Portal Frame Design Example

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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.

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Carter Holt Harvey Limited September 2008 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 4.2 Connection Design 5.0 Mullion Design, Side Wall

5.1 Wind Loading 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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7.1 Wind Loading 7.2 Connection Design

8.0 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

Published by: CHH Woodproducts New Zealand September 2008

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

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Carter Holt Harvey Limited September 2008

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 30.0 m

Building length 60.0 m, consisting of 6 x 10.0 m bays Building Clear Height 6.0 m

Dominant openings 6.0 x 6.0 m in one end and one side wall

Cladding Pierce fixed sheeting of weight 6.0 kg/m2

Region A6, v500 = 45 m/s, v20 = 37 m/s

Terrain Category 3

Directional Multipliers as per AS/NZS 1170.2:2002

This example has been based on relevant current design standards as detailed below: • AS/NZS 1170.0:2002 Structural design actions. Part 0: General principles

• AS/NZS 1170.1:2002 Structural design actions. Part 1: Permanent, imposed and other actions • AS/NZS 1170.2:2002 Structural design actions. Part 2: Wind actions

• NZS 3603:1993 Timber structures standard

• AS 1720.1-1997 Timber structures. Part 1:Design Methods Note: Snow and Earthquake loading have been ignored due to location.

m

kN

w

*

g

=

0 .

25 ×

1 .

6 =

0 .

40 /

Serviceability

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421

5

23

10

39

2

8 9910

40

0

10 2338

384 9910

40 .

0

5

0

1

₆ 2 9 4

Span

or

mm

.

δ

.

δ

G Q

=

∴













×

+

×

=

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.}

kNm M l w M m kN w w * Q G * Q G * Q G * Q G 8 . 9 8 9 . 9 80 . 0 8 . / 80 . 0 40 . 0 5 . 1 17 . 0 2 . 1 5 . 1 2 . 1 2 2 5 . 1 2 . 1 5 . 1 2 . 1 5 . 1 2 . 1 = ∴ × = = = ∴ × + × = + + + +

Check Bending Capacity

The bending capacity of an beam is based on the critical flange stresses due to bending. For composite timber I-beams the bending moment capacity can be based on a lever arm action about the centroid of the flanges 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

kNm

D

A

f

k

Ø

ØM

_bx _t _f 6 1 1

.

10 .

×

−

=

Refer Technical Note 82

where:

(

)

mm D mm A A MPa f k Ø F F t 324 36 360 3060 12 2 288 318 36 90 33 80 . 0 9 . 0 1 2 1 = − = = ∴ × − − × = = = = * 6

6 .

23

10

324 3060

33

8 .

0

9 .

0 M

kNm

ØM

kNm

ØM

bx bx

>

=

∴

×

=

−

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Check Shear Capacity kN v v * Q G * Q G 0 . 4 2 9 . 9 80 . 0 5 . 1 2 . 1 5 . 1 2 . 1 = ∴ × = + +

From table 14, Technical note 82

* 1 1 . 10 6 . 12 8 . 0 6 . 12 . v kN ØV k ØV > = ∴ × = = 2.3 Wind loading

θ_{= 0˚, Lateral wind critical (by inspection)}

a = min(0.2b, 0.2d, h) = 6.0m m kN w w m kN w w c c k k spacing q w* _u _a _l _pe _pi i / 02 . 2 ) 61 . 0 9 . 0 0 . 1 0 . 1 ( 6 . 1 84 . 0 / 63 . 2 ) 61 . 0 9 . 0 5 . 1 0 . 1 ( 6 . 1 84 . 0 ) . . .( . * 2 * 2 * 1 * 1 − = ∴ − × × × × = − = ∴ − × × × × = − = + − + − Calculate weff Calculate Reactions kN R R 84 . 11 000 . 3 63 . 2 955 . 1 02 . 2 * * − − − = ∴ × + × = Calculate Moment kNm M M kNm M M Wu G Wu G Wu Wu 8 . 28 67 . 30 8 9 . 9 17 . 0 9 . 0 7 . 30 0 . 4 0 . 2 02 . 2 2 0 . 3 63 . 2 955 . 4 84 . 11 * 9 . 0 2 * 9 . 0 * 2 * − = ∴ + × × = = ∴ × × − × − × = + − + − − − −

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Carter Holt Harvey Limited September 2008 Calculate w_eff m kN w w eff eff / 50 . 2 9 . 9 8 67 . 30 * 2 * − − = ∴ × = For uplift

m

kN

w

Wu G Wu G

/

35 .

2

50 .

2

17 .

0

9 .

0

* 9 . 0 * 9 . 0 − + +

=

∴

−

×

=

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.

100 0 . 99 10 39 2 8 9910 69 . 1 10 2338 384 9910 69 . 1 5 0 1 / 69 . 1 50 . 2 45 37 6 2 9 4 2 2 Span or mm δ . . . δ m kN w w v v w w w s s u s s = ∴       × × × × + × × × × = = ×       = ×       = − − − −

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 5 . 0 1 . . 1 . 1       = y M EI S E x _{Eq. C1.1, NZS 3603} where: ? 180 2 / 360 10 2338 9 4 = = = × = E x M mm y Nmm EI 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

( )

(

o h

)

ay o y E y y GJ L y D EI M + +               + = . 2 4 2 2 2

π

Eq. C7, NZS 3603 where: mm D mm y mm y_o =360/2=180 _h =−360/2=−180 =360 2 6 2 9

10 1848

10

7 .

