assessment retrofit flexible diaphragms
TRANSCRIPT
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PAPER CLASS & TYPE: GENERAL NON-REFEREED1ME (Dist.), CPEng, MIPENZ,Technical Director, Holmes Consulting Group, Auckland
ABSTRACT
Unreinforced masonry buildings (URM) have historically performed poorly in large earthquakes.
Many URM building failures have been attributed to poor diaphragm performance and
inadequate wall diaphragm connectivity. This paper provides a brief introduction into
URM buildings and their seismic behaviour. It then outlines a design methodology for the
assessment and retrofit of flexible diaphragms in URM buildings with an accompanying design
example. Recommendations for future research are also provided.
Figure 1. Earthquake Damage to Build ings in
Central Napier after the 1931 Earthquake
(Courtesy of Hastings Distric t Council)
1.0 INTRODUCTION
Unreinforced masonry construction was commonly used
in New Zealand between 1880 and the early 1930s.
The poor performance of these buildings in the 1931
Hawkes Bay earthquake resulted in a rapid decline in
popularity and subsequent prohibition of their use in
1965 (Ingham 2008). It is estimated that approximately
3500 unreinforced masonry (URM) buildings still existin New Zealand today (Russell 2008) with a significant
number of those structures still to be strengthened.
This paper provides a brief introduction into URM
buildings and details a design methodology used by the
author to assess and strengthen flexible diaphragms in
existing URM buildings. The design methodology has
been largely adapted from the New Zealand Society for
Earthquake Engineering document Assessment and
Improvement of the Structural Performance of Buildings
in Earthquakes (NZSEE 2006), and ASCE/SEI 41-06
Seismic Rehabilitation of Existing Buildings (ASCE2006).
It has been observed that some people have found
these documents difficult to use and it is hoped that
this paper may assist other Structural Engineers when
they are assessing and retrofitting flexible diaphragms
in URM buildings. The author is interested in receiving
feedback that readers may have with what is proposed
here.
2.0 UNREINFORCED MASONRY BUILDING
CONSTRUCTION
Unreinforced masonry buildings constructed in
New Zealand in the early 1900s typically consist of
unreinforced masonry perimeter and inter-tenancy walls
with timber framed floors and roofs. URM buildings
typically range in height from between 1 to 6 storeys
with 1 or 2 storey structures being the most common.
Unreinforced masonry was primarily used for the
construction of the perimeter and inter-tenancy walls
because of its non combustibility and, for the case of the
exterior walls, its durability when compared with timber.
A DESIGN METHODOLOGY FOR THE ASSESSMENT
AND RETROFIT OF FLEXIBLE DIAPHRAGMS IN
UNREINFORCED MASONRY BUILDINGSBy: Stuart J Oliver1
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Figure 2 illustrates a cross-section of a typical 5 story
high URM building.
Flooring often consists of straight or diagonal timber
sheathing supported on timber joists that typically
span up to approximately 6 meters. Timber joists are
typically supported on either URM walls, or beams thatoften consist of either heavy timber or structural steel
sections. Where supported on masonry walls the joist
and beam ends were often cut diagonally and supported
in pockets in the walls.
Roofing typically consists of corrugated light gauge steel
sheathing supported on timber purlins. Timber trusses
spanning onto either URM walls or beams are typically
provided to support the roof framing. Where supported
on the URM walls the ends of the purlins and trusses
are typically seated into pockets in the walls.
3.0 LATERAL LOAD RESISTING SYSTEM
The lateral load resisting system for URM buildings
consists of both horizontal and vertical lateral load
resisting elements. Floor and roof diaphragms are
the primary horizontal lateral load resisting elements
in typical, unretrofitted, URM buildings. Collectors, ifprovided in retrofitted structures, are also considered to
be horizontal lateral load resisting elements. Vertical
elements typically consist of the URM walls and their
foundations, and new shearwall or braced frames if the
latter are provided as part of a retrofit design.
During a seismic event the tributary horizontal inertia
forces associated with the face loaded walls are
required to be transferred back into the main body
diaphragm via a system of wall anchors and diaphragm
ties. The diaphragm is then required to transfer the
Figure 2. Section Through a Typical 5 Story URM Building
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inertia forces associated with the face loaded walls and
its own seismic mass to the resisting in plane end walls
again via a series of wall anchors.
Unreinforced masonry buildings have historically
performed poorly in large earthquakes. A recent report
by Bruneau (1994) indentified that many of the failuresdue to earthquakes found in URM buildings during the
last 20 years were related to diaphragms and their
connections to the walls. These walls typically account
for approximately 70% 80% of the total seismic mass
of the structure. Failures of this nature that have been
commonly observed include:
Figure 3. Parapet Failure
(reproduced from FEMA, 1998)
Parapet failure. Referring to Figures 3 & 4 this
failure mechanism occurs when face loaded
parapets are not adequately braced back to the
supporting structure.
Wall diaphragm tension tie failure. This failure
mechanism occurs when insufficient connectionis provided between face loaded walls and the
supporting diaphragm (refer Figures 5 & 6).
Wall diaphragm shear failure. As is illustrated
in Figures 7 & 8 this failure mechanism occurs
when insufficient connection is provided between
diaphragm and the stiff in-plane acting end walls.
Figure 4. Parapet Failure Observed at
the 2007 Gisborne Earthquake
Figure 5. Wall diaphragm tension tie failure(from FEMA 1998)
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Figure 6. Wall Diaphragm Tension
Tie Failure Observed at the
2007 Gisborne Earthquake
Figure 7. Wall Diaphragm Shear
Failure (reproduced from FEMA, 1998)
Figure 8. Wall diaphragm
shear failure (reproduced
from FEMA, 1998)
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When assessing and retrofitting URM buildings it is
essential that each of the above failure mechanisms
are considered when assessing the performance of the
existing timber diaphragms.
4.0 FLEXIBLE DIAPHRAGM ANALYSIS
4.1 Analysis Methodology
In the past some engineers have attempted to analyse
URM structures with timber diaphragms assuming rigid
diaphragm behaviour. While most in-plane loaded URM
walls are relatively rigid, timber framed floor diaphragms
are generally not. Referring to Figure 9 diaphragms
are defined as being flexible when the maximum lateral
deflection of the diaphragm along its length (D) is
greater than twice the average interstory drift (W) of
the vertical lateral load resisting elements of the story
immediately below the diaphragm (ASCE 2006).
Figure 9. Diaphragm and Wall Displacement
Terminology (reproduced f rom ASCE, 2006 )
An assumption of rigid diaphragm behaviour may
lead to unconservative assessments of diaphragm
accelerations and inaccurate estimates of load
distribution between lateral load resisting elements. For
URM buildings with single span flexible diaphragms, six
stories or less in height the NZSEE guidelines (NZSEE,
2006) permit a simplified analysis to be undertaken.
This section of the NZSEE guidelines is based on FEMA
356 (FEMA, 2000) the precursor document to ASCE/SEI
41-06 (ASCE, 2006). The simplified analysis assumes
that each diaphragm spans as a simply supported
element between the masonry end walls and permits
the effects of horizontal torsion to be ignored.
Diaphragm deflections are not to exceed 150 mm under
prescribed code seismic loads for this simplified method
of analysis to be applicable.
The simplified analysis assumes that the in-planeloaded masonry end walls are relatively rigid elements
which, as a consequence of their high stiffness, do not
significantly amplify earthquake ground motions. The
dominant mode of response is assumed to be the in-
plane fundamental mode of the diaphragms excited
by the inertia forces associated with the out-of-plane
loaded walls. It is also assumed that the response of
each story is uncoupled from adjacent stories.
