FLAT SLAB DESIGN TO BS8110:PART 1:1997

Similar documents
DESIGN OF SLABS. 3) Based on support or boundary condition: Simply supported, Cantilever slab,

DESIGN OF SLABS. Department of Structures and Materials Engineering Faculty of Civil and Environmental Engineering University Tun Hussein Onn Malaysia

Design Manual to BS8110

ENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P

SLAB DESIGN. Introduction ACI318 Code provides two design procedures for slab systems:

16. Beam-and-Slab Design

Optimum proportions for the design of suspension bridge

Concrete Design to Eurocode 2

Page 1 of Sven Alexander Last revised SB-Produksjon STATICAL CALCULATIONS FOR BCC 250

Detailing of Reinforcment in Concrete Structures

Structural Analysis. EUROCODE 2 Background and Applications

MECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS

Design of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar

SECTION 5 ANALYSIS OF CONTINUOUS SPANS DEVELOPED BY THE PTI EDC-130 EDUCATION COMMITTEE LEAD AUTHOR: BRYAN ALLRED

Eurocode 4: Design of composite steel and concrete structures

SECTION 3 DESIGN OF POST TENSIONED COMPONENTS FOR FLEXURE

Eurocode 2: Design of concrete structures

Chapter 8. Flexural Analysis of T-Beams

Design and Construction of Cantilevered Reinforced Concrete Structures

FOUNDATION DESIGN. Instructional Materials Complementing FEMA 451, Design Examples

APPENDIX C Slab Calculation 2

The following sketches show the plans of the two cases of one-way slabs. The spanning direction in each case is shown by the double headed arrow.

SECTION 3 DESIGN OF POST- TENSIONED COMPONENTS FOR FLEXURE

STRUSOFT EXAMPLES PRE-STRESS 6.4

Two-Way Post-Tensioned Design

ADVANCED SYSTEMS FOR RATIONAL SLAB REINFORCEMENT

HUS-A 6 / HUS-H 6 / HUS-I 6 / HUS-P 6 Screw anchor in precast prestressed hollow core slabs

9.3 Two-way Slabs (Part I)

Prestressed Concrete I-Beam and TxGirder Haunch Design Guide

Design of reinforced concrete columns. Type of columns. Failure of reinforced concrete columns. Short column. Long column

SEISMIC DESIGN. Various building codes consider the following categories for the analysis and design for earthquake loading:

FOOTING DESIGN EXAMPLE

APOLLO SALES LTD SITE SCAFFOLD STEP DESIGN CHECK CALCULATIONS

Chapter - 3 Design of Rectangular Beams and One-way Slabs

MECHANICS OF SOLIDS - BEAMS TUTORIAL 1 STRESSES IN BEAMS DUE TO BENDING. On completion of this tutorial you should be able to do the following.

ADVANTAGES OF STEEL FIBRE REINFORCED CONCRETE IN INDUSTRIAL FLOORS

Reinforced Concrete Design Project Five Story Office Building

HUS-V Screw anchor. HUS-V Screw anchor. Basic loading data (for a single anchor) Mean ultimate resistance

TZ WEDGE ANCHOR FOR CRACKED AND UNCRACKED CONCRETE

Chapter 5 Bridge Deck Slabs. Bridge Engineering 1

A transverse strip of the deck is assumed to support the truck axle loads. Shear and fatigue of the reinforcement need not be investigated.

Basics of Reinforced Concrete Design

Formwork for Concrete

Design Of Reinforced Concrete Structures ii Two-Way Slabs

SLAB DESIGN EXAMPLE. Deck Design (AASHTO LRFD 9.7.1) TYPICAL SECTION. County: Any Hwy: Any Design: BRG Date: 7/2010

Concrete Design Manual

LOAD-CARRYING CAPACITY OF AXIALLY LOADED RODS GLUED-IN PERPENDICULAR TO THE GRAIN

A Case Study Comparing Two Approaches for Applying Area Loads: Tributary Area Loads vs Shell Pressure Loads

Technical handbook Panel Anchoring System

AISI CHEMICAL COMPOSITION LIMITS: Nonresulphurized Carbon Steels

The design process. Design life. Actions on structures. How to design concrete structures using Eurocode 2 2. Getting started