57 Nmm

GJ

Nmm

EI

_y

=

×

=

×

mm

L_ay =9910/4=2478 (Restraint at quarter points)

(

)

(

)

kNm

M

E E

7 .

43

180

2

10 1848

2478

180

4

360

10

7 .

57

6 2 2 2 9

=

∴

+

×

+

























+

×

=

₋

π

1 . 18 180 10 7 . 43 10 2338 . 1 . 1 1 5 . 0 6 9 1 = ∴       × × × × = ⇒ S S Calculate k8 Since 25>S>10 76 . 0 1 . 18 5000 1 1 . 18 0116 . 0 1 . 18 175 . 0 21 . 0 . . . 8 3 2 8 3 4 2 3 2 1 8 = ∴ × + × + × + = + + + = − k k S a S a S a a k Since k8>0.73

kNm

D

A

f

k

Ø

ØM

_bx _t _f 6 1 1

.

10 .

×

−

=

Refer Technical Note 82

where: mm D mm A MPa f k Ø F t 324 3060 33 0 . 1 9 . 0 1 2 1 = = = = =

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Carter Holt Harvey Limited September 2008 * 6

4 .

29

10

324 3060

33

0 .

1

9 .

0 M

kNm

ØM

kNm

ØM

bx bx

>

=

∴

×

=

−

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

.

k

1

.

k

8

.

f

_c

.

A

_f

.

D

1

10

6

kNm

−

×

=

Calculate shear and support reaction for wind load. Considering local pressure factors

Case 1 m kN w w m kN w w w c c k k spacing q w*_i _u _a _l _pe _pi _g / 88 . 1 17 . 0 9 . 0 ) 61 . 0 9 . 0 0 . 1 0 . 1 ( 6 . 1 84 . 0 / 48 . 2 17 . 0 9 . 0 ) 61 . 0 9 . 0 5 . 1 0 . 1 ( 6 . 1 84 . 0 . 9 . 0 ) . . .( . * 2 * 2 * 1 * 1 − = ∴ × + − × × × × = − = ∴ × + − × × × × = + − = + − + − kN R R Wu G Wu G 81 . 11 9 . 6 0 . 6 48 . 2 2 9 . 3 88 . 1 9 . 9 1 * 9 . 0 2 * 9 . 0 − = ∴       × × − + × − = + + Case 2 m kN w w m kN w w w c c k k spacing q w*_i _u _a _l _pe _pi _g / 88 . 1 17 . 0 9 . 0 ) 61 . 0 9 . 0 0 . 1 0 . 1 ( 6 . 1 84 . 0 / 09 . 3 17 . 0 9 . 0 ) 61 . 0 9 . 0 2 0 . 1 ( 6 . 1 84 . 0 . 9 . 0 ) . . .( . * 2 * 2 * 1 * 1 − = ∴ × + − × × × × = − = ∴ × + − × × × × = + − = + − + − kN R R Wu G Wu G 2 . 12 4 . 8 0 . 3 09 . 3 2 9 . 6 88 . 1 9 . 9 1 * 9 . 0 2 * 9 . 0 − = ∴       × × − + × − = + +

Calculate dead & live load combined actions

kN R 4.0 2 9 . 9 8 . 0 * = × =

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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,

kN

k

R

k

R

Max

2 .

12

8 .

0

0 .

4 ,

0 .

1

2 .

12 max

1 * max 1 *

−

=

∴











 −

=

Check shear capacity

Since k₁was taken into account in the calculation of designaction, apply k₁=1.0

* 1 6 . 12 6 . 12 0 . 1 6 . 12 . v kN ØV k ØV > = ∴ × = =

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 Table 3, Technical note 82

k ØQ S* ≤ Eq. 4.1, NZS 3603 k n nkQ Q = . . Eq. 4.2, NZS 3603 k n ØnkQ ØQ = . . . where: kN Q k Ø =0.8 ₁ =1.0 _k =0.526 NZS 3603

k=1.4 since nails are through plywood with flat head nails.

k=1.1 since we are proposing 20 nails per connection Cl. 4.2.2.2(g) NZS 3603 (linear interpolation between 1.3 for 50 nails and 1.0 for 4 nails)

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

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

4 . 19 526 . 0 1 . 1 4 . 1 0 . 1 8 . 0 0 . 1 6 . 12 = ∴ × × × × × = × n n

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.

n ØQ S* ≤ Eq. 4.5, NZS 3603 k n nkQ Q = . . Eq. 4.6, NZS 3603 k n ØnkQ ØQ = . . .

