1) The document discusses the design of laterally supported flexural steel members. It covers topics like conditions for beams to qualify as laterally supported, modes of failure for beams, and design procedures.
2) An example problem is presented showing the design of a simply supported laterally supported beam carrying a factored point load at midspan. The design is carried out selecting an appropriate I-section, checking shear and bending capacity, and verifying against web buckling and crippling.
3) Key steps in the design of laterally supported beams are outlined, including determining loads, selecting section, classification, checking shear and bending strength, and verifying local failures like web buckling and crippling are
The document discusses flat slab construction and design. It begins by defining a flat slab as a reinforced concrete slab without beams that transfers loads directly to supporting columns. It describes various types of flat slabs including simple flat slabs, those with drop panels or column heads, or both. The document outlines design considerations for flat slabs including analyzing column and middle strips, estimating depth, and calculating moments and shear. It also discusses advantages such as reduced height and construction time. In summary, the document provides information on flat slab types, design methodology, and benefits compared to other construction methods.
This document provides an overview of the design of steel structures. It begins with an introduction discussing common steel structures like industrial buildings, warehouses, stadiums and bridges. It explains that pre-engineered steel buildings are popular due to quick construction. The document then discusses the role of civil engineers in designing steel structures to ensure safety, economy and durability. It provides details on the Indian code for steel design, IS 800:2007, and the limit state design approach. The document further discusses types of steel sections, structural members, design loads, design philosophies like working stress method and limit state method. It concludes with explaining partial safety factors for loads and material resistance in limit state design.
The document provides details on the design procedure for beams. It discusses estimating loads, analyzing beams to determine shear forces and bending moments, and designing beams. The design process involves selecting the beam size and shape, calculating the effective span, determining critical moments and shears, selecting reinforcement, and checking requirements such as shear capacity, deflection limits, and development lengths. An example problem demonstrates designing a singly reinforced concrete beam with a span of 5 meters to support a working live load of 25 kN/m.
1. The nominal resisting moment of reinforced concrete beams with compression steel is calculated as the sum of two parts: the moment due to compression concrete and tensile steel, and the moment due to compression steel and tensile steel.
2. The strain in the compression steel is checked to determine if it has yielded, and then the compression stress is calculated.
3. The analysis procedure involves determining the neutral axis location, checking compression steel yield, and calculating section ductility and design moment strength.
This document provides an overview of reinforced concrete slab bridge design. It discusses the types of reinforced concrete bridges, including slab, beam and slab, arch, box girder, cable-stayed, and integral bridges. It also outlines the loads that must be considered in slab bridge design, including truck, other roadway, sidewalk, and impact loads. Finally, it details the design steps for slab and edge beam components, including calculating bending moments from dead and live loads, determining the effective depth, area of main and distributed reinforcement, and designing the edge beam reinforcement.
This document discusses the design of beams. It defines different types of beams like floor beams, girders, lintels, purlins, and rafters. It describes how beams are classified based on their support conditions as simply supported, cantilever, fixed, or continuous beams. Commonly used beam sections include universal beams, compound beams, and composite beams. The document also covers plastic analysis of beams, classification of beam sections, and failure modes of beams.
this slide will clear all the topics and problem related to singly reinforced beam by limit state method, things are explained with diagrams , easy to understand .
OUTLINE
introduction
classification
loads
materials used
Type of reinforcement
RCC
construction methods in RCC
Analysis and design
Detailing
Basic Rules
Site visit
video
The document discusses flat slab construction and design. It begins by defining a flat slab as a reinforced concrete slab without beams that transfers loads directly to supporting columns. It describes various types of flat slabs including simple flat slabs, those with drop panels or column heads, or both. The document outlines design considerations for flat slabs including analyzing column and middle strips, estimating depth, and calculating moments and shear. It also discusses advantages such as reduced height and construction time. In summary, the document provides information on flat slab types, design methodology, and benefits compared to other construction methods.
This document provides an overview of the design of steel structures. It begins with an introduction discussing common steel structures like industrial buildings, warehouses, stadiums and bridges. It explains that pre-engineered steel buildings are popular due to quick construction. The document then discusses the role of civil engineers in designing steel structures to ensure safety, economy and durability. It provides details on the Indian code for steel design, IS 800:2007, and the limit state design approach. The document further discusses types of steel sections, structural members, design loads, design philosophies like working stress method and limit state method. It concludes with explaining partial safety factors for loads and material resistance in limit state design.
The document provides details on the design procedure for beams. It discusses estimating loads, analyzing beams to determine shear forces and bending moments, and designing beams. The design process involves selecting the beam size and shape, calculating the effective span, determining critical moments and shears, selecting reinforcement, and checking requirements such as shear capacity, deflection limits, and development lengths. An example problem demonstrates designing a singly reinforced concrete beam with a span of 5 meters to support a working live load of 25 kN/m.
1. The nominal resisting moment of reinforced concrete beams with compression steel is calculated as the sum of two parts: the moment due to compression concrete and tensile steel, and the moment due to compression steel and tensile steel.
2. The strain in the compression steel is checked to determine if it has yielded, and then the compression stress is calculated.
3. The analysis procedure involves determining the neutral axis location, checking compression steel yield, and calculating section ductility and design moment strength.
This document provides an overview of reinforced concrete slab bridge design. It discusses the types of reinforced concrete bridges, including slab, beam and slab, arch, box girder, cable-stayed, and integral bridges. It also outlines the loads that must be considered in slab bridge design, including truck, other roadway, sidewalk, and impact loads. Finally, it details the design steps for slab and edge beam components, including calculating bending moments from dead and live loads, determining the effective depth, area of main and distributed reinforcement, and designing the edge beam reinforcement.
This document discusses the design of beams. It defines different types of beams like floor beams, girders, lintels, purlins, and rafters. It describes how beams are classified based on their support conditions as simply supported, cantilever, fixed, or continuous beams. Commonly used beam sections include universal beams, compound beams, and composite beams. The document also covers plastic analysis of beams, classification of beam sections, and failure modes of beams.
this slide will clear all the topics and problem related to singly reinforced beam by limit state method, things are explained with diagrams , easy to understand .
OUTLINE
introduction
classification
loads
materials used
Type of reinforcement
RCC
construction methods in RCC
Analysis and design
Detailing
Basic Rules
Site visit
video
ANALYSIS OF FRAMES USING SLOPE DEFLECTION METHODSagar Kaptan
slope deflection equations are applied to solve the statically indeterminate frames without side sway. In frames axial deformations are much smaller than the bending deformations and are neglected in the analysis.
This seminar discusses plastic analysis, which is used to determine the collapse load of structures. It introduces key concepts like plastic hinges, which form at locations of maximum moment and allow large rotations. The plastic section modulus and shape factor are presented as ways to calculate the moment capacity of a fully yielded cross-section. Common collapse mechanisms like simple beams, fixed beams under uniform and point loads, and propped cantilevers are analyzed using the static method of plastic analysis or virtual work method. Determining collapse loads for various structural configurations is demonstrated through examples.
This document introduces different parts of a truss including joints, members, and reactions. It then provides the essential formulas for determining the stability and determinacy of trusses based on the number of joints (j), members (b), and reactive components (r). The remainder of the document works through examples of truss structures, calculating values for b, r, and j, and determining in each case whether the truss is stable/unstable and determinate/indeterminate based on the formulas.
Introduction-Plastic hinge concept-plastic section modulus-shape factor-redistribution of moments-collapse mechanism.
Theorems of plastic analysis - Static/lower bound theorem; Kinematic/upper bound theorem-Plastic analysis of beams and portal frames by equilibrium and mechanism methods.
This document discusses the design of two-way slabs. It defines a two-way slab as having a ratio of long to short spans of less than 2. The main types of two-way slabs described are flat slabs with drop panels, two-way slabs with beams, flat plates, and waffle slabs. The basic steps of two-way slab design are outlined, including choosing the slab type and thickness, the design method, calculating moments, determining reinforcement, and checking shear strength. Two common design methods are described: the direct design method which uses coefficients, and the equivalent frame method which analyzes frames cut between columns.
This document provides an overview of plastic analysis for structural elements. It discusses key concepts like plastic hinges, plastic section modulus, shape factors, and load factors. Plastic analysis is used to determine the ultimate or collapse load of a structure by considering the redistribution of moments that occurs after sections yield. Common failure mechanisms for determinate and indeterminate beams involve the formation of one or more plastic hinges. Methods for plastic analysis include the static/equilibrium method and kinematic/mechanism method. Examples are given for calculating the collapse load of simple structural configurations using these methods.
1) The document discusses the analysis of flanged beam sections like T-beams and L-beams. It covers topics like effective flange width, positive and negative moment regions, and ACI code provisions for estimating effective flange width.
2) Examples are provided for analyzing a T-beam and an L-beam section. This includes calculating the effective flange width, checking steel strain, minimum reinforcement requirements, and computing nominal moments.
3) Reinforcement limitations for flange beams are also outlined, covering requirements for flanges in compression and tension.
information on types of beams, different methods to calculate beam stress, design for shear, analysis for SRB flexure, design for flexure, Design procedure for doubly reinforced beam,
This document provides an overview of the design of steel beams. It discusses various beam types and sections, loads on beams, design considerations for restrained and unrestrained beams. For restrained beams, it covers lateral restraint requirements, section classification, shear capacity, moment capacity under low and high shear, web bearing, buckling, and deflection checks. For unrestrained beams, it discusses lateral torsional buckling, moment and buckling resistance checks. Design procedures and equations for determining effective properties and capacities are also presented.