Guidelines for the Design of Post-Tensioned Floors

Concrete Frame Design Manual

SPECIFICATIONS, LOADS, AND METHODS OF DESIGN

RC Detailing to Eurocode 2

Optimising plate girder design

Hilti HIT-HY 150 MAX with rebar

HOW TO DESIGN CONCRETE STRUCTURES Foundations

Technical Notes 3B - Brick Masonry Section Properties May 1993

Structural Design Criteria & Design Loads on Delhi Metro Rail and Checking of Safety of Radial Joints in Tunnel Lining Segments

5 G R A TINGS ENGINEERING DESIGN MANUAL. MBG Metal Bar Grating METAL BAR GRATING MANUAL MBG METAL BAR GRATING NAAMM

STEEL BUILDINGS IN EUROPE. Multi-Storey Steel Buildings Part 10: Guidance to developers of software for the design of composite beams

METHODS FOR ACHIEVEMENT UNIFORM STRESSES DISTRIBUTION UNDER THE FOUNDATION

How To Build A Luxury Apartment Complex In London

Stresses in Beam (Basic Topics)

Redundant and Robust Structures by Joint Ductility

Rigid and Braced Frames

5 Steel elements. 5.1 Structural design At present there are two British Standards devoted to the design of strucof tural steel elements:

Preliminary steel concrete composite bridge design charts for Eurocodes

Structural welding is a process by which the parts that are to be connected are heated and

Structural Forensic Engineering Case Study : MRR2 Project. Prof. Dr. Azlan Abdul Rahman Forensic Engineering Universiti Teknologi Malaysia

Foreword. This manual provides technical information for designers and contractors to design and plan the construction operation.

MECHANICS OF SOLIDS - BEAMS TUTORIAL TUTORIAL 4 - COMPLEMENTARY SHEAR STRESS

Egcobox Software Manuel Version Max Frank GmbH & Co. KG

TABLE OF CONTENTS. INTRODUCTION... 5 Advance Concrete... 5 Where to find information?... 6 INSTALLATION... 7 STARTING ADVANCE CONCRETE...

Type of Force 1 Axial (tension / compression) Shear. 3 Bending 4 Torsion 5 Images 6 Symbol (+ -)

Session 5D: Benefits of Live Load Testing and Finite Element Modeling in Rating Bridges

Numerical modelling of shear connection between concrete slab and sheeting deck

Reinforced Concrete Slab Design Using the Empirical Method

DESIGN: (SUBSTRUCTURES, SPECIAL STRUCTURES AND MATERIALS) PART 12 BD 31/01 THE DESIGN OF BURIED CONCRETE BOX AND PORTAL FRAME STRUCTURES SUMMARY

Steel and composite bridges in Germany State of the Art

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS

Design MEMO 60 Reinforcement design for TSS 102

USE OF CFRP LAMINATES FOR STRENGTHENING OF REINFORCED CONCRETE CORBELS

SDNS108 Dynamic response of a reinforced concrete slab leaned on 4 with dimensions subjected to a concentrated loading

Design MEMO 54a Reinforcement design for RVK 41

In-situ Load Testing to Evaluate New Repair Techniques

Code of Practice for Structural Use of Concrete 2013

Structural Products. CI/SfB (23) Nh2 P356 August 2008

OPTIMAL DIAGRID ANGLE TO MINIMIZE DRIFT IN HIGH-RISE STEEL BUILDINGS SUBJECTED TO WIND LOADS

Introduction. Eurocodes. Specification. Cost

HVU with HAS/HAS-E rod adhesive anchor

IMPROVING THE STRUT AND TIE METHOD BY INCLUDING THE CONCRETE SOFTENING EFFECT

ETABS. Integrated Building Design Software. Concrete Shear Wall Design Manual. Computers and Structures, Inc. Berkeley, California, USA

Introductory Examples. InfoCAD Program System

Residential Decks. Planning and Development Services Department

Structural Design Calculation For Pergola

EUROPEAN ORGANISATION FOR TECHNICAL APPROVALS

November 20, Heather Sustersic Dear Professor Sustersic,

MILMAN & ASSOCIATES STRUCTURAL CONSULTING ENGINEERS/ PROJECT MANAGERS

Transcription:

Consulting Company for App'd by FLAT SLAB DESIGN TO Slab geometry Span of slab in x-direction; Span of slab in y-direction; Column dimension in x-direction; Column dimension in y-direction; External column dimension in x-direction; External column dimension in y-direction; Edge dimension in x-direction; Edge dimension in y-direction; Effective span of internal bay in x direction; Effective span of internal bay in y direction; Effective span of end bay in x direction; Effective span of end bay in y direction; Span x = 7200 mm Span y = 7200 mm l x = 400 mm l y = 400 mm l x1 = 250 mm l y1 = 250 mm e x = l x1 / 2 = 125 mm e y = l y1 / 2 = 125 mm L x = Span x l x = 6800 mm L y = Span y l y = 6800 mm L x1 = Span x l x / 2 = 7000 mm L y1 = Span y l y / 2 = 7000 mm e x A B C Span x Span x 1 Span y e y l x1 l y1 l x l x 2 l y 3 Span y l y h Slab details Depth of slab; h = 250 mm 1

Consulting Company for App'd by Characteristic strength of concrete; f cu = 35 N/mm 2 Characteristic strength of reinforcement; f y = 500 N/mm 2 Characteristic strength of shear reinforcement; f yv = 500 N/mm 2 Material safety factor; γ m = 1.15 Cover to bottom reinforcement; c = 20 mm Cover to top reinforcement; c = 20 mm Loading details Characteristic dead load; G k = 7.000 kn/m 2 Characteristic imposed load; Q k = 5.000 kn/m 2 Dead load factor; γ G = 1.4 Imposed load factor; γ Q = 1.6 Total ultimate load; N ult = (G k γ G) + (Q k γ Q) = 17.800 kn/m 2 Moment redistribution ratio; β b = 1.0 Ratio of support moments to span moments; i = 1.0 DESIGN SLAB IN THE X-DIRECTION SAGGING MOMENTS End bay A-B Effective span; Midspan moment; Support moment; L = 7000 mm d = 200 mm m = (N ult L 2 ) / (2 (1 + (1 + i)) 2 ) = 74.823 knm/m m = i m = 74.823 knm/m Design reinforcement K = m / (d 2 f cu) = 0.053 z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 187.3 mm Area of reinforcement designed; A s_des = m / (z f y / γ m) = 919 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 919 mm 2 /m Provide 20 dia bars @ 150 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 2094 mm 2 /m PASS - Span reinforcement is OK Check deflection Design service stress; f s = 2 f y A s_req / (3 A s_prov β b) = 146 N/mm 2 Modification factor; k 1 = min(0.55+(477n/mm 2 -f s)/(120 (0.9N/mm 2 +(m/d 2 ))),2) = 1.545 Allowable span to depth ratio; 0.9 26 k 1 = 36.151 Actual span to depth ratio; L / d = 35.000 2

Consulting Company for App'd by PASS - Span to depth ratio is OK Internal bay B-C Effective span; Midspan moment; Support moment; L = 6800 mm d = 202 mm m = (N ult L 2 ) / (2 ( (1 + i) + (1 + i)) 2 ) = 51.442 knm/m m = i m = 51.442 knm/m Design reinforcement K = m / (d 2 f cu) = 0.036 z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 191.9 mm Area of reinforcement designed; A s_des = m / (z f y / γ m) = 617 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 617 mm 2 /m Provide 16 dia bars @ 200 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1005 mm 2 /m PASS - Span reinforcement is OK Check deflection Design service stress; f s = 2 f y A s_req / (3 A s_prov β b) = 204 N/mm 2 Modification factor; k 1 = min(0.55+(477n/mm 2 -f s)/(120 (0.9N/mm 2 +(m/d 2 ))),2) = 1.601 Allowable span to depth ratio; 0.9 26 k 1 = 37.469 Actual span to depth ratio; L / d = 33.663 PASS - Span to depth ratio is OK HOGGING MOMENTS INTERNAL STRIP Penultimate column B3 Consider the reinforcement concentrated in half width strip over the support d = 200 mm Support moment; m = 2 i m = 149.646 knm/m K = m / (d 2 f cu) = 0.107 z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 172.5 mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1996 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1996 mm 2 /m Provide 20 dia bars @ 150 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 2094 mm 2 /m PASS - Support reinforcement is OK 3

