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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 elasticityEmean = 8,800 N/mm²
Second moment of area I = b × h³ / 12 = 2,720,000 mm4
Z to top edge of timberZ = 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 (0.5 kN/m² dead UDL. K3 = 1)

K3 (long term load)

K3 = 1
Dead load (UDL)Fdead,udl = 0.5 kN/m²
Total dead load (dead UDL + joist self weight)Fdead = Fdead,udl + Fjoist = 0.533 kN/m²
Total load per metreF = Fdead × Lspacing = 0.213 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)) = 1.5 mm
Effective span

Leff = Lcl + a = 1 m
Check bending stress
Grade bending stress for C16 (BS5268-2:2002 Table 8)σpar = 5.3 N/mm²
Permissible bending stressσadm = σpar × K3 × K7 × K8 = 6.62 N/mm²
Bending momentM = F × Leff²/ 8 = 0.0267 kNm
Bending stress σ = (M × 106) / Z = 0.467 N/mm²
σ <= σadm ( 0.467 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 stressτadm = τpar × K3 × K8 = 0.737 N/mm²
Shear stressτ = (3 × F × Leff / 2 × 10³) / (2 × b × h) = 0.0443 N/mm²
τ <= τadm ( 0.044 N/mm² <= 0.737 N/mm² ) therefore OK
Check deflection
Permissible deflectionδadm = 0.003 × Leff = 3 mm
Bending deflection δbending = (5 × F × Leff4) / (384 × Emean × I) = 0.117 mm
Shear deflectionδshear = (12 × F × Leff²) / (5 × Emean × b × h) = 0.0161 mm
Total deflectionδtotal = δbending + δshear = 0.133 mm
δtotal <= δadm ( 0.133 mm <= 3.005 mm ) therefore OK

Consider medium term loading (0.5 kN/m² dead UDL + 1.5 kN/m² live UDL. K3 = 1.25)

K3 (medium term load)

K3 = 1.25
Dead load (UDL)Fdead,udl = 0.5 kN/m²
Total dead load (dead UDL + joist self weight)Fdead = Fdead,udl + Fjoist = 0.533 kN/m²
Live load (UDL)Flive,udl = 1.5 kN/m²
Total load per metreF = ( Fdead + Flive,udl ) × Lspacing = 0.813 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)) = 4.6 mm
Effective span

Leff = Lcl + a = 1 m
Check bending stress
Grade bending stress for C16 (BS5268-2:2002 Table 8)σpar = 5.3 N/mm²
Permissible bending stressσadm = σpar × K3 × K7 × K8 = 8.27 N/mm²
Bending momentM = F × Leff²/ 8 = 0.103 kNm
Bending stress σ = (M × 106) / Z = 1.79 N/mm²
σ <= σadm ( 1.795 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 stressτadm = τpar × K3 × K8 = 0.921 N/mm²
Shear stressτ = (3 × F × Leff / 2 × 10³) / (2 × b × h) = 0.17 N/mm²
τ <= τadm ( 0.17 N/mm² <= 0.921 N/mm² ) therefore OK
Check deflection
Permissible deflectionδadm = 0.003 × Leff = 3.01 mm
Bending deflection δbending = (5 × F × Leff4) / (384 × Emean × I) = 0.451 mm
Shear deflectionδshear = (12 × F × Leff²) / (5 × Emean × b × h) = 0.062 mm
Total deflectionδtotal = δbending + δshear = 0.513 mm
δtotal <= δadm ( 0.513 mm <= 3.014 mm ) therefore OK

Consider short term loading (0.5 kN/m² dead UDL + 1.8 kN live point load. K3 = 1.5)

K3 (short term load)

K3 = 1.5
Dead load (UDL)Fdead,udl = 0.5 kN/m²
Total dead load (dead UDL + joist self weight)Fdead = Fdead,udl + Fjoist = 0.533 kN/m²
Live point loadFlive,point = 1.8 kN
Total load per metreF = Fdead × Lspacing = 0.213 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)) = 9.45 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 stressσadm = σpar × K3 × K7 × K8 = 9.92 N/mm²
Bending momentM = (F × Leff²/ 8) + (Flive,point × Leff / 4) = 0.481 kNm
Bending stress σ = (M × 106) / Z = 8.42 N/mm²
σ <= σadm ( 8.422 N/mm² <= 9.924 N/mm² ) therefore OK
Check shear stress
Grade shear stress for C16 (BS5268-2:2002 Table 8)τpar = 0.67 N/mm²
Permissible shear stressτadm = τpar × K3 × K8 = 1.11 N/mm²
Shear stressτ = (3 × (( F × Leff / 2) + Flive,point) × 10³) / (2 × b × h) = 0.793 N/mm²
τ <= τadm ( 0.793 N/mm² <= 1.106 N/mm² ) therefore OK
Check deflection
Permissible deflectionδadm = 0.003 × Leff = 3.03 mm
Bending deflection from uniformly distributed load δbending,udl = (5 × F × Leff4) / (384 × Emean × I) = 0.121 mm
Shear deflection from uniformly distributed loadδshear,udl = (12 × F × Leff²) / (5 × Emean × b × h) = 0.0164 mm
Bending deflection from point load δbending,point = (Flive,point × 10³ × Leff3) / (48 × Emean × I) = 1.61 mm
Shear deflection from point loadδshear,point = (24 × Flive,point × 10³ × Leff) / (5 × Emean × b × h) = 0.275 mm
Total deflectionδtotal = δbending,udl + δshear,udl + δbending,point + δshear,point = 2.03 mm
δtotal <= δadm ( 2.026 mm <= 3.028 mm ) therefore OK

Design summary

PermissibleApplied/ActualUtilisationResult
Long term load shear stress (N/mm²)0.740.046 %OK
Long term load bending stress (N/mm²)6.620.477.1 %OK
Long term load deflection (mm)30.134.4 %OK
Medium term load shear stress (N/mm²)0.920.1718.4 %OK
Medium term load bending stress (N/mm²)8.271.7921.7 %OK
Medium term load deflection (mm)3.010.5117 %OK
Short term load shear stress (N/mm²)1.110.7971.7 %OK
Short term load bending stress (N/mm²)9.928.4284.9 %OK
Short term load deflection (mm)3.032.0366.9 %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 flat roofs consisting of four or more joists.

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