HERA

Innovation
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P O Box 76-134
Manukau City,
New Zealand

Phone: +64-9-262 2885

Fax: +64-9-262 2856
Email: structural@hera.org.nz
Website: www.hera.org.nz

No. 73 April/May 2003

The author(s) of each item in this publication are noted at the The material herein has been the subject of review by a
beginning of the article. number of people. The effort and input of these reviewers is
greatly appreciated.

Introduction
maintaining the larger column size over the top 4
The introduction to this issue covers a number of storeys.
followup matters relating to articles and papers in
The result showed that carrying the larger column
previous issues. Readers' input and suggestions
size over the full height was the more economical
for research topics for the HERA Structural
outcome.
Division is also sought.
However, a fabricator reader makes the following
Following this is a design example, on the design
comments, which are very relevant and may
of a telescopic boom. Such booms are commonly
change the overall outcome in specific
used in portable lifting platforms such as "cherry
circumstances:
pickers". Their simple appearance and ease of
operation belie the complexities involved in their (1) The larger section size has an increased
design, which invokes most of the general bare surface area / metre length of around 20%.
steel design provisions of NZS 3404, as well as the If a required surface coating to the column
delights of combined biaxial bending plus torsion surface for either corrosion protection or fire
plus axial compression! This makes such a design protection is sufficiently expensive, the extra
example perfect for the DCB ! 20% surface area may alter the economics
of the outcome. Given that most multi-
This is followed by three short papers or articles
storey columns will carry some form of
relating to topics of current interest.
coating for fire protection or corrosion
The issue, as usual, concludes with the protection purposes, this should have been
references. included.

Progress on Errata to AS/NZS 2312:2002 In This Issue Page

As advised in the paper on the new corrosion Design Example 73.1: Design of a 3
protection Standard [1] presented on pages 10-15 Telescopic Lifting Boom
of DCB No. 72, there are a number of minor errors
and omissions in this new Standard. These are Restraint issues Relating to Portal 12
currently being rectified by the committee. It is Frame Spine Beams
intended that the changes be published by July Floor Vibration Program NZFl_Vib 1: 22
2003. What the Slab Output Represents

Column Splice Cost Comparison Revisited Design of Multi-Storey Steel 23

Buildings for Satisfactory In-Service
Page 15 of DCB No. 72 presented a short article
Wind Induced Vibration Response:
on a column splice cost comparison between two
Update on DCB No. 66
column options for the top 4 storeys of a multi-
storey building. The choices were between
SPM 0103: Potential Problem and 24
introducing a non-standard splice to step the
Solution
column size down from a 310UC97 to a 250UC73,
or using a standard bolted splice from
References 25
HERA Report R4-100 [2] for the 310UC97 and

HERA Steel Design & Construction Bulletin Page 1 No. 73, April/May 2003
 Charles Clifton's note: This was certainly an Would all DCB readers please make this
omission and should have been included in change in their copy of DCB No. 65.
any comparison made. For spray applied
fire protection or for single coat paint Congratulations to Linus Lim on Achieving his
2
systems, which have an applied cost/m of PhD.
2
under $50/m , [3], the difference would be
unlikely to change the conclusion that Linus Lim, the University of Canterbury fire
maintaining the heavier column section size engineering student whose floor slab panel
over the top 4 levels is the more cost- research has been fundamental to the
effective outcome. However, for solid board development of the second edition of the Slab
or intumescent paint fire protection systems, Panel Method (SPM) fire engineering design
2
which have applied costs/m ranging from procedure, has successfully completed his PhD
2 2
$50/m to over $200/m , the cost saving in degree.
coating on the smaller column might well
change the outcome. His work on membrane action in fire exposed
concrete floor systems has been excellent and
(2) The extra weight and size of the larger Charles Clifton congratulates him, personally and
column may slightly increase transport on behalf of HERA, on an excellent project well
costs, by decreasing the number of lengths completed.
of column that can be put on a truck. Thus
for every three full loads of columns another Linus now joins the Sydney office of a respected
trip will be needed to transport the extra fire engineering consultant, where his talents and
weight. This isn't taken account of in the expertise will be well put to use.
comparison. This would slightly reduce the
margin in favour of maintaining the heavier Research Topics Sought
column size given in the DCB No. 72
comparison. Over the last five years, HERA has received
funding from the Foundation for Research, Science
This reader's key point was that the most cost- and Technology (FRST) for our research
effective option from that comparison is dependent programme into enhanced steel building
on the input assumptions. While that outcome performance in high risk events.
would be typical, it will not always apply and
engineers should not consider that as the definitive This funding has enabled the development of
answer for all future considerations of this type. major new design developments, such as:

Sound advice… • The flange bolted joint (DCB Nos. 58 and

62), for which standard connection details
Corrigenda to the Design of Circular Bolted are currently being developed by Raed Zaki,
Flange Annulus Connection HERA Assistant Structural Engineer

DCB No. 65 contains a design procedure for the • The sliding hinge joint (DCB No. 68) for
design of Circular Bolted Flange Annulus which standard connection details are also
connections. These are bolted connections currently being developed
between lengths of circular columns, where the
two lengths have an annulus plate welded to their • The slab panel method (SPM) of floor
ends, with adjacent plates being bolted via a ring system design for dependable inelastic
of high strength structural bolts. response in severe fires.

A user of the procedure has raised two issues Their current funding is for a programme of
regarding the procedure. These are as follows: research work through to mid-2004. Despite
submitting what the HERA Structural Engineer
Q1: The dimension m2 is defined in equation considers to be our best ever bid to FRST for the
65.4.2 but not shown in that equation. Is the next six years built environment funding round, this
equation correct? bid has been declined. This means that all current
A1: Yes it is. The dimension m2 is used in the government funding of HERA research ceases as
determination of α, from Fig. 65.16. That of mid-2004. As a result, funding for the HERA
variable is then used in equation 65.4.2. Structural Division will decrease by over 50% at
the end of the 2003/2004 financial year. This will
Q2: In section 3.5.2, the equation reference for require a complete re-evaluation of HERA's
*
calculating Ntw is wrong. structural steel activities. We need to undertake
* the following actions:
A2: That is correct; Ntw is calculated from
equation 65.3, not equation 65.13 as stated.

HERA Steel Design & Construction Bulletin Page 2 No. 73, April/May 2003
• Determine which of the current range and
scope of activities undertaken by the HERA
Structural Division the industry would like to
see maintained; then
• Develop funding streams to support these
activities

This includes putting up a research programme for

the next few years of topics that are a high priority
to the industry. We will be seeking input from DCB
readers on all this in the forthcoming year,
however for now we are asking readers to come
back with any research topics that they would like
to see HERA undertake. Please send a brief
description (not more than 100 words suggested)
to Charles Clifton, HERA Structural Engineer:
Email: structural@hera.org.nz

Design Example 73.1: Design of

a Telescopic Lifting Beam
This design example has been prepared by Murray Landon,
Consulting Engineer from Tauranga. It is based on an actual
job, but some details have been changed for confidentiality
reasons and ease of presentation. Some editorial notes have
been added by G Charles Clifton, HERA Structural Engineer.

Introduction and Scope

Fig. 73.1
General details Typical Example of a Telescopic Lifting Boom at
o
Maximum Extension and Raised to 45 from
Fig. 73.1 shows a typical example of a Vertical
telescopic lifting boom. Despite their simplicity
of appearance and operation, they are
not the most straight forward item to design. The design does not cover:
The boom comprises a telescopic beam nested
• Design for other positions: the critical other
into a main beam. RHS members are used for
position to check would be the fully
both the main beam and the telescopic beam.
extended boom raised just off the ground,
The sizes of these two members must be such as giving maximum bending moment but no
to allow nesting to occur, clearances for the plastic axial compression
guides, hydraulic ram to extend and retract the • Design for concentrated load transfer at the
telescopic beam etc. Selection of suitable pairs of supports and other points.
RHS sizes for this can be made using the Design
Capacity Tables for Hollow Sections [4], which
Design Method
covers sizing of nested sections.
Design is to NZS 3404 [5] using the Alternative
Scope of design example Design Method (Working Load Design) of
Appendix P.
This design example covers:
The working load design method is used because
• Design of the boom for the design actions the design loads have been evaluated using the
shown in Fig. 73.2 (but ignoring the self- working load provisions of AS 1418.1 [6].
o
weight), when raised to an angle of 45 and
o
sitting on ground with a slope of 10 to the Use is also made of Formulas for Stress and Strain
horizontal [7] by Roark. Item references are given to the fifth
• The boom is in the fully extended position and sixth editions, which covers the two editions
• The design actions include a primary torque most commonly used by designers.
at the end of the boom of 2.25 kNm.

