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Structural Engineering

The Future of High-Rise Building : Innovations and Challenges

Leave a Comment / Civil Engineering, Construction, Structural Engineering /

The skyline of our cities is continuously evolving, with high-rise buildings playing a significant role in shaping urban environments. As we look towards the future, the design of these towering structures is being influenced by technological advancements, sustainability concerns, and innovative architectural practices. This blog delves into the exciting innovations and challenges that will define the future of high-rise building design. Innovations in High-Rise Building Design 1. Sustainable Building Design Sustainability is at the forefront of modern architecture. As the world grapples with climate change, architects and builders are prioritizing eco-friendly designs. High-rise buildings are increasingly being constructed with sustainable materials and energy-efficient systems. 2. Smart Buildings The integration of technology into building design is creating smarter, more efficient high-rises. Smart buildings utilize interconnected devices and systems to optimize performance and enhance the quality of life for occupants. 3. Innovative Construction Techniques New construction methods are revolutionizing how high-rise buildings are designed and built. 4. Mixed-use Developments The concept of mixed-use developments is gaining traction. These buildings combine residential, commercial, and recreational spaces within a single structure, promoting a live-work-play environment. Challenges in High-Rise Building Design While innovations are paving the way for the future, several challenges need to be addressed to ensure the successful implementation of new high-rise building designs. 1. Sustainability and Environmental Impact Achieving true sustainability in high-rise buildings is a complex task. While sustainable technologies are advancing, their implementation can be costly and challenging. 2. Technological Integration Integrating advanced technologies into high-rise buildings poses several challenges. 3. Urban Density and Infrastructure High-rise buildings are often seen as a solution to urban density, but they also bring their own set of challenges. 4. Safety and Resilience Ensuring the safety and resilience of high-rise buildings is paramount, especially in the face of natural disasters and other emergencies. Case Studies: Future High-Rise Buildings 1. The Edge, Amsterdam The Edge is a prime example of a smart and sustainable high-rise building. Located in Amsterdam, this building is known for its energy efficiency and advanced technological integration. 2. Shanghai Tower, China Shanghai Tower is an iconic high-rise that showcases the potential of innovative construction techniques and sustainable design. 3. Bosco Verticale, Milan Bosco Verticale, or Vertical Forest, in Milan is a unique example of integrating greenery into high-rise buildings. Conclusion The future of high-rise building design is filled with exciting possibilities and significant challenges. Innovations in sustainable practices, smart technologies, and construction techniques are transforming how these structures are designed and built. However, addressing the environmental impact, ensuring technological integration, managing urban density, and guaranteeing safety and resilience are critical to the successful implementation of these innovations. As we move forward, collaboration between architects, engineers, urban planners, and policymakers will be essential to create high-rise buildings that are not only impressive feats of engineering but also sustainable, smart, and socially inclusive spaces that enhance the quality of urban life. The future of our cities depends on the successful evolution of high-rise building design, and the journey has just begun.

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Essential skills for structural engineers

7 Essential Skills for Structural Engineers: Building the Future

Leave a Comment / Education System, Civil Engineering, Construction, Structural Engineering /

Introduction Introduction: Structural engineers play a pivotal role in shaping the infrastructure that supports our daily lives. From towering skyscrapers to robust bridges, their expertise ensures that these structures are safe, efficient, and resilient. To excel in this field, structural engineers must possess a diverse set of skills that go beyond textbook knowledge. In this blog, we will explore the essential skills every structural engineer needs to master to build a successful career and contribute to National & International vibrant landscape. 1. Strong Analytical and Mathematical Skills Structural engineering is deeply rooted in mathematics and physics. Engineers must use these disciplines to analyze the forces acting on structures and ensure they can withstand various loads. Here’s why strong analytical and mathematical skills are crucial: Understanding Complex Calculations: Engineers must perform complex calculations to determine the strength and stability of structures. This involves using principles of calculus, algebra, and geometry to assess load distribution, stress, and strain. Structural Analysis: Analytical skills help engineers evaluate the behavior of structures under different conditions. This includes assessing the impact of external forces such as wind, earthquakes, and traffic loads on bridges and buildings. Precision and Accuracy in Basics: Precision is vital in structural engineering. Even a minor error in calculations can lead to significant structural failures. Engineers must be meticulous in their work to ensure safety and reliability. 2. Proficiency in Design and Drafting Software In today’s digital age, proficiency in design and drafting software is essential for structural engineers. These tools streamline the design process, improve accuracy, and enhance visualization. Key software skills include: Computer-Aided Design (CAD): CAD software, such as AutoCAD and Revit, allows engineers to create detailed 2D and 3D models of structures. These models help in visualizing designs and making necessary adjustments before construction begins. Building Information Modeling (BIM): BIM software, like Revit, enables engineers to create comprehensive digital representations of buildings. BIM integrates various aspects of a project, including architecture, engineering, and construction, promoting collaboration and efficiency. Structural Analysis Software: Tools like STAAD.Pro, ETABS are used for structural analysis and design. These programs help engineers simulate real-world conditions, analyze structural behavior, and optimize designs for safety and efficiency. 3. In-Depth Knowledge of Building Materials A thorough understanding of building materials is essential for structural engineers. Different materials have unique properties that affect their performance and suitability for various applications. Key aspects include: Material Properties: Engineers must be familiar with the mechanical properties of materials such as steel, concrete, wood, and composites. This includes knowledge of strength, elasticity, durability, and thermal properties. Material Selection: Selecting the right material for a project is crucial. Engineers must consider factors like cost, availability, environmental impact, and compatibility with other materials when making decisions. Innovations in Materials: Staying updated on advancements in material science is vital. New materials and technologies can offer improved performance, sustainability, and cost-effectiveness. Engineers should be open to adopting innovative materials that enhance the quality and longevity of structures. 4. Excellent Problem-Solving Abilities Structural engineering is a field that constantly presents challenges and unforeseen issues. Excellent problem-solving abilities are essential to navigate these complexities and find effective solutions. This skill involves: Critical Thinking: Engineers must analyze problems from multiple angles and develop logical, evidence-based solutions. Critical thinking helps in identifying the root cause of issues and devising strategies to address them. Creativity: Innovation often stems from creative problem-solving. Engineers should think outside the box to develop unique solutions that improve efficiency, safety, and aesthetics. Decision-Making: Quick and informed decision-making is crucial in the fast-paced construction industry. Engineers must weigh the pros and cons of different solutions and make timely decisions to keep projects on track. 5. Effective Communication and Teamwork Structural engineering projects are collaborative efforts involving various stakeholders, including architects, contractors, clients, and government officials. Effective communication and teamwork are essential for successful project execution. Key aspects include: Clear Communication: Engineers must convey complex technical information clearly and concisely to non-technical stakeholders. This includes writing reports, presenting findings, and discussing project details with clients and team members. Collaboration: Working as part of a multidisciplinary team requires strong collaboration skills. Engineers must coordinate with architects, contractors, and other professionals to ensure that designs are feasible, cost-effective, and meet project requirements. Negotiation and Conflict Resolution: Negotiation skills are important when dealing with project constraints, budget limitations, and differing opinions. Engineers should be adept at resolving conflicts and finding mutually beneficial solutions. 6. A Solid Understanding of Safety and Compliance Regulations Safety is paramount in structural engineering. Engineers must have a comprehensive understanding of safety standards and compliance regulations to ensure that structures are safe for use. Key areas include: Building Codes and Standards: Familiarity with national and international building codes, such as the Indian Standards (IS), is essential. These codes provide guidelines for design, construction, and maintenance, ensuring safety and quality. Risk Assessment: Engineers must assess potential risks and implement measures to mitigate them. This involves identifying hazards, evaluating their impact, and designing structures to withstand adverse conditions. Environmental and Sustainability Regulations: Incorporating sustainable practices and complying with environmental regulations are increasingly important. Engineers should design eco-friendly structures that minimize environmental impact and promote sustainability. 7. Continuous Learning and Adaptability The field of structural engineering is constantly evolving, driven by technological advancements, new materials, and changing regulations. Continuous learning and adaptability are essential for staying relevant and excelling in this dynamic industry. This involves: Lifelong Learning: Engineers should pursue ongoing education through courses, workshops, and certifications. Staying updated on the latest trends and technologies helps them remain competitive and innovative. Adaptability: The ability to adapt to new technologies, methodologies, and industry changes is crucial. Engineers should be open to learning and integrating new tools and practices into their work. Professional Development: Joining professional organizations, attending conferences, and networking with peers can provide valuable insights and opportunities for growth. Engaging with the engineering community fosters knowledge sharing and professional development. Conclusion: Structural engineering is a multifaceted field that demands a diverse set of skills. From strong analytical abilities and proficiency in

