2.2 structural calculations auckland (avalon block)

Project: Nationwide FLS Upgrades

Avalon Block

Structural Calculations: Auckland

Reference: 246313

Prepared for: Ministry of Education

Revision: 0

20/10/2015

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Project 246313 File Avalon Block - FLS Upgrade Calculations Auckland.docx 20/10/2015 Revision 3

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Report Title Nationwide FLS Upgrades - Avalon Block Structural Calculations: Auckland

Document ID Avalon Block FLS Upgrade Calculations Auckland.docx

Project Number 232603

File Path P:\246313\03 Project Delivery\Design\Avalon Block\Avalon Block FLS Upgrade Calculations Auckland.docx

Client Ministry of Education Client Contact

Rev Date Revision Details/Status Prepared by Author Verifier Approver

0 20 October 2015 Issue to Client PGD PGD CPH JFF

Current Revision 0

Approval

Author Signature Approver Signature

Name Phil Don Name John Finnegan

CPEng Number - CPEng Number 145791

Title Structural Engineer Title Technical Director

FLS Upgrade – Avalon Block Auckland Region - Calculations

SUMMARY 2

1. BUILDING OVERVIEW 3

1.1 Original Structure 3

1.2 FLS Upgrade – Option 1 5

1.3 FLS Upgrade – Option 2 6

2. LATERAL LOADING DEMANDS 7

2.1 Seismic Loading 7

2.2 Wind Loading 17

2.3 Design Loading For Lateral Bracing 22

3. LATERAL BRACING DESIGN 23

3.1 Option 1 23

3.2 Option 2 30

4. ROOF LINTEL DESIGN 34

4.1 Option 1 34

4.2 Option 2 44

5. ROOF DIAPHRAGM CAPACITY CHECK 58

6. FLOOR DIAPHRAGM CAPACITY CHECK 60

7. FOUNDATION CAPACITY CHECK 63

Client: Ministry of Education Date: 18/08/2015

Project/Job: FLS Upgrade Job No: 246313

Subject: Avalon Block - Auckland Sheet No: 2 By: PGD

Summary The Ministry of Education (MOE) wishes to provide schools with standard options to modify their classrooms to allow for a Flexible Learning Space (FLS).

This document provides detailed calculations for the standard options to create a FLS for an Avalon Block. These details are intended to be generic and have been developed with the following site characteristics:

Importance Level 2

Site Subsoil Class D

Location Auckland

This document provides structural calculations for two FLS options set out by the MOE. These are Option 1 and Option 2. The main variances between the two options lie in the footprint the smaller breakout areas have relative to the general flexible learning areas. The FLS options are based around 4 classroom blocks that will be expanded into a larger learning environment that can be altered as required to suit the class type.

For these strengthening schemes, the lateral bracing elements are to have a minimum capacity of 67%NBS.




1. Building Overview

1.1 Original Structure Avalon Blocks were constructed from 1955 to 1962 and feature lightweight timber framing with a lightweight roof. Classrooms feature a suspended timber floor on concrete piles and concrete ring foundations. Cloakrooms and toilets run along the rear of the classrooms and are founded on a concrete slab on grade. Avalon Blocks are characterised by large windows to the front, a mono pitched roof and clerestory windows above a timber beam near the rear of the classrooms. Originally large skylights were present in each classroom. A front view of a typical 3 Classroom Avalon Block is shown in the figure below.

Front View of Typical Avalon Block

The figure below shows a rear view of the clerestory windows and the approximate location of the rear classroom wall. Transverse walls are interrupted by a single door between classrooms.

Rear View of Clerestory Windows in a Typical Avalon Block

Clerestory Windows

Approximate location of rear classroom wall.




12.2m

7.4m

9.7m

2.5m

2.3m

2.9m

1.5m

2.4m Windows

Windows

7.4m

Avalon Block – Plan View Showing Wall Layout

Avalon Block – Side Elevation

Note: The plan shows a typical Avalon classroom. Avalon type classrooms ‘blocks’ consist of classrooms shown in the plan stacked side by side to create a larger Avalon block. The number of classrooms that make up a block varies throughout the country.

The FLS upgrades are based on an Avalon block of 4 classrooms




1.2 FLS Upgrade – Option 1 Option 1 for the FLS upgrade consists of opening up the four classrooms to become one large flexible learning space. At the rear of the classrooms, alterations are made with the existing layout, creating small breakout spaces as well as additional space for bag storage.

To allow for the FLS upgrade, as well as to strengthen the lateral bracing capacity of the structure to meet a minimum capacity of 67%NBS, timber lintels are to be installed across the openings in the classroom walls and new Gib bracing walls are to be installed throughout the structure.

The figure below shows a plan layout with the preliminary structural scheme for the new FLS layout.

FLS Layout and Structural Scheme Option 1




1.3 FLS Upgrade – Option 2 Option 2 for the FLS upgrade varies from Option 1 through creating larger openings and larger small breakout areas. The structural system for this option will require the installation of steel lintels in areas rather than timber lintels (due to the larger spans), as well as the installation of Gib bracing walls similar to Option 1.

The figure below shows a plan layout with the preliminary structural scheme for the new FLS layout.

FLS Layout and Structural Scheme Option 2




2. Lateral Loading Demands

2.1 Seismic Loading

2.1.1 General Parameters

The seismic weight will be calculated for a single classroom and then linearly scaled to provide the approximate seismic weight of four classrooms.

The lateral forces that must be resisted by the elements in a building are a function of the seismic weight and the horizontal acceleration as estimated by the horizontal design action coefficient.

Earthquake actions are calculated as per the equivalent static method, which applies the force due to the weight of the roof and upper half of the walls at roof level and the force due to the total weight of the building at foundation level.

From NZS1170.0, section 4.2.1, the ultimate limit state load combination that relates to earthquake actions is as follows:

𝐸𝐸𝑑𝑑,𝑑𝑑𝑑𝑑𝑑𝑑 = 𝐺𝐺 + 𝐸𝐸𝑢𝑢 + 𝛹𝛹𝑐𝑐𝑄𝑄

𝐺𝐺 = 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐴𝐴𝐴𝐴𝑃𝑃𝐴𝐴𝐴𝐴𝑃𝑃𝐴𝐴

𝐸𝐸𝑢𝑢 = 𝐸𝐸𝑃𝑃𝑃𝑃𝑃𝑃ℎ𝑞𝑞𝑞𝑞𝑃𝑃𝑞𝑞𝑃𝑃 𝐴𝐴𝐴𝐴𝑃𝑃𝐴𝐴𝐴𝐴𝑃𝑃𝐴𝐴

𝑄𝑄 = 𝐼𝐼𝑃𝑃𝐼𝐼𝐴𝐴𝐴𝐴𝑃𝑃𝐼𝐼 𝐴𝐴𝐴𝐴𝑃𝑃𝐴𝐴𝐴𝐴𝑃𝑃𝐴𝐴

𝛹𝛹𝑐𝑐 = 𝐸𝐸𝑃𝑃𝑃𝑃𝑃𝑃ℎ𝑞𝑞𝑞𝑞𝑃𝑃𝑞𝑞𝑃𝑃 𝐶𝐶𝐴𝐴𝑃𝑃𝐶𝐶𝐴𝐴𝑃𝑃𝑃𝑃𝑃𝑃𝐴𝐴𝐴𝐴𝑃𝑃 𝐹𝐹𝑃𝑃𝐴𝐴𝑃𝑃𝐴𝐴𝑃𝑃

Permanent Actions:

Permanent actions include the self-weight of all building elements. These are estimated based on the materials present in each element.

