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You are now beginning the course. Please turn to Topic 1 and work your way through the sessions. Remember to use the chart in the Course Outline if you are unsure about the next activity or section of work to complete. Note: These learning materials are based on the Australian Standards, published by Standards Australia Australia International Inc. You can access these through the RMIT Library at Standards Australia Online. Before you start this part of the course you are expected to have a basic understanding of the concepts of structural design and the limit state design philosophy. Detailed notes on design of concrete structures using AS3600 is given as part of the study guide.
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You are now beginning the course. Please turn to Topic 1 and work your way through the sessions. Remember to use the chart in the Course Outline if you are unsure about the next activity or section of work to complete. Note: These learning materials are based on the Australian Standards, published by Standards Australia Australia International Inc. You can access these through the RMIT Library at Standards Australia Online. Before you start this part of the course you are expected to have a basic understanding of the concepts of structural design and the limit state design philosophy. Detailed notes on design of concrete structures using AS3600 is given as part of the study guide.
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Reinforced concrete has been a most popular structural material for the last few decades. As explained earlier, concrete is strong in compression and weak in tension. Due to low tensile strength, flexural cracks develop at early stages of loading. In order to reduce the formation of cracks, a concentric or an eccentric force can be imposed in the longitudinal direction of the structural element. This imposing longitudinal force is called a prestressing force. This is a compressive force that prestresses the sections along the span of the structural member in advance of the application of gravity and horizontal loads. The prestressing force reduces cracking in the concrete members as this section is considerably stiffer than the equivalent reinforced section. Prestressing induces internal forces, which are of opposite sign to the external loads. Therefore, they significantly reduce deflection. Prestressed members can be smaller in depth than their reinforced concrete members for the same span and loads. The depth of a prestressed member is usually about 70% of the depth of equivalent reinforced concrete member. Although there are savings in using concrete and reinforcements, specialised prestressing materials and their arrangement add to the cost of prestressed structural members. Formwork is more expensive, as the geometry of prestressed sections is complex. Though there are more initial costs involved, if a large number of precast units are constructed, the difference between them and reinforced concrete sections is not very large. Long-term indirect savings like less maintenance cost, better quality and lightweight structures are achieved. In general, when the the span of a reinforced concrete member exceeds 20 m, it is worthwhile to consider the design option of a prestressed concrete member to avoid the risk of excessive cracking and deflection.
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High-strength steel and high-strength concrete provide economical and structurally efficient solutions to the design of prestressed concrete members. High-strength steel is required for the design of these members as large initial strains are suitable for prestressing concrete. Therefore, the use of highstrength steel is a necessity for the construction and design of prestressed members.
Lea earn rn ing Out co com m es Upon successful completion of this Topic you will be able to:
describe the materials used for prestressed concrete
explain the effects of prestress on beam behaviour
determine the transverse forces caused by the tendons
determine the stresses induced by prestress
determine the stresses induced by applied loads
determine the combined stresses induced by prestress and applied loads.
Background Ba ckground Skil ls and Know ledge
Before you start this Topic you are expected to be able to:
list the materials used in partial and fully prestressed concrete members
calculate the different types of loads on structures
design simple reinforced concrete members without prestressing force
draw stress diagrams of members under combined loading.
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Ses ession sion 1 .1: Met hods of Prest r es essing sing A prestressing force is usually imposed to a concrete member by highly tensioned steel like wire, string or bar. Hydraulic jacks are used as a mechanical system for tensioning the steel. The tensioning operation can be before or after the concrete is cast. If the members are tensioned before the concrete is cast, they are called pre-tensioned members. When the members are tensioned after the concrete is cast, they are called posttensioned members. In this session you will learn the procedure for pre-tensioning and post-tensioning concrete members. The prestressing tendons are initially tensioned between fixed supports and anchored. Then the concrete is cast. When the concrete has reached its required strength, the wires are cut or released from the fixed supports. The steel attempts to contract and as a result, the concrete is compressed. In post-tensioning, the concrete is cast around hollow ducts, which are fixed to any required profile. The steel tendons are unstressed in the ducts during the concrete pour. When the concrete has reached its required strength, the tendons are tensioned. For illustration purposes, look at Figures 1.5 and 1.8 of your textbook. Learning Le arning Out come
Upon successful completion of this session, you will be able to list the materials used in partial and fully prestressed concrete members.
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Activit y 1A Methods of Prestressing Read:
Warner, Rangan, Hall and Faulkes; Concrete Structures , sections 1.3 – 1.4 (pp.4 – 13), 3.1, 3.3.1 - 3.3.4, 6.1, 6.2.1 – 6.2.2 Study: Powerpoint presentation: Week 1: Concrete Structures 2.ppt (22.571 2.ppt (22.571 M)6
Ses ession sion 1.2: Calcul at ion of St St r es esse ses s In this session, you will learn to compute the stresses in a prestressed concrete member due to applied loads and induced forces. Imposing a prestressing force, which satisfies the particular conditions of geometry, and loading of a given member is determined from the principles of mechanics and of stress strain relationships. The stress components on a prestressed concrete section caused by the prestress, self-weight and applied external loads are usually calculated by simple beam theory and assuming material behaviour as elastic. In practice, concrete does not behave like a true linearly elastic material. However, elastic computations are useful, if tensile stresses occur at service loads. Therefore, cracking is likely to occur if compressive stresses are excessive, and large time-dependent shortening is expected. The stress calculations on a cross section caused by all the applied loads plus prestress can be calculated separately and summed to obtain the combined stress distribution at any particular load stage. You can consider first the stresses caused by prestress and ignore all the other loads. Finally, all the stresses are added together to get the combined stresses.
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Learning Out comes
Upon successful completion of this session, you will be able to:
determine the transverse forces caused by the tendons
determine the stresses induced by prestress
determine the stresses induced by applied loads
determine the combined stresses induced by prestress and applied loads.
determine the decompression moment
determine the cracking moment.
Activit y 1B Calculat ion of St resses Read and Review : Warner, Rangan, Hall and Faulkes; Concrete Structures , sections 6.1, 6.2
Powerpoint presentation: Week 1: Concrete Structures 2.ppt (22.571 M) Study Example: Warner, Rangan, Hall and Faulkes; Concrete Structures , 6.1 Develop Solutions: Problems 1 and 2 provided separately
When you have completed the problems turn to the back of this Learning Guide to find solutions.
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Sum m ary and Outcom e Checkli st
In this Topic you learnt that prestressed concrete has been a popular structural material for the last few decades. You learnt that, in order to reduce the formation of cracks, a concentric or an eccentric force is imposed in the longitudinal direction of the structural element. This imposing longitudinal force is called a prestressing force. You also learnt to compute the stresses in a prestressed concrete member due to applied loads and induced forces. You learnt how to impose a prestressing force which satisfies the particular conditions of geometry and loading of a given member determined from the principles of mechanics and of stress strain relationships. Tick the box for each statement with which you agree:
I can describe the materials used for prestressed concrete.
I can explain the effects of prestress on beam behaviour.
I can determine transverse forces caused by the tendons.
I can determine the stresses induced by prestress.
I can determine the stresses induced by applied loads.
I can determine the combined stresses induced by prestress and applied loads.
Assessment This Topic will be assessed in the Final Examination and as is relevant to your work in the Major Design Project.
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In this Topic, you will gain knowledge about analysis and design of concrete bridge structures. In this activity you will be directed to various internet sites to introduce yourself to various types of bridges and their mechanisms. You will learn to use the grillage method of analysis to determine design actions on the bridge structure in particular the design actions acting on the bridge deck and supporting partially prestressed concrete girders. In the Activity you will be asked to determine the loads, calculate the bending moments, shear forces and twisting moments. You will also learn to apply the vehicle load on bridges by using the grillage method of analysis. The Major Design Project of prestressed concrete girders for a bridge of given specification will also be introduced in this topic The Project is to be completed over the duration of the total course and you will apply knowledge gained from Topics 1 to 8. Your deliverables will be to:
establish the critical design loads and load combinations for the bridge
model the bridge deck and supporting girders using the Grillage method
determine design actions for Ultimate Strength Design and Serviceability Design of the bridge deck and girders.
design and detail pre stressed concrete girders to satisfy the AS3600 and AS5100.2 requirements considering: o
strength at transfer and stress limits at full service loads
o
losses of pre stress
o
deflections at service loads
o
flexural strength at ultimate design loads
o
shear strength at ultimate strength design loads.
At the end of this Topic you will be able to complete the first 3 bullet points given above.
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Learn ing Out com es Upon successful completion of this Topic you will be able to:
identify and describe different types of bridge structure
explain the grillage method of analysis of bridge structures
idealise a bridge structure into a number of longitudinal and transverse beam elements in a single horizontal plane, rigidly interconnected at nodes
explain the global structural actions of a girder bridge
model the bridge structure for global analysis
explain the effects and concept of interaction between longitudinal and transverse bending
determine the web line for slab supports
determine the vertical deflection of each web line
model the deflection effects by creating torsionally stiff beam elements
determine the loads under the serviceability condition for ultimate strength design.
Background Skil ls and Know ledge
Before you start this Topic you are expected to be able to:
draw SFDs and BMDs from the given shear forces and bending moments
analyse the beam elements by the stiffness method of analysis
analyse the plane frame by the stiffness method of analysis
evaluate the dead and live loads on bridge structures
use the Bridge Code of Practice for vehicle loads.
