Euro-Code 5 Folder: _EC5 Beams

Euro-Code 5
Beams

Half-span roof:
rd

a b

l
System:
Beam length l = 7,00 m
Beam width b = 14,00 cm
Beam height an a ha = 26,00 cm
Pitch α = 6,00 °
Support length t = 10,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = Glulam
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = BS14h
fm,g,k = TAB("wood/EC"; fm.k; SG=SG) = 28,00 N/mm²
fv,g,k = TAB("wood/EC"; fv.k; SG=SG) = 2,70 N/mm²
ft,90,g,k = TAB("wood/EC"; ft,90.k; SG=SG) = 0,45 N/mm²
fc,90,k = TAB("wood/EC"; fc,90.k; SG=SG) = 5,50 N/mm²
E0,mean = TAB("wood/EC"; E0.mean; SG=SG) = 12500,00 N/mm²
γM = 1,30
γS = 1,10

Load:
rd = 12,50 kN/m

Calculation:
fm,g,d = fm,g,k * kmod / γM = 19,38 N/mm²
fv,g,d = fv,g,k * kmod / γM = 1,87 N/mm²
fc,90,d = fc,90,k * kmod / γM = 3,81 N/mm²
ft,90,g,d = ft,90,g,k * kmod / γM = 0,31 N/mm²

Stress resultants for analysis of load bearing capacity:

Md = rd * l² / 8 = 76,56 kNm
Vd = l / 2 * rd = 43,75 kN

hs,req = 150*Vd/(b*fv,g,d) = 250,67 mm

hs,req / (10*ha) = 0,96 < 1

hb = ha+l*100*TAN(α) = 99,57 cm
Euro-Code 5 Folder: _EC5 Beams

Point of maximum stress:

x= l/(1+hb/ha) = 1,45 m
Mx,d = Vd * x - rd * x² * 0,5 = 50,30 kNm
W x,req = 1100 * Mx,d / fm,g,d = 2855,01 cm³
hx = ha+x*100*TAN(α) = 41,24 cm
Wx = b * hx² / 6 = 3968,39 cm³

W x,req / W x = 0,72 < 1

Required area moment of second degree at a deflection of l/300:

Ireq = 3130000 * Md / 1,4 * l / E0,mean = 95853,12 cm4
hm,req = 0,1*(12000*Ireq/b)1/3 = 43,47 cm
hm = (hb-ha)/2+ha = 62,78 cm
hm,req / hm = 0,69 < 1

Analysis of load-bearing capacity:

Shearing stress at footing a:
τa,d = 1,5*Vd*10/(b*ha) = 1,80 N/cm²
τa,d / fv,g,d = 0,96 < 1

Stresses in the edge parallel to the grain:

σm,0,d = 0,001*(1+4*(TAN(α))²)*6*106*Mx,d/(b*hx²) = 13,24 N/mm²
σm,0,d / fm,g,d = 0,68 < 1

Stresses in the edge with grains at an angle:

σm,α,d = 0,001*(1-4*(TAN(α))²)*6*106*Mx,d/(b*hx²) = 12,12 N/mm²
fm,α,d = fm,g,d/(fm,g,d/fc,90,d*(SIN(α))²+(COS(α))²) = 18,55 N/mm²
σm,α,d / fm,α,d = 0,65 < 1

Bearing pressure:
Assumption: l < 150 mm;l1 > 150 mm ; kc,90 according to Tab. 5.1.5
⇒ kc,90 = 1,0
req_A = Vd *10 / ( fc,90,d * kc,90) = 114,83 cm²
req_t = req_A / b = 8,20 cm
req_t / t = 0,82 < 1
Euro-Code 5 Folder: _EC5 Buck

Buck

Buckling lengths:
Euler's crippling load:
l= 5,00 m

lef = 2*l = 10,00 m

lef = l = 5,00 m

lef = 0,7*l = 3,50 m

lef = 0,5*l = 2,50 m

Euro-Code 5 Folder: _EC5 Buck

l
Kr

Sorting grade SG = SEL("Wood/EC"; SG; ) = S10

E0,mean = TAB("Wood/EC"; E0.mean; SG=SG) = 11000,00 N/mm²
Kr = 100000,00 1/N
lef = √(4+(π²*E0,mean)/(l*Kr))
l*√ = 10,27 m
Euro-Code 5 Folder: _EC5 Buck

Roof post:

System:
Post height h = 3,40 m
Cross-sectional width b = 14,00 cm
Cross-sectional thickness d = 14,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90

Strength grade SG= SEL("wood/EC"; SG; ) = S10

ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
fc,0,k = TAB("wood/EC"; fc,0,k; SG=SG) = 21,00 N/mm²
fc,90,k = TAB("wood/EC"; fc,90,k; SG=SG) = 5,00 N/mm²
γM = 1,30

Load at the head:

Vc,d = 50,40 kN

Calculation:
Buckling length lef = h = 3,40 m
A= b *d = 196,00 cm²
i= √(12)*MIN(b;d)
1/√ = 4,04 cm
Degree of slenderness λmin = 100*lef/i = 84,16 > 30
⇒ Buckling proof required!
kc = TAB("wood/ECbuckl"; kc; SG=SG; λ=λ
λmin) = 0,427
σc,0,d = 10*Vc,d / A = 2,57 N/mm²
fc,0,d = fc,0,k * kmod/γM = 14,54 N/mm²
fc,90,d = fc,90,k * kmod/γM = 3,46 N/mm²

Structural verifications:
Buckling: σc,0,d /(kc*fc,0,d) = 0,41 < 1
Lateral pressure: σc,0,d/fc,90,d = 0,74 < 1
Euro-Code 5 Folder: _EC5 Dowel

Dowel
Beam connection:

e3

h
e2
e1
a1 a1 a1
b
t1 t2 t2

Fd
System:
Distance a1 = 4,00 cm
Width b = 3*a1 = 12,00 cm
Distance e1 = 4,00 cm
Distance e2 = 8,50 cm
Distance e3 = 5,50 cm
Thickness t1= 6,00 cm
Thickness t2 = 8,00 cm

Load:
Fd = 40,00 kN

Material:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80

Strength grade SG= SEL("wood/EC"; SG; ) = S10

ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
γM = 1,30

Dowel: S235
Dowel diameter d = SEL("wood/DPin"; d; )/10 = 1,20 cm
fu,k = 360,00 N/mm²
γS = 1,10 N/mm²

Calculation:
My,d = 0.8*fu,k*d³/(6*γS )*103 = 75403,64 Nmm
Post:
fh,1,d = 0.082 * ρk * (1-0.1*d)*kmod /γM = 16,87 N/mm²
Euro-Code 5 Folder: _EC5 Dowel

Support:
k90 = 1.35 + 0.15 * d = 1,53
fh,2,d = 0.082 * ρk * (1-0.1*d)*kmod/( γM*k90) = 11,03 N/mm²

β= fh,2,d / fh,1,d = 0,654

RD1 = fh,1,d * t1 * d * 100 =12146

N
RD2 = 0.5 * fh,1,d * t2 * d * β * 100 =5296
N
RD3 = √(2*β*(1+β)+4*β*(2+ β)*My,d/(fh,1,d*103*d*t1²))-β)
1,1*fh,1,d*10²*t1*d/(2+β )*(√ =5254
N
RD4 = √(2*β/(1+β))* √(2*My,d*fh,1,d*d*10)
1.1*√ =5405
N

RD = 0.001 * MIN(RD1;RD2;RD3;RD4) = 5,25 kN

n= Fd / ( 2 * RD) = 3,8

selected: 4 PIN ∅ 12mm S 235 l = 20 cm

Euro-Code 5 Folder: _EC5 Dowel

Tensile splice Design of strut

t1
t2
t1

dN
System:
Timber thickness t1 = 8,00 cm
Timber thickness t2 = 12,00 cm

Load:
Fsd = 116,00 kN

Materials: NH S 10
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80
Strength grade SG= SEL("wood/EC"; SG; ) = S10
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
γM = 1,30
γS = 1,10
Dowel: DPin ∅ 16mm S 275
Dowel diameter d = SEL("wood/DPin"; d; )/10 = 0,80 cm
fu,k = 360,00 N/mm²

Calculation:
Minimum distances according to Tab.6.6a:
a1 = 7* d = 5,60 cm
a2 = 3*d = 2,40 cm
a3,t = 7* d = 5,60 cm
a4,t = 3*d = 2,40 cm

Design value of bearing stress resistance:

fh,0,d = 0.082 * ρk * (1-0.1*d)*kmod/γM = 17,64 N/mm²

My,d = 1000*0.8*fu,k*d³/(6*γS ) = 22341,82 Nmm

kM = 10*t1 / √( My,d / ( fh,0,d * d *10) ) = 6,36

RD1 = 100* fh,0,d * t1 * d = 11289,60 N

RD2 = 100* 0.5 * fh,0,d * t2 * d = 8467,20 N
RD3 = 100* 0.367 * fh,0,d * t1 * d * ( 2 * √(1 + 3 / kM²) - 1 ) = 4445,08 N
RD4 = 155.6 * fh,0,d * t1 * d / kM = 2762,05 N