Roofs and Floors – Roofs and floors will be assumed to have a weight of 0.4 kPa.

Timber Walls – All timber walls and windows will be assumed to have a weight of 0.3 kPa

Imposed Actions:

Live Floor Loads – Live loads on classroom floors will be taken as 3.0 kPa as per NZS1170.1 Table 3.1.

Live Roof Loads – Live loads on non-accessible roofs will be taken as 0.25 kPa as per NZS1170.1 Table 3.1. This will be ignored in the seismic weight calculations (negligible) but will be included in the lintel design.




2.1.2 Seismic Weight of Classroom Section

Seismic weight will be calculated for the tributary areas of the longitudinal bracing lines. The sum of these areas will be the total seismic weight of the building. Section dimensions are shown below.

Typical Unit Weights

Wall Weight WW = 0.3 kPa Typical for timber framed buildings

Roof Weight RW = 0.4 kPa Typical for timber framed buildings

Floor Weight FW = 0.4 kPa Typical for timber framed buildings

Floor Live Load Q = 3.0 kPa NZS1170.1, Table 3.1

Combination Factor ΨE = 0.3 NZS1170.5, Section 4.2

Longitudinal

Tran

sver

se

Section 1

Section 2

1

2

3

5 6

9 10

Bracing Line 1

Bracing Line 2

4.85m

2.3m

1.25m

7.4m

Section 3

4

Bracing Line 3

7 8 2.55m

1.25m




- Longitudinal Tributary Weight Calculations

Section 1

Wall Length Height Weight 2 7.4 2.4 0.3 × 7.4 × 2.4 = 5.33 kN 3 7.4 3.0 0.3 × 7.4 × 3.0 = 6.66 kN 5 2.3 2.4 0.3 × 2.3 × 2.4 = 1.66 kN 6 2.3 2.4 0.3 × 2.3 × 2.4 = 1.66 kN 7 2.55 3.65 0.3 × 2.55 × 3.65 = 2.79 kN 8 2.55 3.65 0.3 × 2.55 × 3.65 = 2.79 kN

Total WallSec1 = 20.89 kN

Plan Area AreaSec1 = 7.4 m × (2.3m + 2.55 m + 1.25 m)

AreaSec1 = 45.14 m2

Roof Weight RoofSec1 = 0.4 × AreaSec1

RoofSec1 = 18.1 kN

Section 2

Wall Length Height Weight 4 7.4 2.9 0.3 × 7.4 × 2.9 = 6.44 kN 9 4.85 3.25 0.3 × 4.85 × 3.25 = 4.73 kN 10 4.85 3.25 0.3 × 4.85 × 3.25 = 4.73 kN


Plan Area AreaSec2 = 7.4 m × 4.85 m

AreaSec2 = 35.89 m2


RoofSec2 = 14.36 kN

Section 3

Wall Length Height Weight 1 7.4 2.4 0.3 × 7.4 × 2.4 = 5.33 kN


Plan Area AreaSec3 = 7.4 m × 1.25 m

AreaSec3 = 9.25 m2


RoofSec3 = 3.7 kN




- Floor Live Loads

Plan Area AreaFloor = 9.7 m × 7.4 m

AreaFloor = 71.78 m2

Floor Weight GFloor = AreaFloor × FW

GFloor × FW = 71.78 × 0.4

GFloor = 28.71 kN

Area Reduction Factor Ψa = 0.3 + 3 / √(AreaFloor) (Section 3.4.2, NZS 1170.1)

Ψa = 0.65

Floor Live Load QFloor = Ψa × ΨE × Q × AreaFloor

QFloor = 0.65 ×0.3 × 3 × 71.78

QFloor = 41.99 kN

2.5 m Concrete Slab on Grade

9.7 m

7.4 m

Timber Floor on Concrete Piles




- Tributary Weight to Clerestorey Windows

The tributary seismic weight for the critical load path (Line 3) is calculated below. This load path is required to transfer roof loads into the lower ceiling diaphragm, which in turn transfers them to bracing line 1. Demands calculated here are local tributary demands only and have already been included in the global demand calculation.

Wall Weight (WW) 0.3 kPa Typical for timber framed buildings

Roof Weight (RW) 0.4 kPa Typical for timber framed buildings

- Clerestory Windows - Tributary Seismic Weight Calculations

Tributary Weight from Walls Wall Length Height Weight

1 2.55 3.65 0.3 × 2.55 × 3.65 / 2 = 1.40 kN 2 2.55 3.65 0.3 × 2.55 × 3.65 / 2 = 1.40 kN 3 7.4 1.5 0.3 × 7.4 × 1.5 = 3.33 kN

Total WallCW = 6.13 kN

Tributary Weight from Roof AreaCW = 7.4 m × 2.55 m

AreaCW = 18.87 m2

RoofCW = AreaCW × RW

AreaCW × RW = 18.87 × 0.4

RoofCW = 7.55 kN

Longitudinal

Transverse 1 2

Clerestory Windows

Trib Area of Clerestorey Windows

3 4.85m

4.85m

2.55m

2.3m




2.1.3 Horizontal Design Action Coefficients

Seismic Coefficient Variables

Importance Level: Importance Level 2 – Normal Structure

Design Working Life: 50 Years

Fundamental Period: Tn = 0.4s

Site Subsoil Class (Section 3.1.3): D – Deep or Soft Soil

Site Location (Table 3.3): Site Location 7 - Auckland

NZS1170.5 – Section 3.1: Elastic Site Spectra for Horizontal Loading:

C(T) = Ch(T) Z R N(T,D)

Where:

Ch(T) = spectral shape factor from Clause 3.1.2

Z = hazard factor from Clause 3.1.4

R = the return period factor from Clause 3.1.5

N(T,D) = near fault factor from Clause 3.1.6

This gives the following seismic coefficients for various levels of ductility.

Seismic Coefficient (ductility of 3.0 Sp 0.7) CdT3.0 = 0.1274

For the design of ductile strengthening elements (Gib Bracing Walls)

Seismic Coefficient (ductility of 2.5 Sp 0.5) CdT2.5 = 0.1050

For the assessment of existing elements

Seismic Coefficient (ductility of 1.25 Sp 0.925)CdT1.25= 0.3157

For the design and assessment of non-ductile elements




2.1.4 Demand on Wall Bracing Elements

Demand will be calculated as per the equivalent static method given in NZS1170.5. For buildings of this type it is overly conservative to add 8% of the total demand to roof level as per clause 6.2.1.3, so this will not be carried out.

Current building codes require that the lateral bracing elements are able to withstand forces generated at roof level. This includes the weight of the roof and the upper half of the walls.

