The document discusses reinforcement detailing requirements according to Eurocode 2 (EC2). It covers general rules on bar spacing, minimum bend diameters, and anchorage and lapping of bars. For anchorage, it explains how to calculate the basic and design anchorage lengths according to EC2 equations and factors. A worked example calculates the design anchorage length for straight and bent H16 bars in concrete C25/30 with 25mm cover.
This document provides the design of a rectangular water tank with a capacity of 2500 cubic meters. It includes:
1) Design of the roof slab as a flat slab with columns spaced 5 meters apart and a thickness of 240mm.
2) Design of columns with a size of 350mm and reinforcement of 6 bars of 16mm diameter.
3) Design of the vertical walls with a thickness of 230mm at the base reducing to 180mm in the middle. Reinforcement of 16mm diameter bars at 125mm centers is provided.
4) Checks for crack width for the columns and walls show the crack width is less than the permissible 0.2mm.
This document provides structural calculations for the main canopy of a building located in Mumbai. It includes STAAD analysis of the steel structure, material properties, load assumptions, and results of the analysis. Key sections analyzed include the outer MS frame, inner MS frame, supporting MS pipes and tubes. Loads considered are self-weight, wind load, and live load. The analysis checks the steel structure for deflection under these loads.
The document discusses the design of slabs and flat slabs according to EC2. It covers designing for shear in slabs, including punching shear. It discusses detailing requirements for solid slabs. For flat slab design, it provides an example of flexural design. Key aspects covered include:
- Three approaches for designing shear reinforcement in slabs
- Determining shear resistance without shear reinforcement
- Requirements for punching shear including control perimeters and calculating shear stress
- Determining the need for and design of punching shear reinforcement
- Worked example of punching shear design at a column
The document discusses the design of slabs and flat slabs according to EC2. It covers designing for shear in slabs, including punching shear. It discusses detailing requirements for solid slabs. For flat slab design, it provides an example of flexural design. Key aspects covered include:
- Three approaches for designing shear reinforcement in slabs
- Determining shear resistance without shear reinforcement
- Punching shear requirements including control perimeters and shear stress calculation
- Determining required shear reinforcement area for punching shear
- Detailing requirements for solid slabs
The document discusses the design of slabs and flat slabs according to EC2. It covers designing for shear in slabs, including punching shear. It discusses detailing requirements for solid slabs. For flat slab design, it provides an example of flexural design. Key aspects covered include:
- Three approaches for designing shear reinforcement in slabs
- Determining shear resistance without shear reinforcement
- Punching shear requirements including control perimeters and shear stress calculation
- Determining required shear reinforcement area for punching shear
- Detailing requirements for solid slabs
One way slab is designed for an office building room measuring 3.2m x 9.2m. The slab is 150mm thick with 10mm diameter reinforcement bars spaced 230mm centre to centre. It is simply supported on 300mm thick walls and designed to support a 2.5kN/m2 live load. Reinforcement provided meets code requirements for minimum area and spacing. Design checks for cracking, deflection, development length and shear are within code limits.
This document provides a summary of reinforced concrete columns (RCC columns). It defines a column and describes different types of columns based on reinforcement and length. Short columns are less than 12 times the minimum thickness, while long columns are greater than 12 times the thickness. The document outlines preliminary sizing of columns and the functions of tie/spiral reinforcement. It includes design equations for axially loaded columns in working stress design (WSD) and ultimate stress design (USD). Two sample problems are worked through demonstrating column design using both methods.
The document discusses reinforcement detailing requirements according to Eurocode 2 (EC2). It covers general rules on bar spacing, minimum bend diameters, and anchorage and lapping of bars. For anchorage, it explains how to calculate the basic and design anchorage lengths according to EC2 equations and factors. A worked example calculates the design anchorage length for straight and bent H16 bars in concrete C25/30 with 25mm cover.
This document provides the design of a rectangular water tank with a capacity of 2500 cubic meters. It includes:
1) Design of the roof slab as a flat slab with columns spaced 5 meters apart and a thickness of 240mm.
2) Design of columns with a size of 350mm and reinforcement of 6 bars of 16mm diameter.
3) Design of the vertical walls with a thickness of 230mm at the base reducing to 180mm in the middle. Reinforcement of 16mm diameter bars at 125mm centers is provided.
4) Checks for crack width for the columns and walls show the crack width is less than the permissible 0.2mm.
This document provides structural calculations for the main canopy of a building located in Mumbai. It includes STAAD analysis of the steel structure, material properties, load assumptions, and results of the analysis. Key sections analyzed include the outer MS frame, inner MS frame, supporting MS pipes and tubes. Loads considered are self-weight, wind load, and live load. The analysis checks the steel structure for deflection under these loads.
The document discusses the design of slabs and flat slabs according to EC2. It covers designing for shear in slabs, including punching shear. It discusses detailing requirements for solid slabs. For flat slab design, it provides an example of flexural design. Key aspects covered include:
- Three approaches for designing shear reinforcement in slabs
- Determining shear resistance without shear reinforcement
- Requirements for punching shear including control perimeters and calculating shear stress
- Determining the need for and design of punching shear reinforcement
- Worked example of punching shear design at a column
The document discusses the design of slabs and flat slabs according to EC2. It covers designing for shear in slabs, including punching shear. It discusses detailing requirements for solid slabs. For flat slab design, it provides an example of flexural design. Key aspects covered include:
- Three approaches for designing shear reinforcement in slabs
- Determining shear resistance without shear reinforcement
- Punching shear requirements including control perimeters and shear stress calculation
- Determining required shear reinforcement area for punching shear
- Detailing requirements for solid slabs
The document discusses the design of slabs and flat slabs according to EC2. It covers designing for shear in slabs, including punching shear. It discusses detailing requirements for solid slabs. For flat slab design, it provides an example of flexural design. Key aspects covered include:
- Three approaches for designing shear reinforcement in slabs
- Determining shear resistance without shear reinforcement
- Punching shear requirements including control perimeters and shear stress calculation
- Determining required shear reinforcement area for punching shear
- Detailing requirements for solid slabs
One way slab is designed for an office building room measuring 3.2m x 9.2m. The slab is 150mm thick with 10mm diameter reinforcement bars spaced 230mm centre to centre. It is simply supported on 300mm thick walls and designed to support a 2.5kN/m2 live load. Reinforcement provided meets code requirements for minimum area and spacing. Design checks for cracking, deflection, development length and shear are within code limits.
This document provides a summary of reinforced concrete columns (RCC columns). It defines a column and describes different types of columns based on reinforcement and length. Short columns are less than 12 times the minimum thickness, while long columns are greater than 12 times the thickness. The document outlines preliminary sizing of columns and the functions of tie/spiral reinforcement. It includes design equations for axially loaded columns in working stress design (WSD) and ultimate stress design (USD). Two sample problems are worked through demonstrating column design using both methods.
This document provides information on reinforcement detailing according to Eurocode 2 (EC2). It begins with an overview of the structural Eurocodes and the contents of EC2. Key topics covered in more detail include reinforcement properties, minimum cover requirements, crack control, bar spacing, bond stress calculations, and the design of anchorage and lap lengths. Worked examples are provided to demonstrate how to calculate the design anchorage length for tension reinforcement according to the equations and factors specified in EC2. In summary, the document outlines the main requirements for reinforcement detailing in concrete structures as defined by EC2.
check it out: http://goo.gl/vqNk7m
CADmantra Technologies pvt. Ltd. is a CAD Training institute specilized in producing quality and high standard education and training. We are providing a perfact institute for the students intersted in CAD courses CADmantra is established by a group of engineers to devlop good training system in the field of CAD/CAM/CAE, these courses are widely accepted worldwide.
#catiatraining
#ANSYS #CRE-O
#hypermesh
#Automobileworkshops
#enginedevelopment
#autocad
#sketching
The document provides design details for a rectangular concrete tank with three chambers. It discusses load combinations and factors used by the Portland Cement Association (PCA) that differ slightly from American Concrete Institute (ACI) specifications. An interior wall and short exterior wall of the tank are then designed. The interior wall is designed for both vertical and horizontal bending using #8 bars spaced at 6 inches and 8 inches, respectively. The short exterior wall uses a 14 inch thickness with #6 bars at 8 inches for vertical bending to resist a moment of 28,672 lb-ft/ft.
