Portal Frame Design Example

7/24/2019 Portal Frame Design Example

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Project: 30 metre Span LVL Portal Frame Design Date: Sept. 08

For: Carter Holt Harvey Woodproducts NZ Page: 1 / 92

At: Industrial Park, Auckland, New Zealand Designed : C.R

Carter Holt Harvey Limited September 2008

LVL Portal Frame DesignCHH Woodproducts New Zealand

Disclaimer

This design example has been prepared solely to provide guidance and recommendations to suitably qualified engineers and other suitably qualifieddesign professionals for diligent and professional use by them (and no other person) in the calculation of design solutions for LVL portal frame systemsin accordance with currently available New Zealand Standards.

To the best of Carter Holt Harveys knowledge and belief this example has been prepared in accordance with currently available technology andexpertise however good design and construction practice may be affected by factors outside the control of Carter Holt Harvey and beyond the control and

scope of this design example. This example is not intended to be used as the sole recipe, nor is it to be considered the authoritative method, forproducing the relevant design and it is assumed that the relevant designers will employ sound and current engineering knowledge and will take allreasonable care when designing LVL portal frame solutions using this example.

Accordingly, Carter Holt Harvey and its employees, agents and design professionals accept no liability or responsibility whatsoever and howsoeverarising for any losses, damages, costs or expenses (whether direct, indirect and/or consequential) arising from any errors or omissions which may becontained in this example, nor does it accept responsibility to any persons whatsoever for designs prepared in reliance upon this example or any otherinformation contained in this document.


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Table of Contents

1.0 Introduction

2.0 Purlin design2.1 Dead Load2.2 Live load2.3 Wind load2.4 Proposed Purlin Layout2.5 Connection Design2.6 Lateral restraint design2.7 Purlins supporting axial loading

3.0 Portal frame design3.1 Proposed Portal Frame3.2 Serviceability3.3 Strength3.4 Design Actions3.5 Rafter Design

3.5.1 Combined bending and compression3.5.2 Combined bending and tension

3.5.3 Flybrace design3.6 Column Design

3.6.1 Combined bending and compression3.6.2 Combined bending and tension3.6.3 Flybrace design

3.7 Gusset Design3.7.1 Knee Gusset Design3.7.2 Ridge Gusset Design3.7.3 Nail Ring Design

3.7.3.1 Knee Nail Ring Design

3.7.3.2 Ridge Nail Ring Design3.8 Column to Footing Design

4.0 Girt Design, Side Wall4.1 Wind Loading4.2 Connection Design

5.0 Mullion Design, Side Wall5.1 Wind Loading5.2 Connection Design

6.0

Eaves Beam Design6.1 Wind Loading6.2 Connection Design

7.0 Girt Design, End Wall


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7.1

Wind Loading7.2 Connection Design

8.0 Mullion Design, End Wall8.1 Wind Loading8.2 Connection Design

9.0 Longitudinal Bracing Design

10.0 Bibliography

Appendix 1 - Mullion deflection, bending and shear equations

Appendix 2 - 90mm thick hy90 compared with 63mm thick hySPAN

Published by: CHH Woodproducts New ZealandSeptember 2008

Enquires : Free call 0800 808 131Free fax 0800 808 132

Web : www.chhwoodproducts.co.nz/engineerszone


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1.0 Introduction

This design example has been provided as an aid to engineers in the development of design solutions for LVLand I-beam portal frame systems. The development of loading and the design of footings are not covered aspart of this example as their nature is not specific to timber. The design example has been prepared assumingthe building is proposed for Auckland, is within an Industrial Estate, and is subject to the following siteinformation:

Building Span 30.0 mBuilding length 60.0 m, consisting of 6 x 10.0 m baysBuilding Clear Height 6.0 m

Dominant openings 6.0 x 6.0 m in one end and one side wallCladding Pierce fixed sheeting of weight 6.0 kg/m2

Region A6, v500 = 45 m/s, v20= 37 m/sTerrain Category 3Directional Multipliers as per AS/NZS 1170.2:2002

This example has been based on relevant current design standards as detailed below: AS/NZS 1170.0:2002 Structural design actions. Part 0: General principles AS/NZS 1170.1:2002 Structural design actions. Part 1: Permanent, imposed and other actions AS/NZS 1170.2:2002 Structural design actions. Part 2: Wind actions NZS 3603:1993 Timber structures standard AS 1720.1-1997 Timber structures. Part 1:Design Methods

Note: Snow and Earthquake loading have been ignored due to location.

Other Referenced Design Documents: Technical Note 82-07-04 - Limit States Design Information for Specific Engineering Design for New

Zealand Construction. Mitek Specifiers and Users Manual.


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2.0 Purlin Design

Purlin Span 10,000-90 = 9910 mmPurlin Spacing 1600 mm (max.)

Propose HJ360 90 hyJOIST for use as purlin

Typically a hyJOIST purlin roof system becomes cost effective at spans above 6.0 m whilst hySPAN or MSG pinepulins remain cost effective for spans less than 6.0 m.

2.1 Dead load

Assume roof sheeting mass of 6.0 kg/m2plus a miscellaneous load of 1.0 kg/m2

kN/m.w

tself_weigh...

w

g

g

170

1000

8196107

*

*

=

+

=

Serviceability

Deflection of timber i-beams requires the consideration of shear deflection as well as bending deflection.Additional guidance on the calculation of shear deflection can be found in many Timber Design texts and is briefly

discussed in Technical Note 82. Timber components subjected to long term loads such as dead load require theconsideration of creep effects. Table 2.5, NZS 3603:1993 demonstrates the relationship between duration of loadand creep. The k2factor is applied to elastic deflections. LVL products are considered dry at the time of supplyand can be assumed to have a moisture content less than 18%.

Refer Technical Note 82 for Section and Material Properties.

495020

103928

9910170

102338384

9910170502

8384

56

2

9

424

2

2

Spanormm.

.

....

.GA

w.l

.EI

.w.lk

)(k

G

wx

shearbendingT

=

+

=

+=

+=

Serviceability limits for timber purlins are the same as those applied to other building products. For long termdeflection of industrial purlins span/300 or 30.0 mm are deemed acceptable.

2.2 Live load

Live load of 0.25 kPa applied in accordance AS/NZS 1170.1:2002 Table 3.2.

mkNw*g /40.06.125.0 ==

Serviceability


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421523

103928

9910400

102338384

991040.05016

2

9

4

Spanormm.

....

G

Q

=

+

=

Strength

Based on respective k1and load combination factors, combined dead and live load design actions will always bemore critical for design than permanent loads where low roof masses (less than 20 kg/m 2) are applied.

kNmM

lwM

mkNw

w

*

QG

*

QG

*

QG

*

QG

8.9

8

9.980.0

8

.

/80.0

40.05.117.02.1

5.12.1

22

5.12.1

5.12.1

5.12.1

=

==

=

+=

+

+

+

+

Check Bending Capacity

The bending capacity of an I-beam is based on the critical flange stresses due to bending. For composite timber I-beams the bending moment capacity can be based on a lever arm action about the centroid of the flanges withone flange in tension and the other in compression for a single span application. The restraint offered to thecompression flange is instrumental in the capacity of the I-beam. Further guidance on the bending momentcapacities of I-beams may be found in Technical note 82.

Purlin design assumes the use of pierce fixed roof sheeting providing continuous lateral restraint to thetop flange of the purlin. Since compression edge is fully restrained k8=1.0.

So for bending about XX axis

Since k8>0.73

kNmDAfkM ftbx6

11 10....

= Refer Technical Note 82

where:

( )

mmD

mmA

A

MPafk

F

F

t

32436360

3060

122

2883183690

3380.09.0

1

2

1

==

=

=

===

*

6

6.23

103243060338.09.0

MkNmM

kNmM

bx

bx

>=

=


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Check Shear Capacity

kNv

v

*

QG

*

QG

0.4

2

9.980.0

5.12.1

5.12.1

=

=

+

+

From table 14, Technical note 82

*

1

1.10

6.128.06.12.

vkNV

kV

>=

==

2.3 Wind loading

= 0, Lateral wind critical (by inspection)

a = min(0.2b, 0.2d, h) = 6.0m

mkNw

w

mkNw

w

cckkspacingqw pipelau*

i

/02.2

)61.09.00.10.1(6.184.0

/63.2

)61.09.05.10.1(6.184.0

)...(.

