EXPAN Portal Frames Design Guide

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Design Guide Workshops- 2013

Dr Felix Scheibmair, The University of Auckland

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Presentation outline

• Long span roof guide –an introduction

• Timber rivet guide –Introduction and overview

• Designing with timber rivets– Design example

• Quick Connect

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Long span roof guide

• Covers LVL mono and double pitch portal frames and trusses

• Gives preliminary sizing tables to ease design

• Contains examples and design flowcharts

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Long span roof guide content

• Introduction

• Scope

• Mono-pitch portal frame solutions

• Double pitch portal frame solutions

• Truss solutions

• Summary of trends and findings

• Example wind load calculation

• Supporting materials

• Wind pressures and ratios, Flange width selection to minimise waste, box section cross-sectional areas and moment of inertia

• Portal frame design example

• Truss design example

• References

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Scope of guide – Portal frames

• Portals with propped apex

• Spans of 20m – 50m

• Columns heights of 5m, 6m, 7m

• Pinned column bases

• Gable portal frames

• Spans of 15m – 45m

• Column heights of 5m, 6m, 7m


• Cross section and material

• All members box sections of Grade 11 LVL

• Assumptions

• 8m Frame spacing

• 5° roof pitch

• Importance level 2

• Design location: Auckland

• Design life: 50 years

• Pre-cambering equal to self-weight deflection

• Fixed rafter to Column and Apex connections

• Same cross-section throughout for whole portal frame

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Scope of guide – Truss

• Design parameters

• Spans of 50m, 60m, 70m

• Columns clear height of 7m


• Cross section and material

• All members solid sections of Grade 11 LVL

• Assumptions

• Pratt truss

• 8m truss spacing

• Importance level 2

• Design location: Auckland

• Design life: 50 years

• 5° roof pitch

• Pre-cambering equal to self-weight deflection

• Simple pinned connections throughout truss

• All connection requirements to be met using timber rivets

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Propped portal frame – Sample

• This example – Gravity and wind, terrain category 1

• 45mm web thicknesses

Frame Mid‐Column Frame Mid‐Column Frame Mid‐Column Frame Mid‐Column Frame

web size (d x tw) 600 x 45 200 x 45 900 x 45 240 x 45 1000 x 45 300 x 45 1200 x 45 400 x 45 1200 x 45

flange size (bf x tf) 240 x 90 240 x 63 240 x 90 240 x 90 200 x 90 200 x 90 400 x 90 400 x 45 600 x 90

Volume of LVL(m3)



Volume of LVL(m3)



Volume of LVL(m3) 21.

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6.19 9.15 12.77 15.89

20.4

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5.44 8.25 11.53 15.46 21.0

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5.20 7.94 9.33 15.03

Column

Height (m)Box Beam

Portal Frame Span (m)

20 25 30 35 40

• Notes

The volume of LVL shown is the volume required for constructing two spans of rafters and three columns, which gives two spans of mono-pitch portal frames

The units for web size and flange size are millimeters (mm)

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Truss– Sample solution

• This example – 60m span, terrain category 1

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Timber rivet design guide

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Timber rivet design guide

• Timber rivets introduction– What are they and why you should consider designing with them

• Design process – Flowcharts for the design of rivets applied parallel, perpendicular or at an angle to the grain

• Detailing of rivet connections and rivet plates– General notes and important considerations

• Structural design approach– Details on lateral resistance, withdrawal resistance and joint deflection

• Design examples– Examples of truss, base, moment, hanger and shear wall connections

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Timber Rivets

• High load transfer capacity

• Stiff connections due to tight-fit dowel action – no initial slip

• High ductility and the ability to dissipate dynamic loads

• Easy to install and inspect in the field

• No pre-drilling required and no reduction of timber cross section

• Cost effective fastener

• Standard lengths of 40, 65 and 90mm

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Applications

• Connections for long span truss

• Long span beam splices

• Beam to column and column to foundation connections

• Moment connections

• Energy dissipating connections

• Shear wall hold-downs

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Mechanisms of connection failure

• Ductile failure mode– Rivet yielding, embedment failure in timber and localized crushing