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Session 2.1: Types of Br idge St ru ctu res Historically, the two most important materials used in bridge construction were stone and timber. These two materials were common until the nineteenth century. The first reinforced concrete bridge was built in 1875 in France. And the first prestressed concrete bridge was built in 1949 in Belgium. Learning Out come
Upon successful completion of this session you will be able to identify and describe different types of bridge structure In modern days, concrete bridges are generally constructed with partially prestressed girders supporting a deck slab. Full prestressing of girders is often impractical and uneconomical. Research has proved that a combination of non-prestressed and prestressed reinforcement is better than full prestressing. Increased labour costs had led to simplification and rationalisation in bridge construction technology. Arch bridges and other complicated bridge forms are built in exceptional cases. Precast girders help in rapid and economical construction of short-span bridges. Advancements in false work technology proved to be economical for long cast-in-place bridges. For long span bridges, cantilever bridges are preferred over other types of bridges. The first cantilever bridge using precast segments was constructed in early 1960. In England and Australia, segmental bridges were constructed by erecting precast segments onto false work and by prestressing together. This method was used for a 600 m-long flyover in London, in 1961. The Gateway Bridge in Brisbane and its duplicate, now renamed the Sir Leo Hielscher Bridges, on completion of construction in 1986, the 260 m main span of the bridge was a world record for a prestressed concrete free cantilever bridge. It held this record for over 15 years. The box girder is still the largest prestressed concrete, single box in the world, measuring 15 m deep at the pier, with a box width of 12 m and an overall deck width of the 6 lanes of 22 m .
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However, research has now proved that for this much range of span, cable-stayed bridges are more economical and elegant. The Barrios de Luna Bridge in Spain is currently the longest spanning concrete cable-stayed bridge with a main span of 440 m. This bridge was opened to traffic in 1984. The main objectives of bridge design are safety, serviceability, economy and elegance. Safety and serviceability are achieved through the engineer’s analytical and design skills, whereas economy and elegance depend upon research, experience and the creativity of engineers. Design of bridge structures must satisfy the bridge design code of the country in which they are constructed. In general, concrete bridges are divided into two main parts, namely substructure and superstructure. The substructure cost is about 30% and superstructure cost 70% of the bridge’s construction cost. The substructure consists of foundation, piers and abutments. Superstructure costs are divided mainly into construction mechanism, formwork and the cost of materials.
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Activit y 2A Types of Bridge St ruct ures Study:
website: Wikipedia: Types of bridges 7
website: Nova, Superbridge8
Powerpoint presentation: Week 2: Grillage Analysis of Girder Bridges.ppt (650 k)
When studying these websites and others take particular note of: definitions of the various types of bridges explanations of their behaviour under loads descriptions of the types of supports used for bridge structures descriptions of the components of a bridge structure and their function. Develop Solutions: identify various types of bridge structures explain the behaviour of each type of bridge under load describe the types of supports used for bridge structures describe the components of bridge structure and their function.
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http://en.wikipedia.org/wiki/Bridge
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Session 2.2: The Gri llage Met hod and Br idge Desig n Loads In a grillage analysis the structure is idealised into a number of longitudinal and transverse beam elements in a horizontal plane. Beam elements are rigidly connected at nodes. Transverse beams may be orthogonal or skewed with respect to the longitudinal beams. Therefore all types of decks (such as skew, curved or irregular) can be analysed. Learning Out comes
Upon successful completion of this session you will be able to:
explain the grillage method of analysis of bridge structures
idealise a bridge structure into a number of longitudinal and transverse beam elements in a single horizontal plane, rigidly interconnected at nodes
explain the global structural actions of a girder bridge
model the bridge structure for global analysis
explain the effects and concept of interaction between longitudinal and transverse bending
determine the web line for slab supports
determine the vertical deflection of each web line
model the deflection effects by creating torsionally stiff beam elements
determine the loads under the serviceability condition for ultimate strength design.
In a grillage analysis vertical loads are applied at nodes only. Rotations about two horizontal axes, the vertical displacement at each node and the forces (such as bending moment and shear force in beams connected to each node) are determined by matrix analysis.
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The main longitudinal beams are assigned the flexural properties of the full section of each girder including the deck slab. In multi-span girder bridges it is common to consider the slab to be divided into boxes and for the full width of the cantilever to be included in the outer box. This method of idealisation will not provide exact results, but the error is negligible. In the grillage method of analysis the division of the slab into longitudinal elements is not strictly necessary. But the division does help in load distribution. Hence the deck slab can be modelled as transverse elements of rectangular cross-section. The longitudinal edge elements are added to represent the edge beam. These edge beams do not have considerable effect on performance of the bridge structure, but are often helpful in application of load on the cantilever portion. In this method transverse grillage elements can be orthogonal or skewed beams or simply represent a width of slab equal to node spacing when transverse beams are provided the elements should represent the stiffness of the effective transverse member. Torsional rigidity of a member can also be counted in grillage analysis.
Activit y 2B Using Space Gass Space Gass: If you are unfamiliar with Space Gass , turn to the back of this Learning Package and read Using Space Gass for the First Time . Use the guide and the software to become familiar with modelling and analysing structures.
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Bridg e Design Loads
The design of concrete bridge components is governed by the code of practice of a country. The traffic decks and the loads they contain for the design of a bridge superstructure are chosen so that the design is made for the worst possible load case. Loads and design combinations for the bridge design depend upon traffic volume, flood levels, location and aesthetic requirements. In general, economics of a bridge govern the design. The bridge span also plays an important role in determining the cost of bridges. The span of a bridge depends upon site constraints and the type of crossing. If span length variation is allowed, the economical choice will depend on the relative cost of the superstructure and substructure. The initial cost of the bridge also depends upon materials, time to design and construct, methods of construction and fabrication, earthwork and modes of transport used for the bridge. There are some other factors which influence the design of bridges like life-cycle costing, risk, aesthetics and type of bridge. The common type of concrete bridges have prestressed concreted girders and reinforced concrete as deck slab. In this Topic the design loads are calculated in accordance with The Australian Standard for Bridge design Code AS 5100 Part 2 . The Australian Standard AS5100.2 is used as a guideline and considers loads and load effects divided into permanent effects (PE), thermal effects and transient effects. The classification of loads is given in Clause 22.1 of AS5100. For the purpose of this topic and the assignment the permanent effects of structural dead load and any superimposed dead load, and the transient effects of vehicular traffic loads, will only be considered. The dead loads and corresponding load factors are given in Clause 5 of AS5100.2. Road Traffic loads and load factors are given in Clause 6 of AS5100.2. The design life of a bridge is normally assumed 100 years. The bridge has to satisfy ultimate limit states for ultimate strength and serviceability
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To obtain design loads for Serviceability and Ultimate Limit states, the nominal dead loads are multiplied by load factors and the Road traffic loads are multiplied by load factors and dynamic allowances given in the code. If more than one lane in loaded an accompanying load factor is applied to the traffic load as given in the code.
Using Space Gass for Gr ill age Analysis The first step in using Space Gass is to locate its icon on the desktop of your computer, and to click on it.
When you open the software you will have to click on two items that relate to the appropriate use of your student version of Space Gass, and its limitations with respect to the full operating version. The following window will appear
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When you are about to start a new analysis using Space Gass you need to select File, New.
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Before you start any work you should check that the units are appropriate to the problem you are working with. You do this by going to Structure, Units:
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and making any necessary changes in the Units window:
Next, you need to input some headings (Structure, Headings):
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Here are the headings set for this problem via the Headings Input window:
To create the model for grillage analysis the Structures Wizard is used to simulate the bridge deck with one of software’s inbuilt structural system. This is then customised for our particular problem.
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This item is appropriate for our problem (2D grillage)
When you click on the selected structure you will be presented with the detail window – in this case the 2D Grillage. The details for our particular problem have been entered.
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Note that the geometry parameters that you are required to specify are defined in the image at the top left. Note also that the Properties items are defined by the options available at the bottom right. When you have completed the details specifications and clicked on OK, a graph of this item will be inserted into the work area of the main Space Gass window:
You now need to do a series of checks for Node Restraints, Material Properties (especially taking note of Young’s Modulus) and Section Properties
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Enter the section properties for the rectangular slab section and the I shaped prestressed girder here. The Section properties can also be determined by computer. In order to do this you need to use the shape builder function of the software as shown in the next page.
Before analysis can be completed the Loads must be applied to the nodes (Node Loads) and members (Member Concentrated Loads). In this Analysis all the loads are applied as node loads.
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You need to load the Load Case Titles:
You can also define load combinations (Combination Load Cases).
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Now we need to activate Analysis of the problem that we have just defined – both Linear Static Analysis and Non-linear Static Analysis. (Linear Static Analysis only is shown below, but the process is exactly the same for Non-linear Static Analysis. You might need to perform Buckling Analysis also, but this feature is disabled in your student version of the software.)
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When the Analysis components you have selected are completed and you click on OK for each of them, you can check Bending Moment Diagrams, Shear Force Diagrams etc. You can also print out the input and output files, that the maximum of deflections, shear forces and bending moments can then be identified.
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Maximum Ultimate Bending Moment
Maximum Ultimate Shear force
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Activit y 2C The Gril lage Method of Analysis Applied t o Bridges Review:
Powerpoint presentation: Week 2: Grillage Analysis of Girder Bridges.ppt (650 k) Major Design Proj ect : Determine the maximum design actions on the bridge deck and prestressed concrete bridge beams given in the project to be provided by your tutor. Notes: You should continue to work on the major design project throughout the remainder of the course. You can refer to the Assessment schedule for further details on submission etc. This project is to be attempted in groups of three.
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Sum m ary and Outcom e Checkli st
In this Topic you learnt how to use the grillage method of analysis for partially prestressed concrete bridges. You discovered how to determine the loads, calculate the bending moments, shear forces and twisting moments. You also learnt how to apply the vehicle load on bridges by using the grillage method of analysis. Tick the box for each statement with which you agree:
I can identify and describe the different types of bridge structure. I can explain the grillage method of analysis of bridge structures. I can idealise a bridge structure into a number of longitudinal and transverse beam elements in a single horizontal plane, rigidly interconnected at nodes. I can explain the global structural actions of a girder bridge. I can model the bridge structure for global analysis. I can explain the effects and concept of interaction between longitudinal and transverse bending.
I can determine the web line for slab supports.
I can determine the vertical deflection of each web line.
I can model the deflection effects by creating torsionally stiff beam elements. I can determine the loads under the serviceability condition for ultimate strength design.