*

2

*

2

*

1

*

1

=

=

=

=

=

+

+

Calculate weff

Calculate Reactions

kNR

R

84.11

000.363.2955.102.2

*

*

=

+=

Calculate Moment

kNmM

M

kNmM

M

WuG

WuG

Wu

Wu

8.28

67.308

9.917.09.0

7.30

0.40.202.22

0.363.2955.484.11

*

9.0

2

*

9.0

*

2*

=

+

=

=

=

+

+


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Calculate weff

mkNw

w

eff

eff

/50.2

9.9

867.30

*

2

*

=

=

For uplift

mkNw

w

WuG

WuG

/35.2

50.217.09.0

*

9.0

*

9.0

+

+

=

=

Serviceability

To obtain the serviceability wind load the ultimate uniform loads can be factored by the square of the ratioserviceability wind speed to ultimate wind speed.

1000.99

103928

991069.1

102338384

991069.1501

/69.150.245

37

6

2

9

4

2

2

Spanormm

...

mkNw

wv

vw

w

w

s

s

u

ss

=

+

=

=

=

=

The acceptance of serviceability is at the engineers discretion. On the basis of applied local pressure factors andthe instantaneous nature of the wind gust span/100 is deemed acceptable.

Strength

Since the tension flange is fully restrained under uplift actions and the hyJOIST purlin is a composite section, useAppendix C of NZS3603:1993 for stability calculations.

Check Capacity

Calculate S1

5.0

1.

.1.1

=

yM

EIS

E

x Eq. C1.1, NZS 3603

where:

?

1802/360102338 49

=

===

E

x

M

mmyNmmEI Technical Note 82


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Eqn. C7 may be employed due to the continuous restraint offered to the tension flange by the pierce fixed sheeting.A suitably designed lateral restraint system provides intermediate buckling restraint to the purlins.

Calculate Euler Buckling Moment

( )

( )ho

ay

oy

Eyy

GJL

yD

EI

M+

+

+

=.2

4

2

2

2

Eq. C7, NZS 3603

where:

mmDmmymmy ho 3601802/3601802/360 ===== 2629

101848107.57 NmmGJNmmEIy ==

mmLay 24784/9910 == (Restraint at quarter points)

( )

( )

kNmM

M

E

E

7.43

1801802

1018482478

1804

360107.57

6

2

2

2

9

=

+

+

+

=

1.18

180107.43

102338.1.1

1

5.0

6

9

1

=

=

S

S

Calculate k8

Since 25>S>10

76.0

1.185000

11.180116.01.18175.021.0

...

8

32

8

3

4

2

3218

=

+++=

+++=

k

k

SaSaSaak

Since k8>0.73

kNmDAfkM ftbx6

11 10....

= Refer Technical Note 82

where:

mmD

mmA

MPafk

F

t

324

3060

330.19.0

1

2

1

=

=

===


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*

6

4.29

103243060330.19.0

MkNmM

kNmM

bx

bx

>=

=

Note: Where k8< 0.73 the moment capacity becomes a function of the compression flange bucklingrather than the tension flange being critical. The moment capacity equation is altered to represent thiswhere the characteristic tension stress is replaced by the product of the stability factor k8and thecharacteristic compression stress.

ie. kNmDAfkkM fcbx6

181 10.....

=

Calculate shear and support reaction for wind load.

Considering local pressure factors

Case 1

mkNww

mkNw

w

wcckkspacingqw gpipelau*

i

/88.117.09.0)61.09.00.10.1(6.184.0

/48.2

17.09.0)61.09.05.10.1(6.184.0

.9.0)...(.

*

2

*

2

*

1

*

1

=

+=

=

+=

+=

+

+

kNR

R

WuG

WuG

81.11

9.60.648.22

9.388.1

9.9

1

*

9.0

2*

9.0

=

+=

+

+

Case 2

mkNw

w

mkNw

w

wcckkspacingqw gpipelau*

i

/88.1

17.09.0)61.09.00.10.1(6.184.0

/09.3

17.09.0)61.09.020.1(6.184.0

.9.0)...(.

*

2

*

2

*

1

*

1

=

+=

=

+=

+=

+

+

kNR

R

WuG

WuG

2.12

4.80.309.32

9.688.1

9.9

1

*

9.0

2*

9.0

=

+=

+

+

Calculate dead & live load combined actions

kNR 0.42

9.98.0*=

=


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Timber capacity is dependant on the duration of the load in question, this must be taken into account in thedetermination of the critical load case. One method of assessing the critical design load is to remove the durationof load factor,k1, from the capacity equation and divide the load action effect by k1,

kNk

R

k

RMax

2.12

8.0

0.4,

0.1

2.12max

1

*

max

1

*

=

=

Check shear capacity

Since k1 was taken into account in the calculation of designaction, apply k1=1.0

*

1

6.12

6.120.16.12.

vkNV

kV

>=

== Table 14, Technical note 82

Therefore the HJ360 63 hyJOIST is suitable for use as a purlin based on the implied loading ata spacing not exceeding 1600 mm

2.4 Proposed Purlin Layout

2.5 Connection design

Connection of hyJOIST purlins to LVL rafters needs to ensure that the structural integrity of both the hyJOIST purlinand the hySPAN rafter are maintained. Connection to the hyJOIST by nailing through the plywood web providesthe most cost effective method of connection for purlins typically subject to high wind loads (please note this type

of connection is not recommended for i-beams subject to high permanent and/or live loads). Nailing throughplywood allows for nailing close to the end/edge of the plywood. Packing out the web and using proprietary joisthangers can also provide a suitable connection however the cost of the packing, brackets and labour involved canmake this an expensive alternative.


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Purlin connection blocks, or seating blocks as they are sometimes called, have been used in a numberof design situations for connection of C or I beam purlins where the connection block is either screwedor nailed to the rafter and the web of the composite purlin is connected directly to the connectionblock. A purlin connection block is proposed for connection using 2.87 diameter nails through theplywood web and 14g type 17 screws through the connection block to the rafter. Target the connectionfor design shear capacity, Vpsof the purlin.

Note: The selection of a suitable purlin connection block needs to take into account the end and edge distances ofthe fasteners as well as the spacing along and across the grain. The use of 4 x-banded connection block reducesthe tendency of the long band to split, allowing for the spacing of fasteners into the face to be similar along the

grain to across the grain. The orientation of the connection block is important where the plywood web is fixed tothe face of the connection block.

Calculate minimum number of 2.87 FH nails

Joint Group J5 Table 3, Technical note 82

kQS * Eq. 4.1, NZS 3603

kn QknQ ..= Eq. 4.2, NZS 3603

kn QknQ ...=

where:

kNQk k 526.00.18.0 1 === NZS 3603

k=1.4 since nails are through plywood with flat head nails.k=1.1 since we are proposing 20 nails per connection Cl. 4.2.2.2(g) NZS 3603

(linear interpolation between 1.3 for 50 nails and 1.0 for 4 nails)

Other k modification factors are not relevant as timber is dry, nails are in single shear and are nailed into theedge or face of the timber.

From Table 14, Technical Note 82 Vps=12.6.k1

4.19

526.01.14.10.18.00.16.12

=

=

n

n

Say 20/50x2.87 FH nails, nailed through plywood web into purlin connection block

Calculate minimum number of 14g type 17 Hex Head screwsType 17 screws are preferred for timber connection as they are a self drilling screws through the timber.


nQS * Eq. 4.5, NZS 3603

kn QknQ ..= Eq. 4.6, NZS 3603

kn QknQ ...=


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where:

kNQk k 303.30.1*8.0 1 === NZS 3603

Other k modification factors are not relevant as timber is dry, screws are in single shear and are screwed into theedge or face of the timber.*=0.8 is applied as Type 17 screws are as reliable as nails in service.

From Table 14, technical note 82 Vps=12.6.k1

76.4

303.30.18.00.16.12

=

=

n

n

Say 5/100x14g type 17 Hex Head screws, screwed through the purlin connection block into the rafter.