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Long-term-success

Reality about Initial Career Development in Engineering

Leave a Comment / Civil Engineering, Structural Engineering /

While the global job market might be a bit slow, the job market in India is active and full of opportunities. Many young people are not planning their careers for long-term success, and this is something we’ve noticed. In the initial parts of a career, efforts like learning, working on projects, and investing time and money are required to develop the basic foundation required for long-term success. Now a days, young engineering candidates feel that they will get good jobs and a good salary once they are out of college. So they join some crash course in software and put that in their CV, feeling as if the employer is waiting for them. But when these candidates are interviewed, it is seen that they do not have even the most basic knowledge, which they learned in school or in engineering. Due to social media in general, these young candidates are becoming victims of instant gratification and feel that they will get success instantly once they pass out. Really, what they have done in engineering and what they are doing now to develop their knowledge is a big question. But their expectations without their ability are very high. Once they fail in their interview, they blame their fate. So, young people need to think about how they want to build their careers and develop their initial professional development plan. Note that it is not a shortcut plan like a crash course. Students can see successful people and learn from them how they reached that position through hard work, failures, patience, etc. from the beginning. Among the people we interview, a large group are those who are studying hard for exams like MPSC and UPSC or are freshers. These exams are tough, and sometimes people spend 3 to 4 years preparing. But only a few (say 5%) get selected for government jobs, and almost 95% of candidates are now wondering what to do next. Another group of people we interview are those who have studied even more after college, like post-graduation, or who have taken special computer courses. But sometimes, even with extra education, they’re not ready for basic jobs when they apply for them. It’s good to study more, but where and what you study are important. Going to a good college like the IITs or NITs may not be possible for all students. Sometimes, people with only a few skills, like one small piece of software knowledge, end up doing simple jobs in big companies. In many outsourcing companies, clerical or machine-type work is expected of candidates. At the start, these candidates get good pay. But after a while, they don’t get paid much more, and if they want to change jobs, they don’t get it because they only know a few things or have a few skills. These people should place more emphasis in the beginning of their careers on basic concepts, practical application, and the overall knowledge required to become experts in their field to reach a successful and sustainable position in the long term. To have a successful career, you should be good at what you study and know a lot about what you love. Also, you should know how to apply your knowledge in practice. This will help you grow in your career. In the first few years of your job, work at a place that matches what you’re good at, and you have a chance to get overall experience, not clerical experience. Working at the same place for 3–4 years will teach you a lot about your job. The things you learn during this time will guide your career. Get suggestions from the experts in the field. Choosing a job that feels meaningful and enjoyable will lead to growth in your career and a happy work life. It’s important to start early and work comfortably, developing the basic foundation for a successful career journey. Note that a seed takes a lot of time and effort to become a tree and bear fruit. With patience and consistent efforts, fruits will automatically follow. Looking at this kind of scenario in the industry, we have developed the “Initial Career Development Master Course in Structural Engineering” for Civil Engineers, where you will get the all-round knowledge required to become successful in a structural engineering career all over the world. Anil MahadikStructural Consultant and AuditorChartered EngineerTraining Coach

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Industrial steel structures

Unlock your civil engineering potential with Learning Beyond

Leave a Comment / Structural Engineering, Civil Engineering, Construction, Education System /