𝑭𝑭𝒙𝒙 = 𝑪𝑪𝒅𝒅(𝑻𝑻)µ=𝒙𝒙(𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹 𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾 + 𝑼𝑼𝑼𝑼𝑼𝑼𝑾𝑾𝑼𝑼 𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾 𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾)

- Demand on Longitudinal Bracing Lines

The demand on each longitudinal wall bracing element will be calculated from its tributary seismic weight (Section 2.1) and the horizontal design action coefficient for a ductility of µ=2.5 (Sp=0.5) as per the Ministry of Education Guidelines. As mentioned above, multiply the weights calculated by 4 to calculate the demand for a 4-classroom Avalon Block.

Demand on Clerestory Windows

DemCW = CdT2.5 × (WallCW + RoofCW)

DemCW = CdT2.5 × (6.13+7.55) × 4

DemCW = 5.75 kN

Demand on Bracing Line 1

DemBL1 = CdT2.5 × (WallSec1 / 2 + RoofSec1)

DemBL1 = CdT2.5 × (20.89 / 2 + 18.1) × 4

DemBL1 = 11.99 kN



DemBL2 = CdT2.5 × (15.9/2 + 14.36)× 4

DemBL2 = 9.37 kN

Fx






DemBL3 = CdT2.5 × (5.33/2 + 3.70) × 4

DemBL3 = 2.67 kN

The total base shear in the longitudinal direction is therefore equal to:

F_EQLong = DemBL1 + DemBL2 + DemBL3

F_EQLong = 24.03kN

Note: The clerestorey demand is not included in this calculation as the weight in this region is included in Bracing Line 1.

- Transverse Demand

The total transverse demand will be the same as the total demand in the longitudinal direction.

F_EQTrans = F_EQLong

F_EQTrans = 24.03kN

2.2 Wind Loading Determine lateral wind loading demands (in accordance with NZS 1170.2) on the structure to confirm which lateral loading form governs.

2.2.1 Basic Wind Pressure Calculation

Wind Region = A6

Importance Level = 2

Terrain Category = 3

Shielding Multiplier (Ms) = 1.00

Topographical Multiplier(Mt) = 1.00

Vdes SLS = 30.7m/s

Vdes ULS = 37.4m/s

PBasic SLS = 0.57kPa

PBasic ULS = 0.84kPa




These wind parameters have been calculated for a terrain category 3 site with no topographical and shielding factors. These may need to be altered when a design is carried out for a specific Avalon Block with different site parameters.

The spreadsheet calculations to determine the basic wind pressures are shown below.




2.2.2 Design Wind Pressure Calculation

As per Cl 2.4.1 of NZS1170.2, the design wind pressure is calculated by the following equation:

Pdesign = Pbasic x Cfig x Cdyn


Cdyn = 1.0 (Low rise single storey structure with natural frequency above 1Hz)

Determine Cfig for different loading faces and directions:

Cp,i = NA (Internal pressures cancel each other out)

Windward Face:

Cp,e (W) = 0.7 (Table 5.2(A) - h<25m, wind speed doesn’t vary with height)

Leeward Face:

Longitudinal Loading

h = 3.4m (avg)

d = 29m

b = 12m

d/b = 2.4

Cp,e (L)Long = -0.28 (Table 5.2(B) Roof Pitch less than 10 degrees)

Transverse Loading

h = 3.4m (avg)

d = 12m

b = 29m

d/b = 0.4

Cp,e (L)Trans = -0.50 (Table 5.2(B))

See over the page for the Avalon Block dimensions.




Kc,e = 0.8 (Table 5.5 - 3 Effective Surfaces)

Ka Long = 0.9 (Table 5.4 – 25m2 Tributary Area - Conservative)

Ka Trans = 0.8 (Table 5.4 – >100m2 Tributary Area)

Kp = 1.0

Kl = 1.0 (Bracing elements not directly supporting the cladding)

12m

29m




Cfig,e = Cp,e x Ka x Kc,e x Kl x Kp

Cfig,e W = 0.7 x 0.9 x 0.8 x 1.0 x 1.0

= 0.50 (All directions)

Cfig,e L (Long) = -0.28 x 0.9 x 0.8 x 1.0 x1.0

= -0.20

Cfig,e L (Trans) = -0.50 x 0.8 x 0.8 x 1.0 x1.0

= -0.32

Hence, the net Cfig factors are as follows:

Cfig net (Long ) = 0.5+0.20 (Windward and leeward)

= 0.70

Cfig net (Trans) = 0.5+0.32 (Windward and leeward)

= 0.82

The ULS design pressures in both directions are therefore equal to the following:

PdesULS(Long) = 0.70 x 0.84kPa

= 0.59 kPa

PdesULS(Trans) = 0.82 x 0.84kPa

= 0.69 kPa

2.2.3 Wind Shear Forces

F_WindLong = 0.59kPa × 12m × 3.4m × 0.5

F_WindLong = 12.04kN

F_WindTrans = 0.27kPa × (29m × 2.9m × 0.5) + 0.42kPa×(29m × 3.9m × 0.5)

F_WindTrans = 35.10kN




2.3 Design Loading For Lateral Bracing Section 2.1 and 2.2 show the lateral design force calculations for Earthquake and Wind Loading respectively. Compare these demands to determine the critical force to design the lateral bracing to.

Since the majority of the lateral bracing elements in this system will be from new Gib bracing walls. For these elements, it would be more suitable to use a ductility of 3.0 and an Sp factor of 0.7. The seismic forces will therefore have to be scaled up as the seismic coefficient of ductility 3.0 is higher than for a ductility of 2.5 with an Sp factor of 0.5.

2.3.1 Longitudinal Loading

F_EQLongDuct3 = F_EQLong×(CdT3.0/CdT2.5)

F_EQLongDuct3 = 29.16 kN

F_WindLong = 12.04kN

Seismic loading governs in this direction. The wind and seismic demands are similar, therefore design bracing to 100%NBS and use the lower bound bracing capacity specified in the Gib manual for the various Gib bracing walls (capacities usually vary for wind and seismic loading).

2.3.2 Transverse Loading

F_EQTransDuct3 = F_EQTrans×(CdT3.0/CdT2.5)

F_EQTransDuct3 = 29.16kN

F_WindTrans = 35.10 kN

In this direction, wind loading governs. It is therefore to be ensured that the lateral bracing capacity exceeds both the full wind and seismic design forces in this direction.




3. Lateral Bracing Design

3.1 Option 1

3.1.1 Overview/Bracing Scheme

The preliminary bracing scheme is shown in the figure below.

Justify the preliminary structural scheme being adequate to resist lateral loading demand. The critical areas to check are as follows:

• Longitudinal Direction: o New bracing walls on Bracing Line 1-3 o Clerestorey cantilever studs

• Transverse Direction:

o Existing bracing walls on the exterior bracing lines o New Bracing walls on the interior bracing lines

Bracing Line 1

Bracing Line 2

Bracing Line 3

Clerestorey Line

7.3m




3.1.2 Transverse Direction

Internal Bracing Line

Distribute lateral loads based on tributary widths. All walls are spaced at approximately 7.3m centres, which means each of the internal wall lines have a tributary width equal to 25% of the entire building with (29m).