For more complicated URM structures where the
simplified analysis method is not applicable a more
detailed modal response spectrum analysis could be
undertaken using general purpose structural analysis
software. The flexible diaphragms could be modelled
using plane stress elements in this instance. In
some cases, as part of a preliminary design, multi-
span flexible diaphragm structures might be analysed
using the simplified analysis method. In this case the
diaphragm could be treated as a series of independent
simply supported spans. The results of this preliminary
analysis would need to be confirmed by a more rigorous
method in later design phases.
The steps of the simplified analysis procedure
considered when performing a diaphragm assessment
are as follows:
Step 1: For each axis of the building and at each
level calculate the fundamental period of the
diaphragms.
Step 2: Calculate the seismic loads for diaphragms
using the periods determined in Step 1 and theappropriate spectral accelerations calculated
using the New Zealand Loadings Standard,
AS/NZS 1170.5 (SNZ, 2004) as modified by
the NZSEE guidelines.
Step 3: Verify that the existing diaphragms have
adequate strength and stiffness to resist the
required seismic loads and strengthen the
diaphragms if necessary.
Step 4: Calculate the seismic loads generated at the
wall to diaphragm connections.
Step 5: Assess the capacity of the existing wall
diaphragm ties and provide supplementary
wall anchors and sub-diaphragm ties as
required.
Note that the above methodology only considers the
steps of a seismic assessment related to the performance
of diaphragms. A complete building assessment would
also typically include a review of the foundations
and any geological site hazards, vertical lateral load
resisting elements and non-structural components (i.e.
architectural, mechanical and electrical). Details ofsuch a review are outside the scope of this paper and
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readers are referred to the NZSEE guidelines for further
information (NZSEE, 2006).
A detailed description of Steps 1-5 is included in the
following sections.
4.2 Diaphragm Period Determination
Using the NZSEE guidelines the fundamental period,
T1, of the flexible diaphragms can be estimated using
Equation 1.0:
Where: D= maximum in-plane diaphragm deflection
due to a lateral load of 1.0 g, in metres.
For the purposes of this equation thediaphragm shall be considered to remain
elastic under the prescribed elastic load.
ASCE/SEI 41-06 (ASCE, 2006) provides default
expected shear stiffness values, Gd, that can be used to
calculate deflections in existing diaphragms. These are
summarised in Table 1 below. Of the diaphragm types
shown it has been the authors experience that single
straight sheathed diaphragms are the most common.
These diaphragms consist of timber sheathing (typically
150 x 25 boards) laid perpendicular and nailed to the
floor joists with two or more nails at each joist. Shear
D1 3.07T
Eqn. 1
Table 1. Default Strength Values For Exis ting
Timber Diaphragms (ASCE, 2006)
Diaphragm Type Shear Stiffness,
Gd(KN/m)
Yield Strength,
Rn (N/m)
Single Straight Sheathing 350 1750
Double Straight Sheathing Chorded 2600 8750
Unchorded 1200 5850
Single Diagonal Sheathing Chorded 1400 8750
Unchorded 700 6130
Double Sheathing with Chorded 3200 13100
Straight Sheathing or
Flooring Above
Unchorded 1600 9130
Double Diagonal
Sheathing
Chorded 3100 13100
Unchorded 1600 9130
forces perpendicular to the sheathing are resisted by
a nail couple generated at each joist. Shear forces
parallel to the direction of the sheathing are transferred
through the nails in the supporting joists or framing
members below the sheathing joists.
Detailed descriptions of the other diaphragm types listedin Table 1 are given in ASCE/SEI 41-06 (ASCE 2006).
Using the shear stiffness values, Gd, provided in Table 1
the diaphragm stiffness, KD, can be calculated for each
diaphragm in each direction as:
Eqn. 2
Where: b = Diaphragm width.
Gd = Diaphragm shear stiffness from Table 1.
L = Diaphragm span.
The maximum in-plane diaphragm deflection for each of
the existing diaphragms, D, in each direction can then
be calculated using Equations 3:
Eqn. 3
Where: Vu = Diaphragm lateral load.
KD = Diaphragm stiffness.
L
bGK dD
4
D
uD
K
V
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When determining the diaphragm lateral load only
the tributary seismic mass of the out-of-plane loaded
walls and the diaphragm itself should be included. The
seismic mass associated with the in-plane loaded, end
walls should not be included.
If it is found that the existing diaphragms requirestrengthening, and that new plywood sheathing
and/or chord elements are required, the stiffness of
the strengthened diaphragm can be calculated in
accordance with the New Zealand Timber Structures
Standard, NZS 3603 (SNZ, 1993).
4.3 Diaphragm Seismic Load Calculations
For each of the diaphragms being assessed the
diaphragm seismic load, VD, can be calculated as:
VD= C1C3Cd(T1)WD Eqn. 4
Where: C1 = Modification factor to relate
expected inelastic displacements
to those calculated for linear elastic
response. Values recommended in
ASCE/SEI 41-06 (ASCE, 2006) are:
C1 = 1.5 for T1< 0.10 seconds
C1 = 1.0 for T1> Ts Linear interpolation may be used to
calculate intermediate values of C1.
T1 = Fundamental period of the
diaphragm in the direction being
considered.
Ts = Characteristic period of the response
spectrum, defined as the period
associated with the transition from
the constant acceleration segment
to the constant velocity segment of
the response spectrum. In terms of
the AS/NZS 1170 response spectra:
Ts = 0.40 s for Soil Types A, B & C
Ts = 0.5 s for Soil Type D
Ts = 1.0 s for Soil Type E
C3 = Modification factor to account for
P-effects and is a function of the
stability coefficient, i, calculated in
accordance with Section 6.5.2 of
AS/NZS 1170.5 (SNZ, 2004).
C3 = 1.0 when i < 0.1 in all stories,
otherwise,
C3 = 1+5( 0.1)/T1 where is themaximum value of ifor all stories.
Cd(T
1) = Horizontal design coefficient
calculated in accordance with
Section 5.2.1 of AS/NZS 1170.
WD = Effective seismic mass associated
with the fundamental mode of the
diaphragm i.e. tributary seismic
mass of the face loaded walls and
the diaphragm itself.
Equation 4 is a modified version of the NZSEE Equation
4E-8 recommended by the Author. The original
formulation has been simplified by removing modification
factors of C2and Cmwhich are both equal to 1.0 for this
application.
A second departure from the original NZSEE formulation
was to incorporate ductility directly into the equation by
using the AS/NZS 1170 horizontal design coefficient,
Cd(T1), rather than the AS/NZS 1170 elastic site spectral
value, C (T1). It is believed this removes an extra step
in the analysis procedure and is more closely aligned to
conventional New Zealand design office practice.
Care is required when determining an appropriate
structural ductility factor, , for the assessment. Whendetailing is such that the failure mechanism is expected
to be ductile (i.e. nail pull-out) relatively high levels of
ductility can be expected. The New Zealand Timber
Structures Standard (SNZ, 1993) recommends that a
structural ductility factor, , of up to 4.0 can be assumedfor these applications. When other less ductile failure
modes govern the structural ductility factor, , of 1.25may be more appropriate.
ASCE/SEI 41-06 provides some guidance on the
expected structural ductility capacity of existing timber
diaphragms. In this standard component modification
factors (m-factors) to account for the level of expected
ductility at various structural performances limit states
are specified. Table 2 below details the ASCE/SEI
41-06 m-factor applicable to existing timber diaphragms
at the life safety limit state. These are analogous to the
structural ductility factor, , used in AS/NZS 1170.