Once upon a time, in a world filled with towering skyscrapers, intricate bridges, and awe-inspiring infrastructure, there was a group of individuals who possessed the power to shape the very foundations of our society. They were the unsung heroes, the masterminds behind the scenes, The Civil Engineers. At the heart of this tale lies a passion for creating a better world. Our brand, Learning Beyond, was born out of a deep desire to empower aspiring engineers with the knowledge and skills needed to transform their dreams into reality. Civil Engineering At Learning Beyond At Learning Beyond, we understand that every great achievement begins with a solid foundation. Our carefully crafted course is designed to equip you with the fundamental principles of structural engineering, laying the groundwork for your future success. With us, you will learn to analyse and design structures and master the art of sustainable development. We offer more than just theoretical knowledge. Our brand story is one of practicality and hands-on experience. We believe in learning by doing. That’s why our course goes beyond textbooks and classrooms. Through real-life case studies, immersive simulations, and interactive workshops, we bring the world of structural engineering to life. You will have the opportunity to work on real projects, our industry experts will guide you every step of the way. We understand that the journey to becoming a structural engineer can be challenging, but fear not! Learning Beyond community is always there to guid you. At Learning Beyond you are not just a student; you are part of a family. We foster an environment where you can connect with like-minded individuals, share your experiences, and learn from one another. Our instructors are not just teachers; they are mentors who will inspire and guide you as you embark on this incredible adventure. As the story unfolds, you will find yourself gaining the confidence and expertise needed to tackle complex structural engineering problems. You will witness your designs come to life, standing tall and proud against the backdrop of the world you helped shape. And as you reflect on your journey, you will realize that you have become part of a legacy, joining the ranks of the great civil engineers who came before you. So, dear dreamer, are you ready to step into this extraordinary world of structural engineering? Are you ready to turn your dreams into reality? Join us at Learning Beyond, where our brand story intertwines with yours, creating a narrative of innovation, resilience, and boundless possibilities. Together, let’s build a better tomorrow.

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Manual Design Of Doubly Reinforced Continuous Beam

Manual Design Of Doubly Reinforced Continuous Beam

2 Comments / Structural Engineering, Civil Engineering, High Rise Building /

In this blog, you will learn step by step about the design of doubly reinforced beam of first floor of G+5 building procedure using I.S. 456:2000. For daily blogs, subscribe to our blog page and learn complete information about the structural engineering industry Given:- 1) Grade of Concrete = M202) Grade of Steel = Fe5003) Clear Cover to Reinforcement (c) = 25 mm4) Length of Beam :-  B24 & B25 = 4060 mm5) Unit Weight of Concrete = 25 kN/m26) Slab Dimensions :- S3 = (1975 mm X 4060 mm) & S4 = (3050 mm X 4060 mm)7) Floor to Floor Height (H) = 3000 mm Step 1 :- Trial Dimension of The Beam :- Assume the Width of The Beam (b) = 230 mmEffective Depth of Beam (d) = (L/10) to (L/15) = (4290/10) to (4290/15)                                               = 429 mm to 286 mm                                  Take, d = 325 mmAssume, 20 mm Diameter Bars are to be Provided  at a Clear Cover of 25 mm.Therefore,  Overall Depth (D) = 325 + (20/2) + 25 = 360 mm ≈ 375 mm…(Rounded off on Higher Side)Therefore, Effective Depth of Beam Provided (d) = 375 – (20/2) – 25 = 340 mm  Step 2 :- Effective Span (Le) :- Minimum of The Following , i) Le = L + d = 4060 + 340 = 4400 mmii) Le = L + Bearing = 4060 + 230 = 4290 mmTherefore, Effective Span (Le) = 4290 mm Step 3 :- Load Calculations :- Load on Beam B24 & B25 :-i) Super Imposed Dead Load (SIDL) :-Wall Load = Wall Thickness X Floor to Floor Height X Unit Weight of Bricks                   = (0.150 X 3 X 20) = 9 kN/m                  …(No Deduction of Depth of Beam is Made From Floor to Floor Height)ii) Slab Load Transferring on Beam :-Slab S3 :-a) Self Weight of Slab (DL) = D X Unit Weight of Concrete                                             = 0.125 X 25                                             = 3.125 kN/m2b) Live Load (LL) = 2 kN/m2  …[Refer Table No. 1 , Page No. 7 , I.S. 875  (Part 2) : 1987]c) Super Imposed Dead Load (SIDL) =Floor Finish = Wt. of Screeding (50 mm Thk.) + Flooring (10 mm Thk.)                     = (0.05 X 24) + (0.01 X 22)    …[Refer I.S. 875 (Part 1) : 1987]                     = 1.42 kN/m2                     ≈ 1.5 kN/m2Total Load of Slab S3 (w)  = 3.125 + 2 + 1.5 = 6.625 kN/m2 Trapezoidal Load of  Slab S3 is Transferring  on Beam B24 & B25 Because S3 Slab is Two Way Slab Which is Given by,WS3 = [{(w. Lx)/2} X {1- (1/ 3. β2)}]       Where, β = l_y/l_x       = {(6.625 X 1.975 ) / 2} X [1- {1/ 3 X (4.06/1.975)2}]       = 6.02 kN/mSlab S4 :-a) Self Weight of Slab (DL) = D X Unit Weight of Concrete                                             = 0.125 X 25                                             = 3.125 kN/m2b) Live Load (LL) = 2 kN/m2  …[Refer Table No. 1, Page No. 7, I.S. 875  (Part 2) : 1987]c) Super Imposed Dead Load (SIDL) =Floor Finish = Wt. of Screeding (50 mm Thk.) + Flooring (10 mm Thk.)                    = (0.05 X 24) + (0.01 X 22)  … (Refer I.S. 875 (Part 1) : 1987)                    = 1.42kN/m2 ≈ 1.5 kN/m2Total Load of Slab S4 (w)  = 3.125 + 2 + 1.5 = 6.625 kN/m2 Trapezoidal Load of Slab S4 is Transferring  on Beam B24 & B25 Which is Given by,WS4 = [{(w. Lx)/2} X {1- (1/ 3. β2)}]       ={(6.625 X 3.05) / 2} X [1- {1/ 3 X (4.06/3.05)2}]       = 8.2 kN/mTherefore, Total Load of Slabs Transfers on Beam B24 & B25 = WS1 + WS4                                                                                                    = 6.02 + 8.2                                                                                                    = 14.22  kN/miii) Self Weight of Beam (DL) = b X D X Unit Weight of Concrete                                                 = 0.230 X 0.375 X 25                                                 = 2.156 kN/m Total Load on Each Beam B24 & B25 (W) = 9 + 14.22 + 2.156 = 25.376 kN/m                                                                                                      ≈ 25.4 kN/mUltimate Load (Wu) = 25.4 X 1.5 = 38.1 kN/m Step 4 :- Bending Moment (Mu) :- Max. Moment on ContinuousBeam (B24 & B25)  is (M) = W . Le2 / 8                                            = (25.4 X 4.292) / 8                                            = 58.43 kN.mUltimate Moment on ContinuousBeam (B20 & B21)  is (Mu) =58.43X 1.5 = 87.65 kN.m CONTINUOUS BEAM HAVING TWO EQUAL SPAN CARRYING UNIFORMALLY DISTRIBUTED LOAD (UDL) OVER ENTIRE SPAN :- Step 5 :- Limiting Moment of Resistance :- Mulim= 0.133 fck.b.d2Mulim= 0.133 X 20 X 230 X 3402Mulim= 0.133 X 20 X 230 X 3402Mulim= 70724080 NMulim= 70724080 / 106Mulim= 70.72 kN.mAs, Mulim<  Mu Note :- Either We Have to Increase Depth and Design as a Singly Reinforced Beam or if We Are Not Allowed to Increase Depth of Beam Then We Will Design the Beam as a Doubly Reinforced Beam. So We Design Assumed Beam of 230 X 375 mm As Doubly Reinforced Beam. Step 6 :- Balanced Moment (Mu2) :-                                 Equate,  Mu2 = Mu – Mulim Mu2 = 87.65 –  70.72                                Mu2 = 16.93 kN.m Step 7 :- Area of Steel Calculation (Ast) :- [Refer Cl. No. 26.5.1.1 (a) & (b), Page No. 46 & 47,I.S. 456 : 2000] Astmin = As/bd = 0.85/ FyAstmin= As = [(0.85 X 230 X 340) / 500]Astmin= As = 132.94 mm2Astmax = 0.04 b.D = 0.04 X 230 X375          = 3450 mm2 i) Steel in Tensile Zone :-Astreq= Ast1+Ast2 [For kumax & fsc Refer above Tables From, Illustrated  Design of Reinforced Concrete Buildings by S. R. Karve & V. L. Shah] ii) Steel in Compression Zone :-dc / d= (D – d) / d = (375 – 340) / 340 = 0.1By Interpolation,fsc = 412 N/mm2     …Refer Table 4.2.2 Step 8 :- Check For Shear :- Codal Provisions for Shear :- [Refer Cl. No. 26.5.1.6, Page No. 48 & Cl. No. 40.1, 40.3, Page No. 72 of  I.S. 456 : 2000] i) Maximum Shear Force (Vu) :-V = (5.W . Le/ 8)     = [(5 X 25.4 X 4.29) / 8]     = 68.1 kNVu= 68.1 X 1.5 = 102.15 kN ii) Nominal Shear Stress (τv) :-τv = Vu / b.d   