RD = 0.001 * MIN(RD1;RD2;RD3;RD4) = 2,76 kN

required number of dowels n = Fsd/(2*RD) = 21,0

selected: 6 dowels ∅ 16 S235 l=28cm

Euro-Code 5 Folder: _EC5 Dowel

Tension diagonal:
2

ß
slot plate

Ft 1
System:
Rod 1 : 1 * 160mm * 160mm Glulam
Rod 2 : 1 * 160mm * 180mm Glulam
System angle β = 60,0 °
Timber thickness t1 = 16,00 cm
Timber thickness t2 = 16,00 cm
Thickness of slotted plate t3 = 0,80 cm

Construction material CM= SEL("wood/kmod"; CM; ) = Glulam

Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80

Strength grade SG= SEL("wood/EC"; SG; ) = BS11

ρk = TAB("wood/EC"; ρk; SG=SG) = 410,00 kg/m³
γM = 1,30
γS = 1,10

Dowel: DPin ∅ 16mm S 275

Dowel diameter d = SEL("wood/DPin"; d; )/10 = 1,60 cm
fu,k = 430,00 N/mm²

Ft,d = 80,79 kN

Calculation:
Minimum distances of Rod1 according to Tab.6.6a:
a1 = 7* d = 11,20 cm
a2 = 3*d = 4,80 cm
a3,t = 7* d = 11,20 cm
a4,t = 3*d = 4,80 cm

Minimum distances Rod2 according to Tab.6.6a:

a1 = ( 3 + 4 * ABS(COS( β)))* d = 8,00 cm
a2 = 3*d = 4,80 cm
a3,t continuous top chord
a4,t = ( 2 + 2 * ABS(SIN(β))) * d = 5,97 cm

Rod 1 :
fh,1,d = 0.082 * ρk * (1-0.1*d)*kmod /γM = 17,38 N/mm²
Euro-Code 5 Folder: _EC5 Dowel

Rod 2 :
fh,0.d = 0.082 * ρk * (1-0.1*d)*kmod /γM = 17,38 N/mm²
k90 = 1.35 + 0.15 * d = 1,59
fh,2,d = fh,0.d /(k90*(SIN(β))²+(COS(β))²) = 12,05 N/mm²

My,d = 0.8*fu,k*d³*1000/(6*γS) = 213488,48 Nmm

Design value of drift pins in Rod 1:

RD1 = 100* fh,1,d *( t1 / 2 - t3) * d = 20022 N
RD2 = 1.1*fh,1,d*(10*t1/2-10*t3)*d*10*(√√(2+4*My,d/(fh,1,d*10*d*(10*t1/2-10*t3)²))-1) =13437
N
RD3 = 1.5* √(20*My,d*fh,1,d*d) = 16345 N

RD = 0.001 * MIN(RD1;RD2;RD3) = 13,44 kN

n= Ft,d / ( 2 * RD) = 3,0

selected: 4 DPin ∅ 16 S 275 l = 16cm

Design value of drift pins in Rod 2:

RD1 = 100* fh,2,d *( t1 / 2 - t3) * d = 13882 N
RD2 = 1.1*fh,2,d*(10*t1/2-10*t3)*10*d*(√√(2+4*My,d/(fh,2,d*10*d*(10*t1/2-10*t3)²))-1) = 10528 N
RD3 = 15* √(0.2*My,d*fh,2,d*d) = 13610 N

RD = 0.001 * MIN(RD1;RD2;RD3) = 10,53 kN

n= Ft,d / ( 2 * RD) = 3,8

selected: 4 DPin ∅ 16 S 275 l = 16cm

Euro-Code 5 Folder: _EC5 Dowel

Tension diagonal:
2

Ft
System:
Rod 1 : 2 * 60mm * 140mm
Rod 2 : 1 * 100mm * 180mm
System angle β = 60,00 °
Timber thickness t1 = 6,00 cm
Timber thickness t2 = 10,00 cm

Construction material CM= SEL("wood/kmod"; CM; ) = solid wood

Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80

Strength grade SG= SEL("wood/EC"; SG; ) = S7

ρk = TAB("wood/EC"; ρk; SG=SG) = 350,00 kg/m³

γM = 1,30
γS = 1,10
Dowel: DPin ∅ 12mm
Dowel diameter d = SEL("wood/DPin"; d; )/10 = 1,20 cm
fu,k = 360,00 N/mm²

Load:
Ft,d = 40,00 kN

Calculation:
Minimum distances of Rod1 according to Tab.6.6a:
a1 = 7* d = 8,40 cm
a2 = 3*d = 3,60 cm
a3,t = 7* d = 8,40 cm
a4,t = 3*d = 3,60 cm

Minimum distances Rod2 according to Tab.6.6a:

a1 = ( 3 + 4 * ABS(COS(β)))* d = 6,00 cm
a2 = 3*d = 3,60 cm
a3,t continuous top chord
a4,t = ( 2 + 2 * ABS(SIN( β))) * d = 4,48 cm

Rod 1 :
fh,1,d = 0.082 * ρk * (1-0.1*d)*kmod/γM = 15,54 N/mm²
Euro-Code 5 Folder: _EC5 Dowel

Rod 2 :
fh,0.d = 0.082 * ρk * (1-0.1*d)*kmod/γM = 15,54 N/mm²
k90 = 1.35 + 0.15 * d = 1,53
fh,2,d = fh,0.d/(k90*(SIN(β))²+(COS(β))²) = 11,12 N/mm²

β= fh,2,d / fh,1,d = 0,716

My,d = 10 3*0.8*fu,k*d³/(6* γS) = 75403,64 Nmm

RD1 = 100* fh,1,d * t1 * d = 11189 N

RD2 = 100* 0.5 * fh,1,d * t2 * d * β = 6676 N
√(2*β *(1+β)+4*β*(2+β)*My,d/(fh,1,d*103*d*t1²))-β)
RD3 = 1.1*fh,1,d*t1*100*d/(2+β)*(√ = 5026 N
RD4 = 1.1* √(2*β/(1+β))*√√(2*My,d *fh,1,d*10*d) = 5329 N

RD = 0.001 * MIN(RD1;RD2;RD3;RD4) = 5,03 kN

n= Ft,d / ( 2 * RD) = 4,0

selected: 4 DPin ∅ 12 S 235 l = 22cm

Euro-Code 5 Folder: _EC5 Nail

Nail

Nail calculation for each shear joint, thin outside plate:

Thin outside plate
t1
t2
t1

dN
System:
Sheet thickness t1 = 0,15 cm
Timber thickness t2 = 4,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = S13
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
γM = 1,30
γS = 1,10
Nails: 38x100
Nail length lN = 10,00 cm
Nail diameter dN = 0,38 cm
t1 / dN = 0,39 < 0,5
otherwise a different calculation applies
Calculation:
pre-drilled:
fh,k = 0.082 * ρk * (1-0.1*dN ) = 29,98 N/mm²
not pre-drilled::
fh,k = 0.082 * ρk * (10*dN)-0.3 = 20,88 N/mm²

Characteristic value for yield moment of square nails

My,d = 270 * (10*dN)2.6 / γS = 7896,02 Nmm
Characteristic value for yield moment of round nails
My,d = 180 * (10*dN)2.6 / γS = 5264,02 Nmm

fh,d = kmod / γM * fh,k = 14,46 N/mm²

RD1 = 100*0.5* fh,d * t2 * dN = 1098,96 N

RD2 = √(20*My,d*fh,d*dN)
1.1*√ = 836,65 N

RD = 0.001 * MIN(RD1;RD2) = 0,84 kN

Euro-Code 5 Folder: _EC5 Nail

Design of a shear connection:

lr l l
h1 h
2
hi h
n
be h

b2 b b2

System:
Thickness of outer wood d2 = 4,00 cm
Nail distance lr = 6,00 cm
Nail distance be = 16,00 cm
Cross beam width b = 16,00 cm
Cross beam height h = 40,00 cm
Height h1= 24,00 cm
Height h2 = 28,00 cm
Height h3 = 32,00 cm
Height h4 = 36,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = Glulam
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = BS11
Shear and torsion fv,g,k = TAB("wood/EC"; fv,k; SG=SG) = 2,70 N/mm²
Rectangular shear ft,90,g,k = TAB("wood/EC"; ft,90.k; SG=SG) = 0,45 N/mm²
γMH = 1,30

Strength grade SG= SEL("wood/EC"; SG; ) = S10

Rectangular shear ft,0,k = TAB("wood/EC"; ft,0,k; SG=SG) = 14,00 N/mm²

Nails 38x100:
Nail diameter d = 0,38 cm
Nail length lN = 10,00 cm
Nail strength RD = 6000*d² = 866,40 N

γF = 1,50

fv,g,d = kmod*fv,g,k/ γMH = 1,869 N/mm²

ft,90,g,d = kmod*ft,90,g,k/ γMH = 0,312 N/mm²
ft,0,d = kmod*ft,0,k/γMH = 9,692 N/mm²
leff = 10*MIN(d2;12*d;lN-d2) = 40,00 mm

