glulam curved beam design to ec5

P1: PAB/RPW P2: PAB

BLUK117-Porteous October 7, 2007 9:9

242 Structural Timber Design to Eurocode 5

Instantaneous deflection at mid-span of thebeam, uinst

uinst = uinst,b,dl + uinst,b,Q + uinst,s,dl + uinst,s,Q uinst = 18.77 mm

Limitation on deflection at theinstantaneous state – span/300, winst

Table 4.6 (EC5, Table 7.2)

winst = ℓ

300winst = 30 mm

i.e. OKFinal deflection due to permanentactions, ufin.G

(equation (4.41); EC5, equation (2.3))

ufin.G = (uinst,b,dl + uinst,s,dl) · (1 + kdef) ufin.G = 14.82 mm

Final deflection due to variable andquasi-permanent actions, ufin,Q


ufin.Q = (uinst,b,Q + uinst,s,Q) · (1 + ψ2 · kdef) ufin.Q = 9.51 mm

Final deflection due to permanent andquasi-permanent actions, unet.fin

unet.fin = ufin.G + ufin.Q unet.fin = 24.33 mm

Adopt EC5 limitation on deflection –use span/250, wnet.fin

(Table 4.6 (EC5, Table 7.2))

wnet.fin = ℓ

250wnet.fin = 36 mm

i.e. OK

Example 6.7.4 A curved glulam beam with a constant cross-section, 175 mm wide, hav-ing a profile as shown in Figure E6.7.4, and with an effective span of 18.0 m is to beused in the construction of the roof for a school hall. The beam will be laterally supportedalong the full length of the compression edge and there will be no load sharing betweenglulam beams. It is strength class GL 32h in accordance with BE EN 1194:1999, madefrom 30-mm-thick laminations, and will function in service class 2 conditions. For the de-sign loading condition given below, which includes an allowance for the self-weight of thebeam, ignoring SLS requirements, confirm that the beam will comply with the design rulesin EC5. The design loading arises from a combination of permanent and short-term variableloading.

1. Glulam beam geometric propertiesBreadth of each beam, b b = 175 mm

Depth of the beam at the apex, hap hap = 960 mm

Effective span of beam, ℓ ℓ = 18.0 m

Bearing length at each end of the beam,ℓb

ℓb = 200 mm

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Design of Glued Laminated Members 243

Straight section

t = 30 mm

hap = 960 mm

rin = 18 m

β = 9°

Effective span =18 m

Bearing length at each end = 200 mm

Design action = 6 kN/m

Fig. E6.7.4

Angle of slope of the straight memberlengths of the beam, β

β = 9.deg

Inner radius of the curved beam section,rin

rin = 18 m

Thickness of laminations in the beam, t t = 30 mm

Section modulus of the beam about they–y axis at the apex, Wy

Wy =b · hap2

6Wy = 2.69 × 107 mm3

2. Glulam propertiesTable 6.2, homogeneous grade GL 32h

Characteristic bending strength, fm.g.k fm.g.k = 32 N/mm2

Characteristic shear strength, fv.g.k fv.g.k = 3.8 N/mm2

Characteristic bearing strength, fc.90.g.k fc.90.g.k = 3.3 N/mm2

Characteristic compressive strength,fc.0.g.k

fc.0.g.k = 29 N/mm2

Characteristic tensile strengthperpendicular to the grain, ft,90.g.k

ft,90.g.k = 0.5 N/mm2

3. Partial safety factorsTable 2.8 (UKNA to BS EN 1990:2002, Table NA.A1.2(B))) for the ULS

Permanent actions, γG γG = 1.35

Variable actions, γQ γQ = 1.5

Table 2.6 (UKNA to EC5, Table NA.3):

Material factor for glulam at ULS, γM γM = 1.25

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4. ActionsDesign action on the beam, qd qd = 6 kN/m

5. Modification factorsFactor for short-duration loading andservice class 2, kmod.short

(Table 2.4 (EC5, Table 3.1))

kmod.short = 0.9

h = 960 mm

Size factor based on maximum depth= 960 mm, kh

(Table 2.11 (EC5, 3.3))

kh = 1

Bearing factor, kc.90 kc.90 = 1(equation (4.22) (EC5, 6.1.5,(1))

Lateral stability factor for beam, kcrit

(4.5.1.2 (EC5, 6.3.3))kcrit = 1

Factor applied to obtain bending stressin the apex zone, kcurve,b


r = rin = 0.5 · hap

kcurve,b = 1 + 0.35 ·!

hap

r

"+ 0.6 ·

!hap

r

"2

kcurve,b = 1.02

Factor applied to obtain the tensilestress perpendicular to the grain in theapex zone, kcurve,t

kcurve,t = 0.25!