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DESIGN-OF-SINGLY-REINFORCED-BEAM-1

MANUAL DESIGN OF SINGLY REINFORCED BEAM

1 Comment / Structural Engineering, Civil Engineering, Construction, Design, Steel Structure /

In this blog, you will learn step by step about the design of singly reinforced beam of first floor of G+5 building procedure using I.S. 456:2000. For daily blogs, subscribe to our blog page and learn complete information about the structural engineering industry Given :- (SINGLY REINFORCED BEAM) 1) Grade of Concrete = M202) Grade of Steel = Fe5003) Clear Cover to Reinforcement (c) = 25 mm4) Length of Beam = 2660 mm5) Unit Weight of Concrete = 25 kN/m26) Slab Dimensions :- S1 = (2660 mm X 2890 mm) & S2 = (940 mm X 2660 mm)7) Floor to Floor Height (H) = 3000 mm Step 1 :- Trial Dimension of The Beam :- Assume the Width of The Beam (b) = 230 mmEffective Depth of Beam (d) = (L/10) to (L/15) = (2660/10) to (2660/15)                                               = 266 mm to 177.33 mmTake, d = 300 mm  …(Rounded off on Higher Side)Assume, 12 mm Diameter Bars are to be Provided  at a Clear Cover of 25 mm.Therefore, D = 300 + (12/2) + 25 = 331 mm ≈ 375 mm …(Rounded off on Higher Side)Therefore, Effective Depth of Beam Provided (d) = 375 – (12/2) – 25 = 344 mm  Step 2 :- Effective Span (Le) :- Le = L + d = 2660 + 344 = 3004 mm Step 3 :- Load Calculations :- i) Super Imposed Dead Load (SIDL) :-Wall Load = Wall Thickness X Floor to Floor Height X Unit Weight of Bricks                   = (0.150 X 3 X 20) = 9 kN/m                  …(No Deduction of Depth of Beam is Made From Floor to Floor Height)ii) Slab Load Transferring on Beam :-Slab S2 :-a) Self Weight of Slab (DL) = D X Unit Weight of Concrete                                             = 0.125 X 25                                             = 3.125 kN/m2b) Live Load (LL) = 2 kN/m2  …[Refer Table No. 1 , Page No. 7 , I.S. 875  (Part 2) : 1987]c) Super Imposed Dead Load (SIDL) =Floor Finish = Wt. of Screeding (50 mm Thk.) + Flooring (10 mm Thk.)                         + Sunk  Load(325mm Thk.)                     = (0.05 X 24) + (0.01 X 22) + (0.325 X 20)…[Refer I.S. 875 (Part 1) :1987]                     = 7.92 kN/m2                     ≈ 8 kN/m2Total Load of Slab S2 (w)  = 3.125 + 2 + 8 = 13.125 kN/m2 Rectangular Load of Slab S2 is Transferring  on Beam B6 Because S2 Slab is One Way Slab Which is Given by,WS2 = [(w. Lx) / 2] = [(13.125 X 1.17) / 2] = 7.678 kN/mSlab S1 :-a) Self Weight of Slab (DL) = D X Unit Weight of Concrete                                             = 0.125 X 25                                             = 3.125 kN/m2b) Live Load (LL) = 2 kN/m2  …[Refer Table No. 1 , Page No. 7 , I.S. 875  (Part 2) : 1987]c) Super Imposed Dead Load (SIDL) =Floor Finish = Wt. of Screeding (50 mm Thk.) + Flooring (10 mm Thk.)                     = (0.05 X 24) + (0.01 X 22)  …[Refer I.S. 875 (Part 1) : 1987]                     = 1.42kN/m2 ≈ 1.5 kN/m2Total Load of Slab S1 (w)  = 3.125 + 2 + 1.5 = 6.625 kN/m2 Triangular Load of Slab S1 is Transferring  on Beam B6 Because S1 Slab is Two Way Slab Which is Given by,WS1 = [(w. Lx)/3] = [(6.625 X 2.66)/3) = 5.874 kN/mTherefore, Total Load of Slabs Transfering on Beam B6 = WS1 + WS2 = 5.874+7.678                                                                                                            = 13.55  kN/m iii) Self Weight of Beam (DL) = b X D X Unit Weight of Concrete                                                 = 0.230 X 0.375 X 25                                                 = 2.156 kN/mTotal Load on Beam B6 = Wall Load + Slab load + Self Wt. of Beam                                       =  9 + 13.55 + 2.156 = 24.706 kN/mUltimate Load (Wu) = 24.706 X 1.5 = 37.059 kN/m. Step 4 :- Bending Moment (Mu) :- Mu= Wu . Le2/ 8 = 37.059 X 3.0042 / 8 = 41.8 kN.m Step 5 :- Check For Depth :-  Equate   Mumax & Mulim,              Mumax =  Mulim41.8 X 106 = 0.133 X fck X b X dreq2               41.8 X 106 = 0.133 X 20 X 230 X dreq2dreq= 261.38 mm  < 344mm   …(dreq< dprovided)              Therefore,  Safe Step 6 :- Area of Steel Calculations  (Ast) :- [Refer Cl. No. 26.5.1.1 (a) & (b),Page No. 46 & 47,I.S. 456 : 2000] Therefore,  Provide  4-T12  Bars.Therefore, Astprovided = (One Bar Area). (No. of Bars to be Provided)                                 = π/4 X〖 12〗^2  X 4                 Astprovided  = 452.45 mm2                                 >Astmin (134.504 mm2)   …ok                                >Astmax (3450 mm2)        …ok Step 7 :- Check For Shear :- Codal Provisions for Shear :- [Refer Cl. No. 26.5.1.6, Page No. 48 & Cl. No. 40.1, 40.3, Page No. 72 of  I.S. 456 : 2000] i) Maximum Shear Force (Vu) :-Vu= Wu . Le/ 2  ….( For Simply Supported Beam )     =  [(37.059 X 3.004) / 2]     = 55.662 kN ii) Nominal Shear Stress (τv) :-τv = Vu / b.d    = [(55.662 X 103) / (230 X 344)]= 0.7 N/mm2 iii) % of Steel (Pt) :-Pt=  [(100. Astprovided) / b.d]Therefore , Pt = (100 X 452.45) / (230 X 344) = 0.57 % iv) Design Shear Stress (τc) :- (Refer Table No. 19, Page No.73, I.S. 456 : 2000)By Interpolation,τc = [0.48 + {(0.56-0.48) / (0.75-0.5)}X (0.57-0.5)]    = 0.5  N/mm2  < 0.7 N/mm2As, τc < τvWe Need to Design For Shear Reinforcement. Pt τc (For M20) 0.5 0.48 0.57 ? 0.75 0.56 As per Table No. 20 , Page No. 73 , I.S. 456 : 2000Maximum Shear Stress for M20 Grade Concrete is, τcmax= 2.8 N/mm2Therefore,  Beam B6 is Safe in Shear. Step 8 :- Design of Shear Reinforcement :- i) Shear Resisted by Stirrups (Vus) :-Vus= Vu – Vuc ….( For Simply Supported Beam )      = Vu – τc .b.d      = 55.662 X 103 – 0.5 X 230 X 344      = 16102 N      = 16.102 kNProvide 2-Legged 8mm Diameter StirrupsAsv = 2 X π/4 X 64 = 100.544 mm2 ii) Spacing of Stirrups (Sv) :- iii) Check For Stirrups Spacing :-Spacing of Stirrups Should Not be Greater Than Minimum of The Following,Spacing of Stirrups (Sv) = Min. of    i) 0.75d                                                         ii) 300 mmTherefore,

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Manual Design Of Two Way Slab

