Portal Frame Design Example

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30 metre Span LVL Portal Frame Design Carter Holt Harvey Woodproducts NZ Industrial Park, Auckland, New Zealand

Date: Sept. ‘08 Page: 1 / 92 Designed : C.R

LVL Portal Frame Design CHH Woodproducts New Zealand

Disclaimer This design example has been prepared solely to provide guidance and recommendations to suitably qualified engineers and other suitably qualified design professionals for diligent and professional use by them (and no other person) in the calculation of design solutions for LVL portal frame systems in accordance with currently available New Zealand Standards. To the best of Carter Holt Harvey’s knowledge and belief this example has been prepared in accordance with currently available technology and expertise 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, for producing the relevant design and it is assumed that the relevant designers will employ sound and current engineering knowledge and will take all reasonable 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 howsoever arising for any losses, damages, costs or expenses (whether direct, indirect and/or consequential) arising from any errors or omissions which may be contained in this example, nor does it accept responsibility to any persons whatsoever for designs prepared in reliance upon this example or any other information contained in this document. Carter Holt Harvey Limited

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Table of Contents 1.0

Introduction

2.0

Purlin design 2.1 Dead Load 2.2 Live load 2.3 Wind load 2.4 Proposed Purlin Layout 2.5 Connection Design 2.6 Lateral restraint design 2.7 Purlins supporting axial loading

3.0

Portal frame design 3.1 Proposed Portal Frame 3.2 Serviceability 3.3 Strength 3.4 Design Actions 3.5 Rafter Design 3.5.1 Combined bending and compression 3.5.2 Combined bending and tension 3.5.3 Flybrace design 3.6 Column Design 3.6.1 Combined bending and compression 3.6.2 Combined bending and tension 3.6.3 Flybrace design 3.7 Gusset Design 3.7.1 Knee Gusset Design 3.7.2 Ridge Gusset Design 3.7.3 Nail Ring Design 3.7.3.1 Knee Nail Ring Design 3.7.3.2 Ridge Nail Ring Design 3.8 Column to Footing Design

4.0

Girt Design, Side Wall 4.1 Wind Loading Connection Design 4.2

5.0

Mullion Design, Side Wall Wind Loading 5.1 5.2 Connection Design

6.0

Eaves Beam Design 6.1 Wind Loading 6.2 Connection Design

7.0

Girt Design, End Wall

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8.0

7.1 Wind Loading 7.2 Connection Design Mullion Design, End Wall 8.1 Wind Loading 8.2 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

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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 LVL and I-beam portal frame systems. The development of loading and the design of footings are not covered as part of this example as their nature is not specific to timber. The design example has been prepared assuming the building is proposed for Auckland, is within an Industrial Estate, and is subject to the following site information: Building Span Building length Building Clear Height Dominant openings Cladding Region Terrain Category Directional Multipliers

30.0 m 60.0 m, consisting of 6 x 10.0 m bays 6.0 m 6.0 x 6.0 m in one end and one side wall Pierce fixed sheeting of weight 6.0 kg/m2 A6, v500 = 45 m/s, v20 = 37 m/s 3 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. Carter Holt Harvey Limited

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2.0 Purlin Design Purlin Span Purlin Spacing

10,000-90 =

9910 mm 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 pine pulins remain cost effective for spans less than 6.0 m.

2.1 Dead load Assume roof sheeting mass of 6.0 kg/m2 plus a miscellaneous load of 1.0 kg/m2

7.0 × 1.6 × 9.81 + self_weight 1000 ∴ w*g = 0.17 kN/m w*g =

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 the consideration of creep effects. Table 2.5, NZS 3603:1993 demonstrates the relationship between duration of load and creep. The k2 factor is applied to elastic deflections. LVL products are considered dry at the time of supply and can be assumed to have a moisture content less than 18%. Refer Technical Note 82 for Section and Material Properties.

δT = k2(δbending + δshear )  5.w.l 4  5 × 0.17 × 9910 4 w.l 2  0.17 × 99102  δ = k2  + = 2 . 0 . +   9 6  384 × 2338 × 10 8 × 2.39 × 10   384.EI x 8.GAw  ∴ δG = 20.0mm or Span 495 Serviceability limits for timber purlins are the same as those applied to other building products. For long term deflection 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.

w*g = 0.25 × 1.6 = 0.40kN / m Serviceability


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 5 × 0.40 × 99104 0.40 × 99102  δQ = 1.0. + 9 6  384 × 2338 × 10 8 × 2.39 × 10  ∴ δG = 23.5mm or Span 421 Strength Based on respective k1 and load combination factors, combined dead and live load design actions will always be more critical for design than permanent loads where low roof masses (less than 20 kg/m2) are applied.

w1*.2G +1.5Q = 1.2 × 0.17 + 1.5 × 0.40 ∴ w1*.2G +1.5Q = 0.80kN / m

M 1*.2G +1.5Q = ∴ M 1*.2G +1.5Q

w.l 2 0.80 × 9.9 2 = 8 8 = 9.8kNm

Check Bending Capacity The bending capacity of an I-beam is based on the critical flange stresses due to bending. For composite timber Ibeams the bending moment capacity can be based on a lever arm action about the centroid of the flanges with one flange in tension and the other in compression for a single span application. The restraint offered to the compression flange is instrumental in the capacity of the I-beam. Further guidance on the bending moment capacities 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 the top flange of the purlin. Since compression edge is fully restrained k8=1.0. So for bending about XX axis Since k8>0.73

ØM bx = Ø.k1 . f t . A f .D1 × 10 −6 kNm

Refer Technical Note 82

where:

Ø = 0.9 k1 = 0.80 AF = 90 × 36 −

f t = 33MPa

(318 − 288) × 12

∴ AF = 3060mm

2

2

D1 = 360 − 36 = 324mm

ØM bx = 0.9 × 0.8 × 33 × 3060 × 324 × 10−6 kNm ∴ ØM bx = 23.6kNm > M * Carter Holt Harvey Limited

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Check Shear Capacity v1*.2G +1.5Q = ∴ v1*.2G +1.5Q

0.80 × 9.9 2 = 4.0kN

From table 14, Technical note 82

ØV = k1 .12.6 = 0.8 × 12.6 ∴ ØV = 10.1kN > v *

2.3 Wind loading θ = 0˚, Lateral wind critical (by inspection) a = min(0.2b, 0.2d, h) = 6.0m

w*i = qu .spacing .(ka .kl .c pe − c pi ) w1* = 0.84 × 1.6 × (1.0 × 1.5 ×− 0.9 − + 0.61) ∴ w1* = −2.63kN / m w2* = 0.84 × 1.6 × (1.0 × 1.0 ×− 0.9 − + 0.61) ∴ w2* = −2.02kN / m

Calculate weff Calculate Reactions

R* = − 2.02 × 1.955+ − 2.63 × 3.000 ∴ R* = − 11.84kN Calculate Moment −

2.63 × 3.0 2 − M = 11.84 × 4.955 − − 2.02 × 2.0 × 4.0 2 * ∴ M Wu = − 30.7 kNm * Wu

−

0.17 × 9.9 2 − + 30.67 8 = −28.8kNm

M 0*.9G +Wu = 0.9 × ∴ M 0*.9G +Wu Carter Holt Harvey Limited

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Calculate weff −

30.67 × 8 9 .9 2 * ∴ weff = − 2.50kN / m * weff =

For uplift

w0*.9G +Wu = 0.9 × 0.17 − 2.50 ∴ w0*.9G +Wu = − 2.35kN / m Serviceability To obtain the serviceability wind load the ultimate uniform loads can be factored by the square of the ratio serviceability wind speed to ultimate wind speed.

v ws =  s  vu

2

  × ws  2

 37  ws =   × − 2.50= − 1.69kN / m  45   5× − 1.69 × 9910 4 − 1.69 × 9910 2  + δ w = 1.0.  9 8 × 2.39 × ×10 6   384 × 2338 × 10 ∴ δ w = 99.0mm or Span 100 The acceptance of serviceability is at the engineer’s discretion. On the basis of applied local pressure factors and the 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, use Appendix C of NZS3603:1993 for stability calculations.

Check Capacity Calculate S1

1.1.EI x  S1 =    M E .y 

0.5

Eq. C1.1, NZS 3603

where:

EI x = 2338 × 109 Nmm 4 ME = ? Carter Holt Harvey Limited

y = 360 / 2 = 180mm

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 2

2   (EI y ) D + yo2  π  + GJ  4  Lay  ME = (2. yo + yh )

Eq. C7, NZS 3603

where:

yo = 360 / 2 = 180mm 9

EI y = 57.7 × 10 Nmm

yh = −360 / 2 = −180mm D = 360mm 2

GJ = 1848 × 106 Nmm 2

Lay = 9910 / 4 = 2478mm

(Restraint at quarter points) 2

 3602  π  6 57.7 × 10  + 180 2   + 1848 × 10 4 2478    ME = − 2 × 180+ 180 ∴ M E = 43.7 kNm

(

9

)

(

1.1. × 2338 × 109  ⇒ S1 =   6  43.7 × 10 × 180 

)

0.5

∴ S1 = 18.1

Calculate k8 Since 25>S>10

k8 = a1 + a2 .S + a3 .S 2 + a4 .S 3 k8 = 0.21 + 0.175 × 18.1+ − 0.0116 × 18.12 +

1 × 18.13 5000

∴ k8 = 0.76 Since k8>0.73

ØM bx = Ø.k1 . f t . A f .D1 × 10 −6 kNm

Refer Technical Note 82

where:

Ø = 0.9 k1 = 1.0 AF = 3060mm

f t = 33MPa

2

D1 = 324mm Carter Holt Harvey Limited

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ØM bx = 0.9 × 1.0 × 33 × 3060 × 324 × 10−6 kNm ∴ ØM bx = 29.4kNm > M * Note: Where k8 < 0.73 the moment capacity becomes a function of the compression flange buckling rather than the tension flange being critical. The moment capacity equation is altered to represent this where the characteristic tension stress is replaced by the product of the stability factor k8 and the characteristic compression stress. ie. ØM bx = Ø.k1 .k 8 . f c . A f .D1 × 10 −6 kNm Calculate shear and support reaction for wind load. Considering local pressure factors Case 1

w*i = qu .spacing .(ka .kl .c pe − c pi ) + 0.9.wg w1* = 0.84 × 1.6 × (1.0 × 1.5 ×− 0.9 − + 0.61) + 0.9 × 0.17 ∴ w1* = −2.48kN / m w2* = 0.84 × 1.6 × (1.0 × 1.0 ×− 0.9 − + 0.61) + 0.9 × 0.17 ∴ w2* = −1.88kN / m R0*.9G +Wu =

 1  3 .9 2 − 1 . 88 × + −2.48 × 6.0 × 6.9  9 .9  2 

∴ R0*.9G +Wu = −11.81kN Case 2

w*i = qu .spacing .(ka .kl .c pe − c pi ) + 0.9.wg w1* = 0.84 × 1.6 × (1.0 × 2 ×− 0.9 − + 0.61) + 0.9 × 0.17 ∴ w1* = −3.09kN / m w2* = 0.84 × 1.6 × (1.0 × 1.0 ×− 0.9 − + 0.61) + 0.9 × 0.17 ∴ w2* = −1.88kN / m R

* 0.9 G +Wu

 1  6 .9 2 = + −3.09 × 3.0 × 8.4 − 1.88 × 9 .9  2 

∴ R0*.9G +Wu = −12.2kN Calculate dead & live load combined actions

R* =

0 .8 × 9 .9 = 4.0kN 2


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Timber capacity is dependant on the duration of the load in question, this must be taken into account in the determination of the critical load case. One method of assessing the critical design load is to remove the duration of load factor,k1, from the capacity equation and divide the load action effect by k1, * RMax  − 12.2 4.0  = max ,  k1  1 .0 0 .8  * Rmax ∴ = −12.2kN k1

Check shear capacity Since k1 was taken into account in the calculation of design action, apply k1=1.0

ØV = k1.12.6 = 1.0 × 12.6 ∴ ØV = 12.6kN > v*

Table 14, Technical note 82

Therefore the HJ360 63 hyJOIST is suitable for use as a purlin based on the implied loading at a 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 purlin and the hySPAN rafter are maintained. Connection to the hyJOIST by nailing through the plywood web provides the 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 through plywood allows for nailing close to the end/edge of the plywood. Packing out the web and using proprietary joist hangers can also provide a suitable connection however the cost of the packing, brackets and labour involved can make this an expensive alternative.


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Purlin connection blocks, or seating blocks as they are sometimes called, have been used in a number of design situations for connection of C or I beam purlins where the connection block is either screwed or nailed to the rafter and the web of the composite purlin is connected directly to the connection block. A purlin connection block is proposed for connection using Ø2.87 diameter nails through the plywood web and 14g type 17 screws through the connection block to the rafter. Target the connection for design shear capacity, ØVps of the purlin. Note: The selection of a suitable purlin connection block needs to take into account the end and edge distances of the fasteners as well as the spacing along and across the grain. The use of 4 x-banded connection block reduces the 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 to the face of the connection block.