Consulting Company for App'd by Internal column C3 Consider the reinforcement concentrated in half width strip over the support d = 200 mm Support moment; m = 2 i m = 102.884 knm/m K = m / (d 2 f cu) = 0.073 z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 182.1 mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1300 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1300 mm 2 /m Provide 20 dia bars @ 200 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1571 mm 2 /m PASS - Support reinforcement is OK HOGGING MOMENTS EXTERNAL STRIP Penultimate column B1, B2 Consider one and a half bays of negative moment being resisted over the edge and penultimate column Width of span; B = 7200 mm Edge distance; e = 125 mm d = 200 mm Support moment; m = m i (e + B + B / 2) / ((0.5 B) + (0.2 B) + e) = 158.265 knm/m K = m / (d 2 f cu) = 0.113 z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 170.5 mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 2134 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 2134 mm 2 /m Provide 20 dia bars @ 125 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 2513 mm 2 /m PASS - Support reinforcement is OK Internal column C1, C2 Consider one and a half bays of negative moment being resisted over the edge and penultimate column Width of span; B = 7200 mm Edge distance; e = 125 mm d = 200 mm Support moment; m = m i (e + B + B / 2) / ((0.5 B) + (0.2 B) + e) = 108.810 knm/m 4

Consulting Company for App'd by K = m / (d 2 f cu) = 0.078 z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 180.9 mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1383 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1383 mm 2 /m Provide 20 dia bars @ 200 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1571 mm 2 /m PASS - Support reinforcement is OK Corner column A1 d = 206 mm Total load on column; S = ((Span x / 2) + e x) ((Span y / 2) + e y) N ult = 247 kn Area of column head; A = l x l y1 = 0.100 m 2 Support moment; m = S (1 (N ult A / S) 1/3 ) / 2 = 99.639 knm/m K = m / (d 2 f cu) = 0.067 z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 189.3 mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1211 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1211 mm 2 /m Provide 16 dia bars @ 150 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1340 mm 2 /m PASS - Support reinforcement is OK Edge column A2, A3 d = 202 mm Total load on column; S = Span x (Span y / 2 + e y) N ult = 477 kn Area of column head; A = l x1 l y = 0.100 m 2 Support moment; m = S (1 (N ult A / S) 1/3 ) / 5.14 = 78.476 knm/m K = m / (d 2 f cu) = 0.055 z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 188.8 mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 956 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 956 mm 2 /m Provide 16 dia bars @ 175 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1149 mm 2 /m 5

Consulting Company for App'd by PASS - Support reinforcement is OK Between columns 1-2, 2-3 Around the perimeter between the column heads provide a minimum of 50% of the required end span bottom reinforcement. Area of reinforcement required; A s_req = A sx1 / 2 = 1047 mm 2 /m Provide 16 dia bars @ 150 centres - 'U' bars with 1600 mm long legs Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1340 mm 2 /m PASS - Edge reinforcement is OK Distribution reinforcement Provide 12 dia bars @ 300 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 377 mm 2 /m DESIGN SLAB IN THE Y-DIRECTION SAGGING MOMENTS End bay 1-2 Effective span; Midspan moment; Support moment; L = 7000 mm d = 220 mm m = (N ult L 2 ) / (2 (1 + (1 + i)) 2 ) = 74.823 knm/m m = i m = 74.823 knm/m Design reinforcement K = m / (d 2 f cu) = 0.044 z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 208.6 mm Area of reinforcement designed; A s_des = m / (z f y / γ m) = 825 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 825 mm 2 /m Provide 20 dia bars @ 200 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1571 mm 2 /m PASS - Span reinforcement is OK Check deflection Design service stress; f s = 2 f y A s_req / (3 A s_prov β b) = 175 N/mm 2 Modification factor; k 1 = min(0.55+(477n/mm 2 -f s)/(120 (0.9N/mm 2 +(m/d 2 ))),2) = 1.579 Allowable span to depth ratio; 0.9 26 k 1 = 36.942 Actual span to depth ratio; L / d = 31.818 PASS - Span to depth ratio is OK Internal bay 2-3 Effective span; L = 6800 mm 6