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where:

kN Q

k

Ø =0.8* ₁ =1.0 _k =3.303 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 ØV_ps=12.6.k₁

76 . 4 303 . 3 0 . 1 8 . 0 0 . 1 6 . 12 = ∴ × × × = × n n

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

2.6 Lateral restraint design

The lateral restraint system needs to prevent the top and bottom flange of the hyJOIST purlin 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

(

1

)

05 . 0 . . . 34 35 33 * + = r A A n d M k k k F Eq. B9, NZS 3603 where: 0 . 1 33 = k (Wind loading) 4 . 0 34 = k 5 5 , 2 1 22 min 5 , 2 1 min 35 =      + =       + = m k

kNm

M

_A

=

28 .

79

mm d =360

3 =

r

n

(

)

kN F F A A 0 . 2 1 3 360 10 79 . 28 05 . 0 5 4 . 0 0 . 1 6 = ∴ + × × × × × =

Check capacity of lateral restraint – propose 90x45 hyCHORD

kN N

N_c* = _t* =2.0

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 L_ay=1600 mm and L_ax=1600 mm (defined by purlin spacing)

ncx c ØN N* ≤ and ncy c

ØN

N

*

≤

Eq. 3.17, NZS 3603 Minor axis buckling is critical by inspection

A f k k N_ncy = ₁. ₈. _c. Eq. 3.19, NZS 3603 A f k k Ø ØNncy = . 1. 8. c. ∴ where: 2

4050

45

90

45

9 .

0 f

MPa

A

mm

Ø

=

_c

=

×

=

4050 45 9 . 0 × ₁× ₈× × = k k ØN_ncx

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kN k k ØN_ncx =164.0. ₁. ₈. ∴

Calculate k8 for buckling about the minor axis

b L or b L k S 10. ay

3 = whichever is less Eq. 3.15, NZS 3603

6 . 35 45 1600 3 3 = ∴ = S S Since 25>S3>10 23 . 0 6 . 35 5 . 235 . 8 937 . 1 8 5 8 6 = ∴ × = = − k k S a k a Since k1 = 1.0 * 3 . 38 _c ncx kN N ØN = > ∴

Consider tension strength

nt t ØN N* ≤ . Eq. 3.20 NZS 3603 A f k k N_nt = ₁. ₄. _t. Eq. 3.21 NZS 3603 A f k k Ø ØN_nt = . ₁. ₄. _t. ∴ where: 0 . 1 33 9 . 0 = 4 = = f MPa k Ø _t Technical Note 82 2 4050 45 90 mm A= × =

0 .

1

=

k

4050 33 0 . 1 0 . 1 9 . 0 × × × × = nt ØN kN ØN_nt =120.3 ∴

Consider connection between purlins and lateral restraint

Use screws for increased withdrawal capacity for practical purposes

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n ØQ S* ≤ Eq. 4.5, NZS 3603 k n nkQ Q = . . Eq. 4.6, NZS 3603 k n ØnkQ ØQ = . . . where: kN Q k Ø =0.8* 1 =1.0 k =3.303 NZS 3603

Consider Q_k 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

Therefore portion of diameter in penetration = 4.76

Calculate reduction from capacity relating to 7 da Cl. 4.3.2(e), NZS 3603 Reduction factor = 0.68 7 76 . 4 = N Qk =0.68×3.303=2247 ∴ So: 11 . 1 247 . 2 0 . 1 8 . 0 0 . 1 0 . 2 = ∴ × × × = × n n

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

2.7 Purlins subject to axial loads

Purlins in end bays may be subjected to tension and compression forces from braced bays. 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

Deflection

Load Case k₂

Vertical Horizontal

Dead load* 2.0 96.2 mm or span/302 16.2 mm or height/396

Live load 1.0 75.5 mm or span/385 9.6 mm or height/668

Wind loading

Lateral wind₁ 1.0 134.7 mm or span/216 28.4 mm or height/225

Lateral wind2 1.0 74.5 mm or span/390 15.7 mm or height/408

Longitudinal wind1 1.0 108.5 mm or span/268 13.5 mm or height/475

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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

Column Rafter M* N* V* M* N* V* Load Case k1 kN kN kN kN kN kN 1.2G+1.5Q 0.8 -240.0 -84.1 71.5 -268.0 -60.4 50.5 0.9G+Wu - Lat 1.0 271.0 101.0 55.4 293.0 67.3 87.6 1.2G+Wu - Lat 1.0 -276.0 -113.0 62.3 -307.0 -79.2 -95.0

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k1 factored Design Actions Column Rafter M*/k1 N*/k1 V*/k1 M*/k1 N*/k1 V*/k1 Load Case k1 kN kN kN kN kN kN 1.2G+1.5Q 0.8 -300.0 -105.1 89.4 -335.0 -75.5 63.1 0.9G+Wu - Lat 1.0 271.0 101.0 55.4 293.0 67.3 87.6 1.2G+Wu - Lat 1.0 -276.0 -113.0 62.3 -307.0 -79.2 -95.0 3.5 Rafter Design

A check of the capacity of main frame members of a timber portal frame involves a check of combined bending 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

0 .