HERA Steel Design & Construction Bulletin Page 3 No. 73, April/May 2003
 BOOM LAYOUT
∗
F = 7.50 KN
(Inc dynamic factor)

∗
M z = 2.25KNm
(Inc dynamic factor) F

∗ ∗
F z F y
3.30 m

Telescopic Beam

E
1.70 m

4.20 m

Main Beam \ Axes

Y Z

B
1.20 m Hydraulic Ram
A
Angle = 45o
X

Fig. 73.2
o
Layout of Telescopic Boom, When Extended and Raised 45 , Along with
Design Actions and Positions for Design Checks

Boom Layout for Design Example

Section Properties for the Two RHS Members
This is shown in Fig. 73.2.
The members chosen are a 200 x 100 x 5 RHS for
The telescopic beam slides inside the main beam the telescopic beam and a 250 x 150 x 6 RHS for
on plastic guides. These are provided on all four the main beam. Both are Grade C350 to AS1163
sides, to provide the necessary twist restraint. [8].

The ram for the telescopic beam is joined midway Their section properties and nominal capacities
between points D and E. It is assumed that the that are relevant to this design example are given
friction of the slides transfers the compressive in Table 73.1.
force from the telescopic beam to the main beam
at point C; this is also shown in Figure 73.2.

HERA Steel Design & Construction Bulletin Page 4 No. 73, April/May 2003
 Table 73.1
Section Properties and Nominal Capacities ∗ o
F y = F x sin 45 (Approx since on 10 slope)
Telescopic Beam - = 5.30 KN
200 x 100 x 5
-6 4 -6 4
Ix = 14.4 x 10 m Iy = 4.92 x 10 m ∗ o
F z = F x cos 45 (Approx since on 10 slope)
rx = 71.5 mm ry = 41.8 mm = 5.30 KN
-6 4
J = 12.1 x 10 m ∗
F x = F x sin 10 (Ground slopes downwards
kf = 0.925
in positive x direction)
= 1.30 KN
Ns = 911 kN
Mz = 36.0 kNm
Msx = 62.8 kNm Msy = 31.6 kNm R *y ,E = 15.6 KN (Find by taking moments
Vv x = 380 kN Vvy = 189 kN about point D)

Main Beam- 250 x 150 x 6 R *z, E = 5.30 KN

-6 4 -6 4
Ix = 38.4 x 10 m Iy = 17.5 x 10 m
rx = 92.0 mm ry = 62.2 mm
R *y ,D = -10.3 KN (F*
y - R *y ,E )
-6 4
J = 39.0 x 10 m
∗
kf = 0.907 M*x,E = F y x 3.30
= 17.5 KNm
Ns = 1433 kN
Mz = 83.0 kNm ∗
M*y ,E = F x x 3.30
Msx = 131 kNm Msy = 72.9 kNm
Vv x = 577 kN Vvy = 348 kN = 4.30 KNm

M*z,E = 2.25 KNm (design torsion; this is

Notes: applied at the end of the
1. The above are obtained from [4], except that the
nominal capacities for compression, moment and shear boom, in conjunction with
∗
are given in this table, rather than the design capacities. the applied load F which
This is because the alternative design method is being
used. That requires determination of nominal capacity,
is acting through the shear
to which a factor of safety, Ω, is then applied. Table centre of the beam)
P3.3 of [5] gives the required values of Ω, which are
repeated in Table 73.2. Vy* ,E = 5.30 KN
2. The nominal capacity equals the design capacity, as
given by [4], multiplied by (1/φ) or (1/0.9). Vx,* E = 1.30 KN

Table 73.2
Determination of Second-Order Effects
Alternative Design Method Factors of Safety (Ω)
Permissible Strength Factor Of Safety (Ω Ω)
In an elastic analysis to NZS 3404 Clause 4.4, the
For
first requirement is to determine the influence of
Bending 0.60
second-order effects. For the telescopic beam,
Shear 0.62
which is a sway member, this requires the
Axial Forces 0.60 calculation of the elastic buckling load, Noms.
Combined Actions 0.60
The elastic effective length factor for a sway
Note: These are the values from NZS 3404 Table P3.3 that
are relevant to this design example. member with a fixed base is given by NZS 3404
Fig. 4.8.3.2 case number 5, ie. ke = 2.2.
Design Check on Telescopic Beam
However, the telescopic beam is not fixed at point
Critical location for member capacity E, but is continuous past that point to D. It
therefore undergoes rotation at E due to the
This is, by inspection, at the right hand side of applied moment over length DE. This rotation will
point E. increase the deflection at point F.

Design actions in member To allow for this, the deflection at F due to

cantilever plus support rotation must be
Refer to the axes shown in Fig. 73.2 for the signs determined, then converted to an equivalent fixed-
of the forces.

HERA Steel Design & Construction Bulletin Page 5 No. 73, April/May 2003
ended cantilever length EF that would give the Calculation of elastic buckling load factors
same deflection at F. This is to Clause 4.9.2.3;

Cantilever deflection at F, excluding rotation at E Nomsx 409

ëcx = *
= = 77.2
N 5.30
PL3
υ = = 22.0 x 10 −3 m Nomsy
3EI 140
ëcy = *
= = 26.4
N 5.30
P = 5.30 kN
L = 3.30 m Consideration of second-order effects
E = 200 GPa
-6
Ix = 14.4 x 10 m
4 As λcx and λcy ≥ 10, from Clause 4.4.2.2.1 of [5]
second order effects on the telescopic beam
Slope at E member can be neglected.

Bending about the X-axis

M∗x, E 2L
èE = = 3.44 x 10− 3 radians (From
6EI Determining critical cross section along boom
th
Table 3, item 3e, page 103 of [7] or page 107, 6 for bending moment determination
edition)
This involves checking Clause 5.3.3.
M*x,E = 17.5 KNm
M* 17.5
L = 1.70 m At point E; = = 0.28
Msx,E 62.8
Total deflection at E
M* 39.8
At point B; = = 0.30
υ(total ) = υ + èE × 3.30 = 33.4 × 10−3 m Msx,B 131

Hence, in practice, both points need checking as

Equivalent beam length as fixed ended either could be critical.
cantilever
Section moment capacity at E
PL3eq
υ= ΩMSX,E = 0.6 x 62.8 = 37.7 kNm
3EIx Ω = 0.6, from Table 73.2
Msx = 62.8 kNm, from Table 73.1
υ 3EIx
Leq = 3 = 3.79m Member moment capacity for length EF
P
This is determined using Clause 5.6.2 of [5]. The
This equivalent length is then used for Noms segment EF is the relevant segment; this has
calculation in the x and y directions. constant cross section. The end at E has full twist
restraint from the support to the main member; the
Calculation of Nomsx and Nomsy end at F is unrestrained.

π 2 EI x   π 2EI 
N omx = = 409KN   π 2EI   
(k ex L ) 2
Moa,EF = Mo =   2 

y
GJ +  2 w    = 929KNm
L    L e   
-6 4 1   e 
Ix = 14.4 x 10 m (Eqn 5.6.1.1(4))
k ex = 2.2 (case 5, NZS 3404 Fig. 4.8.3.2;
-6 4
allows for curvature in main beam, Iy = 4.92 x 10 m
etc) Le = 3.30 m
L = 3.79 m k t = 1.0 (FU)
k l = 1.0 (Load applied above flange, but
π 2Ely this is taken into account by the
Nomy = = 140kN
(key L )2 torsion moment M*z )
k r = 1.0 (FU)
-6 4
Iy = 4.92 x 10 m
k ey = 2.2 G = 80 GPa
-6 4
L = 3.79 m J = 12.1 x 10 m
Iw = 0 (RHS member; see Clause 5.6.1.4)

HERA Steel Design & Construction Bulletin Page 6 No. 73, April/May 2003
   2  M  M*z 2.30 × 10 -3
 M   τu* = = = 12.4MPa
á s,EF = 0.6  sx  + 3 −  sx  = 1.0 2 × Ae × t 2 × 18.5 × 10 -3 × 5 x 10 -3
  Moa    Moa
 
1 
(Eqn 5.6.1.1(3)) Ae = (b − t)(d − t ) = 18.5 × 10−3 m2
(see section 8.2.4.1 of [9]
Msx,E = 62.8 KNm or Appendix H, Clause
H4 of [5])
αm,E = 1.25 (see Table 5.6.2, case Number 2). b = 100 mm
d = 200 mm
ΩMbx,E = Ωαs αm Msx ≤ ΩMsx t = 5 mm
Vw = Vv x = 380 kN
ΩMbx,E = ΩMsx = 37.7 kNm fy = 350 MPa

As αs αm > 1.0, the segment EF has full lateral *

Veq = 27.7kN < ΩVw = 0.62 x 380 = 236 kN
restraint.
Check interaction of equivalent shear and
(Check on point B is covered in the design of the bending
main member)

Checking x-axis moment adequacy M*x,E 17.5

= = 0.46 < 0.75
ÙM sx 0.6 × 62.8
M*x,E ≤ ΩMbx,E is required
Therefore don’t need to check interaction of
equivalent shear and bending (Clause 5.12.2 of
M*x,E = 17.5 kNm ≤ 37.7 kNm √ OK
NZS 3404 applied to working loads).

