Assessment Retrofit Flexible Diaphragms

8/12/2019 Assessment Retrofit Flexible Diaphragms

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Volume 23 No. 1 April 2010 19

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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20 Journal of the Structural Engineering Society New Zealand Inc.

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


Single Diagonal Sheathing Chorded L/b 3.0 2.0


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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SESOC Journal


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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SESOC Journal


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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SESOC Journal


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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SESOC Journal


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


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


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).


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


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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SESOC Journal


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


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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SESOC Journal


m257.014730

3785D Eqn. 3

s89.0 0.2573.07T1

A.7 Revised Diaphragm Seismic Load Calculation


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


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


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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SESOC Journal


East West Direction


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


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.


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


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


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


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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SESOC Journal


A.12.1 In-Plane Seismic Response


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


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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SESOC Journal


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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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.

Assessment Retrofit Flexible Diaphragms

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