Introduction & under ground water tank problemdhineshkumar002
The document discusses the design of an underground rectangular reinforced concrete water tank. It provides steps for calculating earth pressure, determining member thicknesses, and designing reinforcement for the long walls, short walls, and roof slab. The long walls are designed as vertical cantilevers and the short walls as continuous slabs. Reinforcement is checked for bending and cracking stresses. The example shows calculating load intensities, bending moments, required depths and areas of steel for the tank walls and slab according to code specifications.
The document discusses the double integration method for calculating deflections in beams. It introduces the concept of using Macaulay's notation to write the bending moment expression in beams with point loads as a single equation using square brackets. This allows integrating the differential equation of the beam twice to obtain an expression for the deflection throughout the beam with just two integration constants, avoiding multiple equations that would otherwise be needed. Macaulay's notation makes the double integration method more efficient for problems involving point loads.
Circular slabs are commonly used as roofing elements or covers for circular structures. They experience bending stresses like a saucer with tension on the bottom and compression on top when loaded. Analysis of circular slabs is conducted using plate bending theory in polar coordinates, where bending moments are expressed as radial and tangential components. Common support conditions include simply supported, fixed, or partially fixed edges. Reinforcement is typically provided in a rectangular grid oriented for the maximum of the radial and tangential bending moments, with consideration for sign of moments at edges. An example problem demonstrates design and shear check of a circular slab.
This document provides information on the design of reinforced concrete columns, including:
- Columns transmit loads vertically to foundations and may resist both compression and bending. Common cross-sections are square, circular and rectangular.
- Columns are classified as braced or unbraced depending on lateral stability, and short or slender based on buckling resistance. Short column design considers axial load capacity while slender column design accounts for second-order effects.
- Reinforcement details include minimum longitudinal bar size and spacing and design of lateral ties. Slender column design determines loadings and calculates moments from stiffness, deflection and biaxial bending effects. Design charts are used to select reinforcement for columns under axial and uniaxial
This presentation introduces plastic analysis concepts. It discusses stress-strain curves and the difference between elastic and plastic analysis. Key assumptions of plastic analysis are that plane sections remain plane and the stress-strain relationship is identical in compression and tension. Plastic hinges form where the moment equals the plastic moment. Shape factors determine the plastic modulus for different cross-sections. Methods of plastic analysis include static and kinematic approaches. Failure mechanisms involve forming plastic hinges until collapse. Beam examples and problems are provided to demonstrate plastic analysis methods.
The document discusses the design of columns in concrete structures. It covers several topics related to column design including: member strength and capacity versus section capacity, moment magnification, issues regarding slenderness effects, P-Delta analysis, and effective design considerations. The key steps in column design are outlined, including determining loads, geometry, materials, checking slenderness, computing design moments and capacities, and iterating the design as needed. Factors that influence column capacity such as slenderness, bracing, and effective length and stiffness are also described.
1) Two-way slabs are slabs that require reinforcement in two directions because bending occurs in both the longitudinal and transverse directions when the ratio of longest span to shortest span is less than 2.
2) The document discusses various types of two-way slabs and design methods, focusing on the direct design method (DDM).
3) Using the DDM, the total factored load is first calculated, then the total factored moment is distributed to positive and negative moments. The moments are further distributed to column and middle strips using factors that consider the slab and beam properties.
1) Eccentric connections experience both direct axial forces and bending moments due to eccentric loads. This results in more complex stress distributions compared to concentric connections.
2) For bracket connections with eccentric loads, the direct shear stress and bending stress due to the moment must be calculated and combined using the Pythagorean theorem.
3) For welded joints with eccentric loads, both the direct shear stress and bending stress in the weld must be determined and combined, considering the weld geometry, load magnitude and eccentricity. The resultant stress must satisfy allowable stress criteria.
04-LRFD Concept (Steel Structural Design & Prof. Shehab Mourad)Hossam Shafiq II
The document discusses load and resistance factor design (LRFD) methods for structural design. It provides information on:
1) Types of loads that must be considered in design like dead, live, and environmental loads. Load factors are used to increase calculated loads to account for uncertainties.
2) Resistance factors are used to reduce nominal member strength to account for variability in material strength and dimensions.
3) The LRFD method aims for a 99.7% reliability target where factored resistance must exceed factored loads based on load combinations outlined in codes.
The document discusses design provisions for composite steel girder bridge sections from AASHTO LRFD specifications. It covers elastic stresses in composite sections, effective flange widths, web buckling resistance, and provisions for proportioning sections, determining flexural and shear resistance for compact and non-compact sections in positive and negative flexure. Key points include determining depths in compression, classifying sections as compact or non-compact, and calculating nominal flexural and shear resistances considering factors such as web buckling and lateral torsional buckling.
1. Reinforced masonry working stress design of flexural members uses assumptions including plane sections remaining plane after bending and neglecting all masonry in tension.
2. The balanced condition occurs when the extreme fiber stress in the masonry equals the allowable compressive stress and the tensile stress in reinforcement equals the allowable tensile stress.
3. Shear design of reinforced masonry considers mechanisms such as dowel action and the ability of shear reinforcement to restrict crack growth and resist tensile stresses. Allowable shear stresses depend on the presence of shear reinforcement.
ANALYSIS OF FRAMES USING SLOPE DEFLECTION METHODSagar Kaptan
slope deflection equations are applied to solve the statically indeterminate frames without side sway. In frames axial deformations are much smaller than the bending deformations and are neglected in the analysis.
This seminar discusses plastic analysis, which is used to determine the collapse load of structures. It introduces key concepts like plastic hinges, which form at locations of maximum moment and allow large rotations. The plastic section modulus and shape factor are presented as ways to calculate the moment capacity of a fully yielded cross-section. Common collapse mechanisms like simple beams, fixed beams under uniform and point loads, and propped cantilevers are analyzed using the static method of plastic analysis or virtual work method. Determining collapse loads for various structural configurations is demonstrated through examples.
This document introduces different parts of a truss including joints, members, and reactions. It then provides the essential formulas for determining the stability and determinacy of trusses based on the number of joints (j), members (b), and reactive components (r). The remainder of the document works through examples of truss structures, calculating values for b, r, and j, and determining in each case whether the truss is stable/unstable and determinate/indeterminate based on the formulas.
Introduction-Plastic hinge concept-plastic section modulus-shape factor-redistribution of moments-collapse mechanism.
Theorems of plastic analysis - Static/lower bound theorem; Kinematic/upper bound theorem-Plastic analysis of beams and portal frames by equilibrium and mechanism methods.
This document discusses the design of two-way slabs. It defines a two-way slab as having a ratio of long to short spans of less than 2. The main types of two-way slabs described are flat slabs with drop panels, two-way slabs with beams, flat plates, and waffle slabs. The basic steps of two-way slab design are outlined, including choosing the slab type and thickness, the design method, calculating moments, determining reinforcement, and checking shear strength. Two common design methods are described: the direct design method which uses coefficients, and the equivalent frame method which analyzes frames cut between columns.
This document provides an overview of plastic analysis for structural elements. It discusses key concepts like plastic hinges, plastic section modulus, shape factors, and load factors. Plastic analysis is used to determine the ultimate or collapse load of a structure by considering the redistribution of moments that occurs after sections yield. Common failure mechanisms for determinate and indeterminate beams involve the formation of one or more plastic hinges. Methods for plastic analysis include the static/equilibrium method and kinematic/mechanism method. Examples are given for calculating the collapse load of simple structural configurations using these methods.
1) The document discusses the analysis of flanged beam sections like T-beams and L-beams. It covers topics like effective flange width, positive and negative moment regions, and ACI code provisions for estimating effective flange width.
2) Examples are provided for analyzing a T-beam and an L-beam section. This includes calculating the effective flange width, checking steel strain, minimum reinforcement requirements, and computing nominal moments.
3) Reinforcement limitations for flange beams are also outlined, covering requirements for flanges in compression and tension.
information on types of beams, different methods to calculate beam stress, design for shear, analysis for SRB flexure, design for flexure, Design procedure for doubly reinforced beam,
This document provides an overview of the design of steel beams. It discusses various beam types and sections, loads on beams, design considerations for restrained and unrestrained beams. For restrained beams, it covers lateral restraint requirements, section classification, shear capacity, moment capacity under low and high shear, web bearing, buckling, and deflection checks. For unrestrained beams, it discusses lateral torsional buckling, moment and buckling resistance checks. Design procedures and equations for determining effective properties and capacities are also presented.
Introduction & under ground water tank problemdhineshkumar002
The document discusses the design of an underground rectangular reinforced concrete water tank. It provides steps for calculating earth pressure, determining member thicknesses, and designing reinforcement for the long walls, short walls, and roof slab. The long walls are designed as vertical cantilevers and the short walls as continuous slabs. Reinforcement is checked for bending and cracking stresses. The example shows calculating load intensities, bending moments, required depths and areas of steel for the tank walls and slab according to code specifications.
The document discusses the double integration method for calculating deflections in beams. It introduces the concept of using Macaulay's notation to write the bending moment expression in beams with point loads as a single equation using square brackets. This allows integrating the differential equation of the beam twice to obtain an expression for the deflection throughout the beam with just two integration constants, avoiding multiple equations that would otherwise be needed. Macaulay's notation makes the double integration method more efficient for problems involving point loads.