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Table 2. ASCE/SEI 41-06 m-Factors for the Life Safety
Limit State (ASCE, 2006)
Diaphragm Type Diaphragm
Length/Width
Ratio (L/b)
m-Factor
Single Straight Sheathing Chorded L/b 3.0 2.0
Unchorded L/b 3.0 1.5
Double Straight Sheathing Chorded L/b 3.0 2.0
Unchorded L/b 3.0 1.5
Single Diagonal Sheathing Chorded L/b 3.0 2.0
Unchorded L/b 3.0 1.5
Double Diagonal Chorded L/b 3.5 2.5
Sheathing Unchorded L/b 3.5 2.0
It is noted that the m-factors detailed in Table 2 are
relatively low when compared with the structural ductility
factor, , used for the design of new ductile timberstructures. It is understood that additional research is
currently being undertaken at the University of Auckland
to determine appropriate structural ductility factors that
could be used for the assessment of existing timber
diaphragms in New Zealand.
4.4 Diaphragm Capacity Assessment
When assessing the capacity of diaphragms both
deformation and strength need to be considered.
4.4.1 Deformation Assessment
Equation 3 can be used to determine the expected
diaphragm deflections using the diaphragm seismic
loads calculated in the previous section. Details on how
to determine the diaphragm stiffness were provided
previously in Section 4.2. Diaphragm deflections should
be scaled in accordance with AS/NZS 1170.5 Section
7.2.1 to account for inelastic deformation. P-effects
are already included in the modification factor C3 and
as such AS/NZS 1170.5 Section 7.2.1.2 does not apply.
NZSEE guidelines limit the maximum diaphragm
deflections to 150 mm for this assessment methodology
to be applicable. The Author also recommends a rule
of thumb approach where diaphragm deflections are
limited to less than half the thickness of the supported
out-of-plane URM walls to ensure the diaphragm
deformations do not adversely affect the stability of
these elements. It is hoped that in time research will
provide greater guidance on appropriate diaphragm
deformation limits.
If the existing diaphragms are found to have inadequate
stiffness the diaphragms should be strengthened.
4.4.2 Strength Assessment
NZSEE guidelines recommend that the parabolic
load distribution illustrated in Figure 10 be used when
assessing the capacity of flexible diaphragms. The
parabolic load distribution is intended to emulate
the expected distribution of horizontal inertia forces
developed in the diaphragm.
The parabolic load distribution illustrated in Figure 10
can be expressed as (ASCE, 2006):
2
dd
DE
L
2x1
L
1.5Vw
Eqn. 5
Where: wE = Diaphragm inertia load (kN/m).
VD = Total diaphragm inertia load (kN).
Ld = Diaphragm span (m).
x = Distance from centre line of the
diaphragm (m).
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Figure 10. Recommended Load Distribution
for Flexible Diaphragm Analysis
(reproduced from ASCE 2006)
Similarly the diaphragm shear force and bending
moment distributions can be expressed as:
2d
2
d
DE
L
x
2
3
L
xVV
32
5 dD3d
4D
d
2D
E
LV
2L
xV
4L
x3VM
Eqn. 6
Eqn. 7
Maximum design actions can then be calculated as:
Eqn. 8
Eqn. 9
Default strength values detailed in Table 1 from ASCE/
SEI 41-06 (ASCE, 2006) can be used to assess the shear
capacity of existing diaphragms. Similar values can be
calculated from first principals using the methodology
detailed in Appendix 11B of the NZSEE guidelines or
found from similar references (ICC, 2007). It has been
noted that these values are significantly less than those
recommended in Table 11.1 of the NZSEE guidelines.However it is unknown why such a large discrepancy
exists.
NZS 3603 (SNZ, 1993), the New Zealand Timber
Structures Standard can be used to design diaphragms
strengthened with new plywood sheathing when this is
required. Good references for the design of new plywood
diaphragms include Timber Design Guide (Buchanan,
2007) and Horizontal Timber Diaphragms for Wind and
Earthquake Resistance(Smith et. al., 1986).
Plywood overlays with stapled sheet metal blocking can
also be used to strengthen existing diaphragms whenit is desired to keep the existing timber sheathing for
heritage or other reasons. As illustrated in Figure 11
in this instance light gauge sheet metal strapping is
2
V
V D
maxE,
32
L5VM dDmaxE,
Figure 11. Stapled Sheet Metal Blocking Used as Part of the
Auckland Ar t Gallery Diaphragm Strengthening Works
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provided at plywood sheet edges and staples are used
to fasten the plywood to the blocking. This mitigates
the need for timber blocking at the plywood sheet edges
which can be problematic when the existing timber
sheathing is to remain.
Diaphragm strengthening using this technique iscommon in North America and has recently been used
at the Auckland Art Gallery and Christchurch Arts Centre
Old Arts School as part of seismic refurbishment works.
Formal diaphragm chord elements are often absent
in existing diaphragms. Table 1 provides strength
and stiffness values for both chorded and unchorded
diaphragms. Values given for unchorded diaphragms
have been downgraded acknowledging the reduced
structural performance expected of these elements.
Diaphragm chord elements can be retrofitted into
existing diaphragms relatively easily. Multiple bays of
existing timber joists can be utilised by making them
continuous using nail plates and/or bolts. When new
chord elements are required perpendicular to existing
joists new light gauge metal straps nailed to timber
blocking can be used.
Diaphragm chord elements are typically detailed as
elastically responding elements by following capacity
design procedures or using nominally ductile (i.e. =1.25) design loads.
4.5 Calculation of Wall Diaphragm ConnectionLoads
Both in-plane and out-of-plane connection loads need
to be considered when assessing existing, or retrofitting
new wall-diaphragm ties. Wall-diaphragm ties sustain
predominantly in-plane loading at the diaphragm
connections to the lateral load resisting end walls.
Out-of-plane (i.e. tension and compression) loading of
the ties occurs where they are restraining face loaded
walls.
Concurrency of in-plane and out-of-plane seismic
loading is typically not considered when assessing
existing or designing new wall-diaphragm ties.
4.5.1 Out-Of-Plane Seismic Response
Out-of-plane seismic response of masonry walls in
flexible diaphragm structures is complex and still not
fully understood. Out-of-plane wall-diaphragm tie forces
could be determined by one of the following methods:
i. The connections could be designed using a
capacity design approach such that they have
adequate capacity to resist the maximum
reactions that could be generated by the out-of-
plane response of the wall. Research undertaken
by Blaikie (Blaikie, 2001) has shown that peak
out-of-plane wall inertia force generated in a wall,
Fph,Wall,can be calculated as:
Eqn. 10
Where: 1 = Dynamic magnification factorto account for wall-diaphragm
resonance.
t = Wall thickness (m).
H = Height of wall between floors (m).
WP = Tributary weight of the wall (kN/m).
Pu = Overburden load due to the weight
of the building above (kN/m).
The dynamic magnification factor, 1is a functionof the stiffness of the diaphragm. For single
story buildings with diaphragm periods of greater
than 1.0 seconds 11.3. For similar buildingswith diaphragm periods of less than 0.5 seconds
13.0.
Studies of three storey buildings have shownthat for buildings with a diaphragm period of
0.5 seconds 1 2.0. For similar buildings with
diaphragm periods of 1.0 seconds, 1 was also
found to be a function of the stiffness of the
lateral load resisting end walls. For stiff end
walls (i.e. period of 0.5 seconds) 11.6. When
more flexible end walls are present (i.e. period of
1.0 seconds) 11.2.
ii. The methodology recommended in NZSEE
guidelines (NZSEE, 2006) uses the Parts
Provisions detailed in Section 8 of AS/NZS 1170.It is worth noting that the NZSEE guidelines
recommend assuming a part ductility factor, p,of 1.0 when calculating connection loads. This
appears to contradict AS/NZS 1170 which states
that when considering non-ductile connections a
part ductility factor, p, of 1.25 should be used.
iii. Using the out-of-plane wall anchorage provisions
detailed in ASCE/SEI 41-06 (ASCE, 2006) i.e.:
Eqn. 11
up1 PWH
t2Wallph,F
pSTie WTP dC
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Where: PTie = Wall diaphragm connection
load (kN/m)
= Out-of-plane wall responsecoefficient taken to be 1.2 for
flexible diaphragm structures at
the life safety limit state.
