c/c 0.13 - plasterboard and skim 0.120 2 0.24 Joists 200*50 at 400 c/c 0.15 - lath and plaster 0.25 -

DL= 0.44 studs 100*50 at 400 c/c 0.08 1 0.08 access and storage LL= 0.30 studs 125*50 at 400 c/c 0.09 -

studs 150*50 at 400 c/c 0.11 - Balcony weight No kN/m2 noggins 1 0.02

single ply membrane+underlay 0.040 1 0.04 DL= 0.43 sedum lightweight bauder system 0.450 -

sedum standard system 1.240 - 3*felt system 0.110 1 0.11 W or SW S to E

standing seam zinc 0.100 - basic wind speed Vb 0.0025mm asphalt 0.580 - altitude of site A 200.00 200.00

Millboard composite decking 0.170 1 0.17 plywood 12mm 0.084 - distance to coast D 10.00 10.00plywood 18mm 0.126 1 0.13 max height of building He 7.50 7.50

firring average 50mm 0.038 1 0.04 altitude factor Sa 1.20 1.20foam insulation 150mm 0.048 1 0.05 directional factor Sd 1.00 0.85plasterboard and skim 0.120 1 0.12 seasonal factor Ss 1.00 1.00

lath and plaster 0.250 - probability factor Sp 1.00 1.00joists 100*50 at 400 c/c 0.075 - wind speed Vs 0.00 0.00joists 125*50 at 400 c/c 0.094 - terrain and bldg factor Sb 1.65 1.65joists 150*50 at 400 c/c 0.113 - effective wind speed Ve 0.00 0.00joists 175*50 at 400 c/c 0.131 1 0.13 dynamic pressure qs 0.00 0.00Joists 200*50 at 400 c/c 0.150 - net coefficient Cp 1.20 1.20

DL= 0.78 LL= 4.00 net surface pressure on walls p 0.00 0.00

net surface pressure on evelation area of pitched roofpitch angle 25 deg. - - pitch angle 30 deg. - - pitch angle 35 deg. - - pitch angle 40 deg. - - pitch angle 45 deg. - -

Loadings: 18106

Wind Loading Calculations to BS6399 Part 2 1997

Mexboro 155 (wide) 530mm up to 4.4m spanMexboro 155 (alternate) 420mm up to 4.9m spanMexboro 155 (narrow) 305mm up to 5.7m span

with 75 screed, tiles, insulation, batten, plasterboard

Ballantine Arnold Ltd Page: 2 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Location: balcony joists ────────────────────────

Domestic floor joist ├─┤b ──────────────────── dead+imposed A──┐ ┌─┐ ─┬─ These calculations follow │ │ │d the domestic floor joists ███████████████████████████ └─┘ ─┴─ example by V C Johnson in _▲_ _▲_ A-A TRADA Design Aid DA1 and /// A──┘ ooo BS5268-2:2002.

├─────────── L ───────────┤

The following assumptions are made in these calculations: ■ that the timber has a moisture content of service class 1 or 2 (i.e. K2=1) ■ the floor can adequately distribute any concentrated point load to at least two joists. ■ the centres of joists do not exceed 610 mm ■ that load sharing of the joists can occur & K8 = 1.1.

Effective span of joist L=2.6 m Centres of joists crs=400 mm Dead load including self weight dead=0.78 kN/m² Imposed udl load (on floor) live=4.00 kN/m² Imposed point load (on one joist) PL=1.35 kN Depth of section d=175 mm Width of section b=50 mm

Joist laterally restrained with support restraint as Table 19. Bearing length lb=50 mm Strength class C24 to Table 8.

Grade stresses ────────────── Bending parallel to grain bparg=7.5 N/mm² Shear parallel to grain sparg=0.71 N/mm² Compression perpend to grain cperd=2.4 N/mm² Mean modulus of elasticity Emean=10800 N/mm²

Modification factors ──────────────────── Duration of loading K3=1 Depth factor K7=(300/d)^0.11=(300/175)^0.11 =1.0611 Load sharing (Clause 2.9) K8=1.1 From BS5268-2 Table 18, bearing is < 75 mm from joist end. Bearing Modification factor K4=1.0

Ballantine Arnold Ltd Page: 3 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Permissible stresses ──────────────────── Bending parallel to grain sigmad=K3*K7*K8*bparg =1*1.0611*1.1*7.5 =8.7539 N/mm² Compress perp to grain (no wane) sigcad=K3*K4*K8*cperd=1*1*1.1*2.4 =2.64 N/mm² Shear parallel to grain torad=K3*K8*sparg=1*1.1*0.71 =0.781 N/mm²

Strength ──────── UDL case: UDL per metre run on joist f1=(dead+live)*crs/1000 =(0.78+4)*400/1000 =1.912 kN/m Bending moment M1=f1*L^2*10^6/8=1.912*2.6^2*10^6/8 =1.6156E6 Nmm Point load case with udl (dead only): UDL per metre run on joist (dead) f2=(dead)*crs/1000=(0.78)*400/1000 =0.312 kN/m Bending moment (dead load only) M2=f2*L^2*10^6/8=0.312*2.6^2*10^6/8 =263640 Nmm Bending moment point load M3=PL*L*10^6/4=1.35*2.6*10^6/4 =877500 Nmm Total bending moment M4=M2+M3=263640+877500=1.1411E6 Nmm Worst case moment (udl case governs) M=M1=1.6156E6 Nmm Section modulus Z=b*d^2/6=50*175^2/6=255208 mm³ Bending stress sigma=M/Z=1.6156E6/255208 =6.3307 N/mm² Since sigma < sigmad ( 6.3307 N/mm² < 8.7539 N/mm² ) bending stress is less than permissible stress therefore OK.

Deflection ────────── Second moment of area I=b*d^3/12=50*175^3/12=22.331E6 mm UDL case: Actual deflection including shear del1=5*f1*(L*1000)^4/(384*Emean*I)+12*f1*(L*1000)^2/(5*Emean*b*d) =5*1.912*(2.6*1000)^4/(384*10800*22.331E6)+12*1.912*(2.6*1000)^2 /(5*10800*50*175) =5.0456 mm Point load case: Actual deflection including shear del2=5*f2*(L*1000)^4/(384*Emean*I)+12*f2*(L*1000)^2/(5*Emean*b*d) =5*0.312*(2.6*1000)^4/(384*10800*22.331E6)+12*0.312*(2.6*1000)^2 /(5*10800*50*175) =0.82333 mm del3=PL*1000*(L*1000)^3/(48*Emean*I) =1.35*1000*(2.6*1000)^3/(48*10800*22.331E6) =2.0497 mm del4=del2+del3=0.82333+2.0497=2.873 mm UDL load case governs del=del1=5.0456 mm Limiting deflection Clause 2.10.7 dlim=0.003*L*1000=0.003*2.6*1000 =7.8 mm Deflection does not exceed limit ( 5.0456 mm ≤ 7.8 mm ) therefore OK.

Ballantine Arnold Ltd Page: 4 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Shear and bearing ───────────────── UDL case: Maximum applied shear force V1=f1*L/2=1.912*2.6/2=2.4856 kN Point load case: Maximum applied shear force V2=f2*L/2+PL=0.312*2.6/2+1.35 =1.7556 kN Worst shear (udl case governs) V=V1=2.4856 kN Shear stress tora=3*V*1000/(2*b*d) =3*2.4856*1000/(2*50*175) =0.4261 N/mm² Since tora ≤ torad ( 0.4261 N/mm² ≤ 0.781 N/mm² ) shear stress does not exceed permissible therefore OK. Bearing stress on support sigba=V*1000/(lb*b) =2.4856*1000/(50*50) =0.99424 N/mm² Since sigba ≤ sigcad ( 0.99424 N/mm² ≤ 2.64 N/mm² ) bearing stress does not exceed permissible therefore OK.

Joists: 175 mm x 50 mm @ 400 mm crs Strength class C24 to Table 8. Bending stress 6.3307 N/mm² Permissible bending 8.7539 N/mm² DESIGN Deflection 5.0456 mm SUMMARY Limiting deflection 7.8 mm Shear stress 0.4261 N/mm² Permissible shear 0.781 N/mm² Bearing stress 0.99424 N/mm² Permissible bearing 2.64 N/mm²

Ballantine Arnold Ltd Page: 5 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Location: trimming to lightwell ───────────────────────────────

Timber beam to BS5268-2:2002 ────────────────────────────

███████████████████████████████████████ Simply supported _▲_ _▲_ beam subjected to /// ooo vertical loads. ├──────────────── L ──────────────────┤

Beam span L=2.15 m All loads are positive downwards, reactions are positive upwards, sagging moments are positive.

├───────Lbu───────┤ Gku Dead UDL ├───Lau──┤ Qku Imposed UDL ┌────────┐ ═════════╧════════╧═════ Distances are measured ▀ ▀ from left hand support

Uniformly distributed load 1 of 1 Dist. from left support to start Lau(1)=0 m Distance from left support to end Lbu(1)=2.15 m Dead load (unfactored) Gku(1)=0.98 kN/m Imposed load (unfactored) Qku(1)=5 kN/m

BMs at 20th points, from left to right (sagging is positive) ──────────────────────────────────────────────────────────── 0 0.65651 1.2439 1.7622 2.2114 2.5915 2.9025 3.1443 3.3171 3.4208 3.4553 3.4208 3.3171 3.1443 2.9025 2.5915 2.2114 1.7622 1.2439 0.65651 0

Maximum span bending moment 3.4553 kNm

End shears ────────── Shear force at left hand end 6.4285 kN Shear force at right hand end 6.4285 kN Design shear force Fve'=6.4285

Unfactored dead shear at LHE 1.0535 kN Unfactored imposed shear at LHE 5.375 kN Unfactored dead shear at RHE 1.0535 kN Unfactored imposed shear at RHE 5.375 kN

Ballantine Arnold Ltd Page: 6 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 │y ┌───────┐ ─┬─ Section design parameters │ │ │ ───────────────────────── │ │ │ x ─ •│ │• ─ x │d │ │ │ │ │ │ └───────┘ ─┴─ │y

├── b ──┤

Design bending moment M=3.4553 kNm Design shear force Fve=6.4285 kN Design axial load (+ve comp) Fa=0 kN Depth of section d=175 mm Width of section b=150 mm Eff length for bending about xx Lex=2150 mm Eff length for bending about yy Ley=400 mm Length of bearing lb=50 mm From BS5268-2 Table 18, bearing is < 75 mm from joist end. Bearing Modification factor K4=1.0 Strength class C24 to Table 8. Timber service class adopted tmclass=2 Timber service class modification factor K2=1 as Table 16. Modification factors: Bending parallel to grain K2ben=1.0 Tension parallel to grain K2ten=1.0 Compression parallel to grain K2com=1.0 Compression ┴ to grain K2per=1.0 Shear parallel to grain K2shr=1.0 Mean & min modulus of elasticity K2mod=1.0

Section properties ────────────────── Inertia about xx axis Ix=b*d^3/12=150*175^3/12 =66.992E6 mm Inertia about yy axis Iy=d*b^3/12=175*150^3/12 =49.219E6 mm Area of cross section A=b*d=150*175=26250 mm² Radius of gyration @ x-x axis ix=SQR(Ix/A)=SQR(66.992E6/26250) =50.518 mm Radius of gyration @ y-y axis iy=SQR(Iy/A)=SQR(49.219E6/26250) =43.301 mm

