Report - Purlin Calculator to BS 5268-2:2002 and BS 5268-7.6:1990

Purlin and rafter details

Timber strength class	C16
Clear span of purlin	L_cl = 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	b_s = 38 mm
Rafter depth	h_s = 95 mm
Rafter spacing	s_s = 400 mm

Modification factors

Depth factor

K₇ = (300 / h)^0.11 = 1.11

Purlin section properties

Minimum modulus of elasticity	E_min = 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)	ρ_mean = 370 kg/m³
Purlin self weight per linear metre (g = 9.81 m/s²)	F_j = b × h × ρ_mean × g = 0.031 kN/m
Rafter self weight per linear metre (g = 9.81 m/s²)	F_s = b_s × h_s × ρ_mean × g = 0.013 kN/m

Consider long term loading (1 kN/m² dead UDL. K3 = 1)

K3 (long term load)	K₃ = 1
Dead load (UDL)	F_d = 1 kN/m²
Total load	F = (1.25 × s / s_s (F_d × ( s_s / 1000 ) + F_s ) × cos(α) ) + F_jcos(α) = 2.67 kN/m
Grade bending stress for C16 (BS5268-2:2002 Table 8)	σ_m,par = 5.3 N/mm²
Permissible bending stress	σ_m,adm = σ_m,par × K₃ × K₇ = 5.86 N/mm²
Compression perpendicular to grain for C16 (BS5268-2:2002 Table 8)	σ_c,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 = (L_cl × F / 2) / (σ_c,per × K₃ × b - (F / 2)) = 11 mm
Effective span on slope	L = L_cl + a = 1.01 m
Check bending stress
Bending moment	M = F × L² / 8 = 0.341 kNm
Bending stress	σ_m,a = (M × 10⁶) / Z = 1.97 N/mm²
	σ_m,a <= σ_m,adm ( 1.974 N/mm² <= 5.862 N/mm² ) therefore OK
Check shear stress
Grade shear stress for C16 (BS5268-2:2002 Table 8)	τ_par = 0.67 N/mm²
Permissible shear stress	τ_adm = τ_par × K₃ = 0.67 N/mm²
Shear stress	τ = (3 × F × L / 2) / (2 × b × h) = 0.234 N/mm²
	τ <= τ_adm ( 0.234 N/mm² <= 0.67 N/mm² ) therefore OK
Check deflection
Permissible deflection	δ_adm = 0.003 × L = 3.03 mm
Bending deflection	δ_bending = (5 × F × L⁴) / (384 × E_min × I) = 0.604 mm
Shear deflection	δ_shear = (12 × F × L²) / (5 × E_min × b × h) = 0.131 mm
Total deflection	δ_total = δ_bending + δ_shear = 0.735 mm
	δ_total <= δ_adm ( 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)	F_d = 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)	F_imposed,udl = 1 × ((75 - α) / 45) = 0.889 kN/m²
Total load	F = (1.25 × s / s_s ( F_imposed,udlcos(α) + F_d ) ( s_s / 1000 ) + F_s ) × cos(α) ) + F_jcos(α) = 4.53 kN/m
Grade bending stress for C16 (BS5268-2:2002 Table 8)	σ_m,par = 5.3 N/mm²
Permissible bending stress	σ_m,adm = σ_m,par × K₃ × K₇ = 7.33 N/mm²
Compression perpendicular to grain for C16 (BS5268-2:2002 Table 8)	σ_c,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 = (L_cl × F / 2) / (σ_c,per × K₃ × b - (F / 2)) = 15 mm
Effective span on slope	L = L_cl + a = 1.02 m
Check bending stress
Bending moment	M = F × L² / 8 = 0.584 kNm
Bending stress	σ_m,a = (M × 10⁶) / Z = 3.38 N/mm²
	σ_m,a <= σ_m,adm ( 3.379 N/mm² <= 7.328 N/mm² ) therefore OK
Check shear stress
Grade shear stress for C16 (BS5268-2:2002 Table 8)	τ_par = 0.67 N/mm²
Permissible shear stress	τ_adm = τ_par × K₃ = 0.838 N/mm²
Shear stress	τ = (3 × F × L / 2) / (2 × b × h) = 0.399 N/mm²
	τ <= τ_adm ( 0.399 N/mm² <= 0.838 N/mm² ) therefore OK
Check deflection
Permissible deflection	δ_adm = 0.003 × L = 3.05 mm
Bending deflection	δ_bending = (5 × F × L⁴) / (384 × E_min × I) = 1.04 mm
Shear deflection	δ_shear = (12 × F × L²) / (5 × E_min × b × h) = 0.224 mm
Total deflection	δ_total = δ_bending + δ_shear = 1.27 mm
	δ_total <= δ_adm ( 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

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