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## Purlin and rafter details

 Timber strength class C16 Clear span of purlin Lcl = 1 m Purlin width b = 72 mm Purlin depth h = 120 mm Purlin spacing on slope s = 2500 mm Rafter slope α = 35 degrees Rafter width bs = 38 mm Rafter depth hs = 95 mm Rafter spacing ss = 400 mm

## Modification factors

 Depth factor K7 = (300 / h)0.11 = 1.11

## Purlin section properties

 Minimum modulus of elasticity Emin = 5,800 N/mm² Second moment of area I = b × h³ / 12 = 10,400,000 mm4 Section modulus Z = b × h² / 6 = 173,000 mm³ Average density for C16 grade timber (BS 5268-2:2002 Table 8) ρmean = 370 kg/m³ Purlin self weight per linear metre (g = 9.81 m/s²) Fj = b × h × ρmean × g = 0.031 kN/m Rafter self weight per linear metre (g = 9.81 m/s²) Fs = bs × hs × ρmean × g = 0.013 kN/m

 K3 (long term load) K3 = 1 Dead load (UDL) Fd = 1 kN/m² Total load F = (1.25 × s / ss (Fd × ( ss / 1000 ) + Fs ) × cos(α) ) + Fjcos(α) = 2.67 kN/m Grade bending stress for C16 (BS5268-2:2002 Table 8) σm,par = 5.3 N/mm² Permissible bending stress σm,adm = σm,par × K3 × K7 = 5.86 N/mm² Compression perpendicular to grain for C16 (BS5268-2:2002 Table 8) σc,per = 1.7 N/mm² Notional bearing length(Note from BS 5268-7.5 Clause 4.2: 'The bearing length required at each end of the purlin, calculated in accordance with 5.6, may not be sufficient for practical construction purposes.') a = (Lcl × F / 2) / (σc,per × K3 × b - (F / 2)) = 11 mm Effective span on slope L = Lcl + a = 1.01 m Check bending stress Bending moment M = F × L² / 8 = 0.341 kNm Bending stress σm,a = (M × 106) / Z = 1.97 N/mm² σm,a <= σm,adm ( 1.974 N/mm² <= 5.862 N/mm² ) therefore OK Check shear stress Grade shear stress for C16 (BS5268-2:2002 Table 8) τpar = 0.67 N/mm² Permissible shear stress τadm = τpar × K3 = 0.67 N/mm² Shear stress τ = (3 × F × L / 2) / (2 × b × h) = 0.234 N/mm² τ <= τadm ( 0.234 N/mm² <= 0.67 N/mm² ) therefore OK Check deflection Permissible deflection δadm = 0.003 × L = 3.03 mm Bending deflection δbending = (5 × F × L4) / (384 × Emin × I) = 0.604 mm Shear deflection δshear = (12 × F × L²) / (5 × Emin × b × h) = 0.131 mm Total deflection δtotal = δbending + δshear = 0.735 mm δtotal <= δadm ( 0.735 mm <= 3.033 mm ) therefore OK

## Consider medium term loading (1 kN/m² dead UDL + 1 kN/m² imposed UDL. K3 = 1.25)

 K3 (medium term load) K3 = 1.25 Dead load (UDL) Fd = 1 kN/m² BS 5268-7.5 Clause 4.3: For a roof slope greater than 30° and not exceeding 75°: an imposed load obtained by linear interpolation between the values at 30° roof slope, e.g. 0.75 kN/m2, and zero for a 75° roof slope. Imposed load (UDL) Fimposed,udl = 1 × ((75 - α) / 45) = 0.889 kN/m² Total load F = (1.25 × s / ss ( Fimposed,udlcos(α) + Fd ) ( ss / 1000 ) + Fs ) × cos(α) ) + Fjcos(α) = 4.53 kN/m Grade bending stress for C16 (BS5268-2:2002 Table 8) σm,par = 5.3 N/mm² Permissible bending stress σm,adm = σm,par × K3 × K7 = 7.33 N/mm² Compression perpendicular to grain for C16 (BS5268-2:2002 Table 8) σc,per = 1.7 N/mm² Notional bearing length(Note from BS 5268-7.5 Clause 4.2: 'The bearing length required at each end of the purlin, calculated in accordance with 5.6, may not be sufficient for practical construction purposes.') a = (Lcl × F / 2) / (σc,per × K3 × b - (F / 2)) = 15 mm Effective span on slope L = Lcl + a = 1.02 m Check bending stress Bending moment M = F × L² / 8 = 0.584 kNm Bending stress σm,a = (M × 106) / Z = 3.38 N/mm² σm,a <= σm,adm ( 3.379 N/mm² <= 7.328 N/mm² ) therefore OK Check shear stress Grade shear stress for C16 (BS5268-2:2002 Table 8) τpar = 0.67 N/mm² Permissible shear stress τadm = τpar × K3 = 0.838 N/mm² Shear stress τ = (3 × F × L / 2) / (2 × b × h) = 0.399 N/mm² τ <= τadm ( 0.399 N/mm² <= 0.838 N/mm² ) therefore OK Check deflection Permissible deflection δadm = 0.003 × L = 3.05 mm Bending deflection δbending = (5 × F × L4) / (384 × Emin × I) = 1.04 mm Shear deflection δshear = (12 × F × L²) / (5 × Emin × b × h) = 0.224 mm Total deflection δtotal = δbending + δshear = 1.27 mm δtotal <= δadm ( 1.266 mm <= 3.045 mm ) therefore OK

## Design summary

 Permissible Applied/Actual Utilisation Result Long term load shear stress (N/mm²) 0.67 0.23 35 % OK Long term load bending stress (N/mm²) 5.86 1.97 33.7 % OK Long term load deflection (mm) 3.03 0.73 24.2 % OK Medium term load shear stress (N/mm²) 0.84 0.4 47.7 % OK Medium term load bending stress (N/mm²) 7.33 3.38 46.1 % OK Medium term load deflection (mm) 3.05 1.27 41.6 % OK

## Notes

This design is in accordance with BS 5268-2:2002 Structural use of timber - Part 2: Code of practice for permissible stress design, materials and workmanship and BS 5268-7.6:1990 Structural use of timber - Section 7.6 Purlins supporting rafters.

The major axis of the purlin is perpendicular to the rafter slope.

The horizontal thrust at the eaves is assumed to be transmitted by the ceiling joists to the complimentary rafter.

These calculations conservatively assume that the rafters are continuous over the purlin, however, the rafters may consist of shorter lengths joined at the purlin.

The depth to width ratio of the timber does not exceed 5 and as per the requirements of BS 5268-2 Table 19 there is no risk of buckling under design load provided; The ends are held in position and compression edge held in line, as by direct connection of sheathing, deck or joists.

Timber to be covered, this calculation is not to be used for timber which is fully exposed to the elements.

Wane as allowed in BS 4978:2007 + A2:2017 is permitted.

DEMO