Cd(Ts) = Horizontal design coefficient
calculated in accordance with
Section 5.2.1 of AS/NZS 1170 for
a short period structure.
WP = The tributary weight of the wall
(kN/m).
iv. Use capacity design to detail the wall-diaphragm
connections such that they have adequate
capacity to resist the peak inertia forces that can
be resisted by the diaphragm i.e.:
Eqn. 12
Where: PTie,os= Overstrength wall diaphragm
connection load (kN/m)
2 = Dynamic magnification factorto account for wall-diaphragm
resonance and inelastic modes
of diaphragm response.
VD,os = Diaphragm overstrength shear
capacity (kN).
WD = Effective seismic mass associated
with the fundamental mode of the
diaphragm (kN) (refer Section
4.3).
WP = The tributary weight of the wall
(kN/m).
Research of rigid diaphragm structures with
concrete walls suggests that a dynamic
magnification factor, 2of 2.5 would be appropriatefor such structures (Paulay & Priestley, 1992).
Unfortunately the Author is unaware of any similar
research that has been undertaken to confirm
appropriate values of 2 for flexible diaphragmstructures.
Table 3 below compares the out-of-plane wall forces
calculated using the four design methods detailed above.
The design loads were calculated assuming the building
was founded on a Class C subsoil in Wellington. It was
assumed that the building diaphragm periods were
0.5 seconds, of 4.0 and a diaphragm overstrength,VD,os/WDof 0.23g. Supported out-of-plane loaded walls
were assumed to have H = 3.70 m, t = 350 mm, P u/Wp
= 2.0 and 1= 2.0.
Table 3. Comparison of Out-of-plane Wall
Design Load Calculated Methods
pD
osD,osTie, W
W
VP 2
nalysis Methodology Out-Of-Plane Wall
Design Load
Blaikie 2.27 WpAS/NZS 1170 1.81 WpASCE/SEI 41 -06 0.61 WpDiaphragm Overstrength
Method
0.58 Wp
A parts structural ductility factor, p, of 1.25 was usedand 2was assumed to be 2.5 for the purpose of thisexample. For AS/NZS 1170 and ASCE/SEI 41-06
methods the calculated loads were reduced to 67% of
that which would be used to design a new building as is
often done when evaluating existing structures.
Referring to Table 3 it can be seen that the out-of-
plane wall loads calculated using the AS/NZS 1170
parts provisions are approximately three times that
recommended in ASCE/SEI 41-06. This is a significant
difference. It is also evident from Table 3 that the use
of the Blaikie method can result in large out-of-plane
design loads when compared with the other three
methods, particularly in the lower levels of buildings
where overburden stresses (Pu/Wp) are higher.
Table 3 also illustrates that, when assuming a
dynamic magnification factor, 2 of 2.5, the diaphragmoverstrength method results in out of plane design loads
that are similar to those determined using ASCE/SEI
41-06. Some Structural Engineers are concerned
that using the AS/NZS 1170 parts provisions may be
potentially overly conservative. This is of particular
importance for existing buildings located in regionsof high seismicity as adopting the AS/NZS 1170 parts
provisions can result in unwieldy and costly retrofit
solutions (refer also Section A11.2 of the design example
included in Appendix A).
The validly of the dynamic magnification factor, 2usedin the previous design example needs to be confirmed
by future research before the diaphragm overstrength
method can be adopted. Until this research is completed
it is recommended that the AS/NZS 1170 parts provisions
are used to determine out-of-plane wall forces.
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4.5.2 In-Plane Seismic Response
Capacity design can be used to determine the maximum
in-plane connection load i.e.:
VD,os= osRn Eqn. 13
Where: VD,os= Diaphragm overstrength shear
capacity (kN/m).
os = Diaphragm overstrength factor.
Rn = Diaphragm shear capacity (kN/m).
For those situations when diaphragm overstrengths
cannot be reliably estimated (i.e. plywood diaphragms
when the sheathing has been glued or when significant
non-structural floor finishes are known to exist)
Equation 4 can be used to determine the connection
loads. When using Equation 4 it is recommended that a
nominally ductile (i.e. of 1.25) structural response beassumed.
4.6 Design and Assessment of Wall DiaphragmConnections
4.6.1 Out-Of-Plane Seismic Response
Figures 12 & 13 illustrate typical wall diaphragm ties
for the situation when the floor joists are parallel and
perpendicular to the support masonry wall respectively.
In terms of out-of-plane loading the function of the
diaphragm tie elements are to transfer the horizontal
inertia forces associated with face loaded walls into the
main body of the diaphragm.
Figure 12. Typical Wall Diaphragm Connect ion
with Joists Parallel to the Wall
Figure 13. Typical Wall Diaphragm Connection
with Joists Perpendicular to the Wall
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Poor performance of URM and precast concrete tilt-
up buildings with plywood roof diaphragms in previous
earthquakes (i.e. 1971 San Fernando and 1994
Northridge earthquakes) has resulted in the following
design recommendations (ICBO, 2000):
i. Cross grain bending of boundary joists should beavoided. As is illustrated in Figure 14 cross grain
bending can occur when out-of-plane loaded walls
pull away from the floor diaphragm, when out-of-
plane wall anchor brackets are not provided. In
this case, tension forces that are transferred by
the anchor bolt and resisted by the diaphragm
place the boundary joist in cross grain bending.
Timber typically has low cross grain bending
capacity and in many instances has been found
to be inadequate to resist the necessary seismic
loads in past earthquakes.
ii. Nailing provided at the edges of plywood sheets
should not be considered to be effective in
transferring tensile diaphragm forces between
plywood sheets. This is not recommended as
these nails are in many instances already highly
stressed resisting in-plane diaphragm shear
forces and may have little reserve capacity to
resist tensile diaphragm loading also. In addition
these nails would typically be provided with small
edge distances leaving them prone to pull-through
should they be subject to tensile diaphragm
loading perpendicular to the plywood sheet edge.
iii. Diaphragms shall be provided with continuous
ties or struts between diaphragm chords to
distribute wall out-of-plane anchorage forces into
the diaphragms. This is to ensure that the large
out-of-plane forces generated at diaphragm edges
have a reliable load path back into, and engage,
the main body of the diaphragm. This is analogous
to hanger reinforcing provided in reinforced
concrete beam design. Added chord elements
are permitted to be used to form subdiaphragms
to transmit the anchorage forces to continuous
diaphragm cross-ties. North American Building
Codes (i.e. ACSE 7-05, 2005) limit the maximum
length-to-width ratio of subdiaphragms to 2.5 to 1.
While current URM retrofit practice in New Zealand
typically avoids cross grain bending, there appears
to be less of an awareness regarding the use of nails
to transfer tensile diaphragm forces or the need for
continuous diaphragm cross ties and struts. This is
despite the requirement in AS/NZS 1170.0 (SNZ, 2004)
that floor and roof diaphragms shall be designed to
have ties or struts (where used) able to distribute the
required wall anchorage forces. Subdiaphragms are a
very effective way of providing the necessary continuous
diaphragm cross ties and struts.
The subdiaphragm methodology is a design method
whereby the main diaphragm is broken up into a
number of smaller (sub) diaphragms at the diaphragm
perimeter. The smaller subdiaphragms are designed to
resist the amplified out-of-plane wall anchorage forces
previously described in Section 4.5.1 and span between
diaphragm cross-ties which are typically provided at
approximately 45 m centres.