1

* *

≤













+













ncx c nx x

ØN

N

ØM

M

Eq. 3.23 NZS 3603 0 . 1 * 2 * ≤         +       ncy c nx x ØN N ØM M Eq. 3.24 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

n ØM M* ≤ Eq. 3.3 NZS 3603 Z f k k k k M_n = 1. 4. 5. 8. _b. Eq. 3.4 NZS 3603

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: MPa f k k Ø=0.9 4 = 5 =1.0 _b = 48 Technical Note 82

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Carter Holt Harvey Limited September 2008 83 . 0 900 300 300 167 . 0 11 167 . 0 11 =       = ∴       = k d k Cl. 2.4.6 AS1720.1 3 6 2 2 10 15 . 12 6 90 900 6 . mm Z b d Z × = ∴ × = =

kNm

k

ØM

k

ØM

n n 8 1 6 8 1

.

65 .

435

10

15 .

12

48

83 .

0

0 .

1

0 .

1

9 .

0 =

∴

×

=

Since k1=0.8 kNm k ØM_n =348.52. ₈ 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

5 . 0 1 . . 1 . 1       = y M EI S E x _{Eq. C1} _{NZS 3603}

Since for 900x90 hySPAN

mm y Nmm EI_x 450 2 900 10 17 . 72 12 90 900 13200 12 4 3 = = × = × × = Therefore: 5 . 0 9 1 10 418 . 176       _× = E M S

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.

(21)

Consider Moment Diagram

( )

[

]

0.5 5 .GJ EI L c M _y ay E        = Eq. C3 NZS3603 where: 0 268 0 = =

β

β = ratio of bending moments between buckling restraints

5 . 5 5 = c Table C1 NZS3603 4 9 3 10 71 . 721 12 900 90 13200 Nmm EI_y = × × = ×

Since for rectangular sections:

3

63 .

0

1

3

B

D

B

J



×











×

−

=

Eq. C2 NZS 3603 2 9 3

10

25 .

135

3

90

900

90

63 .

0

1

660 Nmm

GJ



×

=

×











×

−

×

=

Therefore:

[

]

kNm M M E E 68 . 350 10 25 . 135 10 71 . 721 4900 5 . 5 ₉ ₉ 0.5 = ∴ × × ×       = From previous:

(22)

Carter Holt Harvey Limited September 2008 43 . 22 10 68 . 350 10 418 . 176 1 5 . 0 6 9 1 = ∴       × × = S S Since 25>S1>10 56 . 0 43 . 22 5000 1 43 . 22 0116 . 0 43 . 22 175 . 0 21 . 0 . . . 8 3 2 8 3 4 2 3 2 1 8 = ∴ × + × + × + = + + + = − k k S a S a S a a k kNm ØM_n =348.52×0.56=195.2 ∴

ØMn<M* so consider flybrace. The flybrace needs to be located relative to purlin spacing along the rafter but also needs to offer the appropriate level of stability to the rafter. Propose 3rd_{purlin from eave.} Consider moment diagram

( )

[

]

0.5 5 .GJ EI L c M _y ay E        = Eq. C3 NZS 3603 where: 64 . 0 0 . 268 1 . 171 = =

β

82 . 3 5 = c Table C1 NZS 3603 Therefore:

(23)

[

]

kNm M M E E 51 . 685 10 25 . 135 10 71 . 721 1741 82 . 3 ₉ ₉ 0.5 = ∴ × × ×       = From previous: 04 . 16 10 51 . 685 10 418 . 176 1 5 . 0 6 9 1 = ∴       × × = S S Since 25>S1>10 3 4 2 3 2 1 8 a a .S a .S a .S k = + + + Cl C2.10 NZS 3603 86 . 0 04 . 16 5000 1 04 . 16 0116 . 0 04 . 16 175 . 0 21 . 0 8 3 2 8 = ∴ × + × + × + = − k k * 7 . 299 86 . 0 52 . 348 kNm M ØMn = × = > ∴

Check remaining unrestrained section

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

[

]

kNm M M E E 78 . 543 10 25 . 135 10 71 . 721 3160 5 . 5 ₉ ₉ 0.5 = ∴ × × ×       = From previous: 01 . 18 10 78 . 543 10 418 . 176 1 5 . 0 6 9 1 = ∴       × × = S S Since 25>S1>10 3 4 2 3 2 1 8 a a .S a .S a .S k = + + + NZS 3603 Cl C2.10 77 . 0 01 . 18 5000 1 01 . 18 0116 . 0 01 . 18 175 . 0 21 . 0 8 3 2 8 = ∴ × + × + × + = − k k

(24)

Carter Holt Harvey Limited September 2008 * 4 . 268 77 . 0 52 . 348 kNm M ØM_n = × = > ∴

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

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

Calculate Euler Moment

[

]

kNm M M E E 33 . 605 10 25 . 135 10 71 . 721 1600 1 . 3 ₉ ₉ 0.5 = ∴ × × ×       = From previous: 07 . 17 10 33 . 605 10 418 . 176 1 5 . 0 6 9 1 = ∴       × × = S S Since 25>S₁>10 3 4 2 3 2 1 8 a a .S a .S a .S k = + + + Cl C2.10 NZS 3603 81 . 0 07 . 17 5000 1 07 . 17 0116 . 0 07 . 17 175 . 0 21 . 0 8 3 2 8 = ∴ × + × + × + = − k k * 3 . 282 81 . 0 52 . 348 kNm M ØM_n = × = > ∴

Consider column action Major axis buckling XX

(25)

A f k k Nncx = 1. 8. c. Eq. 3.18 NZS 3603 A f k k Ø ØN_ncx = . 1. 8. _c. ∴ where: 2

81000

90

900

45

9 .