Slenderness ─────────── Slenderness ratio @ x-x axis lambdx=Lex/ix=2150/50.518 =42.559 Slenderness ratio @ y-y axis lambdy=Ley/iy=400/43.301 =9.2376 xx axis controls slenderness at lambda=lambdx=42.559

Ballantine Arnold Ltd Page: 7 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Grade stresses ────────────── Compression parallel to grain cparg=7.9 N/mm² Bending parallel to grain bparg=7.5 N/mm² Shear parallel to grain sparg=0.71 N/mm² Compression ┴ to grain cperd=2.4 N/mm² Mean modulus of elasticity Emean=10800 N/mm² Minimum modulus of elasticity Emin=7200 N/mm²

Modification factors ──────────────────── Duration of loading K3=1.0 Depth factor K7=(300/d)^0.11=(300/175)^0.11 =1.0611 Member is not in a load-sharing system as defined by Clause 2.9. Modulus of elasticity mod factor K9=1.21 Member is a trimmer joist or lintel comprising two or more pieces connected together in parallel and acting together to support the loads - as defined in Clause 2.10.11. The minimum modulus of elasticity modified by the factor K9 should be used for deflections E=Emin*K2mod*K9=7200*1*1.21 =8712 N/mm² Load-sharing modification factor K8=1.1 clause 2.10.11 No notches exist at the support K5=1.0

Permissible stresses ────────────────────

Permissible bending stress sigmad=K2ben*K3*K7*K8*bparg =1*1*1.0611*1.1*7.5 =8.7539 N/mm² Shear parallel to grain torad=K2shr*K3*K5*K8*sparg =1*1*1*1.1*0.71 =0.781 N/mm² Compress ┴ to grain (no wane) sigbad=K2per*K3*K4*K8*cperd =1*1*1*1.1*2.4 =2.64 N/mm²

Bending ─────── Applied bending stress sigma=M*1E6*(d/2)/Ix =3.4553*1E6*(175/2)/66.992E6 =4.5131 N/mm² Since sigma ≤ sigmad ( 4.5131 N/mm² ≤ 8.7539 N/mm² ) applied bending stress within permissible.

Check deflection (including shear deflection as per Cl. 2.10.7) ──────────────── Deflection based on E=8712 N/mm² DL deflection without shear dld=dld=0.46717 mm Imposed deflection without shear ild=ild=2.3835 mm Total DL & imposed deflection 2.8507 mm Modulus of rigidity G=E/16=8712/16=544.5 N/mm² Shape factor for rect section KF=1.2

Ballantine Arnold Ltd Page: 8 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Shear area for beam Ay=d*b/KF=175*150/1.2=21875 mm² Total DL & imposed 12.857 kN If total DL & imposed load applied as a UDL, additional deflection due to shear dsu=WT'*L*10^6/(8*Ay*G) =12.857*2.15*10^6/(8*21875*544.5) =0.2901 mm Shear deflection dels=dsu*M/(WT'*L/8) =0.2901*3.4553/(12.857*2.15/8) =0.2901 mm Limiting deflection DELlim=0.003*L*10^3=0.003*2.15*10^3 =6.45 mm Deflection inclusive of shear DEL=dld+ild+dels =0.46717+2.3835+0.2901 =3.1408 mm Since DEL ≤ DELlim ( 3.1408 mm ≤ 6.45 mm ), OK for deflection.

Shear and bearing ───────────────── Shear stress (no notches) tora=3*Fve*1000/(2*b*d) =3*6.4285*1000/(2*150*175) =0.36734 N/mm² Since tora ≤ torad ( 0.36734 N/mm² ≤ 0.781 N/mm² ) shear stress does not exceed permissible therefore OK. Bearing stress on wall plate sigba=Fve*1000/(lb*b) =6.4285*1000/(50*150) =0.85713 N/mm² Since sigba ≤ sigbad ( 0.85713 N/mm² ≤ 2.64 N/mm² ) bearing stress does not exceed permissible therefore OK.

Member: 175 mm x 150 mm Strength class C24 to Table 8. Moisture service class 2 Bending stress 4.5131 N/mm² Permissible bending 8.7539 N/mm² DESIGN Deflection 3.1408 mm SUMMARY Limiting deflection 6.45 mm Shear stress 0.36734 N/mm² Permissible shear 0.781 N/mm² Bearing stress 0.85713 N/mm² Permissible bearing 2.64 N/mm²

Ballantine Arnold Ltd Page: 9 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Location: main breakthrough to function room ────────────────────────────────────────────

Simply supported steel beam ╒═══════════════════════╕ ─────────────────────────── ╘═══════════════════════╛ Calculations are in accordance _▲_ _▲_ with BS5950-1:2000. /// ooo

├────────── L ──────────┤

Beam span L=7.7 m

Section properties ────────────────── 457 x 152 x 52 UB. Dimensions (mm): D=449.8 B=152.4 t=7.6 T=10.9 r=10.2 Properties (cm): Ix=21400 Iy=645 Sx=1100 Sy=133 J=21.4 A=66.6 ry=3.112 rx=17.925

Strength of steel - Clause 3.1.1 ───────────────── The material thickness is 10.9 mm and the steel grade is S 275. Design strength py=275 N/mm² Young's Modulus E=205 kN/mm²

Loading ─────── Dead load factor gamd=1.4 Imposed load factor gami=1.6

All loads are positive downwards, reactions are positive upwards, sagging moments are positive.

├───────Lbu───────┤ Gku Dead UDL ├───Lau──┤ Qku Imposed UDL ┌────────┐ ═════════╧════════╧═════ Distances are measured ▀ ▀ from left hand support. Uniformly distributed load 1 of 1 Dist. from left support to start Lau(1)=0 m Distance from left support to end Lbu(1)=7.7 m Dead load (unfactored) Gku(1)=8.265 kN/m Imposed load (unfactored) Qku(1)=9.3 kN/m

BMs at 40th points, from left to right (sagging is positive) ──────────────────────────────────────────────────────────── 0 19.113 37.247 54.4 70.573 85.765 99.978 113.21 125.46 136.73 147.03 156.34 164.67 172.02 178.39 183.78 188.19 191.62 194.07 195.54 196.03 195.54 194.07 191.62 188.19 183.78 178.39 172.02 164.67 156.34 147.03 136.73 125.46 113.21 99.978 85.765 70.573 54.4 37.247 19.113 0


Ballantine Arnold Ltd Page: 10 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145

End shears ────────── Shear force at left hand end 101.84 kN Shear force at right hand end 101.84 kN Design shear force Fv=101.84 kN


Unfactored dead load deflection 8.6234 mm Unfactored imposed load deflectn 9.7032 mm Total DL & imposed deflection 18.327 mm Span:defln ratio for dead load 892.92 Span:defln ratio for imposed load 793.55 Span:defln ratio for total load 420.15 From Table 8 of BS5950-1:2000, Limiting deflection (brittle) DELlim=L*1000/360=7.7*1000/360 =21.389 mm Since imposed load deflection ≤ DELlim ( 9.7032 mm ≤ 21.389 mm ) deflection within limiting value.

Classification - Clause 3.5.2 ────────────── Classify outstand element of compression flange: Parameter (Table 11 Note b) e=(275/py)^0.5=(275/275)^0.5 =1 Outstand b=B/2=152.4/2=76.2 mm Ratio b'T=b/T=76.2/10.9=6.9908 As b/T ≤ 9e ( 9 ), outstand element of compression flange is classified as Class 1 plastic. Classify web of section: Depth between fillets d=D-2*(T+r)=449.8-2*(10.9+10.2) =407.6 mm Ratio d't=d/t=407.6/7.6=53.632 For sections with neutral axis at mid-depth As d/t ≤ 80e ( 80 ), web is classified as Class 1 plastic.

Shear buckling ────────────── Since d/t ≤ 70e no check for shear buckling is required.

Check section for combined moment and shear ─────────────────────────────────────────── Conservatively the maximum shear is checked with the maximum moment. Shear area Av=t*D=7.6*449.8=3418.5 mm² Shear capacity Pv=0.6*py*Av/10^3=0.6*275*3418.5/10^3 =564.05 kN Design shear force Fv=101.84 kN Elastic modulus Zx=Ix/(D/20)=21400/(449.8/20) =951.53 cm³ Since Fv < 0.6 Pv Moment capacity for compact sec Mc=py*Sx/10^3=275*1100/10^3 =302.5 kNm

Ballantine Arnold Ltd Page: 11 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Limiting moment capacity Mclim=1.2*py*Zx/10^3 =1.2*275*951.53/10^3 =314.01 kNm Since M ≤ Mc ( 196.03 kNm ≤ 302.5 kNm ), applied moment within moment capacity.

UNIVERSAL BEAM 457 x 152 x 52 UB Grade S 275 DESIGN SUMMARY Maximum shear force 101.84 kN Shear capacity 564.05 kN Max. applied moment 196.03 kNm Moment capacity 302.5 kNm Unfactored DL defln 8.6234 mm Unfactored LL defln 9.7032 mm Limiting deflection 21.389 mm ┌ DL shear at LHE 31.82 kN Unfactored │ LL shear at LHE 35.805 kN end shears │ DL shear at RHE 31.82 kN └ LL shear at RHE 35.805 kN

Ballantine Arnold Ltd Page: 12 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Location: main breakthrough to function room with post ──────────────────────────────────────────────────────


├────────── L ──────────┤

Beam span L=7.7/2=3.85 m

Section properties ────────────────── 254 x 102 x 28 UB. Dimensions (mm): D=260.4 B=102.2 t=6.3 T=10 r=7.6 Properties (cm): Ix=4010 Iy=179 Sx=353 Sy=54.8 J=9.57 A=36.1 ry=2.2268 rx=10.539

Strength of steel - Clause 3.1.1 ───────────────── The material thickness is 10 mm and the steel grade is S 275. Design strength py=275 N/mm² Young's Modulus E=205 kN/mm²



├───────Lbu───────┤ Gku Dead UDL ├───Lau──┤ Qku Imposed UDL ┌────────┐ ═════════╧════════╧═════ Distances are measured ▀ ▀ from left hand support. Uniformly distributed load 1 of 1 Dist. from left support to start Lau(1)=0 m Distance from left support to end Lbu(1)=3.85 m Dead load (unfactored) Gku(1)=8.265 kN/m Imposed load (unfactored) Qku(1)=9.3 kN/m







Classification - Clause 3.5.2 ────────────── Classify outstand element of compression flange: Parameter (Table 11 Note b) e=(275/py)^0.5=(275/275)^0.5 =1 Outstand b=B/2=102.2/2=51.1 mm Ratio b'T=b/T=51.1/10=5.11 As b/T ≤ 9e ( 9 ), outstand element of compression flange is classified as Class 1 plastic. Classify web of section: Depth between fillets d=D-2*(T+r)=260.4-2*(10+7.6) =225.2 mm Ratio d't=d/t=225.2/6.3=35.746 For sections with neutral axis at mid-depth As d/t ≤ 80e ( 80 ), web is classified as Class 1 plastic.