hap

r

"kcurve,t = 0.01

(equation (6.25); EC5, equation (6.54))Stressed volume in the apex zone, V(Table 6.6) (Mathcad adjusted to makeit dimensionally correct)

V = β · π · b180.deg

· (h2ap + 2 · rin · hap) V = 0.98 m3

Approximate volume of the beam, Vc

Vc = V + 2 · b · hap

12

−!

rin +hap

2

"· sin(β)

cos(β)Vc = 3.05 m3

Ratio V/Vc

Must be less than 0.67(equation (6.27) (EC5 6.4.3(6))

VVc

= 0.32 less than 0.67 therefore OK

Stress distribution factor for the apexzone, kdis


kdis = 1.4

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Volume factor for the apex zone, kvol

(equation (6.27); EC5, equation (6.51))(Mathcad adjusted to make itdimensionally correct)

kvol =!

0.001 · m3

V

"0.2

kvol = 0.4

Reduction factor due to the curvatureof the laminations, kr

(equation (6.28) and (6.26); EC5,equation (6.49))

rin

t= 600 As the ratio is

greater than 240,kr = 1

Load sharing factor, ksys ksys = 1.0

6. Bending and radial strength in apex zoneAs the beam is of constant cross-section, the critical condition will occur at the apex,under the action of the design loading:

(a) Bending strength condition

Design bending moment, Md Md = qd · ℓ2

8Md = 2.43 × 108 N mm

Design bending stress, σm,0,d

(equation (6.23); EC5, equation (6.42))σm.0.d = kcurve.b · Md

Wyσm.0.d = 9.22 N/mm2

Design bending strength, fm,g,d

fm.g.d =kmod.short · ksys · kh · fm.g.k

γMfm.g.d = 23.04 N/mm2

Design bending strength taking lateraltorsional buckling and laminate effectinto account, fm,r,y.d,

fm.r.y.d = kcrit · kr · fm.g.d fm.r.y.d = 23.04 N/mm2

Bending strength of the glulam beam in apexzone is satisfactory

(b) Radial strength condition(6.4.2.2.2 (EC5. 6.4.3(6))

Design radial tensile stress, σt.90.d

(equation (6.25); EC5, equation(6.54))

σt.90.d = kcurve.t · Md

Wyσt.90.d = 0.12 N/mm2

Design tensile strength perpendicu-lar to the grain, ft.90.d

ft.90.d =kmod.short · ksys · ft.90.g.k

γMft.90.d = 0.36 N/mm2

Design tensile strength perpendicularto the grain taking stress distributionand volume factors into account,ft,r,y.d

(equation (6.27); EC5, equation(6.50))

ft.r.y.d = kdis · kvol · ft.90.d ft.r.y.d = 0.2 N/mm2

Tensile strength of the glulam beam in apexzone is greater than the radial tensile stressand is satisfactory

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7. Shear strengthThe design shear condition due to the design loading:

Design value of the end shearforce perpendicular to the grain –ignoring the reduction permittedin EC5 – see 4.5.2.1, Vd

Vd = qd · ℓ · cos(β)2

Vd = 53.34 kN

Design shear stress, τv.d (adoptingthe full depth of the beam),Equation (6.3)

τv.d = 32

· Vd

b · hapτv.d = 0.48 N/mm2

Design shear strength, fv.g.d

fv.g.d =kmod.short · ksys · fv.g.k

γMfv.g.d = 2.74 N/mm2

Shear strength is satisfactory

8. Bearing strengthThe design bearing condition due to the design loading:

Design value of the end reaction,Reacd

Reacd = Vd Reacd = 53.34 kN

Design bearing stress, σc.β.d

(equation (4.22))σc.β.d = Reacd

b · ℓbσc.β.d = 1.52 N/mm2

Design bearing strength, fc.90.g.d

fc.90.g.d =kmod.short · ksys · fc.90.g.k

γMfc.90.g.d = 2.38 N/mm2

Design compression strengthparallel to the grain, fc.0.g.d

fc.0.g.d =kmod.short · ksys · fc.0.g.k

γMfc.0.g.d = 20.88 N/mm2

Design compression strength at an angle β to thegrain, fc.β.g.d


fc.β.g.d =fc.0.g.d

fc.0.g.d

kc.90 · fc.90.g.d· sin (90.deg − β)2 + cos (90.deg − β)2

fc.β.g.d = 2.43 N/mm2

The bearing stress is less than the bearingstrength; therefore it is acceptable.

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9. Combined shear and tension perpendicular to the grainAs the shear stress at the apex will be zero for the design loading condition, the need to checkthe combined shear and tension condition given in equation (6.30) can be ignored. At anyother positon along the beam the combined stress condition will always be less than 1.

glulam curved beam design to ec5

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