Manual Design Of Two Way Slab

Leave a Comment / Structural Engineering, Civil Engineering, IS Codes /

In this blog, you will learn step by step about the manual design of a two way slab for one floor of the G + 5 building procedure using I.S. 456:2000. For daily blogs, subscribe to our blog page and learn complete information about the structural engineering industry Given :- 1) Grade of Concrete = M202) Grade of Steel = Fe5003) Clear Cover to Reinforcement (c) = 20 mm4) Dimension of Slab =  (3050mm X 4180 mm)5) Assume Beam Support (Bearing) = 230 mm6) Unit Weight of Concrete = 25 kN/m2 Step 1 :- Aspect Ratio (Refer Cl. No. D-1.11, Page No. 90 , I.S. 456:2000) Lx = 3050 mm & Ly = 4180 mm  …(Given)Therefore, Ly/Lx = 4180/3050 = 1.37 < 2  …(Two Way Slab) Step 2 :- Depth of Slab (D) Basic Value (B.V.) = L / D = 26Assume % of Steel = 0.25 %Therefore, Modification Factor (M.F.) = 1.35 …(Refer Fig. 4 , Page No. 38 , I.S.456:2000)Therefore, Effective Depth (d) = SPAN/B.V.X.M.F =  [3050 / (26 X 1.35)] = 86.894 mm Assume Bars to be Provided of Diameter = 8 mm & Provide Clear Cover to Reinforcement = 20 mmTherefore, Overall Depth of Slab (D) = d + c + (8/2) = 86.894 + 20 + 4 = 110.894 ≈ 125 mmTherefore, Effective Depth Provided (dprovided) = D – c – (8/2) = 125 – 20 – 4 = 101 mm Step 3 :- Effective Span (le)                Minimum of,       i) Le = Lx + Bearing = 3050 + 230 = 3280 mm                                          ii) Le = Lx + d = 3050 + 101 = 3151 mm               Therefore,  Effective Span (Le) = 3151 mm Step 4 :- Load Calculations Consider 1 m Strip of Slab,i) Self Weight of Slab (DL) = D X Unit Weight of Concrete                                             = 0.125 X 25                                             = 3.125 kN/m2ii) Live Load (LL)               = 2 kN/m2  …[Refer Table No. 1 , Page No. 7 , I.S. 875  (Part 2) : 1987]iii) Super Imposed Dead Load (SIDL) =      Floor Finish = Wt. of Screeding (50 mm Thk.) + Flooring (10 mm Thk.)                          = (0.05 X 24) + (0.01 X 22) …[Refer I.S. 875 (Part 1) : 1987]                          = 1.42 kN/m2                          ≈ 1.5 kN/m2 Total Load (W) = 3.125 + 2 + 1.5                          = 6.625 kN/m2                          ≈ 6.63 kN/m2Total Ultimate Load (Wu) = 6.63 X 1.5                                           =  9.945 kN/m2                                            ≈ 9.95 kN/m2 Step 5 :- Maximum Design Bending Moment (Mumax) Ly/Lx 1.3 1.37 1.4 -αx 0.065 ? 0.071 +αx 0.049 ? 0.053 Refer Table No. 26, Page No. 90 , I.S. 456:2000 Shorter Span Moments :- 1) -ve Moment at Continuous Edge :- By Interpolation, -αx = 0.065+[{(0.071-0.065) / (1.4-1.3)} X (1.37 – 1.3)]      = 0.0692 Mu = -αx . Wu . le2     = 0.0692 X 9.95 X  3.1512        =  6.836 kN.m 2) +ve Moment at Mid-Span :-By Interpolation,+αx = 0.049+[{(0.053-0.049) / (1.4-1.3)} X (1.37 – 1.3)]      = 0.0518 Mu = +αx . Wu . le2     = 0.0518 X 9.95 X  3.1512       =  5.117 kN.m Longer Span Moments :- 3) -ve Moment at Continuous Edge :-Mu = -αy . Wu . le2     = 0.047 X 9.95 X  3.1512        = 4.64 kN.m 4) +ve Moment at Mid-Span :-Mu = +αy . Wu . le2     = 0.035 X 9.95 X  3.1512       = 3.46 kN.m Therefore, Maximum Design Bending Moment (Mumax) =  6.836 kN.m Step 6 :- Check For Depth               Equate   Mumax & Mulim,              Mumax =  Mulim             6.836 X 106 = 0.133 X fck X b X dreq2              6.836 X 106 = 0.133 X 20 X 1000 X dreq2dreq= 50.7 mm  < 101 mm   …(dreq< dprovided)              Therefore,  Safe Step 7 :- Area of Steel Calculation :- Spacing of Bars :- As per Cl. No. 26.3.3 (b) , Page No. 46 , I.S. 456 : 2000Maximum SpacingBetween Bars          < Minimum of   i) 3d                                                          ii) 300 mmTherefore,Maximum Spacing Between Bars = i) 3 X 101 = 303 mm                                                         ii) 300 mm Note:- Generally We Provide 200 mm Maximum Spacing Between Bars. Spacing of Bars = Area of one bar/ Area of steel * 1000 Along Shorter Span (Lx)  :- 1) At Support :-          = 162.18 mm2Spacing of Bars = (Area of One Bar / Area of Steel) . 1000Consider Bars to Provide are of Diameter 8mm.Therefore, Spacing of Bars = [(π/4) X 82 / 162.18] X 1000 = 3709.98 mmNote:- Generally We Provide 200 mm Maximum Spacing Between Bars. 2) At Mid-Span :-          = 120.1 mm2Spacing of Bars = (Area of One Bar / Area of Steel) . 1000Consider Bars to be Provide are of Diameter 8mm.Therefore, Spacing of Bars = [(π/4) X 82 / 120.1 ] X 1000 = 418.58 mmNote:- Generally We Provide 200 mm Maximum Spacing Between Bars. Along Longer Span (Ly)  :-   d =  dprovided – diameter of bar provided along shorter span = 101 – 8 = 93 mm 1) At Support :-          = 118.53 mm2Spacing of Bars = (Area of One Bar / Area of Steel) . 1000Consider Bars to be Provide are of Diameter 8mm.Therefore, Spacing of Bars = [(π/4) X 82 / 118.53] X 1000 = 424.13mmNote:- Generally We Provide 200 mm Maximum Spacing Between Bars. 2) At Mid-Span :-          = 87.74 mm2Spacing of Bars = (Area of One Bar / Area of Steel) . 1000Consider Bars to be Provide are of Diameter 8mm.Therefore, Spacing of Bars = [(π/4) X 82 / 87.74] X 1000 = 572.97 mmNote:- Generally We Provide 200 mm Maximum Spacing Between Bars. Note:- Calculation of Distribution Steel is Eliminated, Since it is a Two Way Slab. Step 8 :- Check For Deflection :- % of Steel = Pt = 100. Astreq / b.dTherfore, Pt = (100 X 120.1) / (1000 X 101) = 0.11 %Therefore,  Fs = 0.58 X 500 X (120.1 / 251.36) = 138.56  N/mm2% of Steel Provided = Pt = 100. Astprovided  / b.dTherefore , Pt = (100 X 251.36) / (1000 X 101) = 0.248 %M.F. = 2.0Now,

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Manual Design Of One Way slab