Bending about the y-axis The y-axis check is similar and will be OK by
comparison.
M*y ,E ≤ ΩMsy,E is required
Calculation of member compression capacity

M*y ,E = 4.30 kNm ≤ 0.6 x 31.6 = 19.0 kNm √ OK Member compression capacity is calculated for the
length EF.
Calculation of equivalent shear
The actual length EF of 3.3 m is used for this, with
As described in section 8.2.4 of the Structural an effective length factor of 1.0, as this is now
Steelwork Limit State Design Guides Volume 1 [9] design of the member and second-order effects
and also in Commentary Clause C8.5 of have already been determined.
NZS 3404: Part 2 [5], the uniform torsion, M*Z , will
Nc = á cNs
interact with the applied shear, Vy* ,E to produce an
Ncx = 0.895 × 911 = 815 KN
equivalent design shear force.
Ncy = 0.675 × 911 = 615 KN
The second-order influence of on the design M*x
torque must first be determined, as described in αcx = 0.895 From Table 6.3.3(2) of [5]
C8.5.4.2 (c) of [5]. for αb = -0.5 and λnx = 52.5
αcy = 0.675
1 1 αb = -0.5
δz = = = 1.02
M*x,E 17.5  Le   fy 
1- =   (k f )  
1-
Moa,EF 929
ën
-V~~,
 r 
 250 
 
λnx = 52.5
M*z,E = M*z x δ z = 2.25 x 1.02 = 2.30 kNm λny = 89.8
Equivalent shear is now given from C8.5.5.2 of [5] rx = 71.5 mm
as; ry = 41.8 mm
Le = 3.3 m
 τ* V  ke = 1.0
*
Veq = Vy* ,E + u w  = 27.7 KN
 0.60 × f  L = 3.3 m
 y 
kf = 0.925
Vy* ,E = 5.30 KN fy = 350MPa

HERA Steel Design & Construction Bulletin Page 7 No. 73, April/May 2003
Check on member adequacy in compression R *y ,A = - M*Bx / 1 .2 = - 33.2kN
N* ≤ ΩNs
5.30 ≤ 0.6 x 911 = 547 kN √ OK Determination of second-order effects

N* ≤ ΩNcx The same situation applies over length BC of the

5.30 ≤ 0.6 x 463 = 278 kN √ OK main member as applied over length EF of the
telescopic boom, in terms of the need to calculate
N* ≤ ΩNcy an equivalent fixed ended cantilever length BC.
5.30 ≤ 0.6 x 137 = 82.2 kN √ OK
Cantilever deflection at C, excluding rotation
Check combined bending and compression at B

The section capacity check at E uses Clause

PL3 −3
P8.3.4, as applied through Clause 8.3.4.1. υc = = 17.04 × 10 m
3EIx
N* M*x M*y 5.30 17.5 4.30 P = Fy* = 5.30 kN (we are ignoring self
+ + = + +
ÙN s ÙM sx ÙM sy 0.6 × 911 0.6 × 62.8 0.6 × 31.6 weight)
= 0.701 ≤ 1.0 L = 4.20 m
E = 200 GPa
-6 4
The member capacity check on EF uses Clause Ix = 38.4 x 10 m
P8.4.5, as applied through Clause 8.4.5.1.
Slope at B
1.4 1.4
 M*x   M*y   17.5 
1.4
 4.30 
1.4 MB* 2L
  +  = + θB = = 1 .16 × 10 −3 radians
 ÙM   
 ÙM iy   0.6 × 62.1  0.6 × 31.1 
6EI
 cx   
MB* = 22.3 KNm
= 0.475 ≤ 1.0
L = LAB = 1.2 m
Mcx = Mix Full lateral restraint ; see Clause 8.4.5.1
Total deflection at C

 
Mix = Msx 1 −
N*
 = 62.8 × 1 −
5.30 
 = 62.1 KNm υ(total ) = υc + èB × 4.20 = 21.9 ×10 −3 m
 ÙN   0.6 × 815 
 cx 
Equivalent beam length as fixed ended cantilever
 N *   5.30 
Miy = Msy  1 − = 31.6 × 1 −  = 31.1 KNm PL3e
 ÙN   0.6 × 615 
 cy  υ=
3EIx
The design check on the telescopic boom is now
complete. õ 3EIx
Le = 3 = 4.57m
Design Check on Main Member P

The critical cross section for bending is, by This equivalent length is then used for Noms
inspection, over the hydraulic ram at point B. calculation in the x and y directions.

Design actions in member Calculation of Nomsx and Nomsy

M*x,B = F*y x 7.50 The main beam is a combined section and this
= 39.8 KNm needs to be accounted for. The relevant table
∗ from Roark, fifth edition [7] is page 534, table 34,
M*y,B = F x x 7.50 item 1a. From the sixth edition it is on page 670.
= 9.75 KNm
∗ π 2EIx1
M z,B = 2.25 KNm Nomx = K1 × = 170 KN
L2eq, x
∗
V y,B = 5.30 KN
∗
K1 = 0.419 from table 34 of [5]
V x,B = 1.30 KN
Ix2 38.4
R *y ,B = Fy* x (3.3 + 4.2 + 1.2) /1.2 = 38.5 kN = = 2.67 Therefore take as equal to 2.0
Ix1 14.4
(force required at ram for this configuration)
to match tables.

HERA Steel Design & Construction Bulletin Page 8 No. 73, April/May 2003
 a 4.82 Bending about the x-axis
= = 0.58 Therefore take as equal to 0.5
L eq, x 8.36
This involves checking moment capacity at B;
section moment capacity at E and member
P2 0 moment capacity over segment length AF. The
= =0 (There is no increase in
P1 5.30 latter is based on the hydraulic ram not providing
compression load on the effective twist restraint at B, which is probably
larger section) conservative.

L eq = 4.57 + 3.79 = 8.36 m Section moment capacity at B

ΩMsx,B = 0.6 x 131 = 78.6 kNm

K1 x π 2 EÉy 1 Ω = 0.6, from Table 73.2
Nomy = = 39.8KN
(2.2/2 )2 L2eq, y Msx,B = 131 kNm, from Table 73.1 for the
lower section.
The (2.2/2) multiplier on the denominator for Nomy Member moment capacity along length AF
makes allowance for the increase in effective
length for a fixed cantilever specified by NZS 3404
This is determined by using Clause 5.6.2 of [5], in
Fig. 4.8.3.2 Case 5. Given that K1 from Roark conjunction with Clause 5.6.1.1.2 which accounts
already takes account of the cantilever
for the change in cross section at C.
configuration, ie. k=2, with regard to elastic
compression buckling load, the adjustment for
Calculation of non-uniformity factor, αst
NZS 3404 is (2.2/2).
á st = 1.0 - [1.2rr (1 - rs )] = 0.777
K1 = 0.419 from table 34 of [5].

rr = Lr/L = 0.38
Éy 2 17.5
= = 3.56 ⇒ take as equal to 2.0 for tables
Éy 1 4.92 Af m   0.4dm  
rs = 0.6 +  
 = 0.511
Afc   dc  
a 5.4
= = 0.59 ⇒ take as equal to 0.5
L eq, y 9.2 Af m = 0.100 x 0.005 x 2 = 1.00 x 10 m
-3
-3
Af c = 0.150 x 0.006 x 2 = 1.80 x 10 m
P2 0 dm = 200 mm
= =0 dc = 250 mm
P1 5.30
Lr = 3.3 m
L = 1.2 + 4.2 + 3.3 = 8.70 m (Actual
L eq = 1.20 + 4.2 + 3.79 = 9.2 (The distance AB length of main beam AC and length of
is added, because there is no support against telescopic beam EF)
buckling about the y-axis offered by the hydraulic
support at B. The actual length ABC is therefore Calculation of uniform elastic buckling moment,
used). Moa, based on cross-section at B (same as at A)

Calculation of elastic buckling load factors

  π 2El   2 
Moa, u,AF =  y  GJ +  π El w    = 1193 kNm
 2   L2  
ëcx =
N omsx
=
170
= 32.1 1  eL   e   
*
N 5.30 (Eqn 5.6.1.1(4))

N omsy -6 4
39.8 Iy,A = 17.5 x 10 m
ëcy = = = 7.51
N* 5.30 Le = 1.0L = 8.70 m
kt = 1.0 (FU)
As λcx > 10, no need to magnify M*x . kl = 1.0 (Load above flange but taken
into account by moment Mz)
As λcy < 10, must magnify M*y . kr = 1.0 (FU)
G = 80 GPa
-6 4
0.95 J = 39.0 x 10 m
äm = äs = = 1.10 lw = 0 (RHS member; see Clause 5.6.1.4)
 1 
1-  
 λ cy  Reduction in elastic buckling moment for the
segment due to the change in cross-section at C.
M*y ,B = äm M*y ,B first order = 1.10 x 9.75 = 10.73 kNm