This document provides a design example for a reinforced concrete T-beam bridge girder. It includes the design of the deck slab, longitudinal girders, and cross girders. The design uses Courbon's method to calculate live load bending moments and shear forces. Details are given for the design of an interior deck slab panel including reinforcement sizing. Design of the longitudinal girders includes calculating reaction factors and sizing reinforcement to resist bending moments and shear forces from dead and live loads.
DESIGN OF DECK SLAB AND GIRDERS- BRIDGE ENGINEERINGLiyaWilson4
This document provides a design example for a reinforced concrete T-beam bridge girder. It includes the design of the deck slab, longitudinal girders, and cross girders. The design uses Courbon's method to calculate live load bending moments and shear forces. Details are given for the design of an interior deck slab panel including reinforcement sizing. Design of the longitudinal girders includes calculating reaction factors and sizing reinforcement to resist bending moments and shear forces from dead and live loads.
This document discusses the working stress design method for analyzing and designing reinforced concrete beams. It provides equations for determining internal forces, tensile steel ratio, neutral axis depth, and flexural stresses. It also covers topics such as balanced steel ratio, under/over reinforced sections, minimum concrete cover/bar spacing, and designing rectangular and cantilever beams. Doubly reinforced beams are discussed for cases where the cross section dimensions are restricted and the external moment exceeds the section's moment capacity.
This document provides an overview of the design of beams and one-way slabs for flexure, shear, and torsion according to IS 456. It discusses key concepts like requirements for flexural reinforcement, minimum and maximum reinforcement limits, clear cover, deflection control, and selection of member sizes. The document also includes a worked example showing the step-by-step design of a rectangular reinforced concrete beam for flexure. Design checks are performed to check for strength and deflection requirements. Modules for the course will cover analysis and design of beams, one-way slabs, and reinforcement detailing in accordance with limit state design principles and code specifications.
The document summarizes the analysis and design of various foundation types for a seven story building in Nablus city. It describes isolated footings, combined footings, wall footings, mat foundations, and pile foundations. Laboratory test results of soil samples are presented. Loads on each column are calculated. Dimensions, reinforcement details and settlement calculations are provided for each foundation type. Based on the analysis of material quantities, construction costs, and settlement calculations, isolated footings with combined, wall and elevator footings are recommended as the most economical foundation solution.
The document discusses the design of reinforced concrete columns. It provides formulas to calculate the nominal capacity of concentrically loaded columns based on steel ratio and material strengths. Minimum and maximum steel ratios of 1-8% are recommended, with a reasonable range of 1-3%. Clear cover requirements of 40-75mm are outlined depending on soil contact. Tie design considerations include bar diameter, shape, and longitudinal spacing. Spiral reinforcement provides increased ductility and the document discusses formulas for calculating confined concrete strength based on spiral ratio and properties. Minimum spiral ratios and pitch requirements are also provided.
This document provides the calculations to verify the strength of a two-leaf metal swinging gate. It analyzes the dead loads on the gate and designs the main frame, verticals, support columns, and tensioner to support the 1108.58 kg load. The 8x8x1/4 inch steel column and 1/2 inch tension rod designs were verified to withstand the loads using finite element analysis software. The conclusion is that the gate design is sufficient and the tension rod should be included to control deflection of the gate end.
This document provides information on casing design for oil and gas wells. It discusses the reasons for running casing strings, including preventing hole collapse and contamination. It describes the different types of casing strings and how they are classified. Factors involved in casing design are also summarized, such as burst, collapse, and tension pressures. Examples are provided to demonstrate how to calculate collapse pressures under different conditions, including with and without axial tension loads on the casing.
Content;
1. Top spherical dome.
2. Top ring beam.
3. Cylindrical wall.
4. Bottom ring beam.
5. Conical dome.
6. Circular ring beam.
The basics of enticing water tank design and the related components are broadly calculated in this document. The next few documents will demonstrate the design of Intze tank members like column, bracing and foundation. Keep following the updates.....
The document discusses guidelines for detailing reinforcement in concrete structures. It begins by defining detailing as the preparation of working drawings showing the size and location of reinforcement. Good detailing ensures reinforcement and concrete interact efficiently. The document then discusses sources of tension in concrete structures from various loading conditions like bending, shear, and connections. It provides equations from AS3600-2009 for calculating minimum development lengths for reinforcing bars to develop their yield strength based on bar size, concrete strength, and transverse reinforcement. It also discusses lap splice requirements. In summary, the document provides best practice guidelines for detailing reinforcement to efficiently resist loads and control cracking in concrete structures.
This document provides technical design guidelines for steel fiber reinforced concrete floors. It discusses three types of steel fibers and their material properties. It describes EN 14651 beam testing of steel fiber concrete and design values. The document outlines basic design principles for ultimate limit states (ULS) and serviceability limit states (SLS). It discusses resisting and acting forces for ULS. Methods are presented for calculating bending moment and shear force design values from applied loads. Design approaches are given for center point loads and edge loads, including consideration of soil resistance.
This document provides technical design guidelines for steel fiber reinforced concrete floors. It discusses three types of steel fibers and their material properties. It describes EN 14651 beam testing of steel fiber concrete and design values. The document outlines basic design principles for ultimate limit states (ULS) and serviceability limit states (SLS). It discusses resisting and acting forces for ULS. Methods are presented for calculating bending moment and shear from factored loads for various load cases including center point loads and edge loads. Design of soil resistance for punching shear is also covered.
This document provides an analysis and design of a bored pile system. It includes soil properties, pile geometry, load calculations, and checks of capacity including axial, lateral, moment, and settlement. Reinforcement is designed for the pile cap and pile. The pile cap and pile reinforcement are found to be inadequate and require larger diameters.
This document discusses the analysis of singly and doubly reinforced concrete beam sections. It provides definitions and design approaches for singly reinforced, doubly reinforced, and flanged beam sections. The key steps in the design process are outlined, including calculating loads and moments, checking for section type, sizing tension and compression reinforcement, and designing shear reinforcement. Design examples are provided for a singly reinforced and a doubly reinforced concrete beam according to BS 8110 design code standards.
The document provides design formulae for rectangular reinforced concrete beams according to BS8110, including:
1) Choosing beam section dimensions based on span ratios and reinforcement limits
2) Design of singly reinforced beams using moment-depth relationships
3) Design of doubly reinforced beams incorporating compression steel
4) Additional considerations for flanged beam design, accounting for compression in the web.
This document discusses the analysis of singly and doubly reinforced concrete beam sections. It begins by defining singly reinforced sections as having tension reinforcement only, while doubly reinforced sections have reinforcement in both tension and compression zones. Design steps are provided for both section types, including calculating loads, moments, reinforcement areas, and shear reinforcement. Formulas and assumptions used in the design process are also outlined. The goal is for students to learn to properly design reinforced concrete beam sections based on given structural loads and material properties.
This is an overview of my current metallic design and engineering knowledge base built up over my professional career and two MSc degrees : - MSc in Advanced Manufacturing Technology University of Portsmouth graduated 1st May 1998, and MSc in Aircraft Engineering Cranfield University graduated 8th June 2007.
This document provides information on reinforcement detailing according to Eurocode 2 (EC2). It begins with an overview of the structural Eurocodes and the contents of EC2. Key topics covered in more detail include reinforcement properties, minimum cover requirements, crack control, bar spacing, bond stress calculations, and the design of anchorage and lap lengths. Worked examples are provided to demonstrate how to calculate the design anchorage length for tension reinforcement according to the equations and factors specified in EC2. In summary, the document outlines the main requirements for reinforcement detailing in concrete structures as defined by EC2.
check it out: http://goo.gl/vqNk7m
CADmantra Technologies pvt. Ltd. is a CAD Training institute specilized in producing quality and high standard education and training. We are providing a perfact institute for the students intersted in CAD courses CADmantra is established by a group of engineers to devlop good training system in the field of CAD/CAM/CAE, these courses are widely accepted worldwide.
#catiatraining
#ANSYS #CRE-O
#hypermesh
#Automobileworkshops
#enginedevelopment
#autocad
#sketching
The document provides design details for a rectangular concrete tank with three chambers. It discusses load combinations and factors used by the Portland Cement Association (PCA) that differ slightly from American Concrete Institute (ACI) specifications. An interior wall and short exterior wall of the tank are then designed. The interior wall is designed for both vertical and horizontal bending using #8 bars spaced at 6 inches and 8 inches, respectively. The short exterior wall uses a 14 inch thickness with #6 bars at 8 inches for vertical bending to resist a moment of 28,672 lb-ft/ft.