Circular slabs are commonly used as roofing elements or covers for circular structures. They experience bending stresses like a saucer with tension on the bottom and compression on top when loaded. Analysis of circular slabs is conducted using plate bending theory in polar coordinates, where bending moments are expressed as radial and tangential components. Common support conditions include simply supported, fixed, or partially fixed edges. Reinforcement is typically provided in a rectangular grid oriented for the maximum of the radial and tangential bending moments, with consideration for sign of moments at edges. An example problem demonstrates design and shear check of a circular slab.
This document provides information on the design of reinforced concrete columns, including:
- Columns transmit loads vertically to foundations and may resist both compression and bending. Common cross-sections are square, circular and rectangular.
- Columns are classified as braced or unbraced depending on lateral stability, and short or slender based on buckling resistance. Short column design considers axial load capacity while slender column design accounts for second-order effects.
- Reinforcement details include minimum longitudinal bar size and spacing and design of lateral ties. Slender column design determines loadings and calculates moments from stiffness, deflection and biaxial bending effects. Design charts are used to select reinforcement for columns under axial and uniaxial
This presentation introduces plastic analysis concepts. It discusses stress-strain curves and the difference between elastic and plastic analysis. Key assumptions of plastic analysis are that plane sections remain plane and the stress-strain relationship is identical in compression and tension. Plastic hinges form where the moment equals the plastic moment. Shape factors determine the plastic modulus for different cross-sections. Methods of plastic analysis include static and kinematic approaches. Failure mechanisms involve forming plastic hinges until collapse. Beam examples and problems are provided to demonstrate plastic analysis methods.
The document discusses the design of columns in concrete structures. It covers several topics related to column design including: member strength and capacity versus section capacity, moment magnification, issues regarding slenderness effects, P-Delta analysis, and effective design considerations. The key steps in column design are outlined, including determining loads, geometry, materials, checking slenderness, computing design moments and capacities, and iterating the design as needed. Factors that influence column capacity such as slenderness, bracing, and effective length and stiffness are also described.
1) Two-way slabs are slabs that require reinforcement in two directions because bending occurs in both the longitudinal and transverse directions when the ratio of longest span to shortest span is less than 2.
2) The document discusses various types of two-way slabs and design methods, focusing on the direct design method (DDM).
3) Using the DDM, the total factored load is first calculated, then the total factored moment is distributed to positive and negative moments. The moments are further distributed to column and middle strips using factors that consider the slab and beam properties.
1) Eccentric connections experience both direct axial forces and bending moments due to eccentric loads. This results in more complex stress distributions compared to concentric connections.
2) For bracket connections with eccentric loads, the direct shear stress and bending stress due to the moment must be calculated and combined using the Pythagorean theorem.
3) For welded joints with eccentric loads, both the direct shear stress and bending stress in the weld must be determined and combined, considering the weld geometry, load magnitude and eccentricity. The resultant stress must satisfy allowable stress criteria.
04-LRFD Concept (Steel Structural Design & Prof. Shehab Mourad)Hossam Shafiq II
The document discusses load and resistance factor design (LRFD) methods for structural design. It provides information on:
1) Types of loads that must be considered in design like dead, live, and environmental loads. Load factors are used to increase calculated loads to account for uncertainties.
2) Resistance factors are used to reduce nominal member strength to account for variability in material strength and dimensions.
3) The LRFD method aims for a 99.7% reliability target where factored resistance must exceed factored loads based on load combinations outlined in codes.
The document discusses design provisions for composite steel girder bridge sections from AASHTO LRFD specifications. It covers elastic stresses in composite sections, effective flange widths, web buckling resistance, and provisions for proportioning sections, determining flexural and shear resistance for compact and non-compact sections in positive and negative flexure. Key points include determining depths in compression, classifying sections as compact or non-compact, and calculating nominal flexural and shear resistances considering factors such as web buckling and lateral torsional buckling.
1. Reinforced masonry working stress design of flexural members uses assumptions including plane sections remaining plane after bending and neglecting all masonry in tension.
2. The balanced condition occurs when the extreme fiber stress in the masonry equals the allowable compressive stress and the tensile stress in reinforcement equals the allowable tensile stress.
3. Shear design of reinforced masonry considers mechanisms such as dowel action and the ability of shear reinforcement to restrict crack growth and resist tensile stresses. Allowable shear stresses depend on the presence of shear reinforcement.
The origin of the word 'Glulam' comes from the words 'glue' and 'laminated'. Glulam is manufactured by gluing together layers of dimensional lumber or timber boards with structural adhesives to form a structural laminated beam or column. One structural advantage Glulam has over conventional solid timber is that it allows for the manufacture of larger and longer structural members than what could be produced from a single piece of solid timber. An example of a type of structural form that can be constructed from Glulam in buildings is glulam arches.
This document provides instructions for using an anchor bolt schedule. It includes six types of anchor bolts with details like permissible forces, embedment lengths, and plate sizes. Design assumptions are based on working stress method for concrete grades M15 and M20. Examples of typical designs are provided for each bolt type, calculating forces and checking stresses.
Shear Strenth Of Reinforced Concrete Beams Per ACI-318-02Engr Kamran Khan
This document provides a 4 PDH course on the shear strength of reinforced concrete beams per ACI 318-02. It covers topics such as the different modes of failure for beams without shear reinforcement, the shear strength criteria, and calculations for the shear strength provided by concrete. The course content includes introductions to shear stresses in beams, Mohr's circle analysis, beam classifications, and equations for determining nominal shear strength based on the concrete strength and web reinforcement.
This document provides an overview of the design of rectangular reinforced concrete beams that are singly or doubly reinforced. It defines key assumptions in the design process including plane sections remaining plane after bending. It also covers evaluation of design parameters such as moment factors, strength reduction factors, and balanced reinforcement ratios. The design procedures for singly and doubly reinforced beams are described including checking crack width for singly reinforced beams. Figures are also provided to illustrate concepts such as stress distributions and the components of a doubly reinforced beam.
This document discusses the design of circular slabs and slabs with concentrated loads. It begins by introducing circular slabs and their design considerations, such as bending occurring in all directions rather than just two. Reinforcements are provided in two orthogonal directions for simplicity. It then discusses determining bending moments and shear forces in simply supported and fixed edge circular slabs under uniform loads based on formulas. The document also discusses rectangular slabs supported on two edges carrying concentrated loads, including determining the effective load width and formulas to calculate maximum bending moment based on load location. Design examples are provided for both types of slabs.
This document discusses shear stresses and design of shear reinforcement in reinforced concrete beams. It covers key topics such as:
1) The components that resist shear in an RC beam including aggregate interlock, dowel action, and shear reinforcement when used.
2) Design procedures for shear as outlined in ACI 318, including calculating the nominal shear provided by concrete (Vc) and shear reinforcement (Vs), and ensuring the shear capacity exceeds the factored shear demand (Vu).
3) Types and placement of common shear reinforcement such as stirrups. Minimum requirements for stirrups are also covered.
4) An example problem is worked through demonstrating shear design of a simply supported beam section using stirrup
good for engineering students
to get deep knowledge about design of singly reinforced beam by working stress method.
see and learn about rcc structure....................................................
Design of Beam- RCC Singly Reinforced BeamSHAZEBALIKHAN1
Concrete beams are an essential part of civil structures. Learn the design basis, calculations for sizing, tension reinforcement, and shear reinforcement for a concrete beam.
rectangular and section analysis in bending and shearqueripan
The document discusses the design of reinforced concrete beams for bending and shear. It covers the analysis of singly and doubly reinforced rectangular beam sections. Key points covered include the concept of neutral axis, under-reinforced and over-reinforced sections, design of bending reinforcement, design of shear reinforcement including link spacing, and deflection criteria. Worked examples are provided to demonstrate the design of bending and shear reinforcement for rectangular beams.
The document discusses different methods of designing concrete structures, focusing on the limit state method. It describes the limit state method's goal of achieving an acceptable probability that a structure will not become unsuitable for its intended use during its lifetime. The document then discusses stress-strain curves for concrete and steel. It covers stress block parameters and equations for calculating the depth of the neutral axis and moment of resistance for singly reinforced concrete beams. The document concludes by providing examples of analyzing an existing beam section and designing a new beam section.
Study of Steel Moment Resisting Frame with Reduced Beam SectionIJERA Editor
This document summarizes a study on modeling and analyzing a 15-story steel building with and without reduced beam sections (RBS) using time history analysis. The study found that using RBS increased the building's time period by 25% and increased deflections and drifts compared to a building with regular beams. However, base shear was nearly identical between the two buildings. While RBS increased deformations, it also reduced the total steel material needed by about 11.5 tons, providing a cost savings of around $7,480. Therefore, RBS can improve seismic performance by shifting plastic hinging away from beam-column connections while also offering a cost benefit from reduced material.
The document describes the process used by a structural analysis program to design concrete beam flexural reinforcement according to BS 8110-97. The program calculates reinforcement required for flexure and shear. For flexural design, it determines factored moments, calculates reinforcement as a singly or doubly reinforced section, and ensures minimum reinforcement requirements are met. Design is conducted for rectangular beams and T-beams under positive and negative bending.
This document discusses the limit state method for designing reinforced concrete beams. It describes key concepts like limit states, stress-strain curves for concrete and steel, and the parameters used to calculate the depth of the neutral axis and moment of resistance. There are three main types of reinforced concrete beams discussed: singly reinforced, doubly reinforced, and singly or doubly reinforced flanged beams. The document focuses on the design and analysis of singly reinforced beams, providing examples of determining the moment of resistance of a given cross-section, as well as designing a beam to resist a specific bending moment.