Proposed Purlin Connection

2.6 Lateral restraint design

The lateral restraint system needs to prevent the top and bottom flange of the hyJOIST purlin frommoving independently of each other. Many systems are appropriate but may require the fabrication ofspecial components. One of the most effective systems is to use hyJOIST pieces together with ahyCHORD bottom flange restraint and continuous mild steel galvanised strap over the top, as shownbelow.


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Calculate force on lateral restraint

( )1

05.0...

353433

*

+=

r

AA

nd

MkkkF Eq. B9, NZS 3603

where:

0.133=k (Wind loading)

4.034 =k

55,2

122min5,

2

1min

35 =

+=

+=

m

k

kNmMA 79.28=

mmd 360=

3=rn

( )

kNF

F

A

A

0.2

13360

1079.2805.054.00.1

6

=

+

=

Check capacity of lateral restraint propose 90x45 hyCHORD

kNNN tc 0.2**==

Typically a 45 mm thick section is recommended to allow for a 75mm long screw through both the lateral restraintand into the flange of the hyJOIST. Using hyCHORD for the lateral restraint is a good choice given its high strengthand lower cost.

Consider column action

Since Lay=1600 mm and Lax=1600 mm (defined by purlin spacing)

ncxc NN *

and

ncyc NN * Eq. 3.17, NZS 3603

Minor axis buckling is critical by inspection

AfkkN cncy ... 81= Eq. 3.19, NZS 3603

AfkkN cncy .... 81=

where:

240504590459.0 mmAMPaf c ====

4050459.081 = kkNncx


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kNkkNncx ...0.164 81=

Calculate k8for buckling about the minor axis

b

Lor

b

LkS

ay.103 = whichever is less Eq. 3.15, NZS 3603

6.35

45

1600

3

3

=

=

S

S

Since 25>S3>10

23.0

6.355.235

.

8

937.1

8

586

=

=

=

k

k

Sak a

Since k1= 1.0

*3.38 cncx NkNN >=

Consider tension strength

ntt NN .* Eq. 3.20 NZS 3603

AfkkN tnt ... 41= Eq. 3.21 NZS 3603

AfkkN tnt .... 41=

where:

0.1339.04 === kMPaf t Technical Note 82

240504590 mmA ==

0.11 =k

4050330.10.19.0 =ntN

kNNnt 3.120=

Consider connection between purlins and lateral restraint

Use screws for increased withdrawal capacity for practical purposes

Calculate minimum number of 14g type 17 Hex Head screws


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nQS * Eq. 4.5, NZS 3603

kn QknQ ..= Eq. 4.6, NZS 3603

kn QknQ ...=

where:

kNQk k 303.30.1*8.0 1 === NZS 3603


Consider Qkreduction due to the penetration into the receiving member (Purlin/blocking)

Penetration = 75-45 = 30 mm

Since da = 6.3 mm Table 4.5, NZS 3603

Therefore portion of diameter in penetration = 4.76

Calculate reduction from capacity relating to 7 da Cl. 4.3.2(e), NZS 3603

Reduction factor = 68.07

76.4=

NQk 2247303.368.0 ==

So:

11.1

247.20.18.00.10.2

=

=

n

n

Say 2/75x14g type 17 Hex Head screws, screwed through the purlin connection block into the rafter.

2.7 Purlins subject to axial loads

Purlins in end bays may be subjected to tension and compression forces from braced bays. Theseforces need to be considered in the design capacity. Refer to section 9.0, Longitudinal bracing.


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3.0 Portal Frame Design

The following portal frame has been analysed using elastic structural analysis with Microstran. Elasticstructural analysis of a timber portal frame differs little from that applied to steel members except forthe different section and material properties. For solid timber a five percent allowance for sheardeflection is included in the average modulus of elasticity which removes any need for the separateconsideration of shear deflection.

To achieve portal frame action rigid connections need to be made at both the ridge and eave. One ofthe most efficient methods of providing rigid connections is via use of nailed plywood gussets. Theadditional stiffness provided by the knee and ridge gussets is generally ignored in analysis.

3.1 Proposed Portal Frame

Refer Technical Note 82 for Material Properties.

3.2 Serviceability

Serviceability design limits for timber and steel buildings are very similar where the consideration ofcladding and absolute clearances need to be taken into account in the relative stiffness of the frame.Short term duration of loading for wind, live and earthquake loads may be calculated by applying aduration of load factor of 1, hence using the elastic deflection directly from analysis packages. For longterm loads the effects of creep need to be taken into account. NZS 3603 Table 2 defines k2as 2.0 forloading of twelve months or more where the moisture content is less than 18%.

Serviceability 900x90 hySPAN portal frameDeflection

Load Case k2 Vertical HorizontalDead load* 2.0 96.2 mm or span/302 16.2 mm or height/396

Live load 1.0 75.5 mm or span/385 9.6 mm or height/668Wind loadingLateral wind1 1.0 134.7 mm or span/216 28.4 mm or height/225Lateral wind2 1.0 74.5 mm or span/390 15.7 mm or height/408Longitudinal wind1 1.0 108.5 mm or span/268 13.5 mm or height/475Longitudinal wind2 1.0 64.6 mm or span/450 8.1 mm or height/792


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* It is typical to pre-camber the portal by its un-factored deflection (ie. Approx 50 mm in the case)

3.3 Strength

The selection of design moments is important in the design of timber portal frames. The nature of theinteraction of gussets provide specific locations for the selection of critical design actions for thedesign of rafters, columns gussets and nail rings. Hutchings and Bier [2000] provide guidance on thedesign moment locations as shown below.

Location A Rafter design actions at kneeLocation B Column design actionsLocation C Knee gusset design actionsLocation D Gusset to rafter at knee connection actionsLocation E Gusset to column connection actionsLocation F Ridge gusset design actionsLocation G Ridge gusset to rafter design actions

A further check along the rafter is require where the critical design actions may not to be at thegusseted location and should be taken as the maximum along the rafter.

3.4 Design Actions

The consideration of critical design actions also needs to take in account the effect of duration of load factors forcapacity, hence affecting the determination of critical load case. As with steel portal frames the bending momentdiagram should also be taken into account together with the lateral and torsional restraint offered by purlins,

girts and flybraces. The following design actions have been tabled as being of interest, other actions have beendismissed by inspection. The point of contraflexure is within close proximity for each case meaning that thecritical load case can be determined by inspection.

Critical Design Actions

Column Rafter

M* N* V* M* N* V*Load Case k1 kN kN kN kN kN kN

1.2G+1.5Q 0.8 -240.0 -84.1 71.5 -268.0 -60.4 50.5

0.9G+Wu - Lat 1.0 271.0 101.0 55.4 293.0 67.3 87.6

1.2G+Wu - Lat 1.0 -276.0 -113.0 62.3 -307.0 -79.2 -95.0


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k1factored Design Actions

Column Rafter

M*/k1 N*/k1 V*/k1 M*/k1 N*/k1 V*/k1

Load Case k1 kN kN kN kN kN kN

1.2G+1.5Q 0.8 -300.0 -105.1 89.4 -335.0 -75.5 63.1

0.9G+Wu - Lat 1.0 271.0 101.0 55.4 293.0 67.3 87.6

1.2G+Wu - Lat 1.0 -276.0 -113.0 62.3 -307.0 -79.2 -95.0

3.5 Rafter Design

A check of the capacity of main frame members of a timber portal frame involves a check of combined bending andbuckling action, both in plane and out of plane, and a check of combined bending and tension.

3.5.1 Combined bending and compression

Design Criteria

0.1

**

+

ncx

c

nx

x

N

N

M

M Eq. 3.23 NZS 3603

0.1

*2

*

+

ncy

c

nx

x

N

N

M

M Eq. 3.24 NZS 3603


Critical load case - 1.2G+1.5Q

M* = -268.0 kNm Nc* = -60.4 kN V* = 50.5 kN

Consider Bending Moment Capacity

nMM * Eq. 3.3 NZS 3603

ZfkkkkM bn ..... 8541= Eq. 3.4 NZS 3603

For solid sections with member depths greater than 300 mm, apply size factor (k11, AS 1720.1). Forfurther information refer AS1720.1 (Clause 2.4.6) or Technical Note 82.