Mode Im Mode IIIm Mode IV

• Brittle failure mode– Rupture of timber member

• Mixed failure mode– Wood fails before final yielding of the rivet

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Rivet connection design flowcharts

• Load applied parallel to grain

• Load applied perpendicular to grain

• Load applied at angle to grain

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• Determine the design lateral load parallel to the grain

• Set the connection parameters:

Example: Load applied parallel to grain

Nl*

Grain

• Plate thickness (tp)• Rivet length (Lr)• Number of rivet rows (nR)• Number of rivets per row (nC)• Spacing along the grain (a1)• Spacing across the grain (a2)• Loaded end distance (a3t)• Unloaded edge distance (a4c)

Ensure minimum requirements are satisfied such as:

Number of rivets per row nC Loaded end distancea3t [mm]

1-6 75

7-10 100

11-12 125

13-14 150

15-16 175

≥17 200

• Rivet attributes (ultimate tensile strength:1000MPa+, finish: galvanized, dimensions: shank and head characteristics

• Plate dimensions and clearances• Minimum edge and end distances

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• Calculate the design rivet yielding resistance parallel to grain, rQry,l

• Based on two possible ductile failure modes of rivets; (a) single plastic hinge formation, (b) two plastic hinges

• Rivet yielding resistance; ϕrQry,l = ϕrQr,l in which ƒh,0 and Mr,l equal to ƒhy,0 and Mr,y

ϕrQr,l = ϕr k1 k12 kf np nR nC min (Prl,a, Prl,b)

• kf = modification factor for joint position effect (0.9 for edge grain LVL, 1.0 for face grain LVL all allfaces glulam and sawn timber)

• np = number of plates (allows for single sided joints)

Example: Load applied parallel to grain cont’d (a)

(b)

Lp

Lp

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• Calculate the design rivet yielding resistance parallel to grain, rQry,l

• ϕrQr,l = ϕr k1 k12 kf np nR nC min (Prl,a, Prl,b)

• Prl,a = characteristic strength parallel to grain for rivet failure mode (a)

• Prl,b = characteristic strength parallel to grain for rivet failure mode (b)

Example: Load applied parallel to grain cont’d (a)

(b)

Lp

Lp

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• Calculate the design wood block tear-out resistance parallel-to-grain, ϕwQwe,l , based on the wood effective thickness, tefe,l, corresponding to the rivet elastic deformation

• ϕwQw,l =ϕw k1 k12 kf np min (Pw,h, Pw,b, Pw,l)

• ϕw = strength reduction factor for wood failure = 0.7

• Pw,h / Pw,b / Pw,l = characteristic resistance for failure of head / bottom/ side lateral

tensile plane/s, kN

Example: Load applied parallel to grain cont’d

(a) (b) (c)

Qw,l

Grain

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• Is the governing failure mode of the wood block tear-out either mode (b) or (c)?

• If yes, recalculate the wood resistance for the remaining planes to determine whether residual planes can resist higher load by allocating no value to the failed plane(s)

• If no step forward in flowchart

• Is the contribution of the lateral shear planes over 30% of the wood capacity

• If yes, adjustment connection attributes as needed

• If no step forward in flowchart


(b) (c)

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• Is ϕwQwe,l < ϕrQry,l

• If yes,

• Joint design lateral resistance Qs,l = ϕwQwe,l (Brittle failure mode)

• And if, N* ≤ Qs,l then design completed, else adjust as needed and reiterate

• If no, calculate wood block tearoutresistance parallel to grain wQwy,l to determine if mixed mode or ductile mode governs

• Again, recalculate if mode b or c failure and check lateral shear plane contribution below 30%


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• Is ϕwQwy,l < ϕrQry,l

• If yes,

• Joint design lateral resistance Qs,l = ϕrQry,l (Mixed failure mode)


• If no, calculate rivet ultimate resistance parallel to grain rQru,l to determine if mixed mode or ductile mode governs

• Again, recalculate if mode b or c failure and check lateral shear plane contribution below 30%