Shear buckling ────────────── Since d/t ≤ 70e no check for shear buckling is required.

Check section for combined moment and shear ─────────────────────────────────────────── Conservatively the maximum shear is checked with the maximum moment. Shear area Av=t*D=6.3*260.4=1640.5 mm² Shear capacity Pv=0.6*py*Av/10^3=0.6*275*1640.5/10^3 =270.69 kN Design shear force Fv=50.918 kN Elastic modulus Zx=Ix/(D/20)=4010/(260.4/20) =307.99 cm³ Since Fv < 0.6 Pv Moment capacity for compact sec Mc=py*Sx/10^3=275*353/10^3 =97.075 kNm

Ballantine Arnold Ltd Page: 14 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Limiting moment capacity Mclim=1.2*py*Zx/10^3 =1.2*275*307.99/10^3 =101.64 kNm Since M ≤ Mc ( 49.009 kNm ≤ 97.075 kNm ), applied moment within moment capacity.

UNIVERSAL BEAM 254 x 102 x 28 UB Grade S 275 DESIGN SUMMARY Maximum shear force 50.918 kN Shear capacity 270.69 kN Max. applied moment 49.009 kNm Moment capacity 97.075 kNm Unfactored DL defln 2.8762 mm Unfactored LL defln 3.2364 mm Limiting deflection 10.694 mm ┌ DL shear at LHE 15.91 kN Unfactored │ LL shear at LHE 17.903 kN end shears │ DL shear at RHE 15.91 kN └ LL shear at RHE 17.903 kN

Ballantine Arnold Ltd Page: 15 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Location: post to main fnc rm breakthrough ──────────────────────────────────────────

│y ┌─────────┐ ┬ Structural Hollow Section Column │ ┌─────┐ │ │ ──────────────────────────────── │ │ │ │ │ Calculations in accordance with z ─── │ │ │ │ ─── z │D BS5950:Part 1:2000 Sections 4.7 │ │ │ │ │ and 4.8. │ └─────┘ │ │ └─────────┘ ┴ All loads and moments are t ┼─┤ factored. │y

├────B────┤

Factored bending moment axis zz Mz=101*0.05=5.05 kNm Factored shear force y-y direct. Fv=10 kN Axial load (+ve compression) F=101 kN Length of member L=3 m

Steel section properties ──────────────────────── 140 x 140 x 6.3 SHS - Hot finished. Properties (cm): A=33.3 rx=5.44 Zx=141 Sx=166 Ix=984 J=1540 C=206 (where rz=rx, Zz=Zx, Sz=Sx & Iz=Ix) Factored bending moment axis yy My=101*0.05=5.05 kNm


Effective lengths - Clause 4.7.2 ───────────────── Length of member 3000 mm Length between restraints z axis Lz=3000 mm Length between restraints y axis Ly=3000 mm Effective length about z-z axis Lez=Kz*Lz=1*3000=3000 mm Effective length about y-y axis Ley=Ky*Ly=1*3000=3000 mm

Classification - Clause 3.5.2 ────────────── Classify internal element of compression flange: Parameter (Table 12 Note b) e=(275/py)^0.5=(275/275)^0.5 =1 Internal flange size b=B-3*t=140-3*6.3=121.1 mm Web depth d=D-3*t=140-3*6.3=121.1 mm As the section is subject to biaxial bending the classification will be based on the comment in appendix H.3.1 for RHS sections "the maximum axial force in each face should be determined taking account of the moment about the axis parallel to the face".

Ballantine Arnold Ltd Page: 16 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 For the z-z axis ──────────────── Axial force Fc=F+2*My*10^3/(B-t) =101+2*5.05*10^3/(140-6.3) =176.54 kN Ratio - local buckling of web d't=d/t=121.1/6.3=19.222 Ratio r1=Fc*10^3/(2*d*t*py) =176.54*10^3/(2*121.1*6.3*275) =0.42073 Limiting local buckling ratio d'tlim=40*e=40*1=40 As d't < 64e/(1+0.6r1) ( 51.1 ), web is classified as Class 1 plastic.

For the y-y axis ──────────────── Axial force Fc=F+2*Mz*10^3/(D-t) =101+2*5.05*10^3/(140-6.3) =176.54 kN Ratio - local buckling of web d't=b/t=121.1/6.3=19.222 Ratio r1=Fc*10^3/(2*b*t*py) =176.54*10^3/(2*121.1*6.3*275) =0.42073 Limiting local buckling ratio d'tlim=40*e=40*1=40 As d't < 64e/(1+0.6r1) ( 51.1 ), web is classified as Class 1 plastic.

Web shear ───────── Shear area Av=(D/(D+B))*A*100 =(140/(140+140))*33.3*100 =1665 mm² Shear capacity Pv=0.6*py*Av/10^3=0.6*275*1665/10^3 =274.73 kN Shear force Fv=10 kN Since Fv ≤ Pv ( 10 kN ≤ 274.73 kN ), shear force within shear capacity.

Moment capacities ───────────────── Moment capacity for compact sec Mcz=py*Sz/10^3=275*166/10^3 =45.65 kNm Minor axis moment capacity Mcy=Mcz=45.65 kNm Since Mz ≤ Mcz ( 5.05 kNm ≤ 45.65 kNm ) applied moment within moment capacity. Since My ≤ Mcy ( 5.05 kNm ≤ 45.65 kNm ) applied moment within moment capacity.

Local capacity check - Clause 4.8.3.2 ──────────────────── Considering the simplified approach of Clause 4.8.3.2,

F Mz My ──── + ─── + ─── ≤ 1.0 A.py Mcz Mcy

Ballantine Arnold Ltd Page: 17 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Unity factor uf=F*10/(A*py)+Mz/Mcz+My/Mcy =101*10/(33.3*275)+5.05/45.65+5.05 /45.65 =0.33154 The interaction formula is satisfied.

Slenderness - Clause 4.7.2 ─────────── Slenderness (minor axis) lam'y=Ley/(ry*10)=3000/(5.44*10) =55.147 Slenderness (major axis) lam'z=Lez/(rz*10)=3000/(5.44*10) =55.147 Controlling slenderness ratio lambda=55.147

Compressive strength: Perry strut formula from Annex C ────────────────────────────────────────────────────── Limiting slenderness lamb0=0.2*(PI^2*E*1000/py)^0.5 =0.2*(3.1416^2*205*1000/275)^0.5 =17.155 For buckling about y-y axis Robertson cont.for hollow sect. a=2.0 for table 24(a) hot finished Perry factor eta=0.001*a*(lam'y-lamb0) =0.001*2*(55.147-17.155) =0.075984 Euler strength pe=PI^2*E*1000/lam'y^2 =3.1416^2*205*1000/55.147^2 =665.29 N/mm² Factor phi=(py+(eta+1)*pe)/2 =(275+(0.075984+1)*665.29)/2 =495.42 N/mm² Compressive strength pcy=pe*py/(phi+(phi^2-pe*py)^0.5) =665.29*275/(495.42+(495.42^2 -665.29*275)^0.5) =245.45 N/mm² Compression resistance Pcy=A*pcy/10=33.3*245.45/10 =817.34 kN For buckling about z-z axis Robertson cont.for hollow sect. a=2.0 for table 24(a) hot finished Perry factor eta=0.001*a*(lam'z-lamb0) =0.001*2*(55.147-17.155) =0.075984 Euler strength pe=PI^2*E*1000/lam'z^2 =3.1416^2*205*1000/55.147^2 =665.29 N/mm² Factor phi=(py+(eta+1)*pe)/2 =(275+(0.075984+1)*665.29)/2 =495.42 N/mm² Compressive strength pcz=pe*py/(phi+(phi^2-pe*py)^0.5) =665.29*275/(495.42+(495.42^2 -665.29*275)^0.5) =245.45 N/mm² Compression resistance Pcz=A*pcz/10=33.3*245.45/10 =817.34 kN Compressive resistance Pc=817.34 kN Since F ≤ Pc ( 101 kN ≤ 817.34 kN ), design axial compressive load does not exceed compression resistance, section OK.


Equivalent uniform moment factors - Clause 4.8.3.3.4 ───────────────────────────────── For the overall buckling check the equivalent uniform moment factors are conservatively taken as: For lateral torsional buckling mLT=1.0 For major axis flexural bucking mz=1.0 For minor axis flexural buckling my=1.0 For lateral flexural buckling myz=1.0

Buckling resistance - Clause 4.3.6.1 ─────────────────── In accordance with Clause 4.3.6.1 square hollow sections need not be checked separately for resistance to lateral-torsional buckling. Buckling resistance moment Mb=Mcz=45.65 kNm

Member buckling resistance ────────────────────────── Overall buckling check - Clause 4.8.3.3.1 ──────────────────────

F mz.Mz my.My ───── + ──── + ───── ≤ 1.0 Pc py.Zz py.Zy

Unity factor unity=F/Pc+mz*Mz/Mcz+My/Mcy =101/817.34+1*5.05/45.65+5.05 /45.65 =0.34482 The unity relationship is satisfied.

F mLT.Mz my.My ───── + ────── + ───── ≤ 1.0 Pcy Mb py.Zy

Unity factor Unity=F/Pcy+mLT*Mz/Mb+My/Mcy =101/817.34+1*5.05/45.65+5.05 /45.65 =0.34482 The unity relationship is satisfied.

HOT FINISHED In accordance with EN 10210 SQUARE HOLLOW SECTION 140 x 140 x 6.3 SHS Grade S 275 SECTION Section is satisfactory for axial SUMMARY load, and overall buckling check. Axial load 101 kN Compression resistance 817.34 kN Maximum moment z axis 5.05 kNm Moment capacity 45.65 kNm Maximum moment y axis 5.05 kNm Moment capacity 45.65 kNm Local capacity check 0.33154 ≤ 1 Overall buckling checks 0.34482 ≤ 1 0.34482 ≤ 1

Ballantine Arnold Ltd Page: 19 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Location: cut truss purlin ──────────────────────────


├────────── L ──────────┤

Beam span L=3.40 m

Section properties ────────────────── 152 x 152 x 23 UC. Dimensions (mm): D=152.4 B=152.2 t=5.8 T=6.8 r=7.6 Properties (cm): Ix=1250 Iy=400 Sx=182 Sy=80.2 J=4.63 A=29.2 ry=3.7012 rx=6.5428




├─────Lc─────┤ Gkc Dead point load │ Qkc Imposed point load ═════════════════════════ Distances are measured ▀ ▀ from left hand support. Concentrated load 1 of 1 Distance from left support Lc(1)=3.4/2=1.7 m Dead load (unfactored) Gkc(1)=6.54 kN Imposed load (unfactored) Qkc(1)=6.2 kN

├───────Lbu───────┤ Gku Dead UDL ├───Lau──┤ Qku Imposed UDL ┌────────┐ ═════════╧════════╧═════ Distances are measured ▀ ▀ from left hand support. Uniformly distributed load 1 of 2 Dist. from left support to start Lau(1)=0.700 m Distance from left support to end Lbu(1)=2.700 m Dead load (unfactored) Gku(1)=3.945 kN/m Imposed load (unfactored) Qku(1)=2.07 kN/m Uniformly distributed load 2 of 2

Ballantine Arnold Ltd Page: 20 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Dist. from left support to start Lau(2)=0000 m Distance from left support to end Lbu(2)=3.400 m Dead load (unfactored) Gku(2)=0.3 kN/m Imposed load (unfactored) Qku(2)=0 kN/m






Classification - Clause 3.5.2 ────────────── Classify outstand element of compression flange: Parameter (Table 11 Note b) e=(275/py)^0.5=(275/275)^0.5 =1 Outstand b=B/2=152.2/2=76.1 mm Ratio b'T=b/T=76.1/6.8=11.191 As 10e < b/T ≤ 15e ( 15 ), outstand element of compression flange is classified as Class 3 semi-compact. Classify web of section: Depth between fillets d=D-2*(T+r)=152.4-2*(6.8+7.6) =123.6 mm Ratio d't=d/t=123.6/5.8=21.31 For sections with neutral axis at mid-depth As d/t ≤ 80e ( 80 ), web is classified as Class 1 plastic.