Assessment This Topic will be assessed in the mid semester test and Final Examination and is relevant to your work in the Major Design Project.
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In Topic 1 you learnt to compute the stresses in a prestressed concrete member due to applied loads and induced forces. Imposing a prestressing force, which satisfies the particular conditions of geometry and loading of a given member is determined from the principles of mechanics and of stress strain relationships. An alternative procedure, called the equivalent load method, can be used to evaluate the effects of prestress. This method provides the basis for a design technique called load balancing which can be used in the preliminary design of the prestressed concrete structures
Learn ing Out com es When you have successfully completed this Topic you will be able to:
determine all the forces exerted on the concrete member by the prestressing tendon using the equivalent load method calculate stresses in a prestressed concrete beam due to a combination of prestress force, self-weight, permanent load and live load by using the equivalent load method determine the prestressing force to balance a system of externally applied design loads determine an appropriate jacking prestress force for a flexural member and satisfy serviceability conditions at transfer. Background Skil ls and Know ledge
Before you start this Topic you are expected to be able to:
explain the effects of prestress on beam behaviour determine the transverse forces caused by the tendons determine the stresses induced by prestress determine the stresses induced by applied loads determine the combined stresses induced by prestress and applied loads. v2 December 2011
Session 3 .1: Equ iv alent Load Concept In this session, you will learn to calculate the stresses in a prestressed concrete member due to applied loads and induced forces using the equivalent load concept. In the equivalent load method all the forces exerted on the concrete member by the prestressing tendon are determined and used to determine the effects of prestress on the beam. These forces are termed equivalent loads and occur at the anchorages and wherever the cable changes direction. Equivalent loads in common cases such as forces at anchorage, draped cables, and cables with curved profile are determined in the Activity 3A. Learning Out comes
When you have successfully completed this session you will be able to, by using the equivalent load method:
determine all the forces exerted on the concrete member by the prestressing
calculate stresses in a prestressed concrete beam due to a combination of prestress force, self-weight, permanent load and live load.
Activit y 3A Equiv alent Load Concept Read: Warner, Rangan, Hall and Faulkes; Concrete Structures , section 6.3 Study:
Powerpoint presentation: Week 3: Equivalent Load Concept in Prestressed Concrete.ppt (149 k) Study Example: Warner, Rangan, Hall and Faulkes; Concrete Structures , 6.2 Develop Solut ion: Problem 3 provided separately. When you have completed the problem turn to the back of this Learning Guide to find solutions. v2 December 2011
Session 3.2 : Load Balancin g Met hod In this session, you will learn the concept of load balancing, which is a simple development from the equivalent load method of analysis discussed in the previous section. One of the most important uses of prestress in the design of concrete structures is to improve the structural behaviour under service load conditions. Determining the equivalent upward load due to prestress can be useful to determine the required prestress force to balance a given set of applied loads. In load balancing the internal equivalent loads are equal and opposite to a system of externally applied design loads. The external load system is then said to be balanced by the prestress. By choosing an appropriate design load to balance, the designer can determine the cable force and cable profile to be used in a flexural member. The balanced load may be the self-weight of the structure, self-weight plus the permanent load or even the total permanent load plus a proportion of the live load. At the balanced load condition there is no cracking and no deflection hence the serviceability requirements for design are almost automatically satisfied. After the prestressing details have been determined by load balancing, serviceability conditions at transfer and under full working load need to be checked and satisfied. Up to this point we have simply used prestressing force, P as constant throughout length of beam. In reality, loss of prestress force in tendon occurs immediately after tendons are stressed and continue to occur throughout the life of prestressed member. In Topic 5 you will be shown how to calculate loss of prestress at different stages, namely; immediate loss – occurs immediately after transfer (includes friction loss) and deferred loss - time-dependent takes place gradually over time. Load balancing calculations are based on effective prestress, Pe after all losses have occurred. Calculations of conditions immediately after transfer are based on initial prestress force, Pi .
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Learning Out comes
When you have successfully completed this session you will be able to:
determine the prestressing force to balance a system of externally applied design loads determine an appropriate jacking prestress force for a flexural member and satisfy serviceability conditions at transfer.
Activit y 3B Det erm ine Prest ressing Det ails Read:
Warner, Rangan, Hall and Faulkes; Concrete Structures , section 6.4 Review: Powerpoint presentation: Week 3: Equivalent Load Concept in Prestressed Concrete.ppt (149 k) Study Example: Warner, Rangan, Hall and Faulkes; Concrete Structures , 6.3 Develop Solut ion: Problem 4 provided separately. When you have completed the problem turn to the back of this Learning Guide to find solutions.
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Sum mary and Out come Checklist
In this Topic you learnt to compute the stresses in a prestressed concrete member due to applied loads and induced forces using the equivalent load method. You were introduced to the load balancing concept, which is a simple development from the equivalent load method of analysis, to provide means for determining the appropriate prestress details for serviceability design situations. Tick the box for each statement with which you agree:
I can find all the forces exerted on the concrete member by the prestressing tendon (equivalent load method). I can find prestressing details (load balancing method).
Assessment This Topic will be assessed in the mid-semester test and Final Examination and as is relevant to your work in the Major Design Project.
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In all our previous calculations on prestressed concrete beams we’ve used the gross cross sectional area and gross second moment of area of the section. In doing that we assume that the cross section is uncracked (i.e. tension at any point in the section doesn’t exceed cracking stress of the concrete). In normal designs, we allow prestressed concrete beams to be cracked at service loads, although not as severely as in reinforced concrete. Beams which are designed to crack at service loads are known as partially prestressed beams. Most prestressed beams are designed as partially prestressed beams. In this Topic you will learn how to evaluate the post-cracking behaviour of prestressed concrete sections. This situation occurs when the tensile stresses produced by the external moment at a particular section overcomes the compression caused by the prestressing force. As the prestressed section starts to crack the resisting or cracking moment of the section decreases with time due to creep and shrinkage. It is important to note that if the applied moment is greater than the cracking moment, cracking will occur. The post cracking behaviour of cracked sections can be calculated by using simple elastic analysis which is based on the following assumptions:
plane sections remain plane and strain distribution is linear over the depth of the section
a perfect bond exists between the concrete and steel including prestressing
analysis is based on short-term behaviour.
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Background Skil ls and Know ledge
Before you start this Topic you are expected to be able to:
explain the effects of prestress on beam behaviour
determine the transverse forces caused by the tendons
determine the stresses induced by prestress
determine the stresses induced by applied loads
determine the combined stresses induced by prestress and applied loads.
Learn ing Out com es When you have successfully completed this Topic you will be able to:
apply the above assumptions in the analysis of prestressed sections
determine the concrete and steel strains
determine the concrete and steel forces of the cracked cross-section
apply force and moment equilibrium equations
calculate the depth of the neutral axis, dn
calculate the moment corresponding to the sectional strains and stresses.
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Activit y 4A Post Cracking Behaviou r of Prestr essed Concret e Beams in Flexur e Read:
Warner, Rangan, Hall and Faulkes; Concrete Structures , section 6.5 Study: Powerpoint presentation: Week 4: Post Cracking Behaviour of Prestressed Concrete Beams.ppt (111 k) Study Example: Warner, Rangan, Hall and Faulkes; Concrete Structures , 6.4 Develop Solut ion: Problem 5 provided separately When you have completed the problem turn to the back of this Learning Guide to find solutions
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Sum m ary and Outcom e Checkli st In this Topic you learnt to calculate the post cracking behaviour of cracked sections by using simple elastic analysis.
Tick the box for each statement with which you agree:
I can determine the concrete and steel strains. I can calculate the concrete and steel forces of the cracked cross-section. I can apply force and moment equilibrium equations to calculate the depth of the neutral axis, dn and the corresponding sectional moment of the cracked crosssection.
Assessment This Topic will be assessed in the mid-semester test and Final Examination and as is relevant to your work in the Major Design Project.
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It is a well-known fact that the initial prestressing force applied to the concrete undergoes some immediate and time dependent losses. Time dependent losses undergo a progressive process of reduction over a period of significant time. Therefore, it is important to determine the level of prestressing force at each loading stage, from the stage of transfer to the full service loads. The reduction in the prestressing force can be in two stages: 1. Transfer Stage
This is an immediate elastic loss during the construction including elastic shortening of the concrete, anchorage losses and frictional losses. 2. Service Load Stage This is a time-dependent loss, such as creep, shrinkage and those due to temperature effects and steel relaxation. These losses are represented in Figure 5.1.
Figure 5.1: The relationship between immediate and time-dependent losses, copyright RMIT University
An accurate calculation of these losses is not feasible. Different countries’ codes give different empirical recommendations to estimate these losses. A very high degree of accuracy is not desired in determination of losses. Consequently, a lump-sum estimate of losses is more realistic and practical. In this Topic, you will learn to calculate the prestress losses.
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Learn ing Out com es Upon successful completion of this Topic you will be able to:
define and differentiate the types of losses in prestressed concrete structural elements determine elastic deformation losses in pre-tensioned and post-tensioned members determine friction losses along the tendon determine anchorage losses determine creep and shrinkage losses of a prestressed concrete member determine relaxation of steel of a prestressed concrete member. Background Skil ls and Know ledge
Before you start this Topic you are expected to be able to:
calculate the stresses caused by the prestressing force without prestress losses in pre-tensioned members determine the stresses in a prestressed concrete member at transfer stage determine the stresses in a prestressed concrete member under full service load conditions evaluate the effects of a prestressing force on a concrete member. As explained earlier, there are two types of losses. Immediate losses after transfer, such as elastic deformation, friction losses and anchorage losses which are calculated separately for pretensioned and post-tensioned members. Different codes of practice from various countries have given different empirical methods to determine these losses for pre-tensioned and posttensioned members. The other types of losses in prestressed concrete members are time-dependent losses. These losses include shrinkage, creep and relaxation of steel. Creep and shrinkage losses are after transfer in both pre-tensioned and post-tensioned members, whereas relaxation of steel in pre-tensioned members can be before and after transfer. In the case of post-tensioned members, this loss is after transfer of the prestressing force.