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• Is ϕwQwy,l < ϕrQru,l

• If yes,

• Joint design lateral resistance Qs,l = ϕwQwy,l (Mixed failure mode)


• If no, joint design lateral resistance Qs,l = ϕrQru,l (Ductile failure mode)



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Detailing of rivet connections

• Rivets must meet certain criteria for tensile strength, hardness, finish and dimensions

• Plates shall have a cross-section adequate for resisting applied loads and must be more than 3.2mm thick. Finish of plates depends on service conditions

• Rivets must be placed with their major axis aligned with the timber grain and during placement of rivets must commence from the outside inwards in a spiral pattern

• The design guide prescribes minimum spacing requirements for rivet to edge/ end of plate and rivet to edge/ end of timber

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Design examples covered

• Truss connection– Design of LVL truss bottom chord, post and strut connection

• Base connection– Glulam column to foundation connection

• Hanger connection– Connection of a glulam secondary beam to a primary beam

• Shear wall connections– Hold down and floor to wall connection

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Appendices

• Technical Information and specifications

Reference capacity tables

Adjustment factors

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Design Example – Base connection

• Column to foundation connection example, pin joint

• GL10 Radiata Pine glulam

• Load factors taken into account, load combination [0.9G, Wu]

• Two strength limit states of interest: Rivet strength and timber strength

• Efficient design results when difference between capacity of timber and rivets is decreased

• Steel side plate cross-section must be checked for tension and shear loading separately

Design Actions

Np*=50 kN

Nl*=120 kN

500*135

mm

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• Try 65mm long rivets

• Number of rows (parallel to direction of applied load): nR= 8 for tension force and nR= 9 for shear force

• Number of rivets per row: nC= 9 for tension force and nC= 8 for shear force

• Spacing along grain a1= 35mm, across grain a2= 25mm

• End distance: Tension force a3t / shear force a3c,L= 200mm

• Upper end distance, a4c= 163mm (based on member width)

• Side plate thickness tp= 10mm

Connection geometry

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• Qs = Qs,θ

= min (ϕrQru,Ө, Qs,l/cosӨ, Qs,p/sinӨ)

• minimum of design rivet ultimate resistance at angle to grain/ design joint parallel to grain resistance (adjusted for angle to grain)/ design joint perpendicular to grain resistance (adjusted for angle to grain)

• Ө= tan-1(Np*/Nl

*)

• Np*= 50 kN

• Nl* = 120 kN

• Ө= tan-1(50/120) = 22°

Connection lateral resistance

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• Qs= Qs,l

• Qs,l=

ϕwQwe,l if ϕwQwe,l < ϕrQry,l (Brittle mode)

ϕrQry,l if ϕwQwy,l < ϕrQry,l ≤ ϕwQwe,l (Mixed mode)

ϕwQwy,l if ϕrQry,l ≤ ϕwQwy,l ≤ ϕrQru,l (Mixed mode)

ϕrQru,l if ϕrQru,l < ϕwQwy,l (Ductile mode)

• Check if ϕwQwe,l < ϕrQry,l . If so, then Qs,l = ϕwQwe,l (brittle failure mode)

Joint design lateral resistance parallel to the grain

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• ϕrQry,l= ϕrQr,l in which fh,0 and Mr,l equal to fhy,0 and Mry,l

respectively= ϕr k1 k12 kf np nR nC min (Prl,a, Prl,b)

• ϕr = 0.8 (strength reduction failure for ductile failure)

• k1 = 1.0 (for load combination [0.9G, Wu])

• k12= 1.0 (for dry timber)

• kf= 1.0 (for edge grain of glulam)

• np= 2 (number of plates)

• nR= 8 (number of rows of rivets parallel to axial force)

• nC= 9 (number of rows of rivets parallel to shear force)

Rivet capacity corresponding to yielding parallel to grain, ϕrQry,l

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• Xr= 0.87 for glulam

• Jp= 1.0

• fh,0= ƒhy,0 (yielding embedment strength)

• = 71.9ρ (1-0.0024dl) 10-3 for glulam

• ρ= 470 kg/m3 for GL10 glulam

• dl=3.2 mm

• ƒhy,0= 33.5 Mpa

• Lp= 51.8 mm

• Mr,l= Mry,l (yielding capacity of rivet)