Ballantine Arnold Ltd Page: 21 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Shear buckling ────────────── Since d/t ≤ 70e no check for shear buckling is required.

Buckling resistance ─────────────────── Since the beam is subject to possible lateral torsional buckling, the buckling resistance moment Mb is first considered rather than the moment capacity Mc as a guide to selection. A conservative analysis is adopted, taking the worst moment with the greatest effective length.

Effective length - Clause 4.3.5.2 ──────────────── As the point load does not provide restraint to the compression flange, the effective length, Le, is found by considering the restraint conditions at the supports from Table 13 based on the length between the restraints equal to the beam span LT=L=3.4 m Compression flange laterally unrestrained. Both flanges free to rotate on plan. Partial restraint against rotation provided only by pressure of bottom flange onto supports Normal loading conditions therefore Effective length (Table 13) Le=1.2*LT+2*D/1000 =1.2*3.4+2*152.4/1000 =4.3848 m

Clause 4.3.6.6 and Table 18 ─────────────────────────── Equivalent uniform moment factor ──────────────────────────────── The member is loaded between restraints. Moments on beam, all moments are taken as positive. Start m1=0 kNm Quarter point m2=16 kNm Mid-span m3=27.5 kNm Three quarter point m4=16 kNm End m5=0 kNm Equivalent uniform moment factor mLT=0.2+(0.15*m2+0.5*m3+0.15*m4)/M =0.2+(0.15*16+0.5*27.5+0.15*16) /27.424 =0.87643

Resistance to lateral-torsional buckling - Clause 4.3.6 ──────────────────────────────────────── Radius gyration about minor axis ry=SQR(Iy/A)=SQR(400/29.2) =3.7012 cm Slenderness of section lambda=Le/ry*100=4.3848/3.7012*100 =118.47 Elastic modulus Zx=Ix/(D/20)=1250/(152.4/20) =164.04 cm³ Buckling parameter u=(4*Sx^2*(1-Iy/Ix)/(A^2*((D-T) /10)^2))^0.25 =(4*182^2*(1-400/1250)/(29.2^2 *((152.4-6.8)/10)^2))^0.25 =0.84024

Ballantine Arnold Ltd Page: 22 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Torsional index x=0.566*((D-T)/10)*(A/J)^0.5 =0.566*((152.4-6.8)/10)*(29.2 /4.63)^0.5 =20.696 Ratio ratio=lambda/x=118.47/20.696 =5.7244 Slenderness factor v=1/((1+0.05*ratio^2)^0.25) =1/((1+0.05*5.7244^2)^0.25) =0.78463 Ratio βw betaw=Zx/Sx=164.04/182=0.90133 Equivalent slenderness lamLT=u*v*lambda*(betaw)^0.5 =0.84024*0.78463*118.47 *(0.90133)^0.5 =74.152 Limiting slenderness lamlo=0.4*((PI^2*E*10^3)/py)^0.5 =0.4*((3.1416^2*205*10^3) /275)^0.5 =34.31 Perry coefficient etaLT=0.007*(lamLT-lamlo) =0.007*(74.152-34.31) =0.27889 Elastic strength pe=PI^2*E*10^3/(lamLT^2) =3.1416^2*205*10^3/(74.152^2) =367.97 N/mm² Factor phiLT=(py+(etaLT+1)*pe)/2 =(275+(0.27889+1)*367.97)/2 =372.8 N/mm² Factor pey=pe*py=367.97*275=101191 Bending strength pb=(pey)/(phiLT+((phiLT^2-pey)^0.5)) =(101191)/(372.8+((372.8^2 -101191)^0.5)) =178.41 N/mm² Buckling resistance moment Mb=Zx*pb/10^3=164.04*178.41/10^3 =29.267 kNm Equivalent uniform moment factor mLT=0.87643 Since M ≤ Mb/mLT ( 27.424 kNm ≤ 33.393 kNm ), section OK for lateral torsional buckling resistance.

Check section for combined moment and shear ─────────────────────────────────────────── Conservatively the maximum shear is checked with the maximum moment. Shear area Av=t*D=5.8*152.4=883.92 mm² Shear capacity Pv=0.6*py*Av/10^3=0.6*275*883.92/10^3 =145.85 kN Design shear force Fv=19.087 kN Elastic modulus Zx=Ix/(D/20)=1250/(152.4/20) =164.04 cm³ Moment capacity based on elastic modulus of section. Since Fv < 0.6 Pv Mc=py*Zx/10^3=275*164.04/10^3 =45.112 kNm Since M ≤ Mc ( 27.424 kNm ≤ 45.112 kNm ), applied moment within moment capacity. N.B. Buckling resistance is less than moment capacity and therefore buckling resistance controls the design.

Ballantine Arnold Ltd Page: 23 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 UNIVERSAL COLUMN 152 x 152 x 23 UC Grade S 275 DESIGN SUMMARY Maximum shear force 19.087 kN Shear capacity 145.85 kN Max. applied moment 27.424 kNm Moment capacity 45.112 kNm Buckling resistance 29.267 kNm Moment factor (mLT) 0.87643 Resistance (Mb/mLT) 33.393 kNm Unfactored DL defln 4.4427 mm Unfactored LL defln 3.1089 mm Limiting deflection 9.4444 mm ┌ DL shear at LHE 7.725 kN Unfactored │ LL shear at LHE 5.17 kN end shears │ DL shear at RHE 7.725 kN └ LL shear at RHE 5.17 kN

Ballantine Arnold Ltd Page: 24 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Location: post t cut truss purlin ─────────────────────────────────

│y ┌─────────┐ ┬ Structural Hollow Section Column │ ┌─────┐ │ │ ──────────────────────────────── │ │ │ │ │ Calculations in accordance with z ─── │ │ │ │ ─── z │D BS5950:Part 1:2000 Sections 4.7 │ │ │ │ │ and 4.8. │ └─────┘ │ │ └─────────┘ ┴ All loads and moments are t ┼─┤ factored. │y

├────B────┤

Factored bending moment axis zz Mz=19*0.05=0.95 kNm Factored shear force y-y direct. Fv=2.0 kN Axial load (+ve compression) F=19.0 kN Length of member L=2.4 m

Steel section properties ──────────────────────── 90 x 90 x 3.6 SHS - Hot finished. Properties (cm): A=12.3 rx=3.52 Zx=33.8 Sx=39.7 Ix=152 J=237 C=49.7 (where rz=rx, Zz=Zx, Sz=Sx & Iz=Ix) Factored bending moment axis yy My=20*0.05=1 kNm


Effective lengths - Clause 4.7.2 ───────────────── Length of member 2400 mm Length between restraints z axis Lz=2400 mm Length between restraints y axis Ly=2400 mm Effective length about z-z axis Lez=Kz*Lz=1*2400=2400 mm Effective length about y-y axis Ley=Ky*Ly=1*2400=2400 mm

Classification - Clause 3.5.2 ────────────── Classify internal element of compression flange: Parameter (Table 12 Note b) e=(275/py)^0.5=(275/275)^0.5 =1 Internal flange size b=B-3*t=90-3*3.6=79.2 mm Web depth d=D-3*t=90-3*3.6=79.2 mm As the section is subject to biaxial bending the classification will be based on the comment in appendix H.3.1 for RHS sections "the maximum axial force in each face should be determined taking account of the moment about the axis parallel to the face".

Ballantine Arnold Ltd Page: 25 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 For the z-z axis ──────────────── Axial force Fc=F+2*My*10^3/(B-t) =19+2*1*10^3/(90-3.6) =42.148 kN Ratio - local buckling of web d't=d/t=79.2/3.6=22 Ratio r1=Fc*10^3/(2*d*t*py) =42.148*10^3/(2*79.2*3.6*275) =0.26877 Limiting local buckling ratio d'tlim=40*e=40*1=40 As d't < 64e/(1+0.6r1) ( 55.112 ), web is classified as Class 1 plastic.

For the y-y axis ──────────────── Axial force Fc=F+2*Mz*10^3/(D-t) =19+2*0.95*10^3/(90-3.6) =40.991 kN Ratio - local buckling of web d't=b/t=79.2/3.6=22 Ratio r1=Fc*10^3/(2*b*t*py) =40.991*10^3/(2*79.2*3.6*275) =0.26139 Limiting local buckling ratio d'tlim=40*e=40*1=40 As d't < 64e/(1+0.6r1) ( 55.323 ), web is classified as Class 1 plastic.

Web shear ───────── Shear area Av=(D/(D+B))*A*100 =(90/(90+90))*12.3*100 =615 mm² Shear capacity Pv=0.6*py*Av/10^3=0.6*275*615/10^3 =101.48 kN Shear force Fv=2 kN Since Fv ≤ Pv ( 2 kN ≤ 101.48 kN ), shear force within shear capacity.

Moment capacities ───────────────── Moment capacity for compact sec Mcz=py*Sz/10^3=275*39.7/10^3 =10.918 kNm Minor axis moment capacity Mcy=Mcz=10.918 kNm Since Mz ≤ Mcz ( 0.95 kNm ≤ 10.918 kNm ) applied moment within moment capacity. Since My ≤ Mcy ( 1 kNm ≤ 10.918 kNm ) applied moment within moment capacity.

Local capacity check - Clause 4.8.3.2 ──────────────────── Considering the simplified approach of Clause 4.8.3.2,

F Mz My ──── + ─── + ─── ≤ 1.0 A.py Mcz Mcy

Ballantine Arnold Ltd Page: 26 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Unity factor uf=F*10/(A*py)+Mz/Mcz+My/Mcy =19*10/(12.3*275)+0.95/10.918+1 /10.918 =0.23478 The interaction formula is satisfied.