This document provides a design example for a reinforced concrete T-beam bridge girder. It includes the design of the deck slab, longitudinal girders, and cross girders. The design uses Courbon's method to calculate live load bending moments and shear forces. Details are given for the design of an interior deck slab panel including reinforcement sizing. Design of the longitudinal girders includes calculating reaction factors and sizing reinforcement to resist bending moments and shear forces from dead and live loads.
DESIGN OF DECK SLAB AND GIRDERS- BRIDGE ENGINEERINGLiyaWilson4
This document provides a design example for a reinforced concrete T-beam bridge girder. It includes the design of the deck slab, longitudinal girders, and cross girders. The design uses Courbon's method to calculate live load bending moments and shear forces. Details are given for the design of an interior deck slab panel including reinforcement sizing. Design of the longitudinal girders includes calculating reaction factors and sizing reinforcement to resist bending moments and shear forces from dead and live loads.
This document discusses the working stress design method for analyzing and designing reinforced concrete beams. It provides equations for determining internal forces, tensile steel ratio, neutral axis depth, and flexural stresses. It also covers topics such as balanced steel ratio, under/over reinforced sections, minimum concrete cover/bar spacing, and designing rectangular and cantilever beams. Doubly reinforced beams are discussed for cases where the cross section dimensions are restricted and the external moment exceeds the section's moment capacity.
This document provides an overview of the design of beams and one-way slabs for flexure, shear, and torsion according to IS 456. It discusses key concepts like requirements for flexural reinforcement, minimum and maximum reinforcement limits, clear cover, deflection control, and selection of member sizes. The document also includes a worked example showing the step-by-step design of a rectangular reinforced concrete beam for flexure. Design checks are performed to check for strength and deflection requirements. Modules for the course will cover analysis and design of beams, one-way slabs, and reinforcement detailing in accordance with limit state design principles and code specifications.
The document summarizes the analysis and design of various foundation types for a seven story building in Nablus city. It describes isolated footings, combined footings, wall footings, mat foundations, and pile foundations. Laboratory test results of soil samples are presented. Loads on each column are calculated. Dimensions, reinforcement details and settlement calculations are provided for each foundation type. Based on the analysis of material quantities, construction costs, and settlement calculations, isolated footings with combined, wall and elevator footings are recommended as the most economical foundation solution.
The document discusses the design of reinforced concrete columns. It provides formulas to calculate the nominal capacity of concentrically loaded columns based on steel ratio and material strengths. Minimum and maximum steel ratios of 1-8% are recommended, with a reasonable range of 1-3%. Clear cover requirements of 40-75mm are outlined depending on soil contact. Tie design considerations include bar diameter, shape, and longitudinal spacing. Spiral reinforcement provides increased ductility and the document discusses formulas for calculating confined concrete strength based on spiral ratio and properties. Minimum spiral ratios and pitch requirements are also provided.
This document provides the calculations to verify the strength of a two-leaf metal swinging gate. It analyzes the dead loads on the gate and designs the main frame, verticals, support columns, and tensioner to support the 1108.58 kg load. The 8x8x1/4 inch steel column and 1/2 inch tension rod designs were verified to withstand the loads using finite element analysis software. The conclusion is that the gate design is sufficient and the tension rod should be included to control deflection of the gate end.
This document provides information on casing design for oil and gas wells. It discusses the reasons for running casing strings, including preventing hole collapse and contamination. It describes the different types of casing strings and how they are classified. Factors involved in casing design are also summarized, such as burst, collapse, and tension pressures. Examples are provided to demonstrate how to calculate collapse pressures under different conditions, including with and without axial tension loads on the casing.
Content;
1. Top spherical dome.
2. Top ring beam.
3. Cylindrical wall.
4. Bottom ring beam.
5. Conical dome.
6. Circular ring beam.
The basics of enticing water tank design and the related components are broadly calculated in this document. The next few documents will demonstrate the design of Intze tank members like column, bracing and foundation. Keep following the updates.....
The document discusses guidelines for detailing reinforcement in concrete structures. It begins by defining detailing as the preparation of working drawings showing the size and location of reinforcement. Good detailing ensures reinforcement and concrete interact efficiently. The document then discusses sources of tension in concrete structures from various loading conditions like bending, shear, and connections. It provides equations from AS3600-2009 for calculating minimum development lengths for reinforcing bars to develop their yield strength based on bar size, concrete strength, and transverse reinforcement. It also discusses lap splice requirements. In summary, the document provides best practice guidelines for detailing reinforcement to efficiently resist loads and control cracking in concrete structures.
This document provides technical design guidelines for steel fiber reinforced concrete floors. It discusses three types of steel fibers and their material properties. It describes EN 14651 beam testing of steel fiber concrete and design values. The document outlines basic design principles for ultimate limit states (ULS) and serviceability limit states (SLS). It discusses resisting and acting forces for ULS. Methods are presented for calculating bending moment and shear force design values from applied loads. Design approaches are given for center point loads and edge loads, including consideration of soil resistance.
This document provides technical design guidelines for steel fiber reinforced concrete floors. It discusses three types of steel fibers and their material properties. It describes EN 14651 beam testing of steel fiber concrete and design values. The document outlines basic design principles for ultimate limit states (ULS) and serviceability limit states (SLS). It discusses resisting and acting forces for ULS. Methods are presented for calculating bending moment and shear from factored loads for various load cases including center point loads and edge loads. Design of soil resistance for punching shear is also covered.
This document provides an analysis and design of a bored pile system. It includes soil properties, pile geometry, load calculations, and checks of capacity including axial, lateral, moment, and settlement. Reinforcement is designed for the pile cap and pile. The pile cap and pile reinforcement are found to be inadequate and require larger diameters.
This document discusses the analysis of singly and doubly reinforced concrete beam sections. It provides definitions and design approaches for singly reinforced, doubly reinforced, and flanged beam sections. The key steps in the design process are outlined, including calculating loads and moments, checking for section type, sizing tension and compression reinforcement, and designing shear reinforcement. Design examples are provided for a singly reinforced and a doubly reinforced concrete beam according to BS 8110 design code standards.
The document provides design formulae for rectangular reinforced concrete beams according to BS8110, including:
1) Choosing beam section dimensions based on span ratios and reinforcement limits
2) Design of singly reinforced beams using moment-depth relationships
3) Design of doubly reinforced beams incorporating compression steel
4) Additional considerations for flanged beam design, accounting for compression in the web.
This document discusses the analysis of singly and doubly reinforced concrete beam sections. It begins by defining singly reinforced sections as having tension reinforcement only, while doubly reinforced sections have reinforcement in both tension and compression zones. Design steps are provided for both section types, including calculating loads, moments, reinforcement areas, and shear reinforcement. Formulas and assumptions used in the design process are also outlined. The goal is for students to learn to properly design reinforced concrete beam sections based on given structural loads and material properties.
Similar to Masonry Design Example for building with brick materials. (20)
This is an overview of my current metallic design and engineering knowledge base built up over my professional career and two MSc degrees : - MSc in Advanced Manufacturing Technology University of Portsmouth graduated 1st May 1998, and MSc in Aircraft Engineering Cranfield University graduated 8th June 2007.
Covid Management System Project Report.pdfKamal Acharya
CoVID-19 sprang up in Wuhan China in November 2019 and was declared a pandemic by the in January 2020 World Health Organization (WHO). Like the Spanish flu of 1918 that claimed millions of lives, the COVID-19 has caused the demise of thousands with China, Italy, Spain, USA and India having the highest statistics on infection and mortality rates. Regardless of existing sophisticated technologies and medical science, the spread has continued to surge high. With this COVID-19 Management System, organizations can respond virtually to the COVID-19 pandemic and protect, educate and care for citizens in the community in a quick and effective manner. This comprehensive solution not only helps in containing the virus but also proactively empowers both citizens and care providers to minimize the spread of the virus through targeted strategies and education.
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An In-Depth Exploration of Natural Language Processing: Evolution, Applicatio...DharmaBanothu
Natural language processing (NLP) has
recently garnered significant interest for the
computational representation and analysis of human
language. Its applications span multiple domains such
as machine translation, email spam detection,
information extraction, summarization, healthcare,
and question answering. This paper first delineates
four phases by examining various levels of NLP and
components of Natural Language Generation,
followed by a review of the history and progression of
NLP. Subsequently, we delve into the current state of
the art by presenting diverse NLP applications,
contemporary trends, and challenges. Finally, we
discuss some available datasets, models, and
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Masonry Design Example for building with brick materials.
