portal frame design example
TRANSCRIPT
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LVL Portal Frame DesignCHH Woodproducts New Zealand
Disclaimer
This design example has been prepared solely to provide guidance and recommendations to suitably qualified engineers and other suitably qualifieddesign professionals for diligent and professional use by them (and no other person) in the calculation of design solutions for LVL portal frame systemsin accordance with currently available New Zealand Standards.
To the best of Carter Holt Harveys knowledge and belief this example has been prepared in accordance with currently available technology andexpertise however good design and construction practice may be affected by factors outside the control of Carter Holt Harvey and beyond the control and
scope of this design example. This example is not intended to be used as the sole recipe, nor is it to be considered the authoritative method, forproducing the relevant design and it is assumed that the relevant designers will employ sound and current engineering knowledge and will take allreasonable care when designing LVL portal frame solutions using this example.
Accordingly, Carter Holt Harvey and its employees, agents and design professionals accept no liability or responsibility whatsoever and howsoeverarising for any losses, damages, costs or expenses (whether direct, indirect and/or consequential) arising from any errors or omissions which may becontained in this example, nor does it accept responsibility to any persons whatsoever for designs prepared in reliance upon this example or any otherinformation contained in this document.
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Table of Contents
1.0 Introduction
2.0 Purlin design2.1 Dead Load2.2 Live load2.3 Wind load2.4 Proposed Purlin Layout2.5 Connection Design2.6 Lateral restraint design2.7 Purlins supporting axial loading
3.0 Portal frame design3.1 Proposed Portal Frame3.2 Serviceability3.3 Strength3.4 Design Actions3.5 Rafter Design
3.5.1 Combined bending and compression3.5.2 Combined bending and tension
3.5.3 Flybrace design3.6 Column Design
3.6.1 Combined bending and compression3.6.2 Combined bending and tension3.6.3 Flybrace design
3.7 Gusset Design3.7.1 Knee Gusset Design3.7.2 Ridge Gusset Design3.7.3 Nail Ring Design
3.7.3.1 Knee Nail Ring Design
3.7.3.2 Ridge Nail Ring Design3.8 Column to Footing Design
4.0 Girt Design, Side Wall4.1 Wind Loading4.2 Connection Design
5.0 Mullion Design, Side Wall5.1 Wind Loading5.2 Connection Design
6.0
Eaves Beam Design6.1 Wind Loading6.2 Connection Design
7.0 Girt Design, End Wall
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7.1
Wind Loading7.2 Connection Design
8.0 Mullion Design, End Wall8.1 Wind Loading8.2 Connection Design
9.0 Longitudinal Bracing Design
10.0 Bibliography
Appendix 1 - Mullion deflection, bending and shear equations
Appendix 2 - 90mm thick hy90 compared with 63mm thick hySPAN
Published by: CHH Woodproducts New ZealandSeptember 2008
Enquires : Free call 0800 808 131Free fax 0800 808 132
Web : www.chhwoodproducts.co.nz/engineerszone
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1.0 Introduction
This design example has been provided as an aid to engineers in the development of design solutions for LVLand I-beam portal frame systems. The development of loading and the design of footings are not covered aspart of this example as their nature is not specific to timber. The design example has been prepared assumingthe building is proposed for Auckland, is within an Industrial Estate, and is subject to the following siteinformation:
Building Span 30.0 mBuilding length 60.0 m, consisting of 6 x 10.0 m baysBuilding Clear Height 6.0 m
Dominant openings 6.0 x 6.0 m in one end and one side wallCladding Pierce fixed sheeting of weight 6.0 kg/m2
Region A6, v500 = 45 m/s, v20= 37 m/sTerrain Category 3Directional Multipliers as per AS/NZS 1170.2:2002
This example has been based on relevant current design standards as detailed below: AS/NZS 1170.0:2002 Structural design actions. Part 0: General principles AS/NZS 1170.1:2002 Structural design actions. Part 1: Permanent, imposed and other actions AS/NZS 1170.2:2002 Structural design actions. Part 2: Wind actions NZS 3603:1993 Timber structures standard AS 1720.1-1997 Timber structures. Part 1:Design Methods
Note: Snow and Earthquake loading have been ignored due to location.
Other Referenced Design Documents: Technical Note 82-07-04 - Limit States Design Information for Specific Engineering Design for New
Zealand Construction. Mitek Specifiers and Users Manual.
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2.0 Purlin Design
Purlin Span 10,000-90 = 9910 mmPurlin Spacing 1600 mm (max.)
Propose HJ360 90 hyJOIST for use as purlin
Typically a hyJOIST purlin roof system becomes cost effective at spans above 6.0 m whilst hySPAN or MSG pinepulins remain cost effective for spans less than 6.0 m.
2.1 Dead load
Assume roof sheeting mass of 6.0 kg/m2plus a miscellaneous load of 1.0 kg/m2
kN/m.w
tself_weigh...
w
g
g
170
1000
8196107
*
*
=
+
=
Serviceability
Deflection of timber i-beams requires the consideration of shear deflection as well as bending deflection.Additional guidance on the calculation of shear deflection can be found in many Timber Design texts and is briefly
discussed in Technical Note 82. Timber components subjected to long term loads such as dead load require theconsideration of creep effects. Table 2.5, NZS 3603:1993 demonstrates the relationship between duration of loadand creep. The k2factor is applied to elastic deflections. LVL products are considered dry at the time of supplyand can be assumed to have a moisture content less than 18%.
Refer Technical Note 82 for Section and Material Properties.
495020
103928
9910170
102338384
9910170502
8384
56
2
9
424
2
2
Spanormm.
.
....
.GA
w.l
.EI
.w.lk
)(k
G
wx
shearbendingT
=
+
=
+=
+=
Serviceability limits for timber purlins are the same as those applied to other building products. For long termdeflection of industrial purlins span/300 or 30.0 mm are deemed acceptable.
2.2 Live load
Live load of 0.25 kPa applied in accordance AS/NZS 1170.1:2002 Table 3.2.
mkNw*g /40.06.125.0 ==
Serviceability
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421523
103928
9910400
102338384
991040.05016
2
9
4
Spanormm.
....
G
Q
=
+
=
Strength
Based on respective k1and load combination factors, combined dead and live load design actions will always bemore critical for design than permanent loads where low roof masses (less than 20 kg/m 2) are applied.
kNmM
lwM
mkNw
w
*
QG
*
QG
*
QG
*
QG
8.9
8
9.980.0
8
.
/80.0
40.05.117.02.1
5.12.1
22
5.12.1
5.12.1
5.12.1
=
==
=
+=
+
+
+
+
Check Bending Capacity
The bending capacity of an I-beam is based on the critical flange stresses due to bending. For composite timber I-beams the bending moment capacity can be based on a lever arm action about the centroid of the flanges withone flange in tension and the other in compression for a single span application. The restraint offered to thecompression flange is instrumental in the capacity of the I-beam. Further guidance on the bending momentcapacities of I-beams may be found in Technical note 82.
Purlin design assumes the use of pierce fixed roof sheeting providing continuous lateral restraint to thetop flange of the purlin. Since compression edge is fully restrained k8=1.0.
So for bending about XX axis
Since k8>0.73
kNmDAfkM ftbx6
11 10....
= Refer Technical Note 82
where:
( )
mmD
mmA
A
MPafk
F
F
t
32436360
3060
122
2883183690
3380.09.0
1
2
1
==
=
=
===
*
6
6.23
103243060338.09.0
MkNmM
kNmM
bx
bx
>=
=
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Check Shear Capacity
kNv
v
*
QG
*
QG
0.4
2
9.980.0
5.12.1
5.12.1
=
=
+
+
From table 14, Technical note 82
*
1
1.10
6.128.06.12.
vkNV
kV
>=
==
2.3 Wind loading
= 0, Lateral wind critical (by inspection)
a = min(0.2b, 0.2d, h) = 6.0m
mkNw
w
mkNw
w
cckkspacingqw pipelau*
i
/02.2
)61.09.00.10.1(6.184.0
/63.2
)61.09.05.10.1(6.184.0
)...(.
