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Beam details

Beam 56mm x 225mm
5 Layer Glulam
C30 Grade Timber
Timber strength class C30
Service class used 2
Width b = 56 mm
Depth

 h = 225 mm

Span details

Beam clear span Lcl = 1 m
Bearing length Lb = 100 mm
Beam effective span Leff = 1.1 m
Diagrams not to scale

Loading details

Load 1: UDL - Sloping roof, 0° to 30°
Dead load  Fd,1 = 1.15 kN/m² × 8 m = 9.2 kN/m
Imposed load  Fi,1 = 0.75 kN/m² × 8 m = 6 kN/m
Load 2: Partial UDL - Sloping roof, 0° to 30°
Dimension to start of load, A A = 0.2 m
Length of load, B B = 0.4 m
Dead load Fd,2 = 1.15 kN/m² × 7 m = 8.05 kN/m
Imposed load Fi,2 = 0.75 kN/m² × 7 m = 5.25 kN/m
Load 3: Point load - Point load
Distance to point load, A A = 0.8 m
Dead load Fd,3 = 4 kN
Imposed load Fi,3 = 2 kN

Reactions (unfactored)

Dead Imposed Total
Left reaction 7.90 kN 5.00 kN 12.90 kN
Right reaction 9.50 kN 5.70 kN 15.20 kN

Modification factors

Timber service class modification factor K2 as table 16
Bending parallel to grainK2,ben = 1.00
Compression perpendicular to grainK2,per = 1.00
Shear parallel to grainK2,shr = 1.00
Mean & min modulus of elasticityK2,mod = 1.00
Load duration factorK3 = 1.25
Bearing modification factorK4 = 1.00
Depth factor (BS5268-2 clause 2.10.6) K7 = (300 / h)0.11 = 1.03
Load sharing modification factor (BS5268-2 clause 2.10.11)K8 = 1.00
Modulus of elasticity modification factor (BS5268-2 clause 2.9)K9 = 1.00

Glulam modification factors

Bending parallel to grain modification factorK15 = 1.10
Compression perpendicular to grain modification factorK18 = 1.49
Shear parallel to grain modification factorK19 = 1.49
Modulus of elasticity modification factorK20 = 1.00

Modulus of elasticity

Timber minimum modulus of elasticityEmin = 8,200 N/mm²
The minimum modulus of elasticity modified by the factor K9 should be used for deflectionsE = Emin × K2,mod × K9 × K20 = 8,200 N/mm²

Section properties

Area of sectionArea = b × h = 12,600 mm²
Inertia of timber about xx axis Ixx = b × h³ / 12 = 53,200,000 mm4
Z to top edge of timberZ = b × h² / 6 = 472,000 mm³
Average density for C30 grade timber (BS 5268-2:2002 Table 8)ρmean = 460 kg/m³
Self weight (g = 9.81 m/s²)Fself = b × h × Leff × ρmean × g = 62.5 N

Section design parameters

Design bending momentMb = 4,190,000 Nmm
Design shear forceFve = 15,200 N

Check bending stress

Timber grade bending stress parallel to grain (BS5268-2 Table 8)σt,m,g,par = 11 N/mm²
Permissible timber bending stress (factored)σt,m,adm = σt,m,g,par × K2,ben × K3 × K7 × K8 × K15 = 15.6 N/mm²
Maximum bending momentM = 4.19 kNm
Applied bending stress in timberσt,m,max = M / Z = 8.87 N/mm²
 
Pass    σt,m,max <= σt,m,adm ( 8.874 N/mm² <= 15.611 N/mm² ) applied bending stress in timber within permissible
 

Check deflection (including shear deflection as required by clause 2.10.7)

Deflection based on E = 8200 N/mm²
Dead load deflection without shearδd = 0.751 mm
Imposed load deflection without shearδl = 0.463 mm
Total dead & imposed load deflectionδt = 1.21 mm
Modulus of rigidityG = E / 16 = 512 N/mm²
Shape factor for rectangular sectionKF = 1.2
Shear area for beamAy = EA / KF = 10,500 mm²
Total dead & imposed loadWT = 28.1 kN
If total dead & imposed load applied as a UDL, additional deflection due to shear δsu = WT × Leff × 106 / (8 × Ay × G) = 0.718 mm
Shear deflectionδshear = δsu × M / ( WT × Leff / 8 ) = 0.779 mm
Permissible deflectionδadm = 0.003 × Leff × 10³ = 3.3 mm
Total deflection inclusive of shearδmax = δd + δi + δshear = 1.99 mm
 
Pass    δmax <= δadm ( 1.99 mm <= 3.3 mm ), therefore OK for deflection
 

Check shear stress

No notches to occur at the critical shear position.
Timber grade shear stress parallel to grain (BS5268-2 Table 8)τt,g,par = 1.2 N/mm²
Permissible shear parallel to grain (factored)τt,adm = τt,g,par × K2,shr × K3 × K8 × K19 = 2.24 N/mm²
Permissible shear force on timberFt,adm = 2 × τt,adm × b × h / 3 = 18,800 N
Design shear forceFve = 15,200 N
 
Pass    Fve <= Ft,adm ( 15204 N <= 18774 N ) shear capacity of timber is greater than applied shear force, therefore OK
 

Check bearing stress

Timber grade compressive stress perpendicular to grain (BS5268-2 Table 8)σt,c,g,⊥ = 2.2 N/mm²
Permissible compressive stress perpendicular to grain (factored)σt,c,adm = σt,c,g,⊥ × K2,per × K3 × K4 × K8 × K18 = 4.1 N/mm²
Timber bearing stress on supportσt,c,max = Fve / (Lb × b) = 2.71 N/mm²
 
Pass    σt,c,max <= σt,c,adm ( 2.715 N/mm² <= 4.098 N/mm² ) bearing stress is less than permissible timber stress, therefore OK
 

Design summary

PermissibleApplied/ActualUtilisationResult
Shear force (kN)18.815.281 %OK
Bending stress (N/mm²)15.68.8756.8 %OK
Bearing stress (N/mm²)4.12.7166.3 %OK
Deflection (mm)3.31.9960.4 %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.

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.

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