1. - 1 -
Masonry Design Example
Comparisons Using ASD and SD
Richard Bennett, PhD, PE
The University of Tennessee
Chair, 2016 TMS 402/602 Code Committee
2nd Vic Chair, 2022 TMS 402/602 Code Committee
- 2 -
Outline
Introduction: recent masonry code changes
Beams and lintels
Non-bearing walls: out-of-plane
Combined bending and axial force: pilasters
Bearing walls: out-of-plane
Shear walls: in-plane
• Partially Grouted Shear Wall
• Special Reinforced Shear Wall
- 3 -
Reorganization: 2013 TMS 402 Code
Part 1: General
Chapter 1 –
General
Requirements
Chapter 2 –
Notations &
Definitions
Chapter 3 –
Quality &
Construction
Part 2: Design
Requirements
Chapter 4:
General
Analysis &
Design
Chapter 5:
Structural
Elements
Chapter 6:
Reinforcement,
Metal Accessories
& Anchor Bolts
Chapter 7:
Seismic Design
Requirements
Part 3:
Engineered
Design
Methods
Chapter 8:
ASD
Chapter 9:
SD
Chapter 10:
Prestressed
Chapter 11:
AAC
Part 4:
Prescriptive
Design
Methods
Chapter 12:
Veneer
Chapter 13:
Glass Unit
Masonry
Chapter 14:
Partition
Walls
Part 5:
Appendices &
References
Appendix A
– Empirical
Design
Appendix B:
Design of
Masonry infill
Appendix C:
Limit Design
of Masonry
References
- 4 -
2011 vs 2013 Code
Modulus of rupture values increased by 1/3 for all but fully
grouted masonry normal to bed joint
Unit strength tables recalibrated
• Type S mortar, 2000 psi unit strength, f′m = 2000 psi
Shear strength of partially grouted walls: reduction factor of
0.75
2. - 5 -
CMU Unit Strength Table
Net area
compressive
strength of concrete
masonry, psi
Net area compressive strength of ASTM
C90 concrete masonry units, psi (MPa)
Type M or S Mortar Type N Mortar
1,700 --- 1,900
1,900 1,900 2,350
2,000 2,000 2,650
2,250 2,600 3,400
2,500 3,250 4,350
2,750 3,900 ----
TMS 602 Table 2
- 6 -
Partially Grouted Shear Wall Factor
Mean St Dev
Fully grouted
(Davis et al, 2010)
1.16 0.17
Partially grouted
(Minaie et al, 2010)
0.90 0.26
9.3.4.1.2, Equation (9-21)
= 0.75 for partially grouted shear walls
= and 1.0 otherwise
0.90
1.16
0.776
- 7 -
Beam and Lintel Design
Strength Design (Chapter 9)
• = 0.0035 clay masonry
• = 0.0025 concrete masonry
• Masonry stress = 0.8
• Masonry stress acts over = 0.8
• = 0.9 flexure; = 0.8 shear
.
2.25
• Minimum reinf: 1.3
• or 4 3
⁄ ,
• Maximum reinf: 1.5
0.8 0.8
Allowable Stress (Chapter 8)
• Allowable masonry stress: 0.45
• Allowable steel stress
• 32 ksi, Grade 60 steel
⁄ ⁄
2
1 3
⁄
• No min and max reinforcement
• Allowable shear stress
2.25
- 8 -
Beam and Lintel Design
Strength Design (Chapter 9)
1. Determine , depth of compressive
stress block
.
2. Solve for ,
,
.
3. 0.285 ⁄ for CMU
Grade 60 steel:
= 1.5 ksi, 0.00714
= 2 ksi, 0.00952
Allowable Stress (Chapter 8)
1. Assume value of (or ).
Typically 0.85 < < 0.95.
2. Determine a trial value of , .
, /
Choose reinforcement.
3. Determine and ; steel stress
and masonry stress.
4. Compare calculated stresses to
allowable stresses.
5. If masonry stress controls design,
consider other options (such as
change of member size, or
change of ). Reinforcement is
not being used efficiently.
3. - 9 -
ASD: Alternate Design Method
Calculate
3
2 2
2
3
Is ?
For Grade 60 steel,
CMU = 0.312
YES
NO
Compression controls Tension controls
⁄
,
2
1
1
,
1
3
,
2
Iterate. Use 2 as
new guess and repeat.
- 10 -
Comparison of ASD and SD
8 inch CMU
d = 20 in.
- 11 -
Example: Beam
Given: 10 ft. opening; dead load of 1.5 kip/ft; live load of 1.5 kip/ft; 24 in.
high; Grade 60 steel; Type S masonry cement mortar; 8 in. CMU; ’ =
2000 psi
Required: Design beam
Solution:
5.2.1.3: Length of bearing of beams shall be a minimum of 4 in.; typically
assumed to be 8 in.
5.2.1.1.1 Span length of members not built integrally with supports shall be taken
as the clear span plus depth of member, but need not exceed distance between
center of supports.
• Span = 10 ft + 2(4 in.) = 10.67 ft
5.2.1.2 Compression face of beams shall be laterally supported at a maximum
spacing of:
• 32 multiplied by the beam thickness. 32(7.625 in.) = 244 in. = 20.3 ft
• 120 / . 120(7.625 in.)2 / (20 in.) = 349 in. = 29.1 ft
- 12 -
Allowable Stress Design: Flexure
Load
Weight of fully grouted
normal weight: 83 psf
Moment
Determine Assume compression controls
Check if compression controls
Compression
controls
Calculate modular ratio,
1.5 0.083 2ft 1.5 3.17
. .
45.1k ⋅ ft
9.32in.
20in.
0.466 0.312
900
29000ksi
900 2.0ksi
16.1
3 3
. . . ⋅
.
. . .
9.32in.
4. - 13 -
Allowable Stress Design: Flexure
Find ,
Req’d area of steel
Bars placed in bottom U-shaped unit, or knockout bond beam unit.
Use 2 - #9 (As = 2.00 in2)
, 1
1
0.90 9.31in. 7.625in.
16.1 0.90ksi
1
0.466
1
1.94in.
- 14 -
Strength Design: Flexure
Use 2 - #6 (As = 0.88 in2)
= 78.5 k-ft
= 70.6 k-ft
Factored Load
Weight of fully grouted
normal weight: 83 psf
Factored Moment
Find
Depth of equivalent
rectangular stress block
Find ,
Req’d area of steel
. .
62.6k ⋅ ft
1.2 1.6
1.2 1.5 0.083 2ft 1.6 1.5 4.40
.
20in. 20in.
. ⋅
.
.
. . . .
3.78in.
,
. . . . . . .
0.77in.
- 15 -
Check Min and Max Reinforcement
Minimum Reinforcement Check: = 160 psi (parallel to bed joints
in running bond; fully grouted)
Maximum Reinforcement Check: 0.00952
Section modulus
Cracking moment
Check 1.3
. . .
732in.
732in. 160psi
117.1k ⋅ in. 9.76k ⋅ ft
1.3 1.3 9.76k ⋅ ft 12.7k ⋅ ft
78.5k ⋅ ft
0.88in.
7.625in. 20in.
0.00577
- 16 -
Summary: Beams, Flexure
ASD: Allowable tension controls for 0.5 k/ft and 1 k/ft.
Dead Load (k/ft)
(superimposed)
Live Load (k/ft)
Required As (in2)
ASD SD
0.5 0.5
0.34
0.34 ( = 1.5 ksi)
0.26
0.26 ( = 1.5 ksi)
1.0 1.0
0.64
0.65 ( = 1.5 ksi)
0.50
0.52 ( = 1.5 ksi)
1.5 1.5
1.94
5.09 ( = 1.5 ksi)
0.77
0.80 ( = 1.5 ksi)
5. - 17 -
Allowable Stress Design: Shear
Section 8.3.5.4
allows design for
shear at /2 from
face of supports.
Design for DL = 1 k/ft, LL = 1 k/ft
Shear at reaction
/2 from face of support
Design shear force
Shear stress
Allowable masonry
shear stress
Suggest that be
used, not .
.
4in.
.
1.17ft
. . . .