Therefore ZfkkkkkM bn ....... 118541=

where:

MPafkk b 480.19.0 54 ==== Technical Note 82


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83.0900

300

300

167.0

11

167.0

11

=

=

=

k

dk

Cl. 2.4.6 AS1720.1

36

22

1015.12

6

90900

6

.

mmZ

bdZ

=

==

kNmkkM

kkM

n

n

81

681

..65.435

1015.124883.00.10.19.0

=

=

Since k1=0.8

kNmkMn 8.52.348=

Calculate k8The timber structures standard does not talk about critical flange like the steel structures standard however

similar principles apply to the restraint of LVL beams. Guidance is provided for solid sections in Clauses 3.2.5 of

NZS 3603:1993 for end-supported beams with discrete restraint to the compression edge (Cl 3.2.5.2) and tensionedge continuously restrained (Cl 3.2.5.3). Typically these can be useful in the calculation of slenderness of simplebeams and secondary framing however composite sections and members within structural frames requireanalysis using Appendix C of NZS3603:1993 for slenderness calculations.

Consider slenderness equation

5.0

1.

.1.1

=

yM

EIS

E

x Eq. C1 NZS 3603

Since for 900x90 hySPAN

mmy

NmmEIx

4502

900

1017.7212

9090013200

412

3

==

=

=

Therefore:5.0

9

1

10418.176

=

EMS

Calculate Euler moment, ME

Consider compression edge unrestrained from edge of column to point of contraflexure.

Some authors including Milner [1997] have developed theories based on the contribution of lateral restraintoffered to the tension edge by purlins and girts, such theories are beyond the scope of this example.


21/92





Consider Moment Diagram

( )[ ] 5.05 .GJEIL

cM y

ay

E

= Eq. C3 NZS3603

where:

0268

0== = ratio of bending moments between buckling restraints

5.55 =c Table C1 NZS3603

493

1071.72112

9009013200 NmmEIy =

=

Since for rectangular sections:

363.01

3BD

D

BJ

= Eq. C2 NZS 3603

293

1025.1353

90900

900

9063.01660 NmmGJ =

=

Therefore:

[ ]

kNmM

M

E

E

68.350

1025.1351071.7214900

5.5 5.099

=

=

From previous:


22/92





43.22

1068.350

10418.176

1

5.0

6

9

1

=

=

S

S

Since 25>S1>10

56.0

43.225000

143.220116.043.22175.021.0

...

8

32

8

3

4

2

3218

=

+++=

+++=

k

k

SaSaSaak

kNmMn 2.19556.052.348 ==

Mn


23/92





[ ]

kNmM

M

E

E

51.685

1025.1351071.7211741

82.3 5.099

=

=

From previous:

04.16

1051.685

10418.176

1

5.0

6

9

1

=

=

S

S

Since 25>S1>10

3

4

2

3218 ... SaSaSaak +++= Cl C2.10 NZS 3603

86.0

04.165000

104.160116.004.16175.021.0

8

32

8

=

+++=

k

k

*7.29986.052.348 MkNmMn >==

Check remaining unrestrained section

M* = 171.1 kNm Lay = 3160 mm c5= 5.5

Therefore:

[ ]

kNmM

M

E

E

78.543

1025.1351071.7213160

5.5 5.099

=

=

From previous:

01.18

1078.543

10418.176

1

5.0

6

9

1

=

=

S

S

Since 25>S1>10

3

4

2

3218 ... SaSaSaak +++= NZS 3603 Cl C2.10

77.0

01.185000

101.180116.001.18175.021.0

8

32

8

=

+++=

k

k


24/92





*

4.26877.052.348 MkNmMn >==

Consider region along rafter between point of contraflexure and apex along the rafter.

Bending Moment Diagram

Since purlins provide restraint to compression edge, Lay= 1600 mm where c5= 3.1 (moment ratiobetween purlins = 0 (conservative)).

Calculate Euler Moment

[ ]

kNmM

M

E

E

33.605

1025.1351071.7211600

1.3 5.099

=

=

From previous:

07.17

1033.605

10418.176

1

5.0

6

9

1

=

=

S

S

Since 25>S1>10

3

4

2

3218 ... SaSaSaak +++= Cl C2.10 NZS 3603

81.0

07.175000

107.170116.007.17175.021.0

8

32

8

=

+++=

k

k

*

3.28281.052.348 MkNmMn >==


Major axis buckling XX


25/92





AfkkN cncx ... 81= Eq. 3.18 NZS 3603

AfkkN cncx .... 81=

where:

28100090900459.0 mmAMPaf c ====

8100045.9.081 = kkNncx

kNkkNncx ...5.3280 81=

Calculate k8for buckling about the major axis

L=Lax=14221 mm (rafter length from ridge to column)

d

Lor

d

LkS ax

.10

2 = whichever is less NZS 3603 Eq. 3.14

k10= 1.0 (Conservative)

80.15

900

142210.1

2

2

=

=

S

S

Since 25>S2>10

3

4

2

3218 ... SaSaSaak +++= NZS 3603 Cl C2.10

87.0

80.155000

180.150116.080.15175.021.0

8

32

8

=

+++=

k

k

Since k1= 0.8

*1.2278 cncx NkNN >=

Minor axis buckling YY

From previous:

kNkkNncx ...5.3280 81=

Calculate k8for buckling about the minor axis YY

Lay=1600 mm (purlin spacing)


26/92





bLor

bLkS

ay.103= whichever is less Eq. 3.15 NZS 3603

78.17

90

1600

3

3

=

=

S

S

Since 25>S3>103

4

2

3218 ... SaSaSaak +++= Cl C2.10 NZS 3603

78.0

78.175000

178.170116.078.17175.021.0

8

328

=

+++=

k

k

Since k1= 0.8

*03.2047 cncx NkNN >=

Combined actions

0.192.01.2278

4.60

7.299

0.268=

+

Eq. 3.23 NZS 3603

0.183.00.2047

4.60

7.299

0.268 2

=

+

Eq. 3.24 NZS 3603

3.5.2 Combined bending and tension

Design Criteria

0.1

**

+

nnt

t

M

M

N

N Eq. 3.25 NZS 3603


Critical load case - 0.9G+WuLateral wind

M* = 293.0 kNm (at eave)M* = -171.8 kNm (along rafter)Nt* = 69.9 kN V* = 87.6 kN


From previous:

kNmkkMn 81..65.435=

Since k1=1.0, wind gust


27/92





kNmkMn 8.65.435=

Calculate k8


Consider compression edge restrained by purlins at 1600 c/c until point of contraflexure.


( )[ ] 5.05 .GJEIL

cM y

ay

E

= Eq. C3 NZS 3603

where:

60.02.293

2.176== = ratio of bending moments between buckling restraints (purlins)

9.35 =

c Eq. C3 NZS 3603

29

49

1025.135

1071.721

NmmGJ

NmmEIy

=

=

Therefore:

[ ]

kNmM

M

E

E

54.761

1025.1351071.7211600

90.3 5.099

=

=

From previous:


28/92





22.15

1054.761

10418.176

1

5.0

6

9

1

=

=

S

S

Since 25>S1>10

3

4

2

3218 ... SaSaSaak +++= NZS 3603 Cl C2.10

90.0

22.155000

122.150116.022.15175.021.0

8

32

8

=

+++=

k

k

*1.39290.065.435 MkNmMn >==

Check remaining sections between points of contraflexure (ie. Negative moment along the rafter)

Propose flybracing as detailed below

Consider region along rafter between point of contraflexure and apex along the rafter.


29/92






Calculate Euler Moment

Three buckling zones exist for wind uplift, each restrained at strategic purlin locations by flybraces.Consideration of bending moment diagram and restraint locations display.

Region 1 c5 = 5.5, Lay= 5183 mmRegion 2 c5 ~ 3.1, Lay= 2x1600 = 3200 mmRegion 3 c5 = 3.1, Lay= 2x(1050+229) = 2558 mm

Since:

=

ay

EL

cfunctionM 5

Therefore Region 2 is critical buckling region

[ ]

kNmM

M

E

E

66.302

1025.1351071.7213200

1.3 5.099

=

=

From previous:

14.24

1066.302

10418.176

1

5.0

6

9

1

=

=

S

S

Since 25>S1>10

3

4

2

3218

... SaSaSaak +++= Cl C2.10 NZS 3603

49.0

14.245000

114.240116.014.24175.021.0

8

32

8

=

+++=

k

k


30/92





*

5.21349.065.435 MkNmMn >==


ntt NN .* Eq. 3.20 NZS 3603

AfkkN tnt ... 41= Eq. 3.21 NZS 3603

For solid sections with member depths greater than 150 mm, apply k11size factor for tension. Forfurther information refer AS1720.1 (Clause 2.4.6) or Technical Note 82.