= 24900 Nmm

• fax= 11.5ρ dp(1-0.0024dp)10-3

(withdrawal resistance per millimetre penetration)

• = 34.1 N/mm

Determine Prl,a , the characteristic strength parallel to grain for rivet failure mode (a)

, , 24 ,

,2 1

5.3310

With the above parameters Prl,a= 2.86kN

Lp

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• Xr= 0.87 for glulam

• Jp= 1.0

• fh,0= ƒhy,0

• = 71.9ρ (1-0.0024dl) 10-3 for glulam

• ρ= 470 kg/m3 for GL10 glulam

• dl=3.2 mm

• ƒhy,0= 33.5 MPa

• Lp= 51.8 mm

• Mr,l= Mry,l

= 24900 Nmm

• fax= 11.5ρ dp(1-0.0024dp)10-3

= 34.1 N/mm

Determine Prl,b , the characteristic strength parallel to grain for rivet failure mode (b)

, 2 , , 5.3310

With the above parameters Prl,b= 3.13kN

Lp

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Rivet capacity corresponding to yielding parallel to grain, ϕrQry,l

ϕrQry,l = ϕr k1 k12

kf np

nR nC min (Prl,a, Prl,b)

With the above parameters ϕrQry,l = 329.3kN (Yielding mode of failure: a)

• ϕr = 0.8

• k1 = 1.0 for load combination [0.9G, Wu]

• k12= 1.0 for dry timber

• kf= 1.0 for edge grain of glulam

• np= 2

• nR= 8

• nC= 9

• Prl,a= 2.86kN

• Prl,b= 3.13kN

Lp

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• ϕwQwe,l = ϕwQw,l in which tef,l equals to tefe,l

= ϕw np k1 k12 kf min (Pw,h, Pw,b, Pw,l)

• ϕw = 0.7 (strength reduction failure for wood failure)

• Need to calculate characteristic resistance for the head, bottom and lateral failure planes

Wood capacity corresponding to yielding parallel to grain, ϕwQwe,l

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Design Example – Base connectionDetermine Pw,h , characteristic resistance for failure of head tensile plane

Pw,h = Xt ft At,h (1+λ1+λ2) 10-3

With the above parameters Pw,h= 196.5kN, mode (a)

• Xt =1.19 for glulam (tension strength adjustment factor)

• ft =11 MPa for grade GL10 glulam

• At,h =tef,l wc (area of head plane subjected to tensile stress)

• tef,l =tefe,l (Corresponding to rivet elastic deformation, Section 4.1.4.4)

= 44.2 mm

• wc =a2 (nR-1) = 175 mm

• At,h =7736 mm2

• 0.25 1 ,

, ,

.

• ψ = G/E

• G = 670 MPa

• E = 10000 MPa

• ψ = 0.067

• Lc = a1 (nC-1) = 280 mm

• dz = 23.3

• H = 0.54

• As,b = 84000 mm2

• λ1 = 0.573

• λ2 = 0.5 1 ,

,

.

• F = 0.07

• As,l = 42438 mm2

• λ2 = 0.368

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Design Example – Base connectionDetermine Pw,b , characteristic resistance for failure of head tensile plane

Pw,b = (1+λ1-1+λ3) 10-3 x min ,

With the above parameters Pw,b= 180.8kN, mode (c)

• Xs= 0.96 for glulam

• Cb= 0.40

• , , ,

. , ,

• λ3= 0.642

c)

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Determine Pw,l , characteristic resistance for failure of lateral planes

Pw,l = (1+λ2-1+λ3

-1) 10-3 x min , 2 ,

With the above parameters Pw,l= 256.2kN, mode (a)

• Therefore, the wood capacity parallel to grain which corresponds to the rivet elastic deformation, ϕwQwe,l:

• ϕwQwe,l= ϕw np k1 k12 kf min (Pw,h, Pw,b, Pw,l) where min Pw,b governs

• C1= 0.32

With the parameters as calculated ϕwQwe,l = 253.2kN (failure governed by bottom shear plane, mode (c))