*
2
*
2
*
1
*
1
=
=
=
=
=
+
+
Calculate weff
Calculate Reactions
kNR
R
84.11
000.363.2955.102.2
*
*
=
+=
Calculate Moment
kNmM
M
kNmM
M
WuG
WuG
Wu
Wu
8.28
67.308
9.917.09.0
7.30
0.40.202.22
0.363.2955.484.11
*
9.0
2
*
9.0
*
2*
=
+
=
=
=
+
+
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Calculate weff
mkNw
w
eff
eff
/50.2
9.9
867.30
*
2
*
=
=
For uplift
mkNw
w
WuG
WuG
/35.2
50.217.09.0
*
9.0
*
9.0
+
+
=
=
Serviceability
To obtain the serviceability wind load the ultimate uniform loads can be factored by the square of the ratioserviceability wind speed to ultimate wind speed.
1000.99
103928
991069.1
102338384
991069.1501
/69.150.245
37
6
2
9
4
2
2
Spanormm
...
mkNw
wv
vw
w
w
s
s
u
ss
=
+
=
=
=
=
The acceptance of serviceability is at the engineers discretion. On the basis of applied local pressure factors andthe instantaneous nature of the wind gust span/100 is deemed acceptable.
Strength
Since the tension flange is fully restrained under uplift actions and the hyJOIST purlin is a composite section, useAppendix C of NZS3603:1993 for stability calculations.
Check Capacity
Calculate S1
5.0
1.
.1.1
=
yM
EIS
E
x Eq. C1.1, NZS 3603
where:
?
1802/360102338 49
=
===
E
x
M
mmyNmmEI Technical Note 82
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Eqn. C7 may be employed due to the continuous restraint offered to the tension flange by the pierce fixed sheeting.A suitably designed lateral restraint system provides intermediate buckling restraint to the purlins.
Calculate Euler Buckling Moment
( )
( )ho
ay
oy
Eyy
GJL
yD
EI
M+
+
+
=.2
4
2
2
2
Eq. C7, NZS 3603
where:
mmDmmymmy ho 3601802/3601802/360 ===== 2629
101848107.57 NmmGJNmmEIy ==
mmLay 24784/9910 == (Restraint at quarter points)
( )
( )
kNmM
M
E
E
7.43
1801802
1018482478
1804
360107.57
6
2
2
2
9
=
+
+
+
=
1.18
180107.43
102338.1.1
1
5.0
6
9
1
=
=
S
S
Calculate k8
Since 25>S>10
76.0
1.185000
11.180116.01.18175.021.0
...
8
32
8
3
4
2
3218
=
+++=
+++=
k
k
SaSaSaak
Since k8>0.73
kNmDAfkM ftbx6
11 10....
= Refer Technical Note 82
where:
mmD
mmA
MPafk
F
t
324
3060
330.19.0
1
2
1
=
=
===
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*
6
4.29
103243060330.19.0
MkNmM
kNmM
bx
bx
>=
=
Note: Where k8< 0.73 the moment capacity becomes a function of the compression flange bucklingrather than the tension flange being critical. The moment capacity equation is altered to represent thiswhere the characteristic tension stress is replaced by the product of the stability factor k8and thecharacteristic compression stress.
ie. kNmDAfkkM fcbx6
181 10.....
=
Calculate shear and support reaction for wind load.
Considering local pressure factors
Case 1
mkNww
mkNw
w
wcckkspacingqw gpipelau*
i
/88.117.09.0)61.09.00.10.1(6.184.0
/48.2
17.09.0)61.09.05.10.1(6.184.0
.9.0)...(.
*
2
*
2
*
1
*
1
=
+=
=
+=
+=
+
+
kNR
R
WuG
WuG
81.11
9.60.648.22
9.388.1
9.9
1
*
9.0
2*
9.0
=
+=
+
+
Case 2
mkNw
w
mkNw
w
wcckkspacingqw gpipelau*
i
/88.1
17.09.0)61.09.00.10.1(6.184.0
/09.3
17.09.0)61.09.020.1(6.184.0
.9.0)...(.
*
2
*
2
*
1
*
1
=
+=
=
+=
+=
+
+
kNR
R
WuG
WuG
2.12
4.80.309.32
9.688.1
9.9
1
*
9.0
2*
9.0
=
+=
+
+
Calculate dead & live load combined actions
kNR 0.42
9.98.0*=
=
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Timber capacity is dependant on the duration of the load in question, this must be taken into account in thedetermination of the critical load case. One method of assessing the critical design load is to remove the durationof load factor,k1, from the capacity equation and divide the load action effect by k1,
kNk
R
k
RMax
2.12
8.0
0.4,
0.1
2.12max
1
*
max
1
*
=
=
Check shear capacity
Since k1 was taken into account in the calculation of designaction, apply k1=1.0
*
1
6.12
6.120.16.12.
vkNV
kV
>=
== Table 14, Technical note 82
Therefore the HJ360 63 hyJOIST is suitable for use as a purlin based on the implied loading ata spacing not exceeding 1600 mm
2.4 Proposed Purlin Layout
2.5 Connection design
Connection of hyJOIST purlins to LVL rafters needs to ensure that the structural integrity of both the hyJOIST purlinand the hySPAN rafter are maintained. Connection to the hyJOIST by nailing through the plywood web providesthe most cost effective method of connection for purlins typically subject to high wind loads (please note this type
of connection is not recommended for i-beams subject to high permanent and/or live loads). Nailing throughplywood allows for nailing close to the end/edge of the plywood. Packing out the web and using proprietary joisthangers can also provide a suitable connection however the cost of the packing, brackets and labour involved canmake this an expensive alternative.
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Purlin connection blocks, or seating blocks as they are sometimes called, have been used in a numberof design situations for connection of C or I beam purlins where the connection block is either screwedor nailed to the rafter and the web of the composite purlin is connected directly to the connectionblock. A purlin connection block is proposed for connection using 2.87 diameter nails through theplywood web and 14g type 17 screws through the connection block to the rafter. Target the connectionfor design shear capacity, Vpsof the purlin.
Note: The selection of a suitable purlin connection block needs to take into account the end and edge distances ofthe fasteners as well as the spacing along and across the grain. The use of 4 x-banded connection block reducesthe tendency of the long band to split, allowing for the spacing of fasteners into the face to be similar along the
grain to across the grain. The orientation of the connection block is important where the plywood web is fixed tothe face of the connection block.
Calculate minimum number of 2.87 FH nails
Joint Group J5 Table 3, Technical note 82
kQS * Eq. 4.1, NZS 3603
kn QknQ ..= Eq. 4.2, NZS 3603
kn QknQ ...=
where:
kNQk k 526.00.18.0 1 === NZS 3603
k=1.4 since nails are through plywood with flat head nails.k=1.1 since we are proposing 20 nails per connection Cl. 4.2.2.2(g) NZS 3603
(linear interpolation between 1.3 for 50 nails and 1.0 for 4 nails)
Other k modification factors are not relevant as timber is dry, nails are in single shear and are nailed into theedge or face of the timber.
From Table 14, Technical Note 82 Vps=12.6.k1
4.19
526.01.14.10.18.00.16.12
=
=
n
n
Say 20/50x2.87 FH nails, nailed through plywood web into purlin connection block
Calculate minimum number of 14g type 17 Hex Head screwsType 17 screws are preferred for timber connection as they are a self drilling screws through the timber.
Joint Group J4 Table 3, Technical note 82
nQS * Eq. 4.5, NZS 3603
kn QknQ ..= Eq. 4.6, NZS 3603
kn QknQ ...=
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where:
kNQk k 303.30.1*8.0 1 === NZS 3603
Other k modification factors are not relevant as timber is dry, screws are in single shear and are screwed into theedge or face of the timber.*=0.8 is applied as Type 17 screws are as reliable as nails in service.
From Table 14, technical note 82 Vps=12.6.k1
76.4
303.30.18.00.16.12
=
=
n
n
Say 5/100x14g type 17 Hex Head screws, screwed through the purlin connection block into the rafter.
Proposed Purlin Connection
2.6 Lateral restraint design
The lateral restraint system needs to prevent the top and bottom flange of the hyJOIST purlin frommoving independently of each other. Many systems are appropriate but may require the fabrication ofspecial components. One of the most effective systems is to use hyJOIST pieces together with ahyCHORD bottom flange restraint and continuous mild steel galvanised strap over the top, as shownbelow.
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Calculate force on lateral restraint
( )1
05.0...
353433
*
+=
r
AA
nd
MkkkF Eq. B9, NZS 3603
where:
0.133=k (Wind loading)
4.034 =k
55,2
122min5,
2
1min
35 =
+=
+=
m
k
kNmMA 79.28=
mmd 360=
3=rn
( )
kNF
F
A
A
0.2
13360
1079.2805.054.00.1
6
=
+
=
Check capacity of lateral restraint propose 90x45 hyCHORD
kNNN tc 0.2**==
Typically a 45 mm thick section is recommended to allow for a 75mm long screw through both the lateral restraintand into the flange of the hyJOIST. Using hyCHORD for the lateral restraint is a good choice given its high strengthand lower cost.