Consulting Company for App'd by Midspan moment; Support moment; d = 222 mm m = (N ult L 2 ) / (2 ( (1 + i) + (1 + i)) 2 ) = 51.442 knm/m m = i m = 51.442 knm/m Design reinforcement K = m / (d 2 f cu) = 0.030 z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 210.9 mm Area of reinforcement designed; A s_des = m / (z f y / γ m) = 561 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 561 mm 2 /m Provide 16 dia bars @ 200 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1005 mm 2 /m PASS - Span reinforcement is OK Check deflection Design service stress; f s = 2 f y A s_req / (3 A s_prov β b) = 186 N/mm 2 Modification factor; k 1 = min(0.55+(477n/mm 2 -f s)/(120 (0.9N/mm 2 +(m/d 2 ))),2) = 1.798 Allowable span to depth ratio; 0.9 26 k 1 = 42.062 Actual span to depth ratio; L / d = 30.631 PASS - Span to depth ratio is OK HOGGING MOMENTS INTERNAL STRIP Penultimate column C2 Consider the reinforcement concentrated in half width strip over the support d = 220 mm Support moment; m = 2 i m = 149.646 knm/m K = m / (d 2 f cu) = 0.088 z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 195.7 mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1758 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1758 mm 2 /m Provide 20 dia bars @ 150 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 2094 mm 2 /m PASS - Support reinforcement is OK Internal column C3 Consider the reinforcement concentrated in half width strip over the support d = 220 mm 7

Consulting Company for App'd by Support moment; m = 2 i m = 102.884 knm/m K = m / (d 2 f cu) = 0.061 z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 204.0 mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1160 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1160 mm 2 /m Provide 20 dia bars @ 200 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1571 mm 2 /m PASS - Support reinforcement is OK HOGGING MOMENTS EXTERNAL STRIP Penultimate column A2, B2 Consider one and a half bays of negative moment being resisted over the edge and penultimate column Width of span; B = 7200 mm Edge distance; e = 125 mm d = 220 mm Support moment; m = m i (e + B + B / 2) / ((0.5 B) + (0.2 B) + e) = 158.265 knm/m K = m / (d 2 f cu) = 0.093 z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 194.1 mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1875 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1875 mm 2 /m Provide 20 dia bars @ 150 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 2094 mm 2 /m PASS - Support reinforcement is OK Internal column A3, B3 Consider one and a half bays of negative moment being resisted over the edge and penultimate column Width of span; B = 7200 mm Edge distance; e = 125 mm d = 220 mm Support moment; m = m i (e + B + B / 2) / ((0.5 B) + (0.2 B) + e) = 108.810 knm/m K = m / (d 2 f cu) = 0.064 8

Consulting Company for App'd by z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 203.0 mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 1233 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 1233 mm 2 /m Provide 20 dia bars @ 200 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1571 mm 2 /m PASS - Support reinforcement is OK Edge column B1, C1 d = 222 mm Total load on column; S = (Span x / 2 + e x) Span y N ult = 477 kn Area of column head; A = l y1 l x = 0.100 m 2 Support moment; m = S (1 (N ult A / S) 1/3 ) / 5.14 = 78.476 knm/m K = m / (d 2 f cu) = 0.045 z = min((0.5 + (0.25 (K / 0.9))), 0.95) d = 210.1 mm Area of reinforcement required; A s_des = m / (z f y / γ m) = 859 mm 2 /m Area of reinforcement required; A s_req = max(a s_des, A s_min) = 859 mm 2 /m Provide 16 dia bars @ 175 centres Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1149 mm 2 /m PASS - Support reinforcement is OK Between columns A-B, B-C Around the perimeter between the column heads provide a minimum of 50% of the required end span bottom reinforcement. Area of reinforcement required; A s_req = A sy1 / 2 = 785 mm 2 /m Provide 16 dia bars @ 200 centres - 'U' bars with 1600 mm long legs Area of reinforcement provided; A s_prov = π D 2 / (4 s) = 1005 mm 2 /m PASS - Edge reinforcement is OK PUNCHING SHEAR Corner column A1 Design shear transferred to column; V t = ((0.45 Span x) + e x) ((0.45 Span y) + e y) N ult = 202 kn Design effective shear transferred to column; V eff = 1.25 V t = 252 kn Area of tension steel in x-direction; A sx_ten = A scorner = 1340 mm 2 /m Area of tension steel in y-direction; A sy_ten = A scorner = 1340 mm 2 /m Column perimeter; u c = l x1 + l y = 650 mm Average effective depth of reinforcement; d = h c - φ p = 214 mm 9

Consulting Company for App'd by Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = 4.733 N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = 1.811 N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 1292 mm A s_ten = 1731 mm 2 v c = 0.707 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.911 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 119 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 1613 mm A s_ten = 2161 mm 2 v c = 0.707 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.730 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 16 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 1934 mm A s_ten = 2592 mm 2 v c = 0.707 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.609 N/mm 2 Penultimate edge column A2 Design shear transferred to column; Design effective shear transferred to column; v < v c no shear reinforcement required V t = ((0.45 Span x) + e x) (1.05 Span y) N ult = 453 kn V eff = 1.4 V t = 634 kn Area of tension steel in x-direction; A sx_ten = A sx_edge = 1148 mm 2 /m 10