This aspect of timber diaphragm design is often not well
understood in New Zealand and is equally applicable
to the design of new structures with timber diaphragmsand heavy faade elements i.e. concrete masonry
walls, precast concrete and GRC faade panels etc. It
is important that Structural Engineers are familiar with
the subdiaphragm concept and know when it should
be applied. Subdiaphragms will be considered in more
detail in Section 4.6.2 below.
Figure 14. Out-Of-Plane Loading Cross
Grain Bending Failure Mechanism
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Referring to Figures 12 and 13, the tributary out-of-plane
inertia force generated within the walls are transferred
by the tension anchors into the steel brackets; and
then from the steel brackets into the timber joists/
blocking. When joists are orientated perpendicular
to the wall (Figure 13), the joists are able to act as
subdiaphragm ties transferring the out-of-plane forcesinto the subdiaphragm. As is illustrated in Figure 12,
when joists are orientated parallel to the wall, timber
blocking and sheet metal straps can be used to transfer
the out-of-plane forces into the subdiaphragm.
Eccentric loading of the timber joists/blocking by the
steel bracket should be addressed when checking the
adequacy of the existing floor framing. Additional timber
blocking can be used to resist the expected minor axis
bending.
The NZSEE guidelines provide default connectorstrengths that can be used to determine the size and
spacing of tension bolts (NZSEE, 2006). In most
instances insitu testing can be used to justify higher
design values, although consideration of wall cracking
due to out of plane response wall might be considered.
4.6.2 Subdiaphragm Design
Figure 15 illustrates the subdiaphragm concept. Consider
the building subject to an east-west earthquake. In this
instance it is recommended that continuous diaphragm
cross ties are provided between grid lines A and D.
One design solution would be to provide closely spaced(i.e. 500 mm crs), light-gauge sheet metal straps with
timber blocking to act as the diaphragm cross ties/struts
across the width of the diaphragm. However such a
design solution would likely be costly and intrusive to
implement as part of a retrofit.
An alternative design solution using the subdiaphragm
concept would be to utilise the existing beams to act as
the diaphragm cross tie/strut forces and then provide
smaller subdiaphragms that span horizontally between
the diaphragm cross ties/struts.
Consider the length of wall between grid lines D5 and
D6. As shown in Figure 15, a smaller subdiaphragm
could be designed to span between grid lines 5 and 6.
In this instance, when the timber joists are orientated
parallel to the wall, the existing joists are often found
to be adequate to act as the subdiaphragm chords.
The depth of the subdiaphragm is typically increased
until the chord forces are sufficiently reduced that the
existing joists are adequate. Alternatively the existing
joists can be doubled up to increase their capacity. The
sheet metal straps described in the previous section
are used as subdiaphragm cross ties to transfer the
amplified out-of-plane wall anchorage forces (refer
Figure 12) to the rear of the subdiaphragm.
Once the subdiaphragm design is complete, checks
should be made to ensure that the existing beams that
are utilised to act as diaphragm cross diaphragm ties/
struts have adequate axial load capacity, and that their
end connections are sufficient to transfer the necessary
diaphragm cross tie forces.
Referring to Figure 15 a similar strategy can be used for
the north-south earthquake. In this direction the existing
timber joists can be used as subdiaphragm cross-ties
(refer Figure 13). Note that in this direction no beams
are available to act as diaphragm cross ties/struts. As
an alternative, existing floor joists can be doubled up
and made continuous to act as subdiaphragm cross
ties/struts. Existing beams can often be utilised to act
as the subdiaphragm chords.
When designing subdiaphragm elements and cross
diaphragm ties the out-of-plane wall loads calculatedin Section 4.5.1 should be used. Current design office
practice is that these amplified out-of-plane wall forces
are not considered to act concurrently with those
calculated in Section 4.3 for main diaphragm design.
4.6.3 In-Plane Seismic Response
In terms of in-plane loading response the wall
diaphragm ties transfer the tributary horizontal inertia
forces associated with face loaded walls, and the
diaphragms own seismic mass, into the lateral load
resisting in-plane end walls.
Referring to Figures 12 and 13 the inertia forces are
transferred out of the diaphragm via the boundary
joist into the resisting in-plane end walls using grouted
wall anchors. The NZSEE guidelines provide default
connector strengths that can be used to determine the
size and spacing of tension bolts (NZSEE, 2006). In
most instances insitu testing can be also used to justify
higher design values, although consideration of wall
cracking due to out of plane response wall might be
considered.
4.7 Diaphragm Penetrations
Penetrations in diaphragms due to stair openings,
elevators shafts and service risers require special
consideration to ensure that diaphragm performance
is not compromised. Diaphragm penetrations cause
stress concentrations which can lead to poor diaphragm
behaviour if the openings are not addressed in the
diaphragm design.
The shear transfer method (Smith et. al. 1986) is a
simple design method which can be used by Structural
Engineers to determine increased nailing and chord
requirements adjacent to diaphragm openings.
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Figure 15. Subdiaphragm Example
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When diaphragm penetrations occur immediately
adjacent to URM walls structural steel beams can be
used to provide the necessary out-of-plane restraint to
the walls (refer Figure 16). The beams should be tied
back into the diaphragm using subdiaphragm cross ties.
5.0 CONCLUSIONS
URM buildings have historically performed poorly in
large earthquakes. It has been found that many failures
of URM buildings are related to poor performance of the
diaphragms and the wall to diaphragm connections.
Analysis of URM structures with timber floors assuming
rigid diaphragm behaviour may lead to unconservative
assessments of diaphragm accelerations and
inaccurate estimates of load distribution between lateral
load resisting elements.
Many URM structures can be assessed and strengthened
when necessary using a simplified analysis method
contained within the NZSEE guide lines and ASCE/SEI
41-06 as illustrated in this paper.
Figure 16. Typical Out-Of-Plane Wall Support
Detail at Diaphragm Openings
Care is required when detailing the wall to diaphragm
connections to ensure that:
Cross grain bending of boundary joists is avoided.
Eccentric loading of floor framing elements by
out-of-plane wall brackets is considered in thediaphragm design.
Plywood sheet edge nailing is not required to
transfer tensile diaphragm forces.
Diaphragm cross-ties are provided to transfer the
large out-of-plane forces generated at diaphragm
edges back into the main body of the diaphragm.
Stress concentrations due to diaphragm openings
are addressed and the need for supplementary
wall support considered when diaphragm openingsoccur immediately adjacent to URM walls.
Further research into the seismic response of URM
buildings with flexible diaphragms is required to confirm
the following:
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The structural ductility capacity of existing timber
diaphragms.
Acceptable diaphragm deformation limits and the
interaction between the in-plane deformation of
diaphragms and the out-of-plane response of the
supported face loaded walls.
Design loads for the out-of-plane response of
URM walls can be reduced from those currently
specified in the NZSEE recommendations?
Specifically can a capacity design approach be
adopted whereby out-of-plane wall forces are
limited by yielding of the supporting diaphragm i.e.
confirmation of which of the four methods detailed
in Section 4.5.1 is most appropriate for use in New
Zealand?
Many of the principles of subdiaphragm design detailed inthis paper for existing URM buildings are also applicable
to new structures with flexible timber diaphragms and
heavy faade elements. It is important that Structural
Engineers are familiar with the subdiaphragm concept
and know when it should be applied.
6.0 REFERENCES
1. ASCE, ASCE/SEI 41-06 Seismic Rehabilitation
of Existing Buildings, ASCE, Restonm Virgina,
2006.