Calculate minimum number of Ø2.87 FH nails Joint Group

J5


S * ≤ ØQk

Eq. 4.1,

NZS 3603

Qn = n.k .Q k

Eq. 4.2,

NZS 3603

ØQn = Ø.n.k .Qk where:

Ø = 0.8 k1 = 1.0 Qk = 0.526kN k=1.4 since nails are through plywood with flat head nails. k=1.1 since we are proposing 20 nails per connection (linear interpolation between 1.3 for 50 nails and 1.0 for 4 nails)

NZS 3603

Cl. 4.2.2.2(g)

NZS 3603

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

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

12.6 × 1.0 = 0.8 × n × 1.0 × 1.4 × 1.1 × 0.526 ∴ n = 19.4 Say 20/50xØ2.87 FH nails, nailed through plywood web into purlin connection block Calculate minimum number of 14g type 17 Hex Head screws Type 17 screws are preferred for timber connection as they are a self drilling screws through the timber.

Joint Group

J4


S * ≤ ØQn

Eq. 4.5,

NZS 3603

Qn = n.k .Qk

Eq. 4.6,

NZS 3603

ØQn = Ø.n.k .Qk Carter Holt Harvey Limited

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where:

Ø = 0.8 * k1 = 1.0 Qk = 3.303kN

NZS 3603

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

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

12.6 × 1.0 = 0.8 × n × 1.0 × 3.303 ∴ n = 4.76 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 from moving independently of each other. Many systems are appropriate but may require the fabrication of special components. One of the most effective systems is to use hyJOIST pieces together with a hyCHORD bottom flange restraint and continuous mild steel galvanised strap over the top, as shown below.


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Calculate force on lateral restraint *

FA = k33 .k34 .k35 .

0.05M A d (nr + 1)

Eq. B9,

NZS 3603

where:

k33 = 1.0

(Wind loading)

k34 = 0.4

 m +1   22 + 1  k35 = min ,5  = min ,5  = 5  2   2  M A = 28.79kNm d = 360mm nr = 3 FA = 1.0 × 0.4 × 5 ×

0.05 × 28.79 × 106 360(3 + 1)

∴ FA = 2.0kN Check capacity of lateral restraint – propose 90x45 hyCHORD

N c* = N t* = 2.0kN Typically a 45 mm thick section is recommended to allow for a 75mm long screw through both the lateral restraint and into the flange of the hyJOIST. Using hyCHORD for the lateral restraint is a good choice given its high strength and lower cost.

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

N c* ≤ ØN ncx and

N c* ≤ ØN ncy

Eq. 3.17,

NZS 3603

Eq. 3.19,

NZS 3603

Minor axis buckling is critical by inspection

N ncy = k1.k8 . f c . A ∴ ØN ncy = Ø.k1.k8 . f c . A where:

Ø = 0.9

f c = 45MPa

A = 90 × 45 = 4050mm

2

ØN ncx = 0.9 × k1 × k 8 × 45 × 4050 Carter Holt Harvey Limited

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∴ ØN ncx = 164.0.k1 .k 8 .kN Calculate k8 for buckling about the minor axis

S3 =

k10 .L b

or

Lay

Eq. 3.15,

NZS 3603

N t* ≤ Ø.N nt

Eq. 3.20

NZS 3603

N nt = k1.k4 . f t . A

Eq. 3.21

NZS 3603

b

whichever is less

1600 45 ∴ S3 = 35.6 S3 =

Since 25>S3>10

k8 = a5 .S a6 k8 = 235.5 × 35.6 −1.937 ∴ k8 = 0.23 Since k1 = 1.0

∴ ØN ncx = 38.3kN > N c* Consider tension strength

∴ ØN nt = Ø.k1.k4 . f t . A where:

Ø = 0 .9

f t = 33MPa k4 = 1.0

Technical Note 82

A = 90 × 45 = 4050mm 2 k1 = 1.0 ØN nt = 0.9 × 1.0 × 1.0 × 33 × 4050 ∴ ØN nt = 120.3kN 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 Carter Holt Harvey Limited

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Joint Group

J4


S * ≤ ØQn

Eq. 4.5,

NZS 3603

Qn = n.k .Qk

Eq. 4.6,

NZS 3603


Ø = 0.8 * k1 = 1.0 Qk = 3.303kN

NZS 3603


Consider Qk reduction 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

Cl. 4.3.2(e),

NZS 3603

Therefore portion of diameter in penetration = 4.76 Calculate reduction from capacity relating to 7 da Reduction factor =

4.76 = 0.68 7

∴ Qk = 0.68 × 3.303 = 2247 N So:

2.0 × 1.0 = 0.8 × n × 1.0 × 2.247 ∴ n = 1.11 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. These forces 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. Elastic structural analysis of a timber portal frame differs little from that applied to steel members except for the different section and material properties. For solid timber a five percent allowance for shear deflection is included in the average modulus of elasticity which removes any need for the separate consideration of shear deflection. To achieve portal frame action rigid connections need to be made at both the ridge and eave. One of the most efficient methods of providing rigid connections is via use of nailed plywood gussets. The additional 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 of cladding 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 a duration of load factor of 1, hence using the elastic deflection directly from analysis packages. For long term loads the effects of creep need to be taken into account. NZS 3603 Table 2 defines k2 as 2.0 for loading of twelve months or more where the moisture content is less than 18%. Serviceability – 900x90 hySPAN portal frame Load Case

k2

Dead load* Live load Wind loading Lateral wind1 Lateral wind2 Longitudinal wind1 Longitudinal wind2

2.0 1.0


1.0 1.0 1.0 1.0

Deflection Vertical Horizontal 96.2 mm or span/302 16.2 mm or height/396 75.5 mm or span/385 9.6 mm or height/668 134.7 mm or 74.5 mm or 108.5 mm or 64.6 mm or

span/216 span/390 span/268 span/450

28.4 mm or 15.7 mm or 13.5 mm or 8.1 mm or

height/225 height/408 height/475 height/792 September 2008

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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 the interaction of gussets provide specific locations for the selection of critical design actions for the design of rafters, columns gussets and nail rings. Hutchings and Bier [2000] provide guidance on the design moment locations as shown below.

Location A – Rafter design actions at knee Location B – Column design actions Location C – Knee gusset design actions Location D – Gusset to rafter at knee connection actions Location E – Gusset to column connection actions Location F – Ridge gusset design actions Location 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 the gusseted 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 for capacity, hence affecting the determination of critical load case. As with steel portal frames the bending moment diagram 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 been dismissed by inspection. The point of contraflexure is within close proximity for each case meaning that the critical load case can be determined by inspection. Critical Design Actions M* Load Case 1.2G+1.5Q 0.9G+Wu - Lat 1.2G+Wu - Lat Carter Holt Harvey Limited

k1 0.8 1.0 1.0

kN -240.0 271.0 -276.0

Column N* kN -84.1 101.0 -113.0

V* kN 71.5 55.4 62.3

M* kN -268.0 293.0 -307.0

Rafter N* kN -60.4 67.3 -79.2

V* kN 50.5 87.6 -95.0 September 2008

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k1 factored Design Actions Column M*/k1 Load Case 1.2G+1.5Q 0.9G+Wu - Lat 1.2G+Wu - Lat

k1 0.8 1.0 1.0

N*/k1

kN -300.0 271.0 -276.0

kN -105.1 101.0 -113.0

Rafter V*/k1 kN 89.4 55.4 62.3

M*/k1

N*/k1

kN -335.0 293.0 -307.0

kN -75.5 67.3 -79.2

V*/k1 kN 63.1 87.6 -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 and buckling 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

 M x*   N c*    +   ≤ 1.0 ØM ØN nx  ncx    2 *  M x   N c*   ≤ 1 .0   +   ØN  ØM nx  ncy   

Eq. 3.23

NZS 3603

Eq. 3.24

NZS 3603

M * ≤ ØM n

Eq. 3.3

NZS 3603

M n = k1.k 4 .k5 .k8 . f b .Z

Eq. 3.4

NZS 3603

Critical Design Actions Critical load case - 1.2G+1.5Q M* = -268.0 kNm

N c* = -60.4 kN

V* = 50.5 kN

Consider Bending Moment Capacity

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

ØM n = Ø.k1.k4 .k5 .k8 .k11. f b .Z

where:

Ø = 0 .9 k 4 = k 5 = 1 .0


f b = 48MPa

Technical Note 82

September 2008

Project: For: At:



 300  k11 =    d 

0.167

 300  ∴ k11 =    900 

Cl. 2.4.6

0.167

AS1720.1

= 0.83

d 2 .b 900 2 × 90 = 6 6 ∴ Z = 12.15 × 106 mm3 Z=

ØM n = 0.9 × k1 × 1.0 × 1.0 × k 8 × 0.83 × 48 × 12.15 × 10 6 ∴ ØM n = 435.65.k1 .k 8 kNm Since k1=0.8

ØM n = 348.52.k 8 kNm Calculate k8 The 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 tension edge continuously restrained (Cl 3.2.5.3). Typically these can be useful in the calculation of slenderness of simple beams and secondary framing however composite sections and members within structural frames require analysis using Appendix C of NZS3603:1993 for slenderness calculations.

Consider slenderness equation

1.1.EI x  S1 =    M E.y 

0.5

Eq. C1

NZS 3603

Since for 900x90 hySPAN

EI x = 13200 × y=

900 3 × 90 = 72.17 × 1012 Nmm 4 12

900 = 450mm 2

Therefore:

176.418 × 109  S1 =   ME  

0.5

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 restraint offered to the tension edge by purlins and girts, such theories are beyond the scope of this example. Carter Holt Harvey Limited

September 2008

Project: For: At:



Consider Moment Diagram

 c  M E =  5  (EI )y .GJ L   ay 

[

]

0.5

Eq. C3

NZS3603

where:

β=

0 =0 268

β = ratio of bending moments between buckling restraints

c5 = 5.5 EI y = 13200 ×

Table C1

NZS3603

Eq. C2

NZS 3603

90 3 × 900 = 721.71 × 10 9 Nmm 4 12

Since for rectangular sections:

B  D × B3  J = 1 − 0.63 ×  × D 3  90  900 × 90 3  GJ = 660 × 1 − 0.63 × = 135.25 × 10 9 Nmm 2 × 900  3  Therefore:

 5 .5  9 9 ME =   721.71 × 10 × 135.25 × 10  4900  ∴ M E = 350.68kNm

[

]

0.5

From previous:


September 2008

Project: For: At:



176.418 × 109  S1 =  6   350.68 × 10 

0.5

∴ S1 = 22.43 Since 25>S1>10

k8 = a1 + a2 .S + a3 .S 2 + a4 .S 3 k8 = 0.21 + 0.175 × 22.43+ − 0.0116 × 22.432 +

1 × 22.433 5000

∴ k8 = 0.56 ∴ ØM n = 348.52 × 0.56 = 195.2kNm ØMn
Calculate Euler moment, ME

 c  M E =  5  (EI )y .GJ L   ay 

[

]

0.5

Eq. C3

NZS 3603

where:

β=

171.1 = 0.64 268.0

c5 = 3.82

β = ratio of bending moments between buckling restraints Table C1

NZS 3603

Therefore:


September 2008

Project: For: At:



 3.82  9 9 ME =   721.71 × 10 × 135.25 × 10  1741  ∴ M E = 685.51kNm

[

]

0.5

From previous:

176.418 × 109  S1 =  6   685.51 × 10 

0.5

∴ S1 = 16.04 Since 25>S1>10

k 8 = a1 + a 2 .S + a 3 .S 2 + a 4 .S 3

Cl C2.10

k 8 = 0.21 + 0.175 × 16.04+ − 0.0116 × 16.04 2 +

NZS 3603

1 × 16.04 3 5000

∴ k 8 = 0.86 ∴ ØM n = 348.52 × 0.86 = 299.7 kNm > M * Check remaining unrestrained section

M* = 171.1 kNm

Lay = 3160 mm c5 = 5.5

Therefore:

 5 .5  9 9 ME =   721.71 × 10 × 135.25 × 10  3160  ∴ M E = 543.78kNm

[

]

0.5

From previous:

176.418 × 109  S1 =  6   543.78 × 10 

0.5

∴ S1 = 18.01 Since 25>S1>10

k 8 = a1 + a 2 .S + a 3 .S 2 + a 4 .S 3 k 8 = 0.21 + 0.175 × 18.01+ − 0.0116 × 18.012 +

NZS 3603

Cl C2.10

1 × 18.013 5000

∴ k 8 = 0.77


September 2008

Project: For: At:



∴ ØM n = 348.52 × 0.77 = 268.4kNm > M * 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 ratio between purlins = 0 (conservative)). Calculate Euler Moment

 3 .1  9 9 ME =   721.71 × 10 × 135.25 × 10  1600  ∴ M E = 605.33kNm

[

]

0.5

From previous:

176.418 × 109  S1 =  6   605.33 × 10 

0.5

∴ S1 = 17.07 Since 25>S1>10

k 8 = a1 + a 2 .S + a 3 .S 2 + a 4 .S 3 k 8 = 0.21 + 0.175 × 17.07 + − 0.0116 × 17.07 2 +