HERA Steel Design & Construction Bulletin Page 9 No. 73, April/May 2003
Moa,AF = αst x Moa,u,AF = 0.777 x 1,193 = 927 kNm M*Z,B 2.35 × 10 -3
τu* = = = 5.59 MPa
2 × Ae × t 2 × 35.1× 10 -3 × 6.0 × 10 -3
  2  
 M   M sx  
á s,AF = 0.6   sx  + 3  
- 
  = 0.96 Ae = (b − t )(d − t) = 35.1× 10-3 m2
 1  M oa  
  M oa  
  b = 150mm
(Eqn 5.6.1.1 (3)) d = 250mm
t = 6.0mm
Msx = 131 kNm
αm = 1.25 (To allow for self weight and the Vw = Vv x = 577 KN
moment distribution in the beam
due to the telescopic beam fy = 350 MPa
reactions, refer DCB No. 16 for
improved αm ) *
Veq = 20.6 kN << ÙV w = 0.62 x 577 = 358 kN

Mbx = αmαsMs x = 1.25 x 0.96 x 131 = 156 kNm ≤ Msx Check interaction of equivalent shear and
(Eqn 5.6.1.1(1)) bending

Mbx,B = Msx,B = 131 kNm M*x,B 39.8

= = 0.51 < 0.75
ΩMbx,B = 0.6 x 131 = 78.6 kNm ÙM sx 78.6

(As αsαm > 1.0, the segment AF has full lateral Therefore don’t need to consider interaction of
restraint). equivalent shear and bending.

Checking x-axis moment adequacy The y-axis check is similar and will be OK by
comparison.
M*x,B ≤ ÙM bx,B is required
Calculation of member compressive capacity
M*x,B = 39.8 kNm ≤ 78.6 kNm √ OK Member compression capacity is calculated for the
length BF for the x-axis and the length AF for the
Bending about the y-axis y-axis, as the hydraulic ram does not provide
y-axis restraint.
M*y ,B ≤ Ω Ms y B, is required
Both lengths comprise the main member and the
telescopic member, thus requiring design to
M*y ,B = 10.73 kNm (including second - order effects)
Clause 6.3.4.
≤ 0.6 x 72.9 = 43.7 kNm
√ OK This first requires calculation of Nomx and Nomy , so
as to calculate the slenderness ratio.
Calculation of equivalent shear
Calculation of Nomx and Nomy
The second-order influence of Mx* on the design
The relevant table from Roark fifth edition [7]
torque on the main member must first be is Table 34, p.535. For the sixth edition it is on
determined. page 670.
1 Earlier in this main member design, the Nomx and
ä = = 1.04
M* Nomy for the elastic effective lengths for this sway
1- x member were calculated in order to determine the
M oa
second-order effect multipliers. For individual
M*x = 39.8 kNm member design, in accordance with NZS 3404
Moa = 927 kNm Clause 4.8.3.1 (a) (iv), the member effective length
factor is taken as ke = 1.0. This corresponds to a
M*z,B = M*z x äz = 2.25 x 1.04 = 2.35 kNm pin-pin ended member and so means that the item
number from Table 34 of [7] corresponding to the
Equivalent shear is now calculated thus: k e = 1.0 (pin-pin) case must be used. That is case
number 1b.
 τ* V  π 2EIx1
*
= Vy* ,B +  u w  = 20.6 KN Nomx = K1 × = 577 KN
Veq
 0.60 × f
 y

 (ke L eq, x )2
Vy* ,B = 5.30 KN K1 = 1.297 From table 34 of [7], Item 1b.

HERA Steel Design & Construction Bulletin Page 10 No. 73, April/May 2003
Ix2 38.4 Ncy = 0.150 x 911 = 137 kN
= = 2.67 Therefore let equal 2.0 to
Ix1 14.4
αcx = 0.508 for λnx = 113 From Table 6.3.3 (2) of [5]
match tables.
αcy = 0.150 for λny = 223
a 4.2
= = 0.525 Therefore let equal 0.5
L eq, x 7.99 αb = -0.5

P2 0 Check on member adequacy in compression

= =0
P1 5.30
N* ≤ ΩNs
5.30 ≤ 0.6 x 1433 = 860 kN √ OK
k e = 1.0
Leq,x = 4.20 + 3.79 = 7.99 m
N* ≤ ΩNcx
5.30 ≤ 0.6 x 463 = 278 kN √ OK
π2EIy 1
Nomy = K1 × = 149 KN
(ke Leq, y )2 N* ≤ ΩNcy
5.30 ≤ 0.6 x 137 = 82.2 kN √ OK
K1 = 1.297 From table 34 of [1], Item 1b
Check combined bending and compression

Iy 2 17.5 The section capacity check at B uses Clause

= = 3.56 Therefore let equal 2.0 to
Iy 1 4.92 P8.3.4, as applied through Clause 8.3.4.1.
match tables.
N* M*x M*y 5.3 39.8
+ + = +
a 5.4 ÙN s ÙM sx ÙM sy 0.6 × 1433 0.6 × 131
= = 0.588 Therefore let equal 0.5
L 9.19 10.73
+ = 0.767 ≤ 1.0
0.6 × 72.9
P2 0
= =0
P1 5.30 The member capacity check on BF or AF
respectively as appropriate (for the x-axis and the
L eq,y = 5.40 + 3.79 = 9.19 m y-axis) uses Clause P.8.4.5, as applied through
Clause 8.4.5.1.
Calculation of modified member slenderness
1.4 1.4
 M*x   M*y   39.8 
1.4

This is to Clause 6.3.4 (b)   +  = 

 ÙM   ÙM iy   0.6 × 129 
 cx   
1.4
 N   10.73 
ën = 90  s  +  = 0.55 ≤ 1.0
 N om   0.6 × 68.2 

Mcx = Mix Full lateral restraint; see Clause 8.4.5.1

 911 
ënx = 90 x   = 113
 577 
 N* 
 = 131 ×  1 −
5.30 
Mix = Msx 1 −  = 129 KNm
 ÙN   0.6 × 463 
 c 
 911 
ëny = 90 x   = 223
 149   
N*
 = 72.9 × 1 −
5.30 
Miy = Msy  1 −  = 68.2 KNm
 ÙN c   0.6 × 137 
Ns = 911 kN calculated for the smallest cross  
section, which is the telescoping beam.
The design check on the main boom is now
Nomx and Nomy are as calculated above. complete.

Calculation of member compression capacity Conclusion

Nc = αc Ns The telescopic lifting boom shown in Fig. 73.2 can

carry the applied loads when extended and raised
o o
Ncx = 0.508 x 911 = 463 kN 45 and situated on a slope of 10 .

HERA Steel Design & Construction Bulletin Page 11 No. 73, April/May 2003
Restraint Issues Relating to
The use of a spine beam also results in higher
Portal Frame Spine Beams column axial loads, which results in larger column
sizes that may have greater resistance against
This article has been written by G Charles Clifton, HERA
Structural Engineer. accidental impact (but also increase the foundation
loads and possible size of foundation pad or pile
Scope required).

Fig. 73.3 shows a typical example of a spine beam However, the use of a spine beam introduces
in a portal frame building. Spine beams are used restraint issues relating to the spine beam itself,
in propped portal construction to support the portal the rafter it supports and the columns that support
frame in lieu of a column. They carry the vertical it, that must be considered.
loading (either downwards or uplift) in bending
across to supporting columns. This allows the The scope and purpose of this article is to cover
number of supporting columns to be reduced from the general restraint issues relating to portal frame
one every portal frame to one every second or spine beams. It addresses the following topics:
third portal frame, as appropriate. It also means
that an extra beam, typically under the apex, is • restraint of the portal frame rafter
required. • restraint of the spine beam
• restraint of the column
Given the pitch on a portal frame, the clear height • detailed design advice for common
from floor to bottom of spine beam at the apex will configurations of rafter and spine beam
usually be no less than that from floor to underside • some general restraint and design issues.
of rafter at the knees, so the addition of a spine
beam at the apex does not reduce the clear head
height within the building.

Spine Beam

Portal Frame Rafter

Fig. 73.3
Example of Spine Beam in Portal Frame Building

HERA Steel Design & Construction Bulletin Page 12 No. 73, April/May 2003
 L
Portal Frame Rafter
: -,.J . 1i .£»'
■'
TT.’ / Wli ' ••■
:*? r? 1 .'..* jk ; ■ .r .......
Brace
Spine Beam
T~* I -?
r # *

ir'f*' ■ •" •
[ 2• I fI
i

~!