2. Blaikie, E.L., Methodology For The Assessmentof Face Loaded Unreinforced Masonry Walls
Under Seismic Loading, EQC Funded Research
by Opus International Consultants, Project 99/422,
Wellington, New Zealand, 2001.
3. Bruneau, M.,State-Of-The Art Report On Seismic
Performance of Unreinforced Masonary Buildings,
Journal of Structural Engineering, 120(1),1994.
4. Buchanan, A., Timber Design Guide, University
of Canterbury, New Zealand, 2007.
5. ICBO, Guidelines For Seismic Evaluation and
Rehabilitation of Tilt-Up Buildings and Other Rigid
Wall/Flexible Diaphragm Structures International
Conference of Building Officials, 2000.
6. ICC, 2006 International Existing Building Code,
International Code Council, County Club Hills, IL,
2007.
7. Ingham, J.M., The Influence of Earthquakes on
New Zealand Masonry Construction Practice,
14IBMAC Conference, Bondi, Australia, 2008.
8. FEMA, FEMA 306 Evaluation of Earthquake
Damaged Concrete & Masonry Buildings, Applied
Technology Council, Washington D.C., 1998.
9. FEMA, FEMA 356 Prestandard And
Commentary For The Seismic Rehabilitation of
Buildings , ASCE, Washington D.C., 2000.
10. NZSEE, Assessment and Improvement of
the Structural Performance of Buildings in
Earthquakes, NZSEE, Wellington, New Zealand,2006.
11. Paulay, T. & Priestley, M.J.N., Seismic Design
of Reinforced Concrete and Maonry Structures,
John Wiley & Sons Inc, New York, 1992.
12. Russell, A.P. and Ingham, J.M., Trends in the
Architectural Characterisation of Unreinforced
Masonry in New Zealand, 14IBMAC Conference,
Bondi, Australia, 2008.
13. Smith, P.C., Dowrick, D.J. & Dean, J.A., Horizontal
Timber Diaphragms For Wind And EarthquakeResistance, Bulletin of the New Zealand Society
For Earthquake Engineering, Vol. 19, No. 2, 1986.
14. SNZ, NZS 3603:1993 Timber Structures
Standard, Standards New Zealand, Wellington,
New Zealand, 1993.
15. SNZ, ASNZS 1170.5:2004 Structural Design
Actions Part 5 : Earthquake Actions New
Zealand, Standards New Zealand, Wellington,
New Zealand, 2004.
16. ASCE, ASCE 7-05 Minimum Deesign Loads for
Buildings and Other Structures, American Society
of Civil Engineers, Reston, Virginia, USA, 2006.
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APPENDIX A
FLEXIBLE DIAPHRAGM DESIGN EXAMPLE
A.1 Introduction
This design example will detail the assessment and retrofit of the first floor of the five
storey high building previously illustrated in Figures 2 & 15. The building is rectangular
in plan and is approximately 30.4 m long, 18.6 m wide and 22.1 m high.
Unreinforced masonry has been used to construct the perimeter walls. The floors consist
of 20 mm thick straight timber sheathing supported on 300 x 60 mm joists spanning in the
north-south direction. Structural steel beams spanning between cast-iron columns support
the timber joists.
The building is located in Wellington and is founded on Class D subsoil. The intention is
that the building will be seismically strengthened to 67% of that which would be requiredfor a new building at the site.
It is assumed for this example that the material strengths for the existing timber elements
are equal to or greater than that for Radiata Pine No. 1 Framing Grade. The influence of
the floor penetrations on diaphragm behavior has been ignored and the C 3 P-modification factor was assumed to be 1.0 for this design example.
A.2 Building Information
Seismic weights for the first floor were calculated as follows:
Diaphragm self weight = 1098 kN i.e. the self weight of the flooring, joists,
beams, columns, superimposed dead loads
and reduced live load.
Grid Line 1 Wall = 675kN
Grid Line 7 Wall = 673 kN
Grid Line A Wall = 1431 kN
Grid Line D Wall = 1256 kN
The thickness of the walls between Ground Floor and Level 1, and Level 1 and Level 2 is
530 mm. The height of the first floor is 7.4 m above the building base.
A.3 Diaphragm Period Calculation
North South Direction
D1 3.07T Eqn. 1
Need to determine Di.e.
D
u
D
K
V Eqn. 3
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Where:
kN2446g16736751098V u i.e. self weight of the diaphragm and theeast west walls.
L
bG
K
d
D
4
Eqn. 2
m/kN24063509.294
17.4
KD i.e. Gd= 350 kN/m (refer Table 1)
Hence:
m02.12406
2446D
Therefore:
s77.102. 13.07T1
East West Direction
Similarly it can be shown that:
kN3785g1125614311098V u
m/kN8383509.174
29.9
KD
m51.4838
3785D
s72.3T1
A.4 Diaphragm Seismic Load Calculation
North South Direction
VD= C1C3Cd(T1)WD Eqn. 4
Where:
k
STC P1d
CT1 AS/NZS 1170.5 Eqn 5.2(1)
And:
g51.006.10.14.021.1)D,T(ZRNCT h1 TC
7.0pS i.e. assuming = 2.0 capacity
0.4k
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Hence:
18.07.051.0
T1
2.0
Cd
Hence 67% design seismic loading for the north south direction is:
DD W12.0W18.00.10.167.0V d
Hence:
kN295244612.0V d
East West Direction
Similarly it can be shown that:
g28.053.10.14.046.0T1 C
7.0pS
0.2k
10.07.028.0
T1
2.0
Cd
kN253378510.00.10.167.0V d
A.5 Diaphragm Deformation Assessment
Need to ensure that diaphragm deflections do not exceed 150 mm or half the thickness of
the walls (i.e. 530/2 = 265 mm).
North South Direction
m245.02406
2950.2
D
DD
K
V
East West Direction
m604.0838
2530.2
D
DD
K
V
In this instance the diaphragms did not have adequate stiffness and strengthening of the
diaphragm is required.
Consider replacing the straight sheathing with a new F8 Grade 19 mm plywood diaphragm
with 75x3.3 mm edge nailing at 75 mm crs. It is acknowledged that this is a significant
amount of nailing however, as will be seen below in Section A12.2, the shear capacity of
the diaphragm is governed by the subdiaphragm design.
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A.6 Revised Diaphragm Period Calculation
North South Direction
D1 3.07T Eqn. 1
Need to calculate the secant yield stiffness, KD, of the plywood diaphragm.
D
YD
VK
Where: Vy = Diaphragm yield strength (kN)
D = Diaphragm yield displacement (m)
Determine the shear capacity, Qn,of the proposed diaphragm nailing. From NZS 3603
(SNZ, 1993):
kcmd QkknkQ n NZS 3603 Eqn. 4.2
Where: n = Number of nails = 13.3 nails / m
kd = Duration of loading factor = 1.0
km = Multiple nail factor = 1.3
kc = Plywood sheathing with flat head nails factor = 1.4
Qk = Characteristic nail strength from NZS 3603 Table 4.3 = 695 N
Hence:
m/N168706954.13.10.13.13Q n
And;
kN1009168709.292bQ2V n y
From NZS 3603 (SNZ, 1993) the diaphragm yield displacement can be calculated:
321 D NZS 3603 Eqn. 5.2.2
Where: 1 = Flexural deflection of diaphragm (m).2 = Shear deformation of diaphragm (m).3 = Deflection of diaphragm due to nail slip (m).
In this example the flexural diaphragm deflection (1) will be ignored as it is typicallysmall compared with shear and nail slip deformations. This could be confirmed later
during the design once the diaphragm design is complete.