A column is a vertical structural member subjected to compression and bending forces. Short columns fail through crushing or splitting, while slender columns fail through buckling. The document provides examples of calculating required reinforcement area and diameter for a short reinforced concrete column. It also provides examples of calculating the critical buckling load of a rod and determining a suitable universal column section for a given load based on its effective length and slenderness ratio.
1. Tension members are structural members subjected to axial pulling forces that cause elongation. Examples include wire ropes, stayed bridge decks, and bottom chords of trusses.
2. The design strength of a tension member must be greater than the factored tensile force and is limited by either yielding of the gross section, rupture of the critical section, or block shear failure.
3. Design of tension members considers the type of cross section, connections, and calculations of design strength based on yield strength, ultimate strength, and factors of safety. Safety is checked against tension, shear, and block shear failure modes.
The document is a past exam paper for a Civil Engineering course assessing design of reinforced concrete elements. It contains multiple choice and long answer questions testing concepts like assumptions in elastic reinforced concrete theory, types of shear failures, purposes of corner reinforcements, definitions of terms like torsional shear and development length, and differences between short and long columns. It also provides design problems for a reinforced concrete beam and one-way slab requiring calculation of reinforcement areas, spacing, shear checks, and live and dead loads.
This document summarizes design considerations for shear in reinforced concrete structures. It discusses shear strength provided by concrete alone (Vc), shear strength provided by shear reinforcement (Vs), and methods for calculating total shear strength (Vn). It also covers requirements for shear reinforcement spacing and minimum amounts. Design aids are presented for calculating shear capacity of beams, slabs, and members under combined shear and torsion.
Similar to Unit 4 Class Notes -2019 pat..pptx (20)
The document discusses sedimentary rocks, describing their formation from weathering and erosion of pre-existing rocks. Sediment is transported via water or wind and sorted by size into gravel, sand, silt, or clay. Sediment is then deposited and may undergo lithification into sedimentary rock through compaction and cementation. The main types of sedimentary rocks are clastic (formed from fragments), chemical (formed from precipitation), and biochemical (containing organic material). Sedimentary structures provide clues about the environment of deposition, and the interpretation of these rocks can reveal information about past plate tectonic settings and conditions. Sedimentary rocks are an important source of non-metallic and energy resources.
Mineralogy is the study of minerals and their properties. Minerals are naturally occurring inorganic solids with distinct chemical compositions and atomic structures. They commonly form rocks, which make up the Earth's crust. A civil engineer must understand minerals and how their composition affects rock properties and strength. Key physical properties studied in mineralogy include form, color, streak, luster, cleavage, fracture, hardness, and specific gravity. These properties aid in mineral identification. Minerals are also classified as rock-forming or ore-forming based on their chemical groups.
This document discusses dimensional analysis and its applications in fluid mechanics. Dimensional analysis uses dimensions and units to develop dimensionless parameters called Pi terms that relate variables in a system. The Buckingham Pi theorem states that any equation with k variables can be written in terms of k-r independent Pi terms, where r is the minimum number of fundamental dimensions needed to describe the variables. Examples show how to identify the relevant Pi terms for problems and how these terms allow experimental data with different scales to be correlated through a single relationship. Dimensional analysis and similitude are useful for modeling prototypes from scaled down models when the key dimensionless groups match between the two.
This document discusses igneous rocks and the processes involved in their formation. Igneous rocks form from the cooling and solidification of magma. Magma forms from the partial melting of rocks in the crust and upper mantle. The nature of magma and factors like cooling rate, silica content, and dissolved gases determine the texture and composition of the resulting igneous rock. Common igneous rocks include granite, basalt, gabbro, rhyolite and obsidian.
Metamorphic rocks form when rocks are altered by heat, pressure, and chemically active fluids. This process is called metamorphism. There are two main types of metamorphism - contact metamorphism near igneous intrusions which alters rocks locally, and regional metamorphism over large areas from tectonic forces which produces foliated rocks. Metamorphic rocks show different textures based on their grade including slate, phyllite, schist, gneiss, and quartzite, and are important construction materials.
This document provides instructions for conducting a California Bearing Ratio (CBR) test to determine the strength of a soil sample. The CBR test measures the resistance of a soil to penetration by a standard plunger and compares it to a standard material. Key steps include: 1) preparing a remolded or undisturbed soil specimen at optimum moisture content and density; 2) soaking the specimen for 4 days to measure swelling; 3) penetrating the specimen at 1.25mm/min while recording load values; and 4) calculating the CBR value by comparing load values to a standard curve. Proper specimen preparation, soaking, loading procedure, and calculations are necessary to obtain reproducible and valid CBR results for evaluating
This document describes the components and use of a vernier theodolite surveying instrument. It discusses the main parts including the horizontal and vertical circles, telescope, and levels. It explains how to measure horizontal and vertical angles, compute latitudes and departures, and adjust a traverse using Bowditch's rule. The document also discusses sources of errors and provides an example problem to calculate latitudes, departures, and closing error for a traverse.
Contours are imaginary lines on a map connecting points of equal elevation. Contour lines represent contours, with the vertical distance between lines being the contour interval. Contour intervals typically range from 0.2-25m depending on the scale and purpose of the map. Contours are characterized by being equally spaced in flat areas and closer together in steep areas. They are interpolated between known elevation points either by estimation or calculation of slope ratios.
This document contains a question bank with multiple choice and short answer questions related to transportation engineering across 6 units of study. The questions cover topics such as road development plans in India, highway and traffic engineering concepts, geometric design of highways, pavement materials and design, railway engineering, and bridge engineering. This question bank appears to be a study guide for a 4th year transportation engineering class, organized by topic unit for exam preparation.
This document provides an overview of railway engineering basics. It discusses the role of transportation in national development and describes the various modes of transportation systems used in India, including roadways, railways, airways, and waterways. It then focuses on railway engineering, explaining components of the permanent way including rails, sleepers, ballast, and fixtures. It also covers rail gauges and zones used by Indian railways, as well as factors related to rail design and construction such as rail joints, creep, and factors affecting gauge selection.
This document provides an overview of railway engineering basics. It discusses the role of transportation in national development and describes the main modes of transportation in India including roadways, railways, airways, and waterways. It then focuses on railway engineering, explaining the components of the permanent way including rails, sleepers, ballast, and fastenings. It also covers common rail gauges used in India such as broad, meter, and narrow gauges, and discusses factors that affect gauge selection.
railway and bridge engineering ppt.pptxRESHMAFEGADE
This document provides information about railway engineering and permanent way components. It discusses:
1. The different modes of transportation and their merits and demerits, including roadways, railways, airways, and waterways.
2. The classification of Indian railways by gauge, including broad, meter, and narrow gauges.
3. The key components that make up the permanent way of a railway track, such as rails, sleepers, ballast, and fixtures.
4. Different types of rail joints used in railway tracks, including supported, suspended, welded, and staggered joints.
Bitumen is a black or dark colored solid or viscous substance obtained from fractional distillation of crude oil. It is used in highway construction due to its binding and waterproofing qualities. There are different types of bitumen including cutbacks, emulsions, and modified binders. Cutbacks are made by blending bitumen with solvents to reduce viscosity, while emulsions involve suspending bitumen droplets in water using emulsifiers. Modified binders involve adding polymers to bitumen to improve properties like temperature resistance. Pavements can be flexible, using bituminous materials in layers, or rigid, using a solid concrete slab.
Literature Review Basics and Understanding Reference Management.pptxDr Ramhari Poudyal
Three-day training on academic research focuses on analytical tools at United Technical College, supported by the University Grant Commission, Nepal. 24-26 May 2024
Presentation of IEEE Slovenia CIS (Computational Intelligence Society) Chapte...University of Maribor
Slides from talk presenting:
Aleš Zamuda: Presentation of IEEE Slovenia CIS (Computational Intelligence Society) Chapter and Networking.
Presentation at IcETRAN 2024 session:
"Inter-Society Networking Panel GRSS/MTT-S/CIS
Panel Session: Promoting Connection and Cooperation"
IEEE Slovenia GRSS
IEEE Serbia and Montenegro MTT-S
IEEE Slovenia CIS
11TH INTERNATIONAL CONFERENCE ON ELECTRICAL, ELECTRONIC AND COMPUTING ENGINEERING
3-6 June 2024, Niš, Serbia
Electric vehicle and photovoltaic advanced roles in enhancing the financial p...IJECEIAES
Climate change's impact on the planet forced the United Nations and governments to promote green energies and electric transportation. The deployments of photovoltaic (PV) and electric vehicle (EV) systems gained stronger momentum due to their numerous advantages over fossil fuel types. The advantages go beyond sustainability to reach financial support and stability. The work in this paper introduces the hybrid system between PV and EV to support industrial and commercial plants. This paper covers the theoretical framework of the proposed hybrid system including the required equation to complete the cost analysis when PV and EV are present. In addition, the proposed design diagram which sets the priorities and requirements of the system is presented. The proposed approach allows setup to advance their power stability, especially during power outages. The presented information supports researchers and plant owners to complete the necessary analysis while promoting the deployment of clean energy. The result of a case study that represents a dairy milk farmer supports the theoretical works and highlights its advanced benefits to existing plants. The short return on investment of the proposed approach supports the paper's novelty approach for the sustainable electrical system. In addition, the proposed system allows for an isolated power setup without the need for a transmission line which enhances the safety of the electrical network
KuberTENes Birthday Bash Guadalajara - K8sGPT first impressionsVictor Morales
K8sGPT is a tool that analyzes and diagnoses Kubernetes clusters. This presentation was used to share the requirements and dependencies to deploy K8sGPT in a local environment.