’ * a* «
*. i r i

Top of
Column
■
Fig. 73.4
Close up of Spine Beam/Portal Frame/Column Support
HERA Steel Design & Construction Bulletin Page 13 No. 73, April/May 2003
The article is written around the restraint provisions or 20 from [10] applies and the spine beam is not
of NZS 3404 Clauses 5.4 and 6.7 and the use of required to provide restraint to the rafter.
HERA Report R4-92 [10] Restraint Classifications
for Beam Member Moment Capacity Determination Where the rafter is in negative bending going over
to NZS 3404. It covers these items in general the spine beam, the rafter bottom flange is critical.
terms initially, then goes into detail for common In that case, from Connection Detail Nos. 19 or 20,
configurations of rafter and spine beam. the cross section would have less than full twist
restraint (F) without the spine beam and the
Prior to commencing with the first of these topics, additional restraint offered by that beam can be
some definitions of directions and location are used to provide full twist restraint. The restraining
given. These are used repeatedly throughout the system must prevent the rafter from twisting - ie.
article. the bottom flange from moving parallel to the span
of the spine beam.
Definitions of Directions and Location
If the rafter has one or two full depth stiffeners in
These should be read in conjunction with Fig. 73.4, this case (as does the system shown in Fig. 73.3),
which is a view looking along the span of the rafter, then the restraint against twist can be provided
with the spine beam running at right angles through the flexural restraint of the rafter to spine
underneath, across the picture. beam connection and the rafter to purlin
connection. The restraining moments are then
The key directions are: resisted by in-plane bending resistance of the
purlins at the top and the spine beam at the
• parallel to the rafter, which means in the bottom. If there is a typical moment connection
direction of the span of the rafter between rafter and purlin (see Connection Detail
• transverse to the rafter, which means at right 19 from [10]) and at least two bolts between rafter
angles to the span of the rafter and hence and spine beam that can resist in-plane moment in
parallel to the span of the spine beam the spine beam generated by the out-of plane twist
• parallel to the span of the spine beam, which of the supported rafter, then this is a viable
means in the direction of the span of the solution. It is the solution used in Fig. 73.3.
spine beam
• transverse to the span of the spine beam, An alternative is to provide direct lateral restraint to
which means at right angles to the span of the rafter bottom flange via the spine beam. This
the spine beam and hence parallel to the doesn't require the rafter to spine beam or the
span of the rafter rafter to roof connections to provide moment
• the rafter and spine beam are assumed to resistance against rafter twist. However, it puts a
be at right angles in this article lateral restraining force into the spine beam, which
travels along the spine beam and accumulates in
• the top of the column for restraint
accordance with Clause 5.4.3.3 of [5]. This
discussions is taken at the bottom of the
spine beam, as shown in Fig. 73.4. restraining force must be anchored back into the
roof system, or else there could be a restraint
failure caused by the lateral movement of the spine
Restraint of Portal Frame Rafter
beam as a rigid unit, with all rafters connected to
the spine beam rotating in the same direction. In
There are no Connection Details from [10] which
the case of Fig. 73.3, for example, a way of
describe exactly the rafter over spine beam set-up,
providing this anchorage is to connect the spine
however Connection Details 11 and 12 come
beam into an end column in the wind wall, the top
close. The main differences between those details
of which is tied into the roof bracing system in the
and the rafter/spine beam case is that, in the latter:
end bay.
(i) The top of the rafter is typically tied into the
A final alternative offered is to use a fly brace to
roof system, which provides a plane of
the rafter bottom flange, as shown in Connection
inherent lateral restraint
Detail 21 from [10]. In this case, the fly brace plus
spine beam mean that F restraint will be provided,
(ii) The rafter is continuous over the spine
even when the purlin is flexible (see C5.4.2.2 of
beam.
NZS 3404 [5]).
Connection details 19 and 20 are also relevant.
See the cases described under Detailed Advice for
Common Configurations for more detailed
In the case where the rafter is in positive bending
guidance.
going over the spine beam, the rafter top flange
is the Critical Flange (CF - see NZS 3404 [5]
As discussed under Restraint of the Column, later
Clause 5.5 for definition). At points of restraint to
in this article, the column restraining forces parallel
the roofing system, Connection Detail Nos. 19
to the spine beam operate in the same direction as

HERA Steel Design & Construction Bulletin Page 14 No. 73, April/May 2003
 those g
e
negativ nerated by th
e mom e rafter
ent and b
back in
to the must be ottom flange
needs roof pla effectiv un
to ne. Th ely anc der
two sets be designed e restra hored Under
v
of restr to
aint for resist the gre
ining s
ystem snow, d ertical downw
o wnward ar
Restra
ces, no a
t the su ter of these subject s wind) ds loading (
int of S m of the tw to n egative , such dead,
pine B o. suppor
ti n g momen a spine beam live,
eam c olumns t o wil
Fig. 73 the sup
ported and po ver the interm l be
.3 show rafters. s itive m e diate
portal fr s oment
ame ap a system wh T h e momen
under
the oth ex is s ere e v t s
e u ery se
vertical rs being sup pported by a cond involvin
g dead igns will be
stiffnes ported spine b + wind
potentia so by eam, that pro reversed for
lly muc ffered at the a a column. U n der neg duce a
net upli ads
lo
spine b h pex by T
eam. In greater than a colum he C r it ical Fla
ative m
o ment, th
ft .
deflecti order to tha n is
frames,
on at
the ap control t offered by provide nge.
Full se e bottom fla
d if th d , n
ex
govern the spine be between adja al vertical
ferenti e as show either via a s ction restraint ge is the
n ti ff e s hould b
ed
eg. diffe by servicea
am de
sign is cent portal brace a in Connectio ner and
momen e
s s h own in n D e tail 12 t system
rential bility d likely to th e s e F ig fr o
200) or d eflectio e fl system s. 73.3 m
( n limits ection limitati be s is cov and 73 [10] or a fly
relevan frame spacing of (fram ons; ered by
[10].
.4. De
sign of
t lo / e spac Under
Table C ad combinati 250) may ap ing / p o s it iv
2.4.1). on p ly Critical e mom
require There w s (see eg. NZ under the Flange ent, th
e top
d whe il l b e S 4 2 0 3 [11] p r o v ided, b and F flange
suspen re the more s y u ll
ded ce r oof sy tr in g ent lim force fe in spectio section is the
beam iling. s te m it s e d s in n . T restrain
is likely This m s u pports s ystem, to th e rafter h e la t is
suppor
ting co
to be
made
eans th
at the a w via the and thr teral restrainin
ithout th o ugh into g
in Figs
.
lumns,
as
continu
ous ov pine
s e need in-plane rigid the roo
suppor 73.3 and 73 indeed is the er the for furth
er calcu
ity of
the raft f
ting colu .4, and be ers,
mns. pinned am shown See the
c
lation.
at the
end Commo ases describe
n Co d unde
guidanc nfigura r Detail
e. tions e
for m d Advice for
ore d
etailed
R after
Spine b
ea
internal m continous o
column ver
Rc*
R sb* Lateral
restra
RL1* int force

Internal
column

N c*
(a) Prop
ped por
tal fram
e

Rsb*

Proppe
I
d and In
(b) Inte
rmediat
e portal
frame
Lateral
restrain
RL2* t force

termed
ia te Porta
l Frame Fig. 7
HERA S
teel Design
s Showin 3.5
& Cons g Restr
truction aint Fo
Bulletin rces fro
m Spin
e Beam
Page 1 and Co
5 lumn

No. 73,
April/Ma
y 2003
 Propped portal Spine beam
frame Rsb,1*
a below portal
|r b
]i J 1 11
1 Rr,left* Fy r Rr,right *

c
Rsb,2* Restraint to
spine beam
d
e

'L Intermediate
portal frame f>

a,b,c = restraint positions of spine beam

under propped portal frame

d,e,f = restraint positions of spine beam

under intermediate portal frame

Fig. 73.6
In-Plan Restraint to Spine Beam Critical Flange from the Rafter System

Restraint of the Column under Restraint of Portal Frame Rafter, on page 14

above. Whichever solution is adopted to resist
The top of the column, the position of which is these forces needs to be designed for the greatest
shown in Fig. 73.4, must be a point of magnitude restraint force from each source, not
effective restraint, to NZS 3404 [5] Clause 6.7. the sum of the two.
This requires restraining forces determined by
Clause 6.7.2.1 to be resisted about each of the Detailed Advice For Common Configurations
column's principal axes.
Preamble and assumptions
In the direction transverse to the spine beam
(parallel to the rafter), the lateral restraining force This section of the article provides detailed advice
must be transferred by either of design options 1 or on the derivation of restraint forces and their
2 described on pages 18-20 herein. Once this resistance by the overall building system for
restraining force gets into the rafter, it is resisted common spine beam/column configurations.
by the portal frame/roof system as described on
page 17. These configurations are:

In the direction parallel to the spine beam • propped portal frame (Fig. 73.5 (a))
(transverse to the rafter), the lateral restraining • intermediate portal frame (Fig. 73.5 (b))
force must transfer into the rafter and from there to • in-plan restraint to spine beam critical
the roof system. This lateral restraining force is flange from the rafter system (Fig. 73.6).
acting in the same direction as the rafter bottom
flange restraint force generated under negative
moment, solutions for which have been given

HERA Steel Design & Construction Bulletin Page 16 No. 73, April/May 2003
The assumptions made in this section are as Because the roof system provides inherent overall
follows: lateral stability, this force is not required to be
resisted by the portal frame as a lateral force, so it
• The spine beam is continuous over the does not contribute to R*L1 .
supporting column of the propped portal
frames
The column restraint force, R*c , is carried through
• The portal frame building has a normal
bracing system for building stability in into the rafter by the fly braces (or through the
addition to the cladding system (see pages stiffener/column system if design option 2 on
2-4 of DCB No. 50 for more details on what pages 18-20 is used), where it has to be resisted
this requirement entails) by the portal frame, ie.:
• The roof system provides inherent
*
overall lateral stability (see pages 2-4 of RL1 = Rc* (73.3)
DCB No. 50 and/or pages 2.7 to 2.9 of
HERA Report R4-92 [10] for more details).
The restraint force R*L1 can be resisted by the
The details on restraint are as follows: portal frame, in which case it becomes an
additional applied force in the load case generating
Propped portal frame case the design actions on the spine beam and column.
The critical load case for this is typically either
This is shown in Fig. 73.5 (a). In this case the 1.2G + 1.6Q, or 1.2G + 1.2S u, from NZS 4203 [11]
portal frame is providing restraint to the top of the Clause 2.4.3.3.
internal column, requiring a restraint force, R*c ,
However, the fact that the cladding system
given by; provides a stiffening effect - as described on
pages 4-7 of DCB No. 49 and 2-4 of DCB No. 50
Rc* = 0.025 Nc* (73.1) - can be used to share the resistance of this force
between the portal frame and the bracing system
where: in the transverse direction (typically in the gable
walls). The split of this force is in accordance with
N*c = design axial compression on internal
the advice for wind and earthquake given on
column page 7 of DCB No. 49 - ie. for a pinned base portal
frame, 0.5 R*L1 is resisted by the portal frame and
The portal frame is also providing restraint to the
bottom flange of the spine beam, this being the 0.5 R*L1 goes into the roof system and hence to the
critical flange for vertical load cases - ie. the flange gable walls (or other transverse bracing system).
in compression under the negative moment. The
spine beam restraint force at the propped portal, Intermediate portal frame case
R *sb, pp , is given by;
This is shown in Fig. 73.5 (b). In this case, the
spine beam is in positive moment and the top
M*sb,col
pp = 0.025
* flange, which is attached directly to the underside
R sb, (73.2)
(d b - tf )sb of the portal frame, is the critical (compression)
flange. The restraint force, R *sb,ip , is given by:
where:
M*sb, col = design negative moment in spine 0.025 M *sb,ip
ip =
*
R sb, (73.4)
db, tf
beam at the column
= spine beam depth and flange
(d b - tf )sb
thickness
where:
The fly braces shown in that figure must be M*sb,ip = design positive moment in the spine
designed to transfer (R*c + R*sb ) into the rafter, as beam at the intermediate portal frame.
shown in Fig. 73.7, design option 1. The
design of these braces is covered by section This lateral force must be resisted directly by the
2.5.3 of R4-92 [10]. portal frame, thus:

* *
For the reasons given in section 2.5.2 (3) of [10], RL2 = R sb,ip (73.5)
the spine beam restraint force is resisted by
in-plane bending in the rafter (see Fig. 2.3 of [10] *
The restraint force, RL2 , can be resisted solely by
for the distribution of restraint forces in the fly
brace system). the portal frame or split between the portal frame
and the roof/gable wall as described above.

HERA Steel Design & Construction Bulletin Page 17 No. 73, April/May 2003
Accumulated restraint force in roof/gable wall o
equal to R *sb,1 and R *sb,2 for the braces at 45 to
system
the spine beam and rafter, as is shown in
If the decision is made to split the lateral Fig. 73.6.
restraining forces R*L1 and RL2 *
between the portal If the in-plan braces connect into the bottom flange
frames and the roof/gable wall system, then the of the rafter, then this restraint force will tend to
gable wall bracing and the roof braced bay that twist the rafter at that point and must be carried up
feeds forces into the gable wall bracing must be to the purlin line where it is resisted by major axis
designed for accumulated restraint forces in bending in the purlin. If fly braces are used for this
accordance with the parallel restrained members - ie. fly braces from the rafter bottom flange up to
provisions of NZS 3404 [5] Clause 5.4.3.3. The the purlin, the design of these is covered by
restraint forces are those from each portal frame section 2.5.3 and Fig. 2.3 of [10].
(ie. 0.5 R*L1 and 0.5 RL2*
for pinned portal frames)
The other option is to anchor these braces into the
and the upper limit total restraint force into the rafters near the top flange, where the restraint
gable walls or similar is that from one full quantum forces can be effectively anchored into the roof
of restraint force (ie. 0.5 R* for pinned portal system without further consideration. This will
frames) plus six half quanta of restraint forces. eliminate the need for specific rafter twist restraint
In-plan restraint to spine beam critical flange but at the expense of longer and more complex
from the rafter system in-plan bracing members and connection details.

These are the situations shown in Fig. 73.6. In Because there is no net lateral restraint force
these cases, additional restraints are provided, from this in-plan bracing into the plane of the
adjacent to the portal frame/spine beam portal frame, the portal frame lateral restraint
intersection, to reduce the effective length of the forces, R*L1 , and RL2*
, remain as given by
spine beams for lateral buckling (so as to increase equations 73.3 and 73.5, respectively.
the member moment capacity). These restraints
are triangulated back to the portal. Design options 1 and 2 for spine
beam/rafter/column restraint of forces parallel
These restraints may be provided to the spine to the rafter
beam under the propped portal (positions a and c)
or under the intermediate portal (positions d and f). Fig. 73.7 shows two design options for the spine
beam/rafter/column restraint. The first of these
In the first case - ie. at the propped portal - the involves fly braces, which, as previously described,
points of restraint could be in either negative or carry the restraint forces R*sb and R*c up into the
positive moment regions. It is important that the
restraints go to the correct flange of the spine rafter and portal frame/roof system. This option
provides a point of dependable restraint at the top
beam (ie. the critical flange), which will depend on
of the column. It means that the column need not
the sign of the bending moment.
be designed to resist any restraint forces and
In the second case - ie. at the intermediate portal - therefore is designed to resist only N*c and a
the points of restraint will be in positive moment moment in the plane of the spine beam (ie. about
under vertical loading and the restraints would go the column axis transverse to the spine beam)
to the top flange in this instance. (However, for a generated by the eccentric transfer of vertical load
load case involving wind uplift, the moment is requirement of NZS 3404 Clause 4.3.4.2.
reversed although the design actions are likely to
be lower). Design option 1, in contrast, is based on the top
connection transmitting restraint actions due to
The restraints are required to transfer the restraint
( R*sb + R*c ) into the rafter system, although as
force, R*sb , back to points of anchorage. As the
described above only R*c goes into R*L1 , which is
spine beam is continuous under the rafter, the
buckled shape of the spine beam critical flange required to be resisted by the portal frame/roof
over the length of interest (ie. a-b-c or d-e-f) will system.
involve one flange trying to move to the left and the
Design option 2, in contrast, is based on the
other to the right, ie. as shown in Fig. 73.6. Thus
column/spine beam resisting the moment
there will be no net lateral restraint force
generated by the restraint actions shown in
developed along the portal frame.
Fig. 73.8. The peak magnitude of this moment,
The tendency of the spine beam critical flange to which occurs at the top of the column, is given by:
rotate in-plan about points b or e, as appropriate,
must be resisted by the rafters. The transverse *
Mmax =
(R*
sb )
+ Rc* d sb Lcol
(73.6)
forces exerted on the rafter, Rr,* left and R *r,right , are (d sb + Lcol )