Cl C2.10

NZS 3603

1 × 17.07 3 5000

∴ k 8 = 0.81 ∴ ØM n = 348.52 × 0.81 = 282.3kNm > M * Consider column action Major axis buckling XX Carter Holt Harvey Limited

September 2008

Project: For: At:



N ncx = k1.k8 . f c . A

Eq. 3.18

NZS 3603

∴ ØN ncx = Ø.k1.k8 . f c . A where:

Ø = 0.9

f c = 45MPa

A = 900 × 90 = 81000mm

2

ØN ncx = 0.9 × k1 × .k8 × 45 × 81000 ∴ ØN ncx = 3280.5.k1.k8 .kN Calculate k8 for buckling about the major axis L=Lax=14221 mm (rafter length from ridge to column)

S2 =

k10 .L d

Lax whichever is less d

or

k10 = 1.0

NZS 3603

Eq. 3.14

(Conservative)

1.0 × 14221 900 ∴ S 2 = 15.80 S2 =

Since 25>S2>10

k 8 = a1 + a 2 .S + a 3 .S 2 + a 4 .S 3 k 8 = 0.21 + 0.175 × 15.80+ − 0.0116 × 15.80 2 +

NZS 3603

Cl C2.10

1 × 15.80 3 5000

∴ k 8 = 0.87 Since k1 = 0.8

∴ ØN ncx = 2278.1kN > N c* Minor axis buckling YY From previous:

∴ ØN ncx = 3280.5.k1.k8 .kN Calculate k8 for buckling about the minor axis YY Lay=1600 mm (purlin spacing) Carter Holt Harvey Limited

September 2008

Project: For: At:



S3 =

k10 .L b

or

Lay b

whichever is less

Eq. 3.15

NZS 3603

Cl C2.10

NZS 3603

Eq. 3.23

NZS 3603

Eq. 3.24

NZS 3603

Eq. 3.25

NZS 3603

1600 90 ∴ S3 = 17.78 S3 =

Since 25>S3>10

k 8 = a1 + a 2 .S + a 3 .S 2 + a 4 .S 3 k 8 = 0.21 + 0.175 × 17.78+ − 0.0116 × 17.78 2 +

1 × 17.78 3 5000

∴ k 8 = 0.78 Since k1 = 0.8

∴ ØN ncx = 2047.03kN > N c* Combined actions

 268.0   60.4   +  = 0.92 ≤ 1.0  299.7   2278.1  2

 268.0   60.4    +  = 0.83 ≤ 1.0  299.7   2047.0  3.5.2 Combined bending and tension Design Criteria

 N t*   M *   ≤ 1 .0   +     ØN nt   ØM n  Critical Design Actions Critical load case - 0.9G+Wu Lateral wind M* = 293.0 kNm (at eave) M* = -171.8 kNm (along rafter) N t* = 69.9 kN V* = 87.6 kN Consider Bending Moment Capacity From previous:

ØM n = 435.65.k1.k8 kNm Since k =1.0, wind gust

1 Carter Holt Harvey Limited

September 2008

Project: For: At:



ØM n = 435.65.k8kNm Calculate k8 Calculate Euler moment, ME Consider compression edge restrained by purlins at 1600 c/c until point of contraflexure. Bending Moment Diagram

 c  M E =  5  (EI )y .GJ L   ay 

[

]

0.5

Eq. C3

NZS 3603

where:

β=

176.2 = 0.60 293.2

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

c5 = 3.9

Eq. C3

NZS 3603

EI y = 721.71 × 109 Nmm 4 GJ = 135.25 × 109 Nmm 2 Therefore:

 3.90  9 9 ME =   721.71 × 10 × 135.25 × 10  1600  ∴ M E = 761.54kNm

[

]

0.5

From previous:


September 2008

Project: For: At:



176.418 × 10 9  S1 =  6   761.54 × 10 

0.5

∴ S1 = 15.22 Since 25>S1>10

k 8 = a1 + a 2 .S + a 3 .S 2 + a 4 .S 3 k 8 = 0.21 + 0.175 × 15.22+ − 0.0116 × 15.22 2 +

NZS 3603

Cl C2.10

1 × 15.22 3 5000

∴ k 8 = 0.90 ∴ ØM n = 435.65 × 0.90 = 392.1kNm > M * 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.


September 2008

Project: For: At:



Bending Moment Diagram

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 Region 2 Region 3

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

 c5   L   ay 

Since: M E = function

Therefore Region 2 is critical buckling region

 3 .1  9 9 ⇒ ME =   721.71 × 10 × 135.25 × 10  3200  ∴ M E = 302.66kNm

[

]

0.5

From previous:

176.418 × 109  S1 =  6   302.66 × 10 

0.5

∴ S1 = 24.14 Since 25>S1>10

k 8 = a1 + a 2 .S + a 3 .S 2 + a 4 .S 3 k 8 = 0.21 + 0.175 × 24.14+ − 0.0116 × 24.14 2 +

Cl C2.10

NZS 3603

1 × 24.14 3 5000

∴ k 8 = 0.49 Carter Holt Harvey Limited

September 2008

Project: For: At:



∴ ØM n = 435.65 × 0.49 = 213.5kNm > M * Consider tension strength

N t* ≤ Ø.N nt

Eq. 3.20

NZS 3603

N nt = k1.k4 . f t . A

Eq. 3.21

NZS 3603

For solid sections with member depths greater than 150 mm, apply k11 size factor for tension. For further information refer AS1720.1 (Clause 2.4.6) or Technical Note 82.

ØN nt = Ø.k1.k4 .k11. f t . A

Therefore where:

Ø = 0.9 f t = 33MPa k4 = 1.0 A = 900 × 90 = 81000mm k1 = 1.0  150  k11 =    d 

Technical Note 82

0.167

Cl. 2.4.6

AS1720.1

q. 3.25

NZS 3603

V * ≤ ØVn

Eq. 3.3

NZS 3603

Vn = k1.k4 .k5 . f s . AS

Eq. 3.4

NZS 3603

 150  ∴ k11 =    900 

0.167

= 0.74

ØN nt = 0.9 × 1.0 × 1.0 × 0.74 × 33 × 81000 ∴ ØN nt = 1780.2kN Combined actions

 69.9   171.8   +  = 0.84 ≤ 1.0  1780.2   213.5  Calculate Shear Capacity

where:

Ø = 0 .9

k1 = 1.0

k 4 = k 5 = 1 .0

f s = 5.3MPa


Technical Note 82

September 2008

Project: For: At:



AS = 2.b.d / 3 ∴ AS =

2 × 900 × 90 = 54000mm 2 3

Cl 3.2.3.1

NZS 3603

∴ φVS = 0.9 × 1.0 × 1.0 × 5.3 × 54000 ∴ φVS = 257.6kN > V * 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 *

FA = k33 .k34 .k35 .

0.05M A d (nr + 1)

Eq. B9, NZS 3603

where:

k33 = 1.0

(Dead and live loads)

k34 = 0.4  m +1  1+1  ,5  = min ,5  = 1 k 35 = min  2   2  M A = −171.1kNm d = 900mm nr = 1 0.05 × 171.1 × 10 6 F A = 1 .0 × 0 .4 × 1 × 900(1 + 1) ∴ FA = 1.9kN Note: FA is the horizontal force and is shared between two components, one in tension and one in compression. Check capacity of flybrace – propose 90x45 hyCHORD

N c* = N t* = 1.90kN Calculate force in brace

N c* = N t* =


1.90 = 2.7 kN Cos (45)

September 2008

Project: For: At:



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

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

N c* ≤ ØN ncx and

N c* ≤ ØN ncy

Eq. 3.17

NZS 3603

Eq. 3.19

NZS 3603

Eq. 3.15

NZS 3603

Cl C2.10

NZS 3603

Minor axis buckling is critical by inspection

N ncy = k1.k8 . f c . A ∴ ØN ncy = Ø.k1.k8 . f c . A where:

Ø = 0.9

f c = 45MPa

A = 90 × 45 = 4050mm

2

ØN ncx = 0.9 × k1 × k 8 × 45 × 4050 ∴ ØN ncx = 164.03.k1 .k 8 .kN Calculate k8 for buckling about the minor axis

S3 =

k10 .L b

or

Lay b

whichever is less

764 45 ∴ S 3 = 16.98

S3 =

Since 25>S1>10

k 8 = a1 + a 2 .S + a 3 .S 2 + a 4 .S 3 k 8 = 0.21 + 0.175 × 16.98+ − 0.0116 × 16.98 2 +

1 × 16.98 3 5000

∴ k 8 = 0.82 Since k1 = 1.0

∴ ØN ncx = 134.50kN > N c* Carter Holt Harvey Limited

September 2008

Project: For: At:



Consider tension strength

N t* ≤ Ø.N nt

Eq. 3.20

NZS 3603

N nt = k1.k4 . f t . A

Eq. 3.21

NZS 3603

∴ ØN nt = Ø.k1.k4 . f t . A where:

Ø = 0.9 f t = 33MPa k4 = 1.0 A = 90 × 45 = 4050mm k1 = 1.0

Technical Note 82

ØN nt = 0.9 × 1.0 × 1.0 × 33 × 4050 ∴ ØN nt = 120.3kN Consider connection between purlins and rafters and flybrace Screws are required to provide tension connection to rafter/purlin

Calculate minimum number of 14g type 17 Hex Head screws Joint Group

J4


S * ≤ ØQn

Eq. 4.5,

NZS 3603

Qn = n.k .Qk

Eq. 4.6,

NZS 3603


Ø = 0.8 * k1 = 1.0 Qk = 3.303kN

NZS 3603


Consider Qk reduction 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

Cl. 4.3.2(e),

NZS 3603

Therefore portion of diameter in penetration = 4.76 Calculate reduction from capacity relating to 7 da Carter Holt Harvey Limited

September 2008

Project: For: At:



Reduction factor =

4.76 = 0.68 7

∴ Qk = 0.68 × 3.303 = 2247 N So:

ØQn = 0.8 × 1.0 × 2 × 2.247 ∴ ØQn = 4.0kN Consider screws in tension

N * ≤ ØQn

Eq. 4.8,

NZS 3603

Qn = n.k . p.Qk

Eq. 4.9,

NZS 3603

ØQn = Ø.n.k . p.Qk where:

Ø = 0.8 * k1 = 1.0 Qk = 79.5 N / mm

p = 35mm

NZS 3603


ØQn = 0.8 × 2 × 1.0 × 35 × 79.5 ∴ ØQn = 4.45kN Say 2/75x14g type 17 Hex Head screws, screwed through pre-drilled holes in flybrace into rafter and purlin. Proposed flybrace connection


September 2008

Project: For: At:



3.6 Column Design 3.6.1 Combined bending and compression Design Criteria

 M x*   N c*    +   ≤ 1.0  ØM nx   ØN ncx  2  M x*   N c*   ≤ 1 .0   +     ØM nx   ØN ncy 

Eq. 3.23

NZS 3603

Eq. 3.24

NZS 3603

Critical Design Actions Critical load case - 1.2G+1.5Q N c* = -84.1 kN

M* = -240.0 kNm

V* = 71.5 kN

Consider Bending Moment Capacity From previous:

ØM n = 435.65.k1.k8 kNm Since k1=0.8

ØM n = 348.52.k8kNm Calculate k8 For 900x90 hySPAN:

176.418 × 109  S1 =   ME  

0.5

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


September 2008

Project: For: At:



Consider bending moment diagram

 c  M E =  5  (EI )y .GJ L   ay 

[

]

0.5

Eq. C3

NZS 3603

where: Region 1

β=

− 179.3 = 0.59 − 303.0


c5 = 3.92

Eq. C3

NZS 3603

Lay = 2530mm Region 2

β=

0 =0 − 179.3


c5 = 5.5

Eq. C3

NZS 3603

Lay = 3470mm

EI y = 721.71 × 109 Nmm 4 GJ = 135.25 × 109 Nmm 2 Therefore Region 1 is critical:

 3.92  9 9 ME =   721.71 × 10 × 135.25 × 10  2530  ∴ M E = 484.08kNm

[

]

0.5

From previous: Carter Holt Harvey Limited

September 2008

Project: For: At:



176.418 × 10 9  S1 =  6   484.08 × 10 

0.5

∴ S1 = 19.09 Since 25>S1>10

k 8 = a1 + a 2 .S + a 3 .S 2 + a 4 .S 3 k 8 = 0.21 + 0.175 × 19.09+ − 0.0116 × 19.09 2 +

Cl C2.10

NZS 3603

Eq. 3.14

NZS 3603

Fig. 3.5

NZS 3603

1 × 19.09 3 5000

∴ k 8 = 0.71 ∴ ØM n = 348.52 × 0.71 = 247.5kNm Consider column action Major axis buckling XX From previous:

∴ ØN ncx = 3280.5.k1.k8 .kN Calculate k8 for buckling about the major axis L=Lax=6000 mm (column height from rafter to footing)

S2 =

k10 .L d

Lax whichever is less d

or

k10 = 1.0

(conservative)

1.0 × 6000 900 ∴ S 2 = 6.67

S2 =

Since 10

∴ ØN ncx = 3280.5.k1.k8 .kN Carter Holt Harvey Limited

September 2008

Project: For: At:



Calculate k8 for buckling about the minor axis YY Lay=1660 mm (girt spacing)

S3 =

k10 .L b

or

Lay b

whichever is less

Eq. 3.15

NZS 3603

Cl C2.10

NZS 3603

Eq. 3.23

NZS 3603

Eq. 3.24

NZS 3603

Eq. 3.25

NZS 3603

1660 90 ∴ S3 = 18.44

S3 =

Since 25>S3>10

k 8 = a1 + a 2 .S + a 3 .S 2 + a 4 .S 3 k 8 = 0.21 + 0.175 × 18.44+ − 0.0116 × 18.44 2 +

1 × 18.44 3 5000

∴ k 8 = 0.75 Since k1 = 0.8

∴ ØN ncx = 1968.3kN > N c*

Combined actions

 240.0   84.1   +  = 1 .0 ≤ 1 .0  247.5   2624.4  2

 240.0   84.1    +  = 0.98 ≤ 1.0  247.5   1968.3  3.6.2 Combined bending and tension Design Criteria

 N t*   M *   ≤ 1 .0   +   ØM  ØN nt  n    Critical Design Actions Critical load case - 0.9G+Wu Lateral wind M* = 271.0 kNm

N t* = 101.0 kN

V* = 55.4 kN

Consider Bending Moment Capacity From previous, since k1=1.0, wind gust Carter Holt Harvey Limited

September 2008

Project: For: At:



ØM n = 435.65.k8kNm Calculate k8 Calculate Euler moment, ME Consider compression edge restrained by grits at 1660 c/c. Bending Moment Diagram

 c  M E =  5  (EI )y .GJ L   ay 

[

]

0.5

NZS3603

Eq. C3

where:

β=

177.7 = 0.66 271.0

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

c5 = 3.78

NZS3603

Eq. C3

EI y = 721.71 × 109 Nmm 4 GJ = 135.25 × 109 Nmm 2 Therefore:

 3.78  9 9 ME =   721.71 × 10 × 135.25 × 10  1660  ∴ M E = 711.43kNm

[


]

0.5

September 2008

Project: For: At:



From previous:

176.418 × 109  S1 =  6   711.43 × 10 

0.5

∴ S1 = 15.75 Since 25>S1>10

k 8 = a1 + a 2 .S + a 3 .S 2 + a 4 .S 3 k 8 = 0.21 + 0.175 × 15.75+ − 0.0116 × 15.75 2 +

NZS 3603

Cl C2.10

1 × 15.75 3 5000

∴ k 8 = 0.87

∴ ØM n = 435.65 × 0.87 = 379.0kNm > M * Consider tension strength Since k1 = 1.0 , k11 = 0.74

ØN nt = 0.9 × 1.0 × 1.0 × 0.74 × 33 × 81000 ∴ ØN nt = 1780.2kN Combined actions

 101.0   271.0   +  = 0.77 ≤ 1.0  1780.2   379.0 

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


September 2008

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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 connections because the x-band plies help reduce the tendency of the long band plies to split. This allows the nail spacing to be governed by the grain direction of the rafter or column which ever the gusset is being fastened to. Plywood is available in Stress Grade F11 from Carter Holt Harvey in thicknesses up to and including 25 mm. 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 k1 for comparison in the determination of the critical design action. Gusset Design Actions

Load Case 1.35G 1.2G+1.5Q 0.9G+Wu – Lat 1.2G+Wu –Lat 0.9G+Wu – Long 1.2G+Wu –Long

M* kNm -123.0 -324.0 362.0 -382.0 239.1 -295.4

K1 0.6 0.8 1.0 1.0 1.0 1.0

Knee N* kN -31.6 -83.1 102.2 -111.7 64.6 -60.3

V* kN 19.2 50.5 -54.3 -65.2 55.1 -69.7

M* kNm 69.7 183.6 -156.9 171.2 -117.0 161.0

Ridge N* kN -19.0 -50.1 71.1 -75.5 65.9 -56.6

V* kN 2.5 6.6 -5.0 14.5 25.3 7.5

k1 factored Gusset Design Actions Knee M*/k1 Load Case 1.35G 1.2G+1.5Q 0.9G+Wu – Lat 1.2G+Wu –Lat 0.9G+Wu – Long 1.2G+Wu –Long

K1 0.6 0.8 1.0 1.0 1.0 1.0

kNm -205.0 -405.0 362.0 -382.0 239.1 -295.4

N*/k1 kN -52.6 -103.9 102.2 -111.7 64.6 -60.3

Ridge V*/k1 kN 32.0 63.1 -54.3 -65.2 55.1 -69.7

M*/k1 kNm 116.2 229.5 -156.9 171.2 -117.0 161.0

N*/k1 kN -31.7 -62.6 71.1 -75.5 65.9 -56.6

V*/k1 kN 4.2 8.3 -5.0 14.5 25.3 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 a horizontal line across the centroid of the rafter and column intersection as shown below.


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Geometrically, assuming the rafter depth and column depth are equal, the critical section for the knee connection may be calculated by:

Depthcs = D +

L−D D  1 + 1 −  tan θ  2L 

Design Criteria 2

 N c*   M i*   Vi*    +   +   ≤ 1.0  ØN nc   ØM ni   ØVni 

 N t*   ØN nt

  M i*  +    ØM ni

2

  Vi*  +    ØVni

  ≤ 1.0 

Eq. 6.17

NZS 3603

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 in many cases be omitted from calculation such is their effect on sizing. Compression loads are generally past through in bearing and not required for consideration in gusset design.

Critical Design Actions 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] proposes a bilinear stress distribution along the critical section while Hutchings [1987] methodology assumes a triangulated 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 both opening and closing moments of portal frames, and has been used on many portal frame structures. Milner and Crosier [2000] propose a similar calculation based on a triangulated stress distribution but propose an alternate critical section and omit the use of the size factor.


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M i* ≤ ØM ni M ni = k1 .k 8 .k14 .k15 . f pb .

t e .d 6

Eq. 6.9

NZS 3603

Eq. 6.10

NZS 3603

2

Now include size factor - for further information on size factor, k11 refer AS1720.1 (Clause 2.4.6) or Technical Note 82.

ØM ni = Ø.k1.k8 .k11.k14 .k15 . f pb .

Therefore

te .d 2 6

Since the gussets are in pairs:

 t .d 2  ØM ni = 2.Ø.k1.k8 .k11.k14 .k15 . f pb . e  6   Propose 42 mm 4 x-band LVL, where:

Ø = 0 .9 k1 = ? k8 = 1.0

(localised, gusset edges are restrained by gusset stiffeners)

k14 = 1.0 k15 = 1.0

(moisture content < 18%) (only parallel plies are being considered) 0.167

 300  k11 =    d  f b = 48MPa

Cl. 2.4.6

AS1720.1

t e = (42 − (4 × 3.5)) = 28mm

 28 × d 2  ØM ni = 2.0.9 × k1 × 1.0 × 1.0 × 1.0 × k11 × 48 ×  6   ØM n = 403.2 × k1 .k11 .d 2 kNm For 900x90 hySPAN portal frame with 7.5˚ pitch

1200 − 900 900   1 + 1 −  tan (7.5)  2 × 1200  ∴ d = 1177.2mm d = 900 +

 300  k11 =    1177.2  ∴ k11 = 0.80

0.167

Calculate bending moment capacity


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ØM ni = 403.2 × k1 × 0.8 × 1177.2 2 ∴ ØM ni = 447.0 × k1 .kNm Calculate Shear Capacity

V p* ≤ ØVni

Eq. 6.15

NZS 3603

2 Vni = .k1.k8 .k14 .k15 .k18 . f ps .t.d 3

Eq. 6.16

NZS 3603

where:

Ø = 0 .9

f ps = 5.3MPa

Technical Note 82

k1 = ? k14 = 1.0 (moisture content < 15 %) k15 = 1.0 (face grain = 0˚) 2   Vni = 2 0.9 × × k1 × 1.0 × 1.0 × 1.0 × 1.0 × 5.3 × 42 × d  3   ∴Vni = 267.12 × k1 .d .kN Since d = 1177.2mm

∴Vni = 314.45 × k1 .kN Consider tension capacity

N t* ≤ Ø.N nt

Eq. 6.11

NZS 3603

N nt = k1 .k14 .k15 . f pt .t t .d

Eq. 6.12

NZS 3603

∴ ØN nt = Ø.k1 .k14 .k15 . f pt .t e .d where:

Ø = 0 .9

f pt = 33MPa

Technical Note 82

k1 = ? k14 = 1.0 (moisture content < 15 %) k15 = 1.0 (face grain = 0˚) t e = 28mm (parallel plies only) ØN nt = 2[0.9 × k1 × 1.0 × 1.0 × 33 × 28 × d ] ∴ ØN nt = 1663.2.k1 .d .kN Carter Holt Harvey Limited

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Since d = 900mm (use minimum section) - conservative

∴ ØN nt = 1496.9 × k1 .kN Consider compression capacity

N c* ≤ Ø.N nc

Eq. 6.13 NZS 3603

N nc = k1 .k 8 .k14 .k15 . f pc .t e .d

Eq. 6.14 NZS 3603

∴ ØN nc = Ø.k1 .k 8 .k14 .k15 . f pc .t e .d where:

Ø = 0 .9

f pc = 45MPa

Technical Note 82

k1 = ? k14 = 1.0 (moisture content < 15 %) k 8 = 1.0 (localised, gusset edges are restrained by gusset stiffeners) k15 = 1.0 (face grain = 0˚) t e = 28mm (parallel plies only) ØN nc = 2[0.9 × k1 × 1.0 × 1.0 × 1.0 × 45 × 28 × d ] ∴ ØN nc = 2268.0 × k1 .d .kN Since d = 900mm (use minimum section) - conservative

∴ ØN nc = 2041.2 × k1 .kN 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 2

83.1    324.0   50.5   +  +  = 1.07 ≥ 1.0  0.8 × 2041.2   0.8 × 447.0   0.8 × 314.5  It is typical to consider the maximum implied forces on the structure, rather than the applied forces at the specific design location. However if the design criteria is not met then consideration of the implied design actions at the design 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 Carter Holt Harvey Limited

Nc* = -83.1 kN

V* = 50.5 kN

k1=0.8 September 2008

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30 metre Span LVL Portal Frame Design Carter Holt Harvey Woodproducts NZ Industrial Park, Auckland, New Zealand 2

83.1    303.0   50.5   +  +  = 0.97 ≤ 1.0  0.8 × 2041.2   0.8 × 447.0   0.8 × 314.5  Combined bending, compression and shear from Eq. 6.18, NZS 3603:1993 Factor capacities by appropriate duration of load, k1 = 1.0 2

 102.2   362.00   54.3   +  +  = 0.90 ≤ 1.0  1.0 × 1496.9   1.0 × 447.0   1.0 × 314.5 

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 the moment 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 knee gussets. Whilst the ridge gusset may be ‘thinner’ often for consistency of purlin lengths and minimum gusset order quantities it may be preferable to maintain similar gusset thicknesses.

D + L. tan θ Cosθ Depthcs = 0.9.Dgusset Dgusset =

. Design Criteria 2

 N c*   M i*   Vi*    +   +   ≤ 1.0 ØN ØM ØV nc  ni     ni 

 N t*   ØN nt

  M i*  +    ØM ni

Eq. 6.17

NZS 3603

Eq. 6.18

NZS 3603

2

  Vi*   +   ≤ 1.0   ØVni 

Critical Design Actions Critical load case - 1.2G+1.5Q M* = 183.6 kNm

Nc* = -50.1 kN

V* = 6.6 kN

Critical load case - 0.9G+Wu – Lateral wind Carter Holt Harvey Limited

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M* = -156.9 kNm

Nt* = 71.1 kN

V* = -5.0 kN

Consider Bending Moment Capacity From previous, propose 42mm 4 x-band LVL, where:

ØM n = 403.2 × k1 .k11 .d 2 kNm For 900x90 hySPAN portal frame with 7.5˚ pitch

D gusset = ∴ D gusset

900 + 1200 × tan(7.5) Cos (7.5) = 1065.7 mm

d = 0.9 × D gusset = 959.2mm  300  k11 =    959.2  ∴ k11 = 0.82

0.167

Calculate bending moment capacity

ØM ni = 403.2 × k1 × 0.82 × 959.2 2 ∴ ØM ni = 305.5 × k1 .kNm > M * Calculate shear force capacity From previous:

Vni = 267.12 × k1 .d .kN ∴Vni = 256.2 × k1 .kN Calculate Tension Capacity From previous:

∴ ØN nt = 1663.2.k1 .d .kN Since d = 900mm (use minimum section) - conservative

∴ ØN nt = 1496.9 × k1 .kN Calculate Compression Capacity From previous: Carter Holt Harvey Limited

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∴ ØN nc = 2268.0 × k1 .d .kN Since d = 900mm (use minimum section) - conservative

∴ ØN nc = 2041.2 × k1 .kN Combined bending, compression and shear from Eq. 6.17, NZS 3603:1993 Factor capacities by appropriate duration of load, k1 = 0.8 2

50.1 6 .6    183.6     +  +  = 0.63 ≤ 1.0  0.8 × 2041.2   0.8 × 305.5   0.8 × 256.2  Combined bending, compression and shear from Eq. 6.18, NZS 3603:1993 Factor capacities by appropriate duration of load, k1 = 1.0 2

71.1 5 .0    156.9     +  +  = 0.57 ≤ 1.0  1.0 × 2041.2   1.0 × 305.5   1.0 × 256.2  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 on site. 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 critical as it will affect the available spacing and hence number of nails within the group as well as the required penetration into the column/rafter. A staggered nail pattern provides an increased moment capacity 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 combined actions including bending, axial and shear forces. Whilst the bending and axial forces contributions are minor they need to be taken into account. It is normally most efficient to calculate the proportion of force 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 and conservative. For this reason computer packages are often employed to develop design solutions. The following design data have been taken from design capacity tables relating to the corresponding roof pitch and member size. The design methodology, including k factors, from AS1720.1 has been applied to create nail ring capacities 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 to its close relationship between the lateral capacities of nails in testing with CHH’s range of LVL and the published values for joint group JD4. It should be noted that many of the ‘k’ factors used in calculation of connection capacities differ between the standards and it is recommended that for connections these not be mixed and matched. Carter Holt Harvey Limited

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3.7.3.1 Knee nail ring design Critical Design Actions The critical design actions need only be considered in the nail ring design as the effects of stress reversal do not affect the nature of the nail design.