Introduction- e - waste – definition - sources of e-waste– hazardous substances in e-waste - effects of e-waste on environment and human health- need for e-waste management– e-waste handling rules - waste minimization techniques for managing e-waste – recycling of e-waste - disposal treatment methods of e- waste – mechanism of extraction of precious metal from leaching solution-global Scenario of E-waste – E-waste in India- case studies.
Comparative analysis between traditional aquaponics and reconstructed aquapon...bijceesjournal
The aquaponic system of planting is a method that does not require soil usage. It is a method that only needs water, fish, lava rocks (a substitute for soil), and plants. Aquaponic systems are sustainable and environmentally friendly. Its use not only helps to plant in small spaces but also helps reduce artificial chemical use and minimizes excess water use, as aquaponics consumes 90% less water than soil-based gardening. The study applied a descriptive and experimental design to assess and compare conventional and reconstructed aquaponic methods for reproducing tomatoes. The researchers created an observation checklist to determine the significant factors of the study. The study aims to determine the significant difference between traditional aquaponics and reconstructed aquaponics systems propagating tomatoes in terms of height, weight, girth, and number of fruits. The reconstructed aquaponics system’s higher growth yield results in a much more nourished crop than the traditional aquaponics system. It is superior in its number of fruits, height, weight, and girth measurement. Moreover, the reconstructed aquaponics system is proven to eliminate all the hindrances present in the traditional aquaponics system, which are overcrowding of fish, algae growth, pest problems, contaminated water, and dead fish.
Redefining brain tumor segmentation: a cutting-edge convolutional neural netw...IJECEIAES
Medical image analysis has witnessed significant advancements with deep learning techniques. In the domain of brain tumor segmentation, the ability to
precisely delineate tumor boundaries from magnetic resonance imaging (MRI)
scans holds profound implications for diagnosis. This study presents an ensemble convolutional neural network (CNN) with transfer learning, integrating
the state-of-the-art Deeplabv3+ architecture with the ResNet18 backbone. The
model is rigorously trained and evaluated, exhibiting remarkable performance
metrics, including an impressive global accuracy of 99.286%, a high-class accuracy of 82.191%, a mean intersection over union (IoU) of 79.900%, a weighted
IoU of 98.620%, and a Boundary F1 (BF) score of 83.303%. Notably, a detailed comparative analysis with existing methods showcases the superiority of
our proposed model. These findings underscore the model’s competence in precise brain tumor localization, underscoring its potential to revolutionize medical
image analysis and enhance healthcare outcomes. This research paves the way
for future exploration and optimization of advanced CNN models in medical
imaging, emphasizing addressing false positives and resource efficiency.
ACEP Magazine edition 4th launched on 05.06.2024Rahul
This document provides information about the third edition of the magazine "Sthapatya" published by the Association of Civil Engineers (Practicing) Aurangabad. It includes messages from current and past presidents of ACEP, memories and photos from past ACEP events, information on life time achievement awards given by ACEP, and a technical article on concrete maintenance, repairs and strengthening. The document highlights activities of ACEP and provides a technical educational article for members.
A SYSTEMATIC RISK ASSESSMENT APPROACH FOR SECURING THE SMART IRRIGATION SYSTEMSIJNSA Journal
The smart irrigation system represents an innovative approach to optimize water usage in agricultural and landscaping practices. The integration of cutting-edge technologies, including sensors, actuators, and data analysis, empowers this system to provide accurate monitoring and control of irrigation processes by leveraging real-time environmental conditions. The main objective of a smart irrigation system is to optimize water efficiency, minimize expenses, and foster the adoption of sustainable water management methods. This paper conducts a systematic risk assessment by exploring the key components/assets and their functionalities in the smart irrigation system. The crucial role of sensors in gathering data on soil moisture, weather patterns, and plant well-being is emphasized in this system. These sensors enable intelligent decision-making in irrigation scheduling and water distribution, leading to enhanced water efficiency and sustainable water management practices. Actuators enable automated control of irrigation devices, ensuring precise and targeted water delivery to plants. Additionally, the paper addresses the potential threat and vulnerabilities associated with smart irrigation systems. It discusses limitations of the system, such as power constraints and computational capabilities, and calculates the potential security risks. The paper suggests possible risk treatment methods for effective secure system operation. In conclusion, the paper emphasizes the significant benefits of implementing smart irrigation systems, including improved water conservation, increased crop yield, and reduced environmental impact. Additionally, based on the security analysis conducted, the paper recommends the implementation of countermeasures and security approaches to address vulnerabilities and ensure the integrity and reliability of the system. By incorporating these measures, smart irrigation technology can revolutionize water management practices in agriculture, promoting sustainability, resource efficiency, and safeguarding against potential security threats.
4. Introduction :-
Beams are structural element subjected to transverse load in the plane of bending causing bending
moments and shear forces. Symmetrical section about z-z axis (major axis) are economical and
geometrical properties of such section are available in SP 6. The compression flange of the beam can be
laterally supported (restrained) or laterally unsupported (unrestrained) depending upon weather
restrained are provided or not. The beams are designed for maximum BM and checked for maximum SF,
local effect such as vertical buckling and crippling of webs and deflection.
Section 8/PN 52-69/IS 800:2007 shall be followed in the design of such bending members.
4
5. Type of sections :-
Beams can be of different cross sections depending on the span and loadings are shown in below.
5
6. Functional classification of flexural members
6
Flexural members are labeled in different names, such as
Purlins – Member carrying load from sheeting of roof truss
Girt – Member carrying load from side sheeting
Joist – Light member carrying load from floor in a building
Girder – Mostly member carrying heavy loads
Lintel – Member carrying load over window, door opening
Stringer – Mostly member in bridges aligned in the direction of traffic.
Spandrel – Member on periphery
7. LATERALLY SUPPORTED BEAMS
7
Conditions to qualify as laterally supported beam
Section is symmetric about yy axis
Loading is in plane containing yy axis
For sections unsymmetrical about yy axis load acts through shear center
None of its elements i.e. flange and web should buckle until a desired limit state is
achieved
15. 15
Limit states for beams
• Limit state of flexure
• Limit state of shear
• Limit state of bearing
• Limit state of serviceability
16. Modes of failure of Beams:-
A beam transversely loaded in its own plane can attain its full capacity (Plastic moment) only if local and
lateral instabilities are prevented. Local buckling of beam can be due to web crippling and web buckling.
They are avoided by proper dimensioning of the bearing plate and through secondary design checks.
1. Shear strength of Beams Ref. Cl. No. 8.4/PN 59/IS 800:2007
Vd = Design shear strength of web
γm0
= =
Vn Av fyw
√3 γm0
16
18. 3. Local Failures of Flanges :- Ref. Table. No. 2/PN 18/IS 800:2007
The local failure of flanges reduces the plastic moment capacity of the section due to buckling and is avoided
by limiting the outstand to thickness ratios as given in table 2 of IS 800 : 2007.
18
19. 3. Local Failures of Web :- Ref. Table. No. 2/PN 18/IS 800:2007
The web of a beam is thin and can fail locally at supports or where concentrated loads are acting. There are
two types of web failure
1. Web Buckling
2. Web Crippling
19
20. Web Buckling :
20
Q. State and explain the terms with neat sketch :Web buckling Dec. 2014, Dec. 2015.
The web of the beam is thin and can buckle under reaction and concentrated loads with the web behaving
like a short column fixed at the flanges.
For safety against web buckling, the resisting force shall be greater than the reaction or the concentrated load
is dispersed in to web at 450 as shown in fig.
Fwb = (b1 + n1) tw fcd ≥ Reaction, R
Fwb = Resisting force
tw = Thickness of web
fcd = Design compressive stress in web
b1 = Width of bearing plate
n1 = Width of dispersion
For concentrated loads, the dispersion is on both
side and the resisting force can be expressed as,
Fwd = [(b1 + 2 n1 ) tw fcd ] ≥ Concentrated load, P
R
R must be less than fcd(b1+n1)tw
21. Web Crippling :
Q. State and explain the term with neat sketch :Web crippling Dec. 2014, Dec. 2015 May 2017.
Web crippling causes local crushing failure of web due to large bearing stress under reaction at supports or
concentration loads. This occurs due to stress concentration loads. Web crippling is the crushing failure of the
metal at the junction of flange and web. Web crippling causes local buckling of web at the junction of web and
flange. For safety against web crippling, the resisting Force shall be greater than the reaction or the
concentrated load. It will be assumed that the
reaction or concentrated load is dispersed in
to the web with a slope of 1 in 2.5 as shown in fig.
Fwc = Resisting force
tw = Thickness of web
fyw = Yield stress in web
b1 = Width of bearing plate
n2 = Width of dispersion
Fwc =
(b1 + n2) tw fyw
γm0
≥ Reaction,
R
Fwc
=
(b1 + 2 n2) tw fyw
γm0
≥ Concentrated load, P
R
R
R must be less than (b1 + n2) tw fyw / γm0
21
22. Laterally supported beam using single rolled steel section with and without flange plate:
Q. Explain laterally supported beam with suitable sketch. Dec. 2015, Dec. 2016, May 2017.
Beam subjected to BM develop compressive and tensile force and the flange subjected to compressive
forces has the tendency to deflect laterally. This out of plane bending is called lateral bending or buckling of
beam. The lateral bending of beams depends on the effective span between the restrains, minimum moment of
inertial and its presence reduces the plastic moment capacity of the section.