Therefore AfkkkN tnt ..... 1141=

where:


mmA 8100090900 ==

0.11=k

74.0900150

150

167.0

11

167.0

11

=

=

=

k

dk

Cl. 2.4.6 AS1720.1

810003374.00.10.19.0 =ntN

kNNnt 2.1780=

Combined actions

0.184.05.213

8.171

2.1780

9.69

=

+

q. 3.25 NZS 3603

Calculate Shear Capacity

nVV * Eq. 3.3 NZS 3603

Ssn AfkkkV .... 541= Eq. 3.4 NZS 3603

where:

MPafkk

k

s 3.50.1

0.19.0

54

1

===

==

Technical Note 82


31/92





254000

3

9090023/..2

mmA

dbA

S

S

=

=

= Cl 3.2.3.1 NZS 3603

*6.257

540003.50.10.19.0

VkNV

V

S

S

>=

=

Use 900x90 hySPAN as rafter with flybraces to locations as detailed.

3.5.3 Flybrace design

Critical Design Moment at flybrace location M* = -171.1 kNm, where k1=0.8

Calculate force on lateral restraint

( )1

05.0...

353433

*

+=

r

AA

nd

MkkkF Eq. B9, NZS 3603

where:

0.133 =k (Dead and live loads)4.034 =k

15,2

11min5,

2

1min35 =

+=

+=

mk

kNmMA 1.171=

mmd 900=

1=rn

( )kNF

F

A

A

9.1

11900

101.17105.014.00.1

6

=

+

=

Note: FA is the horizontal force and is shared between two components, one in tension and one incompression.

Check capacity of flybrace propose 90x45 hyCHORD

kNNN tc 90.1**==

Calculate force in brace

kNCos

NN tc 7.2)45(

90.1**===


32/92





Typically a 45 mm thick section is recommended to allow for a 75mm long screw through both the flybrace and intothe flange of the hyJOIST. Using hyCHORD for the lateral restraint is a good choice given its high strength andlower cost.


Since Lay=765 mm and Lax=765 mm (defined by brace length)

ncxc NN *

and

ncyc NN * Eq. 3.17 NZS 3603

Minor axis buckling is critical by inspection

AfkkN cncy ... 81= Eq. 3.19 NZS 3603

AfkkN cncy .... 81=

where:

240504590459.0 mmAMPaf c ====

4050459.081 = kkNncx

kNkkNncx ...03.164 81=

Calculate k8for buckling about the minor axis

b

Lor

b

LkS

ay.103= whichever is less Eq. 3.15 NZS 3603

98.16

45

764

3

3

=

=

S

S

Since 25>S1>10

3

4

2

3218 ... SaSaSaak +++= Cl C2.10 NZS 3603

82.0

98.165000

198.160116.098.16175.021.0

8

32

8

=

+++=

k

k

Since k1= 1.0

*50.134 cncx NkNN >=


33/92






ntt NN .* Eq. 3.20 NZS 3603

AfkkN tnt ... 41= Eq. 3.21 NZS 3603

AfkkN tnt .... 41=

where:


mmA 40504590 == 0.1

1=k

4050330.10.19.0 =ntN

kNNnt 3.120=

Consider connection between purlins and rafters and flybrace

Screws are required to provide tension connection to rafter/purlin



nQS * Eq. 4.5, NZS 3603

kn QknQ ..= Eq. 4.6, NZS 3603

kn QknQ ...=

where:

kNQk k 303.30.1*8.0 1 === NZS 3603


Consider Qkreduction due to the penetration into the receiving member (Purlin/blocking)

Penetration = 75-45 = 30 mm

Since da= 6.3 mm Table 4.5, NZS 3603

Therefore portion of diameter in penetration = 4.76

Calculate reduction from capacity relating to 7 da Cl. 4.3.2(e), NZS 3603


34/92





Reduction factor = 68.0776.4 =

NQk 2247303.368.0 ==

So:

kNQ

Q

n

n

0.4

247.220.18.0

=

=

Consider screws in tension

nQN * Eq. 4.8, NZS 3603

kn QpknQ ...= Eq. 4.9, NZS 3603

kn QpknQ ....=

where:

mmpmmNQk k 35/5.790.1*8.0 1 ==== NZS 3603


kNQ

Q

n

n

45.4

5.79350.128.0

=

=

Say 2/75x14g type 17 Hex Head screws, screwed through pre-drilled holes in flybrace into rafter andpurlin.

Proposed flybrace connection


35/92





3.6 Column Design

3.6.1 Combined bending and compression

Design Criteria

0.1

**

+

ncx

c

nx

x

N

N

M

M Eq. 3.23 NZS 3603

0.1

*2

*

+

ncy

c

nx

x

N

N

M

M Eq. 3.24 NZS 3603



M* = -240.0 kNm Nc* = -84.1 kN V* = 71.5 kN


From previous:

kNmkkMn 81..65.435=

Since k1=0.8

kNmkMn 8.52.348=

Calculate k8

For 900x90 hySPAN:

5.09

110418.176

=

EMS


Girts provide tension edge restraint to the outside of the outside of the frame. By inspection from therafter analysis one flybrace is proposed at the middle girt, 3490 mm from the ground.


36/92





Consider bending moment diagram

( )[ ] 5.05 .GJEIL

cM y

ay

E

= Eq. C3 NZS 3603

where:

Region 1

59.00.303

3.179=

= = ratio of bending moments between buckling restraints

92.35=c Eq. C3 NZS 3603

mmLay 2530=

Region 2

03.179

0

== = ratio of bending moments between buckling restraints

5.55 =c Eq. C3 NZS 3603

mmLay 3470=

29

49

1025.135

1071.721

NmmGJ

NmmEIy

=

=

Therefore Region 1 is critical:

[ ]

kNmM

M

E

E

08.484

1025.1351071.7212530

92.3 5.099

=

=

From previous:


37/92





09.19

1008.484

10418.176

1

5.0

6

9

1

=

=

S

S

Since 25>S1>10

3

4

2

3218 ... SaSaSaak +++= Cl C2.10 NZS 3603

71.0

09.19

5000

109.190116.009.19175.021.0

8

32

8

=

+++=

k

k

kNmMn 5.24771.052.348 ==


Major axis buckling XX

From previous:

kNkkNncx ...5.3280 81=

Calculate k8for buckling about the major axis

L=Lax=6000 mm (column height from rafter to footing)

d

Lor

d

LkS ax

.10

2 = whichever is less Eq. 3.14 NZS 3603

k10= 1.0 (conservative) Fig. 3.5 NZS 3603

67.6

900

60000.1

2

2

=

=

S

S

Since 10=

Minor axis buckling YY

From previous:

kNkkNncx ...5.3280 81=


38/92





Calculate k8for buckling about the minor axis YY

Lay=1660 mm (girt spacing)

b

Lor

b

LkS

ay.103 = whichever is less Eq. 3.15 NZS 3603

44.18

90

1660

3

3

=

=

S

S

Since 25>S3>10

3

4

2

3218 ... SaSaSaak +++= Cl C2.10 NZS 3603

75.0

44.185000

144.180116.044.18175.021.0

8

32

8

=

+++=

k

k

Since k1= 0.8

*

3.1968 cncx NkNN >=

Combined actions

0.10.14.2624

1.84

5.247

0.240=

+

Eq. 3.23 NZS 3603

0.198.03.1968

1.84

5.247

0.240 2

=

+

Eq. 3.24 NZS 3603

3.6.2 Combined bending and tension

Design Criteria

0.1

**

+

nnt

t

M

M

N

N Eq. 3.25 NZS 3603


Critical load case - 0.9G+WuLateral wind

M* = 271.0 kNm Nt* = 101.0 kN V* = 55.4 kN


From previous, since k1=1.0, wind gust


39/92





kNmkMn 8.65.435=

Calculate k8


Consider compression edge restrained by grits at 1660 c/c.