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Activit y 5A Types of Losses in Prestressed Concrete (1) Read:
Warner, Rangan, Hall and Faulkes; Concrete Structures , section 10.1, 10.2.1 – 10.2.3, 10.3.1 – 10.3.3 Study: Powerpoint presentation: Week 5: Losses of Prestress.ppt (408 k)
Loss of Prestress Example
(based on Warner worked example 10.1) Losses are to be determined for a simply supported posttensioned beam which spans 25 m and has the cross-section shown in Figure 5.2. The beam carries no permanent load other than its weight. Prestressing steel is contained in two cables each consisting of 18 super grade three wire low relaxation strands of 12.7 mm diameter. For each cable, the area is 1802 mm2. The cables have parabolic profiles with eccentricities of zero at each end and 396 mm at midspan. The concrete strength f c' 45 MPa . Stressing takes place when the concrete is at age 14 days with a strength of 356 MPa. The beam is near a coastal area.
v2 December 2011
pi 1100 MPa
A p 3604 mm 4 two cables As 8000 mm2 A g 570,000 mm 2 I g 3.48 1010 mm4 Gc 29,600 MPa f c' 45 MPa Figure 5.2: Loss of prestress example, copyright RMIT University (Geoff Marchiori)
Eccentricity cable @ mid span 396 mm Stressing takes place @ 14 days lcp 35 MPa …. Beam exposed near – costal area Immediate losses Elastic Loss
The concrete compressive stress at the steel level at mid-span, due to the stressing of one cable is: Pi Pe where Pi pi Ap (one cable) 1100 1802 i A g I g e 1100 1802 1100 1802 396 2 570, 000 3.48 10 10 3.478 8.93 12.4 MPa
ci
The loss of prestress in the first cable when the second cable is stressed is:
p ci
E p Ec
12.4
190,000 80 MPa 29,600
The first cable could thus be stressed to 1180 MPa to allow for concrete elastic compression.
v2 December 2011
Friction loss – Tendon
It is assumed that jacking is carried out at one end only (from Cl 3.4.2.4 AS3600 – 2009). p 0.016 rad / m
(for sheathing with tendons of internal diameter 50 mm ) At 0.20 (zinc-coated flat metal ducts) mid-span9 tot
4h 4 675 279 0.063 rad L 25,000
pa pj e
u tot p Lpa
pj e 0.2 0.0630.01612.5 0.95 pj
At dead end10 tot 2 0.063 0.126 rad pa pj e
0.2 0.126 0.01625
0.91 pj
Anchorage Slip
This is obtained from manufacturer’s information. For example, for VSL multi-strand systems a draw-in loss of 6 mm is given. However, at the critical mid-span section, this loss will be neglected. Deferred losses Shrinkage loss
Design shrinkage strain cs : cs cse csd Cl 3.1.7.2(1) AS3600 - 2009
Autogenous shrinkage strain cse : * cse cse 1.0 e 0.1
t
Cl 3.1.7.2(2)
where t is the time (in days) after setting. Take t = 30 years expected life-span of bridge = 10,950 days
9
L pa 12.5 m
10
L pa 25 m v2 December 2011
* Final autogenous shrinkage strain cse : * cse 0.06 f c' 1.0 50 10 6
Cl 3.1.7.2(4)
0.06 45 1.0 50 10 6 85 10 6 cse 85 10 6 1.0 e
0.110950
85 106
Drying shrinkage strain csd : * csd k1 k 4 csd .b
Cl 3.1.7.2(3)
k1 1.14 figure 3.1.7.2 AS3600 - 2009
for t 10950 days th 2 A g 2 570000 238 mm 2 1500 900 U e k4 0.5 near-coastal area c*sd. b 900 10 6 (Melbourne)
csd 1.14 0.5 900 10 6 513 10 6 The design shrinkage strain is: cs cse csd 85 513 10 598 10 6
6
Loss of prestress due to shrinkage:
p cs
E p
Cl.3.4.3.2
AS3600 – 2009
modification to allow for reinforcement effect 1 15
As 15 8000 1 1.21 A g 570000
p 598 106
190000 93.9 MPa 94 MPa 1.21
Creep loss
Creep strain in concrete cc 0.8 cc ci Ec Cl 3.1.8.1
AS3600 – 2009
ci = sustained stress in concrete at level of tendons
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Assuming that only sustained load is the self-weight = 13.7 kN/m the mid-span bending moment due to sustained load: 13.7 252 Msw 1070 kN / m 8 The stress in concrete at the level of prestressing steel due to prestress and sustained service load.
Figure 5.3: The stress in concrete at the level of prestressing steel, due to prestress and sustained service load, copyright RMIT University (Geoff Marchiori)
Pi Pi e 2 at prestress level A I g g Pi 1802 1100 2 3965 kN The resultant stress at the level of prestressing
3965 10 3 3965 103 3962 1070 106 396 ci 34.8 10 9 34.8 109 570000 24.8 12.2 12.6 MPa Design creep co-efficient: cc cc. b k2 k 3 k 4 k 5 Cl 3.1.8.3 cc .b 2.6
(+b 3.1.8.2) t 30 years t h 238 mm k2 1.143 (Figure 3.1.8.3(A)) k3 0.9 (Figure 3.1.8.3(B)) k4 0.5
k5 1.0 f c' 50 MPa cc 2.6 1.143 0.9 0.5 1.0 1.34
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The creep strain:
ci 12.6 0.8 1.34 29600 0.00046 E c
cc 0.8 cc
Hence loss of prestress due to creep of concrete:
p cc Ep 0.00046 190000 87.4 MPa sum of the losses due to creep and shrinkage: p 94 87.4 181 MPa per cent loss
181 100 16.5% 1100
Tendon stress relaxation
For low relaxation strand take Rb 2% Design relaxation of a tendon R is R k4 k5 k6 Rb Cl 3.3.4.3 AS3600 – 2009
k4 log 5.4 j
1
6
log 5.4 10950
1
6
1.4
pi 1100 0.59 for f p 1870 f pb characteristic minimum breaking strength (Cl 3.3.1) 1870 MPa k5 0.7
k6 1.0 Hence: R 1.4 0.7 1.0 2 1.96% 2% Summary of percentage losses @ mid-span section:
Elastic: 40/1100m
Friction: 5%
Creep and shrinkage: 16.5%
Relaxation: 2%
≈
4%
v2 December 2011
Activit y 5B Types of Losses in Prestressed Concrete (2) Review
Powerpoint presentation: Week 5: Losses of Prestress.ppt (408 k) Develop Solut ion: Problem 6 provided separately. When you have completed the problem turn to the back of this Learning Guide to find solutions.
v2 December 2011
Sum mary and Out come Checklist
In this Topic you learnt that the initial prestressing force applied to the concrete undergoes some immediate and time dependent losses. Time dependent losses undergo a progressive process of reduction over a period of approximately five to six years. Therefore, it is important to determine the level of prestressing force at each loading stage, from the stage of transfer to the full service loads. The reduction in the prestressing force can be in two stages – the transfer stage and the service load stage. Tick the box for each statement with which you agree:
I can define and differentiate the types of losses. I can determine elastic deformation losses in pre-tensioned and post-tensioned members. I can determine friction losses along the tendon. I can determine anchorage losses. I can determine creep and shrinkage losses of a prestressed concrete member. I can determine relaxation of steel of a prestressed concrete member.
Assessment This Topic will be assessed in the Final examination and is relevant to your work in Major Design Project.
v2 December 2011
Serviceability of prestressed concrete members in their deflection and cracking behaviour is an important criterion similar to reinforced concrete sections. Prestressed concrete members are even more sensitive to deflection and post-cracking behaviour requirements, as these beams are more slender than simple reinforced concrete beams.
Learn ing Out com es When you have completed this Topic you will be able to:
list the serviceability requirements for a flexural member
set limits on stress in concrete both at transfer and under full service loads
determine an appropriate prestress force for a flexural member subjected to a given service load
determine eccentricity force for a flexural member
check stress limits
calculate short-term and long-term deflection of a prestressed concrete member under serviceability conditions.
Background Skil ls and Know ledge
Before you start this Topic you are expected to be able to:
determine the combined stresses induced by prestress and applied loads in an uncracked prestressed concrete member
calculate the stresses in a crack prestressed concrete members
determine immediate and deferred losses in prestressed concrete members.
v2 December 2011
Serviceability requirements determine the level of prestress and layout of the tendons in design of theses members. For satisfying these requirements, a reasonably accurate estimate of the magnitude of prestress is needed. There are two critical stages in the design of prestressed concrete for serviceability. The first stage is immediately after the prestressing force is transferred to the concrete. This force is usually designated as Pi. The second critical stage is after the time dependent losses have occurred and the member is under full service loads. At this stage, the magnitude of the prestressing force is at a minimum value and applied loads are at a maximum value. The prestressing force at this stage is usually designated Pe. At both of these stages, serviceability requirements of the member should be satisfied. Strength requirements at each stage are also satisfied. However, it is worthwhile to note the level of prestress, and the amount and distribution of steel are determined by serviceability rather than strength requirements
Activit y 6A Design of a Prestr essed Concret e Member under Service Loads (1) Read:
Warner, Rangan, Hall and Faulkes; Concrete Structures , sections 8.1 - 8.4, 15.1, 15.3.1 - 15.3.3, 15.4.4, 15.5 and 15.5.2. Study: Powerpoint presentation: Week 6: Design of Prestressed Concrete Beams for Serviceability.ppt (205 k)
v2 December 2011
Serviceability, Deflections and Crack Control Example
(based on Warner worked example 8.7) The beam shown in Figure 6.1 spans 25 m and supports a live load of 14.8 kN/m in addition to self-weight.