11.56k
11.56k
. .
.
9.02k
.
. . .
59.2psi
2.25 2.25 2000psi 50.3psi
- 18 -
Allowable Stress Design: Shear
Use #3 stirrups
Check max
shear stress
Req’d steel stress
Determine for
a spacing of 8 in.
59.2psi
OK
Determine so that no shear reinforcement would be required.
Use a 32 in. deep beam if possible;
will slightly increase dead load.
2 2 2000psi 89.4psi
59.2psi 50.2psi 8.9psi
0.5 ⇒
,
.
. . . . .
. .
0.034in.
.
. . .
23.5in.
- 19 -
Strength Design: Shear
Requirement for shear at
/2 from face of support
is in ASD, not SD, but
assume it applies
Design for DL = 1 k/ft, LL = 1 k/ft
Shear at
reaction
/2 from face of support
Design shear force
Design masonry
shear strength
Suggest that be used,
not to find
.
4in.
.
1.17ft
16.00k
. .
.
12.50k
. . . . . .
16.00k
2.25
0.8 2.25 7.625in. 20in. 2000psi 12.28k
- 20 -
Strength Design: Shear
Use #3 stirrups
Check max
Req’d
Determine for
a spacing of 8 in.
> 12.50k OK
Determine so that no shear reinforcement would be required.
Use inverted
bond beam to get
slightly greater .
4
0.9 4 7.625in. 20in. 2000psi 21.82k
. .
.
0.28k
0.5 ⇒
,
.
. .
. .
0.004in.
.
.
. . . .
20.4in.
6. - 21 -
Non-Bearing Partially Grouted Walls
Out-of-Plane
s
b’
b
d
a
t
f
As
= effective compressive width per bar = min{ , 6 , 72 in} (5.1.2)
A. Neutral axis in flange:
a. Almost always the case
b. Design for solid section
B. Neutral axis in web
a. Design as a T-beam section
C. Often design based on a 1 ft width
Minimum reinforcement:
No requirements
Maximum reinforcement:
ASD: none
SD: Same as for beams
- 22 -
Non-Bearing Partially Grouted Walls
Out-of-Plane
Spacing
(inches)
Steel Area in2/ft
#3 #4 #5 #6
8 0.16 0.30 0.46 0.66
16 0.082 0.15 0.23 0.33
24 0.055 0.10 0.16 0.22
32 0.041 0.075 0.12 0.16
40 0.033 0.060 0.093 0.13
48 0.028 0.050 0.078 0.11
56 0.024 0.043 0.066 0.094
64 0.021 0.038 0.058 0.082
72 0.018 0.033 0.052 0.073
80 0.016 0.030 0.046 0.066
88 0.015 0.027 0.042 0.060
96 0.014 0.025 0.039 0.055
104 0.013 0.023 0.036 0.051
112 0.012 0.021 0.033 0.047
120 0.011 0.020 0.031 0.044
- 23 -
Example: Partially Grouted Wall ASD
Given: 8 in CMU wall; 16 ft high; Grade 60 steel, = 2000 psi; Lateral wind load
of = 30 psf (factored)
Required: Reinforcing (place in center of wall)
Solution:
= 0.45(2000psi) = 900 psi = 1.80 x 106 psi
= 32000 psi = 29 x 106 psi
= / = 16.1
Moment
Determine Assume compression controls
Check if compression
controls
Tension controls
3 3
. . . . .
⋅ .
.
. 0.346in.
. .
. .
0.091 0.312
.
.
6912
⋅ .
576
⋅
- 24 -
Example: Partially Grouted Wall ASD
Equation / Value Iteration 1 Iteration 2 Iteration 3
(in.) 0.346 0.699 0.709
0.091 0.183 0.183
, (in.2) 0.0584 0.0603 0.0603
,
(in.) 0.0784 0.0810 0.0810
2 (in.) 0.699 0.709 0.709
Use # 4 @ 40 inches (As = 0.060in2/ft)
(close enough)
7. - 25 -
Example: Partially Grouted Wall SD
Given: 8 in CMU wall; 16 ft high; Grade 60 steel, = 2000 psi; Lateral wind
load of = 30 psf (factored)
Required: Reinforcing (place in center of wall)
Solution:
Use #4 @ 40 inches (As = 0.060in2/ft)
Factored
Moment
Find
Solve as solid
section
Find
,
.
11520
⋅ .
960
⋅
,
. . .
.
. .
0.0.0573
.
.
3.81in. 3.81in.
.
⋅
.
.
. .
. 0.179in.
- 26 -
ASD vs SD: Flexural Members
ASD calibrated to SD for wind and seismic loads
When a significant portion of load is dead load, ASD will
require more steel than SD
When allowable masonry stress controls in ASD, designs are
inefficient
Advantage to SD, which is reason concrete design rapidly
switched to SD about 50 years ago
- 27 -
ASD: Combined Bending and
Axial Load Design Method
Calculate
Is ?
For Grade 60 steel,
CMU = 0.312
YES
Iterate. Use 2 as new
guess and repeat.
NO
Compression controls Tension controls
⁄
3
2 2
2 2
⁄
3
,
2
1
1
2 3
,
′
1
3
,
2
- 28 -
ASD: Combined Bending and
Axial Load Design Method
If tension controls; determine from cubic equation.
6 2 2 2
0
,
1
2
1
Determination of .
0.45
0.45
32ksi
29000ksi
900
0.45
0.45
32
29000
900
0.312
For clay masonry, 700 , 0.368
8. - 29 -
Strength Design: Combined
Bending and Axial Load
Calculate
Is ?
For CMU, Grade 60 steel
0.547
YES
Compression controls
Tension controls
NO
2 2
⁄
0.8 0.8
,
0.8 ⁄
,
0.8 ⁄
- 30 -
Example: Pilaster Design ASD
Given: Nominal 16 in. wide x 16 in. deep CMU pilaster; = 2000 psi; Grade 60
bar in each corner, center of cell; Effective height = 24 ft; Dead load of 9.6 kips and
snow load of 9.6 kips act at an eccentricity of 5.8 in. (2 in. inside of face); Factored
wind load of 26 psf (pressure and suction) and uplift of 8.1 kips ( = 5.8 in.);
Pilasters spaced at 16 ft on center; Wall is assumed to span horizontally between
pilasters; No ties.
Required: Determine required reinforcing using allowable stress design.
Solution:
Vertical
Spanning
= 11.8 in
x
Load
= 5.8 in 2.0 in
= 1800 ksi
= 16.1
Lateral Load
= 0.6(26psf)(16ft) = 250 lb/ft
Inside
- 31 -
Example: Pilaster Design ASD
Weight of pilaster:
Weight of fully grouted 8 in wall (lightweight units) is 75 psf. Pilaster is
like a double thick wall. Weight is 2(75psf)(16in)(1ft/12in) = 200 lb/ft
Usually the load combination with smallest axial load and largest lateral
load controls. Try load combination of 0.6D + 0.6W to determine
required reinforcement and then check other load combinations.
The location and value of maximum moment can be determined from:
If 0 or ,
= moment at top
measured down from top of pilaster
2 8 2
2
- 32 -
Example: Pilaster Design ASD
0.6D + 0.6W
Top of pilaster
Find axial force at this point. Include weight of pilaster (200 lb/ft).
Design for = 2.3 k, = 218 k-in.
Location of
maximum moment
0.6 9.6k 0.6 8.1k 0.9k
0.9k 5.8in. 5.2k ⋅ in.
. . ⋅ .
.
143.1in.
0.9k 0.6 0.20 143.1in.
.
2.3k
. ⋅ . .
.
. . ⋅ .
.
.
.
218k ⋅ in.
Maximum
moment
10. - 37 -
Example: Pilaster Design SD
Given: Nominal 16 in. wide x 16 in. deep CMU pilaster; = 2000 psi; Grade 60
bar in each corner, center of cell; Effective height = 24 ft; Dead load of 9.6 kips and
snow load of 9.6 kips act at an eccentricity of 5.8 in. (2 in. inside of face); Factored
wind load of 26 psf (pressure and suction) and uplift of 8.1 kips ( = 5.8 in.);
Pilasters spaced at 16 ft on center; Wall is assumed to span horizontally between
pilasters; No ties.
Required: Determine required reinforcing using strength design.