Hence, the internal transverse wall lines have the following demand:

F_TransIntWall = (1/4) × F_WindTrans (Wind Loading Governs)

F_TransIntWall = 8.78kN

A large opening is to be placed in the centre of the transverse walls. Strengthen the transverse wall lines up by re-lining with a suitable Gib bracing material.

*Note: Part of Wall 2 will be placed in the higher stud area to assist with load transfer between blocks. On some internal gridlines, Wall 2 and 3 will be one continuous wall. Treat as separate walls (worst case).

Wall 1 Design

Assume 80% of the demand is transferred to the higher stud area, and the remaining 20% is distributed to the lower stud area. This is conservative as the lower stud area has greater than 20% of the total building area.

The lateral loading demand is therefore equal to 7.02kN and 1.76kN in the two areas respectively.

Try lining Wall 1 with Gib GS2. Gib GS2 has a bracing capacity of 85Bu/m (4.25kN/m) (lower bound). Wall 1 is 2.7m long and has an average height of 3.1m.

Hence, the capacity of Wall 1 is equal to the following:

FnW1 = (2.4m/3.1m)×4.25kN/m×2.7m

FnW1 = 8.88kN

The capacity of this wall exceeds the demand, therefore there is no need to transfer seismic load across to the lower stud area.

8.78kN

2.7m 2.6m 2.4m

Wall 1 Wall 2* Wall 3*

7.02kN 1.76kN




Wall 2 and 3 Design

Line Wall 2 with Gib GS2 and Wall 3 with Gib GS1 (worst case). Gib GS2 has a capacity of 85Bu/m (4.25kN/m – worst case) and GS1 has a capacity of 60Bu/m (3kN/m – worst case). Take Wall 2 as having a length of 2.0m (ignore 600mm in the high stud area).

FnW23 = 4.25kN/m×2.0m+3.0kN/m×2.4m

FnW23 = 15.70kN

> 1.76kN OK

This is greater than the total demand of 1.76kN hence it is adequate to line these walls as Gib GS2.

External Bracing Line

The external bracing line has half the tributary width of an internal wall line, therefore:

F_TransExtWall = F_TransIntWall/2

F_TransExtWall = 4.39kN

Transverse end walls are shown to have diagonal timber cross bracing.

External Wall Bracing Capacity

Wall Length L = 7.4 m

Average Wall Height H = 3.4 m

Height Multiplier fhm = 1.0 MOE Guidelines V2 (2>H/L)

Bracing Capacity per metre Bu = 2.5 kN/m Diagonal Timber Cross Bracing

Capacity of External Transverse Wall CapTranE = L × Bu × fhm

CapTranE = 7.4 × 2.5 × 1.0

CapTranE = 18.5 kN

> F_TransExtWall OK

No further strengthening is therefore required on the external transverse bracing walls.

7.4m




3.1.3 Longitudinal Direction

Bracing Line 1 and 3

Utilise new Gib bracing walls on both bracing lines to carry the total demand of Bracing Line 1 and 3 combined.

On bracing Line 3, remove a glazing in selected areas and replace with GS1 timber framed walls. On bracing line 1, utilise a selection of the wall as Gib GS1 or GS2.

As was found in 2.3.1), seismic loading and wind loading are similar, therefore design bracing to resist 100%NBS.

F_BL13Duct3 = (DemBL1 + DemBL3) × (CdT3.0/CdT2.5)

F_BL13Duct3 = 17.79kN

Gib GS1 has a bracing capacity of 60Bu/m (3kN/m). The wall height along this bracing line is equal to 2.4m, therefore this capacity will not need to be reduced further.

The total length of wall required is equal to:

LwallreqdBL13Duct3 = F_BL13Duct3/3

LwallreqdBL13Duct3 = 5.93m

Bracing Line 3 has a total wall length of approximately 5m without requiring the removal of any additional windows. Bracing Line 1 has a total wall length of approximately 18.5m.

Both these bracing lines can therefore be lined with Gib standard (GS1).




Clerestorey Area

Internal window studs sit on a 16” by 5” beam as shown below. The only continuous members of this line are the 5” by 3” timber members on each side of the classrooms. Longitudinal loads will cause these continuous timber members to bend about their strong axis. The capacity of the window studs in bending will be reduced by a factor of Cd(T)µ=2.5 / Cd(T)µ=1.25 to reflect the brittle failure mode.

As the only bracing elements are in the transverse bracing walls, this will give 5 continuous studs between 4 classrooms.

Cantilever Timber Studs Parameters

Stud breadth b = 3in

b = 76.20mm Nominal Dimensions

Stud depth d = 5 in

d = 127.00mm Nominal Dimensions

Bending Strength fb = 22 MPa Estimated Bending Strength

Number of Studs n = 5 5 studs between 4 classrooms

k factors k1 = 1 Earthquake Load

k4 = 1.2 3 members

k5 = 1.0 no grid system

Earthquake Actions

Window Studs

16” by 5” Beam

Skew Nails

Continuous 5” by 3” member




k8 factor

Unrestrained Length L = 0.6m Distance between noggs

L/b = 7.87

k8 factor k8 = 1.0 (NZS3603:1993 Table 2.8)

Max Lateral Bracing Capacity

Lever arm L = 1.0 m

Section Modulus Zx = (b×d2)/6

Zx = 204838mm3

Moment Capacity Mn = k1 × k4 × k5 × k8 × fb × Zx

Mn = 5.41 kNm

Reduced Moment Capacity MnRed = (0.1050/0.3157) × Mn

MnRed = 1.80 kNm

Max Roof Force CapCW = n × MnRed / L

CapCW = 9.00kN

%NBS Calculation

%NBS NBSCW = CapCW / DemCW

NBSCW = 156.64 %

This is greater than the minimum bracing requirement of 67%NBS. No further strengthening is required in this area.




Bracing Line 2

Re-line the four existing walls with Ecoply EP1. These walls are approximately 0.9m in length and 2.9m high. EP1 has a bracing capacity of 105 Bu/m (5.25kN/m – Wind Loading – Worst Case).

Bracing Line 2 has the following demand at ductility 3.0:

F_BL2Duct3 = DemBL2 × (CdT3.0/CdT2.5)


Bracing Line 2 has the following capacity:

FnBL2 = 5.25kN/m×(2.4m/2.9m)×(0.9m×4)

FnBL2 = 15.64kN

FnBL2 > F_BL2Duct3 OK

There will also be some additional lateral bracing capacity due to cantilever action in the timber studs in the glazing areas.

Minimum Bracing Line Capacity Check

Clause 5.4.7 in NZS 3604:2011 states that no bracing line shall have a bracing capacity less than the greater of 100 BUs or 50% of the total bracing demand divided by the number of bracing lines. Check bracing lines 1-3 meet this criteria.