Slenderness - Clause 4.7.2 ─────────── Slenderness (minor axis) lam'y=Ley/(ry*10)=2400/(3.52*10) =68.182 Slenderness (major axis) lam'z=Lez/(rz*10)=2400/(3.52*10) =68.182 Controlling slenderness ratio lambda=68.182

Compressive strength: Perry strut formula from Annex C ────────────────────────────────────────────────────── Limiting slenderness lamb0=0.2*(PI^2*E*1000/py)^0.5 =0.2*(3.1416^2*205*1000/275)^0.5 =17.155 For buckling about y-y axis Robertson cont.for hollow sect. a=2.0 for table 24(a) hot finished Perry factor eta=0.001*a*(lam'y-lamb0) =0.001*2*(68.182-17.155) =0.10205 Euler strength pe=PI^2*E*1000/lam'y^2 =3.1416^2*205*1000/68.182^2 =435.23 N/mm² Factor phi=(py+(eta+1)*pe)/2 =(275+(0.10205+1)*435.23)/2 =377.32 N/mm² Compressive strength pcy=pe*py/(phi+(phi^2-pe*py)^0.5) =435.23*275/(377.32+(377.32^2 -435.23*275)^0.5) =226.71 N/mm² Compression resistance Pcy=A*pcy/10=12.3*226.71/10 =278.85 kN For buckling about z-z axis Robertson cont.for hollow sect. a=2.0 for table 24(a) hot finished Perry factor eta=0.001*a*(lam'z-lamb0) =0.001*2*(68.182-17.155) =0.10205 Euler strength pe=PI^2*E*1000/lam'z^2 =3.1416^2*205*1000/68.182^2 =435.23 N/mm² Factor phi=(py+(eta+1)*pe)/2 =(275+(0.10205+1)*435.23)/2 =377.32 N/mm² Compressive strength pcz=pe*py/(phi+(phi^2-pe*py)^0.5) =435.23*275/(377.32+(377.32^2 -435.23*275)^0.5) =226.71 N/mm² Compression resistance Pcz=A*pcz/10=12.3*226.71/10 =278.85 kN Compressive resistance Pc=278.85 kN Since F ≤ Pc ( 19 kN ≤ 278.85 kN ), design axial compressive load does not exceed compression resistance, section OK.


Equivalent uniform moment factors - Clause 4.8.3.3.4 ───────────────────────────────── For the overall buckling check the equivalent uniform moment factors are conservatively taken as: For lateral torsional buckling mLT=1.0 For major axis flexural bucking mz=1.0 For minor axis flexural buckling my=1.0 For lateral flexural buckling myz=1.0

Buckling resistance - Clause 4.3.6.1 ─────────────────── In accordance with Clause 4.3.6.1 square hollow sections need not be checked separately for resistance to lateral-torsional buckling. Buckling resistance moment Mb=Mcz=10.918 kNm

Member buckling resistance ────────────────────────── Overall buckling check - Clause 4.8.3.3.1 ──────────────────────

F mz.Mz my.My ───── + ──── + ───── ≤ 1.0 Pc py.Zz py.Zy

Unity factor unity=F/Pc+mz*Mz/Mcz+My/Mcy =19/278.85+1*0.95/10.918+1/10.918 =0.24675 The unity relationship is satisfied.

F mLT.Mz my.My ───── + ────── + ───── ≤ 1.0 Pcy Mb py.Zy

Unity factor Unity=F/Pcy+mLT*Mz/Mb+My/Mcy =19/278.85+1*0.95/10.918+1/10.918 =0.24675 The unity relationship is satisfied.

HOT FINISHED In accordance with EN 10210 SQUARE HOLLOW SECTION 90 x 90 x 3.6 SHS Grade S 275 SECTION Section is satisfactory for axial SUMMARY load, and overall buckling check. Axial load 19 kN Compression resistance 278.85 kN Maximum moment z axis 0.95 kNm Moment capacity 10.918 kNm Maximum moment y axis 1 kNm Moment capacity 10.918 kNm Local capacity check 0.23478 ≤ 1 Overall buckling checks 0.24675 ≤ 1 0.24675 ≤ 1

Ballantine Arnold Ltd Page: 28 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Location: cut rafter ────────────────────

b ├───────────────────────┤│o Top of rafter is pinned and restrained ─┬─ ▄█◄│o against horizontal movement, but │ ▄█▀▀2 │o unrestrained against vertical │ ▄█▀▀ movement. │ ▄█▀▀ │c ▄█▀▀ │ ▄█▀▀ │ ▄█▀▀ │ 1▄█▀▀ ─┴─_▲_ Bottom of rafter is pinned and restrained /// against horizontal movement.

Rafter projection onto horizontal b=3.0 m Rafter projection onto vertical c=1.8 m Elastic modulus E=8.8E6 kN/m² Moment of inertia of rafter I=4.46E-5 m Slope of roof alpha=ATN(c/b)=ATN(1.8/3) =0.54042 radians alpha'=DEG(alpha)=30.964 degrees Length of rafter r=SQR(b^2+c^2)=SQR(3^2+1.8^2) =3.4986 m Loads and reactions are positive in the directions shown.

│ │ │ │ │ │ │ │ │ │ │ │ 2 w2 ▄█◄───H2 ▄█▀▀ ▄█▀▀ Load is UDL applied ▄█▀▀ vertically on plan. ▄█▀▀ ▄█▀▀ H1 ▄█▀▀ ───►▄█▀▀ 1 │V1

Plan load per unit length w2=0.58 kN/m Total load on rafter W=w2*b=0.58*3=1.74 kN Vertical reaction at 1 (+ve up) V1=W=1.74 kN Maximum moment (+ve sagging) M=W*COS(alpha)*r/8 =1.74*COS(0.54042)*3.4986/8 =0.6525 kNm Max defln in rafter (+ve is sag) D1=5*W*COS(alpha)*r^3/(384*E*I) =5*1.74*COS(0.54042)*3.4986^3/(384 *8.8E6*44.6E-6) =0.0021197 m Span/deflection ratio s'd=r/D1=3.4986/0.0021197 =1650.5 Deflection/span ratio d's=D1/r=0.0021197/3.4986 =0.60587E-3

Ballantine Arnold Ltd Page: 29 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Horizontal react at 1 (+ve right) H1=W*b/(2*c)=1.74*3/(2*1.8) =1.45 kN Horizontal react at 2 (+ve left) H2=1.45 kN Axial force at 1 (+ve compress.) F1=V1*SIN(alpha)+H1*COS(alpha) =1.74*SIN(0.54042)+1.45*COS(0.54042) =2.1386 kN Shear force at 1 S1=V1*COS(alpha)-H1*SIN(alpha) =1.74*COS(0.54042)-1.45*SIN(0.54042) =0.74602 kN

Ballantine Arnold Ltd Page: 30 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Location: FLOOR BEAMS OF CUT TRUSSES ────────────────────────────────────

Timber beam to BS5268-2:2002 ────────────────────────────




Uniformly distributed load 1 of 1 Dist. from left support to start Lau(1)=0 m Distance from left support to end Lbu(1)=4.50 m Dead load (unfactored) Gku(1)=0.200 kN/m Imposed load (unfactored) Qku(1)=0.800 kN/m






├── b ──┤

Design bending moment M=2.5313 kNm Design shear force Fve=2.25 kN Design axial load (+ve comp) Fa=0 kN Depth of section d=225 mm Width of section b=47. mm Eff length for bending about xx Lex=4500 mm Eff length for bending about yy Ley=000 mm Length of bearing lb=50 mm From BS5268-2 Table 18, bearing is < 75 mm from joist end. Bearing Modification factor K4=1.0 Strength class C24 to Table 8. Timber service class adopted tmclass=1 Timber service class modification factor K2=1 as Table 16. Modification factors: Bending parallel to grain K2ben=1.0 Tension parallel to grain K2ten=1.0 Compression parallel to grain K2com=1.0 Compression ┴ to grain K2per=1.0 Shear parallel to grain K2shr=1.0 Mean & min modulus of elasticity K2mod=1.0

Section properties ────────────────── Inertia about xx axis Ix=b*d^3/12=47*225^3/12=44.613E6 mm Inertia about yy axis Iy=d*b^3/12=225*47^3/12=1.9467E6 mm Area of cross section A=b*d=47*225=10575 mm² Radius of gyration @ x-x axis ix=SQR(Ix/A)=SQR(44.613E6/10575) =64.952 mm Radius of gyration @ y-y axis iy=SQR(Iy/A)=SQR(1.9467E6/10575) =13.568 mm

Slenderness ─────────── Slenderness ratio @ x-x axis lambdx=Lex/ix=4500/64.952 =69.282 Slenderness ratio @ y-y axis lambdy=Ley/iy=0/13.568=0 xx axis controls slenderness at lambda=lambdx=69.282


Modification factors ──────────────────── Duration of loading K3=1 Depth factor K7=(300/d)^0.11=(300/225)^0.11 =1.0322 Member is in a load-sharing system as defined by Clause 2.9. Load-sharing modification factor K8=1.1 The mean modulus of elasticity should be used to calculate deflections and displacements under both dead and imposed loads. Modulus of elasticity E=Emean*K2mod=10800*1=10800 N/mm² No notches exist at the support K5=1.0


Permissible bending stress sigmad=K2ben*K3*K7*K8*bparg =1*1*1.0322*1.1*7.5 =8.5152 N/mm² Shear parallel to grain torad=K2shr*K3*K5*K8*sparg =1*1*1*1.1*0.71 =0.781 N/mm² Compress ┴ to grain (no wane) sigbad=K2per*K3*K4*K8*cperd =1*1*1*1.1*2.4 =2.64 N/mm²


Check deflection (including shear deflection as per Cl. 2.10.7) ──────────────── Deflection based on E=10800 N/mm² DL deflection without shear dld=dld=2.2163 mm Imposed deflection without shear ild=ild=8.8652 mm Total DL & imposed deflection 11.082 mm Modulus of rigidity G=E/16=10800/16=675 N/mm² Shape factor for rect section KF=1.2 Shear area for beam Ay=d*b/KF=225*47/1.2=8812.5 mm² Total DL & imposed 4.5 kN If total DL & imposed load applied as a UDL, additional deflection due to shear dsu=WT'*L*10^6/(8*Ay*G) =4.5*4.5*10^6/(8*8812.5*675) =0.42553 mm

Ballantine Arnold Ltd Page: 33 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Shear deflection dels=dsu*M/(WT'*L/8) =0.42553*2.5313/(4.5*4.5/8) =0.42553 mm Limiting deflection DELlim=0.003*L*10^3=0.003*4.5*10^3 =13.5 mm Deflection inclusive of shear DEL=dld+ild+dels =2.2163+8.8652+0.42553 =11.507 mm Since DEL ≤ DELlim ( 11.507 mm ≤ 13.5 mm ), OK for deflection.