0 f

MPa

A

mm

Ø

=

_c

=

×

=

81000 45 . 9 . 0 × 1× 8× × = k k ØN_ncx kN k k ØNncx =3280.5. 1. 8. ∴

Calculate k8 for buckling about the major axis

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

d L or d L k S 10. ax

2 = whichever is less NZS 3603 Eq. 3.14

k10 = 1.0 (Conservative) 80 . 15 900 14221 0 . 1 2 2 = ∴ × = S S Since 25>S2>10 3 4 2 3 2 1 8 a a .S a .S a .S k = + + + NZS 3603 Cl C2.10 87 . 0 80 . 15 5000 1 80 . 15 0116 . 0 80 . 15 175 . 0 21 . 0 8 3 2 8 = ∴ × + × + × + = − k k Since k1 = 0.8 * 1 . 2278 _c ncx kN N ØN = > ∴

Minor axis buckling YY From previous: kN k k ØN_ncx =3280.5. ₁. ₈. ∴

Calculate k8 for buckling about the minor axis YY Lay=1600 mm (purlin spacing)

(26)

Carter Holt Harvey Limited September 2008 b L or b L k S 10. ay

3 = whichever is less Eq. 3.15 NZS 3603

78 . 17 90 1600 3 3 = ∴ = S S Since 25>S3>10 3 4 2 3 2 1 8 a a .S a .S a .S k = + + + Cl C2.10 NZS 3603 78 . 0 78 . 17 5000 1 78 . 17 0116 . 0 78 . 17 175 . 0 21 . 0 8 3 2 8 = ∴ × + × + × + = − k k Since k₁ = 0.8 * 03 . 2047 _c ncx kN N ØN = > ∴ Combined actions 0 . 1 92 . 0 1 . 2278 4 . 60 7 . 299 0 . 268 ≤ =       +       Eq. 3.23 NZS 3603 0 . 1 83 . 0 0 . 2047 4 . 60 7 . 299 0 . 268 2 ≤ =       +       Eq. 3.24 NZS 3603

3.5.2 Combined bending and tension Design Criteria 0 . 1 * * ≤         +       n nt t ØM M ØN N Eq. 3.25 NZS 3603

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:

kNm k k ØM_n =435.65. 1. 8

(27)

kNm k ØM_n =435.65. 8

Calculate k₈

Calculate Euler moment, M_E

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

( )

[

]

0.5 5 .GJ EI L c M _y ay E        = Eq. C3 NZS 3603 where: 60 . 0 2 . 293 2 . 176 = =

β

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

9 . 3 5 = c Eq. C3 NZS 3603 2 9 4 9

10

25 .

135

10

71 .

721 Nmm

GJ

Nmm

EI

_y

×

=

×

=

Therefore:

[

]

(28)

Carter Holt Harvey Limited September 2008 22 . 15 10 54 . 761 10 418 . 176 1 5 . 0 6 9 1 = ∴       × × = S S Since 25>S1>10 3 4 2 3 2 1 8 a a .S a .S a .S k = + + + NZS 3603 Cl C2.10 90 . 0 22 . 15 5000 1 22 . 15 0116 . 0 22 . 15 175 . 0 21 . 0 8 3 2 8 = ∴ × + × + × + = − k k * 1 . 392 90 . 0 65 . 435 kNm M ØM_n = × = > ∴

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

Propose flybracing as detailed below

(29)

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 c₅= 5.5, L_ay = 5183 mm Region 2 c5 ~ 3.1, Lay = 2x1600 = 3200 mm Region 3 c₅= 3.1, L_ay = 2x(1050+229) = 2558 mm Since: _       = ay E L c function M 5

Therefore Region 2 is critical buckling region

[

]

kNm M M E E 66 . 302 10 25 . 135 10 71 . 721 3200 1 . 3 ₉ ₉ 0.5 = ∴ × × ×       = ⇒ From previous: 14 . 24 10 66 . 302 10 418 . 176 1 5 . 0 6 9 1 = ∴       × × = S S Since 25>S1>10 3 4 2 3 2 1 8 a a .S a .S a .S k = + + + Cl C2.10 NZS 3603 49 . 0 14 . 24 5000 1 14 . 24 0116 . 0 14 . 24 175 . 0 21 . 0 8 3 2 8 = ∴ × + × + × + = − k k

(30)

Carter Holt Harvey Limited September 2008 * 5 . 213 49 . 0 65 . 435 kNm M ØM_n = × = > ∴

Consider tension strength

nt t ØN N* ≤ . Eq. 3.20 NZS 3603 A f k k N_nt = ₁. ₄. _t. 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.

Therefore ØN_nt =Ø.k1.k4.k11.f_t.A where: 0 . 1 33 9 . 0 = ₄ = = f MPa k Ø _t Technical Note 82 mm A=900×90=81000

0 .