Figure 6.1: Serviceability, deflections and crack control worked example, copyright RMIT University (Geoff Marchiori)
Section properties f cp' 35 MPa
f c' 45 MPa
pe 854 MPa
pi 1100 MPa
A p 3640 mm2
As 8000 mm2
A g 570,000 mm2 I g 3.48 10 10 mm4 Ec 30, 000 MPa
Ep 190, 000 MPa
1. Check adequacy of beam at transfer using Cl 8.1.6.2. Initial prestress force: Pi pi Ap 1100 3640 4004 kN
At transfer, beam self-weight needs to be considered. Self-weight of beam: wsw 24 0.57 13.68 kN / m
v2 December 2011
Self-weight of bending moment: wsw l 2 13.68 252 Msw 1069 kNm 8 8
Figure 6.2: Checking adequacy of beam at transfer, copyright RMIT University (Geoff Marchiori) 3 4004 10 3 4004 10 396 621 1069 10 6 621 b 570000 3.48 1010 3.48 1010 7.02 28.3 19.1 16.22 MPa 0.5 f cp
OK as 0.5 f cp 17.5 3 4004 10 3 4004 10 396 279 1069 10 6 279 t 570000 3.48 1010 3.48 1010 7.02 12.7 8.57 2.89 MPa 0.5 f cp
OK as 0.5 f cp 17.5 Check whether the section is cracked under full service load. Mservice
252 l2 wa wl 13.68 14.8 2225 kNm 8 8
wa wsw
Bottom fibre stress: P Pey b Mservice y b 2225 106 621 b 7.02 28.3 3.48 1010 A I g I g
4.38 MPa (tension) Tensile cracking stress of concrete: f r' 0.6 f c' 0.6 45 4.02 MPa b 4.34 4.02
Therefore beam is cracked under service loads.
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2. Calculate deflection of beam Since section is cracked under full service load, I cr is needed for the calculation of deflections.
properties of equivalent cracked section: dn I cr To find dn we equate moments of area above and below neutral axis using transformed section. Assume dn is in the flange. E p
Es Ec Ec 190 APT 3640 23050 mm2 30 200 AST 8000 53330 mm2 30 n
or
Figure 6.3: Calculating deflection of beam, copyright RMIT University (Geoff Marchiori)
moments @ N.A. ( Ay 0 ): 1500dn2 23050 675 dn 53330 815 dn 2 750 dn2 76380dn 59 106 0 dn 234 mm Therefore assumption dn in flange is correct. Determine second moment of area of cracked section I cr , about neutral axis ( I Ay 2 ):
1500 234 3 2 2 I cr 23050 675 234 53330 815 234 12 28.9 10 9 mm4
M Icf I cr I g I cr cr M s
3
v2 December 2011
To determine cracking moment, first calculate stress in bottom fibre due to prestress alone: b
Mcr
P Peyb 7.02 28.3 35.3 MPa A I
bp
f cf' I g y
wservice 28.5 kN
35.3 0.6
45 3.48 1010 621
2204 kNm
Mservice 2225 kNm 3
2204 10 4 I ef 28.9 10 3.48 10 28.9 10 3.46 10 mm 2225 Effective prestress, Pe pe Ap 854 3640 3109 kN 9
10
9
Equivalent upward load due to prestress @ service load: h 3109 0.396 15.8 kN / m 8 l2 252 5 28.5 15.8 25000 4 5 wl 2 service 62.2 mm 384 EI ef 384 30000 3.46 1010 W p 8Pe
allowable limit
l 25000 100 mm 62.2 OK 250 250
Activit y 6B Design of a Prestr essed Concret e Member under Serv ice Loads Review:
Powerpoint presentation: Week 6: Design of Prestressed Concrete Beams for Serviceability.ppt (205 k) Study Example: Warner, Rangan, Hall and Faulkes; Concrete Structures , 8.5, 8.6 and 15.2. Develop Solutions: Problem 7 provided separately
When you have completed these problems turn to the back of this Learning Guide to find solutions. v2 December 2011
Sum mary and Out come Checklist
In this Topic you learnt to calculate the post cracking behaviour of cracked sections by using simple elastic analysis. Tick the box for each statement with which you agree:
I can list the serviceability requirements for a flexural member. I can set limits on stress in concrete both at transfer and under full service loads.
I can determine prestress force for a flexural member.
I can determine eccentricity force for a flexural member.
I can check stress limits.
I can calculate short-term and long-term deflections of a prestressed concrete member
Assessment
This Topic will be assessed in the Final Examination and as is relevant to your work in the Major Design Project.
v2 December 2011
v2 December 2011
The satisfaction of concrete and steel stress limits at service load does not guarantee adequate strength and safety of the structural member. It is important to consider the non-linear behaviour of the member in the over-load range to ensure structural safety of the member. This is possible only by calculating the ultimate capacity of a member, which provides the sufficient margin between the service load and the ultimate load. The ultimate strength of a cross-section in bending is calculated by flexural strength theory, which involves consideration of the strength of both the concrete and steel in tension and compression. In this Topic, you will learn to predict the ultimate flexural strength of a prestressed concrete member with singly and doubly reinforced concrete sections. The ultimate bending moment of a section is determined and design requirements for the strength limit state must be satisfied. In addition to the satisfaction of strength limit state, a measure of the ductility of each section must be determined. Ductile members undergo large deformations before failure. Therefore, design for ductility is very important in determining the ultimate strength of a prestressed concrete member. The determination of the ultimate bending moment of a prestressed concrete section uses almost similar assumptions to those being used in simple non-prestressed reinforced concrete sections. They are:
strain variation along the cross-section is linear, therefore plane sections remain plane
concrete carries compression only. Tensile strength of the concrete is ignored
stresses in concrete, non-prestressed steel and prestressed steel are obtained from actual stress-strain relationships.
v2 December 2011
Learn ing Out com es When you have successfully completed this Topic you will be able to: Upon successful completion of this Topic you will be able to:
state the assumptions made in determination of ultimate flexural strength of a prestressed concrete member
determine the flexural behaviour of a beam member under ultimate load conditions
determine the strains in non-prestressed steel and in the concrete
select an initially appropriate location of a neutral axis
determine the ultimate strain and stress in prestressing steel
verify and relocate the position of the neutral axis
determine the ultimate moment capacity of a prestressed reinforced concrete section.
Background Skil ls and Know ledge
Before you start this Topic you are expected to be able to:
use the concept of a rectangular stress block
determine the combined stresses induced by prestress and applied loads at transfer
calculate the combined stresses induced by prestress and applied loads at service loads
calculate the overall prestress losses including immediate and deferred
perform deflection checks of a prestressed concrete member
v2 December 2011
Activit y 7A Ultim ate Flexur al St rengt h of Prestressed Concrete Beams Read:
Warner, Rangan, Hall and Faulkes; Concrete Structures , sections 7.2, 7.3, 7.4, and 7.6 Study: Powerpoint presentation: Week 7: Ultimate Strength of Prestressed Concrete Beams.ppt (181 k) Study Worked Examples: Warner, Rangan, Hall and Faulkes; Concrete Structures , 7.11, 7.12, 7.13, 7.16 and 7.19. Develop Solutions: Problem 8 provided separately. When you have completed these problems turn to the back of this Learning Guide for solutions.
v2 December 2011
Sum m ary and Outcom e Checkli st
In this Topic, you learnt how to predict the ultimate flexural strength of a prestressed concrete member .You learnt how the ultimate bending moment of a section is determined and how the design requirements for the strength limit state are satisfied. In addition to the satisfaction of strength limit state, a measure of the ductility of each section must be determined. Ductile members undergo large deformations before failure. Therefore, design for ductility is very important in determining the ultimate strength of a prestressed concrete member. Tick the box for each statement with which you agree:
I can state the prestressed concrete member assumptions necessary to determine ultimate strength. I can determine the flexural behaviour of a beam member under ultimate load conditions. I can determine the strains in a non-prestressed steel and in the concrete. I can determine the ultimate strain and stress in prestressing steel I can determine the ultimate moment capacity Mu of a prestressed reinforced concrete section.
Assessment
This Topic will be assessed in the Final Examination and as is relevant to your work in the Major Design Project.
v2 December 2011
This Topic covers the procedure for the design of prestressed concrete sections to resist shear forces resulting from applied loads. As concrete has lower strength in tension and higher strength in compression, shear design is very important in all types of concrete structures.
Learn ing Out com es Upon successful completion of this Topic you will be able to:
state the concept of shear in beams
determine the shear strength of a prestressed concrete beam
analyse and design for flexure-shear cracking
analyse and design for web-shear cracking
satisfy the design requirements to avoid shear failure in beams.
Background Skil ls and Know ledge
Before you start this Topic you are expected to be able to:
determine the shear stresses in a beam without reinforcement
calculate the principal stresses in a beam section without reinforcement
apply Mohr’s Circle of Stresses in a simple homogenous single material beam section
design shear reinforcement in a concrete beam without prestressing steel as per current code of practice
v2 December 2011
The behaviour of prestressed concrete beams in shear failure is different from flexural behaviour. The shear failure occurs suddenly and the diagonal cracks that cause failure are wider than the flexural cracks. Both shear and torsional forces produce shear stresses in prestressed concrete members. This shear stress can result in principal tensile stresses at the critical section and consequently exceed the tensile strength of the concrete. In general, the shearing stresses in beams are not caused by direct shear or pure torsion, but by a combination of applied loads and moments. This will cause diagonal tension or flexural shear stresses in the member. There are some special applications like corbels or brackets or balcony cantilever slabs where direct shear or pure torsion is applied. As explained above, in common structures, shear stresses cause diagonal tension in the concrete and, ultimately, inclined cracking. Traditional reinforcement in the form of hoops or stirrups or ties is used to carry tensile forces in the webs of prestressed concrete beams to resist diagonal cracks. A local shear failure at columns or under concentrated loads can occur in slabs and footings. The punching shear failure is very common in flat slabs and flat plates. In this Topic, you will learn to provide adequate reinforcement to resist shear and torsional forces. It is a well-established fact that the longitudinal compression induced by prestress delays the formation of diagonal cracks. Mohr’s Circle of Stresses can be used to examine the effects of prestress on the formation and direction of inclined cracks. Therefore, introduction of prestress increases the shear strength of reinforced concrete beams. Normally, prestressed sections often have thin webs, and their thicknesses are governed by shear strength considerations.