Calculation of maximum design value:

from nail analysis F90,d,N = 0.002 * 16 * RD = 27,72 kN
from tension bar F90,d,Z = 0.2 / γF * d2 *lN * ft,0,d = 51,69 kN
Euro-Code 5 Folder: _EC5 Nail

detailed structural verification:

be / h = 0,40 < 0,5
⇒ detailed structural verification required
bef = 2 * leff = 80,00 mm
c= 4/3 *√ √(be/h * (1 - be/h)³) = 0,392
lr,ef = 10*√ √(lr²+ ( c * h)² ) = 167,89 mm
Aef = lr,ef *bef = 13431,20 mm²
kr = 0.25 *((1) + (h1 /h2)² + (h1 /h3)² + (h1 /h4)²) = 0,685
η= 1 - 3 * (be/h)² + 2 * (be/h)³ = 0,648

F90,d = 0.001 / ( η * kr ) * 13 * Aef0.8 * ft,90,g,d = 18,34 kN

Decisive F90,d,max = MIN(F90,d;F90,d,N;F90,d,Z) = 18,34 kN

Euro-Code 5 Folder: _EC5 Nail

Nail calculation for each shear joint for two different kinds of wood:
two unequal kinds of wood
t1
t2
t1

< dN
System:
Timber thickness t1 = 4,00 cm
Timber thickness t2 = 4,00 cm
t1< t2
Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG1= SEL("wood/EC"; SG; ) = S13
ρk1 = TAB("wood/EC"; ρk; SG=SG1) = 380,00 kg/m³
Strength grade SG2= SEL("wood/EC"; SG; ) = MS13
ρk2 = TAB("wood/EC"; ρk; SG=SG2) = 400,00 kg/m³
γM = 1,30
γS = 1,10
Nails: 38x100
Nail length lN = 10,00 cm
Nail diameter dN = 0,38 cm

Calculation:
pre-drilled:
fh,k1 = 0.082 * ρk1 * (1-0.1*dN) = 29,98 N/mm²
not pre-drilled::
fh,k1 = 0.082 * ρk1 * (10*dN)-0.3 = 20,88 N/mm²

pre-drilled:
fh,k2 = 0.082 * ρk2 * (1-0.1*dN ) = 31,55 N/mm²
not pre-drilled::
fh,k2 = 0.082 * ρk2 * (10*dN)-0.3 = 21,98 N/mm²

Characteristic value for yield moment of square nails

My,d = 270 * (10*dN)2.6 / γS = 7896,02 Nmm
Characteristic value for yield moment of round nails
My,d = 180 * (10*dN)2.6 / γS = 5264,02 Nmm

fh,d1 = kmod / γM * fh,k1 = 14,46 N/mm²

fh,d2 = kmod / γM * fh,k2 = 15,22 N/mm²

β= fh,d1 / fh,d2 = 0,95

RD1 = 100* fh,d1 * t1 * dN = 2197,92 N

RD2 = 100* 0.5 * fh,d1 * t2 * dN * β = 1044,01 N
√(2*β*(1+β)+4* β*(2+β)*My,d/(fh,d1*103*dN*t1 ²))-β)
RD3 = 110*fh,d1*t1*dN /(2+β)*(√ = 935,89 N
RD4 = √(2*β/(1+β))* √(20*My,d*fh,d1*dN)
1.1*√ = 825,85 N

RD = 0.001 * MIN(RD1;RD2;RD3;RD4) = 0,83 kN

Euro-Code 5 Folder: _EC5 Nail

Nail calculation for each shear joint:

Thick outside plate
t1
t2
t1

dN
System:
Sheet thickness t1 = 0,50 cm
Timber thickness t2 = 4,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = S13
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
γM = 1,30
γS = 1,10
Nails: 38x100
Nail length lN = 10,00 cm
Nail diameter dN = 0,38 cm
otherwise a different calculation applies

Calculation:
pre-drilled:
fh,k = 0.082 * ρk * (1-0.1*dN ) = 29,98 N/mm²
not pre-drilled::
fh,k = 0.082 * ρk * (10*dN)-0.3 = 20,88 N/mm²

Characteristic value for yield moment of square nails

My,d = 270 * (10*dN)2.6 / γS = 7896,02 Nmm
Characteristic value for yield moment of round nails
My,d = 180 * (10*dN)2.6 / γS = 5264,02 Nmm

fh,d = kmod / γM * fh,k = 14,46 N/mm²

RD1 = 100*0.5* fh,d * t2 * dN = 1098,96 N

RD2 = √(20*My,d*fh,d*dN)
1.5*√ = 1140,88 N

RD = 0.001 * MIN(RD1;RD2) = 1,10 kN

Euro-Code 5 Folder: _EC5 Nail

Nail calculation for each shear joint, plate inside:

Plate inside
t1
t2
t1

dN
System:
Timber thickness t1 = 4,00 cm
Sheet thickness t2 = 1,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = S13
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
γM = 1,30
γS = 1,10
Nails: 38x100
Nail length lN = 10,00 cm
Nail diameter dN = 0,38 cm

Calculation:
pre-drilled:
fh,k = 0.082 * ρk * (1-0.1*dN ) = 29,98 N/mm²
not pre-drilled::
fh,k = 0.082 * ρk * (10*dN)-0.3 = 20,88 N/mm²

Characteristic value for yield moment of square nails

My,d = 270 * (10*dN)2.6 / γS = 7896,02 Nmm
Characteristic value for yield moment of round nails
My,d = 180 * (10*dN)2.6 / γS = 5264,02 Nmm

fh,d = kmod / γM * fh,k = 14,46 N/mm²

RD1 = 100* fh,d * t1 * dN = 2197,92 N

RD2 = √(2+4*My,d/(fh,d*103*dN*t1²))-1)
110*fh,d *t1*dN*(√ = 1200,38 N
RD3 = √(20*My,d*fh,d*dN)
1.5*√ = 1140,88 N

RD = 0.001 * MIN(RD1;RD2;RD3) = 1,14 kN

Euro-Code 5 Folder: _EC5 Nail

Nail calculation for each shear joint:

t1
t2
t1

dN
System:
Timber thickness t1 = 4,00 cm
Timber thickness t2 = 4,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = S13
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
γM = 1,30
γS = 1,10
Nails: 38x100
Nail length lN = 10,00 cm
Nail diameter dN = 0,38 cm

Calculation:
pre-drilled:
fh,k = 0.082 * ρk * (1-0.1*dN ) = 29,98 N/mm²
not pre-drilled::
fh,k = 0.082 * ρk * (10*dN)-0.3 = 20,88 N/mm²

Characteristic value for yield moment of square nails

My,d = 270 * (10*dN)2.6 / γS = 7896,02 Nmm
Characteristic value for yield moment of round nails
My,d = 180 * (10*dN)2.6 / γS = 5264,02 Nmm

fh,d = kmod / γM * fh,k = 14,46 N/mm²

kM = 10*MIN(t1;t2) / √( My,d / ( fh,d *10* dN ) ) = 4,09

RD1 = 100* fh,d * t1 * dN = 2197,92 N

RD2 = 100* 0.5 * fh,d * t2 * dN = 1098,96 N
RD3 = 100*0.367*fh,d *t1*dN*(2*√ √(1+3/kM²)-1) = 945,34 N
RD4 = 155.6 * fh,d * t1 * dN / kM = 836,18 N

RD = 0.001 * MIN(RD1;RD2;RD3;RD4) = 0,84 kN

Euro-Code 5 Folder: _EC5 Nail

Determination of required number of nails

t1
t2
t1

System:
Timber thickness t1 = 4,00 cm
Timber thickness t2 = 4,00 cm

Materials:
Construction material CM = SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD = SEL("wood/kmod"; CLD; ) = permanent
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,60
Strength grade SG= SEL("wood/EC"; SG; ) = S10
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
γM = 1,30
γS = 1,10
Nails: 42x110
Nail length lN = 12,00 cm
Nail diameter dN = 0,42 cm

Load:
Fsd = 16,00 kN

Calculation:
Minimum timber thickness adhered to according to 6.3.1.2(11):
(MAX(7*dN;(13*dN -30)*ρk/400))/(MIN(t1;t2)) = 0,73 < 1
Minimum driving depth according to 6.3.1.2(4):
(8*dN)/MIN(t1;t2) = 0,84 < 1
Minimum nail distances according to Tab.6.3.1.2:
a1 = 10* dN = 4,20 cm
a2 = 5 * dN = 2,10 cm
a3,t = 15* dN = 6,30 cm
a4,t = 5 * dN = 2,10 cm

pre-drilled fh,k = 0.082 * ρk * (1-0.1*dN ) = 29,85 N/mm²

not pre-drilled fh,k = 0.082 * ρk * (10*dN) -0.3 = 20,26 N/mm²
fh,d = kmod / γM * fh,k = 9,35 N/mm²