Total Demand FLong = F_BL2Duct3+F_BL13Duct3

FLong = 29.16kN

Minimum allowable capacity per bracing line

Capmin = (FLong/3)×0.5

Capmin = 4.86kN

This is less than 100Bus (5kN). Each bracing line is to have a minimum of 5kN.

Bracing line 1 and 3 will have a minimum capacity of 15kN each and Bracing Line 2 has a capacity of 18.6kN. These gridlines therefore meet the minimum bracing requirements.




3.2 Option 2

3.2.1 Overview/Bracing Scheme

The preliminary bracing scheme is shown in the figure below.

Justify the preliminary structural scheme being adequate to resist lateral loading demand. The critical areas to check are as follows:

• Longitudinal Direction: o New bracing walls on Bracing Line 1-3 o Clerestorey cantilever studs

• Transverse Direction:

o Existing bracing walls on the exterior bracing lines o New Bracing walls on the interior bracing lines

Bracing Line 1

Bracing Line 2

Bracing Line 3

Clerestorey Line

7.3m

1.8m




3.2.2 Transverse Direction

Internal Bracing Line

Distribute lateral loads based on tributary widths. All walls are spaced at approximately 7.3m centres, which means each of the internal wall lines have a tributary width equal to 25% of the entire building with (29m).

Hence, the internal transverse wall lines have the following demand:

F_TransIntWall = (1/4) × F_WindTrans (Wind Loading Governs)

F_TransIntWall = 8.78kN

A large opening is to be placed in the centre of the transverse walls. Strengthen the transverse wall lines up by re-lining with a suitable Gib bracing material.

*Note: Part of Wall 2 will be placed in the higher stud area to assist with load transfer between blocks. On some internal gridlines, Wall 2 and 3 will be one continuous wall, elevation shown is the worst case with an opening between these walls.

Wall 1 Design

Assume 80% of the demand is transferred to the higher stud area, and the remaining 20% is distributed to the lower stud area. This is conservative as the lower stud area has greater than 20% of the total building area.

The lateral loading demand is therefore equal to 7.02kN and 1.76kN in the two areas respectively.

Try lining Wall 1 with Gib GS2. Gib GS2 has a bracing capacity of 85Bu/m (4.25kN/m) (Lower Bound). Wall 1 is 1.8m long and has an average height of 3.1m.

Hence, the capacity of Wall 1 is equal to the following:

FnW1 = (2.4m/3.1m)×4.25kN/m×1.8m

FnW1 = 5.92kN

Load Transfer Design

Need to transfer the following force from the high stud area:

F_Transfer = 7.02kN-5.92kN

F_Transfer = 1.10kN

8.78kN

1.8m 2.2m 2.2m

Wall 1 Wall 2* Wall 3*

7.02kN 1.76kN




This will be transferred through the segment of Wall 2 that is present in the high stud area.

Transfer the load through more closely spaced screw fixings between the timber top chord of the wall and the wall cladding. Gib typically specifies a minimum of 6 gauge screws to fix around the sheet perimeters.

From NZS 3603:1993, 6 gauge screws have a characteristic strength of 0.854kN (Table 4.5). This correlates to a design strength of 0.60kN (Φ=0.7).

Based on this, a total of 2 screws will be required to carry the additional 1.1kN.

Wall 2 runs an allowable distance of 0.6m into the high level stud area. Space these screw fixings at 50mm centres across this 600mm length (approximately 12 screws provided).

Wall 2 and 3 Design

Line Wall 2 with Gib GS2 and Wall 3 with Gib GS1 (worst case). Gib GS2 has a capacity of 85Bu/m (4.25kN/m – worst case) and GS1 has a capacity of 60Bu/m (3kN/m – worst case). Take Wall 2 as having a length of 1.6m (ignore 600mm in the high stud area as this is being utilised for the transfer connection).

FnW23 = 4.25kN/m×1.6m+3.0kN/m×2.2m

FnW23 = 13.40kN

> 2.86kN OK

This is greater than the total demand of 2.86kN (1.1kN+1.76kN), hence it is adequate to line these walls as Gib GS2 and GS1.

External Bracing Line

As was found in 3.1.2, the existing timber framed wall with diagonal timber framing will be adequate.

3.2.3 Longitudinal Direction

Bracing Line 1 and 3

Utilise new Gib bracing walls on both bracing lines to carry the total demand of Bracing Line 1 and 3 combined.

On bracing Line 3, remove a glazing in selected areas and replace with GS1 timber framed walls. On bracing line 1, utilise a selection of the wall as Gib GS2.

As was found in 2.3.1), seismic loading and wind loading are similar, therefore design bracing to resist 100%NBS.

F_BL13Duct3 = (DemBL1 + DemBL3) × (CdT3.0/CdT2.5)


Gib GS1 has a bracing capacity of 60Bu/m (3kN/m). The wall height along this bracing line is equal to 2.4m, therefore this capacity will not need to be reduced further.

The total length of wall required is equal to:

LwallreqdBL13 = F_BL13Duct3/3

LwallreqdBL13 = 5.93m

Bracing Line 3 has a total wall length of approximately 12m without requiring the removal of any additional windows. Bracing Line 1 has over 20m.

Both these bracing lines can therefore be lined with Gib standard (GS1).




Clerestorey Area

From 3.1.3) it was shown that the clerestorey area is adequate to resist seismic loading over 100%NBS

Bracing Line 2

As was done in Option 1, utilise the existing wall space by re-lining these walls with Ecoply EP1.




4. Roof Lintel Design

4.1 Option 1

4.1.1 Lintel Layout

Three lintel types are to be designed to support the roof purlins in the areas where walls have been taking out. The locations of the lintels are shown in the plan below. These lintels are to be labelled Lintel 1, Lintel 2 and Lintel 3.

4.1.2 Design Loads

Load Combinations

Check the lintels under the following load combinations from NZS 1170.0

ULS:

1.35G 1.2G+1.5Q 1.2G+Wu+ψcQ 0.9G+wu

SLS:

G+ψsQ G+ψlQ G+ψsQ+Ws G+ψlQ+Ws

Lintel 1

Lintel 3 Lintel 2

Rafter Span Direction





Gravity Loading:

G = 0.4kPa (Roof – As was used for seismic weight calculations)

Q = 0.25kPa (Non Accessible Roof)

Wind Loading:

From 2.2.1), the following basic pressures were calculated:


PBasic SLS = 0.57kPa

Determine the design wind pressures on the lintels. Conservatively take the worst case wind loading for all lintels.

As per Cl 2.4.1 of NZS1170.2, the design wind pressure is calculated by the following equation:

Pdesign = Pbasic x Cfig x Cdyn

Cdyn = 1.0 (Low rise single storey structure with natural frequency above 1Hz)

Determine Cfig for different loading faces and directions:

Internal:

Cp,i = -0.3, 0 (All walls equally permeable)

Roof:

The roof slope is less than 10 degrees across the entire structure.

Longitudinal Loading

h = 3.4m (avg)

d = 29m

h/d = 0.12

Cp,e (R)Long = -0.9, 0.2 (Table 5.3(A))

Transverse Loading

h = 3.4m (avg)

d = 12m

h/d = 0.28

Cp,e (R)Trans = -0.9, 0.2 (Table 5.3(A))

The effective pressures will be the same for wind loading in both directions. Design lintels separately for uplift case and downwards case.