Consider column action
Since Lay=1600 mm and Lax=1600 mm (defined by purlin spacing)
ncxc NN *
and
ncyc NN * Eq. 3.17, NZS 3603
Minor axis buckling is critical by inspection
AfkkN cncy ... 81= Eq. 3.19, NZS 3603
AfkkN cncy .... 81=
where:
240504590459.0 mmAMPaf c ====
4050459.081 = kkNncx
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kNkkNncx ...0.164 81=
Calculate k8for buckling about the minor axis
b
Lor
b
LkS
ay.103 = whichever is less Eq. 3.15, NZS 3603
6.35
45
1600
3
3
=
=
S
S
Since 25>S3>10
23.0
6.355.235
.
8
937.1
8
586
=
=
=
k
k
Sak a
Since k1= 1.0
*3.38 cncx NkNN >=
Consider tension strength
ntt NN .* Eq. 3.20 NZS 3603
AfkkN tnt ... 41= Eq. 3.21 NZS 3603
AfkkN tnt .... 41=
where:
0.1339.04 === kMPaf t Technical Note 82
240504590 mmA ==
0.11 =k
4050330.10.19.0 =ntN
kNNnt 3.120=
Consider connection between purlins and lateral restraint
Use screws for increased withdrawal capacity for practical purposes
Calculate minimum number of 14g type 17 Hex Head screws
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Joint Group J4 Table 3, Technical note 82
nQS * Eq. 4.5, NZS 3603
kn QknQ ..= Eq. 4.6, NZS 3603
kn QknQ ...=
where:
kNQk k 303.30.1*8.0 1 === NZS 3603
Other k modification factors are not relevant as timber is dry, screws are in single shear and are screwed into theedge or face of the timber.*=0.8 is applied as Type 17 screws are as reliable as nails in service.
Consider Qkreduction due to the penetration into the receiving member (Purlin/blocking)
Penetration = 75-45 = 30 mm
Since da = 6.3 mm Table 4.5, NZS 3603
Therefore portion of diameter in penetration = 4.76
Calculate reduction from capacity relating to 7 da Cl. 4.3.2(e), NZS 3603
Reduction factor = 68.07
76.4=
NQk 2247303.368.0 ==
So:
11.1
247.20.18.00.10.2
=
=
n
n
Say 2/75x14g type 17 Hex Head screws, screwed through the purlin connection block into the rafter.
2.7 Purlins subject to axial loads
Purlins in end bays may be subjected to tension and compression forces from braced bays. Theseforces need to be considered in the design capacity. Refer to section 9.0, Longitudinal bracing.
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3.0 Portal Frame Design
The following portal frame has been analysed using elastic structural analysis with Microstran. Elasticstructural analysis of a timber portal frame differs little from that applied to steel members except forthe different section and material properties. For solid timber a five percent allowance for sheardeflection is included in the average modulus of elasticity which removes any need for the separateconsideration of shear deflection.
To achieve portal frame action rigid connections need to be made at both the ridge and eave. One ofthe most efficient methods of providing rigid connections is via use of nailed plywood gussets. Theadditional stiffness provided by the knee and ridge gussets is generally ignored in analysis.
3.1 Proposed Portal Frame
Refer Technical Note 82 for Material Properties.
3.2 Serviceability
Serviceability design limits for timber and steel buildings are very similar where the consideration ofcladding and absolute clearances need to be taken into account in the relative stiffness of the frame.Short term duration of loading for wind, live and earthquake loads may be calculated by applying aduration of load factor of 1, hence using the elastic deflection directly from analysis packages. For longterm loads the effects of creep need to be taken into account. NZS 3603 Table 2 defines k2as 2.0 forloading of twelve months or more where the moisture content is less than 18%.
Serviceability 900x90 hySPAN portal frameDeflection
Load Case k2 Vertical HorizontalDead load* 2.0 96.2 mm or span/302 16.2 mm or height/396
Live load 1.0 75.5 mm or span/385 9.6 mm or height/668Wind loadingLateral wind1 1.0 134.7 mm or span/216 28.4 mm or height/225Lateral wind2 1.0 74.5 mm or span/390 15.7 mm or height/408Longitudinal wind1 1.0 108.5 mm or span/268 13.5 mm or height/475Longitudinal wind2 1.0 64.6 mm or span/450 8.1 mm or height/792
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* It is typical to pre-camber the portal by its un-factored deflection (ie. Approx 50 mm in the case)
3.3 Strength
The selection of design moments is important in the design of timber portal frames. The nature of theinteraction of gussets provide specific locations for the selection of critical design actions for thedesign of rafters, columns gussets and nail rings. Hutchings and Bier [2000] provide guidance on thedesign moment locations as shown below.
Location A Rafter design actions at kneeLocation B Column design actionsLocation C Knee gusset design actionsLocation D Gusset to rafter at knee connection actionsLocation E Gusset to column connection actionsLocation F Ridge gusset design actionsLocation G Ridge gusset to rafter design actions
A further check along the rafter is require where the critical design actions may not to be at thegusseted location and should be taken as the maximum along the rafter.
3.4 Design Actions
The consideration of critical design actions also needs to take in account the effect of duration of load factors forcapacity, hence affecting the determination of critical load case. As with steel portal frames the bending momentdiagram should also be taken into account together with the lateral and torsional restraint offered by purlins,
girts and flybraces. The following design actions have been tabled as being of interest, other actions have beendismissed by inspection. The point of contraflexure is within close proximity for each case meaning that thecritical load case can be determined by inspection.
Critical Design Actions
Column Rafter
M* N* V* M* N* V*Load Case k1 kN kN kN kN kN kN
1.2G+1.5Q 0.8 -240.0 -84.1 71.5 -268.0 -60.4 50.5
0.9G+Wu - Lat 1.0 271.0 101.0 55.4 293.0 67.3 87.6
1.2G+Wu - Lat 1.0 -276.0 -113.0 62.3 -307.0 -79.2 -95.0
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k1factored Design Actions
Column Rafter
M*/k1 N*/k1 V*/k1 M*/k1 N*/k1 V*/k1
Load Case k1 kN kN kN kN kN kN
1.2G+1.5Q 0.8 -300.0 -105.1 89.4 -335.0 -75.5 63.1
0.9G+Wu - Lat 1.0 271.0 101.0 55.4 293.0 67.3 87.6
1.2G+Wu - Lat 1.0 -276.0 -113.0 62.3 -307.0 -79.2 -95.0
3.5 Rafter Design
A check of the capacity of main frame members of a timber portal frame involves a check of combined bending andbuckling action, both in plane and out of plane, and a check of combined bending and tension.
3.5.1 Combined bending and compression
Design Criteria
0.1
**
+
ncx
c
nx
x
N
N
M
M Eq. 3.23 NZS 3603
0.1
*2
*
+
ncy
c
nx
x
N
N
M
M Eq. 3.24 NZS 3603
Critical Design Actions
Critical load case - 1.2G+1.5Q
M* = -268.0 kNm Nc* = -60.4 kN V* = 50.5 kN
Consider Bending Moment Capacity
nMM * Eq. 3.3 NZS 3603
ZfkkkkM bn ..... 8541= Eq. 3.4 NZS 3603
For solid sections with member depths greater than 300 mm, apply size factor (k11, AS 1720.1). Forfurther information refer AS1720.1 (Clause 2.4.6) or Technical Note 82.
Therefore ZfkkkkkM bn ....... 118541=
where:
MPafkk b 480.19.0 54 ==== Technical Note 82
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83.0900
300
300
167.0
11
167.0
11
=
=
=
k
dk
Cl. 2.4.6 AS1720.1
36
22
1015.12
6
90900
6
.
mmZ
bdZ
=
==
kNmkkM
kkM
n
n
81
681
..65.435
1015.124883.00.10.19.0
=
=
Since k1=0.8
kNmkMn 8.52.348=
Calculate k8The timber structures standard does not talk about critical flange like the steel structures standard however
similar principles apply to the restraint of LVL beams. Guidance is provided for solid sections in Clauses 3.2.5 of
NZS 3603:1993 for end-supported beams with discrete restraint to the compression edge (Cl 3.2.5.2) and tensionedge continuously restrained (Cl 3.2.5.3). Typically these can be useful in the calculation of slenderness of simplebeams and secondary framing however composite sections and members within structural frames requireanalysis using Appendix C of NZS3603:1993 for slenderness calculations.