( )[ ] 5.05 .GJEIL

cM y

ay

E

= NZS3603 Eq. C3

where:

66.00.271

7.177 == = ratio of bending moments between buckling restraints (grits)

78.35=c NZS3603 Eq. C3

29

49

1025.135

1071.721

NmmGJ

NmmEIy

=

=

Therefore:

[ ]

kNmM

M

E

E

43.711

1025.1351071.7211660

78.3 5.099

=

=


40/92





From previous:

75.15

1043.711

10418.176

1

5.0

6

9

1

=

=

S

S

Since 25>S1>10

3

4

2

3218 ... SaSaSaak +++= NZS 3603 Cl C2.10

87.0

75.155000175.150116.075.15175.021.0

8

328

=

+++=

k

k

*0.37987.065.435 MkNmMn >==


Since 0.11=k , 74.0

11=k

810003374.00.10.19.0 =ntN

kNNnt 2.1780=

Combined actions

0.177.00.379

0.271

2.1780

0.101=

+

Use 900x90 hySPAN as column with flybraces to locations as detailed.


41/92





3.7 Gusset Design

The knee and ridge connections of an LVL portal frame can be completed by using a plywood gusset.Plywood gussets allow an ease of fabrication and can be readily fixed using machine driven nails.Plywood or minimum 4 x-band gussets are recommended for use in heavily nailed rigid connectionsbecause the x-band plies help reduce the tendency of the long band plies to split. This allows the nailspacing to be governed by the grain direction of the rafter or column which ever the gusset is beingfastened to.

Plywood is available in Stress Grade F11 from Carter Holt Harvey in thicknesses up to and including 25mm. For thicknesses over 25 mm required for large span portal frames CHH have developed 4 x-band

hySPAN sheets (2400x1200) in a 42mm thickness allowing 28 mm (8 plies) of parallel plies.

Design actions can be factored by the duration of load factor k1for comparison in the determination ofthe critical design action.

Gusset Design Actions

Knee Ridge

M* N* V* M* N* V*

Load Case K1 kNm kN kN kNm kN kN

1.35G 0.6 -123.0 -31.6 19.2 69.7 -19.0 2.5

1.2G+1.5Q 0.8 -324.0 -83.1 50.5 183.6 -50.1 6.6

0.9G+Wu Lat 1.0 362.0 102.2 -54.3 -156.9 71.1 -5.01.2G+Wu Lat 1.0 -382.0 -111.7 -65.2 171.2 -75.5 14.5

0.9G+Wu Long 1.0 239.1 64.6 55.1 -117.0 65.9 25.3

1.2G+Wu Long 1.0 -295.4 -60.3 -69.7 161.0 -56.6 7.5

k1factored Gusset Design Actions

Knee Ridge

M*/k1 N*/k1 V*/k1 M*/k1 N*/k1 V*/k1

Load Case K1 kNm kN kN kNm kN kN

1.35G 0.6 -205.0 -52.6 32.0 116.2 -31.7 4.2

1.2G+1.5Q 0.8 -405.0 -103.9 63.1 229.5 -62.6 8.3

0.9G+Wu Lat 1.0 362.0 102.2 -54.3 -156.9 71.1 -5.01.2G+Wu Lat 1.0 -382.0 -111.7 -65.2 171.2 -75.5 14.5

0.9G+Wu Long 1.0 239.1 64.6 55.1 -117.0 65.9 25.3

1.2G+Wu Long 1.0 -295.4 -60.3 -69.7 161.0 -56.6 7.5

3.7.1 Knee gusset design

The capacity of a plywood gusset is based on the critical depth at which the gusset bends, which is ahorizontal line across the centroid of the rafter and column intersection as shown below.


42/92





Geometrically, assuming the rafter depthand column depth are equal, the criticalsection for the knee connection may becalculated by:

tan2

11

+

+=

L

D

DLDDepthcs

Design Criteria

0.1

*2

**

+

+

ni

i

ni

i

nc

c

V

V

M

M

N

N Eq. 6.17 NZS 3603

0.1

*2**

+

+

ni

i

ni

i

nt

t

V

V

M

M

N

N Eq. 6.18 NZS 3603

It is typical that the design shear and tension action effects have little influence on the size of a gusset and can inmany cases be omitted from calculation such is their effect on sizing. Compression loads are generally pastthrough in bearing and not required for consideration in gusset design.


Load case - 1.2G+1.5Q (Combined bending, compression and shear)

M* = -324.0 kNm Nc* = -83.1 kN V* = 50.5 kN k1=0.8

Load case - 0.9G+Wu (Lateral wind) - (Combined bending, tension and shear)

M* = 362.0 kNm Nt* = 102.2 kN V* = 54.3 kN k1=1.0

Consider bending moment capacity

Many authors have proposed methods of calculating the capacity of plywood gussets. Batchelor [1984] proposesa bilinear stress distribution along the critical section while Hutchings [1987] methodology assumes atriangulated stress distribution across the critical section and recommends the application of a size factor.

Hutchings [1987] methodology is applied in this example. This methodology is suitable for application to bothopening and closing moments of portal frames, and has been used on many portal frame structures. Milner andCrosier [2000] propose a similar calculation based on a triangulated stress distribution but propose an alternatecritical section and omit the use of the size factor.


43/92





nii MM *

Eq. 6.9 NZS 3603

6

......

2

151481

dtfkkkkM epbni = Eq. 6.10 NZS 3603

Now include size factor - for further information on size factor, k11refer AS1720.1 (Clause 2.4.6) or Technical Note82.

Therefore6

........

2

15141181

dtfkkkkkM epbni =

Since the gussets are in pairs:

=

6

.........2

2

15141181

dtfkkkkkM epbni

Propose 42 mm 4 x-band LVL, where:

9.0=

?1=k

0.18=k (localised, gusset edges are restrained by gusset stiffeners)

0.114 =k (moisture content < 18%)

0.115 =k (only parallel plies are being considered)167.0

11

300

=

dk Cl. 2.4.6 AS1720.1

MPafb 48=

( )( ) mmte 285.3442 ==

=

6

28480.10.10.19.0.2

2

111

dkkMni

kNmdkkMn 2111 ..2.403 =

For 900x90 hySPAN portal frame with 7.5 pitch

( )

80.02.1177

300

2.1177

5.7tan12002

90011

9001200900

11

167.0

11

=

=

=

+

+=

k

k

mmd

d

Calculate bending moment capacity


44/92





kNmkM

kM

ni

ni

.0.447

2.11778.02.403

1

21

=

=


nip VV * Eq. 6.15 NZS 3603

dtfkkkkkV psni ........3

218151481

= Eq. 6.16 NZS 3603

where:

MPaf ps 3.59.0 == Technical Note 82

?1=k

0.114=k (moisture content < 15 %)

0.115 =k (face grain = 0)

= dkVni 423.50.10.10.10.1

3

29.02 1

kNdkVni ..12.267 1=

Since mmd 2.1177=

kNkVni .45.314 1=

Consider tension capacity

ntt NN .* Eq. 6.11 NZS 3603

dtfkkkN tptnt ..... 15141= Eq. 6.12 NZS 3603

dtfkkkN eptnt ...... 15141=

where:

MPaf pt 339.0 == Technical Note 82

?1=k

0.114 =k (moisture content < 15 %)

0.115 =k (face grain = 0)

mmte 28= (parallel plies only)

[ ]dkNnt = 28330.10.19.02 1

kNdkNnt ...2.1663 1=


45/92





Since mmd 900= (use minimum section) - conservative

kNkNnt .9.1496 1=

Consider compression capacity

ncc NN .* Eq. 6.13 NZS 3603

dtfkkkkN epcnc ...... 151481= Eq. 6.14 NZS 3603

dtfkkkkN epcnc ....... 151481=

where:

MPaf pc 459.0 == Technical Note 82

?1=k

0.114=k (moisture content < 15 %)

0.18 =k (localised, gusset edges are restrained by gusset stiffeners)

0.115 =

k (face grain = 0)mmte 28= (parallel plies only)

[ ]dkNnc = 28450.10.10.19.02 1

kNdkNnc ..0.2268 1=


kNkNnc .2.2041 1=

Consider Combined Actions

Combined bending, compression and shear from Eq. 6.17, NZS 3603:1993

Factor capacities by appropriate duration of load, k1= 0.8

0.107.15.3148.0

5.50

0.4478.0

0.324

2.20418.0

1.83 2

=

+

+

It is typical to consider the maximum implied forces on the structure, rather than the applied forces at the specificdesign location. However if the design criteria is not met then consideration of the implied design actions at thedesign location may be required. Therefore consider moment and shear forces at critical stress line for analysis.