v2 December 2011
Activit y 8A Design ing Prest ressed Concret e Beams for Shear ( 1) Read:
Warner, Rangan, Hall and Faulkes; Concrete Structures , sections 12.4, 12.5 and 12.6 Study: Powerpoint presentation: Week 8: Shear Strength of Concrete Prestressed Beams.ppt (166 k)
Shear Design Exam ple
(based on Warner worked example 12.2)
A post-tensioned beam with the section shown spans 30 m between simple supports and carries a live load of 17 kN/m. The beam is prestressed by a parabolic cable with an eccentricity at midspan of 524 mm and zero eccentricity at the ends. The cable comprises of 35 strands of 12.7 mm diameter with A p 3500 mm2 , in a 112 mm diameter duct. Owing to friction the prestress force varies in an approximately linear manner from 3880 kN at ends to 3650 kN at midspan. Assume f c' 40 MPa and f sy 400 MPa . Calculate the shear reinforcement for the beam. A g 488400 mm2 I g 83.76 10 9 mm4 wself weight 0.4884 m 2 25 kN / m3 12.2 kN / m w * 40.8 kN / m take effective depth, d 1140 m
v2 December 2011
Figure 8.1: Shear design example, copyright RMIT University (Geoff Marchiori)
The bending moment and shear force at distance x from the left support are: 30 612 kN 2 30 V * 40.8 40.8x 2 612 40.8x kN x M * 612 x 40.8x 2 2 612 x 20.4 x kNm
RL 40.8
+'ve
+'ve
Figure 8.2, Bending moment and shear force, copyright RMIT University (Geoff Marchiori)
v2 December 2011
Design at critical section near support. Location from the support (Cl 8.2.4) of the critical section for design: Error! Objects cannot be created from editing field codes.
do 1220
112 24 1140 mm 2
At the critical section, d 1.14 m the cable force is 3863 kN ; i.e. prestress force variation due to friction loss:
@ x 0 P 3880 kN x 15 P 3656 kN
Figure 8.3, Prestress force variation due to friction loss, copyright RMIT University (Geoff Marchiori)
Linear variation:
3880 3656 3880 x x 3863 kN 15 m 1.14 m Eccentricity of cable at x = 1.14 m:
Figure 8.4: Eccentricity of cable, copyright RMIT University (Geoff Marchiori)
x x 2 1140 1140 2 y 4h 4 524 77 m L L 30000 30000 e 77 mm The inclination of the cable is:
v2 December 2011
4
h x 4 524 1140 1 2 1 2 0.065 rad 30000 L L 30000
The vertical component of the cable force: Pv P sin 3863 0.065 251 kN
Figure 8.5: Vertical component of cable force, copyright RMIT University (Geoff Marchiori)
The effective web width: bv bw 0.5 dd
d
d
Cl 8.2.6
sum of the diameters of the grouted ducts
bv 200 0.5 112 144 mm At this section: V * 612 40.8 1.14 566 kN 2
M * 612 1.14 20.4 1.14 671 kNm Shear carried by concrete: Web-shear cracking precedes flexural-shear cracking the shear carried by the concrete, V t , is therefore that shear for which the principal tension at some point in the web reaches critical value 0.33 f c' . Near the support the moment is small and the
principal tension at the centroidal axis is greater than at the webflange junction. At the centroid ( y 0 ), the horizontal normal stress is:
P A
3863 103 488400
7.91 MPa
v2 December 2011
The vertical normal stress is assumed to be zero (to calculate principal stresses is then taken as +ve if tensile and negative if compressive). The sheer stress is:
Vt Q I bw
where Q is the first moment of area above (or below) the centroid.
Figure 8.6: Shear stress, copyright RMIT University (Geoff Marchiori)
230 Q 760 230 552 2 1 560 75 552 230 25 2 2 2
552 230 200 552 230 2 92.99 106 mm3 then V t 92.99 106 7.71 10 6 Vt MPa 9 83.76 10 144 The principal stress, , at the centroid is: 1
2 0.5 2 0.5 v 2
2
791 0.5 7.71 10 6 V t 7.91 0.5 The critical value of tension is 0.33 f c' 0.33 40 2.11 MPa , so set 1 2.11 MPa 2
2
2.11 7.91 0.5 7.71 10 6 V t 7.91 0.5
Vt 596 kN v2 December 2011
Adding the vertical component of the cable force: Vuc Vt Pv 598 251 847 kN
0.5 Vuc 0.5 0.7 847 296 kN V*566
therefore it is necessary to design shear reinforcement. Calculate V u min Cl 8.2.9 Vu min Vuc 0.10 f c' bv do Vuc 0.6 b v do
847 103 0.10 40 144 1140 847 103 0.63 144 1140 950 kN v uc 0.6bv d0 Is Asv ,min sufficient?
Since: 0.5Vuc V * V u min 296 566 665 provide minimum shear reinforcement only. Shear reinforcement Cl 8.2.8: ' Asv min 0.06 f c bv 0.35 bv s f sy f sy
Asv min 0.06 40 144 0.35 144 0.137 0.126 mm2 / mm With s 400 400 Asv min 0.137 mm2 / mm s Y12 stirrups Asv 220 mm2 s 1606 mm
The maximum spacing is 500 mm Therefore provide Y12 – 500 cts Design near quarter span (x = 7 m)
For beams with uniformly distributed loads the region near the quarter-span point is often critical for flexure-shear cracking at x =7m Cable force = 3773 kN
7000 7000 2 e 4 524 375 mm 30000 30000
v2 December 2011
Cable slope:
4 524 7000 1 2 30000 0.0373 rad 30000
Bottom reinforcement 3Y20 bars Ast 930 mm2 Design forces: V * 612 40.8 7 326 kN 2
M * 612 7 20.4 7 3284 kNm Note that, since at this section the cable is located in the flange and not in the web, bv bw 200 mm Shear carried by concrete:
P P e y b I g I g ; i.e. M dec M dec bp Ig yb A g y b 3773 103 375 668 83.76 109 3 M dec 3773 10 83.76 10 9 668 M dec 2384 106 Nmm 2384 kNm M dec 2384 Vdec 237 kN M * 3284 V * 326 Pv 3773 0.0373 141 kN do 1000 1140 1 1.1 1.6 0.506 1.1 1000 take 1 1.1, 2 3 1.0 f cv 40 MPa 1 1.1 1.6
Cl 8.2.7.1
1
Ast Ap 3 Vuc 1 2 3 bv do f cv Vo Pv b d v o
Cl 8.2.7.2
930 3500 V uc 1.1 1 1 200 1140 40 3 200 1140 1
1
3
237 103 141 10 3 231 237 141 609 kN 0.5 Vuc 0.5 0.7 609 213 kN V *
326 kN
v2 December 2011
Therefore it is necessary to design shear reinforcement. Calculate V u ,min
CI 8.2.9
Vu ,min Vuc 0.10 f c' bv do Vuc 0.6 b v do
Vuc 0.10 40 200 1140 V uc 0.6 200 1140 V uc 0.63 200 1140 609 10 3 0.63 200 1140 753 kN Is Asv ,min … sufficient? Vu ,min 0.7 753 527 kN V *
326 kN
Therefore we require only the minimum reinforcement. Shear reinforcement: The stirrup requirements are as for the previously considered section. Provide Y12 – 500 spacings. Other sections can be completed using spreadsheet.
Activit y 8A Design ing Prest ressed Concret e Beams for Shear ( 2) Review:
Powerpoint presentation: Week 8: Shear Strength of Concrete Prestressed Beams.ppt (166 k) Develop Solutions: Problem 9 provided separately. When you have completed these problems turn to the back of this Learning Guide for solutions.
v2 December 2011
Sum mary and Out come Checklist
In this Topic you learnt how to provide adequate reinforcement to resist shear and torsional forces in prestressed concrete beams. Tick the box for each statement with which you agree:
I can stat e the concept of shear in beams.
I can determine the shear strength of a prestressed concrete beam.
I can analyse and design for flexure shear cracks.
I can analyse and design for web shear cracks.
I can satisfy the design requirements to avoid shear failure in beams.
Assessment This Topic will be assessed in the Final Examination and as is relevant to your work in the Major Design Project.
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In commercial and multistorey buildings the supported floor systems are usually constructed of reinforced concrete that has been cast in place. Two-way slabs and plates are those panels in which the dimensional ratio of length to width is less than two. In general, frame systems comprise conventional slab-beamcolumn combinations for a concrete structure. Flat plates or slabs are those floors which a supported directly on columns without beams. Flat plates or slabs are supported directly over the columns, therefore the slabs span in two directions and provide two-way action to the columns. As there are no beams, the bottom surface of the floor is flat. Flat slabs have some advantages over the conventional beam-slab floor system. The main advantage of a flat slab floor is a smaller height for the overall building. Also, the flat slab formwork is simpler and aesthetically elegant. In general, flat slabs have drop panels and flat plates are without drop panels. The flat slabs deflect along the lines joining adjacent columns, and the interior of the panel deflects further relative to these columns. In practice, two commonly used methods of analysis are the direct design or simplified approach method and the equivalent frame method. Both the direct design and the equivalent frame method are based on the concept of the equivalent frame, except that the direct design method has limited applications. The direct design method is suitable for gravity load only, where the frame method can cope with horizontal loading as well. In this Topic the simplified approach method will be discussed. The learning outcomes of this topic will introduce you to analysis and design of reinforced concrete structures which are non-standard and will facilitate understanding of sophisticated analysis methods developed to cater for these.