Knee, Critical load case - 1.2G+1.5Q M* = -324.0 kNm N c* = -83.1 kN # As per Table 2.7, AS1720.1

V* = 50.5 kN

k1=0.77#

The methodology proposed for the calculation of nail group capacity for combined bending, axial and shear force involves the following steps: 1. Calculate moment capacity of nail rings in accordance with AS1720.1. AS1720.1 provides a capacity calculation for transfer of in plane moments through nailed moment ring such that:

 i =n  r φM j = φ .k1 .k13 .k14 .k16 .k17 .rmax .Qk .∑  i  i =1  rmax 

3  2     

AS1720.1

Eq. 4.2(4)

where: n = number of fasteners Qk= characteristic strength of fastener ri= distance to the ith fastener from the centroid of the fastener group rmax= the maximum value of ri Ø = 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 (Ø3.15 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 nail rings connecting gusset pairs is:

  i=n  r  φM = 2.φ .k1 .k13 .k14 .k16 .k17 .rmax .Qk .∑  i  i =1  rmax  

         3 2

2. Calculate remaining portion of nail capacity after bending actions have been considered. 3. a. φQn = φ .k1 .k13 .k14 .k16 .k17 .Qk b. φN axial / shear


 M*   × φQn × n = 1 −  φM  September 2008

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

 * N axial / shear = max 

(N ) + (v ) , (N ) + (v ) * 2 c

* 2

* 2 t

* 2

  

Engineering Bulletin 2 – Rigid Moment Connection Details can be used for selection of the moment ring capacity 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

and

φM = 0.77 × 454.14 ∴ φM = 349.7 kNm φQn = 0.855kN

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



M* 

 × φQn × n φN axial / shear = 1 − φM 

   324.0  φN axial / shear = 1 −  × 0.855 × (684 × 2 )  349.7  ∴ φN axial / shear = 86.0kN Calculate vectorial sum of axial and shear force, divided by k1 for direct comparison


(N ) + (v ) , (N ) + (v ) * 2 c

* 2

* 2 t

* 2

  

  83.1  2  50.5  2  102.2  2  54.3  2  *   N axial = max + ,   +   / shear   0.77   0.77   1 .0   1 .0    * N axial / shear = max (126.3,115.7 ) * ∴ N axial / shear = 126.3kN * Since N axial / shear > φN either add an additional nail ring or adjust nail size. Try using a Ø3.33 nail. Using Table 3 from Engineering Bulletin 2 the capacity of the nail rings can be factored proportionally to the Characteristic Capacity of the nail laterally loaded in single shear.

Ø3.33/Ø3.15 factor =

898 = 1.11 810

Therefore:

φM = 0.77 × 454.14 × 1.11

and Carter Holt Harvey Limited

∴ φM = 388.15kNm ≥ M * φQn = 0.855 × 1.11 ∴ φQn = 0.949kN September 2008

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φN axial / shear φN axial / shear

 M*   × φQn × n = 1 −  φM  324.0   = 1 −  × 0.949 × (684 × 2 )  388.15 

∴ φN axial / shear = 214.6kN > N * Use nine nail rings of Ø3.33 x75 FH nails to pattern as marked.

Proposed Nail Ring

3.7.3.2 Ridge nail ring design Critical Design Actions Critical load case - 1.2G+1.5Q M* = 183.6 kNm Carter Holt Harvey Limited

Nc* = -50.1 kN

V* = 6.6 kN September 2008

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Engineering Bulletin 2 – Rigid Moment Connection Details can be used for selection of the moment ring capacity 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 Ø3.33 nails in the knee connection, apply same nail size in the ridge, therefore apply 1.11 factor from previous to apply nail ring capacities from Table ##.

φM = 0.77 × 265.96 × 1.11 and

∴ φM = 227.3kNm ≥ M * φQn = 0.949kN

From previous




M* 

 × φQn × n φN axial / shear = 1 − φM 

   183.6  φN axial / shear = 1 −  × 0.949 × (344 × 2 )  227.3  ∴ φN axial / shear = 125.52kN Calculate vectorial sum of axial and shear force, divided by k1 for direct comparison


(N ) + (v ) , (N ) + (v ) * 2 c

* 2

* 2 t

* 2

  

  50.1  2  6.6  2  71.1  2  5.0  2 *   N axial = + max ,   +  / shear   0.77   0.77  1 .0   1 .0    * N axial / shear = max (65.6,71.1)

   

* ∴ N axial / shear = 65.6 kN < φN

Use four nail rings of Ø3.33 x75 FH nails to pattern as marked.


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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 fixed directly to the LVL columns. A similar design philosophy is applied to the design and specification of hold down anchors and base plates as would normally be applied to steel where the buckling of the plate under tension needs to be considered. The connection of the base brackets to the column could be achieved using nails, screws or bolts. Nails are typically not recommended of base plates in larger structures because of the number of nails required combined with the fact they would need to be hand driven through holes in plates. Bolts can be used and are good to aid in the transfer of bracing loads across the column. Screws are ideal for most base bracket connections due to their ease of application. It is important that screw patterns are staggered for both sides of the column so that splitting of the LVL does not occur. Again reactions are factored to take into consideration duration of load factors.

Consider Design Reactions PF1 Rx Load Case 1.35G 1.2G+1.5Q 0.9G+Wu (Lateral) 1.2G+Wu (Lateral) 0.9G+Wu (Long) 1.2G+Wu (Long)

k1 0.6 0.8 1.0 1.0 1.0 1.0

kN 19.19 50.53 -58.60 -53.96 -24.61 52.50

Ry

(Rx2+Ry2)0.5

kN 35.82 86.87 -99.28 115.53 -78.67 102.30

Angle

kN 40.6 100.5 115.3 127.5 82.4 115.0

61.8 59.8 59.4 65.0 72.6 62.8

k1 adjusted values PF1 Load Case 1.35G 1.2G+1.5Q 0.9G+Wu (Lateral) 1.2G+Wu (Lateral) 0.9G+Wu (Long) 1.2G+Wu (Long)

k1 0.6 0.8 1.0 1.0 1.0 1.0

Rx kN 31.98 63.16 -58.60 -53.96 -24.61 52.50

Ry kN 59.70 108.59 -99.28 115.53 -78.67 102.30

(Rx2+Ry2)0.5 kN 67.7 125.6 115.3 127.5 82.4 115.0

Angle 61.8 59.8 59.4 65.0 72.6 62.8

It is typical in Timber structures to provide a moisture barrier at the base of the columns to eliminate the column from getting wet and staying wet during the construction period. This can be typically achieved by using H3.2 treated 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 loads need be considered.

Calculate minimum number of 14g type 17 Hex Head screws Joint Group

J4


S * ≤ ØQn

Eq. 4.5,

NZS 3603

Qn = n.k .Qk

Eq. 4.6,

NZS 3603


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Ø = 0.8 * k1 = 1.0 Qk = 3.303kN

k = 1.25

NZS 3603

Other ‘k’ modification factors are not relevant as timber is dry, screws are in single shear and are screwed through close fitting steel plates into the edge or face of the timber. *Ø=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.

115.3 = 0.8 × 1.0 × n × 1.25 × 3.303 ∴ n = 34.9 Say 48/14gx50 type 17 Hex Head screws, screwed through base plate sides into column.

Proposed Connection


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4.0 Girt Design, side wall Girt Span Girt Spacing

(10,000-(90+65))/2 = 4923 mm 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 and negative wind pressures.

θ = 0˚, Lateral wind qu=0.84 kPa

Case 1 Case 2

cp,e= +0.7, cp,i= -0.56, kL= 1.25 cp,e= -0.3, cp,i= +0.61

θ = 90˚, Longitudinal wind qu=0.76 kPa

Case 1

cp,e= -0.65, cp,i= +0.54, kL= 1.5

Calculate design loading w*i = qu .spacing .(k a .kl .c pe − c pi ) w1* = 0.84 × 1.66 × (1.25 ×+ 0.7 − − 0.56) ∴ w1* = +1.93kN / m w2* = 0.76 ×1.66 × (1.5 ×− 0.65 − + 0.54) ∴ w2* = −1.84kN / m

Serviceability Refer Technical Note 82 for Section and Material Properties. 2

 37  ws =   × + 1.93= + 1.30kN / m  45   5.w.l 4  δ w = k 2 .   384.EI x 


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 5× + 1.30 × 4923 4  δ w = 1.0. 9   384 × 283 × 10  ∴ δ w = 35.3mm or Span 140 Strength Check capacity for positive wind pressures w.l 2 +1.93 × 4.92 = 8 8 * ∴ M = 5.79kNm M* =

Consider shear and support reaction for wind load

w.l +1.93 × 4.9 = 2 2 * * ∴ N = V = 4.73kN V* =

Calculate Bending Moment Capacity

M * ≤ ØM n

Eq. 3.3 NZS 3603

M n = k1.k 4 .k5 .k8 . f b .Z

Eq. 3.4 NZS 3603

where:

Ø = 0 .9

k1 = 1.0

k 4 = k 5 = 1 .0

f b = 48MPa

Technical Note 82

ØM n = 0.9 × 1.0 × 1.0 × 1.0 × k8 × 48 × 271 × 103 ∴ ØM n = 11.7.k8kNm Continuous restraint to compression edge via pierce fixed sheeting, therefore k8=1.0

ØM = 11.7 kNm > M * Calculate Shear Capacity

V * ≤ ØVn

Eq. 3.3 NZS 3603

Vn = k1.k4 .k5 . f s . AS

Eq. 3.4 NZS 3603

where:


September 2008

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Ø = 0 .9

k1 = 1.0

k 4 = k 5 = 1 .0 AS = 2.b.d / 3

f s = 5.3MPa

∴ AS =

Technical Note 82

Cl 3.2.3.1 NZS 3603

2 × 190 × 45 = 5700mm 2 3

∴VS = 0.9 × 1.0 × 1.0 × 5.3 × 5700 ∴VS = 27.2kN > V * Consider negative wind pressures. − w.l 2 1.84 × 4.9 2 M = = 8 8 * ∴ M = −5.5kNm

*

Calculate bending moment capacity From previous:

ØM n = 11.7.k8 kNm Calculate k8 Continuous lateral restraint is provided to the tension edge via pierce fixed sheeting. Calculate S1

d b 190 S! = 3 × = 12.67 45

S1 = 3.

Eq. 3.6

NZS 3603

Since 25>S1>10

k8 = a1 + a2 .S + a3 .S 2 + a4 .S 3 k8 = 0.21 + 0.175 × 12.67 + − 0.0116 × 12.67 2 +

1 × 12.673 5000

∴ k8 = 0.97 ∴ ØM = 11.7 × 0.97 ØM = 11.3kNm > M *

4.2 Connection design Carter Holt Harvey Limited

September 2008

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Connection of hyCHORD girts is easiest performed using proprietary brackets and screws or nails. The proposed bracket 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 we recommend a minimum of 10/Ø3.15x35 FH nails ie. 5/Ø3.15 nails per tab.

Check Capacity Joint Group

J5


S * ≤ ØQn

Eq. 4.1,

NZS 3603

Qn = n.k .Qk

Eq. 4.2,

NZS 3603


Ø = 0.8 k1 = 1.0 Qk = 0.631kN

NZS 3603

n = 10 k=1.25 since nails are through steel side plates < 3.0 mm thickness.

Other ‘k’ modifaction factors are not relevant as timber is dry, nails are in single shear and are nailed into the edge or face of the timber.

ØQn = 0.8 × 10 × 1.0 × 1.25 × 0.631 ∴ Qn = 6.3kN > N * Also confirm Characteristic Strength Capacity of Bracket.

