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Beam details

Timber strength class C24
Service class of timber 2
Timber width bt = 47 mm
Timber depth

 ht = 120 mm
Steel grade Grade 50
Steel width bs = 10 mm
Steel depth

 hs = 100 mm
Bolt diameter 12 mm

Span details

Beam clear span Lcl = 1 m
Bearing length Lb = 100 mm
Beam effective span Leff = Lcl + (2 × (Lb / 2)) = 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 - Timber Dormer Wall
Dimension to start of load, A A = 0.2 m
Length of load, B B = 0.4 m
Dead load Fd,2 = 1 kN/m² × 8 m = 8 kN/m
Imposed load Fi,2 = 0.75 kN/m² × 8 m = 6 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 7.93 kN 5.17 kN 13.10 kN
Right reaction 9.53 kN 5.83 kN 15.36 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
From BS5268-2 Table 18, bearing is < 75mm from joist end.
Bearing modification factorK4 = 1.00
Depth factor K7 = (300 / h)0.11 = 1.11
Load sharing modification factor (BS5268-2 clause 2.10.11)K8 = 1.10
Modulus of elasticity modification factor (BS5268-2 clause 2.9)K9 = 1.14

Modular ratio of steel to timber

Timber minimum modulus of elasticityEmin = 7,200 N/mm²
Modulus of elasticity for grade 50 steelEst = 205,000 N/mm²
The minimum modulus of elasticity modified by the factor K9 should be used for deflectionsE = Emin × K2,mod × K9 = 8,210 N/mm²
Modular ratioMR = Est / E = 25

Section properties

Equivalent timber Area of sectionEA = No. steel pieces × MR × bs × ds + No. timber pieces × (bt × dt) = 36,300 mm²
Inertia of timber about xx axis It = No. timber members × bt × dt³ / 12 = 13,500,000 mm4
Modified Inertia of steel about xx axis Is = No. steel plates × MR × bs × ds³ / 12 = 20,800,000 mm4
Total Inertia about xx axis in equivalent timber Ixx = It + Is = 34,300,000 mm4
Distance to edge of steelYs = ds / 2 = 50 mm
Distance to edge of timberYt = dt / 2 = 60 mm
Extreme fibre is timber sectionYc = Yt = 60 mm
Dist of centroid to steel edgeYn = Ys = 50 mm
Z to top edge of timberZc = Ixx / Yc = 572,000 mm³
Average density for C24 grade timber (BS 5268-2:2002 Table 8)ρmean = 420 kg/m³
Self weight of timber (g = 9.81 m/s²)Fself, timber = (bt × ht × ρmean) × Leff × g = 51.1 N
Self weight of steel (g = 9.81 m/s²)Fself, steel = (bs × hs × ρsteel) × Leff × g = 84.7 N
Total self weight (g = 9.81 m/s²)Fself = 136 N

Section design parameters

Design bending momentMb = 4,260,000 Nmm
Design shear forceFve = 15,400 N

Check bending stress

Strains in the timber and steel are the same at the same distance from the neutral axis. Since Yt > Ys ( 60 mm > 50 mm ) the modified steel stress will be less than that of the timber as a ratio of the distances Ys and Yt.
Timber grade bending stress parallel to grain (BS5268-2 Table 8)σt,m,g,par = 7.5 N/mm²
Permissible timber bending stress (factored)σt,m,adm = σt,m,g,par × K2,ben × K3 × K7 × K8 = 11.4 N/mm²
Maximum bending momentM = 4.26 kNm
Applied bending stress in timberσt,m,max = M/Zc = 7.44 N/mm²
 
Pass    σt,m,max <= σt,m,adm ( 7.444 N/mm² <= 11.406 N/mm² ) applied bending stress in timber within permissible
 
Permissible steel stress from table 2 BS449, grade 50 steel, with a fully restrained section assumed in design, steel plate less than or equal to 40 mm thickσs,adm = 230 N/mm²
Applied bending stress in steelσs,m,max = σt,m,max × MR × Yn / Yc = 155 N/mm²
 
Pass    σs,m,max <= σs,adm ( 154.927 N/mm² <= 230 N/mm² ) applied bending stress in steel within permissible
 