Characteristic value for yield moment of square nails

My,d = 270 * (10*dN )2.6 / γS = 10242,81 Nmm
Characteristic value for yield moment of round nails
My,d = 180 * (10*dN)2.6 / γS = 6828,54 Nmm

kM = 10*MIN(t1 ;t2) / √( My,d / ( fh,d *10* dN ) ) = 3,03

Euro-Code 5 Folder: _EC5 Nail

RD1 = 100* fh,d * t1 * dN = 1570,80 N

RD2 = 100* 0.5 * fh,d * t2 * dN = 785,40 N
RD3 = 36.7 * fh,d * t1 * dN * ( 2 * √(1 + 3 / kM ²) - 1 ) = 751,57 N
RD4 = 155.6 * fh,d * t1 * dN / kM = 806,66 N

RD = 0.001 * MIN(RD1 ;RD2;RD3;RD4) = 0,75 kN

required number of nails n = Fsd/(2*RD) = 10,7

selected: 2x6 nails 42x120 DIN 1151

Euro-Code 5 Folder: _EC5 Nail

Determination of required number of nails

t1
t2
t1
System:
Timber thickness t1 = 4,00 cm
Timber thickness t2 = 4,00 cm
Timber width b = 8,00 cm

Materials:
Construction material CM = SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD = SEL("wood/kmod"; CLD; ) = permanent
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,60
Strength grade SG= SEL("wood/EC"; SG; ) = S10
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
γM = 1,30
γS = 1,10
Nails: 42x110
Nail length lN = 6,50 cm
Nail diameter dN = 0,28 cm

Load:
Fsd = 16,00 kN

Calculation:
t3 = (lN - t1) = 2,50 cm
Checking the lapped joint length according to 6.3.1.2(10):
4 * dN / (t2 - t3) = 0,75 < 1
Minimum timber thickness adhered to according to 6.3.1.2(11):
(MAX(7*dN;(13*dN-30)*ρk/400))/(MIN(t1;t2 )) = 0,49 < 1
Minimum driving depth according to 6.3.1.2(4):
(8*dN)/MIN(t1;t2;t3) = 0,90 < 1
Minimum nail distances according to Tab.6.3.1.2:
a1 = 10* dN = 2,80 cm
a2 = 5 * dN = 1,40 cm
a3,t = 15* dN = 4,20 cm
a4,t = 5 * dN = 1,40 cm

pre-drilled fh,k = 0.082 * ρk * (1-0.1*dN ) = 30,29 N/mm²

not pre-drilled fh,k = 0.082 * ρk * (10*dN) -0.3 = 22,88 N/mm²
fh,d = kmod / γM * fh,k = 10,56 N/mm²

Characteristic value for yield moment of square nails

My,d = 270 * (10*dN)2.6 / γS = 3569,29 Nmm
Characteristic value for yield moment of round nails
My,d = 180 * (10*dN)2.6 / γS = 2379,53 Nmm

kt = MAX(t1;t2;t3)/MIN(t1;t2;t3) = 1,60
kM = 10*MIN(t1;t2;t3) / √( My,d / ( fh,d * 10*dN ) ) = 2,79
Euro-Code 5 Folder: _EC5 Nail

RD1 = 100* fh,d * MIN(t1;t2;t3) * dN = 739,20 N

RD2 = 100*0.5*fh,d *MIN(t1;t2;t3)*dN*(√ √( 3 * kt² + 2 * kt + 3 ) - kt -1 ) = 416,02 N
RD3 = 36.7 * fh,d * MIN(t1;t2;t3) * dN * ( 2 * √(1 + 3 / kM²) - 1 ) = 367,34 N
RD4 = 155.6 * fh,d * MIN(t1;t2;t3) * dN / kM = 412,26 N

RD = 0.001 * MIN(RD1;RD2;RD3;RD4) = 0,37 kN

required number of nails n = Fsd/RD = 43,2

selected: 2x3x8=48 nails 28x65 DIN 1151

Euro-Code 5 Folder: _EC5 Nail

Attachment of roof boarding/sheathing

Fld,d
Fax,d

t1

t2

System:
Timber thickness t1 = 2,80 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Strength grade SG= SEL("wood/EC"; SG; ) = S10
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
γM = 1,30
γS = 1,10
Nails: 38x100
Nail length lN = 7,50 cm
Nail diameter dN = 0,40 cm

Load:
Wind suction Fax,d = 0,55 kN CLD: short term
Roof shear Load scheme g Fla,d1 = 0,25 kN CLD: permanent
Roof shear Load scheme g+s Fla,d2 = 0,45 kN CLD: short term

Calculation:
Load-bearing capacity under shear load:
pre-drilled fh,k = 0.082 * ρk * (1-0.1*dN ) = 29,91 N/mm²
not pre-drilled fh,k = 0.082 * ρk * (10*dN )-0.3 = 20,56 N/mm²

Characteristic value for yield moment of square nails

My,d = 270 * (10*dN)2.6 / γS = 9022,50 Nmm
Characteristic value for yield moment of round nails
My,d = 180 * (10*dN)2.6 / γS = 6015,00 Nmm

decisive load scheme:

Class of load duration CLD= SEL("wood/kmod"; CLD; ) = permanent
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod1 = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,60
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod2 = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90

Fla,d1*kmod2/(Fla,d2*kmod1) = 0,83 < 1

⇒ Load scheme g+s decisive
Euro-Code 5 Folder: _EC5 Nail

fh,d = kmod2 / γM * fh,k =14,23

N/mm²
t2 = (lN - t1) = 4,70 mm
kt = MAX(t1;t2)/MIN(t1;t2) = 1,68
kM = 10*MIN(t1;t2) / √( My,d / ( fh,d *10* dN ) ) = 2,72
RD1 = 100* fh,d * t1 * dN = 1593,76 N
RD2 = 100* 0.5 * fh,d * t2 * dN = 1337,62 N
RD3 = 100*0.367*fh,d * t1 * dN * ( 2 * √(1 + 3 / kM²) - 1 ) = 801,95 N
RD4 = 155.6 * fh,d * t1 * dN / kM = 911,72 N
RD = 0.001 * MIN(RD1 ;RD2;RD3;RD4) = 0,80 kN

Fla,d2/RD = 0,56 < 1

Pullout capacity:
according to DIN 1052-2 Section 6.3.1
f1,d = 40*10 -6*ρk²*kmod2/γM = 4,00 N/mm²
f2,d = 600*10-6 *ρk²*kmod2/γγM =59,98
N/mm²
Rd1 = 100* f1,d * dN * t2 = 752,00 N
Rd2 = 100* f2,d * dN² = 959,68 N
Rd = MIN(Rd1 ;Rd2)/1000 = 0,75 kN

Fax,d/Rd = 0,73 < 1

Combined loads:

(Fax,d/Rd)²+(Fla,d2/RD)² = 0,85 < 1

Euro-Code 5 Folder: _EC5 Nail

Nail calculation for a shear joint (thick plate)

Wood-metal connection with plate t1 > 0.5 d

t1
t2

dN
System:
Sheet thickness t1 = 0,50 cm
Timber thickness t2 = 4,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = S13
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
γM = 1,30
γS = 1,10
Nails: 38x100
Nail length lN = 10,00 cm
Nail diameter dN = 0,38 cm

Calculation:
pre-drilled:
fh,k = 0.082 * ρk * (1-0.1*dN ) = 29,98 N/mm²
not pre-drilled::
fh,k = 0.082 * ρk * (10*dN)-0.3 = 20,88 N/mm²

Characteristic value for yield moment of square nails

My,d = 270 * (10*dN)2.6 / γS = 7896,02 Nmm
Characteristic value for yield moment of round nails
My,d = 180 * (10*dN)2.6 / γS = 5264,02 Nmm

fh,d = kmod / γM * fh,k = 14,46 N/mm²

RD1 = √(2+4*My,d/(fh,d*10 3*dN*t2²))-1)

100*1.1*fh,d*t2*dN*(√ = 1200 N
RD2 = √(20*My,d*fh,d*dN)
1.5*√ = 1141 N
RD3 = 100*fh,d *t2*dN = 2198 N

RD = 0.001 * MIN(RD1;RD2;RD3) = 1,14 kN

Euro-Code 5 Folder: _EC5 Nail

Nail calculation for a shear joint (thin plate) :

Wood-metal connection with plate t1 > 0.5 d

t1
t2

dN
System:
Sheet thickness t1 = 0,15 cm
Timber thickness t2 = 4,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = S13
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
γM = 1,30
γS = 1,10
Nails: 38x100
Nail length lN = 10,00 cm
Nail diameter dN = 0,38 cm
t1 / dN = 0,39 < 0,5
otherwise a different calculation applies
Calculation:
pre-drilled:
fh,k = 0.082 * ρk * (1-0.1*dN ) = 29,98 N/mm²
not pre-drilled::
fh,k = 0.082 * ρk * (10*dN)-0.3 = 20,88 N/mm²