Design Actions at critical stress line, Load case 1.2G+1.5Q

M* = -303.0 kNm Nc* = -83.1 kN V* = 50.5 kN k1=0.8


46/92





0.197.05.3148.0

5.50

0.4478.0

0.303

2.20418.0

1.83 2=

+

+



0.190.05.3140.1

3.54

0.4470.1

00.362

9.14960.1

2.102 2

=

+

+

3.7.2 Ridge Gusset Design

The design of the ridge gusset is similar to the knee gusset where the design capacity is based on themoment resistance offered by the ridge gusset section. Typically a mitre type joint is considered.Hutchings [1989] proposes a 0.9 factor be applied to the critical section as defined below.

Savings in design and fabrication can be made by keeping the distance L constant across the ridge and the kneegussets. Whilst the ridge gusset may be thinner often for consistency of purlin lengths and minimum gussetorder quantities it may be preferable to maintain similar gusset thicknesses.

.

gussetcs

gusset

DDepth

LCos

DD

.9.0

tan.

=

+=

Design Criteria

0.1

*2

**

+

+

ni

i

ni

i

nc

c

V

V

M

M

N

N Eq. 6.17 NZS 3603

0.1

*2

**

+

+

ni

i

ni

i

nt

t

V

V

M

M

N

N Eq. 6.18 NZS 3603



M* = 183.6 kNm Nc* = -50.1 kN V* = 6.6 kN

Critical load case - 0.9G+Wu Lateral wind


47/92





M* = -156.9 kNm Nt* = 71.1 kN V* = -5.0 kN


From previous, propose 42mm 4 x-band LVL, where:

kNmdkkMn2

111 ..2.403 =

For 900x90 hySPAN portal frame with 7.5 pitch

82.0

2.959

300

2.9599.0

7.1065

)5.7tan(1200)5.7(

900

11

167.0

11

=

=

==

=

+=

k

k

mmDd

mmD

CosD

gusset

gusset

gusset


*

1

2

1

.5.305

2.95982.02.403

MkNmkM

kM

ni

ni

>=

=

Calculate shear force capacity

From previous:

kNkV

kNdkV

ni

ni

.2.256

..12.267

1

1

=

=

Calculate Tension Capacity

From previous:

kNdkNnt ...2.1663 1=


kNkNnt .9.1496 1=

Calculate Compression Capacity

From previous:


48/92





kNdkNnc ..0.2268 1=


kNkNnc .2.2041 1=



0.163.02.2568.0

6.65.3058.0

6.1832.20418.0

1.50

2

=

+

+



0.157.02.2560.1

0.5

5.3050.1

9.156

2.20410.1

1.71 2

=

+

+

Use 42 mm 4 x-Band hySPAN as both knee and ridge gusset pairs

3.7.3 Nail ring design

The design of the nail ring is important because more than half of the nailing needs to be performed onsite. It is also important to consider end and edge distances together with allowable nail spacings(both along and across the grain) for the chosen fasteners. Selection of the nail diameter is also criticalas it will affect the available spacing and hence number of nails within the group as well as therequired penetration into the column/rafter. A staggered nail pattern provides an increased momentcapacity by maximising the lever arm action about the nail group centroid.

The design of nail groups associated with rigid moment connections are often subjected to combinedactions including bending, axial and shear forces. Whilst the bending and axial forces contributionsare minor they need to be taken into account. It is normally most efficient to calculate the proportion offorce remaining in the nails after the contribution to the design moment affect is taken out.

The complexity of calculations for the nail ring mean hand calculations can be time consuming andconservative. For this reason computer packages are often employed to develop design solutions. Thefollowing design data have been taken from design capacity tables relating to the corresponding roofpitch and member size.

The design methodology, including k factors, from AS1720.1 has been applied to create nail ringcapacities for a number of section sizes and gusset widths. These tables can be found in Engineering

Bulletin No.2, Rigid Moment Connections using CHH veneer based products. AS1720.1 was used due toits close relationship between the lateral capacities of nails in testing with CHHs range of LVL and thepublished values for joint group JD4. It should be noted that many of the k factors used in calculationof connection capacities differ between the standards and it is recommended that for connectionsthese not be mixed and matched.


49/92





3.7.3.1 Knee nail ring design


The critical design actions need only be considered in the nail ring design as the effects of stress reversal do notaffect the nature of the nail design.

Knee, Critical load case - 1.2G+1.5Q

M* = -324.0 kNm Nc* = -83.1 kN V* = 50.5 kN k1=0.77#

#As per Table 2.7, AS1720.1

The methodology proposed for the calculation of nail group capacity for combined bending, axial andshear force involves the following steps:

1. Calculate moment capacity of nail rings in accordance with AS1720.1. AS1720.1 provides acapacity calculation for transfer of in plane moments through nailed moment ring such that:

=

=

=

2

3

1 max

max171614131 ........

ni

i

ikj

r

rQrkkkkkM AS1720.1 Eq. 4.2(4)

where:

n = number of fastenersQk= characteristic strength of fastenerri= distance to the i

thfastener from the centroid of the fastener grouprmax= the maximum value of ri = capacity factor (0.8 - nails used with primary elements in structures other than houses)k1= duration of load factor (Clause 2.4.11, AS1720.1)k13= 1.0 (nails in side grain)k14= 1.0 (nails in single shear)k16= 1.1 (nails driven through plywood gussets)k17= multiple nail factor for resisting in plane moments (AS1720.1 Table 4.3(B))

Qk= 810 N (3.15 nail, JD4 strength group, AS1720.1 Table 4.1 (B))

Since nail rings will be applied through gusset pairs the total moment resistance offered by nailrings connecting gusset pairs is:

=

=

=

2

3

1 max

max171614131 .........2

ni

i

ik

r

rQrkkkkkM

2. Calculate remaining portion of nail capacity after bending actions have been considered.

3.

a. kn QkkkkkQ ...... 171614131 =

b. nQM

MN nshearaxial

=

*

/ 1


50/92





4.

Calculate vectorial sum of the combined axial and shear forces for comparison with remainingcapacity. These forces are assumed to be evenly distributed over the nail group.

( ) ( ) ( ) ( )

++= 2*2*2*2**

/ ,max vNvNN tcshearaxial

Engineering Bulletin 2 Rigid Moment Connection Details can be used for selection of the moment ringcapacity for the nail ring to suit the 7.5 roof pitch and 1200 mm wide gusset as drawn above.

From Table 50, Engineering Bulletin 2 for nine (9) nail rings

kNmMM

7.34914.45477.0

=

=

and kNQn 855.0=

Calculate remaining nail group capacity after resistance to moment has been calculated.

( )

kNN

N

nQM

MN

shearaxial

shearaxial

nshearaxial

0.86

2684855.0

7.349

0.3241

1

/

/

*

/

=

=

=

Calculate vectorial sum of axial and shear force, divided by k1for direct comparison

( ) ( ) ( ) ( )

( )

kNN

N

N

vNvNN

shearaxial

shearaxial

shearaxial

tcshearaxial

3.126

7.115,3.126max

0.1

3.54

0.1

2.102,

77.0

5.50

77.0

1.83max

,max

*

/

*

/

2222

*

/

2*2*2*2**

/

=

=

+

+

=

++=

Since NN shearaxial >*

/either add an additional nail ring or adjust nail size. Try using a 3.33 nail.

Using Table 3 from Engineering Bulletin 2 the capacity of the nail rings can be factored proportionally tothe Characteristic Capacity of the nail laterally loaded in single shear.