Consulting Company for App'd by Area of tension steel in y-direction; A sy_ten = A sy1e = 2094 mm 2 /m Column perimeter; Average effective depth of reinforcement; u c = (2 l x1)+ l y = 900 mm d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = 4.733 N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = 3.292 N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 2184 mm A s_ten = 3588 mm 2 v c = 0.757 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 1.356 N/mm 2 1.6 v c < v <= 2 v c Shear reinforcement required at perimeter; A sv_req = 5 ((0.7 v) - v c) u d / (0.95 f yv) = 947 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 2826 mm A s_ten = 4628 mm 2 v c = 0.756 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 1.048 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 372 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 3468 mm A s_ten = 5669 mm 2 v c = 0.756 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.854 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 154 mm 2 11

Consulting Company for App'd by Shear reinforcement at a perimeter of 3.75d - (803 mm) u = u c + (2 (k x k y) k d) = 4110 mm A s_ten = 6710 mm 2 v c = 0.755 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.721 N/mm 2 Internal edge column A3 Design shear transferred to column; Design effective shear transferred to column; v < v c no shear reinforcement required V t = ((0.45 Span x) + e x) Span y N ult = 431 kn V eff = 1.4 V t = 604 kn Area of tension steel in x-direction; A sx_ten = A sx_edge = 1148 mm 2 /m Area of tension steel in y-direction; A sy_ten = A sye = 1570 mm 2 /m Column perimeter; Average effective depth of reinforcement; u c = (2 l x1)+ l y = 900 mm d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = 4.733 N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = 3.135 N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 2184 mm A s_ten = 2989 mm 2 v c = 0.712 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 1.292 N/mm 2 1.6 v c < v <= 2 v c Shear reinforcement required at perimeter; A sv_req = 5 ((0.7 v) - v c) u d / (0.95 f yv) = 945 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 2826 mm A s_ten = 3862 mm 2 v c = 0.712 N/mm 2 12

Consulting Company for App'd by Nominal design shear stress at perimeter; v = V eff / (u d) = 0.998 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 365 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 3468 mm A s_ten = 4734 mm 2 v c = 0.712 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.814 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 159 mm 2 Shear reinforcement at a perimeter of 3.75d - (803 mm) u = u c + (2 (k x k y) k d) = 4110 mm A s_ten = 5607 mm 2 v c = 0.711 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.686 N/mm 2 Penultimate edge column B1 Design shear transferred to column; Design effective shear transferred to column; v < v c no shear reinforcement required V t = (1.05 Span x) ((0.45 Span y) + e y) N ult = 453 kn V eff = 1.4 V t = 634 kn Area of tension steel in x-direction; A sx_ten = A sx1e = 2513 mm 2 /m Area of tension steel in y-direction; A sy_ten = A sy_edge = 1148 mm 2 /m Column perimeter; Average effective depth of reinforcement; u c = l x + (2 l y1) = 900 mm d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = 4.733 N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = 3.292 N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 2184 mm A s_ten = 4066 mm 2 13

Consulting Company for App'd by v c = 0.789 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 1.356 N/mm 2 1.6 v c < v <= 2 v c Shear reinforcement required at perimeter; A sv_req = 5 ((0.7 v) - v c) u d / (0.95 f yv) = 789 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 2826 mm A s_ten = 5241 mm 2 v c = 0.788 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 1.048 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 331 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 3468 mm A s_ten = 6416 mm 2 v c = 0.788 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.854 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 104 mm 2 Shear reinforcement at a perimeter of 3.75d - (803 mm) u = u c + (2 (k x k y) k d) = 4110 mm A s_ten = 7592 mm 2 v c = 0.787 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.721 N/mm 2 v < v c no shear reinforcement required 14

Consulting Company for App'd by Penultimate central column B2 Design shear transferred to column; Design effective shear transferred to column; V t = (1.05 Span x) (1.05 Span y) N ult = 1017 kn V eff = 1.15 V t = 1170 kn Area of tension steel in x-direction; A sx_ten = A sx1e = 2513 mm 2 /m Area of tension steel in y-direction; A sy_ten = A sy1e = 2094 mm 2 /m Column perimeter; Average effective depth of reinforcement; u c = 2 (l x + l y) = 1600 mm d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = 4.733 N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = 3.417 N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 4168 mm A s_ten = 9601 mm 2 v c = 0.847 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 1.312 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 872 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 5452 mm A s_ten = 12559 mm 2 v c = 0.847 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 1.003 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 382 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 6736 mm A s_ten = 15516 mm 2 15