Ka = 0.9 (Tributary area approximately 25m2)

Kc,e = 0.8 (4 Effective Surfaces Table 5.5)

Kp = 1.0

Kl = 1.0 (Lintels not directly supporting the cladding)

12m

29m




Cfig,e = Cp,e x Ka x Kc,e x Kl x Kp

Cfig,e Up = -0.9 x 0.9 x 0.8 x 1.0 x 1.0

= -0.648

Cfig,e Down = 0.2 x 0.9 x 0.8 x 1.0 x1.0

= 0.144

Cfig,i = Cp,i x Kc,i

Cfig,i = -0.3 x 0.8

= -0.240 (Downwards)

Hence, the net Cfig factors are as follows:

Cfig net (Up) = -0.648

Cfig net (Down) = 0.144+0.240

= 0.384

The ULS and SLS design pressures in both directions are therefore equal to the following:

PdesULS(Up) = -0.648 x 0.84kPa

= -0.544 kPa

PdesULS(Down) = 0.384 x 0.84kPa

= 0.322 kPa

PdesSLS(Up) = -0.648 x 0.57kPa

= -0.369 kPa

PdesSLS(Down) = 0.384 x 0.57kPa

= 0.219 kPa

*Note: These design wind pressures are lower than that calculated for the Wellington region design. The same lintel members will be used for the Auckland design. The Wellington design loads therefore govern. The lintel design calculations shown on the pages to follow are for the Wellington region. Refer to the Wellington calculations for the wind load derivation.




4.1.3 Lintel 1 Design

Span = 3.7m

Tributary Width = 7.5m (Approximately)

2x290x45 timber members are adequate for this lintel. See the spreadsheets below for the lintel design calculations.





Span = 2.5m


By inspection 2x290x45 timber members are suitable here also.


Span = 2.0m


By inspection 2x290x45 timber members are suitable here also.

4.1.6 Lintel Fixings and Stud Supports

From the lintel calculations shown above, the critical reactions are as follows:

N*t = -4.5kN (Uplift – 0.9G+Wu)

N*c = 12.4kN (Compression – 1.2G+Wu)

Design a lintel fixing to resist these loads.

Use a standard Lumberlok lintel fixing detail. The critical uplift demand is 4.5kN, therefore use a standard 7.5kN detail.

These details show a double stud as the lintel support. Check a double 90x45 stud is adequate to support the reaction loads.




Tensile Capacity (In accordance with NZS 3603:1993):

φNtstud = φ x k1 x k4 x ft x A

φ = 0.80

k1 = 1.00 (Wind Loading Critical)

k4 = 1.14 (Two Support Elements)

ft = 6.00 MPa (SG8 Timber)

A = 90mm x 45mm x 2

= 8100mm2

Hence: φNtstud = 0.8 x 1.0 x 1.14 x 6.00 x 8100 x 10-3

= 44.3kN

> N*t OK

Compression Capacity (In accordance with NZS 3603:1993):

φNcstud = φ x k1 x k8 x fc x A

φ = 0.80

k1 = 0.60 (Critical gravity demand similar to critical wind demand)

fc = 18.00 MPa (SG8 Timber)

Lax = 3500mm (Approximate worst case – adjacent to clerestorey area)

Lay = 1000mm (Conservative – Dwang spacing in wall the studs are built in)

b = 90 mm

d = 90 mm

S2 = 3500/90 (Lax/d)

= 38.88 Critical

S3 = 1000/90 (Lay/b)

= 11.11

Hence: k8 = 0.203 (Table 2.8 – NZS3603:1993)

A = 8100 mm2

Hence: φNcCol = 0.8 x 0.6 x 0.203 x 18.0 x 8100 x 10-3

= 14.21 kN

> N*c OK

A double 90x45 stud is therefore adequate to support the lintel loads.




4.1.7 Jack Stud Framing

It is likely that there will be some clearance between the top of the lintel and the roof level for Lintel 1. As a result, jack stud framing will be required between the roof and the top of the lintel. The maximum jack stud framing height is expected to be no greater than 1000mm.

Try 90x45 jack studs at 600mm. From the Lintel 1 design, the critical uplift wind UDL was calculated to be -2.43kN/m and the critical downwards UDL was calculated to be 6.71kN/m. This correlates to an axial tension load of 1.46kN and an axial compression load of 4.03kN in the jack studs. Check the studs can support these loads.



φ = 0.80


k4 = 1.00 (One Element)


A = 90mm x 45mm

= 4050mm2


= 19.44kN

> N*t OK



φ = 0.80



Lax = 1000mm (Approximate worst case)

Lay = 1000mm (Approximate worst case)

b = 45 mm

d = 90 mm

S2 = 1000/90 (Lax/d)

= 11.11

S3 = 1000/45 (Lay/b)

= 22.22 Critical

Hence: k8 = 0.577 (Table 2.8 – NZS3603:1993)

A = 4050 mm2




Hence: φNcCol = 0.8 x 0.6 x 0.577 x 18.0 x 4050 x 10-3

= 20.19 kN

> N*c OK

90x45 jack studs at 600 centres are therefore adequate to support the roof loads. By inspection it will be adequate to fix these jack studs to the lintels with Lumberlok CPC80 fixings as the uplift loads are minor.




4.2 Option 2

4.2.1 Lintel Layout

4.2.2 Design Loads

The demands will be the same as those shown in 4.1.2) for Option 1. As discussed in 4.1.2) the lintel design will remain the same for all regions, therefore the Wellington design loads govern as the wind loads are higher. The lintel design calculations shown on the pages to follow are for the Wellington region. Refer to the Wellington calculations for the wind load derivation.


Span = 4.6m


A 200 PFC is suitable for this lintel. See spreadsheets below for lintel capacity calculations.

Lintel 1



Lintel 2

Lintel 3

Lintel 4





Span = 7.5m

Tributary Width = 2.6m (Worst Case)

A 200PFC is suitable here also (see spreadsheets below).





Span = 5.2m


2x290x45 timber members are suitable here (see spreadsheets below).





Span = 3.0m


By inspection, 2x290x45 timber members will be suitable here also.

4.2.7 Lintel Fixings and Stud Supports

Steel PFC Lintels

From the lintel calculations shown above, the critical reactions are as follows for the PFC Lintels:

Lintel 1

N*t = -5.9kN (Uplift – 0.9G+Wu)

N*c = 15.4kN (Compression – 1.2G+Wu)

Lintel 2

N*t = -3.3kN x 2

= -6.6kN (Uplift – 0.9G+Wu – PFC Either Side of Stud)

N*c = 8.4kN x 2

= 16.8kN (Compression – 1.2G+Wu – PFC Either Side of Stud)




Design a lintel fixing to resist these loads.

Lintel 1

Utilise the following standard detail shown below with a triple stud and three M12 bolts. Check the stud is adequate to resist tension and compression loads and the bolts have adequate shear capacity to resist the uplift loading reactions.