Ballantine Arnold Ltd Page: 34 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Location: Dormer roof joists ────────────────────────────

Timber beam to BS5268-2:2002 ────────────────────────────




Uniformly distributed load 1 of 1 Dist. from left support to start Lau(1)=0 m Distance from left support to end Lbu(1)=4.3 m Dead load (unfactored) Gku(1)=0.5*0.4=0.2 kN/m Imposed load (unfactored) Qku(1)=000 kN/m




Unfactored dead shear at LHE 0.43 kN Unfactored imposed shear at LHE 0 kN Unfactored dead shear at RHE 0.43 kN Unfactored imposed shear at RHE 0 kN


├── b ──┤

Design bending moment M=0.46225 kNm Design shear force Fve=0.43 kN Design axial load (+ve comp) Fa=0 kN Depth of section d=150 mm Width of section b=47 mm Eff length for bending about xx Lex=4300 mm Eff length for bending about yy Ley=0 mm Length of bearing lb=50 mm From BS5268-2 Table 18, bearing is < 75 mm from joist end. Bearing Modification factor K4=1.0 Strength class C16 to Table 8. Timber service class adopted tmclass=2 Timber service class modification factor K2=1 as Table 16. Modification factors: Bending parallel to grain K2ben=1.0 Tension parallel to grain K2ten=1.0 Compression parallel to grain K2com=1.0 Compression ┴ to grain K2per=1.0 Shear parallel to grain K2shr=1.0 Mean & min modulus of elasticity K2mod=1.0

Section properties ────────────────── Inertia about xx axis Ix=b*d^3/12=47*150^3/12=13.219E6 mm Inertia about yy axis Iy=d*b^3/12=150*47^3/12=1.2978E6 mm Area of cross section A=b*d=47*150=7050 mm² Radius of gyration @ x-x axis ix=SQR(Ix/A)=SQR(13.219E6/7050) =43.301 mm Radius of gyration @ y-y axis iy=SQR(Iy/A)=SQR(1.2978E6/7050) =13.568 mm

Slenderness ─────────── Slenderness ratio @ x-x axis lambdx=Lex/ix=4300/43.301 =99.304 Slenderness ratio @ y-y axis lambdy=Ley/iy=0/13.568=0 xx axis controls slenderness at lambda=lambdx=99.304


Modification factors ──────────────────── Duration of loading K3=1.25 Depth factor K7=(300/d)^0.11=(300/150)^0.11 =1.0792 Member is in a load-sharing system as defined by Clause 2.9. Load-sharing modification factor K8=1.1 The mean modulus of elasticity should be used to calculate deflections and displacements under both dead and imposed loads. Modulus of elasticity E=Emean*K2mod=8800*1=8800 N/mm² No notches exist at the support K5=1.0


Permissible bending stress sigmad=K2ben*K3*K7*K8*bparg =1*1.25*1.0792*1.1*5.3 =7.8649 N/mm² Shear parallel to grain torad=K2shr*K3*K5*K8*sparg =1*1.25*1*1.1*0.67 =0.92125 N/mm² Compress ┴ to grain (no wane) sigbad=K2per*K3*K4*K8*cperd =1*1.25*1*1.1*2.2 =3.025 N/mm²


Check deflection (including shear deflection as per Cl. 2.10.7) ──────────────── Deflection based on E=8800 N/mm² DL deflection without shear dld=dld=7.6537 mm Imposed deflection without shear ild=ild=0 mm Total DL & imposed deflection 7.6537 mm Modulus of rigidity G=E/16=8800/16=550 N/mm² Shape factor for rect section KF=1.2 Shear area for beam Ay=d*b/KF=150*47/1.2=5875 mm² Total DL & imposed 0.86 kN If total DL & imposed load applied as a UDL, additional deflection due to shear dsu=WT'*L*10^6/(8*Ay*G) =0.86*4.3*10^6/(8*5875*550) =0.14306 mm

Ballantine Arnold Ltd Page: 37 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Shear deflection dels=dsu*M/(WT'*L/8) =0.14306*0.46225/(0.86*4.3/8) =0.14306 mm Limiting deflection DELlim=0.003*L*10^3=0.003*4.3*10^3 =12.9 mm Deflection inclusive of shear DEL=dld+ild+dels=7.6537+0+0.14306 =7.7967 mm Since DEL ≤ DELlim ( 7.7967 mm ≤ 12.9 mm ), OK for deflection.



Ballantine Arnold Ltd Page: 38 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Location: ridge beam ────────────────────

Simply supported glulam beam to BS5268-2:2002 ─────────────────────────────────────────────




Uniformly distributed load 1 of 1 Dist. from left support to start Lau(1)=0 m Distance from left support to end Lbu(1)=6.6 m Dead load (unfactored) Gku(1)=1.58+.35=1.93 kN/m Imposed load (unfactored) Qku(1)=1.50 kN/m





Ballantine Arnold Ltd Page: 39 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Rectangular glulam member │y ───────────────────────── ┌───────┐ ─┬─ with horizontal laminates ├───────┤ │ ───────────────────────── ├───────┤ │ Member laterally restrained x ─── ├───────┤ ─── x │d ─────────────────────────── ─┬─├───────┤ │ in bending tp─┴─├───────┤ │ ────────── └───────┘ ─┴─ │y ├── b ──┤ Mx=M'*10^6=18.676*10^6=18.676E6 Nmm V=Fve'*1000=11.319*1000=11319 N Design axial load (+ve compress) F'=0 kN F=F'*1000=0*1000=0 N Number of laminates in section np=9 Thickness of laminates in section tp=45 mm Width of section b=140 mm Depth of section d=tp*np=45*9=405 mm Glulam beam laterally restrained with support restraint as Table 19.

Grade stresses ────────────── For horizontally glue laminated members the grade stresses should be taken as the product of the strength class stresses given in Table 8 for the relevant grade and species and the modification factors of Table 24. Chosen species is a Softwood. Strength class C24 to Table 8.

Section properties ────────────────── Largest section dimension h=d=405 mm Inertia about zz axis Iz=b*d^3/12=140*405^3/12 =775.02E6 mm Inertia about yy axis Iy=d*b^3/12=405*140^3/12 =92.61E6 mm Area of cross section A=b*d=140*405=56700 mm² Radius of gyration about zz axis iz=SQR(Iz/A)=SQR(775.02E6/56700) =116.91 mm Radius of gyration about yy axis iy=SQR(Iy/A)=SQR(92.61E6/56700) =40.415 mm

Grade stresses ────────────── Compression parallel to grain cparg=7.9 N/mm² Bending parallel to grain bparg=7.5 N/mm² Shear parallel to grain sparg=0.71 N/mm² Compression ┴ to grain cperd=2.4 N/mm² Mean modulus of elasticity Emean=10800 N/mm² Minimum modulus of elasticity Emin=7200 N/mm²

Ballantine Arnold Ltd Page: 40 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Modification factors ──────────────────── Timber service class adopted tmclass=3 Timber stress grades and moduli adjusted as table 16. Modification factors: Bending parallel to grain K2ben=0.8 Tension parallel to grain K2ten=0.8 Compression parallel to grain K2com=0.6 Compression ┴ to grain K2per=0.6 Shear parallel to grain K2shr=0.9 Mean & min modulus of elasticity K2mod=0.8 Duration of loading K3=1.25 Depth factor K7=0.81*(d^2+92300)/(d^2+56800) =0.81*(405^2+92300)/(405^2+56800) =0.94022 Member is not in a load-sharing system as defined by Clause 2.9 Load-sharing modification factor K8=1.0 Modification factor K15 K15=1.41 Modification factor K19 K19=2.34 Modification factor K20 K20=1.07

Permissible stresses ──────────────────── Permissible bending stress sigmad=K2ben*K3*K7*K8*K15*bparg =0.8*1.25*0.94022*1*1.41*7.5 =9.9428 N/mm² Depth of notch at support notch=100 mm he=d-notch=405-100=305 mm Notch is on bottom edge K5=he/d=305/405=0.75309 Permissible shear stress torad=K2shr*K3*K5*K8*K19*sparg =0.9*1.25*0.75309*1*2.34*0.71 =1.4076 N/mm²

Bending stress ────────────── Applied bending stress sigma=Mx*(d/2)/Iz =18.676E6*(405/2)/775.02E6 =4.8798 N/mm² Since sigma ≤ sigmad ( 4.8798 N/mm² ≤ 9.9428 N/mm² ) applied bending stress within permissible.

Shear stress ──────────── Shear stress at notch tora=3*V/(2*b*he)=3*11319/(2*140*305) =0.39762 N/mm² Since tora ≤ torad ( 0.39762 N/mm² ≤ 1.4076 N/mm² ) shear stress does not exceed permissible therefore OK.

Ballantine Arnold Ltd Page: 41 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Check for deflection ──────────────────── (including shear defln as reqd by Clause 2.10.7) Deflection based on modified Edef=Emean*K2mod*K20=10800*0.8*1.07 =9244.8 N/mm² DL deflection without shear dld=6.6552 mm Imposed deflection without shear ild=5.1725 mm Total DL & imposed deflection 11.828 mm Modulus of rigidity G=Edef/16=9244.8/16=577.8 N/mm² Shape factor for Rect section KF=1.2 Shear area for beam Ay=d*b/KF=405*140/1.2=47250 mm² Total DL & imposed 22.638 kN If total DL & imposed load applied as a UDL, additional deflection due to shear dsu=WT'*L*10^6/(8*Ay*G) =22.638*6.6*10^6/(8*47250*577.8) =0.68409 mm Shear deflection dels=dsu*M'/(WT'*L/8) =0.68409*18.676/(22.638*6.6/8) =0.68409 mm Dead + Imposed + Shear deflection DEL=dld+ild+dels =6.6552+5.1725+0.68409 =12.512 mm Imposed + shear deflection DELis=ild+dels=5.1725+0.68409 =5.8565 mm Limiting deflection DELlim=0.003*L*10^3=0.003*6.6*10^3 =19.8 mm Beam is not used as part of a domestic floor and is not limited to 14 mm deflection. Since DEL ≤ DELlim ( 12.512 mm ≤ 19.8 mm ), OK for deflection. Precamber of beam unnecessary.

Laminate Thickness 45 mm DESIGN Number of laminates 9 SUMMARY Depth of section 405 mm Width of section 140 mm Beam notched on bottom edge.

Strength class C24 to Table 8. Moisture service class 3 Applied bending stress 4.8798 N/mm² Permiss bending stress 9.9428 N/mm² Applied shear stress 0.39762 N/mm² Permiss shear stress 1.4076 N/mm² Dead+imposed+shear Defl 12.512 mm Limiting deflection 19.8 mm No precamber of beam is necessary

Ballantine Arnold Ltd Page: 42 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Location: eaves beam ────────────────────

Simply supported glulam beam to BS5268-2:2002 ─────────────────────────────────────────────




Uniformly distributed load 1 of 1 Dist. from left support to start Lau(1)=0 m Distance from left support to end Lbu(1)=5.7 m Dead load (unfactored) Gku(1)=0.78+.3=1.08 kN/m Imposed load (unfactored) Qku(1)=0.75 kN/m





Ballantine Arnold Ltd Page: 43 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Rectangular glulam member │y ───────────────────────── ┌───────┐ ─┬─ with horizontal laminates ├───────┤ │ ───────────────────────── ├───────┤ │ Member laterally restrained x ─── ├───────┤ ─── x │d ─────────────────────────── ─┬─├───────┤ │ in bending tp─┴─├───────┤ │ ────────── └───────┘ ─┴─ │y ├── b ──┤ Mx=M'*10^6=7.4321*10^6=7.4321E6 Nmm V=Fve'*1000=5.2155*1000=5215.5 N Design axial load (+ve compress) F'=0 kN F=F'*1000=0*1000=0 N Number of laminates in section np=6 Thickness of laminates in section tp=45 mm Width of section b=140 mm Depth of section d=tp*np=45*6=270 mm Glulam beam laterally restrained with support restraint as Table 19.