Manual Design Of One Way Slab

3 Comments / Structural Engineering, Civil Engineering, Design /

In this blog you will learn about Design of one way slab of first floor of G + 5 building procedure using I.S. 456:2000. Learn Complete Structural Engineering Industry Requirements Practically With Our Daily Blogs Series Given:- 1) Grade of Concrete = M202) Grade of Steel = Fe5003) Clear Cover to Reinforcement (c) = 20 mm4) Dimension of Slab =  (940mm X 2660 mm)5) Assume Beam Support (Bearing) = 230 mm6) Unit Weight of Concrete = 25 kN/m2 Step 1 :-Aspect Ratio (Refer Cl. No. D-1.11, Page No. 90 , I.S. 456:2000)Lx = 940 mm & Ly = 2660 mm  …(Given)Therefore, Ly/Lx = 2660/940 = 2.82 > 2   …(One Way Slab) Step 2 :-Trial Depth of Slab (D) Basic Value (B.V.) = L / D = 20Assume % of Steel = 0.25 %Therefore, Modification Factor (M.F.) = 1.35 …(Refer Fig. 4 , Page No. 38 I.S.456:2000)Therefore, Effective Depth (d) = SPAN/B.V.X.N.F. =  [940/ (20 X 1.35)] = 34.81 mmAssume Bars to be Provided of Diameter = 8 mm & Provide Clear Cover to Reinforcement = 20 mm Therefore, Overall Depth of Slab (D) = d + c + (8/2) = 34.81 + 20 + 4 = 58.81 mmAssume , Overall Depth of Slab (D)  = 125 mmTherefore, Effective Depth Provided (dprovided) = D – c – (8/2) = 125 – 20 – 4 = 101 mm Step 3 :-Effective Span (le)                Minimum of,       i) Le = Lx + Bearing = 940 + 230 = 1170 mm                                          ii) Le = Lx + d = 940 + 101 = 1041 mm                Therefore,  Effective Span (Le) = 1041 mm Step 4 :- Load Calculations Consider 1 m Strip of Slab,i) Self Weight of Slab (DL) = D X Unit Weight of Concrete                                             = 0.125 X 25                                             = 3.125 kN/m2ii) Live Load (LL) = 2 kN/m2  …[Refer Table No. 1 , Page No. 7, I.S. 875  (Part 2) : 1987]iii) Super Imposed Dead Load (SIDL) =Floor Finish = Wt. of Screeding (50 mm Thk.) + Flooring (10 mm Thk.) + Sunk  Load(325 mm Thk.)                     = (0.05 X 24) + (0.01 X 22) + (0.325 X 20)…[Refer I.S. 875 (Part 1) :1987]                     = 7.92 kN/m2                     ≈ 8 kN/m2Total Load (W) = 3.125 + 2 + 8                          = 13.125 kN/m2Total Ultimate Load (Wu) = 13.125X 1.5 ≈ 19.7 kN/m2 Step 5 :- Maximum Design Bending Moment (Mumax) Refer Table No. 26, Page No. 90 , I.S. 456:2000 Span Moment :- Mumax= Wu . le2 / 8Mumax = 19.7 X  1.0412 / 8 Mumax =  2.668 kN.mTherefore ,Mumax= 2.668 kN.m Step 6 :- Check For Depth               Equate   Mumax & Mulim,              Mumax =  Mulim               2.668 X 106 = 0.133 X fck X b X dreq2               2.668 X 106 = 0.133 X 20 X 1000 X dreq2dreq= 31.67 mm  < 101mm   …(dreq< dprovided)              Therefore,  Safe Step 7 :- Area of Steel Calculation :- Spacing of Bars :- As per Cl. No. 26.3.3 (b) , Page No. 46 , I.S. 456 : 2000Maximum SpacingBetween Bars          < Minimum of   i) 3d                                                          ii) 300 mm Therefore,Maximum Spacing Between Bars = i) 3 X 101 = 303 mm                                                         ii) 300 mmNote:- Generally We Provide 200 mm Maximum Spacing Between Bars.         = 61.7 mm2Spacing of Bars = (Area of One Bar / Area of Steel) . 1000Consider Bars to Provide are of Diameter 8mm.Therefore, Spacing of Bars = [(π/4) X 82 / 61.7] X 1000 = 814.78 mmNote:- Generally We Provide 200 mm Maximum Spacing Between Bars. Therefore, Provide 8 mm Diameter Bars at 200 mm c/c. Step 8 :- Distribution Steel Calculation :- Dst = 0.12 % of the Gross Area of Slab     …(For Fe500)Therefore, Dst= (0.12/100) X Ag                  Dst= (0.12/100) . b . D  …(Where, b = 1000 mm &                                                      D = Overall Depth of Slab)                  Dst= (0.12/100) X 1000 X 125                       = 150 mm2Spacing of Bars = (Area of One Bar / Area of Steel) . 1000Consider Bars to be Provide are of Diameter 8 mm.Therefore, Spacing of Bars = [(π/4) X 82 / 150] X 1000 = 335.15 mmMaximum SpacingBetween Bars          < Minimum of   i) 5d  = 5 X 101 = 505 mm                                                          ii) 450 mmNote:- Generally We Provide 200 mm Maximum Spacing Between Bars. Therefore, Provide 8 mm Diameter Bars at 200 mm c/c. Step 9 :- Check For Deflection :- % of Steel = Pt = 100. Astreq / b.dTherefore, Pt = (100 X 61.7) / (1000 X 101) = 0.061 % Therefore,  Fs = 0.58 X 500 X (61.7 / 251.36) = 71.18  N/mm2% of Steel Provided = Pt = 100. Astprovided  / b.dTherefore , Pt = (100 X 251.36) / (1000 X 101) = 0.248 %M.F. = 2.0    As, dreq < dprovided    …Safe for Deflection. REINFORCEMENT DETAILS :-

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Manual Design Of Footing Of G+5 Building Using I.S. 456:2000