HERA Steel Design & Construction Bulletin Page 18 No. 73, April/May 2003
 ult of
requirements. As a res
.8. strength and stiffness mm en tary
as defined in Fig. 73 studies that are ref
erenced from Co
where all variables are by the 2 [5] , the
able to be developed Clause C5.4.3.1 of
NZS 3404 Part
That moment must be m of rem ov ed .
ss-section at the botto requirement has be en
stiffened spine beam cro the sp ine stiffness ca se wh ere the
nnection between However, this is on
ly the
the spine beam, the co n its elf . Th e made of steel or involv
e
d the colum nts are
beam and column an restraining ele me b or
acts in a plane tra ns ve rse to nforced concrete sla
moment in the column other anchorage into a rei to pro vid e the
- ie. about the column's structural system.
The ke y is
the spine beam the ge or
ment ge ne rat ed thr ou gh to "transfer to anchora
principal axis to the mo en oft strength requirements tio ns .
vertical load requirem design restraint ac
eccentric transfer of t the reaction points" [5] the
.4.2. This means tha
NZS 3404 Clause 4.3 axial
biaxial bending plu s d on design actions
column is subject to a co lum n Restraint actions base
n option 2. For
compression for desig Cla us e 8.1 .5, required by Clauses 5.4
.3.1,
with NZS 3404 The restraint actions
section that complies city to design
will be section capa .2 are based on the
the critical cond itio n 5.4.3.2, 5.4.3.3 or 6.7 load
er capacities. For
Clause 8.3.4.2. actions, not the memb s me ans
ing earthquake thi
combinations not includ tra int is
nsider mber requiring res
General Issues to Co that, where the me
, the restrainin g for ce s to be
oversized for strength the
uirements s tha n those generated by
Restraint stiffness req resisted are les ca pa cit y.
mpression
had member moment or co
/AS1250 prior to 1992
Editions of NZS 3404 to me et bo th
ining systems
requirements for restra
Rafter

Stiffeners

Rsb*
Fly braces Rc*

Design Option 1

JLyiL

Rafter

Top Connection

Stiffeners
Rsb*
Bottom Connection Rc*

Design Option 2

Fig. 73.7
er/Column Restraint
2 for Spine Beam/Raft
Design Options 1 and
No. 73, April/May 2003
Page 19
nstruction Bulletin
HERA Steel Design & Co
 ≈ (Rsb* + R c*)
777777777777

dsb
Rsb* + Rc*
M*max
Lcol
Fig. 73.8
Restraint Force Application and Moments for Design Option 2
Load height classification for spine beam Judicious use of beam web stiffeners
supports to rafter
The designer should ascertain carefully where
Where the spine beam supports the rafter, the stiffeners to the beam web are needed in a
rafter load is applied to the spine beam via the rafter/spine beam system and only specify them
spine beam top flange. The structural system where required.
transferring this load is laterally restrained, so that,
for a spine beam with a vertical axis of As described in DCB No. 52, pp. 11-13, stiffeners
symmetry, k1 = 1.0 applies from the NOTE (1) to to the (spine) beam over the supporting
Table 5.6.3 (2). In most instances, such a support intermediate columns will almost always be
is also the end of a segment, both ends of which needed.
are restrained, hence k1 = 1.0 from Table 5.6.3 (2)
irrespective of the load height position. However, stiffeners in either the spine beam or the
rafter over the locations where the rafter is
Bolts fully tensioned supported by the spine beam alone may not be
necessary, especially where fly braces are also
The high strength structural bolts used in and used on the spine beam, and these applications
around the spine beam should be fully tensioned to should be checked more closely.
increase the rigidity of the overall system.
Where slotted or oversize holes are used, these
bolts must be fully tensioned; see NOTE to Clause
5.4.2.1 and 5.4.2.2 of NZS 3404 [5].
HERA Steel Design & Construction Bulletin Page 20 No. 73, April/May 2003
 NZFl-Vib1 Programmed by: Y. Khwaounjoo, HERA Date: 1 May 2002
Version: 1

Project: Example 2: HiBond slab (Location 2)

Designed by: YRK

ββ = 0.05 Slab Span (m) = 2.75

Beam/Joist Span 1 (m) = 9.00 Beam/Joist Span 2 (m) = 9.00
Girder 1 Span 1 (m) = 8.25 Girder 1 Span 2 (m) = 8.25

Vibration analysis based on the recommendation of AISC/CISC Design Guide 11- Floor Vibration
Due
to Human Activity, 1997, Murray et al., NZS3101:Part1:1995 and NZS 3404:Part 1:1997.

E. FINAL RESULT: FOR SLAB SPAN OF 2.75m

Fundamental Peak
Member frequency (Hz) Acceleration (% g)
Slab 18.18 0.02
Beam/Joist 8.18 0.10
Girder 1 12.04 0.05
Girder 2

Combined Floor 6.767 0.199

Limiting Combined Floor Acceleration (% g) = 0.50

2.5
Peak Acceleration (% g)

1.5

0.5 Upper Limit

Actual Value
0 J L J L

1 10 100
Frequency (Hz)

Fig. 73.9
Example of Output from Floor Vibration Program

Floor response against limiting criteria

Fig. 73.9
Example of Output from Floor Vibration Program

HERA Steel Design & Construction Bulletin Page 21 No. 73, April/May 2003
Floor Vibration Program interpreted appropriately. Guidance on this is now
given.
NZFl_Vib 1: What the Slab
Output Represents
Applying The Results from NZFl_Vib1
This article is written by Yadav Khwaounjoo, the developer of
the program NZFl_Vib 1, with additional material by G Charles Fig. 73.9 on page 21 shows an example of the
Clifton, HERA Structural Engineer. output. As previously stated, this is for Design
Example 2 from the User's Manual [12], in which
General all the dimensions of the floor system are shown.

The program NZFl_Vib 1 is a program for the The output given in the Table E: Final Result of
analysis of floor vibration of floor systems Fig. 73.9 is for the slab, beam/joist, girder 1, girder
comprising steel beams supporting a concrete 2 (none in this instance) and combined floor.
slab. It is a spreadsheet based program and
comes with a comprehensive users manual, HERA The first of these, the slab output, is for the slab on
Report R4-112 [12]. rigid supports as determined by [13]. This would
apply to the slab close to supporting columns, for
That manual contains seven worked design example. This value does not consider combined
examples. Fig. 73.9 shows the output from action effects of the slab with the supporting
Design Example No. 2, which comprises a system and should be considered in isolation from
concrete slab on Hibond decking, supported on a the rest of the items in that table.
network of secondary and primary beams.
The rest of these items are determined using [13]
The program implements two published and relate to each floor system component in turn,
procedures for floor vibration assessment. The followed by combined actions.
first of these is the AISC Design Guide Series 11
[13] and the second is the Applied Technology The graphical output is that for the combined floor
Council Design Guide 1 [14]. system.

The reason for using both procedures is because Fig. 73.9 shows a design example where the slab
neither covers the full range of floor systems built spans only 2.75m onto supporting beams
in New Zealand for which vibration checks are (secondary beams), which span onto supporting
required. The first and most widely used girders (primary beams), which are supported by
procedure [13] covers slabs supported on beams, columns. Fig. 73.10 then shows the output for the
where the response of the slab and the response same slab thickness but for the slab span
of the beams will contribute to the overall vibration increased to 5m between secondary beams. The
response. example is not a realistic practical solution,
because in this instance the size of the beams
The second procedure [14] covers the vibration of would need to be increased, as would the slab
the slab when spanning between stiff supports, thickness, however these have been deliberately
such as walls, where the vibration response of the kept the same to illustrate the affects of changing
floor is solely dependent on the slab. only the slab span.

A brief overview of the FOR SLAB SPAN OF 5 m

E. FINAL RESULT:
scope and coverage
of these two floor
vibration design Fundamental Peak
procedures is given in Member frequency (Hz) Acceleration (% g)
DCB No. 56, pp. 25- Slab 4.54 0.70
27. Beam/Joist 6.01 0.19
Girder 1 12.24 0.05
Because NZFl_Vib 1 Girder 2
incorporates and
presents results from Combined Floor 5.462 0.257
both procedures and
these procedures Limiting Combined Floor Acceleration (% g) = 0.50
treat the vibration
response of the slab
differently, the input and output needs to be
Fig. 73.10
appropriate for the support conditions (eg. slab on
Result for the Slab Span Doubled to 5m
walls or slab on beams) and needs to be read and

HERA Steel Design & Construction Bulletin Page 22 No. 73, April/May 2003
 The effect on the slab alone is considerable; the These responses occur at different regions of the
peak acceleration for the slab alone close to rigid floor and, for a complying floor system, all peak
supports is now critical; which would be the case in accelerations must be below the limiting floor
practice if this system were built. acceleration criteria. As previously mentioned, the
upper limit line shown in the graphical output of
Increasing the slab span also affects the beam and NZFl_Vib 1 gives the limiting floor acceleration as
girder stiffnesses and participating masses. In this a function of frequency that each component of the
case, doubling the slab span decreases the beam floor system must meet. The combined floor
frequency and increases the peak acceleration, response is plotted on that graph. However, the
because the increase in participating mass is slab response is not shown there and must be
greater than the increase in stiffness. However, it checked seperately. In the case of the slab shown
slightly increases the girder frequency, although in Fig. 73.11, the limiting acceleration (for this type
this does not change the calculated girder peak of occupancy) for the slab with a frequency of
acceleration. 7.5 Hz is 0.5% g, so the slab is satisfactory.