1

=

k

74 . 0 900 150 150 167 . 0 11 167 . 0 11 =       = ∴       = k d k Cl. 2.4.6 AS1720.1 81000 33 74 . 0 0 . 1 0 . 1 9 . 0 × × × × × = nt ØN kN ØN_nt =1780.2 ∴ Combined actions 0 . 1 84 . 0 5 . 213 8 . 171 2 . 1780 9 . 69 ≤ =       +       q. 3.25 NZS 3603

Calculate Shear Capacity

n ØV V* ≤ Eq. 3.3 NZS 3603 S s n k k k f A V = ₁. ₄. ₅. . Eq. 3.4 NZS 3603 where: MPa f k k k Ø s 5.3 0 . 1 0 . 1 9 . 0 5 4 1 = = = = = Technical Note 82

(31)

2 54000 3 90 900 2 3 / . . 2 mm A d b A S S = × × = ∴ = Cl 3.2.3.1 NZS 3603 *

6 .

257 54000

3 .

5

0 .

1

0 .

1

9 .

0 V

kN

V

S S

>

=

∴

×

=

∴

φ

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

3.5.3 Flybrace design

Critical Design Moment at flybrace location M* = -171.1 kNm, where k1=0.8 Calculate force on lateral restraint

(

1

)

05 . 0 . . . 34 35 33 * + = r A A n d M k k k F Eq. B9, NZS 3603 where: 0 . 1 33 =

k (Dead and live loads)

4 . 0 34 = k 1 5 , 2 1 1 min 5 , 2 1 min 35 =      + =       + = m k

kNm

M

_A

=

−

171 .

1

mm d =900

1 =

r

n

(

)

kN F F A A 9 . 1 1 1 900 10 1 . 171 05 . 0 1 4 . 0 0 . 1 6 = ∴ + × × × × × =

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

Check capacity of flybrace – propose 90x45 hyCHORD

kN N

N_c* _t* 1.90 = =

Calculate force in brace

kN Cos N N_c _t 2.7 ) 45 ( 90 . 1 * * = = =

(32)

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) ncx c ØN N* ≤ and ncy c

ØN

N

*

≤

Eq. 3.17 NZS 3603 Minor axis buckling is critical by inspection

A f k k N_ncy = 1. 8. _c. Eq. 3.19 NZS 3603 A f k k Ø ØN_ncy = . 1. 8. _c. ∴ where: 2

4050

45

90

45

9 .

0 f

MPa

A

mm

Ø

=

_c

=

×

=

4050 45 9 . 0 × ₁× ₈× × = k k ØN_ncx kN k k ØN_ncx =164.03. 1. 8. ∴

Calculate k₈ for buckling about the minor axis

b L or b L k S 10. ay

98 . 16 45 764 3 3 = ∴ = S S Since 25>S1>10 3 4 2 3 2 1 8 a a .S a .S a .S k = + + + Cl C2.10 NZS 3603 82 . 0 98 . 16 5000 1 98 . 16 0116 . 0 98 . 16 175 . 0 21 . 0 8 3 2 8 = ∴ × + × + × + = − k k Since k1 = 1.0 * 50 . 134 _c ncx kN N ØN = > ∴

(33)

Consider tension strength nt t ØN N* . ≤ Eq. 3.20 NZS 3603 A f k k N_nt = 1. 4. _t. Eq. 3.21 NZS 3603 A f k k Ø ØN_nt = . ₁. ₄. _t. ∴ where: 0 . 1 33 9 . 0 = ₄ = = f MPa k Ø _t Technical Note 82 mm A=90×45=4050

0 .

1

=

k

4050 33 0 . 1 0 . 1 9 . 0 × × × × = nt ØN kN ØN_nt =120.3 ∴

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

n ØQ S* ≤ Eq. 4.5, NZS 3603 k n nkQ Q = . . Eq. 4.6, NZS 3603 k n ØnkQ ØQ = . . . where: kN Q k Ø =0.8* ₁ =1.0 _k =3.303 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

Therefore portion of diameter in penetration = 4.76

(34)

Carter Holt Harvey Limited September 2008 Reduction factor = 0.68 7 76 . 4 = N Q_k =0.68×3.303=2247 ∴ So: kN ØQ ØQ n n 0 . 4 247 . 2 2 0 . 1 8 . 0 = ∴ × × × =

Consider screws in tension

n ØQ N* ≤ Eq. 4.8, NZS 3603 k n nk pQ Q = . . . Eq. 4.9, NZS 3603 k n Ønk pQ ØQ = . . . . where: mm p mm N Q k Ø=0.8* ₁ =1.0 _k =79.5 / =35 NZS 3603

kN ØQ ØQ n n 45 . 4 5 . 79 35 0 . 1 2 8 . 0 = ∴ × × × × =

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

(35)

3.6 Column Design

3.6.1 Combined bending and compression Design Criteria

0 .