Grade stresses ────────────── For horizontally glue laminated members the grade stresses should be taken as the product of the strength class stresses given in Table 8 for the relevant grade and species and the modification factors of Table 24. Chosen species is a Softwood. Strength class C24 to Table 8.

Section properties ────────────────── Largest section dimension h=d=270 mm Inertia about zz axis Iz=b*d^3/12=140*270^3/12 =229.64E6 mm Inertia about yy axis Iy=d*b^3/12=270*140^3/12 =61.74E6 mm Area of cross section A=b*d=140*270=37800 mm² Radius of gyration about zz axis iz=SQR(Iz/A)=SQR(229.64E6/37800) =77.942 mm Radius of gyration about yy axis iy=SQR(Iy/A)=SQR(61.74E6/37800) =40.415 mm

Grade stresses ────────────── Compression parallel to grain cparg=7.9 N/mm² Bending parallel to grain bparg=7.5 N/mm² Shear parallel to grain sparg=0.71 N/mm² Compression ┴ to grain cperd=2.4 N/mm² Mean modulus of elasticity Emean=10800 N/mm² Minimum modulus of elasticity Emin=7200 N/mm²

Ballantine Arnold Ltd Page: 44 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Modification factors ──────────────────── Timber service class adopted tmclass=3 Timber stress grades and moduli adjusted as table 16. Modification factors: Bending parallel to grain K2ben=0.8 Tension parallel to grain K2ten=0.8 Compression parallel to grain K2com=0.6 Compression ┴ to grain K2per=0.6 Shear parallel to grain K2shr=0.9 Mean & min modulus of elasticity K2mod=0.8 Duration of loading K3=1.25 Depth factor K7=(300/d)^0.11=(300/270)^0.11 =1.0117 Member is not in a load-sharing system as defined by Clause 2.9 Load-sharing modification factor K8=1.0 Modification factor K15 K15=1.365 Modification factor K19 K19=2.34 Modification factor K20 K20=1.07

Permissible stresses ──────────────────── Permissible bending stress sigmad=K2ben*K3*K7*K8*K15*bparg =0.8*1.25*1.0117*1*1.365*7.5 =10.357 N/mm² Depth of notch at support notch=100 mm he=d-notch=270-100=170 mm Notch is on bottom edge K5=he/d=170/270=0.62963 Permissible shear stress torad=K2shr*K3*K5*K8*K19*sparg =0.9*1.25*0.62963*1*2.34*0.71 =1.1768 N/mm²

Bending stress ────────────── Applied bending stress sigma=Mx*(d/2)/Iz =7.4321E6*(270/2)/229.64E6 =4.3692 N/mm² Since sigma ≤ sigmad ( 4.3692 N/mm² ≤ 10.357 N/mm² ) applied bending stress within permissible.

Shear stress ──────────── Shear stress at notch tora=3*V/(2*b*he) =3*5215.5/(2*140*170) =0.32871 N/mm² Since tora ≤ torad ( 0.32871 N/mm² ≤ 1.1768 N/mm² ) shear stress does not exceed permissible therefore OK.

Ballantine Arnold Ltd Page: 45 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 Check for deflection ──────────────────── (including shear defln as reqd by Clause 2.10.7) Deflection based on modified Edef=Emean*K2mod*K20=10800*0.8*1.07 =9244.8 N/mm² DL deflection without shear dld=6.9924 mm Imposed deflection without shear ild=4.8558 mm Total DL & imposed deflection 11.848 mm Modulus of rigidity G=Edef/16=9244.8/16=577.8 N/mm² Shape factor for Rect section KF=1.2 Shear area for beam Ay=d*b/KF=270*140/1.2=31500 mm² Total DL & imposed 10.431 kN If total DL & imposed load applied as a UDL, additional deflection due to shear dsu=WT'*L*10^6/(8*Ay*G) =10.431*5.7*10^6/(8*31500*577.8) =0.40834 mm Shear deflection dels=dsu*M'/(WT'*L/8) =0.40834*7.4321/(10.431*5.7/8) =0.40834 mm Dead + Imposed + Shear deflection DEL=dld+ild+dels =6.9924+4.8558+0.40834 =12.257 mm Imposed + shear deflection DELis=ild+dels=4.8558+0.40834 =5.2642 mm Limiting deflection DELlim=0.003*L*10^3=0.003*5.7*10^3 =17.1 mm Beam is not used as part of a domestic floor and is not limited to 14 mm deflection. Since DEL ≤ DELlim ( 12.257 mm ≤ 17.1 mm ), OK for deflection. Precamber of beam unnecessary.

Laminate Thickness 45 mm DESIGN Number of laminates 6 SUMMARY Depth of section 270 mm Width of section 140 mm Beam notched on bottom edge.

Strength class C24 to Table 8. Moisture service class 3 Applied bending stress 4.3692 N/mm² Permiss bending stress 10.357 N/mm² Applied shear stress 0.32871 N/mm² Permiss shear stress 1.1768 N/mm² Dead+imposed+shear Defl 12.257 mm Limiting deflection 17.1 mm No precamber of beam is necessary

Ballantine Arnold Ltd Page: 46 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 * Location: oak frame truss * * Data prepared with SCALE proforma 850 * PRINT DATA, RESULTS FROM 46 TYPE PLANE FRAME METHOD ELASTIC NODES NUMBER OF JOINTS 3 NUMBER OF MEMBERS 3 NUMBER OF SUPPORTS 2 NUMBER OF LOADINGS 1 NUMBER OF SEGMENTS 2 TRACE * * Global and local axes ┌── * ───────────────────── │ ┌+┐ ──► local y * local x └►│z│ * Global Y │ │ │ * │ ┌─┐ │ │ Local z points out * │ │ │ └─┘ of the paper, * │ │ │ │ positive moment is * │ local │z│◄┐ anti-clockwise. * │ y ◄── └+┘ │ local x * Global Z out│ ──┘ * of paper └────────────────────────────────► Global X * * Except in the special case of a continuous beam, the members of a * structure do not all run along the global axis X; they may be inclin- * ed to the the global axis and each may run in a different direction. * In the diagram above, the left member is pointing upwards, the right * downwards. To fix the orientation of each member relative to axes X, * Y and Z, every member is assumed to carry a local set of axes denoted * x,y,z with axis x running from the START of the member (at the local * origin) to the END of the member. Movements along x,y - and * anticlockwise about z - are considered positive, as depicted above. * * * Positive │ SHEAR Positive │ SHEAR * Positive │ ┌────────────────────┐ │ Positive * AXIAL ─────► ┌─ │start end│ ┌─ ─────► AXIAL (tension) * (compression) │ └────────────────────┘ │ * └──► Positive MOMENT └──► Positive MOMENT * * When dealing with the joints of a structure (their coordinates, * conditions of support, loads etc.) we refer to global axes X, Y and * Z. When dealing with a member (its section properties, member loads * etc.) we refer to its local axes x, y and z. For quick reference * the effects of positive results for forces acting on the ends of * members are depicted above. THINK OF THE JOINTS AS APPLYING FORCES * TO THE MEMBER ENDS. * * All units are kN & m and combinations thereof. JOINT COORDINATES 1 0 0 SUPPORT 2 2.5 1.1 3 5 0 SUPPORT JOINT RELEASES 1 MOMENT Z

Ballantine Arnold Ltd Page: 47 Made by: CJA Date: 05.04.18 Ref No: C702 ─────────────────────────────────────────────────────────────────────── Office: 6145 3 MOMENT Z MEMBER INCIDENCES 1 1 2 2 2 3 3 1 3 MEMBER RELEASES 1 START MOMENT Z 2 START MOMENT Z END MOMENT Z CONSTANTS E 9E6 ALL G 562500 ALL MEMBER PROPERTIES 1 RECTANGLE DY 0.15 DZ 0.075 2 RECTANGLE DY 0.15 DZ 0.075 3 RECTANGLE DY 0.15 DZ 0.15 LOADING CASE 1 JOINT LOADS 2 FORCE X 0 Y -11.3 MEMBER LOADS 3 FORCE Y PROJECTED UNIFORM W -0.25 SOLVE

LOADING CASE 1 JOINT DISPLACEMENTS JOINT X DISPLACEMENT Y DISPLACEMENT Z ROTATION 1 0.000000000 0.000000000 -0.003429355 2 0.000000000 -0.000939672 -0.000314903 3 0.000000000 0.000000000 0.003429355 4 0.000000000 -0.000469836 -0.000314903 5 0.000000000 -0.000469836 0.000314903 6 0.000000000 -0.005432442 0.000000000

LOADING CASE 1 MEMBER FORCES AT START OF FIRST SEGMENT AND ENDS OF ALL SEGMENTS MEMBER JOINT AXIAL FORCE SHEAR FORCE BENDING MOMENT 1 1 14.0290 0.0000 0.0000 4 -14.0290 0.0000 0.0000 2 -14.0290 0.0000 0.0000 2 2 14.0290 0.0000 0.0000 5 -14.0290 0.0000 0.0000 3 -14.0290 0.0000 0.0000 3 1 0.0000 0.6250 0.0000 6 0.0000 0.0000 0.7813 3 0.0000 0.6250 0.0000

LOADING CASE 1 SUPPORT REACTIONS JOINT X FORCE Y FORCE Z MOMENT 1 12.8409 6.2750 0.0000 3 -12.8409 6.2750 0.0000

EQUILIBRIUM CHECK SUM OF FORCES REACTION

FORCES IN DIRECTION X 0.0000 0.0000 FORCES IN DIRECTION Y -12.5500 12.5500 MOMENTS ABOUT AXIS Z -31.3750 31.3750


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Modbury Recreation Ground Balcony / First Floor Plan Showing Roof structure above Sketch 18106 - sk01 Rev A April 2018 NTS

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Existing Attic trusses over at 400 crs 220*47 bottom chord and rafters 95*47 purlin posts

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Existing double trusses trim around existing roof lights

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Partner Principal rafter to carry new eaves beam M10 Bolts 600 crs Alt Top / Bottom

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Partner Principal rafter to carry new eaves beam

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Roof laid over existing on lay boards

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152*152*23 UC

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90*3.6SHS

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90*3.6SHS

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100*3.6 SHS galvanised fully welded wind frame

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dormer roof joists150*50 C16 @ 400 c/c

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Braces to elevation truss and eaves beam

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Blue - Canopy structure over

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Red - Dormer and existing truss roof structure

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Ridge Beam 405*140C24 Glulam

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Eaves Beam 270*140C24 Glulam

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Rafters 150*50 C24@400 c/c

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Oak or C24 Glulam elevation Frame see sketches SK04 and SK05

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Modbury Recreation Ground Ground Floor Plan Showing balcony / First Floor structure above Sketch 18106 - sk02 Rev A April 2018 NTS

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Standard Guage Cavity Lintel, e.g Catnic CG 90/100

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Existing Attic trusses over at 400 crs 220*47 bottom chord and rafters 95*47 purlin posts

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Balcony joists 175*50 C24 at 400 crs

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Triple joists to trim around lightwell

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Triple joists to trim around lightwell

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140*100 precast concrete lintels

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Twin 457*152*52UB

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At each end of twin beam 2 x 140*6.3SHS Posts 250*250*15 base plates 4mm f/w all round to SSH 4 X M16 HD bolts

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Ensure bottom chords of cut trusses bear on the internal walls, wedge if necessary

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Partner floor joists of the cut trusses with 220*47 C16, M10 bolts at 600 crs alternating top to bottom of member. Over full length of truss

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90*3.6SHS above fixed to block wall

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Modbury Recreation Ground Ground Floor Plan Showing Foundation Layout Sketch 18106 - sk03 Rev A April 2018 NTS

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600*250 Strip foundation

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150mm RC 30 Ground Bearing slab A393 mesh T&B and 40 cover on min 150 compacted hardcore

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450*250 strip foundation

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Depth of foundation to match existing and to be agreed with the Building Inspector

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1200*1200*300 RC30 Pad foundations A393 mesh 75 cover to bottom

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Lateral support to walls at all floor levels to be as B' Regs Part A Diagram 15 Lateral restraint to gable walls at roof level to be as B' Regs Part A Diagram 161200mm*30mm*5mm restraint straps at 2000mm max c/c throughout.

