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## Joist details

 Timber strength class C16 Joist width b = 38 mm Joist depth h = 95 mm Joist spacing s = 400 mm Clear span Lcl = 1 m

Diagrams not to scale

## Modification factors

 Depth factor K7 = (300 / h)0.11 = 1.13 Load sharing modification factor (BS5268-2 clause 2.10.11) K8 = 1.10

## Section properties

 Mean modulus of elasticity Emean = 8,800 N/mm² Second moment of area I = b × h³ / 12 = 2,720,000 mm4 Z to top edge of timber Z = b × h² / 6 = 57,200 mm³ Average density for C16 grade timber (BS 5268-2:2002 Table 8) ρmean = 370 kg/m³ Joist self weight per square metre (g = 9.81 m/s²) Fjoist = b × h × ρmean × g / s = 0.033 kN/m²

## Consider long term loading (0.75 kN/m² dead UDL + 1.5 kN/m² live UDL. K3 = 1)

 K3 (long term load) K3 = 1 Dead load (UDL) Fdead,udl = 0.75 kN/m² Total dead load (dead UDL + joist self weight) Fdead = Fdead,udl + Fjoist = 0.783 kN/m² Live load (UDL) Flive,udl = 1.5 kN/m² Total load per metre F = ( Fdead + Flive,udl ) × Lspacing = 0.913 kN/m Compression perpendicular to grain for C16 (BS5268-2:2002 Table 8) σc,per = 1.7 N/mm² Notional bearing length(Note from BS 5268-7.1 Clause 4.2: 'The bearing length required at each end of the joist, calculated in accordance with 5.5, may not be sufficient for practical construction purposes.') a = (Lcl × F / 2) / (σc,per × K3 × K8 × b - (F / 2)) = 6.47 mm Effective span Leff = Lcl + a = 1.01 m Check bending stress Grade bending stress for C16 (BS5268-2:2002 Table 8) σpar = 5.3 N/mm² Permissible bending stress σadm = σpar × K3 × K7 × K8 = 6.62 N/mm² Bending moment M = F × Leff²/ 8 = 0.116 kNm Bending stress σ = (M × 106) / Z = 2.02 N/mm² σ <= σadm ( 2.023 N/mm² <= 6.616 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 × K8 = 0.737 N/mm² Shear stress τ = (3 × F × Leff / 2 × 10³) / (2 × b × h) = 0.191 N/mm² τ <= τadm ( 0.191 N/mm² <= 0.737 N/mm² ) therefore OK Check deflection Permissible deflection δadm = 0.003 × Leff (or max of 14mm) = 3.02 mm Bending deflection δbending = (5 × F × Leff4) / (384 × Emean × I) = 0.511 mm Shear deflection δshear = (12 × F × Leff²) / (5 × Emean × b × h) = 0.0699 mm Total deflection δtotal = δbending + δshear = 0.581 mm δtotal <= δadm ( 0.581 mm <= 3.019 mm ) therefore OK

## Consider medium term loading (0.75 kN/m² dead UDL + 1.4 kN live point load. K3 = 1.25)

 K3 (medium term load) K3 = 1.25 Dead load (UDL) Fdead,udl = 0.75 kN/m² Total dead load (dead UDL + joist self weight) Fdead = Fdead,udl + Fjoist = 0.783 kN/m² Live point load Flive,point = 1.4 kN Total load per metre F = Fdead × Lspacing = 0.313 kN/m Compression perpendicular to grain for C16 (BS5268-2:2002 Table 8) σc,per = 1.7 N/mm² Notional bearing length(Note from BS 5268-7.1 Clause 4.2: 'The bearing length required at each end of the joist, calculated in accordance with 5.5, may not be sufficient for practical construction purposes.') a = ((Lcl × F / 2) + (Flive,point / 2)) / (σc,per × K3 × K8 × b - (F / 2)) = 9.66 mm Effective span Leff = Lcl + a = 1.01 m Check bending stress Grade bending stress for C16 (BS5268-2:2002 Table 8) σpar = 5.3 N/mm² Permissible bending stress σadm = σpar × K3 × K7 × K8 = 8.27 N/mm² Bending moment M = (F × Leff²/ 8) + (Flive,point × Leff / 4) = 0.393 kNm Bending stress σ = (M × 106) / Z = 6.88 N/mm² σ <= σadm ( 6.881 N/mm² <= 8.27 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 × K8 = 0.921 N/mm² Shear stress τ = (3 × (( F × Leff / 2) + Flive,point) × 10³) / (2 × b × h) = 0.647 N/mm² τ <= τadm ( 0.647 N/mm² <= 0.921 N/mm² ) therefore OK Check deflection Permissible deflection δadm = 0.003 × Leff (or max of 14mm) = 3.03 mm Bending deflection from uniformly distributed load δbending,udl = (5 × F × Leff4) / (384 × Emean × I) = 0.177 mm Shear deflection from uniformly distributed load δshear,udl = (12 × F × Leff²) / (5 × Emean × b × h) = 0.0241 mm Bending deflection from point load δbending,point = (Flive,point × 10³ × Leff3) / (48 × Emean × I) = 1.26 mm Shear deflection from point load δshear,point = (24 × Flive,point × 10³ × Leff) / (5 × Emean × b × h) = 0.214 mm Total deflection δtotal = δbending,udl + δshear,udl + δbending,point + δshear,point = 1.67 mm δtotal <= δadm ( 1.671 mm <= 3.029 mm ) therefore OK

## Design summary

 Permissible Applied/Actual Utilisation Result Long term load shear stress (N/mm²) 0.74 0.19 25.9 % OK Long term load bending stress (N/mm²) 6.62 2.02 30.6 % OK Long term load deflection (mm) 3.02 0.58 19.2 % OK Medium term load shear stress (N/mm²) 0.92 0.65 70.3 % OK Medium term load bending stress (N/mm²) 8.27 6.88 83.2 % OK Medium term load deflection (mm) 3.03 1.67 55.2 % 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.

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

The depth to width ratio of the timber does not exceed 6 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.

Notches are not to exceed 0.125 of the depth of the joist and are to be located between 0.07 and 0.25 of the span from the support.

Holes drilled at the neutral axis (centre line) are not to exceed 0.25 of the depth of a joist and are not to be less than than three diameters (centre to centre) apart and are to be located between 0.25 and 0.4 of the span from the support.

Strutting or blocking between joists is required as follows;
- Joist spans up to 2.5m - no strutting required
- Joist spans 2.5m to 4.5m - provide strutting at mid-span
- Joist spans over 4.5m - provide 2 rows of strutting at third span positions

Solid timber strutting to be at least 38mm thick and be a minimum of three quarters of the joist depth.

Proprietary strutting to be used in accordance with manufacturers recommendations.

The joists are to have a minimum end bearing of 40mm.

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

These calculations are only applicable for floors consisting of four or more joists.

DEMO