ØQ = 18.0kN > N *

Mitek Literature

The 190x45 hyCHORD girt to span 4.9 m at maximum 1660 mm spacing is adequate to support the design load. Proposed Girt Layout


September 2008

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5.0 Mullion design, side wall The mullion is best calculated as a vertical member supporting a series of point loads that share a common spacing, which is typical of mullions. Standard beam formulae have been adapted to best provide accurate but easy to calculate equations. Refer Appendix 1 for beam equations, where n is the number of girts.

Consider side wall mullion Mullion Span

6.4 m

Girt Spacing

1660 mm

Propose 300x63 hySPAN for use as side wall mullion.

5.1 Wind loading Girts provide lateral restraint to the compression edge for positive pressures and to the tension edge for negative wind pressures. Girt reactions have been recalculated excluding the local pressure factors as the mullion does not directly support the cladding.

θ = 0˚, Lateral wind Girt loading (positive pressure)

P*= +8.61 kN

θ = 90˚, Longitudinal wind Girt loading (negative pressure)

P*= -7.35 kN

Serviceability Apply maximum wind pressure for serviceability 2

 37  Ps =   × + 8.61= + 5.82kN  45  n=4  1 P.L3 4  δw = k 2 . .n.3 − 1 + 2  192.EI  2  n 


Appendix 1

September 2008

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 1 5820 × 6400 3 4  δ w = 1 .0 x × 4 × 3 − 1 + 2  6 192 × 13200 × 142 x10  2  4  ∴ δ w = 30.3mm or Span 211

Strength Calculate bending moment and shear Positive wind pressures + P.l 8.61 × 6.4 M = n. = 4× 8 8 * ∴ M = 27.55kNm *

Negative wind pressures − 7.35 × 6.4 P.l = 4× 8 8 * − ∴ M = 23.5kNm

M * = n.

Consider shear and support reaction for wind load + P 8.61 = 4× 2 2 * * + ∴ N = V = 17.2kN

V * = (n − 1).

Calculate Bending Moment Capacity Positive Pressure. Calculate k8 Compression edge restrained by girts at 1660 mm spacing. 0.5 2 L    ay .  d  S1 = 1.35 − 1    b   b    

0.5

1660  300 2  0.5    S1 = 1.35 − 1    63   63    

Eq. 3.5

NZS 3603

0.5

∴ S1 = 14.95 Since 25>S1>10


September 2008

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k8 = a1 + a2 .S + a3.S 2 + a4 .S 3 k8 = 0.21 + 0.175 × 14.95+ − 0.0116 × 14.952 +

1 × 14.953 5000

∴ k8 = 0.90 M * ≤ ØM n

Eq. 3.3

NZS 3603

M n = k1.k 4 .k5 .k8 . f b .Z

Eq. 3.4

NZS 3603

where:

Ø = 0 .9

k 1 = 1 .0

k 4 = k 5 = 1 .0

f b = 48MPa

Technical Note 82

ØM n = 0.9 × 1.0 × 1.0 × 1.0 × k8 × 48 × 945 × 103 ∴ ØM n = 40.82.k8kNm Since k8=0.90

ØM = 36.74kNm > M * Negative Pressure. Calculate k8 Tension edge restrained by girts at 1660 mm spacing. Consider Stability equation for Discrete Restraint to Tension Edge from AS 1720.1.

d  S1 =   b

1.35

 Lay   d

  

0.5

Eq. 3.2(5)

AS 1720.1-2008

M * ≤ ØM n

Eq. 3.3

NZS 3603

M n = k1.k 4 .k5 .k8 . f b .Z

Eq. 3.4

NZS 3603

1.35

 300   1660  S1 =      63   300  ∴ S1 = 19.34

0.5

Since 25>S1>10

k 8 = a1 + a 2 .S + a3 .S 2 + a 4 .S 3 k 8 = 0.21 + 0.175 × 19.34+ − 0.0116 × 19.34 2 +

1 × 19.34 3 5000

∴ k 8 = 0.70


September 2008

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where:

Ø = 0 .9

k 1 = 1 .0

k 4 = k 5 = 1 .0

f b = 48MPa

Technical Note 82


ØM = 28.6kNm > M * Calculate Shear Capacity

V * ≤ ØVn

Eq. 3.3

NZS 3603

Vn = k1.k4 .k5 . f s . AS

Eq. 3.4

NZS 3603

where:

Ø = 0 .9

k1 = 1.0

k 4 = k 5 = 1 .0 AS = 2.b.d / 3

f s = 5.3MPa

∴ AS =

Technical Note 82

Cl 3.2.3.1

2 × 300 × 63 = 12600mm 2 3

NZS 3603

∴VS = 0.9 × 1.0 × 1.0 × 5.3 × 12600 ∴VS = 60.1kN > V * 5.2 Connection design Two different connections are required for the mullion. Connection to the ground is proposed using Mitek CF2x brackets whilst the connection to the eaves beam can be performed using two Mitek N21 Diagonal Cleats. The design capacities expressed in the Mitek literature are based on fully nailing out the holes. We can calculate a reduced number of fasteners for ease of installation whilst maintaining the structural integrity. It is recommended that the reduced number of fasteners are evenly distributed across the tab/bracket area. Propose use of proprietary Ø3.15x35 FH nails for connection to mullion. Joint Group

J5

Table 3, Technical Note 82

S * ≤ ØQn

Eq. 4.1,

NZS 3603

Qn = n.k .Qk

Eq. 4.2,

NZS 3603


September 2008

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Ø = 0.8 k1 = 1.0 Qk = 0.631kN

NZS 3603

n=? k=1.25 since nails are through steel side plates < 3.0 mm thickness.


ØQn = 0.8 × n × 1.0 × 1.25 × 0.631 ØQn = n × 0.631kN ∴n =

17.22 = 27.25 0.631

Say 15/Ø3.15x35 FH nails per tab, per bracket. Stagger nails. Also check Characteristic Strength Capacity of Bracket. One pair of Mitek N21 Diagonal Cleats.

ØQ = 40.0kN > N *

Mitek Literature

One pair of CF2x brackets with 2/M12 Chemical anchors. Load in shear only.

ØQ = 48.0kN > N *

Mitek Literature

Proposed Mullion Connection


September 2008

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September 2008

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6.0 Eaves beam design Eaves Beam Span Mullion location

10,000-(90+(2x41)) = 9828 mm Mid-span (critical)

Propose 400x63 hySPAN for use as eaves beam

6.1 Wind loading Positive pressure P*= +17.22 kN, w*=+0.97kN/m Negative pressure P*= -14.7 kN, w*=-0.69 kN/m Serviceability Positive wind pressure critical for serviceability 2

 37  Ps* =   ×+ 17.22= + 11.64kN  45  2

 37  w =   ×+ 0.97 = + 0.66kN / m  45  * s

 5×+ 0.66 × 98284 +11640 × 98283  δw = 1.0. +  9 48 × 4435 ×109   384 × 4435 × 10 ∴ δw = 70.0mm or Span 140

Strength Check capacity for positive wind pressures w.l 2 P.L + 0.97 × 9.8 2 17.22 × 9.8 + = + 8 4 8 4 * ∴ M = 53.8kNm M* =

Consider shear and support reaction for wind load Carter Holt Harvey Limited

September 2008

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w.l P + 0.97 × 9.8 17.22 + = + 2 2 2 2 * * ∴ N = V = 13.4kN V* =

Calculate Bending Moment Capacity

M * ≤ ØM n

Eq. 3.3

NZS 3603

M n = k1.k 4 .k5 .k8 . f b .Z

Eq. 3.4

NZS 3603

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

ØM n = Ø.k1.k4 .k5 .k8 .k11. f b .Z

Therefore where:

Ø = 0 .9

k 1 = 1 .0

k 4 = k 5 = 1 .0

f b = 48MPa

 300  k11 =    d 

Technical Note 82

0.167

 300  ∴ k11 =    400 

Cl. 2.4.6

0.167

AS1720.1

= 0.95

ØM n = 0.9 × 1.0 × 1.0 × 1.0 × k 8 × 0.95 × 48 × 1680 × 10 3 ∴ ØM n = 68.9.k 8 kNm Continuous restraint to compression edge via pierce fixed sheeting, therefore k8=1.0


V * ≤ ØVn

Eq. 3.3

NZS 3603

Vn = k1.k4 .k5 . f s . AS

Eq. 3.4

NZS 3603

where:

Ø = 0 .9

k1 = 1.0

k 4 = k 5 = 1 .0 AS = 2.b.d / 3

f s = 5.3MPa

∴ AS =

2 × 400 × 63 = 16800mm 2 3


Technical Note 82

Cl 3.2.3.1

NZS 3603

September 2008

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∴VS = 0.9 × 1.0 × 1.0 × 5.3 × 16800 ∴VS = 80.1kN > V * Consider negative wind pressures. w.l 2 P.L − 0.69 × 9.8 2 − 14.7 × 9.8 + = + 8 4 8 4 * − ∴ M = 44.3kNm M* =

Calculate bending moment capacity From previous:

ØM n = 68.94.k8kNm Calculate k8 Continuous lateral restraint is provided to the tension edge via pierce fixed sheeting. Calculate S1

S1 = 3.

d b

S! = 3 ×

Eq. 3.6 NZS 3603

400 = 19.04 63

Since 25>S1>10

k8 = a1 + a2 .S + a3 .S 2 + a4 .S 3 k8 = 0.21 + 0.175 × 19.04+ − 0.0116 × 19.042 +

1 × 19.043 5000

∴ k8 = 0.72 ∴ ØM = 68.94 × 0.72 ØM = 49.6kNm > M *

6.2 Connection design Connection of the eaves beam to the column needs to provide both torsional restraint as well as adequate fastening for the horizontal wind loads. It is proposed to use a combination of a pair of Mitek MS1430 split joist hanger together with a Mitek N21 Diagonal Cleat. Calculate load taken by split joist hangers using 6/ 14g type 17 Hex Head screws per member Carter Holt Harvey Limited

September 2008

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Joint Group

J4


S * ≤ ØQn

Eq. 4.5,

NZS 3603

Qn = n.k .Qk

Eq. 4.6,

NZS 3603


Ø = 0.8 * k1 = 1.0 Qk = 3.303kN

NZS 3603

k=1.25 since nails are through steel side plates < 3.0 mm thickness. Other ‘k’ modification factors are not relevant as timber is dry and screws are in single shear. *Ø=0.8 is applied as Type 17 screws are as reliable as nails in service.

ØQn = 0.8 × 6 × 1.0 × 1.25 × 3.303 ∴ ØQn = 18.2kN Consider use of Diagonal Cleat to provide stability and additional support Joint Group

J5


S * ≤ ØQn

Eq. 4.1,

NZS 3603

Qn = n.k .Qk

Eq. 4.2,

NZS 3603


Ø = 0.8 k1 = 1.0 Qk = 0.631kN n = 10

NZS 3603

k=1.25 since nails are through steel side plates < 3.0 mm thickness. Other ‘k’ modification factors are not relevant as timber is dry, nails are in single shear and are nailed into the edge or face of the timber.

ØQn = 0.8 × 10 × 1.0 × 1.25 × 0.631 ∴ Qn = 6.3kN Consider design Reaction, N* = 13.8 kN

ØQT = 18.2 + 6.3 ∴ QT = 24.5kN > N * Proposed Connection Carter Holt Harvey Limited

September 2008

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Also confirm Characteristic Strength Capacity of Bracket. N21 Diagonal Cleat, one only

ØQ = 20.0kN

Mitek Literature

Pair of MS1430 Split Joist Hangers

ØQ = 29.6kN

Mitek Literature

Total capacity of brackets

ØQ = 20.0 + 29.6 = 49.6kN > N *

7.0 Girt Design, end wall Girt Span Girt Spacing

6000 mm 1660 mm

Propose 240x45 hySPAN for use as end wall girt

7.1 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 and negative wind pressures.

θ = 0˚, Lateral wind qu=0.84 kPa

Case 1 Case 2

cp,e= -0.65, cp,i= +0.61, kL= 1.5, over 6.0 m cp,e= -0.65, cp,i= +0.61, kL= 2.0, over 3.0 m



Case 1

cp,e= +0.7, cp,i= -0.65, kL= 1.25, over 3.0 m

September 2008

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Calculate design loading Case 2 is critical by inspection

w*i = qu .spacing .(ka .kl .c pe − c pi ) w1* = 0.84 × 1.66 × (2.0×− 0.65 − + 0.61) ∴ w1* = −2.66kN / m w2* = 0.84 × 1.66 × (1.0×− 0.65 − + 0.61) ∴ w2* = −1.76kN / m

Calculate weff Calculate Reactions

R* = − 2.66 × 1.5+ − 1.76 × 1.5 ∴ R* = − 6.63kN Calculate Moment − * M Wu = − 6.63 × 3.0 −

2.66 × 1.5 2 − − 1.76 × 1.5 × 2.25 2

* ∴ M Wu = − 10.96kNm

Calculate weff −

10.96 × 8 6 .0 2 * ∴ weff = − 2.44kN / m * eff

w

=

Calculate critical reaction and shear Calculate Reactions

R* =

 1 − 3 .0 2 − 2 . 66 × + 1.76 × 3.0 × 4.5  6 .0  2 

∴ R* = − 7.31kN

Serviceability Refer Technical Note 82 for Section and Material Properties.