Solution:
Vertical
Spanning
= 11.8 in
x
Load
= 5.8 in 2.0 in
Lateral Load
= (26psf)(16ft) = 416 lb/ft
Inside
- 38 -
Example: Pilaster Design SD
0.9D + 1.0W
At top of pilaster:
Find axial force at this point. Include weight of pilaster.
Find location of
maximum moment
0.9 9.6k 1.0 8.1k 0.54k
0.54k 5.8in. 3.1k ⋅ in.
. . ⋅ .
.
143.7in.
,
. ⋅ . .
.
. . ⋅ .
.
.
.
361k ⋅ in.
Maximum
moment
Design for = 2.7 kips, = 361 k-in.
0.54k 0.9 0.20 143.7in.
.
2.69k
- 39 -
Example: Pilaster Design SD
Determine required area of steel (assuming one layer)
Determine
Determine ,
Try 2 - #5, 4 total, one in each cell
⁄
.
11.8in. 11.8in.
. . . . . ⋅ .
⁄
. . . . .
1.50in.
,
. ⁄
. . . . . . . .
⁄
0.57in.
- 40 -
Example: Pilaster Design SD
11. - 41 -
Example: Pilaster Design
- 42 -
ASD vs SD: Pilaster Design
Similar behavior to before
• ASD and SD close when allowable tension stress controls
• ASD more conservative when allowable masonry stress
controls
• Less reinforcement required with SD due to small dead load
factor
SD design easier as steel has generally yielded
Advantage to SD
- 43 -
Design: Bearing Walls – OOP Loads
Strength Design (Chapter 9)
Allowable Stress
(Chapter 8)
• No second-order
analysis required
• No maximum
reinforcement limits
• Use previous design
procedure
• Second order analysis required
• Slender wall procedure
• Moment magnification
• Second-order analysis
• Need to check maximum
reinforcement limits
• Need to check deflections
- 44 -
Strength Design Procedure
• Assumes simple support conditions.
• Assumes midheight moment is approximately maximum moment
• Assumes uniform load over entire height
• Valid only for the following conditions:
• 0.05 No height limit
• 0.20 height limited by 30
Moment:
factored floor load
factored wall load
Deflection:
12. - 45 -
Strength Design Procedure
Solve simultaneous linear equations:
- 46 -
Strength Design Procedure
Deflection Limit
Calculated using allowable stress load combinations
Cracking Moment:
⁄
⁄
Cracked moment of inertia:
Depth to neutral axis:
.
0.007
- 47 -
Example: Wind Loads ASD
18
ft
32
psf
D = 500 lb/ft
Lr = 400 lb/ft
W = -360 lb/ft
2.67
ft
Given: 8 in. CMU wall; Grade 60 steel; Type S masonry
cement mortar; = 2000 psi; roof forces act on 3 in.
wide bearing plate at edge of wall.
Required: Reinforcement
Solution:
Estimate reinforcement
~
. .
0.778
Assume = 0.95
, = 0.080 in.2/ft
Try #5 @ 48 in. (0.078 in.2/ft)
Cross-section
of top of wall
Determine eccentricity
e = 7.625in/2 – 1.0 in.
= 2.81 in.
wall weight is 38 psf for
48 in. grout spacing
- 48 -
Example: Wind Loads ASD
Load Comb. (kip/ft) (kip/ft) (k-ft/ft) (k-ft/ft) , (in2)
0.6D+0.6W 0.084 0.364 -0.049 0.753 0.068
D+0.6W 0.284 0.751 -0.002 0.777 0.059
Check 0.6D+0.6W
Use #5 @ 48 in. (0.078 in.2/ft)
Although close to #5 @ 56 in. (0.066in.2/ft), a wider spacing also reduces wall weight
Find force at top of wall
Find force at
midheight
Find moment
at top of wall
Find moment
at midheight
0.6 0.5 0.6 0.36 0.084
0.084 0.6 0.040ksf 2.67ft 9ft 0.364
0.084
.
ft 0.6 0.032ksf
.
0.049
⋅
. . .
⋅
0.753
⋅
13. - 49 -
Example: Wind Loads ASD
Sample Calculations: 0.6D+0.6W
1. = 0.312; = 1.19in.
2. Assume masonry controls.
Determine .
Since 0.478 in. < 1.18 in.
tension controls.
3. Iterate to find , .
Equation / Value Iteration 1 Iteration 2
(in.) 0.457 0.791
2
⁄ 3
⁄ (k-ft/ft) 0.1110 0.1076
, (in.2/ft) 0.0658 0.0682
,
(in.) 0.1036
2 (in.) 0.791
3
⁄
3
. . . . .
⋅ .
.
. 0.457in.
For centered reinforcement, 2 0
⁄
- 50 -
Example: Wind Loads SD
18
ft
32
psf
D = 500 lb/ft
Lr = 400 lb/ft
W = -360 lb/ft
2.67
ft
Given: 8 in. CMU wall; Grade 60 steel; Type S masonry
cement mortar; = 2000 psi; roof forces act on 3 in.
wide bearing plate at edge of wall.
Required: Reinforcement
Solution:
Estimate reinforcement
~
.
1.30
= 0.24 in.
, = 0.078 in.2/ft
Try #5 @ 48 in. (0.078 in2/ft)
Cross-section
of top of wall
Determine eccentricity
e = 7.625in/2 – 1.0 in.
= 2.81 in.
- 51 -
Example: Wind Loads SD
Summary of Strength Design Load Combination Axial Forces
(wall weight is 38 psf for 48 in. grout spacing)
Load Combination
(kip/ft) (kip/ft) (kip/ft)
0.9D+1.0W
0.9(0.5)+1.0(-0.36) =
0.090
0.9(0.038)(2.67+9) =
0.399
0.489
1.2D+1.0W+0.5Lr
1.2(0.5)+1.0(-0.36)
+0.5(0.4) = 0.440
1.2(0.038)(2.67+9) =
0.532
0.972
= Factored floor load; just eccentrically applied load
= Factored wall load; includes wall and parapet weight, found
at mid-height of wall between supports (9 ft from bottom)
- 52 -
Example: Wind Loads SD
Modulus of rupture: use linear interpolation between no grout and full grout
Ungrouted (Type S masonry cement): 51 psi
Fully grouted (Type S masonry cement): 153 psi
Cracking moment, :
Commentary allows inclusion of axial load (9.3.5.4.4)
Use minimum axial load (once wall has cracked, it has cracked)
Wall properties determined from NCMA TEK 14-1B Section
Properties of Concrete Masonry Walls
51psi 153psi 68psi
⁄
⁄
.
.
.
.
. .⁄
6.97
⋅ .
0.581
⋅
14. - 53 -
Example: Wind Loads SD
Cracked moment of inertia,
Modular ratio,
Depth to
neutral axis,
16.1
.
.
.
.
.
. 0.334in.
16.1 0.0775
. .
3.812in. 0.334in.
.
. .
16.8
.
- 54 -
Example: Wind Loads SD
Find is the moment at the top support of the wall. It includes
eccentric axial load and wind load from parapet.
= factored moment at top of wall
= factored out-of-plane load on parapet
= height of parapet
0.090
.
ft
. .
0.093
⋅
- 55 -
Example: Wind Loads SD
. .
⋅ .
⋅
.
.
.
.
.
.
.
. .
1.309
⋅
- 56 -
Example: Wind Loads SD
Compare to capacity (interaction diagram)
Depth of stress
block,
Design
moment,
⁄
.
.
.
. .
.
. 0.270in.
0.9
.
.
0.0775
.
60ksi 3.812in.
. .
17.19
⋅ .
1.432
⋅
,
16. - 61 -
Bearing Walls: SD Summary
• Factored Loads ,
• Axial load: include wall weight
• Moment: include second-order effects
• Procedures
• Slender wall
• Moment magnification
• Second-order analysis
• Cross-section properties: , ,
• Material properties: , , ,
• Capacity ,
• Interaction diagram
• Need to only obtain point(s) of interest
• Maximum reinforcement
• Shear
- 62 -
ASD vs SD: Bearing Walls
ASD and SD require approximately the same amount of
reinforcement for reasonably lightly loaded walls
SD requires a second-order analysis and a check of
deflections
Advantage to ASD
• Interestingly OOP loading is where designers often use SD
- 63 -
Shear Walls: Shear
Strength Design (Chapter 9)
0.8
4.0 1.75 0.25
0.5
6 0.25
4 1.0
Interpolate for 0.25 1.0
5 2
Allowable Stress (Chapter 8)
4.0 1.75 0.25
4.0 1.75 0.25
Special Reinforced Shear Walls
0.5
3 ⁄ 0.25
2 ⁄ 1.0
Interpolate for 0.25 ⁄ 1.0
5 2
- 64 -
Shear Walls: Maximum Reinforcement
Strength Design (Chapter 9)
Allowable Stress (Chapter 8)
Special reinforced shear walls having
• ⁄ 1 and
• 0.05
• Provide boundary elements
• Boundary elements not
required in certain cases
• OR
• Limit reinforcement
• Based on strain condition
2
17. - 65 -
SD Maximum Reinforcement
Design with Boundary Elements?