Consider slenderness equation
5.0
1.
.1.1
=
yM
EIS
E
x Eq. C1 NZS 3603
Since for 900x90 hySPAN
mmy
NmmEIx
4502
900
1017.7212
9090013200
412
3
==
=
=
Therefore:5.0
9
1
10418.176
=
EMS
Calculate Euler moment, ME
Consider compression edge unrestrained from edge of column to point of contraflexure.
Some authors including Milner [1997] have developed theories based on the contribution of lateral restraintoffered to the tension edge by purlins and girts, such theories are beyond the scope of this example.
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Consider Moment Diagram
( )[ ] 5.05 .GJEIL
cM y
ay
E
= Eq. C3 NZS3603
where:
0268
0== = ratio of bending moments between buckling restraints
5.55 =c Table C1 NZS3603
493
1071.72112
9009013200 NmmEIy =
=
Since for rectangular sections:
363.01
3BD
D
BJ
= Eq. C2 NZS 3603
293
1025.1353
90900
900
9063.01660 NmmGJ =
=
Therefore:
[ ]
kNmM
M
E
E
68.350
1025.1351071.7214900
5.5 5.099
=
=
From previous:
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43.22
1068.350
10418.176
1
5.0
6
9
1
=
=
S
S
Since 25>S1>10
56.0
43.225000
143.220116.043.22175.021.0
...
8
32
8
3
4
2
3218
=
+++=
+++=
k
k
SaSaSaak
kNmMn 2.19556.052.348 ==
Mn
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[ ]
kNmM
M
E
E
51.685
1025.1351071.7211741
82.3 5.099
=
=
From previous:
04.16
1051.685
10418.176
1
5.0
6
9
1
=
=
S
S
Since 25>S1>10
3
4
2
3218 ... SaSaSaak +++= Cl C2.10 NZS 3603
86.0
04.165000
104.160116.004.16175.021.0
8
32
8
=
+++=
k
k
*7.29986.052.348 MkNmMn >==
Check remaining unrestrained section
M* = 171.1 kNm Lay = 3160 mm c5= 5.5
Therefore:
[ ]
kNmM
M
E
E
78.543
1025.1351071.7213160
5.5 5.099
=
=
From previous:
01.18
1078.543
10418.176
1
5.0
6
9
1
=
=
S
S
Since 25>S1>10
3
4
2
3218 ... SaSaSaak +++= NZS 3603 Cl C2.10
77.0
01.185000
101.180116.001.18175.021.0
8
32
8
=
+++=
k
k
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*
4.26877.052.348 MkNmMn >==
Consider region along rafter between point of contraflexure and apex along the rafter.
Bending Moment Diagram
Since purlins provide restraint to compression edge, Lay= 1600 mm where c5= 3.1 (moment ratiobetween purlins = 0 (conservative)).
Calculate Euler Moment
[ ]
kNmM
M
E
E
33.605
1025.1351071.7211600
1.3 5.099
=
=
From previous:
07.17
1033.605
10418.176
1
5.0
6
9
1
=
=
S
S
Since 25>S1>10
3
4
2
3218 ... SaSaSaak +++= Cl C2.10 NZS 3603
81.0
07.175000
107.170116.007.17175.021.0
8
32
8
=
+++=
k
k
*
3.28281.052.348 MkNmMn >==
Consider column action
Major axis buckling XX
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AfkkN cncx ... 81= Eq. 3.18 NZS 3603
AfkkN cncx .... 81=
where:
28100090900459.0 mmAMPaf c ====
8100045.9.081 = kkNncx
kNkkNncx ...5.3280 81=
Calculate k8for buckling about the major axis
L=Lax=14221 mm (rafter length from ridge to column)
d
Lor
d
LkS ax
.10
2 = whichever is less NZS 3603 Eq. 3.14
k10= 1.0 (Conservative)
80.15
900
142210.1
2
2
=
=
S
S
Since 25>S2>10
3
4
2
3218 ... SaSaSaak +++= NZS 3603 Cl C2.10
87.0
80.155000
180.150116.080.15175.021.0
8
32
8
=
+++=
k
k
Since k1= 0.8
*1.2278 cncx NkNN >=
Minor axis buckling YY
From previous:
kNkkNncx ...5.3280 81=
Calculate k8for buckling about the minor axis YY
Lay=1600 mm (purlin spacing)
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bLor
bLkS
ay.103= whichever is less Eq. 3.15 NZS 3603
78.17
90
1600
3
3
=
=
S
S
Since 25>S3>103
4
2
3218 ... SaSaSaak +++= Cl C2.10 NZS 3603
78.0
78.175000
178.170116.078.17175.021.0
8
328
=
+++=
k
k
Since k1= 0.8
*03.2047 cncx NkNN >=
Combined actions
0.192.01.2278
4.60
7.299
0.268=
+
Eq. 3.23 NZS 3603
0.183.00.2047
4.60
7.299
0.268 2
=
+
Eq. 3.24 NZS 3603
3.5.2 Combined bending and tension
Design Criteria
0.1
**
+
nnt
t
M
M
N
N Eq. 3.25 NZS 3603
Critical Design Actions
Critical load case - 0.9G+WuLateral wind
M* = 293.0 kNm (at eave)M* = -171.8 kNm (along rafter)Nt* = 69.9 kN V* = 87.6 kN
Consider Bending Moment Capacity
From previous:
kNmkkMn 81..65.435=
Since k1=1.0, wind gust
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kNmkMn 8.65.435=
Calculate k8
Calculate Euler moment, ME
Consider compression edge restrained by purlins at 1600 c/c until point of contraflexure.
Bending Moment Diagram
( )[ ] 5.05 .GJEIL
cM y
ay
E
= Eq. C3 NZS 3603
where:
60.02.293
2.176== = ratio of bending moments between buckling restraints (purlins)
9.35 =
c Eq. C3 NZS 3603
29
49
1025.135
1071.721
NmmGJ
NmmEIy
=
=
Therefore:
[ ]
kNmM
M
E
E
54.761
1025.1351071.7211600
90.3 5.099
=
=
From previous:
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22.15
1054.761
10418.176
1
5.0
6
9
1
=
=
S
S
Since 25>S1>10
3
4
2
3218 ... SaSaSaak +++= NZS 3603 Cl C2.10
90.0
22.155000
122.150116.022.15175.021.0
8
32
8
=
+++=
k
k
*1.39290.065.435 MkNmMn >==
Check remaining sections between points of contraflexure (ie. Negative moment along the rafter)
Propose flybracing as detailed below
Consider region along rafter between point of contraflexure and apex along the rafter.
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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, Lay= 5183 mmRegion 2 c5 ~ 3.1, Lay= 2x1600 = 3200 mmRegion 3 c5 = 3.1, Lay= 2x(1050+229) = 2558 mm
Since:
=
ay
EL
cfunctionM 5
Therefore Region 2 is critical buckling region
[ ]
kNmM
M
E
E
66.302
1025.1351071.7213200
1.3 5.099
=
=
From previous:
14.24
1066.302
10418.176
1
5.0
6
9
1
=
=
S
S
Since 25>S1>10
3
4
2
3218
... SaSaSaak +++= Cl C2.10 NZS 3603
49.0
14.245000
114.240116.014.24175.021.0
8
32
8
=
+++=
k
k
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*
5.21349.065.435 MkNmMn >==
Consider tension strength
ntt NN .* Eq. 3.20 NZS 3603
AfkkN tnt ... 41= Eq. 3.21 NZS 3603
For solid sections with member depths greater than 150 mm, apply k11size factor for tension. Forfurther information refer AS1720.1 (Clause 2.4.6) or Technical Note 82.
Therefore AfkkkN tnt ..... 1141=
where:
0.1339.04 === kMPaf t Technical Note 82
mmA 8100090900 ==
0.11=k
74.0900150
150
167.0
11
167.0
11
=
=
=
k
dk
Cl. 2.4.6 AS1720.1
810003374.00.10.19.0 =ntN
kNNnt 2.1780=
Combined actions
0.184.05.213
8.171
2.1780
9.69
=
+
q. 3.25 NZS 3603
Calculate Shear Capacity
nVV * Eq. 3.3 NZS 3603
Ssn AfkkkV .... 541= Eq. 3.4 NZS 3603
where:
MPafkk
k
s 3.50.1
0.19.0
54
1
===
==
Technical Note 82
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254000
3
9090023/..2
mmA
dbA
S
S
=
=
= Cl 3.2.3.1 NZS 3603
*6.257
540003.50.10.19.0
VkNV
V
S
S
>=
=
Use 900x90 hySPAN as rafter with flybraces to locations as detailed.