Beams where lateral buckling of the compression flange are prevented are called laterally restrained
beams. Such continuous lateral supports are provided in two ways
i. The compression flange is connected to an
RC slab throughout by shear connectors
ii. External lateral supports are provided at closer
interval to the compression flange so that it as
good continuous lateral support Ref cl no 8.2.1
Design of such laterally supported beams are
carried out using clauses 8.2.1.2, 8.2.1.3,8.2.1.5,
8.4.8.4.1,8.4.1.1,8.4.2.1, and 5.6.1.
22
23. Design steps for laterally supported beams
1. Determine maximum factored shear force ‘V’ and bending moment ‘M’
for a given loading and support condition
2. Selection of cross section
Find
p
Z required =
M.γm0
fy
Take βb = 1 for assuming plastic or compact section
Using Annex H IS 800 : 2007 plastic properties of section. Select required section.
Or
e
Z required =
𝑍𝑝
23
𝑠ℎ𝑎𝑝𝑒 factor (𝑠)
Shape factor
i. I and channel section – 1.11 to 1.15
ii. Rectangular section – 1.5
iii. Circular section – 1.69
iv. Angle and T – section – 1.8
v. Square or diamond section – 2
∴ Zp provided > Z p required
24. 3. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Using Table no 2 / PN 18 / IS 800 : 2007
4. Design Shear Strength ( vd ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007
Vn
γm0
Av fy
√3 γm0
Vd = = > V ∴ ok & Safe
5. Design bending strength (Md )
i. If V 𝑘 0.6 Vd or V < 0.6 Vd
Ref. cl no. 8.2.1.2 & 8.2.1.3 / PN 53 / IS 800 : 2007
then Md = <
βb .fy ZP 1.2 Ze .fy
.
γm0 γm0
........for simply supported beams
<
1.5 ZP . fy
………….for cantilever beams
γm0
βb = 1 for Plastic and compact section
Where
24
25. ii. If V > 0.6 Vd
then Md = Mdv
Where Md = Design bending strength under high shear.
𝜙 = Md - β (Md - Mfd
γm0
) ≤
1.2 Ze .fy
………For plastic and compact section
Ze .fy
………For semi compact section
γm0
If Md > M ∴ ok & Safe
25
26. 6. Checks
a) Web buckling Ref. cl no 8.7.3.1 / PN 67 / IS 800:2007
Fwb = (b1 + n1) tw fcd ≥ Reaction, R
Fwd = [(b1 + 2 n1 ) tw fcd ] ≥ Concentrated load, P ∴ ok & Safe
Where fcd = is calculated using slenderness ratio
0.7 d
r𝑦
= 2.425 ×
d
t𝑤
and table 9 c
b) Web Crippling Ref. cl no 8.7.4 / PN 67 / IS 800:2007
Fwc =
(b1 + n2) tw fyw
γm0
≥ Reaction,
R
Fwc =
(b1 + 2 n2) tw fyw
γm0
c) Deflection Limit
26
≥ Concentrated load, P ∴ ok & Safe
Ref. Table no 6 / PN 31 / IS 800:2007
∴ ok & Safe
δ actual < δ Limiting
27. Design of laterally supported beams
Example 1 A Simply supported beam of effective span 4 m carries a factored point load of 350 kN mid span.
The section is laterally supported throughout the span. Design cross section using I-section.
SPPU-Dec. 2010, 10 marks
1. Determine maximum factored shear force ‘V’ and bending moment ‘M’
Maximum Shear force
V = Reaction = W
= 175 kN-m
2
Maximum Bending Moment
M = WL
4
= 350 kN-m
2. Selection of cross section
p
Z required =
M.γm0
fy
=
350 ×10 6
×1.10
250
= 1540 × 103 mm3
Or
𝑍
Ze required = 𝑝
=
𝑠ℎ𝑎𝑝𝑒 factor (𝑠)
Select an ISLB 500 @ 75 kg/m
1540 ×10 3
27
1.14
= 1350.8 × 103 mm3
(Using annex ‘H’ IS 800:2007)
d= 500 mm, bf = 180 mm, tf = 14.1 mm, tw = 9.2 mm,
Zp = 1773.7 × 103 mm3 > Zp required , Ze = 1545. 2 × 103 mm3
Izz = 38549 × 104 mm4 ,
28. 3. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 180 /2
14.1
= 6.38 < 9.4 ∈
∴ Flange is plastic
Web = b
tw
=
h1
tw
= 430.2
9.2
= 46.76 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
28
29. 4. Design Shear Strength ( vd ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007
Vn
γm0
Av fy
√3 γm0
Vd = = > V
=
Av fy
√3 γm0
=
500 × 9.2 × 250
√3 × 1.10
= 603.3 kN > V = 175 kN-m
∴ ok & Safe
29
30. 5. Design bending strength (Md ) Ref. cl no. 8.2.1.2 & 8.2.1.3 / PN 53 / IS 800 : 2007
i. If V 𝑘 0.6 Vd or V < 0.6 Vd
0.6 Vd = 361.8 kN
∴ V < 0.6 Vd ∴ Md = .
βb .fy ZP
γm0
Md =
βb .fy ZP
.
γm0
=
1×1773.7 ×10 3
× 250
1.10
=403.11 kN –m > M = 350 kN-m
∴ ok & Safe
30
31. 6. Checks
a) Web buckling
fwb = (b1 + n1) tw fcd ≥ Reaction, R
Ref. cl no 8.7.3.1 / PN 67 / IS 800:2007
fwb = [(b1 + 2 n1 ) tw fcd ] ≥ Concentrated load, P
Where fcd = is calculated using slenderness ratio
0.7 d
r𝑦
= 2.425 ×
d
t𝑤 9.2
= 2.425 × 500
=131.78
Buckling Class ‘c’
By interpolation
cd
f = 74.3 − 74.3−66.2
× (131.78 − 130)
140−130
= 72.85 N/mm2
fcd
At Concentrated load
∴ fwb = (180 + 2 × 250 ) ×9.2 ×72.85
= 455.74 kN > P = 350 kN
At Reaction ‘R’
∴ fwb = (180 + 250 ) ×9.2 ×72.85
= 288.19 kN > R = 175 kN
∴ ok & Safe
∴ ok & Safe
31
32. b) Web Crippling Ref. cl no 8.7.4 / PN 67 / IS 800:2007
Fwc
At concentrated load ‘W’
=
(b1 + 2 n2) tw fyw
γm0
=
(180 +2 ×2.5 ×14.1 × 9.2 ×250
1.10
> Concentrated load, P = 350 kN
= 523.77 kN
c) Deflection Limit
∴ ok & Safe
Ref. Table no 6 / PN 31 / IS 800:2007
working load = 350
1.5
= 233.33 kN
δ actual
WL3
=
48 EI
=
233.33 ×103
×40003
48 ×2×105
× 38549 × 10 4
= 4.03 mm
δ Limiting
= span
300
= 4000
300
32
= 13.33 mm
∴ δ actual δ Limiting
< ∴ ok & Safe
Provide an ISLB 500 @ 75 kg/m beam. … … … … … … … … … … … … … … … … . . A n s
33. Design of laterally supported beams
Example 2 Design a laterally supported simply supported beam of 7 m effective span. It carries a load 0f 250
kN which is uniformly distributed load over the whole span. In addition the beam carries a point load of 100 kN
at mid span. SPPU-Dec. 2010, 10 marks
Solution :
Given : UDL = 250 kN, UDL = 250
= 35.71 kN/m
7
Point load at centre = 100 kN
1. Determine maximum factored shear force ‘V’ and bending moment ‘M’
Maximum Shear force
V = Reaction = 1.5 [ 35.71× 7
+ 100
] = 262.48 kN
2 2
Maximum Bending Moment
2
M = 1.5 [ 35.71 × 7
8 4
+ 100 × 7
] = 590.81 kN-m
2. Selection of cross section
p
Z required =
M.γm0
fy
=
590.81 ×10 6
×1.10
33
250
= 2598.55 × 103 mm3
Select an ISLB 600 @ 99.5 kg/m (Using annex ‘H’ IS 800:2007)
d= 600 mm, bf = 210 mm, tf = 15.5 mm, tw = 10.5 mm,
Zp = 2798.56 × 103 mm3 > Zp required , Ze = 2428.9 × 103 mm3
Izz = 72867.6 × 104 mm4 ,
34. 3. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 210 /2
15.5
= 6.77 < 9.4 ∈
∴ Flange is plastic
Web = d
tw
=
h1
tw
= 520
10.5
= 49.52 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
34
35. 4. Design Shear Strength ( vd ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007
Vn
γm0
Av fy
√3 γm0
Vd = = > V
=
Av fy
√3 γm0
=
600 × 10.5 × 250
√3 × 1.10
= 826.66 kN > V = 262.48 kN-m
∴ ok & Safe
35
36. 5. Design bending strength (Md ) Ref. cl no. 8.2.1.2 & 8.2.1.3 / PN 53 / IS 800 : 2007
i. If V 𝑘 0.6 Vd or V < 0.6 Vd
0.6 Vd = 495.99 kN
∴ V < 0.6 Vd ∴ Md = .
βb .fy ZP
γm0
Md =
βb .fy ZP
.