Consulting Company for App'd by v c = 0.847 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.812 N/mm 2 Internal central column B3 Design shear transferred to column; Design effective shear transferred to column; v < v c no shear reinforcement required V t = (1.05 Span x) Span y N ult = 969 kn V eff = 1.15 V t = 1114 kn Area of tension steel in x-direction; A sx_ten = A sx1i = 2094 mm 2 /m Area of tension steel in y-direction; A sy_ten = A sye = 1570 mm 2 /m Column perimeter; Average effective depth of reinforcement; u c = 2 (l x + l y) = 1600 mm d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = 4.733 N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = 3.254 N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 4168 mm A s_ten = 7636 mm 2 v c = 0.785 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 1.249 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 872 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 5452 mm A s_ten = 9988 mm 2 v c = 0.785 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.955 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 418 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 6736 mm 16

Consulting Company for App'd by A s_ten = 12340 mm 2 v c = 0.785 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.773 N/mm 2 Internal edge column C1 Design shear transferred to column; Design effective shear transferred to column; v < v c no shear reinforcement required V t = Span x ((0.45 Span y) + e y) N ult = 431 kn V eff = 1.4 V t = 604 kn Area of tension steel in x-direction; A sx_ten = A sxe = 1570 mm 2 /m Area of tension steel in y-direction; A sy_ten = A sy_edge = 1148 mm 2 /m Column perimeter; u c = l x + (2 l y1) = 900 mm (Library item: Flat slab shear map C1) Average effective depth of reinforcement; d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = 4.733 N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = 3.135 N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 2184 mm A s_ten = 2989 mm 2 v c = 0.712 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 1.292 N/mm 2 1.6 v c < v <= 2 v c Shear reinforcement required at perimeter; A sv_req = 5 ((0.7 v) - v c) u d / (0.95 f yv) = 945 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 2826 mm A s_ten = 3862 mm 2 v c = 0.712 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.998 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 365 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) 17

Consulting Company for App'd by u = u c + (2 (k x k y) k d) = 3468 mm A s_ten = 4734 mm 2 v c = 0.712 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.814 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 159 mm 2 Shear reinforcement at a perimeter of 3.75d - (803 mm) u = u c + (2 (k x k y) k d) = 4110 mm A s_ten = 5607 mm 2 v c = 0.711 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.686 N/mm 2 Internal central column C2 Design shear transferred to column; Design effective shear transferred to column; v < v c no shear reinforcement required V t = Span x (1.05 Span y) N ult = 969 kn V eff = 1.15 V t = 1114 kn Area of tension steel in x-direction; A sx_ten = A sxe = 1570 mm 2 /m Area of tension steel in y-direction; A sy_ten = A sy1i = 2094 mm 2 /m Column perimeter; Average effective depth of reinforcement; u c = 2 (l x + l y) = 1600 mm d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = 4.733 N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = 3.254 N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 4168 mm A s_ten = 7636 mm 2 v c = 0.785 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 1.249 N/mm 2 18

Consulting Company for App'd by Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 872 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 5452 mm A s_ten = 9988 mm 2 v c = 0.785 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.955 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 418 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 6736 mm A s_ten = 12340 mm 2 v c = 0.785 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.773 N/mm 2 Internal column C3 Design shear transferred to column; Design effective shear transferred to column; v < v c no shear reinforcement required V t = Span x Span y N ult = 923 kn V eff = 1.15 V t = 1061 kn Area of tension steel in x-direction; A sx_ten = A sxi = 1570 mm 2 /m Area of tension steel in y-direction; A sy_ten = A syi = 1570 mm 2 /m Column perimeter; Average effective depth of reinforcement; u c = 2 (l x + l y) = 1600 mm d = h c - φ p = 214 mm Maximum allowable shear stress; v max = min(0.8 (f cu), 5) = 4.733 N/mm 2 Design shear stress at column perimeter; v 0 = V eff / (u c d) = 3.099 N/mm 2 Shear reinforcement at a perimeter of 1.50d - (321 mm) PASS - Maximum concrete shear stress not exceeded at column perimeter u = u c + (2 (k x k y) k d) = 4168 mm A s_ten = 6544 mm 2 19