3.5.3 Flybrace design
Critical Design Moment at flybrace location M* = -171.1 kNm, where k1=0.8
Calculate force on lateral restraint
( )1
05.0...
353433
*
+=
r
AA
nd
MkkkF Eq. B9, NZS 3603
where:
0.133 =k (Dead and live loads)4.034 =k
15,2
11min5,
2
1min35 =
+=
+=
mk
kNmMA 1.171=
mmd 900=
1=rn
( )kNF
F
A
A
9.1
11900
101.17105.014.00.1
6
=
+
=
Note: FA is the horizontal force and is shared between two components, one in tension and one incompression.
Check capacity of flybrace propose 90x45 hyCHORD
kNNN tc 90.1**==
Calculate force in brace
kNCos
NN tc 7.2)45(
90.1**===
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Typically a 45 mm thick section is recommended to allow for a 75mm long screw through both the flybrace and intothe flange of the hyJOIST. Using hyCHORD for the lateral restraint is a good choice given its high strength andlower cost.
Consider column action
Since Lay=765 mm and Lax=765 mm (defined by brace length)
ncxc NN *
and
ncyc NN * Eq. 3.17 NZS 3603
Minor axis buckling is critical by inspection
AfkkN cncy ... 81= Eq. 3.19 NZS 3603
AfkkN cncy .... 81=
where:
240504590459.0 mmAMPaf c ====
4050459.081 = kkNncx
kNkkNncx ...03.164 81=
Calculate k8for buckling about the minor axis
b
Lor
b
LkS
ay.103= whichever is less Eq. 3.15 NZS 3603
98.16
45
764
3
3
=
=
S
S
Since 25>S1>10
3
4
2
3218 ... SaSaSaak +++= Cl C2.10 NZS 3603
82.0
98.165000
198.160116.098.16175.021.0
8
32
8
=
+++=
k
k
Since k1= 1.0
*50.134 cncx NkNN >=
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Consider tension strength
ntt NN .* Eq. 3.20 NZS 3603
AfkkN tnt ... 41= Eq. 3.21 NZS 3603
AfkkN tnt .... 41=
where:
0.1339.04 === kMPaf t Technical Note 82
mmA 40504590 == 0.1
1=k
4050330.10.19.0 =ntN
kNNnt 3.120=
Consider connection between purlins and rafters and flybrace
Screws are required to provide tension connection to rafter/purlin
Calculate minimum number of 14g type 17 Hex Head screws
Joint Group J4 Table 3, Technical note 82
nQS * Eq. 4.5, NZS 3603
kn QknQ ..= Eq. 4.6, NZS 3603
kn QknQ ...=
where:
kNQk k 303.30.1*8.0 1 === NZS 3603
Other k modification factors are not relevant as timber is dry, screws are in single shear and are screwed into theedge or face of the timber.*=0.8 is applied as Type 17 screws are as reliable as nails in service.
Consider Qkreduction due to the penetration into the receiving member (Purlin/blocking)
Penetration = 75-45 = 30 mm
Since da= 6.3 mm Table 4.5, NZS 3603
Therefore portion of diameter in penetration = 4.76
Calculate reduction from capacity relating to 7 da Cl. 4.3.2(e), NZS 3603
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Reduction factor = 68.0776.4 =
NQk 2247303.368.0 ==
So:
kNQ
Q
n
n
0.4
247.220.18.0
=
=
Consider screws in tension
nQN * Eq. 4.8, NZS 3603
kn QpknQ ...= Eq. 4.9, NZS 3603
kn QpknQ ....=
where:
mmpmmNQk k 35/5.790.1*8.0 1 ==== NZS 3603
Other k modification factors are not relevant as timber is dry, screws are in single shear and are screwed into theedge or face of the timber.*=0.8 is applied as Type 17 screws are as reliable as nails in service.
kNQ
Q
n
n
45.4
5.79350.128.0
=
=
Say 2/75x14g type 17 Hex Head screws, screwed through pre-drilled holes in flybrace into rafter andpurlin.
Proposed flybrace connection
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3.6 Column Design
3.6.1 Combined bending and compression
Design Criteria
0.1
**
+
ncx
c
nx
x
N
N
M
M Eq. 3.23 NZS 3603
0.1
*2
*
+
ncy
c
nx
x
N
N
M
M Eq. 3.24 NZS 3603
Critical Design Actions
Critical load case - 1.2G+1.5Q
M* = -240.0 kNm Nc* = -84.1 kN V* = 71.5 kN
Consider Bending Moment Capacity
From previous:
kNmkkMn 81..65.435=
Since k1=0.8
kNmkMn 8.52.348=
Calculate k8
For 900x90 hySPAN:
5.09
110418.176
=
EMS
Calculate Euler moment, ME
Girts provide tension edge restraint to the outside of the outside of the frame. By inspection from therafter analysis one flybrace is proposed at the middle girt, 3490 mm from the ground.
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Consider bending moment diagram
( )[ ] 5.05 .GJEIL
cM y
ay
E
= Eq. C3 NZS 3603
where:
Region 1
59.00.303
3.179=
= = ratio of bending moments between buckling restraints
92.35=c Eq. C3 NZS 3603
mmLay 2530=
Region 2
03.179
0
== = ratio of bending moments between buckling restraints
5.55 =c Eq. C3 NZS 3603
mmLay 3470=
29
49
1025.135
1071.721
NmmGJ
NmmEIy
=
=
Therefore Region 1 is critical:
[ ]
kNmM
M
E
E
08.484
1025.1351071.7212530
92.3 5.099
=
=
From previous:
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09.19
1008.484
10418.176
1
5.0
6
9
1
=
=
S
S
Since 25>S1>10
3
4
2
3218 ... SaSaSaak +++= Cl C2.10 NZS 3603
71.0
09.19
5000
109.190116.009.19175.021.0
8
32
8
=
+++=
k
k
kNmMn 5.24771.052.348 ==
Consider column action
Major axis buckling XX
From previous:
kNkkNncx ...5.3280 81=
Calculate k8for buckling about the major axis
L=Lax=6000 mm (column height from rafter to footing)
d
Lor
d
LkS ax
.10
2 = whichever is less Eq. 3.14 NZS 3603
k10= 1.0 (conservative) Fig. 3.5 NZS 3603
67.6
900
60000.1
2
2
=
=
S
S
Since 10=
Minor axis buckling YY
From previous:
kNkkNncx ...5.3280 81=
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Calculate k8for buckling about the minor axis YY
Lay=1660 mm (girt spacing)
b
Lor
b
LkS
ay.103 = whichever is less Eq. 3.15 NZS 3603
44.18
90
1660
3
3
=
=
S
S
Since 25>S3>10
3
4
2
3218 ... SaSaSaak +++= Cl C2.10 NZS 3603
75.0
44.185000
144.180116.044.18175.021.0
8
32
8
=
+++=
k
k
Since k1= 0.8
*
3.1968 cncx NkNN >=
Combined actions
0.10.14.2624
1.84
5.247
0.240=
+
Eq. 3.23 NZS 3603
0.198.03.1968
1.84
5.247
0.240 2
=
+
Eq. 3.24 NZS 3603
3.6.2 Combined bending and tension
Design Criteria
0.1
**
+
nnt
t
M
M
N
N Eq. 3.25 NZS 3603
Critical Design Actions
Critical load case - 0.9G+WuLateral wind
M* = 271.0 kNm Nt* = 101.0 kN V* = 55.4 kN
Consider Bending Moment Capacity
From previous, since k1=1.0, wind gust
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kNmkMn 8.65.435=
Calculate k8
Calculate Euler moment, ME
Consider compression edge restrained by grits at 1660 c/c.