September 2008

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30 metre Span LVL Portal Frame Design Carter Holt Harvey Woodproducts NZ Industrial Park, Auckland, New Zealand 2

 37  ws =   ×− 2.44= − 1.62kN / m  45   5.w.l 4  δw = k2.   384.EI x   5×− 1.62 × 6000 4  δw = 1.0. 9   384 × 684 × 10  ∴ δw = 40.0mm or

Span 150

Strength Calculate Bending Moment Capacity

M * ≤ ØM n

Eq. 3.3

NZS 3603

M n = k1.k 4 .k5 .k8 . f b .Z

Eq. 3.4

NZS 3603

where:

Ø = 0 .9

k 1 = 1 .0

k 4 = k 5 = 1 .0

f b = 48MPa

Technical Note 82

ØM n = 0.9 × 1.0 × 1.0 × 1.0 × k 8 × 48 × 432 × 10 3 ∴ ØM n = 18.7.k 8 kNm Calculate k8 Since negative pressures produce a higher moment than positive pressure for this case and continuous restraint is offered, the tension edge restraint will produce a less stable option, hence we only need to consider capacity for negative pressure in this case.

Continuous lateral restraint is provided to the tension edge via pierce fixed sheeting. Calculate S1

S1 = 3.

d b

S1 = 3 ×

Eq. 3.6

NZS 3603

240 = 16.0 45

Since 25>S1>10


September 2008

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k8 = a1 + a2 .S + a3 .S 2 + a4 .S 3 k8 = 0.21 + 0.175 × 16.0+ − 0.0116 × 16.02 +

1 × 16.03 5000

∴ k8 = 0.86 ∴ ØM = 18.67 × 0.86 ØM = 16.1kNm > M * Calculate Shear Capacity

V * ≤ ØVn

Eq. 3.3

NZS 3603

Vn = k1.k4 .k5 . f s . AS

Eq. 3.4

NZS 3603

where:

Ø = 0 .9

k1 = 1.0

k 4 = k 5 = 1 .0 AS = 2.b.d / 3

f s = 5.3MPa

∴ AS =

2 × 240 × 45 = 7200mm 2 3

Technical Note 82

Cl 3.2.3.1

NZS 3603

∴ VS = 0.9 × 1.0 × 1.0 × 5.3 × 7200 ∴ VS = 34.34kN > V * 7.2 Connection design Propose JH 47x190 to suit 240x45 hySPAN so that the same bracket can be used for both side and end walls (Depth of bracket is 79.2 % of girt depth so suitable).

Consider 12/Ø3.15x35 FH nails. Joint Group

J5


S * ≤ ØQn

Eq. 4.1,

NZS 3603

Qn = n.k .Qk

Eq. 4.2,

NZS 3603


Ø = 0.8 k1 = 1.0 Qk = 0.631kN n = 12

NZS 3603

k=1.25 since nails are through steel side plates < 3.0 mm thickness. Carter Holt Harvey Limited

September 2008

Project: For: At:




ØQn = 0.8 × 12 × 1.0 × 1.25 × 0.631 ∴ Qn = 7.6kN > N * Also confirm Characteristic Strength Capacity of Bracket.

ØQ = 27.0kN > N *

Mitek Literature

The 240x45 hySPAN girt to span 6.3 m at maximum 1660 mm spacing is adequate to support the design load.


September 2008

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8.0 Mullion design, end wall Wind posts may be designed for buildings where smaller frames are used in the end walls. Wind posts would be required to resist axial loads from the frames as well as horizontal wind loads. A reduced frame was not used in this example as its analysis is similar to other portal frame components where special attention is paid to the bending moment diagram and restraint offered by purlins. Additional end wall bracing may also be required to limit the sway achieved by the reduced section frame. End wall mullions are typically symmetrical about the ridge line to allow for repetition of detail and order lengths. Depending on spans and quantities individual calculations of mullion sizes may have cost benefits. In this case given there are only four mullions per end wall we will design the mullion that has the maximum span. End wall mullions can be detailed to fix to the inside or outside of the end wall frame. Connection to the outside of the frame is less complicated because the purlins do not create clashes with proposed mullion locations.

Consider end wall mullion Mullion Span

8.2 m

Girt Spacing

1660 mm

Propose 400x63 hySPAN for use as end wall mullion.

8.1 Wind loading Girts provide lateral restraint to the compression edge for positive pressures and to the tension edge for negative wind pressures. Girt loads have been recalculated excluding local pressure factors.

θ = 0˚, Lateral wind Girt loading (negative pressure)

P*= -10.56 kN

θ = 90˚, Longitudinal wind Carter Holt Harvey Limited

September 2008

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Girt loading (positive pressure)

P*= +10.20 kN

Serviceability Apply maximum wind pressure for serviceability 2

 37  Ps =   × + 10.56= − 7.14kN  45  n=5 δw = k 2 .

P.L3 192.EI

1  1  1   .n − .3 − 1 − 2  n   2  n  

7140 × 8200 3 1  1  1  δ w = 1 .0 x × 5 −  × 3 − 1 − 2 6 192 × 13200 × 336 x10  5   2  5 ∴ δ w = 55.9mm or Span 147

Appendix 1   

Strength Calculate bending moment and shear Positive wind pressures + P.l 10.20 × 8.2 2 M = (n − 1). = (5 − 1)× 8.n 8×5 * + ∴ M = 50.2kNm *

2

Negative wind pressures − P.l 10.56 × 8.2 = (5 2 − 1)× 8.n 8×5 * − ∴ M = 52.0kNm

M * = (n 2 − 1).

Consider shear and support reaction for wind load − P 10.56 = (6 − 1) × 2 2 * * ∴ N = V = 26.4kN

V * = (n − 1).

Calculate Bending Moment Capacity Positive Pressure. Calculate k8 Compression edge restrained by girts at 1660 mm spacing. Carter Holt Harvey Limited

September 2008

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30 metre Span LVL Portal Frame Design Carter Holt Harvey Woodproducts NZ Industrial Park, Auckland, New Zealand 0.5 2 L    ay .  d  S1 = 1.35 − 1    b   b    

0.5

Eq. 3.5

NZS 3603

M * ≤ ØM n

Eq. 3.3

NZS 3603

M n = k1.k 4 .k5 .k8 . f b .Z

Eq. 3.4

NZS 3603

1660  400 2  0.5    S1 = 1.35 − 1    63   63    

0.5

∴ S1 = 17.35 Since 25>S1>10

k8 = a1 + a2 .S + a3 .S 2 + a4 .S 3 k8 = 0.21 + 0.175 × 17.35+ − 0.0116 × 17.352 +

1 × 17.353 5000

∴ k8 = 0.80

where:

Ø = 0 .9

k 1 = 1 .0

k 4 = k 5 = 1 .0

f b = 48MPa

Technical Note 82

ØM n = 0.9 × 1.0 × 1.0 × 1.0 × k 8 × 48 × 1680 × 10 3 ∴ ØM n = 72.6.k 8 kNm Since k8=0.80

ØM = 58.1kNm > M * Negative Pressure. Calculate k8 Tension edge restrained by girts at 1660 mm spacing. Consider Stability equation for Discrete Restraint to Tension Edge from AS 1720.1.

d  S1 =   b

1.35

 Lay   d

1.35

  

0.5

 400   1660  S1 =      63   400  ∴ S1 = 24.69 Carter Holt Harvey Limited

Eq. 3.2(5)

AS 1720.1

0.5

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Since 25>S1>10

k 8 = a1 + a 2 .S + a 3 .S 2 + a 4 .S 3 k 8 = 0.21 + 0.175 × 24.69+ − 0.0116 × 24.69 2 +

1 × 24.69 3 5000

∴ k 8 = 0.47 From previous:


ØM = 34.11kNm < M *

Not sufficient

The effect of beam slenderness have reduced the capacity of this section such that it is not suitable to support the required load. In this type of situation it is an opportunity to select a thicker, lower strength and cost section such as 400x90 hy90 to replace the 400x63 hySPAN. Whilst hy90 has a lower Characteristic Bending Strength the additional thickness means that a 63mm hySPAN and 90mm hy90 compare favourably with each other as a direct strength and stiffness comparison (refer Appendix 2). The fact that the hy90 section in question has a lower depth to breadth ratio means it is more stable, and hence may be suitable for the end wall mullion.

Try 400x90 hy90 as end wall mullion: 1.35

 400   1660  S1 =      90   400  ∴ S1 = 15.26

0.5

Since 25>S1>10

k8 = a1 + a2 .S + a3 .S 2 + a4 .S 3 k8 = 0.21 + 0.175 × 15.26+ − 0.0116 × 15.262 +

1 × 15.263 5000

∴ k8 = 0.89 For hy90:

Ø = 0 .9

k1 = 1.0

k 4 = k 5 = 1 .0

f b = 35MPa

Technical Note 82

So:


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ØM n = 0.9 × 1.0 × 1.0 × 1.0 × k8 × 35 ×

400 2 × 90 6

∴ ØM n = 75.60.k8 kNm Since k8=0.89


V * ≤ ØVn

Eq. 3.3

NZS 3603

Vn = k1.k4 .k5 . f s . AS

Eq. 3.4

NZS 3603

where:

Ø = 0 .9

k1 = 1.0

k 4 = k 5 = 1 .0

f s = 5.3MPa

Technical Note 82

Consider reduced section at notch for shear capacity, notch to match 240 deep girt.

AS = 2.b.d / 3 ∴ AS =

2 × 240 × 90 = 14400mm 2 3

Cl 3.2.3.1

NZS 3603

∴VS = 0.9 × 1.0 × 1.0 × 5.3 × 14400 ∴VS = 68.7kN > V * Consider notching of mullion top. The mullion notch to accommodate the rafter needs to be checked. The notch will only fracture due to an opening moment which would be caused by a positive wind pressure. Since the rafter acts as a support for the mullion the moment at the notch is zero, however the shear force needs to be considered as per equation 3.7, NZS 3603.

M* = 0 kNm, V* = N* = 26.4 kN

V * + 1 .2

M* ≤ 1.5Ø.k1.k4 .k5 .k7 . f s . Asn dn

Eq. 3.7, NZS 3603

where:

Ø = 0 .9

k1 = 1.0

k 4 = k 5 = 1 .0

f s = 5.3MPa

Technical Note 82

It is best practice to always make the notch slope as long as possible to limit the stress concentration at the notch location. Carter Holt Harvey Limited

September 2008

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Since:

a = (400 − 240) = 160mm bn = 4 × 160 = 640mm a ≥ 0.1.d 2 .2 k7 = 0.25 d 2 .2 ∴ k7 = = 0.49 4000.25 Asn = 2.b.d n / 3

Table 3.1 NZS 3603

Cl 3.2.6 NZS 3603

2 × 240 × 90 = 14400mm 2 3 26.4kN ≤ 50.5kN ∴ AS =

8.2 Connection design Connection to the ground is proposed using Mitek CF2x brackets, similar to the side wall mullion. The connection to the rafter requires tension capable fixings. One system using proprietary brackets involves the use of three Mitek concealed purlin cleats (CP80) together with one suitable length cyclone tie. Consider system capacity CT1200 Cyclone Tie, wrapped around mullion with 5 nails per end.

ØQ = 13.0kN *

Mitek Literatue

3/ Concealed purlin cleats (CP80), fixed with 8/Ø3.15x30 FH nails and 4/14gx35 type 17 screws per bracket

ØQ = 1.5 × 16.0 = 24.0kN ∴ ØQ = 13.0 + 24.0 = 37.0kN > N

Mitek Literatue *

Mitek Literatue

Proposed Connection


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9.0 Longitudinal Bracing Portal Frame action takes care of the lateral loads however Longitudinal Bracing is required to transfer the longitudinal forces to the ground. For spans up to 20m strap bracing is a cheap and easy way to achieve the required longitudinal bracing requirements. For spans above 20m the installation time required for strap bracing can outweigh any advantages of the low costs of materials. Traditional threaded rod may be used for bracing or timber braces can be developed. Timber (LVL) braces can be used for loading in tension and compression and can be suitably fixed to purlins and girts to reduce buckling lengths. Both Mitek and Pryda have proprietary brackets suitable for connection to timber frames. Matching bracing points with mullion/wind post loads is theoretically the best option but is nearly always impractical for larger structures. Essentially the braced bays will act as a truss transferring the loads through the purlins between braced bays. This can be easily modelled using structural analysis packages.


cp,e (Windward) = +0.7, cp,e (Leeward)= -0.3

Calculate design loading Calculate end wall force

1  heave + hridge  b . Area = . 2 2 2 1  6.789 + 8.764  30 Area = . . 2 2  2 Area = 58.3m 2 / side * Pend _ wall = q u . A.(c p ,e (W ) − c p ,e ( L ) ) * + − Pend _ wall = 0.76 × 58.3 ×( 0.7 − 0.3) * ∴ Pend _ wall = 44.3kN

Consider friction due to sheeting profile

d 60.0 = = 7 .7 h 7.79 d 60.0 = = 2 .0 b 30.0

Cl 5.5 AS/NZS 1170.2:2002

Since both building ratios exceed 4 the friction force acting on the building needs to be resisted. Calculate Friction Force Calculate Load Area for friction

Area = (b + 2h)(d − 4h) Carter Holt Harvey Limited

AS/NZS 1170.2

Eqn. 5.5(a) September 2008

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(30.0 + 2 × 7.8)(60 − 4 × 7.8 ) 2 ∴ Area = 656.6m 2 / side Area =

For large buildings the area may be broken up into its roof and wall contribution to take advantage of the lower loads to be resisted however for this example the loads do not reach levels where any significant advantage may be gained.

c f = 0.04 * friction

P

- Ribs across the wind direction

AS/NZS 1170.2

Table 5.9

= qu . A friction .c f

* Pfriction = 0.76 × 656.6 × 0.04 * ∴ Pfriction = 20.0kN

Total load to be resisted by bracing per side * * PTotal = Pend* _ wall + Pfriction * PTotal = 44.3 + 20.0 * ∴ PTotal = 64.3kN

For long buildings it is good practice to have a bracing bay each end of the building rather than relying on the building to transfer all bracing loads from one end to the other, for this example we propose two braced bays.