Characteristic value for yield moment of square nails

My,d = 270 * (10*dN)2.6 / γS = 7896,02 Nmm
Characteristic value for yield moment of round nails
My,d = 180 * (10*dN)2.6 / γS = 5264,02 Nmm

fh,d = kmod / γM * fh,k = 14,46 N/mm²

RD1 = 100* ( √(2)-1)*fh,d * t2 * dN = 910,4 N

RD2 = √(20*My,d*fh,d*dN)
1.1*√ = 836,6 N

RD = 0.001 * MIN(RD1;RD2) = 0,84 kN

Euro-Code 5 Folder: _EC5 Nail

Nail calculation for a shear joint:

t1
t2

dN
System:
Timber thickness t1 = 4,00 cm
Timber thickness t2 = 4,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = S10
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
γM = 1,30
γS = 1,10
Nails: 38x100
Nail length lN = 10,00 cm
Nail diameter dN = 0,38 cm

Calculation:
pre-drilled:
fh,k = 0.08 * ρk * (1-0.1*dN) = 29,24 N/mm²
not pre-drilled::
fh,k = 0.082 * ρk * (10*dN)-0.3 = 20,88 N/mm²

Characteristic value for yield moment of square nails

My,d = 270 * (10*dN)2.6 / γS = 7896,02 Nmm
Characteristic value for yield moment of round nails
My,d = 180 * (dN*10)2.6 / γS = 5264,02 Nmm

fh,d = kmod / γM * fh,k = 14,46 N/mm²

kt = MAX(t1;t2)/MIN(t1;t2) = 1,00
kM = 10*MIN(t1 ;t2) / √( My,d / ( fh,d *10* dN ) ) = 4,09

RD1 = 100* fh,d * MIN(t1;t2) * dN = 2197,92 N

RD2 = √(3*kt²+2*kt+3)-kt-1)
100* 0.5*fh,d *MIN(t1;t2)*dN*(√ = 910,41 N
RD3 = 100* 0.367 * fh,d * MIN(t1;t2) * dN * ( 2 * √(1 + 3 / kM²) - 1 ) = 945,34 N
RD4 = 100* 1.556 * fh,d * MIN(t1;t2) * dN / kM = 836,18 N

RD = 0.001 * MIN(RD1;RD2;RD3;RD4) = 0,84 kN

Euro-Code 5 Folder: _EC5 Nail

Tie rod connection

h2
b t 1 t 2 t1
F
d
α

s h1

System:
Timber thickness t1 = 3,80 cm
Timber thickness t2 = 3,80 cm
Timber height h1 = 14,00 cm
Timber height h2 = 10,00 cm
Nail distance s = 3,50 cm
Nail distance b = 2,50 cm
Nail rows a = 3
Nail columns v = 3

Load:
Fd = 15 kN

Materials:
Construction material CM = SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD = SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80
Strength grade SG= SEL("wood/EC"; SG; ) = S10
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
γM = 1,30
γS = 1,10
Nails: 42x110
Nail length lN = 11,00 cm
Nail diameter dN = 0,42 cm

Calculation:
t3 = (lN - t1 - t2 ) = 3,40 cm
pre-drilled:
fh,k = 0.082 * ρk * (1-0.1*dN) = 29,85 N/mm²
not pre-drilled::Imposed Load:
fh,k = 0.082 * ρk * (10*dN)-0.3 = 20,26 N/mm²
Characteristic value for yield moment of square nails
My,d = 270 * (10*dN)2.6 / γS = 10242,81 Nmm
Characteristic value for yield moment of round nails
My,d = 180 * (10*dN)2.6 / γS = 6828,54 Nmm

fh,d = kmod / γM * fh,k = 12,47 N/mm²

kt = MAX(t1;t2)/MIN(t1;t2;t3) = 1,12
kM = 10*MIN(t1;t2;t3) / √( My,d / ( fh,d *10* dN ) ) = 2,98
Euro-Code 5 Folder: _EC5 Nail

RD1 = 100* fh,d * MIN(t1;t2;t3) * dN = 1780,72 N

RD2 = 100* 0.5 * fh,d * MIN(t1;t2;t3) * dN = 890,36 N
RD3 = 36.7 * fh,d * MIN(t1;t2;t3) * dN * ( 2 * √(1 + 3 / kM²) - 1 ) = 858,26 N
RD4 = 155.6 * fh,d * MIN(t1;t2;t3) * dN / kM = 929,80 N

RD = 0.001 * MIN(RD1;RD2;RD3;RD4) = 0,86 kN

required number of nails n = Fd/(2*RD) = 8,7

selected: 3x3=9 nails 42x110 DIN 1151

Euro-Code 5 Folder: _EC5 Nail

Wind brace connection:

a4 a2 a4
h

h
ß
F
t s b
System:
Beam thickness h = 16,00 cm
Driving depth t = 11,40 cm
Sheet thickness s = 0,60 cm
Angle of tensile force β = 45,00 °
Tensile force Fd = 23,54 kN

Materials: Post
Construction material CM = SEL("wood/kmod"; CM; ) = Glulam
Class of load duration CLD = SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = BS14h
ρk = TAB("wood/EC"; ρk; SG=SG) = 410,00 kg/m³
γM = 1,30
Screws: 4 Sr ∅ 12*120 DIN 571
Screw diameter d = 1,20 cm
fu,k = 300,00 N/mm²
γS = 1,10

Minimum distances Rod1 according to Tab.6.23:

a1 = 7* d = 8,40 cm
a2 = 4*d = 4,80 cm
a4,t = 3*d = 3,60 cm

Design value of bearing stress resistance:

fh,d = 0.082 * ρk * (1-0.1*d)*kmod /γM = 20,48 N/mm²
0.8/d = 0,67 < 1
⇒ calculation as for pins:
(0.6*(t+s)+4*d)/t = 1,05 > 1
⇒ def = 9 *d = 10,80 mm
My,d = 0.8*fu,k *def³/(6*γS) = 45807,71 Nmm
s/(d/2) = 1,00 < 1
⇒ thin plate
RD1 = ( √(2)-1)*fh,d*(t-1.5*d)*d*100 = 9772,52 N
RD2 = √(20* My,d * fh,d *d)
1.1 *√ = 5219,54 N

RD = 0.001 * MIN(RD1;RD2) = 5,22 kN

Euro-Code 5 Folder: _EC5 Nail

Design value for pullout:

f3,d = kmod/γM*(1.5+6*d)*√ √(ρk) = 121,96 N/mm²
Rd = f3,d * (0.6*(t+s)-d)*0.01 = 7,32 kN

Combined loads:

(Fd*SIN(β)/(4*Rd))²+(Fd*COS(β)/(4*RD))² = 0,96 < 1

Euro-Code 5 Folder: _EC5 Shrinkage

Shrinkage
Shrinkage and swelling of wood:

b
System:
The only appreciable stress is generated ⊥ to the fibre.
Width b = 32,00 cm
Height h = 12,00 cm
Coefficient for the degree of shrinkage and swelling β90 = 0,24 %

Strength grade SG= SEL("wood/EC"; SG; ) = S10

E90,mean = TAB("wood/EC"; E90.mean; SG=SG) = 370,00 N/mm²

Timber moisture before w1 = 10,00 %

Timber moisture after w2 = 15,00 %

Stress:
σc,90= E90,mean*β90*(w2-w1)/200 = 2,22 N/mm²

Total force:
Fσ,90 = σc,90 * b * h = 852,48 N
Euro-Code 5 Folder: _EC5 spec dowel

spec dowel
Diagonal tie connection with special dowels:

a2

h
a2

ß
a

h1

d1 d d1
System:
Distance a = 8,00 cm
Distance a2 = 9,00 cm
System angle β = 40,00 °
Beam height h = 30,00 cm
Beam thickness d = 16,00 cm
Strut height h1 = 18,00 cm
Strut thickness d1 = 8,00 cm
Number of rows nz = 2
Tensile force Ft,d = 165,00 kN

Construction material CM= SEL("wood/kmod"; CM; ) = solid wood

Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80

Transition to DIN 1052-2

Fd = Ft,d/1.35 = 122,22 kN
Selection of dowel: Split ring connector Type D according to DIN 1052-2, Table 4, 6 or 7 and
8
Dowel ∅ 65 - A with 6 * M12 + 12 * washers 58/6
Number of dowels selected per cut n = 6
per.Nc = 11,00 kN
Dowel height hc = 2,70 cm
Dowel diameter dc = 6,50 cm
Minimum wood dimensions of the beam from b/a = 110/40
Minimum wood dimensions of the strut from b/a = 100/40
Distance a1 = 14,00 cm
a2.1 = dc+hc/2 = 7,85 cm
a3,t = 14,00 cm
a4,strut = 10/2 = 5,00 cm
a4,beam = 11/2 = 5,50 cm
l= a2/SIN(β) = 14,00 cm
a1/l = 1,00 < 1
a2.1/a2 = 0,87 < 1
a4,beam/(h/2-a2 ) = 0,92 < 1
a4,strut/((h1-a)/2) = 1,00 < 1

nef = MIN( 2 + ( 1- n/nz / 20) * ( n/nz - 2 );6) = 2,85

Structural verification:
Fd/(2*nz*nef*per.Nc) = 0,97 < 1
Euro-Code 5 Folder: _EC5 spec dowel