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Learn ing Out com es Upon successful completion of this Topic you will be able to:
state the assumptions and applications based on a simplified or direct design method for two-way flat slabs
divide a floor in both directions as design strips
calculate the total moments in the design strip at the supports and at mid-span
estimate how these moments are distributed across the width of the design strip
repeat the calculations of the design strip moments and their distribution across the width in the other direction
determine the dimensions of the design strip
calculate the total static moment
calculate total end moments and span moments
determine design moment factors for an end span
determine the proportion of design moments assigned to the column strip
state applications based on the idealised frame method or determine the design moments in interior and edge strips in both x and y directions
design and layout the reinforcement.
Background Skil ls and Know ledge
Before you start this Topic you are expected to be able to:
explain basic concepts of reinforced concrete
describe flexural behaviour of reinforced concrete beams
calculate loads, shear forces and bending moments for common reinforced concrete structural elements of rectangular section
evaluate ultimate moment capacity for a reinforced member
explain the concept of limit state design for reinforced concrete structures
evaluate dead, live and wind loads on a given structure. v2 December 2011
Session 9.1: Sim pli fi ed Met hod of Analysis In this session you will learn to design the flat slab or flat plate floor system with the direct design, or simplified method of analysis. The applications of this method are limited as this method is more approximate and has the following limitations:
The successive spans must be fairly uniform in each direction - they should not differ by more than one third of the longer span.
A minimum of three continuous spans is necessary in each direction.
All loads shall be due to gravity only, and uniformly distributed over the entire panel. The live load should not exceed three quarters of the dead load.
It is worth noting that you will find the majority of normal floor systems satisfy these conditions Learning Out comes
Upon successful completion of this session, you will be able to:
determine whether the slab geometry and loading permits the use of the simplified method
select slab depth to satisfy shear and deflection requirements. If the slab thickness chosen is not adequate, then revise the dimensions or strengthen the slab-column or both
divide the structure into equivalent design frames bound by centre lines of panels on each side of a line of columns
compute the total statical moment
select the distribution factors of the negative and positive moments to the exterior and interior columns and spans
distribute the equivalent frame moments to the column and middle strips
determine whether the trial slab depth is adequate for shear
design the flexural reinforcement
select the size and spacing of the reinforcement to satisfy the crack control requirements.
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Activit y 9A Sim plif ied Meth od of Analysis – Flat Slabs (1) Read:
Warner, Rangan, Hall and Faulkes; Concrete Structures , sections 17.1-17.4 and 18.1-18.3 Study: Powerpoint presentation: Week 9: Introduction to Flat Slab Analysis and Design.ppt (1.419 M)
Analysis of Flat Slabs Exam ple (Simp lif ied Meth od)
(based on Warner worked example 19.3) A two way flat slab supporting a retail shopping mall has a repetitive floor plan with 400 mm square columns placed in a rectangular grid of 6.5 m 6 m. Live load is 5 kPa. Drop panels are 2.2 m 2 m. Projection of the drop panel below the slab is 75 mm. Slab thickness is 250 mm. Concrete specific weight = 24 3 kN/m . The slab comprises three continuous spans in each direction ( x and y). For both an internal design strip and an edge design strip, calculate the total moments in both the x and y directions. Divide the floor system into column strips and middle strips for both the internal design strip and the edge design strip. Determine the moment distribution across the width of the design strips in both the x and y directions.
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Figure 9.1, Two way flat slab, copyright RMIT University (Rebecca Gravina)
Geometry check: Ratio longer span /shorter span < 2; i.e. 6.5/6 = 1.08 < 2. There are more than 2 spans in one direction. Live load is less than three quarters of dead load. Hence simplified method is acceptable. Design load Fd 1.2G 1.5Q 1.2 24 0.25 1.5 5 14.7 kN / m 2
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X direction
Figure 9.2, Design load – x direction, copyright RMIT University (Geoff Marchiori)
Interior design strip Fd Lt L2o Total static moment M o 8
Cl 6.10.4.2
Fd 14.7 kN / m 2 Lt = width of design strip = 6.0 m L = span of slab in x-direction from centre to centre of columns = 6.5 m For this slab with drop panels asup
400 75 275 mm 2
Lo L minus 7 times the sum of the values of dsup at each end of the span
6500 - 2 0.7 275 6115 mm Therefore: 2
14.7 6 6.115 M o 412 kNm 8
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Edge design strip strip Fd Lt L2o Total static moment M o 8 Fd 14.7 kN / m 2 Lt 3.0 0.2 3.2 m Lo 6.115 m 2
14.7 3.2 6.115 M o 220 kNm 8 Distribution design strip bending bending moments End Span – design moment
Table 6.10.4.3 (A) of AS3600
ve moment at exterior support 0.25 M o ve midspan moment 0.50 M o ve moment at interior support 0.75 M o Interior span – design moment
Table 6.10.4.3 (B) of AS3600
ve moment at support 0.65 M o ve midspan moment 0.35 M o Bending moment distribution factors between column and middle strips (Table 6.9.5.3 of AS3600) AS3600)
ve moment at interior support 0.6 to 1.0, say 0.75 to column strip, 0.25 to middle strip ve moment at an exterior support 0.75 to 1.0, say 0.75 to column strip, 0.25 to middle strip ve moment in all spans 0.5 to 1.0, say 0.6 to column strip, 0.4 to middle strip Interior design strip M o 412 kNm
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End Span
ve support moment, column strip 0.25 M o 0.75 77 kNm ve interior support moment, column strip 0.75 Mo 0.75 232 kNm ve moment midspan, column strip 0.5 Mo 0.6 124 kNm ve exterior support moment, half middle strip 0.25 M o 0.25 13 kNm 2 ve interior support moment, half middle strip 0.75 M o 0.25 39 kNm 2 ve moment midspan, half middle strip 0.5 M o 0.4 41 kNm 2 Interior span
ve support moment, column strip 0.65 M o 0.75 211 kNm ve midspan moment, column strip 0.35 M o 0.6 87 kNm 0.65 M o 0.25 ve support moment, half middle strip 2 34 kNm 0.35 M o 0.4 ve midspan moment, half middle strip 2 29 kNm Edge Design Strip M o 220 kNm
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End Span
ve exterior support moment, column strip 0.25 Mo 0.75 41 kNm ve interior support moment, column strip 0.75 Mo 0.75 124 kNm ve interior span moment, column strip 0.5 M o 0.6 66 kNm ve exterior support moment, middle strip 0.25 Mo 0.25 14 kNm ve interior support moment, middle strip 0.75 Mo 0.25 41 kNm ve moment midspan, middle strip 0.5 M o 0.4 44 kNm Interior Span
ve support moment, column strip 0.65 M o 0.75 107 kNm ve midspan moment, column strip 0.35 M o 0.6 46 kNm ve support moment, middle strip 0.65 M o 0.25 36 kNm ve midspan moment, middle strip 0.35 M o 0.4 30 kNm Y direction
Figure 9.3, Design load – y direction, copyright RMIT University (Geoff Marchiori)
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Interior design strip Total static moment M 0 Lt 6.5 m
L 6.0 m
Lo 6000 2 0.7 275 5615 m 2
14.7 6.5 5.615 M o 377 kNm 8 Edge design strip Total static moment M 0 6.5 0.2 3.45m 2 Lo 5.615 m Lt
2
14.7 3.45 5.615 M o 200 kNm 8
Activit y 9B Sim plif ied Meth od of Analysis – Flat Slabs (2) Review: Powerpoint presentation: Week 9: Introduction to Flat Slab Analysis and Design.ppt (1.419 M) Develop Solutions: Problem 10 provided separately When you have completed these problems turn to the back of this Learning Guide to find solutions.
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Session 9.2: Design of Flat Slabs for Flexu r e and Defl ect ion Cont rol An essential criteria in the design of slabs is adequate control of deflections at service loads. A minimum thickness of slab needs to be selected at the preliminary design stage and later minor adjustments are made during the final design stage to ensure calculated deflections are within allowable limits. Learning Out comes
Upon successful completion of this session, you will be able to:
determine whether the trial slab depth is adequate for satisfy deflection requirements
design the flexural reinforcement
select the size and spacing of the reinforcement to satisfy the crack control requirements
Design codes provide rules for allowable span to depth ratios. Deflection in slabs can be due to long-term creep and shrinkage of the concrete. The code procedure must be followed to comply with deflection requirements. Once an appropriate thickness of slab has been selected, the design moments and shears of a flat slab are obtained using either the simplified method given in codes or an idealised frame method. In this topic as previously mentioned we will be using the simplified method. The reinforcement required to resists these moments is obtained be evaluating the tensile reinforcement area Ast or tensile reinforcement ratio , to satisfy that the design moment M* is less than or equal to the ultimate moment capacity of the section, M u. The tensile reinforcement ratio should be not less than minimum requirements. Minimum centre to centre spacing of bars must also comply with detailing requirements given in codes for adequate control of flexure cracks at service loads.
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Activit y 9B Design of Flat Slabs for Flexur e and Defl ection Contr ol ( 1) Read:
Warner, Rangan, Hall and Faulkes; Concrete Structures , sections 19.1,19.4 Review: Powerpoint presentation: Week 9: Introduction to Flat Slab Analysis and Design.ppt (1.419 M) Study Worked Examples: Warner, Rangan, Hall and Faulkes; Concrete Structures , example 19.3 Develop Solutions: Problem 11 below this Activity Box. When you have completed these problems turn to the back of this Learning guide to find solutions.