No
Is
1 OR 3 AND 3
AND
0.10 : Geometrically symmetrical walls
0.05 : Geometrically unsymmetrical walls
No boundary
elements
required
Design boundary
elements per TMS
402 Section 9.3.6.6.2
Design with TMS 402 Section 9.3.3.2.
Area of flexural tensile reinforcement ≤ area
required to maintain axial equilibrium under
the following conditions
A strain gradient corresponding to in
masonry and in tensile reinforcement
Axial forces from loading combination D +
0.75L + 0.525QE .
Compression reinforcement, with or
without lateral restraining reinforcement,
can be included.
Is ⁄ 1 ?
= 1.5
Ordinary reinforced walls: = 1.5
Intermediate reinforced walls: = 3
Special reinforced walls: = 4
Yes No
Yes
Yes No
- 66 -
Shear Walls: Shear Capacity Design
Special Reinforced Shear Walls
Strength Design (Chapter 9)
Allowable Stress (Chapter 8)
• Seismic design load required
to be increased by 1.5 for
shear
• Masonry shear stress
reduced for special walls
• Design shear strength, ,
greater than shear
corresponding to 1.25
(increases shear at least
1.39 times)
• Except need not be
greater than 2.5 . (doubles
shear)
- 67 -
ASD: Distributed Reinforcement
Calculate
Is ?
YES
Solve for ,
∗
NO
Compression controls
6
1
3 3
,
∗
1
2
1
2
1
Determine from the quadratic equation
1
3 3 6
6
0
,
∗
1
2 1
1
1
2
1
Tension controls
∗
= distributed reinforcement (in.2/ft)
- 68 -
SD: Distributed Reinforcement
Assume tension controls
2 2
⁄
0.8
,
0.8 ⁄
,
∗ ,
0.65
Approximate required
distributed reinforcement
Approximate 0.9
18. - 69 -
Example: Shear Wall ASD
Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type
S mortar; ′ = 2000psi; superimposed dead load of 1 kip/ft. In-plane
seismic load of 50 kips. 0.5 (just less than 0.5)
Required: Design the shear wall; ordinary reinforced shear wall
Solution: Check using 0.6D+0.7E load combination.
• 0.7 50k 10ft 350k ⋅ ft 4200k ⋅ ft
• Axial load,
• Need to know weight of wall to determine .
• Need to know reinforcement spacing to determine wall weight
• Estimate wall weight as 45 psf
• Wall weight: 45psf 10ft 16ft 7.2k
• 1 k ft
⁄ 16ft 7.2k 23.2k
• 0.6 0.7 0.2 0.53 0.53 23.2k 12.3k
- 70 -
Example: Shear Wall ASD
6
1
3 3
4200k ⋅ in. 12.3k
192in.
6
1
3
192in. 0.90ksi 7.625in. 12.3k
192in.
3
0.0550
Calculate ; for preliminary design purposes use full thickness of wall
Since tension controls. Solve quadratic equation.
1
3 3 6 6
0
1
3
192in. 32ksi
7.625in.
16.11
12.3k
192in.
3
4200k ⋅ in. 12.3k
192in.
6
4200k ⋅ in. 12.3k
192in.
6
0
0.147
- 71 -
Example: Shear Wall ASD
Calculate required area of reinforcement
,
∗
1
2 1
1
1
2
1
1
2
0.147 192in. 7.625in. 32ksi
0.147
1 0.147
1
16.11
12.3k
1
2
1 0.147 32ksi 192in.
0.00934
in.
in.
0.112
in.
ft
Try #5 @ 32 in. (0.120in.2/ft)
Due to module; use 40 in. (0.093in.2/ft) for interior bars
32 in. 32 in.
40 in. 40 in. 40 in.
- 72 -
Example: Shear Wall ASD
Stressed to 93% of
allowable
Used equivalent thickness = 3.57 in.
(40 in. grout spacing)
19. - 73 -
Example: Shear Wall ASD
Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel,
′ =2000psi; #5 at 32in. ends; #5 @ 40in. interior; superimposed dead
load of 1 kip/ft. In-plane seismic load of 50 kips.
Required: Design for shear
Solution: Net area is shown below
2 1.25in. 192in. 6 8in. 7.625in. 2.5in. 726in.
- 74 -
Example: Shear Wall ASD
Shear Stress:
OK
120in.
192in.
0.625
Shear Span:
0.7 50k
726in.
48.2psi
Max Shear:
,
2
3
5 2
2
3
5 2 0.625 2000psi 0.75 83.8psi
Masonry Shear:
1
2
4 1.75 0.25
1
2
4 1.75 0.625 2000psi 0.25
12300lb
726in.
0.75
51.9psi
OK
- 75 -
Example: Shear Walls SD
Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type
S mortar; ′ = 2000 psi; superimposed dead load of 1 kip/ft. In-plane
seismic load of 50 kips. 0.5 (just less than 0.5)
Required: Design the shear wall; ordinary reinforced shear wall
Solution: Check using 0.9D+1.0E load combination.
• 50k 10ft 500k ⋅ ft
• Axial load,
• Need to know weight of wall to determine .
• Need to know reinforcement spacing to determine wall weight
• Estimate wall weight as 45 psf
• Wall weight: 45psf 10ft 16ft 7.2k
• 1 k ft
⁄ 16ft 7.2k 23.2k
• 0.9 0.2 0.80 0.80 23.2k 18.6k
- 76 -
Example: Shear Walls SD
,
0.8 ⁄
0.8 2ksi 7.625in. 3.96in. 18.6k 0.9
⁄
60ksi
0.460in.
, depth
of stress
block
, ,
area of
steel
Estimate d 0.9 0.9 192in. 173in.
,
∗
,
dist. steel
,
∗ ,
0.65
0.460in.
0.65 192in.
12in.
ft
0.044 in. ft
⁄
Try #4 @ 48 in. (0.050in.2/ft)
2 2
⁄
0.8
173in. 173in.
2 18.6k 173in. 192in. 2 6000k ⋅ in.
⁄
0.9 0.8 2000psi 7.625in.
3.96in.
20. - 77 -
Example: Shear Walls SD
-100
0
100
200
300
400
500
600
700
800
0 500 1000 1500 2000 2500
Axial
(kip)
Moment (kip-ft)
Design Strength
Factored Loads Balanced Point
Stressed to 94% of
design strength
Used equivalent thickness = 3.39 in.
(48 in. grout spacing)
- 78 -
Example: Comparison to ASD
‐100
0
100
200
300
400
500
0 200 400 600 800 1000 1200 1400
Axial
(kip)
Moment (k‐ft)
0.7 SD: #4 @ 48 in.
ASD: #5 @ 32,40 in.
SD: 5 - #4
ASD: 6 - #5
- 79 -
Example: Shear Walls SD
Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel,
′ =2000 psi; #4 at 48in.; superimposed dead load of 1 kip/ft. In-plane
seismic load of 50 kips.
Required: Design for shear
Solution: Net area is shown below
2 1.25in. 192in. 5 8in. 7.625in. 2.5in. 685in.
- 80 -
Example: Shear Walls SD
OK
120in.
192in.
0.625
Shear Span:
Max Shear:
,
4
3
5 2
0.8
4
3
5 2 0.625 685in. 2000psi 0.75 91.9kip
Masonry
Shear:
4 1.75 0.25
0.8 4 1.75 0.625 685in. 2000psi 0.25 18600lb 0.75
56.2kip
OK
21. - 81 -
Example: Shear Walls SD
• Calculate axial force based on = 83.8 in.
• Include compression reinforcement
• = 323 kips
• Assume a live load of 1 k/ft
• D + 0.75L + 0.525QE = (1k/ft + 0.75(1k/ft))16ft = 28 kips
Section 9.3.3.2 Maximum Reinforcement
Since / 1, strain gradient is based on 1.5 .