γm0
=
1×2798.56 ×10 3
× 250
1.10
= 636.04 kN –m > M = 590.81 kN-m
∴ ok & Safe
36
37. 6. Checks
a) Web buckling
fwb = (b1 + n1) tw fcd ≥ Reaction, R
Ref. cl no 8.7.3.1 / PN 67 / IS 800:2007
fwd = [(b1 + 2 n1 ) tw fcd ] ≥ Concentrated load, P
Where fcd = is calculated using slenderness ratio
0.7 d
r𝑦
= 2.425 ×
d
t𝑤 10.5
= 2.425 × 600
=138.57
Buckling Class ‘c’
By interpolation
cd
f = 74.3 − 74.3−66.2
× (138.57 − 130)
140−130
= 67.35 N/mm2
fcd
At Reaction ‘R’
∴ fwb = (210 + 300) ×10.5 × 67.35
= 361.46 kN > R = 262.48 kN
∴ ok & Safe
37
38. b) Web Crippling
At concentrated load ‘W’
Ref. cl no 8.7.4 / PN 67 / IS 800:2007
Fwc
=
(b1 + 2 n2) tw fyw
γm0
=
(210 +2 ×2.5 ×15.5 × 10.5 ×250
1.10
> R = 262.48 kN
= 593.60 kN
c) Deflection Limit
∴ ok & Safe
Ref. Table no 6 / PN 31 / IS 800:2007
δ actual
=
WL3
+ 5 WL4
48 EI 348 EI
= 12.56 mm
= span
= 7000
δ Limiting 300 300
38
= 23.33 mm
∴ δ actual δ Limiting
< ∴ ok & Safe
Provide an ISLB 600 @ 99.5 kg/m beam. … … … … … … … … … … … … … … … … . . A n s
39. Design of laterally supported beams
Example 3 An ISLB 600 @ 99.5 kg/m has been used a simply supported beam over 7.2 m span. Determine the
safe uniformly distributed load ‘W’ so that the beam can carry in flexure. Assuming compressive flange is
restrained throughout the span against lateral buckling and Fe 410 steel.
SPPU-Dec. 2010, 10 marks
Solution :
1. Sectional properties
39
ISLB 600 @ 99.5 kg/m (Using annex ‘H’ IS 800:2007)
d= 600 mm, bf = 210 mm, tf = 15.5 mm,
Zp = 2798.56 × 103 mm3,
tw = 10.5 mm,
Ze = 2428.9 × 103 mm3, h1 = 520.2 mm
40. 2. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 210 /2
14.1
= 6.77 < 9.4 ∈
∴ Flange is plastic
Web = b
tw
=
h1
tw
= 520.2
10.5
= 49.54 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
40
41. 3. Maximum Bending Moment M (factored)
M =
WL2
=
8
1.5 × W × 72002
8
= 9.72 W
4. Bending Strength of section
Md = P y .
γm0
=
βb . Z f 1× 2798.56 ×10 3
× 250
1.10
= 636.03 kN-m
5. UDL ‘W’ equating Md and factored ‘M’
9.72 W = 636.03 kN-m
W = 65.43 kN/m
Now, self weight of section
=
99.5 × 9.81
41
1000
= 0.976 kN/m
Hence, UDL W = 65.43 – 0.976 = 64.45 kN/m
∴ W = 64.45 kN/m … … … … … … … … … A n s
42. Design of laterally supported beams
Example 4 Design a laterally supported beam of effective span 6 m for the following data :
Maximum bending moment M = 150 kN-m
Maximum Shear force V = 210 kN.
Grade of steel Fe 410
Solution :
Given :
SPPU-Dec. 2011, 13 marks, May 2014, 15 marks.
1. Determine maximum factored shear force ‘V’ and bending moment ‘M’
Maximum Shear force = 210 kN
Maximum Bending Moment = 150 kN-m
2. Selection of cross section
p
Z required =
M.γm0
fy
=
150 ×10 6
×1.10
42
250
= 660 × 103 mm3
Select an ISLB 350 @ 49.5 kg/m (Using annex ‘H’ IS 800:2007)
d= 350 mm, bf = 165 mm, tf = 11.4 mm, tw = 7.4 mm, h1 = 288.3 mm ,
Zp = 851.11 × 103 mm3 > Zp required.
43. 3. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 165/2
11.4
= 7.23 < 9.4 ∈
∴ Flange is plastic
Web = d
tw
=
h1
tw
= 288.3
7.4
= 38.95 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
43
44. 4. Design Shear Strength ( vd ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007
Vn
γm0
Av fy
√3 γm0
Vd = = > V
=
Av fy
√3 γm0
=
350 × 7.4 × 250
√3 × 1.10
= 339.84 kN > V = 262.48 kN-m
∴ ok & Safe
44
46. Laterally Unsupported Beam :-
Q. Explain laterally Unsupported beam with suitable sketch. Dec. 2016, May 2017.
Beam subjected to BM develop compressive and tensile force and the flange subjected to compressive
forces has the tendency to deflect laterally. This out of plane bending is called lateral bending or buckling of
beam. The lateral bending of beams depends on the effective span between the restrains, minimum moment of
inertial and its presence reduces the plastic moment capacity of the section.
The value of Mcr can be calculated using the equation given in cl. 8.2.2.1 / PN 54 / IS 800 : 2007
The design of bending compressive strength can be calculated using a set of equation as specified in cl 8.2.2
(Table 13 a and 13 b / PN 54 to 57 / IS 800:2007.
The design of laterally unsupported consist of selecting a section based on the plastic sectional modulus
and checking for its shear capacity & deflection.
46
47. Modes of Failure of Beam :-
Q. Explain modes of failure of beam with suitable sketches. Dec. 2016.
i. Bending Failure : Bending failure may be done due to fracture of tension flange and due to crashing of
compression flange.
ii. Shear failure : It occurs when buckling of web of beam near location of high shear failure.
iii. Torsional buckling: Due to combined moment on the beam buckle about both axis called torsional
buckling.
iv. Lateral buckling : due to large span of beams deflection takes place beyond limits.
v. Local failure of Web : due to heavy shear in some regions of beam web fails in crippling or buckling.
47
48. Design steps for laterally Unsupported beams
1. Determine maximum factored shear force ‘V’ and bending moment ‘M’
for a given loading and support condition
2. Selection of cross section
Find
p
Z required =
Md
βb . fbd
Take βb = 1 for assuming plastic or compact section
Assume, fcd = 120 to 140 N/mm2 for I section
Using Annex H IS 800 : 2007 plastic properties of section. Select required section.
Or
e
Z required =
Zp
48
shape factor (s)
Shape factor
i. I and channel section – 1.11 to 1.15
ii. Rectangular section – 1.5
iii. Circular section – 1.69
iv. Angle and T – section – 1.8
v. Square or diamond section – 2
∴ Zp provided > Z p required
49. 3. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Using Table no 2 / PN 18 / IS 800 : 2007
4. Effective length of beam Ref. 8.3.1 / Table No 15 / PN 58 / IS 800 : 2007
Depending upon support condition, LLT calculated using Table – 15.
5. Design Shear Strength ( vd ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007
Vn
γm0
Av fy
49
√3 γm0
Vd = = > V
∴ ok & Safe
6. Design bending strength (Md )
Md = Zp . βb . fbd
Where, Zp = plastic sectional moduli
fbd = design bending compressive stress
Md > M
∴ ok & Safe
Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007
50. 6. Design bending strength (Md ) Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007
7. Check : Deflection Limit
50
Ref. Table no 6 / PN 31 / IS 800:2007
<
δ actual δ Limiting
∴ ok & Safe
51. Examples on Laterally Unsupported Beams
Example 1 Design a suitable I section for and simply supported beam of span 5 m carrying a dead load of 20
kN/m and imposed load of 40 kN/m. The beam is laterally unsupported throughout the span. Take fy = 250 MPa.
SPPU- Dec. 2010, 15 Marks
Solution : Factored Load = 1.5 ×(20 + 40) = 90 kN/m
1. Determine maximum factored shear force ‘V’ and bending moment ‘M’
i. Maximum shear force :
V = WL
= 90 × 5000
= 225 kN
2 2
ii. Maximum bending force:
8 2
2 2
M = WL
= 90 × 5000
= 281.25 kN-m
2. Selection of cross section
Assuming fbd = 130 N/mm2
p
Z required =
Md
βb . fbd
6
= 281.25 ×10
130
51
= 2163.46 × 103 mm3
Select ISWB 500 @ 95.2 kg/m
d= 500 mm, bf = 250 mm, tf = 14.7 mm,
Zp = 2351.35 × 103 mm3 > Zp required,
tw = 9.9 mm, rmin = 49.6 mm,
= 52290.9 × 104 mm4
Ixx
52. 3. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 250/2
14.7
= 8.50 < 9.4 ∈
∴ Flange is plastic
Web = d
tw
=
h1
tw
= 431
9.9
= 43.53 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
52
53. 4. Effective length of beam
Effective length KL = 5000 mm.
Ref. 8.3.1 / Table No 15 / PN 58 / IS 800 : 2007
As end are restrained against torsion but compression flange is laterally unsupported.