Therefore the horizontal load to be transferred in each braced bay is 32.15 kN. Consider Roof Bracing Layout (Wall Bracing similar except tension only)

Calculate force in roof brace

Rbrace =

32.15 = 40.3kN Cos37.1

Calculate force in wall brace

Rbrace =

32.15 = 38.2kN Cos32.6


September 2008

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Design for critical load, 40.3 kN Propose 2/90x45 hyCHORD braces, separated with 7mm plywood strips to form a spaced column. Calculate Buckling capacity

N ncx = k1.k8 . f c . A

NZS3603

Eq. 3.18

∴ ØN ncx = Ø.k1.k8 . f c . A where:

Ø = 0 .9

f c = 45MPa

A = 2 × 90 × 35 = 6300mm

Technical Note 82

ØN ncx = 0.9 × k1 × .k8 × 45 × 6300 ∴ ØN ncx = 255.2 × k1 .k8 .kN Minor axis critical for buckling by inspection Minor axis buckling YY Calculate k8 for buckling about the minor axis YY

Lay = ∴ Lay

Purlin _ spacing Sinα 1600 = = 2652mm Sin37.1

This component is considered to be a spaced column. Clause E4 of AS1720.1 details stability equations for spaced columns. This procedure can be used to determine the slenderness coefficient for this member.

S 5 = 0.3g13 .g 28 .L.

A I

Eq. E4(5)

AS1720.1

Table 3.2 Table E5

AS1720.1 AS1720.1

where: g13=1.0 g28=1.0 A=2x90x35=6300 mm2

90 12 = 3.421 × 10 3 mm 4

(

)

I yy = 77 3 − 7 3 × ∴ I yy


September 2008

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∴ S 5 = 0.3 × 1.0 × 1.0 × 2652 ×

6300 3.421 × 10 3

∴ S 5 = 34.15 Since S5>25

k 8 = a5 .S a6 k 8 = 235.5 × 34.15 −1.937

Eq. 3.18

NZS3603

∴ k 8 = 0.25 Since k1 = 1.0

∴ ØN ncx = 64.37 kN > N c* Consider connection, propose proprietary Mitek B145 bracket system

Consider 2/M16 Bolts For three member system

be = 2 × 35 = 70mm

Table 4.9

NZS3603

Qskl = 2.Qkl

Table 4.9

NZS3603

Qkl = min(k11 . f cj .d a2 ,0.5 × be . f cj .d a )

Cl 4.4.2(a)(i)

NZS3603

Apply joint group J3 for bolts parallel to the grain

Technical Note 82

Since load is parallel to the grain


September 2008

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∴ Qkl = min(2.0 × 45.2 × 16.0 2 ,0.5 × 70 × 45.2 × 16.0) ∴ Qkl = min(23.14,25.31) ∴ Qkl = 23.14kN Calculate Bolt system capacity

N * ≤ ØQn

Eq. 4.16

NZS3603

Qn = n.k1 .k12 .k13 .Qsk

Eq. 4.17

NZS3603

N * ≤ ØQn

Eq. 4.18

NZS3603

Qn = f pj . Aw

Eq. 4.19

NZS3603

ØQn = Ø.n.k1 .k12 .k13 .Qk where:

Ø = 0.7 n = 2 Qsk = 2 × 23.14 = 46.28kN k1 = 1.0 k12 = 1.0 k13 = 1.0 Therefore:

ØQn = 0.7 × 2 × 1.0 × 1.0 × 1.0 × 46.28 ∴ ØQn = 64.8kN > N * Consider bracket bolts in tension

ØQn = Ø.n. f pj . Aw where:

Ø = 0 .7 n = 2

f pj = 14.5MPa

Aw = 65 × 65 − π × 18 2

NZS 3603

∴ Aw = 3206.7 mm 2 Therefore:

ØQn = 0.7 × 2 × 14.5 × 3206.7 ∴ ØQn = 65.1kN > N * Check colt capacity for fin plate connection to cleat. Propose M16 G8.8 bolt, including threads ØV = 59.3 kN > N*


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Consider Wall Bracing Connection

9.1 Purlins subject to axial loads Purlins in end bays may be subjected to tension and compression forces from braced bays. These forces need to be considered in the design capacity. The load to be transferred through the purlin system in both tension and compression is relative to the force in the brace. This force can be calculated as a designer would for a steel building. Critical Design Actions Critical load case - 0.9G+Wu (from section 2) M* = -28.8 kNm

N c* = N t* = 9.64 kN

ØM = 29.4 kNm

Consider column action of purlins subject to axial force due to bracing loads. Major axis buckling XX

N ncx = k1.k8 . f c . A

Eq. 3.18

NZS3603

∴ ØN ncx = Ø.k1.k8 . f c . A Take only the flange area into account. Remember to include for the penetration of the web into the flange. Technical Note 82 includes guidance on the calculation of hyJOIST section properties. Carter Holt Harvey Limited

September 2008

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where:

Ø = 0 .9

f c = 45MPa

t.(hr − hw )   A = 2. B.h f −  2   9 × (318 − 288)   2 A = 2. 90 × 36 −  = 6210mm 2   6 EA = 13200 × 6210 = 81.97 × 10 N ØN ncx = 0.9 × k1 × .k8 × 45 × 6210 ∴ ØN ncx = 251.50.k1.k8 .kN Calculate k8 for buckling about the major axis L=Lax=9910 mm (Purlin length)

 0.823(EA)  S3 =   PE  

0.5

Eq. D2

(

)

 0.823 81.97 × 106  S3 =   PE    67.46 × 106  ∴ S3 =   PE  

NZS3603

0.5

0.5

Calculate Euler buckling load.

PE = PE =

π 2 (EI )y L2E

π 2 × 2338 × 109 9910 2

∴ PE = 235150 N Calculate S3

 67.46 × 106  ∴ S3 =    235150 

0.5

= 16.93

Since 25>S1>10a


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k8 = a1 + a2 .S + a3.S 2 + a4 .S 3 k8 = 0.21 + 0.175 × 16.93+ − 0.0116 × 16.932 +

1 × 16.933 5000

∴ k8 = 0.82 Since k1 = 1.0


∴ ØN ncx = 251.5.k1 .k 8 .kN Calculate k8 for buckling about the minor axis YY Lay=3303 mm (lateral restraint spacing) Calculate Euler buckling load. Design assumes that top edge of purlin is continuously restrained by roof sheeting and bottom flange is effectively restrained by lateral restraint. 2

   2  (EI )y  π   d + yo2  + GJ  Lay   4  PE = (EI )× + (EI )y y o ( y o + 2. y e ) + (EA) (EA)

Eq. D3

NZS3603

2

2   π   360 57.7 × 109 ×  + 180 2  + 1848 × 106    2478   4  PE = 9 2338 × 10 57.7 × 109 180(180 + 2 × 0 ) + + 81.97 × 106 81.97 × 106 ∴ PE = 127582 N

(

) (

)

Calculate S3

 67.46 × 106  ∴ S3 =    127582 

0.5

= 22.99

Since 25>S1>10

k 8 = a1 + a 2 .S + a3 .S 2 + a 4 .S 3 k 8 = 0.21 + 0.175 × 22.99+ − 0.0116 × 22.99 2 +

1 × 22.99 3 5000

∴ k 8 = 0.53 Carter Holt Harvey Limited

September 2008

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Since k1 = 1.0

∴ ØN ncy = 133.3kN > N c* Combined actions

 28.8   9.6   +  = 1.02 ≤ 1.0  29.4   205.8 

Eq. 3.23

NZS3603

Eq. 3.24

NZS3603

2

 28.8   9.6    +  = 1.03 ≤ 1.0  29.4   133.3 

Note that the combined actions are 3 % over, however both axial and bending moment capacities are based on the flange area of the hyJOIST, although listed at 36 mm the hyJOIST flanges have a minimum thickness of 38 mm, therefore increasing the capacity by 3 %.

Consider tension strength

N t* ≤ Ø.N nt

Eq. 3.20

NZS 3603

N nt = k1.k4 . f t . A

Eq. 3.21

NZS 3603

∴ ØN nt = Ø.k1.k4 . f t . A Since the section depth of the individual components is less then 150, size effect factor k11 can be ignored.

where:

Ø = 0 .9

f t = 33MPa k4 = 1.0

Technical Note 82

k1 = 1.0 ØN nt = 0.9 × 1.0 × 1.0 × 33 × 6210 ∴ ØN nt = 184.44kN Combined actions

 28.8   9.6   +  = 1.03 ≤ 1.0  29.4   184.4 

Eq. 3.25

NZS 3603

Note that the combined actions are 3 % over, however both axial and bending moment capacities are based on the flange area of the hyJOIST, although listed at 36 mm the hyJOIST flanges have a minimum thickness of 38 mm, therefore increasing the capacity by 3 %.

The HJ360 90 hyJOIST are suitable for the imposed combined actions from longitudinal winds wind loads and bracing. Carter Holt Harvey Limited

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10.0 References 1. CHH Woodproducts New Zealand, Technical Note 82-07-04, Limit States Design Information for Specific Engineering Design for New Zealand Construction. 2. CHH Woodproducts New Zealand, Engineering Bulletin No. 2 –Rigid Moment Connections using CHH veneer based products. 3. Batchelor, M.L. (1984), Improved Plywood Gussets for Timber Portal Frames, Proceedings of the Pacific Timber Engineering Conference, Auckland 1984, Paper No. 185B. 4. Hutchings B.F (1989), Moment Joist Design, Design, Construct and Detailing in Timber Conference, 15-17 May, 1992, Timber Development Association (NSW) Ltd. 5. Hutchings B.F and Bier H (2000), Timber Engineering Design Made More Accessible, www.chhwoodproducts.co.nz/engineerszone 6. Milner H.R (1987), The Design and Construction of Timber Portal Frames, Chisolm Institute of Technology 7. Milner H.P and Crozier D.A (2000), Structural Design of Timber Portal Frame Buildings, Engineers Australia Pty Ltd. 8. National Association of Forest Industries, Timber Datafile SS1, Timber Portal Frames, National Association of Forest Industries 9. Standards Australia, AS 1720.1-1997 Timber Structures, Part 1: Design methods 10. Standards New Zealand, NZS 3603:1993 Timber structures standard


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Appendix 1 - Mullion deflection, bending and shear equations The following design action equations have been provided from commonly available equations for a series of evenly spaced point loads as idealised for a mullion in service. Due to the nature of loading of mullions ‘n’ in the equations is the number of girts supported by the mullion. It is assumed that the loads applied by girts at each location are equal. Where n is odd:

1  1  1   n − n .3 − 2 1 − n 2      2 n − 1 .P.L M max = 8.n n . P R* = 2

∂=

P.L3 192.EI

(

)

Where n is even:

 1 P.L3 4  .n.3 − 1 + 2  192.EI  2  n  P.L M max = n. 8 n.P R* = 2

∂=

Note: The reaction equation differs slightly from the conventional reaction equation for a series of point loads supported by a simply supported beam. This is to take into account the fact that a girt is located at the base of the mullion.


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Appendix 2 – 90mm thick hy90 compared with 63mm thick hySPAN The following serviceability and strength (bending moment) comparisons between a 63 mm hySPAN section and a 90mm thick hy90 have been provided to illustrate the relative similarities between the sections. The Characteristic Properties have been taken from “Limit States Design Information” Technical Note 82-07-04. Serviceability

EI = E.

B.D 12

Strength 3

90.D 3 = 71250.D 3 12 63.D 3 = 13200. = 69300.D 3 12 > EI hySPAN

EI hy 90 = 9500. EI hySPAN ∴ EI hy 90

B.D 2 6 90.D 2 ØM hy 90 = Ø.k1−8 .35. = 525.D 2 6 63.D 2 ØM hySPAN = Ø.k1−8 .48. = 504.D 2 6 ∴ ØM hy 90 > ØM hySPAN ØM = Ø.k1−8 . f b

As can be seen above, in both serviceability and strength limit states, an equivalent depth 90mm thick hy90 exhibits structural properties exceeding those of a 63mm hySPAN. Please note that these comparisons do not take into account the effects of lateral stability.


September 2008

Portal Frame Design Example

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