Bending Moment Diagram
( )[ ] 5.05 .GJEIL
cM y
ay
E
= NZS3603 Eq. C3
where:
66.00.271
7.177 == = ratio of bending moments between buckling restraints (grits)
78.35=c NZS3603 Eq. C3
29
49
1025.135
1071.721
NmmGJ
NmmEIy
=
=
Therefore:
[ ]
kNmM
M
E
E
43.711
1025.1351071.7211660
78.3 5.099
=
=
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From previous:
75.15
1043.711
10418.176
1
5.0
6
9
1
=
=
S
S
Since 25>S1>10
3
4
2
3218 ... SaSaSaak +++= NZS 3603 Cl C2.10
87.0
75.155000175.150116.075.15175.021.0
8
328
=
+++=
k
k
*0.37987.065.435 MkNmMn >==
Consider tension strength
Since 0.11=k , 74.0
11=k
810003374.00.10.19.0 =ntN
kNNnt 2.1780=
Combined actions
0.177.00.379
0.271
2.1780
0.101=
+
Use 900x90 hySPAN as column with flybraces to locations as detailed.
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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 connectionsbecause the x-band plies help reduce the tendency of the long band plies to split. This allows the nailspacing to be governed by the grain direction of the rafter or column which ever the gusset is beingfastened to.
Plywood is available in Stress Grade F11 from Carter Holt Harvey in thicknesses up to and including 25mm. For thicknesses over 25 mm required for large span portal frames CHH have developed 4 x-band
hySPAN sheets (2400x1200) in a 42mm thickness allowing 28 mm (8 plies) of parallel plies.
Design actions can be factored by the duration of load factor k1for comparison in the determination ofthe critical design action.
Gusset Design Actions
Knee Ridge
M* N* V* M* N* V*
Load Case K1 kNm kN kN kNm kN kN
1.35G 0.6 -123.0 -31.6 19.2 69.7 -19.0 2.5
1.2G+1.5Q 0.8 -324.0 -83.1 50.5 183.6 -50.1 6.6
0.9G+Wu Lat 1.0 362.0 102.2 -54.3 -156.9 71.1 -5.01.2G+Wu Lat 1.0 -382.0 -111.7 -65.2 171.2 -75.5 14.5
0.9G+Wu Long 1.0 239.1 64.6 55.1 -117.0 65.9 25.3
1.2G+Wu Long 1.0 -295.4 -60.3 -69.7 161.0 -56.6 7.5
k1factored Gusset Design Actions
Knee Ridge
M*/k1 N*/k1 V*/k1 M*/k1 N*/k1 V*/k1
Load Case K1 kNm kN kN kNm kN kN
1.35G 0.6 -205.0 -52.6 32.0 116.2 -31.7 4.2
1.2G+1.5Q 0.8 -405.0 -103.9 63.1 229.5 -62.6 8.3
0.9G+Wu Lat 1.0 362.0 102.2 -54.3 -156.9 71.1 -5.01.2G+Wu Lat 1.0 -382.0 -111.7 -65.2 171.2 -75.5 14.5
0.9G+Wu Long 1.0 239.1 64.6 55.1 -117.0 65.9 25.3
1.2G+Wu Long 1.0 -295.4 -60.3 -69.7 161.0 -56.6 7.5
3.7.1 Knee gusset design
The capacity of a plywood gusset is based on the critical depth at which the gusset bends, which is ahorizontal line across the centroid of the rafter and column intersection as shown below.
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Geometrically, assuming the rafter depthand column depth are equal, the criticalsection for the knee connection may becalculated by:
tan2
11
+
+=
L
D
DLDDepthcs
Design Criteria
0.1
*2
**
+
+
ni
i
ni
i
nc
c
V
V
M
M
N
N Eq. 6.17 NZS 3603
0.1
*2**
+
+
ni
i
ni
i
nt
t
V
V
M
M
N
N Eq. 6.18 NZS 3603
It is typical that the design shear and tension action effects have little influence on the size of a gusset and can inmany cases be omitted from calculation such is their effect on sizing. Compression loads are generally pastthrough in bearing and not required for consideration in gusset design.
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] proposesa bilinear stress distribution along the critical section while Hutchings [1987] methodology assumes atriangulated stress distribution across the critical section and recommends the application of a size factor.
Hutchings [1987] methodology is applied in this example. This methodology is suitable for application to bothopening and closing moments of portal frames, and has been used on many portal frame structures. Milner andCrosier [2000] propose a similar calculation based on a triangulated stress distribution but propose an alternatecritical section and omit the use of the size factor.
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nii MM *
Eq. 6.9 NZS 3603
6
......
2
151481
dtfkkkkM epbni = Eq. 6.10 NZS 3603
Now include size factor - for further information on size factor, k11refer AS1720.1 (Clause 2.4.6) or Technical Note82.
Therefore6
........
2
15141181
dtfkkkkkM epbni =
Since the gussets are in pairs:
=
6
.........2
2
15141181
dtfkkkkkM epbni
Propose 42 mm 4 x-band LVL, where:
9.0=
?1=k
0.18=k (localised, gusset edges are restrained by gusset stiffeners)
0.114 =k (moisture content < 18%)
0.115 =k (only parallel plies are being considered)167.0
11
300
=
dk Cl. 2.4.6 AS1720.1
MPafb 48=
( )( ) mmte 285.3442 ==
=
6
28480.10.10.19.0.2
2
111
dkkMni
kNmdkkMn 2111 ..2.403 =
For 900x90 hySPAN portal frame with 7.5 pitch
( )
80.02.1177
300
2.1177
5.7tan12002
90011
9001200900
11
167.0
11
=
=
=
+
+=
k
k
mmd
d
Calculate bending moment capacity
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kNmkM
kM
ni
ni
.0.447
2.11778.02.403
1
21
=
=
Calculate Shear Capacity
nip VV * Eq. 6.15 NZS 3603
dtfkkkkkV psni ........3
218151481
= Eq. 6.16 NZS 3603
where:
MPaf ps 3.59.0 == Technical Note 82
?1=k
0.114=k (moisture content < 15 %)
0.115 =k (face grain = 0)
= dkVni 423.50.10.10.10.1
3
29.02 1
kNdkVni ..12.267 1=
Since mmd 2.1177=
kNkVni .45.314 1=
Consider tension capacity
ntt NN .* Eq. 6.11 NZS 3603
dtfkkkN tptnt ..... 15141= Eq. 6.12 NZS 3603
dtfkkkN eptnt ...... 15141=
where:
MPaf pt 339.0 == Technical Note 82
?1=k
0.114 =k (moisture content < 15 %)
0.115 =k (face grain = 0)
mmte 28= (parallel plies only)
[ ]dkNnt = 28330.10.19.02 1
kNdkNnt ...2.1663 1=
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Since mmd 900= (use minimum section) - conservative
kNkNnt .9.1496 1=
Consider compression capacity
ncc NN .* Eq. 6.13 NZS 3603
dtfkkkkN epcnc ...... 151481= Eq. 6.14 NZS 3603
dtfkkkkN epcnc ....... 151481=
where:
MPaf pc 459.0 == Technical Note 82
?1=k
0.114=k (moisture content < 15 %)
0.18 =k (localised, gusset edges are restrained by gusset stiffeners)
0.115 =
k (face grain = 0)mmte 28= (parallel plies only)
[ ]dkNnc = 28450.10.10.19.02 1
kNdkNnc ..0.2268 1=
Since mmd 900= (use minimum section) - conservative
kNkNnc .2.2041 1=
Consider Combined Actions
Combined bending, compression and shear from Eq. 6.17, NZS 3603:1993
Factor capacities by appropriate duration of load, k1= 0.8
0.107.15.3148.0
5.50
0.4478.0
0.324
2.20418.0
1.83 2
=
+
+
It is typical to consider the maximum implied forces on the structure, rather than the applied forces at the specificdesign location. However if the design criteria is not met then consideration of the implied design actions at thedesign location may be required. Therefore consider moment and shear forces at critical stress line for analysis.
Design Actions at critical stress line, Load case 1.2G+1.5Q
M* = -303.0 kNm Nc* = -83.1 kN V* = 50.5 kN k1=0.8
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0.197.05.3148.0
5.50
0.4478.0
0.303
2.20418.0
1.83 2=
+
+
Combined bending, compression and shear from Eq. 6.18, NZS 3603:1993
Factor capacities by appropriate duration of load, k1= 1.0
0.190.05.3140.1
3.54
0.4470.1
00.362
9.14960.1
2.102 2
=
+
+
3.7.2 Ridge Gusset Design
The design of the ridge gusset is similar to the knee gusset where the design capacity is based on themoment resistance offered by the ridge gusset section. Typically a mitre type joint is considered.Hutchings [1989] proposes a 0.9 factor be applied to the critical section as defined below.
Savings in design and fabrication can be made by keeping the distance L constant across the ridge and the kneegussets. Whilst the ridge gusset may be thinner often for consistency of purlin lengths and minimum gussetorder quantities it may be preferable to maintain similar gusset thicknesses.