Tensile splice with special dowels:

a1 a1
a a a

t1
t2
t1
System:
Distance a = 22,00 cm
Distance a1 = 6,00 cm
Thickness t1= 8,00 cm
Thickness t2 = 10,00 cm
Cut surfaces nz = 2
Tensile force Ft,d = 95,00 kN

Construction material CM= SEL("wood/kmod"; CM; ) = solid wood

Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80

Transition to DIN 1052-2

Fd = Ft,d/1.35 = 70,370 kN
Selection of dowel: Split ring connector Type A according to DIN 1052-2, Table 4, 6 or 7
Dowel ∅ 95 - A with 2 * M12 + 4 * washers 58/6
Number of dowels selected per cut n = 3
zul.Nc = 17,00 kN
bmin= 12,00 cm
amin= 22,00 cm
amin4= bmin/2 = 6,00 cm
a/amin = 1,00 < 1
a1/amin4 = 1,00 < 1

nef = MIN(2 + ( 1- n / 20) * ( n - 2 );6) = 2,85

Structural verification:
Fd/(nz*nef*zul.Nc ) = 0,73 < 1
Euro-Code 5 Folder: _EC5 Stress analysis

Stress analysis
Arched beam:

Fd

t α

d
System:
Cutting depth t = 12,00 cm
Support length d = 24,00 cm
Beam width b = 16,00 cm
Support angle α = 55,00 °

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = Glulam
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80
Strength grade SG= SEL("wood/EC"; SG; ) = BS14k
ρk = TAB("wood/EC"; ρk; SG=SG) = 410,00 kg/m³
Rectangular shear fc,0,k = TAB("wood/EC"; fc,0.k; SG=SG) = 27,50 N/mm²
Rectangular shear fc,90,k = TAB("wood/EC"; fc,90.k; SG=SG) = 5,50 N/mm²
γM = 1,30

Load:
Fd = 300,00 kN

Calculation:
fc,0,d = fc,0,k * kmod / γM = 16,92 N/mm²
fc,90,d = fc,90,k * kmod / γM = 3,38 N/mm²
Vertical supporting force:
β= 90 - α = 35,00 °
kc,ββ = 1 /((fc,0,d/fc,90,d)*(SIN( β))²+(COS(β
β ))²) = 0,431
Vd,max = 100* b*d*kc,β*fc,0,d*0.001 = 280,03 kN
Horizontal supporting force:
kc,αα= α))²+(COS(α
1 /((fc,0,d/fc,90,d)*(SIN(α α))²) = 0,271
Hd,max = 100* b*d*kc,α*fc,0,d*0.001 = 176,08 kN

Fd,h = Fd * COS(αα) = 172,07 kN

Fd,v = α)
Fd * SIN(α = 245,75 kN

Structural verifications:
Fd,h / Hd,max = 0,98 < 1
Fd,v / Vd,max = 0,88 < 1
Euro-Code 5 Folder: _EC5 Stress analysis

Centre purlin with 2-axis deflection:

Fd

b
Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = S10
fm,k = TAB("wood/EC"; fm.k; SG=SG) = 24,00 N/mm²
γM = 1,30

Load:
Loading moments as a result of the sloping load
My,d = 20,00 kNm
Mz,d = 5,00 kNm

Calculation:
fm,d = fm,k * kmod / γM = 16,62 N/mm²
Approximation W y,req = 1000*(My,d + Mz,d) / fm,d = 1504,21 cm³

selected: Cross section:

b= 16,00 cm
h= 24,00 cm
Wy = b * h² / 6 = 1536,00 cm³
Wz = h * b² / 6 = 1024,00 cm³

Structural verification:
km = 0,70 for rectangular cross section
otherwise km = 1.0
(My,d/W y*1000+km*Mz,d/W z*1000)/fm,d = 0,99 < 1
Euro-Code 5 Folder: _EC5 Stress analysis

Compression member connection

1 1

Fd
System:
Cross-sectional width in cut 1-1 b = 8,00 cm
Cross-sectional in cut 1-1 h = 16,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80
Strength grade SG= SEL("wood/EC"; SG; ) = S10
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
Rectangular shear fc,0,k = TAB("wood/EC"; fc,0.k; SG=SG) = 21,00 N/mm²
γM = 1,30

Load:
Fd = 112,00 kN

Calculation:
A= b*h = 128,00 cm²
fc,0.d = fc,0,k * kmod / γM = 12,92 N/mm²

σc,0.d = Fd * 10 / A = 8,75 N/mm²

Structural verification:
σc,0.d / fc,0.d = 0,68 < 1
Euro-Code 5 Folder: _EC5 Stress analysis

Structural analysis of deflection in two directions and pressure:

qd
qd α

Nd
h

l b
System:
Beam width b = 18,00 cm
Beam height h = 24,00 cm
Beam length l = 3,50 m
Load angle α = 75,00 °

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80
Strength grade SG= SEL("wood/EC"; SG; ) = S10
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
Rectangular shear fc,0,k = TAB("wood/EC"; fc,0.k; SG=SG) = 21,00 N/mm²
fm,k = TAB("wood/EC"; fm.k; SG=SG) = 24,00 N/mm²
γM = 1,30

Load:
qd = 13,21 kN/m
Nd = 94,50 kN

Calculation:
qd,y = qd * SIN(α) = 12,76 kN/m
qd,z = qd * COS(α) = 3,42 kN/m
My,d = qd,y * l² / 8 = 19,54 Nmm
Mz,d = qd,z * l² / 8 = 5,24 Nmm

fc,0,d = fc,0,k * kmod / γM = 12,92 N/mm²

fm,d = fm,k * kmod / γM = 14,77 N/mm²

Wy = b * h² / 6 = 1728,00 cm³
Wz = h * b² / 6 = 1296,00 cm³
A= b*h/1 = 432,00 cm²

km = 0,70 for rectangular cross section

otherwise km = 1.0
σc,0,d = 10 * Nd / A = 2,19 N/mm²

Structural verifications:
(σc,0,d/fc,0,d)²+(My,d/W y*1000+km*Mz,d/W z*1000)/fm,d = 0,99 < 1
(σc,0,d/fc,0,d)²+(km*My,d/W y*1000+Mz,d/W z*1000)/fm,d = 0,84 < 1
Euro-Code 5 Folder: _EC5 Stress analysis

Structural analysis of deflection in two directions and tension:

qd qd
α

Nd h

l b
System:
Beam width b = 18,00 cm
Beam height h = 24,00 cm
Beam length l = 3,50 m
Load angle α = 75,00 m

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80
Strength grade SG= SEL("wood/EC"; SG; ) = S10
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
ft,0,k = TAB("wood/EC"; ft.0.k; SG=SG) = 14,00 N/mm²
fm,k = TAB("wood/EC"; fm.k; SG=SG) = 24,00 N/mm²
γM = 1,30

Load:
qd = 10,21 kN/m
Nd = 47,50 kN

Calculation:
qd,y = qd * SIN(α) = 9,86 kN/m
qd,z = qd * COS(α) = 2,64 kN/m
My,d = qd,y * l² / 8 = 15,10 Nmm
Mz,d = qd,z * l² / 8 = 4,04 Nmm

ft,0,d = ft,0,k * kmod / γM = 8,62 N/mm²

fm,d = fm,k * kmod / γM = 14,77 N/mm²

Wy = b * h² / 6 = 1728,00 cm³
Wz = h * b² / 6 = 1296,00 cm³
A= b*h/1 = 432,00 cm²

km = 0,70 for rectangular cross section

otherwise km = 1.0

Structural verification:
10*Nd/A/ft,0,d+(My,d/W y*1000+km*Mz,d/W z*1000)/fm,d = 0,87 < 1
Euro-Code 5 Folder: _EC5 Stress analysis

End support of a laminated timber beam

Vd
d
System:
Width of laminated timber beam b = 18,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = Glulam
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80
Strength grade SG= SEL("wood/EC"; SG; ) = BS11
Rectangular shear fc,90,k = TAB("wood/EC"; fc,90.k; SG=SG) = 5,50 N/mm²
fm,k = TAB("wood/EC"; fm.k; SG=SG) = 24,00 N/mm²
γM = 1,30

Load:
Vd = 98,00 kN

Calculation:
fc,90,d = fc,90,k * kmod / γM = 3,38 N/mm²
Assumption: l > 150 mm otherwise kc,90 according to Tab. 5.1.5
⇒ kc,90 = 1,00
erf_A = Vd *10 / ( fc,90,d * kc,90) = 289,94 cm²

erf_d = erf_A / b = 16,11 cm

sel. d = 17,00 cm

Structural verification:
erf_d / d = 0,95 < 1
Euro-Code 5 Folder: _EC5 Stress analysis

Stress analysis of shear force and torsion:

Td
h

Vd b
System:
Support width b = 30,00 cm
Support depth h = 14,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = Glulam
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = BS16k
fv,g,k = TAB("wood/EC"; fv.k; SG=SG) = 2,70 N/mm²
γM = 1,30

Load:
relevant shear force Vd = 18,00 kN
relevant torsional moment Td = 1,74 kNm

Calculation:
A= b*h = 420,00 cm²
η= 1+0.6/(b/h) = 1,28
τtor,d = 3000*Td*η/(b*h²) ?? = 1,14 N/mm²
τv,d = 30*Vd/(2*A) = 0,64 N/mm²
fv,d = kmod * fv,g,k / γM = 1,87 N/mm²

Structural verifications:
Shear: τv,d / fv,d = 0,34 < 1
Torsion: τtor,d / fv,d = 0,61 < 1
Combination: τtor,d / fv,d+(τv,d / fv,d)² ?? = 0,73 < 1
Euro-Code 5 Folder: _EC5 Stress analysis

Stud:

a l l1 l
Fc,d Fc,d

h
System:
Length of wood beam projection a = 6,00 cm
Support length l = 6,00 cm
Support depth d = 12,00 cm
clear distance between studs l1 = 34,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = S10
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
fc,90,k = TAB("wood/EC"; fc.90.k; SG=SG) = 5,00 N/mm²
fm,k = TAB("wood/EC"; fm.k; SG=SG) = 24,00 N/mm²
γM = 1,30

Load:
Fc,d = 20,00 kN

Calculation:
fc,90.d = kmod * fc,90,k / γM = 3,46 N/mm²
for: 15 / l1 = 0,44 < 1
and: l / 15 = 0,40 < 1
⇒according to Tab. 5.1.5 kc,90 = 1+a*(150-10*l)/1700 = 1,32
A= l*d = 72,00 cm²
σc,90.d = 10 * Fc,d / A = 2,78 N/mm²

Structural verification:
σc,90.d / ( kc,90 * fc,90.d ) = 0,61 < 1
Euro-Code 5 Folder: _EC5 System

System

Acceptable force of a double shoulder joint:

β /2
β /2 N1 N2
1
α
t v1 t v2
h

l v1
l v2
System:
Beam height h = 22,00 cm
Beam width b = 14,00 cm
Length of wood beam projection lv1 = 22,00 cm
Length of wood beam projection lv2 = 30,00 cm
Shoulder depth tv1 = 3,50 cm
Shoulder depth tv = 4,50 cm
Angle α = 45,00 °
Comply with structural rules for shoulder depth and length of wood beam projection.
Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80
Strength grade SG= SEL("wood/EC"; SG; ) = S13
ρk = TAB("wood/EC"; ρk; SG=SG) = 380,00 kg/m³
fv,k = TAB("wood/EC"; fv.k; SG=SG) = 2,50 N/mm²
fc,0,k = TAB("wood/EC"; fc,0.k; SG=SG) = 23,00 N/mm²
fc,90,k = TAB("wood/EC"; fc.90.k; SG=SG) = 5,00 N/mm²
γM = 1,30

Load:
Nd = 86,00 kN

Calculation:
fv,d = kmod * fv,k / γM = 1,54 N/mm²
fc,0,d = kmod * fc,0,k / γM = 14,15 N/mm²
fc,90,d = kmod * fc,90,k / γM = 3,08 N/mm²

with approximation:
kF = 1 / ((fc,0,d/fc,90,d)*(SIN(α))²*COS(α)+COS(α)³) = 0,51
ks = 4 / ((fc,0,d/fc,90,d)*(SIN(α))²+(COS(α))²+2*COS(α)+1) = 0,77

lv1,min = 10*Nd/2 *COS(α)/(b * fv,d) = 14,10 cm

lv1,1 = 8* tv = 36,00 cm
lv1,min / lv1,1 = 0,39 < 1
lv1,min / lv1 = 0,64 < 1
Euro-Code 5 Folder: _EC5 System

lv2,min = 10*Nd/2 *COS(α)/(b * fv,d) = 14,10 cm

lv2,1 = 8* tv = 36,00 cm
lv2,min / lv2,1 = 0,39 < 1
lv2,min / lv2 = 0,47 < 1

RF,d = b * tv * fc,0,d * kF / 10 = 45,46 kN

RS,d = b * tv * fc,0,d * ks / 10 = 68,64 kN

detailed calculation:
kc,α = 1 / ((fc,0,d/fc,90,d) * (SIN(α))² + (COS(α))² ) = 0,358
RF,d = kc,α * fc,0,d * b * tv / COS(α)/10 = 45,13 kN

kc,α = 1 / ((fc,0,d/fc,90,d) * (SIN(α/2))² + (COS(α/2))² ) = 0,655

fc,0,5α,d = kc,α * fc,0,d = 9,27 N/mm²

RS,d = fc,0,5α,d * α * b / ( (COS(α/2))²)/100 = 68,42 kN

Structural verification:
Nd / (RF,d+RS,d) = 0,76 < 1
Euro-Code 5 Folder: _EC5 System

Acceptable force of a face staggered joint:

N
β /2
β /2 α
N1

h tv

lv
System:
Beam height h = 22,00 cm
Beam width b = 14,00 cm
Length of wood beam projection lv = 22,00 cm
Shoulder depth tv = 4,50 cm
Angle α = 45,00 °

Comply with structural rules for shoulder depth and length of wood beam projection.
Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80
Strength grade SG= SEL("wood/EC"; SG; ) = S13
fv,k = TAB("wood/EC"; fv.k; SG=SG) = 2,50 N/mm²
fc,0,k = TAB("wood/EC"; fc.0.k; SG=SG) = 23,00 N/mm²
fc,90,k = TAB("wood/EC"; fc.90.k; SG=SG) = 5,00 N/mm²
γM = 1,30

Load:
Nd = 63,00 kN

Calculation:
fv,d = kmod * fv,k / γM = 1,54 N/mm²
fc,0,d = kmod * fc,0,k / γM = 14,15 N/mm²
fc,90,d = kmod * fc,90,k / γM = 3,08 N/mm²

with approximation:
ks = 4 / ((fc,0,d/fc,90,d)*(SIN(α))²+(COS(α))²+2*COS(α)+1) = 0,77
lv,min = 10*Nd *COS(α)/(b * fv,d) = 20,66 cm
lv,1 = 8* tv = 36,00 cm
lv,min / lv,1 = 0,57 < 1
lv,min / lv = 0,94 < 1

RS,d = b * tv * fc,0,d * ks / 10 = 68,64 kN

detailed calculation:
kc,α = 1 / ((fc,0,d/fc,90,d) * (SIN(α/2))² + (COS(α/2))² ) = 0,655
fc,0,5α,d = kc,α * fc,0,d = 9,27 N/mm²

RS,d = fc,0,5α,d * α * b / ( (COS(α/2))²)/100 = 68,42 kN

Structural verification:
Nd / RS,d = 0,92 < 1
Euro-Code 5 Folder: _EC5 System

Acceptable force of a heel staggered joint:

N2
α

h tv

lv
System:
Beam height h = 22,00 cm
Beam width b = 14,00 cm
Length of wood beam projection lv = 22,00 cm
Shoulder depth tv = 4,50 cm
Angle α = 45,00 °
Comply with structural rules for shoulder depth and length of wood beam projection.
Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80
Strength grade SG= SEL("wood/EC"; SG; ) = S13
fv,k = TAB("wood/EC"; fv.k; SG=SG) = 2,50 N/mm²
fc,0,k = TAB("wood/EC"; fc.0.k; SG=SG) = 23,00 N/mm²
fc,90,k = TAB("wood/EC"; fc.90.k; SG=SG) = 5,00 N/mm²
γM = 1,30

Load:
Nd = 43,00 kN

Calculation:
fv,d = kmod * fv,k / γM = 1,54 N/mm²
fc,0,d = kmod * fc,0,k / γM = 14,15 N/mm²
fc,90,d = kmod * fc,90,k / γM = 3,08 N/mm²

with approximation:
kF = 1 / ((fc,0,d/fc,90,d)*(SIN(α))²*COS(α)+COS(α)³) = 0,51
lv,min = 10*Nd *COS(α)/(b * fv,d) = 14,10 cm
lv,1 = 8* tv = 36,00 cm
lv,min / lv,1 = 0,39 < 1

lv,min / lv = 0,64 < 1

RF,d = b * tv * fc,0,d * kF / 10 = 45,46 kN

detailed calculation:
kc,α = 1 / ((fc,0,d/fc,90,d) * (SIN(α))² + (COS(α))² ) = 0,358
Rd = kc,α * fc,0,d * b * tv / COS(α)/10 = 45,13 kN

Structural verification:
N d / Rd = 0,95 < 1
Euro-Code 5 Folder: _EC5 System

Connection of a binding joist to a support:

s1 s

h1
c

h1
c

h1
Schlitzblech

b1b1
System:
Centroidal distance s = 16,00 cm
Centroidal distance s1 = 5,00 cm
Dowel distance b1 = 5,00 cm
Dowel distance h1 = 8,00 cm
Thickness of slotted plate m = 1,00 cm
Width of binding joist b = 16,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = Glulam
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80
Strength grade SG= SEL("wood/EC"; SG; ) = BS11
ρk = TAB("wood/EC"; ρk; SG=SG) = 410,00 kg/m³
γM = 1,30
γS = 1,10