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Design of Flat Slab f or Flexure Exam ple and deflection contr ol
(based on Warner worked example 19.3) A two way flat slab supporting a retail shopping mall has a repetitive floor plan with 400 mm square columns placed in a rectangular grid of 6.5 m x 6 m. Live load is 5 kPa. Drop panels are 2.2 m x 2m. Projection of the drop panel below the slab is 75 mm. Slab thickness is 250 mm. f c' 32 MPa ,
f sy 500 MPa
Determine the flexural reinforcement required in the slab using the simplified method given in AS3600.
check minimum slab thickness Lef d
/ Lef 1000Ec k3 k 4 F def
1
3
Assume:
overall slab thickness = D = 250 mm
d effective depth D cover
diameter main steel bars 16 mm
diameter main bars 2
Therefore: d 250 20
16 222 mm 2
Lef effective span = lesser of Ln D or L Take longer span L = 6.5 m
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Ln clear span = 6500 400 6100 mm Ln D 6100 250 6350 mm Lef 6350 mm k3 1.05 drop panels k4 2.1 Ec 2400 1.5 0.043 32 28600 MPa
Cl 3.1.2
f c' 32 MPa f sy 500 MPa 1 250 1.0 kcs g s kcs 1 q
/ Lef Fdef
S 0.7
- short term - long term L 0.4 g dead load = 0.25 24 6 kPa q live load 5 kPa kcs 2.0 Fdef 1 2 6 0.7 2 0.4 5 18 7.5 25.5 kPa
kN /m2
check:
1 1000 28600 6350 1.05 2.1 250 222 25.5 28.6 36.4 OK
1
3
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Reinforcement design At each location, the various strips of slab are designed as rectangular shallow beams. The moments in the x-direction are larger than those in the ydirection. Hence bars in the x–direction will be place closer to the slab surface. We will use N16 main bars. Effective depth at x-bars ds 250 20 8 222 mm Effective depth at y-bars ds 250 20 16 8 206 mm Note at the first interior columns, the reinforcement is designed for the greater of the moments shown, that is, between end span and interior span. Each middle strip is designed to resist the sum of the moments assigned to its two adjoining halves.
Minimum reinforcement ratio (Cl 9.1.1) for slabs supported on columns: pmin
' 2 Ast D f ct . f 0.24 bd d f sy
Take b = 1000 mm, x-direction ds 222 mm : D 250 mm f ct' . f 0.6 f c' 3.39 MPa f sy 500 MPa 2
Ast 250 3.39 pmin 0.24 0.0021 bd 222 500 Ast ,min 0.0021 1000 222 466.2 mm2 / m To determine spacing over 1000 mm width: Area 1N16 bar 201 mm2 1000 Spacing 201 431 mm 466.2
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Check minimum spacing (CI 9.4.1) Minimum spacing is the lesser of: 2.0Ds or 300 mm 2.0 227 or 300 mm Therefore adopt 300 mm minimum spacing: Minimum reinforcement N16 - 300 cts 1000 Ast ,min 201 670 mm2 / m 300
Minimum moment: Mu f sy Ast .min 0.85d
0.8 500 670 0.85 222 10 6 51 kNm / m For all moments less than 51 kNm/m (width) provide minimum reinforcement. Reinforcement in x - direction (d = 222 mm) Edge design strip Column strip – width = 1.7 m
Figure 9.5, Edge design strip, copyright RMIT University (Geoff Marchiori)
Provide minimum reinforcement N16 – 300 cts @ -ve moment exterior support - endspan M *
41 kNm 24 kNm / m 1.7 m
+ve moment endspan M * 66 kN 46 kNm +ve moment interior span M * 27 kNm / m 1.7 m since M * Mu ,min 51 kNm / m
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However provide more than minimum reinforcement: @ -ve moment interior support (adopt the larger value) 124 kNm M * 73 kNm / m 1.7 m Proportion tensile reinforcement required: M * Mu f sy Ast 0.85d 7.3 106 0.8 500 Ast 0.85 222
Ast 967 mm2 / m Spacing required: b A (1N16 bar) Ast 1000 201 207 200 mm 967
s
Therefore provide N16 – 200 cts at negative moment interior support. 1 1 middle strip edge design middle strip interior design 2 2 total width 1.5 m 1.5 m 3.0 m
Figure 9.6, Spacing required, copyright RMIT University (Geoff Marchiori)
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Since all M * in middle strip are less than M u ,min provide minimum reinforcement in –ve M and +ve M regions, N16 300cts Interior design strip Column strip width = 3 m
Figure 9.7, Interior design strip, copyright RMIT University (Geoff Marchiori)
Provide minimum reinforcement N16 – 300 cts in regions where M * Mu ,min (51 kNm / m) . Only need to provide more than minimum reinforcement at interior support, -ve moment. M * 77 kNm / m M u f sy Ast 0.85d
77 106 0.8 500 Ast 0.85 222 Ast 1020 mm2 / m Spacing required: b Ab (1N16 bar) Ast 1000 201 197 175 mm (round down to nearest 25 mm) 1020
s
Therefore provide N16 – 175 cts at -ve moment interior support. (½ middle strip – interior design)/(width = 1.5 m)
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Figure 9.8: Spacing required, copyright RMIT University (Geoff Marchiori)
Since all M * in middle strip are less than M u ,min provide minimum reinforcement in +ve and –‘ve moment regions; i.e. N16 – 300 cts Reinforcement in y – direction, d = 206 mm Minimum reinforcement = N16 – 300 cts Ast ,min 670 mm2 / m Minimum moment M u f sy Ast ,min 0.85d
0.8 500 670 0.85 206 47 kNm / m For all moments less than 47 kNm/m provide minimum reinforcement. Continue to follow procedure as given for x-direction in previous pages.
Activit y 9D Design of Flat Slabs for Flexur e and Defl ection Contr ol ( 2) Review:
Powerpoint presentation: Week 9: Introduction to Flat Slab Analysis and Design.ppt (1.419 M) Develop Solutions: Problem 11 provided separately When you have completed these problems turn to the back of this Learning guide to find solutions. v2 December 2011
Session 9 .3: Desig n of Flat Slabs f or Shear Normally slabs are not likely to fail in shear. But the region of a flat plate or flat slab under a column is subjected to a combination of severe bending moment and shear forces. The slab is just like a thin, wide beam. Flat slabs could fail in a beamtype flexural shear or punching shear. In this session you will learn to design flat slabs for shear failure. Learning Out comes
Upon successful completion of this session, you will be able to:
determine ultimate shear strength
calculate design shear force
calculate the total unbalanced moment transmitted directly through flexure
calculate the design bending moment
design numbers and spacing of reinforcements to resist shear
determine short-term and long-term loads
determine slab depth requirements for deflection in column-beam strips
determine slab depth requirements for deflection in the middle beam strip
revise the slab or drop panel depth if required
Two types of shear can be induced in flat slabs.
Punching shear results from force transfer from column to slab. The critical section for punching shear is d/2 outside the column periphery.
Torsional shear results from the transfer of bending moments between the slab and column.
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Shear Design ar ound Colum n Exam ple
(based on Warner worked example 19.4) Design the region around an interior column and an exterior in the flat slab floor system of lecture example Week 10 – flexure.
Figure 9.10, Two way flat slab design for shear, copyright RMIT University (Rebecca Gravina)
G – dead load = 6 kPa Q – live load = 5 kPa Design load: Fd 1.2G 1.5Q 1.2 6 1.5 5 14.7 kN / m 2 Determine V * V * 6.5 6 14.7 573 kN
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Determine M v*
Cl 6.10.4.4
M v* unbalanced bending moment at interior support within width of the design strip F M v* M M ' From week 9 – lecture example: M interior support, end span, design strip 0.75 412 309 kNm ' Minterior support, interior span, design strip 0.65 412 268 kNm
M v* M M ' 309 268 41 kNm AS3600 requires that M v* at an interior support should not be less than the value given in: 2 2 M v* 0.06 1.2 g 0.75q Lt Lo 1.2 gLt L'o M v* 0.06 1.2 6 0.75 5 6 6.115 2 1.2 6 6 6.115 2 M v* 50.48 kNm
Check whether V * and M v* can be resisted by the concrete (Cl 9.2.4). since M v* is not zero V uo V u uM v* 1.0 8V * ad om From figure 9.2.1 (B) do ,min 2 2 as the average of dx and dy
a b column width take dom i.e. dom
292 276 2
284 mm
Note slab thickness in vicinity of columns = 250 + 75 = 325 mm Therefore a = b = 400 + 284 = 684 mm
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From figure 9.2.1(A):
U = length of critical shear perimeter = (684 + 684) x 2 = 2736 mm Figure 9.11: Critical shear perimeter, copyright RMIT University (Geoff Marchiori)
V uo = ultimate shear strength of a slab with no moment transfer; i.e. M v* 0 ) Cl 9.2.3 (assume there is no shear head) Vuo udom f cv 0.3 c ud om f cv as c 0
2 ' where f cv 0.17 1 f c 0.34 f c' h f c' 32 MPa 400 Cl 9.2.1.4 1 h 400 2 f cv 0.17 1 32 0.34 32 2.89 1.92 1 adopt f cv 1.92 MPa Vu 0 udom f cv 2736 284 1.92 1492 kN
V u 0 1492 10 3 V u uM v* 2736 50.5 106 1.0 1.0 18 573 103 684 284 18V A* d0 m 1492 103 1297 kN 1 0.15 check: V * 0.7Vu 573 908 OK
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Design around an edge column Determine V *
6 V * 6.5 14.7 2 287 kN
Figure 9.12: Design around an edge column, copyright RMIT University (Geoff Marchiori)
Determine M v* M v* 13 77 13 103 kN Check whether V * and M v* can be resisted by the concrete Cl 9.2.4
dom 284 mm 284 a 400 542 mm 2 b 400 284 684 mm u 2 a b 2 542 684 1768 mm f cv 1.92 MPa h 1
Figure 9.13, Checking whether V * and M v* can be resisted by the concrete, copyright RMIT University (Geoff Marchiori)
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Vuo udom f cv 1768 284 1.92 964 kN V uo
Vu
964 10 3
638 kN uM v* 1768 103 10 6 1 * 1 8 287 103 542 284 8 V ad om Vu 447 kN V * (103)
Vv* and M v* can be resisted by concrete alone
Activit y 9E Design of Flat Slabs for Shear ( 2) Review: Powerpoint presentation: Week 9: Introduction to Flat Slab Analysis and Design.ppt (1.419 M) Study Worked Examples:
Warner, Rangan, Hall and Faulkes; Concrete Structures , example 19.3 - 19.7 Develop Solutions: Problem 12 provided separately When you have completed these problems turn to the back of this Learning guide to find solutions.
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