Vertical restraint to flat and shallow-pitched roofs to be as B' Regs Part A Diagram 161000mm*30mm*5mm restraint straps at 2000mm max c/c throughout.

Restraint Straps

timber packer bolted to web of beamwith 10mm through-bolts at 400c/c

Simpson JHA 270/50 joist hanger

Ballantine Arnold LtdStandard Steelwork Details

Unless shown otherwise on the architects drawings, all external steelwork and steelwork built into the outer leaf of cavity walls is to be hot dip galvanised to BS EN 1461. Internal steelwork to be shot blasted and coated with 2c red oxide primer. All site cuts and welds to be protected with zinc-rich galvafroid paint. All steelwork to be fire protected with intumescent paint or 2 layers plasterboard with taped jointsto architects details Protection to Steelwork

Twin Beams Bolted Together

16mm grade 8.8 bolt sleeved in 40*40*3 SHS at both ends of beam and at 600mm c/c

to suit cavity50

8mm web plate 4mm continuous fillet weld all round

2No 16mm grade 8.8 bolts with 30mm edge distance to 18mm holes

10

Beam to Beam Connections

Padstones

Unless shown otherwide on the drawing, beams are to extend by a minimum of 150mm onto supporting walls and are to sit on concrete padstones cut from 450mm long * 100mm wide * 140mm deep precast concrete lintel sections.

twin beams to bear on a common padstone250*150 deep * full width of the wallunless shown otherwise on the drawings

outer leaf beam to be galvanised

Beam to Post Connection

4No M12 grade 8.8 bolts to 250*120*8 top plate with 4mmfillet weld to SHS post

250*120*10 base plate4No M12 resin boltsinto concrete padstoneor onto top flange ofsteel beam below

shot fire starter ties at 300 c/c where posts are built into masonry walls, or shot fire nails to studs at 300 c/c for timber.

Fabricated steel to be to BS-EN1090 and CE-marked EXC2class 1 for dwellings up to 4 storeys; class 2b for non domestic buildings up to 4 storys

Joist Connection for Beam Restraint

Web Plate Connection Angle Cleat Connection

70*70*8 RSA cleat2No M16 Grade 8.8bolts to web of eachbeam30mm edge distanceto 18mm holes

top and bottom flanges cut back as required

All Beams are to be Connected Together At or Near to Supports

8mm end plates with 6mm fillet weldsand 4No M12 grade 8.8 bolts

Detail shown is for beams over a common postSimilar Detail is required for beams over a common padstone

4No M12 grade 8.8 bolts to 250*120*8 end plates with 4mm fillet weld to SHS post

Posts to be fully restrained by joists, noggins and packers at each floor level

Posts Connected End-to-End

These are standard details for typical installations and may not all apply to a particular project

.nominally 300mm or as shown on drwg.

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joist hanger

timber packer bolted to web of beamwith 10mm through-bolts at 600c/c

Lateral support to walls at all floor levels to be as B' Regs Part A Diagram 15 Lateral restraint to gable walls at roof level to be as B' Regs Part A Diagram 161200mm*30mm*5mm restraint straps at 2000mm max c/c throughout.

Vertical restraint to flat and shallow-pitched roofs to be as B' Regs Part A Diagram 161000mm*30mm*5mm restraint straps at 2000mm max c/c throughout.

Restraint Strapsjoist hanger

C24 grade timber wallplate 50mmthick bolted to masonry wall withM10 Rawlbolts at 400 c/c for floorsand at 600 c/c for roofs unlessshown otherwise on the drawings

Ballantine Arnold LtdStandard Timber Details

Joist Connectionto Steel Beam

Alternative Connections to Steel Ridge Beams

50*50 collar nailed to rafters

joist with max 50mm notch skew nailed to 100*38timber bearer boltedto top flange beamM10 at 600 c/c

50*50 collar nailed to rafters

joist with max 50mm notch skew nailed to 50*50timber bearer through-bolted to web of beam M10 at 600 c/c

Wallplate to Masonry Wall

External Timber Stud WallsRender or timber cladding on battens over breathable membrane all to architect specification over 9mm OSB3 nailed with 65mm* 3mm nails at 150 c/c to 140*38 or 150*50 C16 studs at 400 c/c bearing on treated 140*50 or 150*50 C24 treated sole plate fixed down with 120mm long M8 concrete screws or bolts at 600 c/c over DPC. Twin 150*50 C16 lintels on double cripple studs to openings where not shown otherwise on the drawings.

Internal Loadbearing Timber Stud Walls9mm OSB3 nailed with 65mm* 3mm nails at 150 c/c to 100*50 or 89*38 C16 studs at 400 c/c bearing on treated 100*50 or 89*50 C24 treated sole plate fixed down with 120mm long M8 concrete screws or bolts at 600 c/c over DPC. Twin 150*50 C16 lintels on double cripple studs to openings where not shown otherwise on the drawings.

Internal Non-loadbearing Timber Stud Walls100*50 or 89*38 C16 studs at 400 c/c bearing on treated 100*50 or 89*38 C16 sole plate fixed down with 120mm long M8 concrete screws or bolts at 600 c/c over DPC. Twin 150*50 C16 lintels on double cripple studs to openings where not shown otherwise on the drawings.

Timber construction details all as Trada publication 'Timber Frame Construction'

rafter notched over 50*38 C24 timber bearer bearer fixed to ridge beam with 75*5 screws at 150c/c or 100*4.2 nails at 100 c/c

glulam ridge beam

Glulam Beam Details

Glulam beam-to-beam connections to be formed using steel hangers fabricated to the required geometry. Hangers are to be fabricated from 5mm plate hooked fully over the top of the carrier beam and with a level 50mm bearing shelf to support the secondary member.

Where a concealed joint is required use Simpson Strongtie BTALU240 with 44No 60*4 nails and 6No 12mm dowels :- capacity is limited to18kN for 90mm beam; 23kN for 115mm beam; 30kN for 140 beam widths


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Ballantine Arnold LtdFoundations: Standard Details

600

150

225

Depth of foundations are to be confirmed on site with the Building Inspector. Minimum depth in clays and shrinkable soils of 900mm and 500mm other soils unless noted otherewise on drawingsThe suitability of the formation is to be confirmed on site with the Building Inspector (the design is based on an allowable bearing pressure of 100kN/m2). Soft spots and pockets to be removed and backfilled with lean mix concrete. Foundations are to be founded on natural undisturbed ground and NOT to be formed on placed fill

Cavity Wall Single leaf wall

225

450

Dwarf wall for B&B / suspended floor

225

450

Beam and Block Flooring

Min 150mm void

50 Gen 1 Oversite

For stone facing increase the foundation width by the width of the stonework so asto maintain 150mm shoulder each face

Strip and pad foundation concrete to be Grade Gen 1 (10N) to BS 8500-2 in non aggressive soils and compacted when placing. Strip footing shall be locally widened for piers

Trench fill foundations of equivalent width may be used as an acceptable alternative to strip foundations and are recommended > 1.2m below GL

Isolated posts to bear on 440*440 7N dense concrete block foundation piers on 600*600*225 concrete pad foundations unless indicated otherwise on the drawings

Where necessary foundations can be stepped to suit the ground levels, steps to be in accordance with the Building Regulations,

Excavations shall be kept drained of surface and ground water at all times. All mass concrete poured foundations are to be poured on the same day as excavation of the ground, otherwise a concrete blinding concrete shall be used at the the base of the excavation

L

ST

Min Overlap L = Twice height of step (normally 450mm) or Thickness of foundation or 300mm, whichever the greater

S should not be greater than T, i.e 225mm normally

E.g for a 225mm thick Strip:- for 225 step, L = 450 for 440 step L= 880 Stepped foundation :

Trench FillGen 1 (10 N) concrete

225

Ground Bearing Slabs:

Screed / Insulation / DPM / DPC all in accordance with the architects details

100mm Fibre reinforced RC 30 Mix (polypropylene fibres to specialist supplier design, typically 4 to 6kg/m3)

ScreedInsulation

100mm sand blindingMinimum 150mm compacted clean inert granular fill, compacted in maximum 150mm layers to a maximum depth of 600mm

Level and compacted natural / undisturbed ground

For ground bearing slabs that bear onto foundation walls the slab shall be 150mm RC30 with A393 mesh, 50 cover to the bottom and a minimum 500mm laps between sheets. All on 150mm compacted hardcore

215mm , 7N dense concreteblocks laid flat(1: 3 to 4 mortar mix)

max

500

Min 1000


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Ballantine Arnold Ltd Civil Structural and Marine Engineering Consultants

Function Room Knock-Through Beam: Propping Method Statement

Modbury Recreation Ground Hall Job 18106.

Overall propping load (from beam shear calculation) = 203 kN

Total number of accrows at 15kN/each = 14.No.

Deflection to be induced into new beam by driving shims 8 mm (from DL deflection calculation)

Needle beams and accrow props are to be installed at equal intervals at a maximum spacing of 1.2m so as to install at least the minimum number of accrows stated above. Strongboy type props are not to be used as they have a very small load capacity. Needle beams to be 100*100*10 SHS where 1 or 2 accrows are used at each end. Needle beams to be 152*152*23 UC where 3 accrows are used at each end.

1. Install needle beams and accrow props as above. Props are to bear onto an adequate bearing surface capable of 1.5 tons per accrow.

2. Break out masonry and cast concrete padstones to support new beams at each end.

3. Break out one face of the wall. 4. Install steel beams to support the broken-out face of the wall and bolt

together if required. 5. Slate pack and drive shims between the top of the beam and the underside of

the masonry wall above so as to pick up the load of the masonry above and to induce a deflection into the beam as stated above.

6. Repeat items 3,4,5 for the other face of the wall. 7. Make good masonry and fully pack between the top of the beam and

underside of the wall with dry-pack 3:1 mortar. 8. Remove props and make good after 48 hours of dry-pack mortar.

This information is given in good faith as a guide to the contractor. The contractor should check and verify propping requirements to suit his particular methods and is to remain liable for all aspects of the temporary works scheme.

Documents

· /// A──┘ ooo BS5268-2:2002. ├─────────── L ───────────┤ The following assumptions are made in these calculations: that the