Consulting Company for App'd by v c = 0.746 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 1.190 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 834 mm 2 Shear reinforcement at a perimeter of 2.25d - (482 mm) u = u c + (2 (k x k y) k d) = 5452 mm A s_ten = 8560 mm 2 v c = 0.746 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.910 N/mm 2 Shear reinforcement required at perimeter; A sv_req = (v - v c) u d / (0.95 f yv) = 403 mm 2 Shear reinforcement at a perimeter of 3.00d - (642 mm) u = u c + (2 (k x k y) k d) = 6736 mm A s_ten = 10576 mm 2 v c = 0.746 N/mm 2 Nominal design shear stress at perimeter; v = V eff / (u d) = 0.736 N/mm 2 CURTAILMENT OF REINFORCEMENT Internal column v < v c no shear reinforcement required Radius of circular yield line; r = (l x l y / π) 1/2 (1.05 Span x 1.05 Span y / (l x l y)) 1/3 = 1601 mm Minimum curtailment length in x-direction; Minimum curtailment length in y-direction; Corner column l int_x = Max(r + 12 D, 0.25 Span x) = 1841 mm l int_y = Max(r + 12 D, 0.25 Span y) = 1841 mm Radius of yield line; r = (l x1 l y / π) 1/2 ((0.45 Span x + e x) (0.45 Span y + e y)/ (l x1 l y)) 1/3 Minimum curtailment length in x-direction; r = 863 mm l corner_x = Max(r + 12 D, 0.2 Span x) = 1440 mm 20

Consulting Company for App'd by Minimum curtailment length in y-direction; l corner_y = Max(r + 12 D, 0.2 Span y) = 1440 mm Edge columns Radius of yield line in x-direction; r = (l x1 l y / π) 1/2 ((0.45 Span x + e x) (1.05 Span y) / (l x1 l y)) 1/3 Minimum curtailment length in x-direction; r = 1130 mm l edge_x = Max(r + 12 D, 0.2 Span x) = 1440 mm Radius of yield line in y-direction; r = (l x l y1 / π) 1/2 ((0.45 Span y + e y) (1.05 Span x) / (l x l y1)) 1/3 Minimum curtailment length in y-direction; r = 1130 mm l edge_y = Max(r + 12 D, 0.2 Span y) = 1440 mm 21

Consulting Company for App'd by A B C e x 1 e y l y1 n l x1 n Span x s r e Span x s r f s 0.2 x Span y c x 2 Span y Span y q p j q d a j e x b l f ly 0.5 x Span y p g h 3 k k m q l x 0.5 x Span x 0.2 x Span x When the effective span in the x direction, L x, is greater than the effective span in the y direction, L y, the reinforcement in the outer layer is assumed to be that in the x direction otherwise it is assumed to be that in the y direction. 22

Consulting Company for App'd by REINFORCEMENT KEY a = 20 dia bars @ 150 centres - (2094 mm 2 /m); b = 16 dia bars @ 200 centres - (1005 mm 2 /m) c = 20 dia bars @ 200 centres - (1570 mm 2 /m); d = 16 dia bars @ 200 centres - (1005 mm 2 /m) e = 20 dia bars @ 125 centres - (2513 mm 2 /m); f = 20 dia bars @ 200 centres - (1570 mm 2 /m) g = 20 dia bars @ 150 centres - (2094 mm 2 /m); h = 20 dia bars @ 200 centres - (1570 mm 2 /m) j = 20 dia bars @ 150 centres - (2094 mm 2 /m); k = 20 dia bars @ 200 centres - (1570 mm 2 /m) l = 20 dia bars @ 150 centres - (2094 mm 2 /m); m = 20 dia bars @ 200 centres - (1570 mm 2 /m) n = 16 dia bars @ 150 centres - (1340 mm 2 /m) p = 16 dia bars @ 175 centres - (1148 mm 2 /m); q = 16 dia bars @ 150 centres - (1340 mm 2 /m) r = 16 dia bars @ 175 centres - (1148 mm 2 /m); s = 16 dia bars @ 200 centres - (1005 mm 2 /m) Distribution bars = 12 dia bars @ 300 centres - (377 mm 2 /m) Shear reinforcement is required - Refer to output above for details. 23