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

Deflection based on E = 8208 N/mm²
Dead load deflection without shearδd = 1.16 mm
Imposed load deflection without shearδl = 0.743 mm
Total dead & imposed load deflectionδt = 1.91 mm
Modulus of rigidityG = E / 16 = 513 N/mm²
Shape factor for rectangular sectionKF = 1.2
Shear area for beamAy = EA / KF = 30,200 mm²
Total dead and imposed loadW = 28.5 kN
If total dead & imposed load applied as a UDL, additional deflection due to shear δsu = W × Leff × 106/(8 × Ay × G) = 0.252 mm
Shear deflectionδshear = δsu × Mb/(W × L/8) = 0.275 mm
Permissible deflectionδadm = 0.003 × L × 10³ = 3.3 mm
Total deflection inclusive of shearδmax = δd + δl + δshear = 2.18 mm
 
Pass    δmax <= δadm ( 2.18 mm <= 3.3 mm ), therefore OK for deflection
 

Check shear stress

No notches to occur at the critical shear position.
Design shear forceFve = 15,400 N
Timber grade shear stress parallel to grain (BS5268-2 Table 8)τt,g,par = 0.71 N/mm²
Permissible shear parallel to grain (factored)τt,adm = τt,g,par × K2,shr × K3 × K8 = 0.976 N/mm²
Permissible shear force on timberFt,adm = 2 × τt,adm × No. timber members × bt × dt / 3 = 7,340 N
Since Fve > Ft,adm ( 15355 N > 7341 N ) shear capacity of timber alone exceeds applied shear force. The steel plates are required to extend over the support and act with the timber to resist the applied shear force.
Permissible shear stress on steelτs,adm = 100 N/mm²
Permissible shear force on steelFs,adm = 200,000 N
Total shear capacity of sectionFtotal,adm = Ft,adm + Fs,adm = 207,000 N
 
Pass    Fve <= Ftotal,adm ( 15355 N <= 207341 N ) applied shear force is less than the shear capacity of combined steel and timber, therefore OK
 

Check bearing stress

Timber grade compressive stress perpendicular to grain (BS5268-2 Table 8)σt,c,g,⊥ = 1.9 N/mm²
Permissible compressive stress perpendicular to grain (factored)σt,c,adm = σt,c,g,⊥ × K2,per × K3 × K4 × K8 = 2.61 N/mm²
Timber bearing stress on supportσt,c,max = Fve/(Lb × No. timber members × bt) = 1.63 N/mm²
 
Pass    σt,c,max <= σt,c,adm ( 1.634 N/mm² <= 2.613 N/mm² ) bearing stress is less than permissible timber stress, therefore OK
 

Bolts between timber and steel plates along beam

Total load on beamW = total left reaction + total right reaction = 28.5 kN
Maximum reaction at bearingRmax = 15.4 kN
Basic bolt shear capacity (tables 69-73)Vcapacity = 2.23 kN
Number of interfacesNinterfaces = 2
Minimum number of bolts required at bearings calcNbolts, bearings, calc = Rmax / (Ninterfaces × Vcapacity) = 3.44
Limiting bolt spacingSlimit = min(2.5 × ht, 600) = 300 mm
Minimum number of bolts along length of beamNbolts_along_length = W / (Ninterfaces × Vcapacity) = 6.38

Bolts to be staggered along the length of the beam, alternately set ht/4 = 30 mm above and below the centre line.
Bolts are to be spaced at 300 mm maximum centres.
Bolts at the supports are to be located 0.5 x the bearing length (Lb) = 50 mm from the inner edge of the support.
Minimum end and edge distances in the timber are to be 4 x bolt diameter = 48 mm.
Under point loads, provide an additional bolt at the centre line of the beam.

Design summary

PermissibleApplied/ActualUtilisationResult
Shear force (kN)20715.47.4 %OK
Timber bending stress (N/mm²)11.47.4465.3 %OK
Steel bending stress (N/mm²)23015567.4 %OK
Bearing stress (N/mm²)2.611.6362.5 %OK
Deflection (mm)3.32.1866.1 %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 BS449 - Specification for the use of structural steel in building.

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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