4 Comments / Structural Engineering, Civil Engineering, Design, footing, IS Codes, Structure /

In this blog you will learn Manual Design Of Footing of G+5 Building Using I.S. 456:2000 with the help of example. Step by step guid to clear conceptLearn Complete Structural Engineering Indusry Requirements Practically With Our Daily Blogs Series Given: Load Calculation (Step 1): Load From Column      = 1500 kN Self Weight of Footing = Assume 10% of the column load                                     = (10/100) X 1500                                     = 150 kN  ———————————————————–Total Weight (W)         = 1650 kN Area of Footing (A) (Step 2): Area of Footing (A) = Total Weight (W) / Net Soil Bearing Capcity (S.B.C.)= 1650 / 300= 5.5 m2Take Equal Projections From Both Sides, 5.5 = (0.6 + 2x) . (0.3 + 2x)5.5 = 0.18 +1.2x + 0.6x + 4×25.5 = 0.18 +1.8x + 4×2Therefore, by Solving Above Equation,x = 0.95 mTake Projection of 1 m on Both Side. Therfore, Size of Footing = (0.3+ 2 X 1 ) & (0.6 + 2 X 1) = 2.3 X 2.6 mTherefore Area of Footing Provided = 2.3 X 2.6 = 5.98 m2 Upward Soil Pressure (Pau) (Step 3): Pau = (Factored Load of Column / Actual Area of Footing Provided)     = 1.5 X 1500 / 5.98Pau= 376.25 kN/m2 Bending Moment (B.M.) (Step 4): The Critical Section for B.M. Along the Column Face as Shown in the Figure 2, From X-axis:      = (2.6 – 0.6) / 2     = 1 mConsider  a Strip of 1m.Therefore, Mux = [Pau X 1 X (Cx2/2)]                         = [376.25 X 1 X (12 /2)]                 Mux = 188.125  kN.m From Y-axis:      = (2.3 – 0.3) / 2     = 1 mTherefore, Muy = [Pau X 1 X (Cy2/2)]                         = [376.25 X 1 X (12 /2)]                  Muy = 188.125  kN.m Depth of Footing Required (dreq) (Step 5): Equate The Maximum Bending Moment (Max. of Mux & Muy) With Limiting Moment of Resistance ,Consider a Strip of 1m.Mumax= Mulim188.125 X 106 = 0.133 fck.b.dreq2188.125 X 106 = 0.133 X 20 X 1000 X dreq2dreq = 265.94 mmProvide 12 mm Diameter Bars at a Clear Cover of 50 mm.Therefore, Overall Depth (D) = 265.94 + 50 + (12/2)                                                = 321.94 mm ≈ 600 mm …(rounded -off on higher side) Overall Depth (D) = 450 mmTherefore, Effective Depth Provided (dprovided) =  600 – 50 – (12/2) = 544 mm Area of Steel Calculation (Ast) (Step 6): Astx Along X-Direction: Astx = 826.8 mm2Provide 12 mm Diameter Bars.Spacing = (π/4 × 12² ÷ 826.8) ×100                        = 136.8 mm ≈ 125 mmProvide 12 mm Diameter Bars @ 125 mm c/c Asty Along Y-Direction: Asty = 826.8 mm2Provide 12 mm Diameter Bars.Spacing = (π/4 × 12² ÷ 826.8) ×100                        = 136.8 mm ≈ 100 mmProvide 12 mm Diameter Bars @ 100 mm c/c Check For One Way Shear (Step 7): The Critical Section for One Way Shear is Taken at a Distance of ‘d’ From Column Periphery. The Offset x = Cx – d = 1 – 0.544 = 0.456 mSimilarly,  y = Cy – d  = 1 – 0.544 = 0.456 mDepth of Footing at Critical Section (d’) = 544 mm Shear Force Along X-Direction :-Vux = x . L . Pau = 0.456 X 1 X 376.25 = 171.57 kN Shear Force Along Y-Direction :-Vuy = y . B . Pau = 0.456 X 1 X 376.25 = 171.57 kN Vumax = 209.2 kN Nominal Shear Stress (τv): τv  = Vumax/ b’.d’ = [(171.57 X 103) / (1000 X 544)]  = 0.32 N/mm2 % of Steel (Pt): Pt = 100 Astmax / b’d’ = [(100 X 1130) / (1000 X 544)] = 0.21 % Design Shear Stress (τc): τc =  0.32 N/mm2  > 0.32 N/mm2As, τc > τv    … Safe in One Way Shear Pt τc  ( For M20) <0.15 0.28 0.21 ? 0.25 0.36 Refer Table No. 19, Page No.73, I.S. 456 : 2000 Check For Two Way Shear (Step 8): The Critical Section for Two Way Shear is Taken at a Distance of ‘d/2’ From Column Periphery. Depth of Footing at Critical Section (D’) = 750 mmEffective Depth at Critical Section = d’ = 600 – 50 – (12/2 ) = 544 mmb’ = 2 . Perimeter of Critical Section    = 2 X (1144 + 844)    = 3976 mm Shear Force on Shaded Area is Given by,Vu = (Footing Area – Critical Section Area) . Upward Soil PressureVu = [(2.6 X 2.3) – (1.144 X 0.844)] X 376.25Vu = 1886.7 kN Nominal Shear Stress (τv): τv  = Vumax/ b’. d’ = [(1886.7 X 103) / (3976 X 544)]  = 0.87 N/mm2 Design Shear Stress (τc’): τc‘= ks. (0.25√fck )Where,  ks= Minimum of,    i) 1                                            ii) 0.5 + βc = 0.5 + 0.5 = 1                                             Where, βc = (Breadth / Depth)column = 300 / 600 = 0.5Therefore, ks= 1 Therefore,   τc‘= 1 X (0.25√20) = 1.118 N/mm2   >  0.87 N/mm2As, τc > τv    … Safe in Two Way Shear

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Which skills do structural engineer required?

1 Comment / Structural Engineering, Civil Engineering, Construction, Education System, Steel Structure /

Structural engineering is an approximate science. It has been seen that even practicing structural engineers are also finding difficulty in understanding structural behaviour and its consequences from analysis, design, drawing till site construction phase. Most of the times the assumption part is missed during design of structure considering life cycle of the structure. Skill set for civil engineers Now a day’s structural engineers are using sophisticated softwares like ETABS, STAAD, RAM etc. While using those softwares they depend on software fully which may lead to blunders if the results are not verified by preliminary manual calculations and behaviour understanding. In practice the problems are different and difficult to define. This requires overall structural understanding and critical thinking process. Classical understanding to decide form, material and load path of structure is also required. Understanding this requirement of structural engineers we have a knowledge sharing course which will help to bridge up the understanding gap.

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