Gbt8
WL2 NZS 3603 Eqn. 5.2.5
Where: W = 1009 kN
L = 17.4 m
G = 455 MPa
b = 29.9 m
T = 19.2 mm
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Hence:
mm4.82.19109.294558
104.171010093
33
2
And;
2
mea1 n3 NZS 3603 Eqn. 5.2.6
Where: a = 2.0
m = 8
en = 1.4 mm
Hence:
mm8.16
2
4.1821
3
And;
mm258.164.80 D
Therefore:
m/kN40360025.0
1009
KD
Hence the elastic deflection of the plywood sheathed diaphragm and a 1g lateral load can
be calculated as:
m061.040360
2446
D Eqn. 3
Therefore:
s43.0 0.0613.07T1
East West Direction
Similarly it can be shown that:
kN604168709.172V y
mm9.142.19104.174558
109.29106043
33
2
mm3.26
2
4.1255.01
3
mm413.269.140 D
m/kN14730041.0
604
KD
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m257.014730
3785D Eqn. 3
s89.0 0.2573.07T1
A.7 Revised Diaphragm Seismic Load Calculation
North South Direction
Using the revised diaphragm period calculated above:
g20.10.10.14.00.3)D,T(ZRNCT h1 TC
g30.07.020.1
T1
2.84
Cd
Note the above assumes the nailed plywood diaphragm will be designed to have a ductility
displacement capacity, , of 4.0. Hence the revised 67% design seismic loading for thenorth south direction is:
kN535246630.00.19.167.0V d
East West Direction
Similarly it can be shown that:
g84.00.10.14.011.2T1 C
15.07.084.0
T1
4.0
Cd
kN379378515.00.10.167.0V d
A.8 Revised Diaphragm Shear Strength Assessment
North South Direction
m/N89509.292
535
2b
VV DmaxE, Eqn. 8
m/kN13500168708.0Rn Refer Section A.6
Hence the proposed diaphragm nailing has adequate shear capacity. Determine the actual
ductility required, Act. Because T1 is less than 0.7 s consideration of AS/NZS 1170 theproportionality is not linear and consideration of k is required..
85.1
9.2950.132
24467.02.10.109.167.0
bR2
WSTCCC67.0k
n
p131
Act,
HenceAct can be calculated as 2.4.
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East West Direction
Similarly it can be shown that:
m/N108904.172
379
2b
VV DmaxE,
m/N13500Rn (refer above).
Again proposed diaphragm nailing is adequate. Because T1 is greater than 0.7s Actcansimply be calculated as:
2.34.175.132
3794
bR2
V
n
,Eact
A.9 Revised Diaphragm Deformation Assessment
North South Direction
mm3240360
5354.2Act
D
DD
K
V
East West Direction
mm8214730
3792.3Act
D
DD
K
V
The diaphragm deflections in both directions are less than the maximum permitted of150 mm. Hence the proposed diaphragm has adequate stiffness.
A.10 Diaphragm Chord Design
Diaphragm chords will be designed to remain elastic using capacity design.
North South Direction
Overstrength shear capacity of the diaphragm nailing, Qos, can be calculated as:
nosos,D QV Eqn. 13
Where: os = Nail overstrength factor taken as 1.6Qn = Characteristic shear capacity diaphragm nailing
= 16.87 kN/m (refer Section A.6)
Hence:
m/kN0.2787.166.1V os,D
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Hence the diaphragm overstrength factor in the north-south direction, os,NS, can becalculated as:
02.3535
9.290.272
V
bV2
D
os,D
NS,
os
Hence the maximum bending moment, Mos,max, generated in the diaphragm at overstrength
can be calculated as:
kNm439032
4.1753502.35L
32
V5M
Dos,NSmaxos, Eqn. 9
The maximum overstrength diaphragm chord forces can then be calculated as:
kN1479.29
4390M max,os b
P chordxos,
This design load should be used to design the diaphragm chord.
East West Direction
Similarly it can be shown that:
48.2379
4.170.272NS,
os
kNm439032
9.2937948.25 maxos,M
kN2524.17
4390chordxos,P
A.11 Calculation of Wall Diaphragm Connection Loads
A.11.1 In-Plane Seismic Response
The diaphragm overstrength capacity, VD,os, was calculated previously in Section A.11 as16.87 kN/m.
A.11.2 Out-Of-Plane Seismic Response
North South Direction
The same design will be used for grid line 1 and grid line 7 walls. From Section A.2 the
tributary self weight of the critical grid line 1 wall was 675 kN i.e. 37.7 kN/m.
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Using the AS/NZS 1170 part provisions, assuming a part structural ductility factor, p, of1.25, hiof 7.4 m, and a part risk factor, Rp, of 1.0 the out-of-plane wall loads, Fph, can be
calculated as 0.67 x 1.70 = 1.13 g.
Hence the out-of-plane wall load for the north-south seismic response, Fph,NS Wall can be
calculated as:
m/kN7.427.3713.1WF wall,iph ph,NSF
East West Direction
Similarly it can be shown that:
m/kN9.479.29
1431wall,i W
m/kN1.549.4713.1WF wall,iph EWph,F
Out of interest compare the AS/NZS 1170 parts loads with the overstrength diaphragm
acceleration in each direction i.e.
g66.02449
53502.3
W
VV
D
NS,D
NS,
osos
g25.0
3785
37948.2
W
VV
D
NS,D
EW,
os
os
Hence the AS/NZS 1170 out-of-plane parts load are 1.7 and 4.5 times greater than the
overstrength diaphragm acceleration and in the north south and east west directions
respectively. Note that as will be seen in Section A12.2 the diaphragm nailing was
governed by the AS/NZS 1170 parts loads. If the diaphragm was designed without
consideration of the AS/NZS 1170 parts loads it can be shown that the overstrength
diaphragm accelerations would be approximately 1/3 of that detailed above. This equates
to the AS/NZS 1170 out-of-plane parts load that would be 5 and 12 times greater than the
overstrength diaphragm accelerations and in the north south and east west directions
respectively. It seems improbable that this could occur. This demonstrates that further
research into this area is warranted.
A.12 Design of Wall Diaphragm Connections
For this example it will be assumed that no existing wall-diaphragm anchors exist and that
new anchors will be provided as part of the proposed retrofit.
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A.12.1 In-Plane Seismic Response
North South Direction
From Section A12.1 above VD, os = 16.87 kN/m. From Table 10.B2 of the NZSEE
guidelines the default shear capacity, Qn, of an M16 anchor bolt grouted at least 200 mminto a masonry wall can be taken as 9.0 kN.
Hence the max anchor spacing can be calculated as:
m53.087.16
0.90.1
V
Qs
os,D
,n
max
Note that a strength reduction factor, , of 1.0 was used in the above equation inaccordance with typical capacity design procedures. The existing joists are at 450 mm
hence the new wall anchors could be provided at each joist location i.e. 450 mm centres.
East West Direction
The same in-plane diaphragm load occurs in the east west direction because the nail
spacing is the same. Hence new wall anchors at 500 mm centres will be adequate in this
direction also.
A.12.2 Out-Of-Plane Seismic Response
North South Direction
Determine Wall Anchor Spacing:
From Section A11.1 above Fph, NS = 42.7 kN/m. From Table 10.B2 of the NZSEE
guidelines the default tension capacity, Pn, of an M16 anchor bolt grouted 50 mm less than
the thickness of the wall can be taken as 11.0 kN.
Hence the max anchor spacing can be calculated as:
m18.07.42
0.117.0
F
Ps
NSWall,ph
nmax
This is not a practicable anchor spacing. Consider the required anchor tension capacity if
an anchor is provided at each joist location.
kN4.277.0
7.4245.0FsP
NSWall,phActn
Given that the anchor is to be embedded in a 470 mm thick masonry this required capacity
may be achievable although it will need to be confirmed by insitu anchor testing.