Note: Wood failure mode (c) governs, therefore ϕwQwe,l must be recalculated for the remaining planes to determine whether then residual head tensile plane and lateral shear planes can resist higher load: λ1= 0, λ3

-1= 0

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Again, determine Pw,h and Pw,l

Pw,h= Xt ft At,h (1+λ1+λ2) 10-3, mode (a)

Using the parameters selected earlier Pw,h= 138.5kN, mode (a)

Pw,l= (1+λ2-1+λ3

-1) 10-3 x min , ,mode

2 , ,mode

Using the parameters selected earlier Pw,l= 180.6kN, mode (a)

Recalculate wood capacity parallel to grain based on rivet elastic deformation ϕwQwe,l

ϕwQwe,l= ϕw np k1 k12 kf min (Pw,h, Pw,l) = 0.721.01.01.0 min(138.5, 180.6)= 193.5 kN (failure governed by head tensile plane, mode (a))

Timber failure mode (a) governs failure, therefore no recalculation is required. The residual head tensile plane and lateral shear planes resist lower load of 193.5 kNcompared to timber capacity of 253.2 kN when considering all resisting planes. Thus, ϕwQwe,l = 253.2 kN.

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Design Example – Base connectionCheck that the contribution of the lateral shear planes is less than 30% of the total joint capacity

(1+λ2-1+λ3

-1)-1 < 0.3

(1+0.368-1+0.642-1) -1 < 0.3

0.190 < 0.3

This portion of design is governed by brittle failure mode

OK, connection does not need to be redesigned

Check if Qs,l = ϕwQwe,l (brittle failure mode)

If ϕwQwe,l< ϕrQry,l

253.2< 329.3 OK

Qs,l= ϕwQwe,l

= 253.2 kN

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• Similar procedure to parallel to grain

• Failure modes for timber perpendicular to grain

Joint design lateral resistance perpendicular to grain

tef,ptef,pb

(a) (b)

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Ultimate rivet capacity at an angle of 22 degrees to the grain , ϕrQru,Ө

ϕrQru,Ө = , ,

, ,

ϕrQru,Ө = . .

. ° . °

The rivet ultimate capacity at angle angle of 22 degrees to the grain ϕrQru,Ө= 376.9 kN

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Finally, check joint ultimate lateral resistance

N*≤ Qs

∗ ∗2 ∗2

∗ 120 50N*= 130.0 kN

Qs= Qs,θ

= min (Qs,l/cosӨ, Qs,p/sinӨ, ϕrQru,Ө)= min (253.2/cos22°, 63.2/sin22°, 376.9)= min (273.1, 168.7, 376.9) = 168.7 kN

130.0 ≤ 168.7, OK (Connection mode of failure: Brittle wood splitting perpendicular-to-grain)

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Finally, check joint ultimate lateral resistance

Adopt 65 mm long rivets with an array of 8 rows by 9 columns, spacing 35 mm by 25 mm along and across the grain, and 200 mm end distance

Configuration:

8*9, 35*25 mm

End distance:

200 mm

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If ductile behaviour at the connection ultimate capacity is desirable

• The failure mode of the connection can be improved from brittle/mixed to ductile by increasing the wood resistance with larger rivet spacing across and along the grain.

• Spreadsheets can be used to speed up the computation process, and once set up, adjustments in spacing, end and edge distances and capacities for a range of rivet lengths can be evaluated relatively quickly.

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Timber Rivet Connections Design Guide

• Detailed description of rivet characteristics and uses

• A stand alone document which aims to widely disseminate information on timber rivet connections

• All phases of design explained in detail

• Comprehensive design examples and easy to follow design flow charts

Key Points

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Quick Connect

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Quick Connect

• An expedient moment connection for portal frame buildings

• Allows for offsite fabrication

• Versatile system –hybrid material connections

• Does not govern design of primary members

• Can use spreadsheets for designs

Key Points

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Quick Connect

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More Information needed?

• Contact EWPA

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Thank you

Documents

EXPAN Portal Frames Design Guide