Load at tie point:

Vc,d = 34,60 kN
Mc,d = 0,01*Vc,d * (s1+s) = 7,27 kNm

Geometric values:
r1 = √(b1²+(h1/2)²) = 6,4 cm
r2 = √(b1²+(h1*1,5)²) = 13,0 cm
αM = ATAN(b1/(h1*1,5)) = 22,62 °

Calculation:
Drift pin forces:
as the result of shear force FV,d = Vc,d / 8 = 4,33 kN
as the result of moment FM,d = 100*Mc,d*r2/(4*(r1²+r2²)) = 11,25 kN

Components:
FM2,d,V = FM,d * SIN(αM) = 4,33 kN
FM2,d,H = FM,d * COS(αM) = 10,38 kN

Fd,max = √((FM2,d,V+FV,d)²+FM2,d,H²) = 13,52 kN

α= ATAN((FM2,d,V+FV,d)/FM2,d,H) = 39,84 °
Euro-Code 5 Folder: _EC5 System

selected: Dowel DPin ∅ 16mm

d= 1,60 cm
fu,k = 360,0 N/mm²

My,d = 0,8*fu,k*d³/(6*γS) = 178,7 kNmm

fh,0,d = 0,082 * ρk * (1-0,1*d)*kmod/γM = 17,38 N/mm²
k90 = 1,35 + 0,15 * d = 1,59
fh,1,d = fh,0,d/(k90*(SIN(α))²+(COS(α))²) = 13,99 N/mm²
t1 = (b-m)/2 = 7,50 cm

RD1 = 100*fh,1,d * t1 * d = 16788 N

RD2 = 110*fh,1,d* t1*d*(√√(2+4*My,d/(fh,1,d*d* t1²))-1) = 11124,52 N
RD3 = √(2*My,d*fh,1,d*d)
150*√ = 13416,44 N

RD = 0,001 * MIN(RD1;RD2;RD3) = 11,12 kN

Structural verification:
Fd,max / ( 2 * RD) = 0,61 < 1
Euro-Code 5 Folder: _EC5 System

Joist hanger:
B'

F1,d
hN
h H
H'

s
h1
h1

System:
Height of principal beam hH = 32,00 cm
Height of short-tie beam hN = 24,00 cm
Joist hanger distance h1 = 6,00 cm
Nail distance s = 15,20 cm
Load:
F1,d = 12,00 kN

Materials:
Principal beam: 120mm*380mm
Construction material CM= SEL("wood/kmod"; CM; ) = Glulam
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80
Strength grade SG= SEL("wood/EC"; SG; ) = BS11
ρk = TAB("wood/EC"; ρk; SG=SG) = 410,00 kg/m³
ft,90,g,k = TAB("wood/EC"; ft.90.k; SG=SG) = 0,45 N/mm²
γM = 1,30
Short-tie beam: 140mm*240mm NH S 10

selected: GH joist hanger 04 140*160 with 14 RNä 4.0x60

Horizontal distance between centroidal axis of nails B' = 18,60 cm
Vertical Nail distance H' = 12,00 cm
Joist hanger thickness db = 0,20 cm
Nail diameter dN = 0,40 cm
Nail length lN = 6,00 cm
Number of nails nN = 14
According to manufacturer's table R0.d = 16,40 kN
f= 1/(1-0.93*(s+h1)/hH) = 2,61
tef = MIN(12*dN ;lN-db) = 4,80 cm
ft,90.d = kmod /γM*ft,90,g,k = 0,28 N/mm²

Rt,90.d = 0.0055*f*(10*tef )0.8*(10*hH+4*√

√(100*B'*H'))0.8*ft,90.d = 20,86 kN
Rd = MIN(Rt,90.d ;R0.d) = 16,40 kN

Structural verification:
F1,d/Rd = 0,73 < 1
Euro-Code 5 Folder: _EC5 System

Notch joist

qd

he h
ε

d Fd
x l
System:
Cross-sectional width b = 14,00 cm
Support length d = 18,00 cm
Cross sectional height h = 88,00 cm
Cross sectional height at footing he = 78,00 cm
Notch distance x = 13,00 cm
Notch length l = 20,00 cm
Beam length l1 = 850,00 cm

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = Glulam
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 2
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = BS14k
ρk = TAB("wood/EC"; ρk; SG=SG) = 410,00 kg/m³
fv,g,k = TAB("wood/EC"; fv.k; SG=SG) = 2,70 N/mm²
fc,90,k = TAB("wood/EC"; fc.90.k; SG=SG) = 5,50 N/mm²
for solid wood kn = IF(CM="Glulam";6,5;5) = 6,50
γM = 1,30

Load:
Supporting force Fd = 95,30 kN
Line load qd = 2,80 kN/m

Calculation for splintering:

Maximum design shear force:
for uniformly distributed load kr = 1-2*h/l1 = 0,79
Fd,b = k r * Fd = 75,29 kN
for individual load kF =(0.5-h/l1)*e/h with e the distance of the force from footing k
and Fb,d =kF * Fd
fv,d = kmod * fv,g,k / γM = 1,87 N/mm²
l=MAX(l;0.01)
ε= ATAN((h - he )/l) = 26,57 °
kε = 1+(1.1*(TAN(ε ))-1.5)/(√
√(h*10)) = 1,10
α= he / h = 0,89
k90 = √(h*10)*(√
kn/(√ √(1/α−α²)))
√(α*(1-α))+0,8*(x/h)*√ = 0,575
kv = MIN(1;k90*kε) = 0,63

Structural verification:
(1500*Fd,b/(100*b*he))/(kv*fv,d) = 0,88 < 1
Euro-Code 5 Folder: _EC5 System

Calculation of bearing pressure:

fc,90,d = fc,90,k * kmod / γM = 3,81 N/mm²
Assumption: Support length t > 150 mm otherwise kc,90 according to Tab. 5.1.5
⇒ kc,90 = 1,0
erf_A = Fd *10 / ( fc,90,d * kc,90) = 250,13 N/mm²
erf_d = erf_A / b = 17,87 mm

Structural verification:
erf_d / d = 0,99 < 1
Euro-Code 5 Folder: _EC5 System

Rail post:
Hd

h1
h3h2
h

System:
Post distance l = 1,70 m
Rail height h1 = 100,00 cm
Dowel distance h2 = 5,00 cm
Dowel distance h3 = 23,00 cm
γQ = 1,50

Load: according to DIN 1055-3

H' = 1,00 kN/m
Hd = γQ * l * H' = 2,55 kN/m

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = Glulam
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = short term
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,90
Strength grade SG= SEL("wood/EC"; SG; ) = BS11
ft,90.g,k = TAB("wood/EC"; ft,90.k; SG=SG) = 0,45 N/mm²
γM = 1,30

Calculation:
Mc,d = 0,01*Hd*(h1+h2+h3/2) = 2,97 kNm
F1,d = Hd/2+Mc,d/(h3*0,01) = 14,19 kN
F2,d = Mc,d/(h3*0,01)-Hd/2 = 11,64 kN

Transition to DIN 1052-2:

F1 = MAX(F1,d;F2,d)/1,4 = 10,14 kN

selected: 2 DPin ∅ 85 - D

mit zul_Nc = 14,50 kN

Structural verification:
F1/ zul_Nc = 0,70 < 1
Euro-Code 5 Folder: _EC5 System

Suspender:

t t

Ft,d
α
Fc,d Fc,d
System:
Height of diagonal brace h = 14,00 cm
Width of diagonal brace b = 14,00 cm
Height of suspender hH = 18,00 cm
Width of suspender bH = 14,00 cm
Shoulder depth tv = 3,00 cm
Diameter of pin d = 1,20 cm
Angle α = 45,00 °
Comply with structural rules for shoulder depth and length of wood beam projection.

Materials:
Construction material CM= SEL("wood/kmod"; CM; ) = solid wood
Class of load duration CLD= SEL("wood/kmod"; CLD; ) = normal
Utility class UC= SEL("wood/kmod"; UC; ) = 1
⇒ kmod = TAB("wood/kmod"; kmod; CM=CM; CLD=CLD; UC=UC) = 0,80
Strength grade SG= SEL("wood/EC"; SG; ) = S13
ft,0,k = TAB("wood/EC"; ft.0.k; SG=SG) = 18,00 N/mm²
γM = 1,30

Calculation:
Aw,v = 2*b*tv = 84,00 cm²
Aw,bo = (d+1)*(hH-2*tv) = 26,40 cm²
Aw = bH*hH-MAX(Aw,bo;Aw,v) = 168,00 cm²
ft,0,d = kmod * ft,0,k / γM = 11,08 N/mm²
Maximum force to be accepted by the construction:
Ft,d,max = ft,0,d*Aw*0,1 = 186,14 kN