Check a triple 90x45 stud is adequate to support the reaction loads.



φ = 0.80


k4 = 1.20 (Three Support Elements)


A = 90mm x 45mm x 3

= 12150mm2


= 70.00kN

> N*t OK






φ = 0.80



Lax = 3500mm (Approximate worst case – adjacent to clerestorey area)


b = 90 mm

d = 135 mm

S2 = 3500/90 (Lax/d)

= 38.88 Critical

S3 = 1000/135 (Lay/b)

= 7.41

Hence: k8 = 0.203 (Table 2.8 – NZS3603:1993)

A = 12150 mm2

Hence: φNcCol = 0.8 x 0.6 x 0.203 x 18.0 x 12150x 10-3

= 21.31 kN

> N*c OK

A triple 90x45 stud is therefore adequate to support the lintel loads. At the base of the studs utilise a similar base fixing to that shown in Option 1. As a triple stud is used, this will require nailing on both faces of the studs to ensure all three members are adequately fixed together.

Bolt Shear Capacity

φQnbolt = φ x n x k1 x k12 x k13 x Qsk

φ = 0.7

n = 2 bolts (worst case)

k1 = 1.0 (wind loading)

k12 = 1.0 (dry timber)

k13 = 1.0 (less than 4 fasteners)

Qsk = 10.4kN (M12 bolts, be = 90mm)

Hence: φQnbolt = 0.7 x 2 x 1.0 x 1.0 x 1.0 x 10.4

= 14.56kN

> N*t OK

2 M12 bolts are adequate.




Lintel 2

Utilise the detail shown below where the PFC runs through the stud wall. Check to see a double 90x45 stud adequate to resist compression and tension loads.



φ = 0.80


k4 = 1.14 (Two Support Elements)


A = 90mm x 45mm x 2

= 8100mm2


= 44.4kN

> N*t OK



φ = 0.80



Lax = 2400mm


b = 45 mm




d = 90 mm

S2 = 2400/90 (Lax/d)

= 26.66 Critical

S3 = 1000/45 (Lay/b)

= 22.22

Hence: k8 = 0.576 (Table 2.8 – NZS3603:1993)

A = 8100 mm2

Hence: φNcCol = 0.8 x 0.6 x 0.576 x 18.0 x 8100x 10-3

= 40.31 kN

> N*c OK

A double 90x45 stud is therefore adequate to support the lintel loads. At the base of the studs utilise a similar base fixing to that shown in Option 1. This will be adapted slightly for the two studs being separated in this situation.

Bolt Shear Capacity

φQnbolt = φ x n x k1 x k12 x k13 x Qsk

φ = 0.7

n = 1 bolt (worst case)

k1 = 1.0 (wind loading)

k12 = 1.0 (dry timber)

k13 = 1.0 (less than 4 fasteners)

Qsk = 10.4kN x 2 (M12 bolts, be = 90mm, double shear)

Hence: φQnbolt = 0.7 x 1.0 x 1.0 x 1.0 x 1.0 x 10.4 x 2

= 14.56kN

> N*t OK

1 x M12 bolt is adequate.

PFC Splice Detail

As the PFC will run continuous through the wall, utilise a splice detail for constructability. Place the splice 1.0m from the wall edge. Splice with two bolts inside the web of the PFC similar to the figure below.

Design for the expected loading demands 1.0m in from the edge of the support.




The lintels span 7.5m between the supports and have the following critical demands:

M*max = 15.7kNm (2.24kN/m UDL)

V*max = 8.4kN

The demands at the splice are therefore equal to the following:

M*Splice = 7.3kNm

V*Splice = 6.2kN

Space the bolts 120mm apart. The critically loaded bolt will be top bolt as it will be subject to both tension and shear. The critical bolt demands are therefore equal to the following: N*Bolt = 7.3kNm/0.120m = 61kN V*Bolt = 6.2kN/2 = 3.1kN Try M16 bolts (8.8/S)

φNf = 104kN

φVfn = 59.3kN

Check combined actions

(N*/φNf)2+(V*/φVfn)2 = (3.1kN/59.3kN)2+(61kN/104kN)2

= 0.347

< 1.0 OK

Try a 12mm, Grade 300 base plate.

There will be a 28mm lever arm between the bolt and the inside of the PFC flange.

M*plate = 61kN x 0.028m

= 1.71kNm

φMnplate = φFySy

φ = 0.9

Fy = 310 MPa

Sy = Bt2/4




= 36B

The minimum plate length (B) required for the capacity to exceed 1.71kNm is therefore equal to 170mm. By yield line theory this will be able to be achieved.

Timber Lintels

The timber lintel fixings will be the same as that shown in Option 1. The uplift and compression demands are less than that of the Option 1 lintels.

4.2.8 Jack Stud Framing

The UDL loads on Lintel 1 are expected to be similar to those in Option 1, therefore by inspection, 90x45 studs at 600 centres will be adequate.

= Yield Length (Exceeds 170mm)




5. Roof Diaphragm Capacity Check The roof diaphragm of the typical Avalon Block is braced by diagonal timber sarking The diaphragm will be treated similar to a wall with a bracing capacity of 84Bu/m (4.2kN/m), the same as a diagonal timber sarked wall as per the Ministry of Education Guidelines.

*Note: The roof diaphragm was found to have the capacity to resist seismic loading greater than 67%NBS for the Wellington Region. The seismic forces in the Wellington Region (Soil Class C) are greater than the Auckland Region (Soil Class D). The roof diaphragm will therefore be adequate in Auckland also.

Below are the roof diaphragm capacity calculations for the Wellington loading case. Refer to the Wellington design for the load derivation.

.

F*=7.3kN/4.85m per Classroom (Ductility 2.5, Sp 0.5) See 2.1.4

F*=(70.59kN/8)/3.7m (Ductility 2.5, Sp 0.5) See 2.3

Bracing Line 2

Bracing Line 1

Clerestorey Line




Diaphragm Capacity (Longitudinal)

Length L = 7.4 m

Width h = 4.85 m

Bracing Capacity Bu = 4.2 kN/m MOE Guidelines V2

Height Modification Factor fmh = 1.0 MOE Guidelines V2 (H/L < 2)

Longitudinal Diaphragm Capacity CapDL = L × h × fmh

CapDL = 7.4 × 4.2 × 1.0

CapDL = 31.08 kN > 7.30kN ⟹ 100% NBS

By inspection, the diaphragm will be adequate to resist ductility 1.25 demands also.

Diaphragm Capacity (Transverse)

Length L = 9.7 m

Height h = 3.7 m

Bracing Capacity Bu = 4.2 kN/m MOE Guidelines V2

Height Modification Factor fmh = 1.0 MOE Guidelines V2 (H/L < 2)

Longitudinal Diaphragm Capacity CapDT = L × h × fmh

CapDT = 9.7 × 4.2 × 1.0

CapDT = 40.74 kN > 8.8kN ⟹ 100% NBS

By inspection, the diaphragm will be adequate to resist ductility 1.25 demands also.