.
gussetcs
gusset
DDepth
LCos
DD
.9.0
tan.
=
+=
Design Criteria
0.1
*2
**
+
+
ni
i
ni
i
nc
c
V
V
M
M
N
N Eq. 6.17 NZS 3603
0.1
*2
**
+
+
ni
i
ni
i
nt
t
V
V
M
M
N
N Eq. 6.18 NZS 3603
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
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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:
kNmdkkMn2
111 ..2.403 =
For 900x90 hySPAN portal frame with 7.5 pitch
82.0
2.959
300
2.9599.0
7.1065
)5.7tan(1200)5.7(
900
11
167.0
11
=
=
==
=
+=
k
k
mmDd
mmD
CosD
gusset
gusset
gusset
Calculate bending moment capacity
*
1
2
1
.5.305
2.95982.02.403
MkNmkM
kM
ni
ni
>=
=
Calculate shear force capacity
From previous:
kNkV
kNdkV
ni
ni
.2.256
..12.267
1
1
=
=
Calculate Tension Capacity
From previous:
kNdkNnt ...2.1663 1=
Since mmd 900= (use minimum section) - conservative
kNkNnt .9.1496 1=
Calculate Compression Capacity
From previous:
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kNdkNnc ..0.2268 1=
Since mmd 900= (use minimum section) - conservative
kNkNnc .2.2041 1=
Combined bending, compression and shear from Eq. 6.17, NZS 3603:1993
Factor capacities by appropriate duration of load, k1= 0.8
0.163.02.2568.0
6.65.3058.0
6.1832.20418.0
1.50
2
=
+
+
Combined bending, compression and shear from Eq. 6.18, NZS 3603:1993
Factor capacities by appropriate duration of load, k1= 1.0
0.157.02.2560.1
0.5
5.3050.1
9.156
2.20410.1
1.71 2
=
+
+
Use 42 mm 4 x-Band hySPAN as both knee and ridge gusset pairs
3.7.3 Nail ring design
The design of the nail ring is important because more than half of the nailing needs to be performed onsite. It is also important to consider end and edge distances together with allowable nail spacings(both along and across the grain) for the chosen fasteners. Selection of the nail diameter is also criticalas it will affect the available spacing and hence number of nails within the group as well as therequired penetration into the column/rafter. A staggered nail pattern provides an increased momentcapacity by maximising the lever arm action about the nail group centroid.
The design of nail groups associated with rigid moment connections are often subjected to combinedactions including bending, axial and shear forces. Whilst the bending and axial forces contributionsare minor they need to be taken into account. It is normally most efficient to calculate the proportion offorce remaining in the nails after the contribution to the design moment affect is taken out.
The complexity of calculations for the nail ring mean hand calculations can be time consuming andconservative. For this reason computer packages are often employed to develop design solutions. Thefollowing design data have been taken from design capacity tables relating to the corresponding roofpitch and member size.
The design methodology, including k factors, from AS1720.1 has been applied to create nail ringcapacities for a number of section sizes and gusset widths. These tables can be found in Engineering
Bulletin No.2, Rigid Moment Connections using CHH veneer based products. AS1720.1 was used due toits close relationship between the lateral capacities of nails in testing with CHHs range of LVL and thepublished values for joint group JD4. It should be noted that many of the k factors used in calculationof connection capacities differ between the standards and it is recommended that for connectionsthese not be mixed and matched.
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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 notaffect the nature of the nail design.
Knee, Critical load case - 1.2G+1.5Q
M* = -324.0 kNm Nc* = -83.1 kN V* = 50.5 kN k1=0.77#
#As per Table 2.7, AS1720.1
The methodology proposed for the calculation of nail group capacity for combined bending, axial andshear force involves the following steps:
1. Calculate moment capacity of nail rings in accordance with AS1720.1. AS1720.1 provides acapacity calculation for transfer of in plane moments through nailed moment ring such that:
=
=
=
2
3
1 max
max171614131 ........
ni
i
ikj
r
rQrkkkkkM AS1720.1 Eq. 4.2(4)
where:
n = number of fastenersQk= characteristic strength of fastenerri= distance to the i
thfastener from the centroid of the fastener grouprmax= the maximum value of ri = capacity factor (0.8 - nails used with primary elements in structures other than houses)k1= duration of load factor (Clause 2.4.11, AS1720.1)k13= 1.0 (nails in side grain)k14= 1.0 (nails in single shear)k16= 1.1 (nails driven through plywood gussets)k17= multiple nail factor for resisting in plane moments (AS1720.1 Table 4.3(B))
Qk= 810 N (3.15 nail, JD4 strength group, AS1720.1 Table 4.1 (B))
Since nail rings will be applied through gusset pairs the total moment resistance offered by nailrings connecting gusset pairs is:
=
=
=
2
3
1 max
max171614131 .........2
ni
i
ik
r
rQrkkkkkM
2. Calculate remaining portion of nail capacity after bending actions have been considered.
3.
a. kn QkkkkkQ ...... 171614131 =
b. nQM
MN nshearaxial
=
*
/ 1
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4.
Calculate vectorial sum of the combined axial and shear forces for comparison with remainingcapacity. These forces are assumed to be evenly distributed over the nail group.
( ) ( ) ( ) ( )
++= 2*2*2*2**
/ ,max vNvNN tcshearaxial
Engineering Bulletin 2 Rigid Moment Connection Details can be used for selection of the moment ringcapacity for the nail ring to suit the 7.5 roof pitch and 1200 mm wide gusset as drawn above.
From Table 50, Engineering Bulletin 2 for nine (9) nail rings
kNmMM
7.34914.45477.0
=
=
and kNQn 855.0=
Calculate remaining nail group capacity after resistance to moment has been calculated.
( )
kNN
N
nQM
MN
shearaxial
shearaxial
nshearaxial
0.86
2684855.0
7.349
0.3241
1
/
/
*
/
=
=
=
Calculate vectorial sum of axial and shear force, divided by k1for direct comparison
( ) ( ) ( ) ( )
( )
kNN
N
N
vNvNN
shearaxial
shearaxial
shearaxial
tcshearaxial
3.126
7.115,3.126max
0.1
3.54
0.1
2.102,
77.0
5.50
77.0
1.83max
,max
*
/
*
/
2222
*
/
2*2*2*2**
/
=
=
+
+
=
++=
Since NN shearaxial >*
/either add an additional nail ring or adjust nail size. Try using a 3.33 nail.
Using Table 3 from Engineering Bulletin 2 the capacity of the nail rings can be factored proportionally tothe Characteristic Capacity of the nail laterally loaded in single shear.
3.33/3.15 factor = 11.1810
898=
Therefore:
*15.388
11.114.45477.0
MkNmM
M
=
=
andkNQ
Q
n
n
949.0
11.1855.0
=
=
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Calculate remaining nail group capacity after resistance to moment has been calculated.
( )
*
/
/
*
/
6.214
2684949.015.388
0.3241
1
NkNN
N
nQM
MN
shearaxial
shearaxial
nshearaxial
>=
=
=
Use nine nail rings of 3.33 x75 FH nails to pattern as marked.
Proposed Nail Ring
3.7.3.2 Ridge nail ring design
Critical Design Actions
Critical load case - 1.2G+1.5Q
M* = 183.6 kNm Nc* = -50.1 kN V* = 6.6 kN
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Engineering Bulletin 2 Rigid Moment Connection Details can be used for selection of the moment ringcapacity for the nail ring to suit the 7.5 roof pitch and 1200 mm wide gusset as drawn above.
From Table 50, Engineering Bulletin 2 apply four (4) nail rings. Since we are using 3.33 nails in theknee connection, apply same nail size in the ridge, therefore apply 1.11 factor from previous to applynail ring capacities from Table ##.
*3.227
11.196.26577.0
MkNmM
M
=
=
and kNQn 949.0= From previous
Calculate remaining nail group capacity after resistance to moment has been calculated.
( )
kNN
N
nQM
MN
shearaxial
shearaxial
nshearaxial
52.125
2344949.03.227
6.1831
1
/
/
*
/
=
=
=
Calculate vectorial sum of axial and shear force, divided by k1for direct comparison
( ) ( ) ( ) ( )
( )
NkNN
N
N
vNvNN
shearaxial
shearaxial
shearaxial
tcshearaxial
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Project: 30 metre Span LVL Portal Frame Design Date: Sept. 08
For: Carter Holt Harvey Woodproducts NZ Page: 53 / 92
At: Industrial Park, Auckland, New Zealand Designed : C.R
Carter Holt Harvey Limited September 2008
3.8 Column to footing connection design
Connection of portal frame columns to footings can be achieved by base brackets that are suitably sized and fixeddirectly to the LVL columns. A similar design philosophy is applied to the design and specification of hold downanchors and base plates as would normally be applied to steel where the buckling of the plate under tensionneeds to be considered.