OK
= 0.446(188in.) = 83.8 in.
Strain c/d, CMU c/d, Clay
1.5 0.446 0.530
3 0.287 0.360
4 0.232 0.297
- 82 -
Example: Shear Walls SD
Shear reinforcement requirements in Strength Design
• Except at wall intersections, the end of a horizontal reinforcing bar needed to
satisfy shear strength requirements shall be bent around the edge vertical
reinforcing bar with a 180-degree hook.
• At wall intersections, horizontal reinforcing bars needed to satisfy shear
strength requirements shall be bent around the edge vertical reinforcing bar
with a 90-degree standard hook and shall extend horizontally into the
intersecting wall a minimum distance at least equal to the development length.
- 83 -
Example: Shear Wall ASD
Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type
S mortar; ′ = 2000 psi; superimposed dead load of 1 kip/ft. In-plane
seismic load of 50 kips. 0.5
Required: Design the shear wall; special reinforced shear wall
Solution:
• Flexural reinforcement remains the same (although ASCE 7 allows a
load factor of 0.9 for ASD and special shear walls)
• Design for 1.5V, or 1.5(0.7)(50 kips) = 52.5 kips (Section 7.3.2.6.1.2)
• 52.5k 726in.
⁄ 72.3psi
• Maximum 83.8psi OK
- 84 -
Example: Shear Wall ASD
Use #5 @ 16 in.
Use #5 bars in
bond beams.
Determine
spacing.
Required steel
stress
Masonry Shear:
1
4
4 1.75 0.25
1
4
4 1.75 0.625 2000psi 0.25
12300lb
726in.
36.7psi
,
72.3psi
0.75
36.7psi 59.7psi
0.5 ⇒
0.5
,
0.5 0.31in. 32000psi 192in.
59.7psi 726in.
21.9in.
22. - 85 -
Example: Shear Wall ASD
Masonry Shear:
1
4
4 1.75 0.25
1
4
4 1.75 0.625 2000psi 0.25
15300lb
1464in.
35.1psi
Due to closely spaced bond beams, fully grout wall.
Shear Stress:
52.5k
1464in.
35.9psi
7.625in. 192in. 1464in.
Shear Area:
81psf 10ft 16ft 13.0k
Wall weight:
Dead load: 1 k ft
⁄ 16ft 13.0k 29.0k
Axial load: 0.53 0.53 29.0k 15.3k
- 86 -
Example: Shear Wall ASD
Use #4 bars in
bond beams.
Determine
spacing.
Required steel
stress
,
35.9psi
1.0
35.1psi 0.8psi
0.5 ⇒
0.5
,
0.5 0.20in. 32000psi 192in.
0.8psi 1464in.
524in.
Spacing determined by prescriptive requirements
Maximum Spacing Requirements (7.3.2.6)
• minimum{ one-third length, one-third height, 48 in. }
min
192 .
3
,
120 .
3
, 48 . min 64 . , 40 . , 48 . 40 .
- 87 -
Example: Shear Wall ASD
Use #5 @ 40 in.
Prescriptive Reinforcement Requirements (7.3.2.6)
• 0.0007 in each direction
• 0.002
Vertical: ρ 6 0.31 . 2 1464 . 2
⁄ 0.00127 OK
Horizontal: ρ , max 0.002 0.00127, 0.0007 0.00073
Determine bar size for 40 in. spacing
, 0.00073 7.625 . 40 . 0.22 .
An alternate is #4 @ 32 in. 0.00082
- 88 -
Example: Shear Wall ASD
Section 8.3.4.4 Maximum Reinforcement
Requirements only apply to special reinforced shear walls
No need to check maximum reinforcement since only need to check if:
• / 1 and / 0.625
• 0.05 ′
• 0.05(2000psi)(1464in2) = 146 kips
• Assume a live load of 1 k/ft
• 1.0 0.7 0.2 1.0 0.7 0.2 0.5 29k 16k 47.0k
OK
23. - 89 -
Example: Shear Wall ASD
If we needed to check maximum reinforcing, do as follows.
OK
For distributed reinforcement, the reinforcement ratio is obtained
as the total area of tension reinforcement divided by . For
Assume 5 out of 6 bars in tension.
2
16.1 2ksi
2 60ksi 16.1
60ksi
2ksi
0.00582
5 0.31in.
7.625in. 188in.
0.00108
- 90 -
Example: Shear Walls SD
Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type
S mortar; ′ = 2000 psi; superimposed dead load of 1 kip/ft. In-plane
seismic load of 50 kips. 0.5
Required: Design the shear wall; special reinforced shear wall
Solution:
• Flexural reinforcement remains the same
• Shear capacity design: Design shear strength, , greater than shear
corresponding to 1.25
- 91 -
Example: Shear Walls SD
• For #4 @ 48 in., =18.6 k; = 552 k-ft; = 613 k-ft
• 1.25 = 766 k-ft;
• Design for shear of 76.6 kips
• But wait, need to check load combination of 1.2D + 1.0E
• = [1.2 + 0.2( )] = 1.3 = 30.1 k, = 709 k-ft ,
• 1.25 = 886 k-ft
• Design for shear of 88.6 kips
• But wait, Section 7.3.2.6 has maximum spacing requirements:
• min{1/3 length of wall , 1/3 height of wall, 48 in.} = 40 in.
• Decrease spacing to 40 in.
• = 778 k-ft, 1.25 = 972 k-ft
• Design for shear of 97.2 kips
- 92 -
Example: Shear Walls SD
• Bottom line: any change in wall will change , which will change
design requirement
• Often easier to just use = 2.5 , or = 2.5 = 2.0 .
• Design for shear of 100 kips
24. - 93 -
Example: Shear Walls SD
Shear Area:
(6 - #4 bars)
2 1.25in. 192in. 6 8in. 7.625in. 2.5in. 726in.
Max Shear:
,
4
3
5 2
0.8
4
3
5 2 0.625 726in. 2000psi 0.75 97.4kip
Options:
• Design for shear from 1.25 = 97.2 kips
• Increase to 2100 psi, , 99.8 kips
• requires a unit strength of 2250 psi
• Grout more cells
- 94 -
Example: Shear Walls SD
Masonry
Shear:
4 1.75 0.25
4 1.75 0.625 726in. 2000psi 0.25 18600lb
99.0kip
Required
Steel
Strength: ,
100k
0.8 0.75
99.0k 67.7k
Determine
spacing:
Use #5 bars
0.5 ⇒
0.5
,
0.5 0.31in. 60ksi 192in.
67.7k
26.4in.
Use #5 at 24 in. o.c.
- 95 -
Example: Shear Walls SD
Maximum reinforcement
• = 0.446(188in.) = 83.8 in.
• Calculate axial force based on = 83.8 in. (#4 @ 40 in.)
• Since 1
⁄ , = 1.5
• Include compression reinforcement
• = 298 kips
• Assume a live load of 1 k/ft
• D + 0.75L + 0.525QE = 23.2kip + 0.75(16 kip) = 35.2 kips
OK
- 96 -
Example: Shear Walls SD
• Section 9.3.6.5: Maximum reinforcement provisions of 9.3.3.5 do not
apply if designed by this section (boundary elements)
• Special boundary elements not required if:
OK
0.10 geometrically symmetrical sections
0.05 geometrically unsymmetrical sections
AND
OR AND
For our wall, 1
⁄
0.10 0.10 2ksi 1464in. 293k
1 3
3
25. - 97 -
Example: Shear Walls SD
• Prescriptive Reinforcement Requirements
• 0.0007 in each direction
• 0.002 total
• Vertical: 6(0.20in2)/1464in2 = 0.00082
• Horizontal: 5(0.31in2)/[120in(7.625in)] = 0.00169
• Total = 0.00082 + 0.00169 = 0.00251 OK
- 98 -
ASD vs SD: Shear Walls
SD provides significant savings in overturning steel
• Most, if not all, steel in SD is stressed to fy. Stress in steel in
ASD varies linearly
SD generally requires slightly less shear steel
Advantage to SD
• But, maximum reinforcement requirements can sometimes be
hard to meet; designers then switch to ASD, which requires
more steel. This makes no sense.
- 99 -
ASD vs SD: Final Outcome
Strength
Design
Allowable
Stress
Design
- 100 -
Thank you
Questions
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