The FLS upgrades will not alter the strength of the roof diaphragm as no structural elements that make up the diaphragm are being removed. It will also not alter the distance the diaphragm is required to span between bracing elements. Based on this, the roof diaphragm will still be adequate to resist 100%NBS once the FLS alterations are made.




6. Floor Diaphragm Capacity Check Timber floors feature 6” by 1” boards double nailed to floor joists. Floor diaphragms will have the same capacity in each direction as the ceiling diaphragms assessed above. Due to the presence of concrete foundations beneath all walls the floor will only be required to transfer forces from its own weight and the live loads applied directly to it.

*Note: The floor diaphragm was found to have the capacity to resist seismic loading greater than 67%NBS for the Wellington Region. The seismic forces in the Wellington Region (Soil Class C) are greater than the Auckland Region (Soil Class D). The floor diaphragm will therefore be adequate in Auckland also.

Below are the floor diaphragm capacity calculations for the Wellington loading case. Refer to the Wellington design for the load derivation.

4.85 m

3.7 m 3.7 m

4.85 m

DemFloorT = 13.84kN/3.7m (Ductility 2.5, Sp 0.5)

DemFloorL = 13.86kN/4.85m (Ductility 2.5, Sp 0.5)




Longitudinal Capacity Check

The floor has a similar weight to the roof diaphragm. The seismic load in the floor diaphragm is therefore equal to the roof diaphragm load, plus the additional load generated due to the live load on the floor.

Plan Area AreaFloorL = 9.7 m × 3.7 m

AreaFloorL = 35.89 m2

Area Reduction Factor Ψa = 0.3 + 3 / √(AreaFloorL) (Section 3.4.2, NZS 1170.1)

Ψa = 0.8

Floor Live Load QFloorL = Ψa × ΨE × Q × AreaFloorL = 0.8 × 0.3 × 3 × 35.89

QFloorL = 25.84 kN

Total Demand DemFloorL = Cd(T)µ=2.5(WDFloorL + QFloorL) = 7.3kN + (0.254 × 25.84)

DemFloorL = 13.86kN

%NBS

%NBS Floor Diaphragm - Longitudinal NBSFloorL = CapDL / DemFloorL

NBSFloorL = 31.08 / 13.86

NBSFloorL = >100%NBS

As this is a floor diaphragm, it would be best for damage to be minimised. Scale up the demands to ductility 1.25. This will result in the seismic demands increasing by a factor of 3.0.

%NBSrevised


NBSFloorL = 31.08 / (13.86x3)

NBSFloorL = 75%NBS>67%NBS OK




Transverse Floor Diaphragm Demand

The transverse wind demands were found to be higher than the seismic demands for lateral bracing. The seismic demands will be critical in this situation, particularly if a ductility of 1.25 is used as the live load in the floor will be included and the demand will be scaled up for a lower ductility.

From 2.1.4, the transverse seismic demand was calculated to be 58.18kN (ductility 2.5, Sp 0.5). This correlates to a maximum transverse diaphragm shear of 7.27kN. Add the additional seismic demand generated due to the live load on the floor.

Plan Area AreaFloorT = 7.4 m × 4.85 m

AreaFloorT = 35.89 m2

Area Reduction Factor Ψa = 0.3 + 3 / √(AreaFloorT) (Section 3.4.2, NZS 1170.1)

Ψa = 0.8

Floor Live Load QFloorT = Ψa × ΨE × Q × AreaFloorT = 0.8 × 0.3 × 3 × 35.89

QFloorT = 25.84 kN

Total Demand DemFloorT = Cd(T)µ=2.5(WDFloorT + QFloorT) = 7.27 + (0.254 × 25.84)

DemFloorT = 13.84 kN

%NBS

%NBS Floor Diaphragm - Transverse NBSFloorT = CapDT / DemFloorT

NBSFloorT = 40.74 / 13.84

NBSFloorT = >100%NBS

When considering as ductility 1.25.

%NBSrevised


NBSFloorL = 40.74 / (13.84x3)

NBSFloorL = 98%NBS>67%NBS OK




7. Foundation Capacity Check Concrete foundation walls are present beneath all classroom walls as shown below.

*Note: The foundations were found to have the capacity to resist seismic loading greater than 67%NBS for the Wellington Region. The seismic forces in the Wellington Region (Soil Class C) are greater than the Auckland Region (Soil Class D). The foundations will therefore be adequate in Auckland also.

Below are the foundation capacity calculations for the Wellington loading case. Refer to the Wellington design for the load derivation.

Original plans do not detail the bottom plate to foundation wall connection. The Avalon Blocks that have been inspected typically have 12mm hold-downs at approximately 3ft centres. Both bolts and bent reinforcing rod have been observed for this purpose. The following calculations assume that hold-downs have been inspected by a qualified person and found to be present and in reasonable condition.

The foundations and hold-downs will be subject to seismic forces resulting from the full dead load of the walls, dead load of the roof and dead and live load of the floor. The foundations are expected to remain undamaged during a seismic event, so demands corresponding to a ductility of 1.25 will be used.




Demand at Foundation Level

The demand on the wall will be calculated as shown below using a ductility of µ=1.25:

𝑭𝑭𝒙𝒙 = 𝑪𝑪𝒅𝒅(𝑻𝑻)µ=𝒙𝒙(𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹 𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾 + 𝑻𝑻𝑹𝑹𝑾𝑾𝑾𝑾𝑾𝑾 𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾 𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾+ 𝑭𝑭𝑾𝑾𝑹𝑹𝑹𝑹𝑼𝑼 𝑫𝑫𝑾𝑾𝑾𝑾𝒅𝒅 𝑳𝑳𝑹𝑹𝑾𝑾𝒅𝒅𝑳𝑳 + 𝑭𝑭𝑾𝑾𝑹𝑹𝑹𝑹𝑼𝑼 𝑳𝑳𝑾𝑾𝑳𝑳𝑾𝑾 𝑳𝑳𝑹𝑹𝑾𝑾𝒅𝒅𝑳𝑳)

Tributary Weight TribWeight = WallSec1 + RoofSec1 + WallSec2 + RoofSec2 + GFloor + QLFloor

Demand (Foundations) DemFound = Cd(T)µ=1.25 × (TribWeight) = 0.7641 × (139.95)

DemFound = 106.94 kN

12mm bolt in single shear in 45mm bottom plate – parallel to grain

Effective Timber Thickness be = 90 mm Table 28.4 of Timber Design Guide

Bolt diameter d = 12 mm

Bolt - Characteristic Strength Qu = min (72.2 × d2, 18 × d × be) Cl 4.4.2 NZS3603:1993

Qu = 10.4 kN

Transverse direction will be critical. Assume bolts are present at 3 ft (0.9 m) centres.

Wall Length L = 7.4 m

Spacing s = 0.9 m

Number of bolts n = L/s = 8

Capacity per transverse wall Ctw = n × Qu = 83.2 kN

Total Capacity TCtw = 2 × Ctw = 166.4 kN

%NBS NBS = TCtw / DemFound = 166.4 / 106.94

NBS = >100%NBS

Fx

2.2 structural calculations auckland (avalon block)

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