5. Design Shear Strength ( vd ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007
Vn
Vd = =
γm0 √3 γ
Av fy
m0
= 649.51 kN > V = 225 kN
=
500×9.9×250
√3 ×1.1
∴ ok & Safe
6. Design bending strength (Md )
Md = Zp . βb . fbd
Where, Zp = plastic sectional moduli
fbd = design bending compressive stress
Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007
KL = 5000
rmin
49.6
500
53
14.7
= 100.86, h = = 34.01
tf
54. From Table no14 / PN 57 / IS 800 :2007
For
kL
rmin
= 100.86,
h
t𝐹
= 34.01
By interpolation
35−30
A = 270.9 − 270.9 −257.7
× (34.01 − 30) = 260.31 N/mm2
35−30
B = 231.1 − 231.1 −219.3
× (34.01 − 30)
cr, b
f = 260.31 − 260.31 −221.56
× (100.86 − 100)
= 221.56 N/mm2
= 256.97 N/mm2
110−100
= 256.97 N/mm2
fcr, b
54
55. Now, for rolled section αLT = 0.21
fbd = ?
Cl No. 8. 2. 2 / PN 54 / IS 800:2007
Ref. Table no 13a / PN 55 / IS 800:2007
By interpolation
bd
f = 152 +(
300−250
163.6 −152.3
)× (256.57 − 250)
fbd = 153.48 N/mm2
Md= Zp . βb . fbd = 1 × 2351.35 ×103 × 153.48
= 360.88 kN-m > M = 281.25
∴ ok & Safe
55
56. 7) Check : Deflection Limit Ref. Table no 6 / PN 31 / IS 800:2007
δ actual
5 W L4
=
384 EI
=
5 ×60 ×50004
384× 2×105
× 52290.9 ×104
= 4.66 mm
δ Limiting =
span
300
= 5000
300
56
= 16.67 mm
<
δ actual δ Limiting
∴ ok & Safe
Hence, Provide ISWB 500 @ 95.2 kg/m. … … … … … … … … … … … … … … … … … A n s .
57. Examples on Laterally Unsupported Beams
Example 2 A floor beam in a building has a span of 6 m. it is simply supported over supports and carries a
uniformly distributed load of 40 kN/m, inclusive of self weight. Design the beam if the compression flange is
unrestrained throughout the span against lateral buckling and Fe 410 steel. SPPU- Dec. 2011, 15 Mark
Solution : Factored Load = 1.5 × 40 = 60 kN/m
1. Determine maximum factored shear force ‘V’ and bending moment ‘M’
i. Maximum shear force :
V = WL
= 60 × 6000
= 180 kN
2 2
ii. Maximum bending force:
8 2
2 2
M = WL
= 60 ×6000
= 270 kN-m
2. Selection of cross section
Assuming fbd = 130 N/mm2
p
Z required =
Md
βb . fbd
6
= 281.25 ×10
130
57
= 2163.46 × 103 mm3
Select ISWB 500 @ 95.2 kg/m
d= 500 mm, bf = 250 mm, tf = 14.7 mm,
Zp = 2351.35 × 103 mm3 > Zp required,
tw = 9.9 mm, rmin = 49.6 mm,
= 52290.9 × 104 mm4
Ixx
58. 3. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 250/2
14.7
= 8.50 < 9.4 ∈
∴ Flange is plastic
Web = d
tw
=
h1
tw
= 431
9.9
= 43.53 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
58
59. 4. Effective length of beam
Effective length KL = 6000 mm.
Ref. 8.3.1 / Table No 15 / PN 58 / IS 800 : 2007
As end are restrained against torsion but compression flange is laterally unsupported.
5. Design Shear Strength ( vd ) Ref. cl no. 8.4 / PN 59 / IS 800 : 2007
Vn
Vd = =
γm0 √3 γ
Av fy
m0
= 649.51 kN > V = 225 kN
=
500×9.9×250
√3 ×1.1
∴ ok & Safe
6. Design bending strength (Md )
Md = Zp . βb . fbd
Where, Zp = plastic sectional moduli
fbd = design bending compressive stress
Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007
KL = 6000
rmin
49.6
500
59
14.7
= 120.96, h = = 34.01
tf
60. From Table no14 / PN 57 / IS 800 :2007
For
kL
rmin
= 120.96 ,
h
t𝐹
= 34.01
By interpolation
A = 202.4 −
35−30
202.4 −190.1
× (34.01 − 30) = 192.53 N/mm2
35−30
B = 179.0 − 179.0 −167.1
× (34.01 − 30)
cr, b
f = 192.53 − 192.53 −169.45
× (120.96 − 120)
= 169.45 N/mm2
= 190.31 N/mm2
130−120
= 190.31 N/mm2
fcr, b
60
61. Now, for rolled section αLT = 0.21
fbd = ?
Cl No. 8. 2. 2 / PN 54 / IS 800:2007
Ref. Table no 13a / PN 55 / IS 800:2007
By interpolation
bd
f = 106.8 +( 134.1 −106.8
) × (190.31 − 150)
200−150
128.80 N/mm2
fbd =
Md= Zp . βb . fbd = 1 × 2351.35 ×103 × 128.80
= 302.85 kN-m > M = 281.25
∴ ok & Safe
61
62. 7) Check : Deflection Limit Ref. Table no 6 / PN 31 / IS 800:2007
δ actual =
5 W L4
384 EI
=
5 ×40 ×60004
384× 2×105
× 52290.9 ×104
= 6.45 mm
δ Limiting =
span
300
= 6000
300
62
= 20 mm
<
δ actual δ Limiting
∴ ok & Safe
Hence, Provide ISWB 500 @ 95.2 kg/m. … … … … … … … … … … … … … … … … … A n s .
63. Examples on Laterally Unsupported Beams
Example 3 Determine the safe uniformly load that the beam ISLB 600 @ 99.5 kg/m has been used as a simply
supported over 7.2 m span. The compression flange of beam is not restrained against lateral buckling. At the
ends beam is fully restrained in torsion but both the flange are free to warp at the ends.
SPPU- Dec. 2012, 15 Marks, May 2016, Dec. 2016, 16 Marks
Solution :
1. Properties of ISLB 600 @ 99.5 kg/m
A = 126.69 ×102 mm2, D = 600 mm,
63
bf = 210 mm,
tw = 10.5 mm,
h1= 520.2 mm,
Ze = 2428.9 × 103 mm3
tf = 15.5 mm,
rmin = 37.9 mm,
Zp = 2798.56 × 103 mm3,
Ixx = 52290.9 × 104 mm4
64. 2. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 210/2
15.5
= 6.77 < 9.4 ∈
∴ Flange is plastic
Web = d
tw
=
h1
tw
= 520.2
10.5
= 49.54 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
64
66. Now, for rolled section αLT = 0.21
for fcr, b = 92.17 N/mm2 fbd = ?
Cl No. 8. 2. 2 / PN 54 / IS 800:2007
Ref. Table no 13a / PN 55 / IS 800:2007
By interpolation
bd
f = 63.6 +( 70.5 −63.6
) × (89.50 − 80)
90−80
70.15 N/mm2
fbd =
Md= Zp . βb . fbd = 1 × 2798.56 ×103 × 70.15
= 196.32 kN-m
66
67. 4) Safe uniformly distributed load (w)
M = Md
8
wL2
= 196.32
w = 30.30
1.5
67
Safe UDL w = 30.30
= 20.2 kN/m Including self weight.
w = 20.2 kN … … … … … … … … … … … … … … … … … A n s .
68. Examples on Laterally Unsupported Beams
Example 4 A simply supported beam of effective spam 8 m carries uniformly distributed load w kN/m
throughout the span. The compression flange is laterally unsupported throughout the span. Determine intensity
of uniformly distributed load ‘w’ so that ISMB 400 @ 61.6 kg/m provided for beam can carry safely.
SPPU- May 2017, 10 Marks
Solution :
1. Properties of ISMB 400 @ 61.6 kg/m
A = 7846 mm2, d= 400 mm,
68
bf = 140 mm,
tw = 8.9 mm,
h1= 334.4 mm,
tf = 16 mm,
rmin = 28.2 mm,
Zp = 1176.18 × 103 mm3,
69. 2. Classification of Section Ref. Table no 2 / PN 18 / IS 800 : 2007
Flange =
b = bf /2
tf tf
= 140/2
15.5
= 4.375 < 9.4 ∈
∴ Flange is plastic
Web = d
tw
=
h1
tw
= 334.4
8.9
= 37.57 < 84 ∈
∴ Web is plastic.
∴ Section is Plastic
69
70. 3. Design bending strength (Md )
Md = Zp . βb . fbd
From Table no14 / PN 57 / IS 800 :2007
Ref. cl no. 8.2.2/ PN 54 / IS 800 : 2007
= 283.68,
KL = 8000 h = 400
rmin 28.2 tf 16
= 25.
By interpolation
cr, b 290−280
f = 74.7 − 74.7 −71.8
× (283.68 − 280) = 73.63 N/mm2
fcr, b = 73..63 N/mm2
70
71. Now, for rolled section αLT = 0.21
for fcr, b = 73.63 N/mm2 fbd = ?
Cl No. 8. 2. 2 / PN 54 / IS 800:2007
Ref. Table no 13a / PN 55 / IS 800:2007
By interpolation
bd
f = 56.8 +( 63..6 −56.8
) × (73.63 − 70)
80−70
59.27 N/mm2
fbd =
Md= Zp . βb . fbd = 1 × 1176.18 ×103 × 59.27
= 69.71 kN-m
71
72. 4) Safe uniformly distributed load (w)
M = Md
8
wL2
= 69.71
w = 8.71
1.5
72
Safe UDL w = 8.71
= 5.80 kN/m Including self weight.
w = 5.80 kN … … … … … … … … … … … … … … … … … A n s .