The connection of the base brackets to the column could be achieved using nails, screws or bolts. Nails aretypically not recommended of base plates in larger structures because of the number of nails required combinedwith the fact they would need to be hand driven through holes in plates. Bolts can be used and are good to aid inthe transfer of bracing loads across the column. Screws are ideal for most base bracket connections due to theirease of application. It is important that screw patterns are staggered for both sides of the column so that splittingof the LVL does not occur.
Again reactions are factored to take into consideration duration of load factors.
Consider Design Reactions
PF1
Rx Ry (Rx2+Ry
2)0.5 Angle
Load Case k1 kN kN kN
1.35G 0.6 19.19 35.82 40.6 61.8
1.2G+1.5Q 0.8 50.53 86.87 100.5 59.8
0.9G+Wu (Lateral) 1.0 -58.60 -99.28 115.3 59.4
1.2G+Wu (Lateral) 1.0 -53.96 115.53 127.5 65.0
0.9G+Wu (Long) 1.0 -24.61 -78.67 82.4 72.6
1.2G+Wu (Long) 1.0 52.50 102.30 115.0 62.8
k1 adjusted values
PF1
Rx Ry (Rx2+Ry
2)0.5 Angle
Load Case k1 kN kN kN
1.35G 0.6 31.98 59.70 67.7 61.8
1.2G+1.5Q 0.8 63.16 108.59 125.6 59.8
0.9G+Wu (Lateral) 1.0 -58.60 -99.28 115.3 59.41.2G+Wu (Lateral) 1.0 -53.96 115.53 127.5 65.0
0.9G+Wu (Long) 1.0 -24.61 -78.67 82.4 72.6
1.2G+Wu (Long) 1.0 52.50 102.30 115.0 62.8
It is typical in Timber structures to provide a moisture barrier at the base of the columns to eliminate the columnfrom getting wet and staying wet during the construction period. This can be typically achieved by using H3.2treated Plywood and melthoid at both the LVL column end and ground as detailed in the structural drawings.Downwards loads may be considered to be taken out in bearing so for the design of connections only uplift loadsneed be considered.
Calculate minimum number of 14g type 17 Hex Head screws
Joint Group J4 Table 3, Technical note 82
nQS * Eq. 4.5, NZS 3603
kn QknQ ..= Eq. 4.6, NZS 3603
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7/24/2019 Portal Frame Design Example
54/92
Project: 30 metre Span LVL Portal Frame Design Date: Sept. 08
For: Carter Holt Harvey Woodproducts NZ Page: 54 / 92
At: Industrial Park, Auckland, New Zealand Designed : C.R
Carter Holt Harvey Limited September 2008
kn QknQ ...=
where:
25.1303.30.1*8.01
==== kkNQk k NZS 3603
Other k modification factors are not relevant as timber is dry, screws are in single shear and are screwed throughclose fitting steel plates into the edge or face of the timber.*=0.8 is applied as Type 17 screws are as reliable as nails in service.
Since critical design reaction is 115.3 kN, calculate minimum number of 14g screws.
9.34
303.325.10.18.03.115
=
=
n
n
Say 48/14gx50 type 17 Hex Head screws, screwed through base plate sides into column.
Proposed Connection
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7/24/2019 Portal Frame Design Example
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Project: 30 metre Span LVL Portal Frame Design Date: Sept. 08
For: Carter Holt Harvey Woodproducts NZ Page: 55 / 92
At: Industrial Park, Auckland, New Zealand Designed : C.R
Carter Holt Harvey Limited September 2008
4.0 Girt Design, side wall
Girt Span (10,000-(90+65))/2 = 4923 mmGirt Spacing 1660 mm
Propose 190x45 hyCHORD for use as side wall girt
4.1 Wind loading
The capacity of solid timber girts is also dependant on the nature of lateral tortional buckling restraint and the
critical edge to which the loading and restraint is provided. It is therefore important to consider both positive andnegative wind pressures.
= 0, Lateral wind
qu=0.84 kPa Case 1 cp,e=+0.7, cp,i=
-0.56, kL= 1.25Case 2 cp,e=
-0.3, cp,i=+0.61
= 90, Longitudinal wind
qu=0.76 kPa Case 1 cp,e=-0.65, cp,i=
+0.54, kL= 1.5
Calculate design loading
mkNw
w
mkNw
w
cckkspacingqw pipelau*
i
/84.1
)54.065.05.1(66.176.0
/93.1
)56.07.025.1(66.184.0
)...(.
*
2
*
2
*
1
*
1
=
=
+=
=
=
+
+
Serviceability
Refer Technical Note 82 for Section and Material Properties.
=
=
=
++
x
w
s
EI
lw.k
mkNw
.384
..5
/30.193.145
37
4
2
2
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7/24/2019 Portal Frame Design Example
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Project: 30 metre Span LVL Portal Frame Design Date: Sept. 08
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At: Industrial Park, Auckland, New Zealand Designed : C.R
Carter Holt Harvey Limited September 2008
1403.35
10283384
492330.15019
4
Spanormm
..
w
w
=
=
+
Strength
Check capacity for positive wind pressures
kNmM
lwM
79.58
9.493.1
8
.
*
22
*
=
==
+
Consider shear and support reaction for wind load
kNVN
lwV
73.4
2
9.493.1
2
.
**
*
==
==
+
Calculate Bending Moment Capacity
nMM * Eq. 3.3 NZS 3603
ZfkkkkM bn ..... 8541= Eq. 3.4 NZS 3603
where:
MPafkk
k
b 480.1
0.19.0
54
1
===
== Technical Note 82
kNmkM
kM
n
n
8
38
.7.11
10271480.10.10.19.0
=
=
Continuous restraint to compression edge via pierce fixed sheeting, therefore k8=1.0
*7.11 MkNmM >=
Calculate Shear Capacity
nVV * Eq. 3.3 NZS 3603
Ssn AfkkkV .... 541= Eq. 3.4 NZS 3603
where:
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Project: 30 metre Span LVL Portal Frame Design Date: Sept. 08
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At: Industrial Park, Auckland, New Zealand Designed : C.R
Carter Holt Harvey Limited September 2008
MPafkkk
s 3.50.10.19.0
54
1
===
== Technical Note 82
25700
3
451902
3/..2
mmA
dbA
S
S
=
=
=
Cl 3.2.3.1 NZS 3603
*2.27
57003.50.10.19.0
VkNV
V
S
S
>=
=
Consider negative wind pressures.
kNmM
lwM
5.5
8
9.484.1
8
.
*
22*
=
==
Calculate bending moment capacity
From previous:
kNmkMn 8.7.11=
Calculate k8
Continuous lateral restraint is provided to the tension edge via pierce fixed sheeting.
Calculate S1
b
dS .31 = Eq. 3.6 NZS 3603
67.1245
190
3! ==S
Since 25>S1>10
97.0
67.125000
167.120116.067.12175.021.0
...
8
32
8
3
4
2
3218
=
+++=
+++=
k
k
SaSaSaak
*3.11
97.07.11MkNmM
M>=
=
4.2 Connection design
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7/24/2019 Portal Frame Design Example
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Project: 30 metre Span LVL Portal Frame Design Date: Sept. 08
For: Carter Holt Harvey Woodproducts NZ Page: 58 / 92
At: Industrial Park, Auckland, New Zealand Designed : C.R
Carter Holt Harvey Limited September 2008
Connection of hyCHORD girts is easiest performed using proprietary brackets and screws or nails. The proposedbracket is manufactured by Mitek. It is important to ensure that the depth of proprietary brackets is at least 60%of the depth for beams up to 50 mm thick. Propose JH47x190 to suit 190x45 hyCHORD. It is typical to apply a
practical minimum number of nails for bracket and beam stability, for members around 190 mm deep werecommend a minimum of 10/3.15x35 FH nails ie. 5/3.15 nails per tab.
Check Capacity
Joint Group J5 Table 3, Technical note 82
nQS * Eq. 4.1, NZS 3603
kn QknQ ..=
Eq. 4.2, NZS 3603
kn QknQ ...=
where:
10
6