Modelling the Effect of Stiff Supports

Design of Multi-Storey Steel
For a slab spanning directly onto stiff supports, eg. Buildings for Satisfactory In-
walls, type NR into the Sec. Beam/Joist and the
Pri. Beam 1 boxes. The slab response is then Service Wind-Induced
calculated in accordance with [14], which is the Acceleration Response: Update
appropriate procedure in this instance (ie. with no
supporting beams).
on DCB No. 66
This update is written by G Charles Clifton, HERA Structural
However, when the slab spans onto a mixture of Engineer.
stiff supports and supporting beams, then the
details of these beams must be entered and the Background
calculation is undertaken to [13]. Fig. 73.11 shows
the results for a 150 mm thick slab on Hi-bond, DCB No. 66, pp. 1-10, February 2002 contains a
spanning from a wall onto a 530UB82 supporting procedure for the design of multi-storey steel
beam - ie. one end of the deck span is supported buildings for satisfactory in-service wind induced
by the wall, the other runs over the supporting acceleration response. This procedure is based
beam. on a preliminary design technique developed by
Cenek et. al. [15] and the
FOR SLAB SPANNING 5m FROM WALL ONTO Joint Australian/New Zealand
E. FINAL RESULT: A SUPPORTING BEAM wind loadings standard,
AS/NZS 1170.2:2002 [16].
Fundamental Peak
Member frequency (Hz) Acceleration (% g) The DCB paper also gives the
Slab 7.50 0.48 background to this topic, a
Beam/Joist 8.09 0.05 commentary to the procedure
Girder 1 and a worked design
Girder 2 example.

Between then and now, the

Combined Floor 5.632 0.241 following developments have
occurred which call for an
Limiting Combined Floor Acceleration (% g) = 0.50 update on the DCB No. 66
procedure:
Fig. 73.11
Result for 5 m Span Slab on Hi-Bond, 150 mm • AS/NZS 1170.2 [16] has been published (it
Thick, Spanning from Wall to over a Supporting was in draft when that paper was written)
Beam
• Some errors and ambiguities have been
In this instance, what the results show is: noted in that paper

• the response of the slab alone close to the • Clark Hyland has undertaken further
supporting wall (which in this instance has research into the comparison of design
the highest peak acceleration value) provisions for calculating the peak
• the response of the supporting beam, and acceleration at the top of the building and
• the combined response of the slab and found a more accurate expression for this.
supporting beam, where the slab Details are given in session 1 of [17] and
contribution is that away from the supporting briefly discussed on the next page.
wall.

HERA Steel Design & Construction Bulletin Page 23 No. 73, April/May 2003
These aspects are all covered in the update SPM0103: Potential Problem
details, which are as given below.
and Solution
Details of Update
This short article is written by G Charles Clifton, HERA
Structural Engineer.
(1) The building mass to be used, mo, should be
expressed in kg/m height and is the average
The second edition of the Slab Panel Method has
building mass/unit height over the top
been published in DCB No. 71. It comes with a
one-third of the building, when applying the
computer program, SPM0103, which allows
provisions of DCB No. 66 or Appendix G of
designers to rapidly design the slab panels.
AS/NZS 1170.2 [16]
The program is available as a single executable
(2) In the example of application given on pages
file entitled SPM0103. It is accompanied by a
7,8 of DCB No. 66, mo is also expressed as
sample calculation file, being the design
tonnes/m height when making the level 1
example presented in section 9, pages 12-14 of
assessment. For the equation used in that
1.3 DCB No. 71.
assessment, h / mo > 1.6, this change from
kilogrammes to tonnes is correct. Equation
To date, no one to whom the program has been
G1 from [16] is the same equation, but with
sent has reported any problems with its
the 1.6 changed to 0.0016 to allow mo in
installation. One firm, however, has reported a
kg/m height to be used.
problem with trying to run new jobs. What
happens is that they can input the data on the
(3) The design acceleration and the
input screen, however the program does not
acceleration limit must be in the same units.
perform the calculations when the calculate button
The amax from equation 66.2, DCB No. 66, is
is pressed, displaying instead a run time error.
calculated in m/s/s and converted to milli-g
through the (1000/9.81) factor at the end
The SPM0103 version of the program is written as
of the equation. The equation given in
a design tool, however the coding for the program
Fig. 66.3 is equation 66.7 without the
actually contains code for the design version and
(1000/9.81) adjustment factor. Its units are
for a research version, which was developed for
therefore m/s/s, although the vertical axis of
HERA research use and allows more variables to
that figure is given in milli-g
be altered than are required for design. In the
design version, the coding for the research version
(4) Also in equation 66.7, fo = fundamental
is supposed to be switched off. However, what
frequency, not fundamental period as stated
has happened, in this instance, is that modules of
the research version have been activated by the
(5) Note the frequency scale in Fig. 66.2 is in
design example and have tried to run in
steps of 20, not 2!
conjunction with modules of the design version,
causing incompatibility problems and a run-time
(6) It is recommended that readers use
error to develop. Why we don't know.
Equation G3(1) from AS/NZS 1170.2 [16]
instead of equation 66.2 from the
However, if a user encounters this problem, the
DCB No. 66 paper to calculate the
solution is straight forward. It is to overwrite the
(cross-wind) design maximum acceleration
data, in the design example supplied, with the
at the top of the building. This is the
design data for the case under consideration, do
critical wind acceleration for design. This
the calculation, then save as under a new
recommendation comes from Hyland's
filename. That approach has worked in the two
comparison study, which shows Equation
instances where this problem has occurred.
G3(1) giving the best match in cross-wind
acceleration with experiment of all the
If any users have encountered this problem,
procedures available. This equation gives
please advise Charles Clifton, email address:
the acceleration in m/s/s directly, which is
structural@hera.org.nz
then compared with alimit from equation 66.7,
taking account of the units. Equation G3(1)
is more complex to use than the Cenek et.
al. equation 66.2, but gives more accurate
answers. See details from Session 1 of [17]
for more information. When using Equation
G3(1), be careful in determining the value
of Cf s to use from Clause 6.3.2.3 of
AS/ZNS 1170.2 and make sure the
appropriate option for the building's shape
and proportions is being used.

HERA Steel Design & Construction Bulletin Page 24 No. 73, April/May 2003
References
13. Murray, TM et. al.; Floor Vibration due to
1. AS/NZS 2312:2002, Guide to the Protection
Human Activity; American Institute of Steel
of Structural Steel Against Atmospheric
Construction, 1997, Steel Design Guide
Corrosion by the Use of Protective Coatings;
Series 11.
Standards New Zealand, Wellington.
14. Allen, DE et. al.; Mnimising Floor Vibration;
2. Hyland C; Structural Steelwork Connections
Applied Technology Council, Redwood City,
Guide Incorporating Amendment No. 1;
USA, 1999, ATC Design Guide; 1.
HERA, Manukau City, New Zealand,
1999/2001, HERA Report R4-100.
15. Cenek, P et. al.; Designing for Dynamic
Serviceability Under Wind Loading; Recent
3. Hyland, C; Structural Steelwork Estimating
Advances in Wind Engineering, Volume 1,
Guide; HERA, Manukau City, 1998, HERA
TF Sun (Editor); Pergamon Press, 1989, pp.
Report R4-96.
399-406.

16. AS/NZS 1170.2: 2002, Structural Design

4. Design Capacity Tables for Structural Steel,
Actions Part 2; Wind Actions; Standards
Second Edition, Volume 2: Hollow Sections,
New Zealand, Wellington.
Australian Institute of Steel Construction,
Sydney, Australia, 1999.
17. Hyland, C et.al; Notes Prepared for the Steel
Structures Seminar 2003; HERA, Manukau
5. NZS 3404: 1997, plus Amendment No. 1:
City, New Zealand, 2003, HERA Report
2001, Steel Structures Standard; Standards
R4-119.
New Zealand, Wellington, New Zealand.

6. AS 1418.1 : 2002 (Plus Amendment No1 :

1997), Cranes ( Including Hoists and
Winches) Part 1 : General Requirements;
Standards Australia, Sydney, Australia.

7. Roark, RJ and Young, W C; Formulas for

Stress and Strain (Sixth Edition); McGraw-
Hill International, Tokyo, Japan, (1975).

8. AS 1163:1991, Structural Steel Hollow

Sections; Standards Australia, Sydney,
Australia.

9. Clifton, GC; Structural Steelwork Limit State

Design Guides Volume 1; HERA, Manukau
City, 1994, HERA Report R4-80.

10. Clifton, GC; Restraint Classifications for

Beam Member Moment Capacity
Determination to NZS 3404:1997; HERA,
Manukau City, 1997, HERA Report R4-92.

11. NZS 4203:1992, General Structural Design

and Design Loadings for Buildings;
Standards New Zealand, Wellington, New
Zealand.

12. Khwaounjoo, YR; Report and User’s Manual

for NZFl_Vib 1 Program (Program for the
Analysis of Floor Vibration); HERA,
Manukau City, New Zealand, 2002, HERA
Report R4-112.

HERA Steel Design & Construction Bulletin Page 25 No. 73, April/May 2003