1

* *

≤













+













ncx c nx x

ØN

N

ØM

M

Eq. 3.23 NZS 3603 0 . 1 * 2 * ≤         +       ncy c nx x ØN N ØM M Eq. 3.24 NZS 3603

Critical Design Actions Critical load case - 1.2G+1.5Q

M* = -240.0 kNm N_c* = -84.1 kN V* = 71.5 kN Consider Bending Moment Capacity

From previous: kNm k k ØM_n =435.65. 1. 8 Since k₁=0.8 kNm k ØM_n =348.52. 8 Calculate k8 For 900x90 hySPAN: 5 . 0 9 1 10 418 . 176       _× = E M S

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.

(36)

Consider bending moment diagram

( )

[

]

0.5 5 _EI _._GJ L c M _y ay E        = Eq. C3 NZS 3603 where: Region 1 59 . 0 0 . 303 3 . 179 = − − =

β

92 . 3 5 = c Eq. C3 NZS 3603 mm L_ay =2530 Region 2 0 3 . 179 0 = − =

β

5 . 5 5 = c Eq. C3 NZS 3603 mm L_ay =3470 2 9 4 9

10

25 .

135

10

71 .

721 Nmm

GJ

Nmm

EI

_y

×

=

×

=

Therefore Region 1 is critical:

[

]

(37)

09 . 19 10 08 . 484 10 418 . 176 1 5 . 0 6 9 1 = ∴       × × = S S Since 25>S1>10 3 4 2 3 2 1 8 a a .S a .S a .S k = + + + Cl C2.10 NZS 3603 71 . 0 09 . 19 5000 1 09 . 19 0116 . 0 09 . 19 175 . 0 21 . 0 8 3 2 8 = ∴ × + × + × + = − k k kNm ØM_n =348.52×0.71=247.5 ∴

Consider column action Major axis buckling XX From previous: kN k k ØN_ncx =3280.5. ₁. ₈. ∴

Calculate k₈ for buckling about the major axis

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

d L or d L k S 10. ax

k₁₀ = 1.0 (conservative) Fig. 3.5 NZS 3603 67 . 6 900 6000 0 . 1 2 2 = ∴ × = S S Since 10<S2 k8 =1.0and k1 = 0.8 * 4 . 2624 _c ncx kN N ØN = > ∴

Minor axis buckling YY From previous: kN k k ØN_ncx =3280.5. ₁. ₈. ∴

(38)

Calculate k₈ for buckling about the minor axis YY L_ay=1660 mm (girt spacing) b L or b L k S 10. ay

44 . 18 90 1660 3 3 = ∴ = S S Since 25>S3>10 3 4 2 3 2 1 8 a a .S a .S a .S k = + + + Cl C2.10 NZS 3603 75 . 0 44 . 18 5000 1 44 . 18 0116 . 0 44 . 18 175 . 0 21 . 0 8 3 2 8 = ∴ × + × + × + = − k k Since k1 = 0.8 * 3 . 1968 _c ncx kN N ØN = > ∴ Combined actions 0 . 1 0 . 1 4 . 2624 1 . 84 5 . 247 0 . 240 ≤ =       +       Eq. 3.23 NZS 3603 0 . 1 98 . 0 3 . 1968 1 . 84 5 . 247 0 . 240 2 ≤ =       +       Eq. 3.24 NZS 3603

3.6.2 Combined bending and tension Design Criteria 0 . 1 * * ≤         +       n nt t ØM M ØN N Eq. 3.25 NZS 3603

Critical load case - 0.9G+W_u Lateral wind

M* = 271.0 kNm N t* = 101.0 kN V* = 55.4 kN Consider Bending Moment Capacity

(39)

kNm k ØM_n =435.65. 8

Calculate k₈

Calculate Euler moment, M_E

Consider compression edge restrained by grits at 1660 c/c. Bending Moment Diagram

( )

[

]

0.5 5 _EI _._GJ L c M _y ay E        = NZS3603 Eq. C3 where: 66 . 0 0 . 271 7 . 177 = =

β

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

78 . 3 5 = c NZS3603 Eq. C3 2 9 4 9

10

25 .

135

10

71 .

721 Nmm

GJ

Nmm

EI

_y

×

=

×

=

Therefore:

[

]

kNm M M E E 43 . 711 10 25 . 135 10 71 . 721 1660 78 . 3 ₉ ₉ 0.5 = ∴ × × ×       =

(40)

Carter Holt Harvey Limited September 2008 From previous: 75 . 15 10 43 . 711 10 418 . 176 1 5 . 0 6 9 1 = ∴       × × = S S Since 25>S₁>10 3 4 2 3 2 1 8 a a .S a .S a .S k = + + + NZS 3603 Cl C2.10 87 . 0 75 . 15 5000 1 75 . 15 0116 . 0 75 . 15 175 . 0 21 . 0 8 3 2 8 = ∴ × + × + × + = − k k * 0 . 379 87 . 0 65 . 435 kNm M ØM_n = × = > ∴

Consider tension strength Since

k

₁

=

1 .

0

,

k

₁₁

=

0 .