3.33/3.15 factor = 11.1810

898=

Therefore:

*15.388

11.114.45477.0

MkNmM

M

=

=

andkNQ

Q

n

n

949.0

11.1855.0

=

=


51/92






( )

*

/

/

*

/

6.214

2684949.015.388

0.3241

1

NkNN

N

nQM

MN

shearaxial

shearaxial

nshearaxial

>=

=

=

Use nine nail rings of 3.33 x75 FH nails to pattern as marked.

Proposed Nail Ring

3.7.3.2 Ridge nail ring design



M* = 183.6 kNm Nc* = -50.1 kN V* = 6.6 kN


52/92





Engineering Bulletin 2 Rigid Moment Connection Details can be used for selection of the moment ringcapacity for the nail ring to suit the 7.5 roof pitch and 1200 mm wide gusset as drawn above.

From Table 50, Engineering Bulletin 2 apply four (4) nail rings. Since we are using 3.33 nails in theknee connection, apply same nail size in the ridge, therefore apply 1.11 factor from previous to applynail ring capacities from Table ##.

*3.227

11.196.26577.0

MkNmM

M

=

=

and kNQn 949.0= From previous


( )

kNN

N

nQM

MN

shearaxial

shearaxial

nshearaxial

52.125

2344949.03.227

6.1831

1

/

/

*

/

=

=

=

Calculate vectorial sum of axial and shear force, divided by k1for direct comparison

( ) ( ) ( ) ( )

( )

NkNN

N

N

vNvNN

shearaxial

shearaxial

shearaxial

tcshearaxial


53/92





3.8 Column to footing connection design

Connection of portal frame columns to footings can be achieved by base brackets that are suitably sized and fixeddirectly to the LVL columns. A similar design philosophy is applied to the design and specification of hold downanchors and base plates as would normally be applied to steel where the buckling of the plate under tensionneeds to be considered.

The connection of the base brackets to the column could be achieved using nails, screws or bolts. Nails aretypically not recommended of base plates in larger structures because of the number of nails required combinedwith the fact they would need to be hand driven through holes in plates. Bolts can be used and are good to aid inthe transfer of bracing loads across the column. Screws are ideal for most base bracket connections due to theirease of application. It is important that screw patterns are staggered for both sides of the column so that splittingof the LVL does not occur.

Again reactions are factored to take into consideration duration of load factors.

Consider Design Reactions

PF1

Rx Ry (Rx2+Ry

2)0.5 Angle

Load Case k1 kN kN kN

1.35G 0.6 19.19 35.82 40.6 61.8

1.2G+1.5Q 0.8 50.53 86.87 100.5 59.8

0.9G+Wu (Lateral) 1.0 -58.60 -99.28 115.3 59.4

1.2G+Wu (Lateral) 1.0 -53.96 115.53 127.5 65.0

0.9G+Wu (Long) 1.0 -24.61 -78.67 82.4 72.6

1.2G+Wu (Long) 1.0 52.50 102.30 115.0 62.8

k1 adjusted values

PF1

Rx Ry (Rx2+Ry

2)0.5 Angle

Load Case k1 kN kN kN

1.35G 0.6 31.98 59.70 67.7 61.8

1.2G+1.5Q 0.8 63.16 108.59 125.6 59.8

0.9G+Wu (Lateral) 1.0 -58.60 -99.28 115.3 59.41.2G+Wu (Lateral) 1.0 -53.96 115.53 127.5 65.0

0.9G+Wu (Long) 1.0 -24.61 -78.67 82.4 72.6

1.2G+Wu (Long) 1.0 52.50 102.30 115.0 62.8

It is typical in Timber structures to provide a moisture barrier at the base of the columns to eliminate the columnfrom getting wet and staying wet during the construction period. This can be typically achieved by using H3.2treated Plywood and melthoid at both the LVL column end and ground as detailed in the structural drawings.Downwards loads may be considered to be taken out in bearing so for the design of connections only uplift loadsneed be considered.



nQS * Eq. 4.5, NZS 3603

kn QknQ ..= Eq. 4.6, NZS 3603


54/92





kn QknQ ...=

where:

25.1303.30.1*8.01

==== kkNQk k NZS 3603

Other k modification factors are not relevant as timber is dry, screws are in single shear and are screwed throughclose fitting steel plates into the edge or face of the timber.*=0.8 is applied as Type 17 screws are as reliable as nails in service.

Since critical design reaction is 115.3 kN, calculate minimum number of 14g screws.

9.34

303.325.10.18.03.115

=

=

n

n

Say 48/14gx50 type 17 Hex Head screws, screwed through base plate sides into column.

Proposed Connection


55/92





4.0 Girt Design, side wall

Girt Span (10,000-(90+65))/2 = 4923 mmGirt Spacing 1660 mm

Propose 190x45 hyCHORD for use as side wall girt

4.1 Wind loading

The capacity of solid timber girts is also dependant on the nature of lateral tortional buckling restraint and the

critical edge to which the loading and restraint is provided. It is therefore important to consider both positive andnegative wind pressures.

= 0, Lateral wind

qu=0.84 kPa Case 1 cp,e=+0.7, cp,i=

-0.56, kL= 1.25Case 2 cp,e=

-0.3, cp,i=+0.61

= 90, Longitudinal wind

qu=0.76 kPa Case 1 cp,e=-0.65, cp,i=

+0.54, kL= 1.5

Calculate design loading

mkNw

w

mkNw

w

cckkspacingqw pipelau*

i

/84.1

)54.065.05.1(66.176.0

/93.1

)56.07.025.1(66.184.0

)...(.

*

2

*

2

*

1

*

1

=

=

+=

=

=

+

+

Serviceability

Refer Technical Note 82 for Section and Material Properties.

=

=

=

++

x

w

s

EI

lw.k

mkNw

.384

..5

/30.193.145

37

4

2

2


56/92





1403.35

10283384

492330.15019

4

Spanormm

..

w

w

=

=

+

Strength

Check capacity for positive wind pressures

kNmM

lwM

79.58

9.493.1

8

.

*

22

*

=

==

+

Consider shear and support reaction for wind load

kNVN

lwV

73.4

2

9.493.1

2

.

**

*

==

==

+

Calculate Bending Moment Capacity

nMM * Eq. 3.3 NZS 3603

ZfkkkkM bn ..... 8541= Eq. 3.4 NZS 3603

where:

MPafkk

k

b 480.1

0.19.0

54

1

===

== Technical Note 82

kNmkM

kM

n

n

8

38

.7.11

10271480.10.10.19.0

=

=

Continuous restraint to compression edge via pierce fixed sheeting, therefore k8=1.0

*7.11 MkNmM >=


nVV * Eq. 3.3 NZS 3603

Ssn AfkkkV .... 541= Eq. 3.4 NZS 3603

where:


57/92





MPafkkk

s 3.50.10.19.0

54

1

===

== Technical Note 82

25700

3

451902

3/..2

mmA

dbA

S

S

=

=

=

Cl 3.2.3.1 NZS 3603

*2.27

57003.50.10.19.0

VkNV

V

S

S

>=

=

Consider negative wind pressures.

kNmM

lwM

5.5

8

9.484.1

8

.

*

22*

=

==


From previous:

kNmkMn 8.7.11=

Calculate k8

Continuous lateral restraint is provided to the tension edge via pierce fixed sheeting.

Calculate S1

b

dS .31 = Eq. 3.6 NZS 3603

67.1245

190

3! ==S

Since 25>S1>10

97.0

67.125000

167.120116.067.12175.021.0

...

8

32

8

3

4

2

3218

=

+++=

+++=

k

k

SaSaSaak

*3.11

97.07.11MkNmM

M>=

=

4.2 Connection design


58/92





Connection of hyCHORD girts is easiest performed using proprietary brackets and screws or nails. The proposedbracket is manufactured by Mitek. It is important to ensure that the depth of proprietary brackets is at least 60%of the depth for beams up to 50 mm thick. Propose JH47x190 to suit 190x45 hyCHORD. It is typical to apply a

practical minimum number of nails for bracket and beam stability, for members around 190 mm deep werecommend a minimum of 10/3.15x35 FH nails ie. 5/3.15 nails per tab.

Check Capacity


nQS * Eq. 4.1, NZS 3603

kn QknQ ..=

Eq. 4.2, NZS 3603

kn QknQ ...=

where:

10

6