Referring to Figure 15 consider the design of the subdiaphragm. In this example it will
provide 1 bay deep subdiaphragms i.e. between gridlines 1 and 2 to support the grid line 1
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46 Journal of the Structural Engineering Society New Zealand Inc.
wall, and between 6 and 7 to support the grid line 7 wall. Diaphragm cross-ties will be
provided on grid lines A, B, C and D.
Check Subdiaphragm Shear:
m/kN6.1352
9.56.22b2
LFVsd
sdNS,phmax,E
Note that in the above equation the AS/NZS 1170 parts loads used was that calculated
assuming a part structural ductility factor, p, of 3.0 ( i.e. 0.67 x 0.90 g) because in thisinstance the required diaphragm nailing has been determined. From Section A.8
Rn= 13.5 kN/m hence the subdiaphragm has adequate shear strength.
Check Subdiaphragm Chords:
kN9.379.48
9.57.42
b8
LF
P
2
sd
2NS,ph
max,E
sd
The capacity of the existing beams along grid lines 2 and 6 will need to be checked to
ensure that they have adequate capacity to resist this axial load. Also the necessary
detailing will have to be provided to ensure the beams are adequately connected to the
subdiaphragm i.e. provide new ribbon plates if necessary.
Referring to section A.10 the loads used to design the primary diaphragm chords on grid
lines 1 and 7 were greater than the subdiaphragm loads calculated above and as such will
be adequate to act as subdiaphragm chords.
Check Subdiaphragm Cross-Ties:
In this instance the existing 300x 60 joists will act as the subdiaphragm cross ties. Using
NZS 3603 it can be shown that the existing timber joists have adequate capacity to resist
the additional out-of-plane wall axial loads. A steel bracket bolted to the existing joist can
be detailed to transfer the load from the wall anchor into the existing joist.
Adequate nailing is required to transfer the subdiaphragm cross-tie loads into the new
plywood diaphragm. The design load can be calculated as:
m/kN92.39.4
7.4245.0
b
FsVsd
NS,phActE
Hence 75 x 3.33 nails at 200 centres would be adequate i.e. Qn= 827 N/nail).
Design Diaphragm Cross-Ties:
The diaphragm cross-tie force can be calculated as:
1i,sdi,sdNS,phmax,E bbF5.0P
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Volume 23 No. 1 April 2010 47
Where bds,i and bds,i+1 are the adjacent subdiaphragm spans. Hence on grid line B the
diaphragm cross-tie force is:
kN248)9.572.5(7.425.0bbF5.0P 1i,sdi,sdNS,phmax,E
It can be shown that this load can be resisted by doubling up an existing joist line adjacent
grid lineB. Nail plates could be detailed to provide a continuous load path between grid
lines 1 & 7. Similarly the grid line A diaphragm cross-tie force can be calculated as:
kN122)072.5(7.425.0bbF5.0P 1i,sdi,sdNS,phmax,E
Note that this load is less than the diaphragm chord force of 252 kN calculated previously
in Section A.10 for the east-west diaphragm response. Hence the capacity of the
diaphragm chord is adequate. The design grid line C & D cross ties will be similar to that
already described above.
East West Direction
Determine Wall Anchor Spacing:
From Section A11.1 above Fph, EW= 54.1 kN/m. Referring to Figure 12 in this direction
the spacing of the wall anchors is governed by the capacity of the light gauge metal strap
subdiaphragm cross-ties. Consider a proprietary metal strap with an axial load capacity,
Rn, of 14.8 kN. Hence the max anchor spacing can be calculated as:
m274.01.54
8.14
F
Rs
NSWall,ph
nmax
Hence provide wall anchors and subdiaphragm cross-ties at 275 mm centres. It is
acknowledged that this is a very close spacing. The anchor spacing could be increased by
using stronger, custom light gauge metal straps or alternatively using lower out-of-plane
design loads. The latter may be possible in the future pending improved design
methodologies.
Note that insitu anchor testing will still be required to confirm the required anchor
capacity.
Consider the design of the subdiaphragm. In this example it will provide 1 bay deepsubdiaphragm i.e. between gridlines A and B to support the grid line A wall, and between
C and D to support the grid line D wall. Diaphragm cross-ties will be provided on grid
lines 1, 2, 3, 4, 5, 6 and 7.
Check Subdiaphragm Shear:
m/kN3.127.52
9.46.28
b2
LFV
sd
sdNS,phmax,E
From Section A.8 Rn= 13.5 kN/m hence the subdiaphragm has adequate shear strength.
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Check Subdiaphragm Chords:
kN0.287.58
9.41.54
b8
LFP
2
sd
2NS,ph
max,Esd
It can be shown that the existing joists have adequate capacity to resist this load.
Detail Subdiaphragm Cross-Ties:
Referring to Figure A.1 below care is required to ensure that the 95 mm vertical
eccentricity between the anchor bracket and the light gauge metal strap subdiaphragm
cross-ties is addressed in the design. Two rows of joists have been blocked-out in order to
keep the stabilising vertical shears, V1, to a manageable level.
Figure A.1 Subdiaphragm Tie Force Distribution
Knowing that the existing 300 x 60 joists are at 450 mm centres the stabilising vertical
shear force, V1, to be transferred through the new timber blocking can be calculated as:
kN56.14502
958.14
s2
eFV
joist
ph
1
Skew nails could be used to secure the timber blocking to the existing joists. Referring to
Figure A.1 the capacity of the right most joist to resist this additional seismic load (i.e.
1.56/0.275 = 5.68 kN/m) would need to be verified.
First consider force F1, the load to be transferred
kN41.4
30019
958.14
dt
eFF
joistply
ph
1
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Too much food?
For those of you who watch what you eat, here's the final word on nutrition and health. It's a reliefto know the truth after all those conflicting nutritional studies.
1. The Japanese eat very little fat and suffer fewer heart attacks than us.
2. The Mexicans eat a lot of fat and suffer fewer heart attacks than us.
3. The Chinese drink very little red wine and suffer fewer heart attacks than us.
4. The Italians drink a lot of red wine and suffer fewer heart attacks than us.
5. The Germans drink a lot of beer and eat lots of sausages and fats and suffer fewer heartattacks than us.
CONCLUSION:
Eat and drink what you like.Speaking English is apparently what kills you.
Hence a standard proprietary 6 kN light metal gauge metal brace strap would be
adequate. Force F2 can then be calculated as:
kN4.1041.48.14FFF 1ph2
The proposed proprietary light gauge metal strap is pre-punched with 2 rows of 3.15diameter holes at 32 mm crs. It can be shown that 2 rows of 3.15 x 75 nails at 32 mm crs
are adequate. Wood screws can be used over the balance of the sub diaphragm tie length
to transfer the out-of-plane wall loads into the subdiaphragm. The capacity of the wood
screws, Rn,would need to be greater than:
m/kN60.27.5
8.14
b
FR
sd
ph
n
Design Diaphragm Cross-Ties:
Typical the diaphragm cross-tie force is:
kN265)9.49.4(1.545.0bbF5.0P 1i,sdi,sdNS,phmax,E
It can be shown that this load can be resisted by the existing steel beams. The capacity of
steel beam end connections should checked to ensure that a continuous cross-tie has been
provided across the width of the diaphragm. Similarly the grid line 1 and 7 diaphragm
cross-tie force can be calculated as:
kN132)092.4(1.545.0bbF5.0P 1i,sdi,sdNS,phmax,E
This load is less than the diaphragm chord force of 151 kN calculated previously in
Section A.10 for the north-south diaphragm response. Hence the diaphragm chord
designed previously will also be adequate to act as the grid line 1 & 7 diaphragm cross-tie.