74

81000 33 74 . 0 0 . 1 0 . 1 9 . 0 × × × × × = nt ØN kN ØN_nt =1780.2 ∴ Combined actions 0 . 1 77 . 0 0 . 379 0 . 271 2 . 1780 0 . 101 ≤ =       +      

(41)

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

Knee Ridge M* N* V* M* N* V* Load Case K1 kNm kN kN kNm kN kN 1.35G 0.6 -123.0 -31.6 19.2 69.7 -19.0 2.5 1.2G+1.5Q 0.8 -324.0 -83.1 50.5 183.6 -50.1 6.6 0.9G+Wu – Lat 1.0 362.0 102.2 -54.3 -156.9 71.1 -5.0 1.2G+Wu –Lat 1.0 -382.0 -111.7 -65.2 171.2 -75.5 14.5 0.9G+Wu – Long 1.0 239.1 64.6 55.1 -117.0 65.9 25.3 1.2G+Wu –Long 1.0 -295.4 -60.3 -69.7 161.0 -56.6 7.5 k1 factored Gusset Design Actions

Knee Ridge M*/k1 N*/k1 V*/k1 M*/k1 N*/k1 V*/k1 Load Case K1 kNm kN kN kNm kN kN 1.35G 0.6 -205.0 -52.6 32.0 116.2 -31.7 4.2 1.2G+1.5Q 0.8 -405.0 -103.9 63.1 229.5 -62.6 8.3 0.9G+Wu – Lat 1.0 362.0 102.2 -54.3 -156.9 71.1 -5.0 1.2G+Wu –Lat 1.0 -382.0 -111.7 -65.2 171.2 -75.5 14.5 0.9G+Wu – Long 1.0 239.1 64.6 55.1 -117.0 65.9 25.3 1.2G+Wu –Long 1.0 -295.4 -60.3 -69.7 161.0 -56.6 7.5

3.7.1 Knee gusset design

The capacity of a plywood gusset is based on the critical depth at which the gusset bends, which is 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:

θ

tan 2 1 1       − + − + = L D D L D Depth_cs Design Criteria 0 . 1 * 2 * * ≤       +       +       ni i ni i nc c ØV V ØM M ØN N Eq. 6.17 NZS 3603

0 .

1

* 2 * *

≤













+













+













ni i ni i nt t

ØV

V

ØM

M

ØN

N

Eq. 6.18 NZS 3603

It is typical that the design shear and tension action effects have little influence on the size of a gusset and can 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.

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 N_t* = 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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ni i ØM M* ≤ Eq. 6.9 NZS 3603 6 . . . . . . 2 15 14 8 1 d t f k k k k M e pb ni = Eq. 6.10 NZS 3603

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

82. Therefore 6 . . . . . . . . 2 15 14 11 8 1 d t f k k k k k Ø ØM e pb ni =

Since the gussets are in pairs:













=

6 .

.

2

2 15 14 11 8 1

d

t

f

k

Ø

ØM

e pb ni

Propose 42 mm 4 x-band LVL, where:

9 . 0 = Ø

?

1

=

k

0 . 1 8 =

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

0 .

1

14

=

k

(moisture content < 18%) 0 . 1 15 =

k (only parallel plies are being considered)

167 . 0 11 300       = d k Cl. 2.4.6 AS1720.1 MPa f_b =48

(

)

(

)

mm t_e = 42− 4×3.5 =28













_×

×

=

6

28

48

0 .

1

0 .

1

0 .

1

9 .

0 .

2

2 11 1

d

k

ØM

_ni kNm d k k ØM_n 2 11 1. . 2 . 403 × =

For 900x90 hySPAN portal frame with 7.5˚ pitch

( )

80 . 0 2 . 1177 300 2 . 1177 5 . 7 tan 1200 2 900 1 1 900 1200 900 11 167 . 0 11 = ∴       = = ∴       × − + − + = k k mm d d

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kNm

k

ØM

k

ØM

ni ni

.

0 .

447

2 .

1177

8 .

0

2 .

403

1 2 1

×

=

∴

×

=

Calculate Shear Capacity

ni p

ØV

V

*

≤

Eq. 6.15 NZS 3603 d t f k k k k k V_ni . . . _ps.. 3 2 18 15 14 8 1 = Eq. 6.16 NZS 3603 where: MPa f Ø =0.9 _ps =5.3 Technical Note 82

?

1

=

k

0 .

1

14

=

k

(moisture content < 15 %) 0 . 1 15 = k (face grain = 0˚)     × × × × × × × × × = k d Vni 1.0 1.0 1.0 1.0 5.3 42 3 2 9 . 0 2 ₁ kN d k Vni =267.12× 1. . ∴ Since d =1177.2mm kN k V_ni =314.45× ₁. ∴

Consider tension capacity

nt t ØN N* ≤ . Eq. 6.11 NZS 3603 d t f k k k N_nt = 1. 14. 15. _pt. _t. Eq. 6.12 NZS 3603 d t f k k k Ø ØN_nt = . ₁. ₁₄. ₁₅. _pt. _e. ∴ where: MPa f Ø =0.9 pt =33 Technical Note 82

?

1

=

k

0 .

1

14

=

k

(moisture content < 15 %) 0 . 1 15 = k (face grain = 0˚) mm

t_e =28 (parallel plies only)

[

k d

]

ØN_nt =20.9× 1×1.0×1.0×33×28× kN d k ØN_nt =1663.2. 1. . ∴