Concrete and Steel in Construction PDF

CONCRETE AND STEEL IN CONSTRUCTION Principles of Reinforced Concrete JRPLUCENA by Robert Lucena LEARNING OUTCOMES 1. Discuss the use and properties of concrete and steel in constructing structures What we...

CONCRETE AND STEEL IN CONSTRUCTION Principles of Reinforced Concrete JRPLUCENA by Robert Lucena LEARNING OUTCOMES 1. Discuss the use and properties of concrete and steel in constructing structures What we will do 1. Discussion I will discuss concepts, theories, information, etc. related to the topic to attain objectives 1 to 2 2. Open Forum and Student Evaluation Let’s see how much you learn and what are your ideas to share with others 3. Problem Solving Let’s apply what we learn! What we will do 2. Open Forum and Student Evaluation Let’s see how much you learn and what are your ideas to share with others 3. Problem Solving Let’s apply what we learn! 4. Lecturer Evaluation Answer the evaluation form so I can improve my lecture  Inside our toolbox Calculators Notes and pen References  1. Discuss the use and properties of concrete Argument and steel in constructing structures Should the use of reinforced concrete structures be prioritized over alternative sustainable building materials in modern urban construction, considering both environmental impact and structural longevity?  1. Discuss the use and properties of concrete Concrete in Construction and steel in constructing structures Concrete is a mixture of sand, gravel, crushed rock, or other aggregates held together in a rocklike mass with a paste of cement and water. Aggregates Binder  1. Discuss the use and properties of concrete Concrete in Construction and steel in constructing structures Assessment (item 1) Advantages Disadvantages List 4 and justify with your group List 4 and justify with your group  1. Discuss the use and properties of concrete Concrete in Construction and steel in constructing structures Advantages Disadvantages Considerable compressive strength Low tensile strength Resistance to fire and water Additional cost for formworks Rigidity Heavy members Economic Quality control Castable to various shapes Minimal need of highly skilled workers  1. Discuss the use and properties of concrete Concrete in Construction and steel in constructing structures Design Codes ACI 318-11 Building Code Requirements NSCP 2015 for Structural Concrete  1. Discuss the use and properties of concrete Concrete in Construction and steel in constructing structures Assessment (item 2) Types of Portland Cement List 5 and discuss in the group the applicability  1. Discuss the use and properties of concrete Concrete in Construction and steel in constructing structures Types of Portland Cement Type I - The common, all-purpose cement used for general construction work. Type II - A modified cement that has a lower heat of hydration than does Type I cement and that can withstand some exposure to sulfate attack. Type III - A high-early-strength cement that will produce in the first 24 hours a concrete with a strength about twice that of Type I cement. This cement does have a much higher heat of hydration. Type IV - A low-heat cement that produces a concrete which generates heat very slowly. It is used for very large concrete structures. Type V - A cement used for concretes that are to be exposed to high concentrations of sulfate.  1. Discuss the use and properties of concrete Concrete in Construction and steel in constructing structures The compressive strength of concrete, 𝑓𝑐 ′ , is determined by testing to failure 28-day-old 6-in. diameter by 12-in. concrete cylinders 𝑓𝑐 ′ ranges from 17 MPa (2,500 psi) to 138 MPa (20,000 psi) Ultimate strength Concrete fails at at strain of 0.002 0.003 strain  1. Discuss the use and properties of concrete Concrete in Construction and steel in constructing structures Assessment (item 3) How can you obtain the modulus of elasticity of concrete given the stress- strain curve (chat in messenger use PM)  1. Discuss the use and properties of concrete Concrete in Construction and steel in constructing structures Static modulus of elasticity, 𝑬𝒄 (a) The initial modulus is the slope of the stress–strain diagram at the origin of the curve. (b) The tangent modulus is the slope of a tangent to the curve at some point along the curve—for instance, at 50% of the ultimate strength of the concrete. (c) The slope of a line drawn from the origin to a point on the curve somewhere between 25% and 50% of its ultimate compressive strength is referred to as a secant modulus. (d) Another modulus, called the apparent modulus or the long- term modulus, is determined by using the stresses and strains obtained after the load has been applied for a certain length of time.  1. Discuss the use and properties of concrete Concrete in Construction and steel in constructing structures Static modulus of elasticity, 𝑬𝒄 𝐸𝑐 = 𝑤𝑐 1.5 (0.043) 𝑓𝑐 ′ with 𝑤𝑐 varying from 1500 to 2500 𝑘𝑔/𝑚3 and 𝑓𝑐 ′ in MPa Simplified (normal concrete having a density of 2, 320 𝑘𝑔/𝑚3 ) 𝐸𝑐 = 4700 𝑓𝑐 ′  1. Discuss the use and properties of concrete Concrete in Construction and steel in constructing structures Tensile Strength of Concrete The tensile strength of concrete varies from about 8% to 15% of its compressive strength. A major reason for this small strength is the fact that concrete is filled with fine cracks. ASTM C78 Standard Test Method for Flexural Strength of Concrete (Using Simple Beam with Third-Point Loading) Flexural tensile strength, 𝑓𝑟 , is defined as the modulus of rupture ′ 𝒇𝒓 = 𝟎. 𝟕 𝒇𝒄 𝑴𝑷𝒂  1. Discuss the use and properties of concrete Concrete in Construction and steel in constructing structures Effect of Aggregates Flexural tensile strength, 𝑓𝑟 , is defined as the modulus of rupture ′ 𝒇𝒓 = 𝟎. 𝟕𝝀 𝒇𝒄 𝑴𝑷𝒂 λ depends on the aggregate that is replaced with lightweight material. If only the coarse aggregate is replaced (sand-lightweight concrete), λ = 0.85. If the sand is also replaced with lightweight material (all- lightweight concrete), λ = 0.75.  1. Discuss the use and properties of concrete Concrete in Construction and steel in constructing structures Effect of Aggregates Flexural tensile strength, 𝑓𝑟 , is defined as the modulus of rupture ′ 𝒇𝒓 = 𝟎. 𝟕𝝀 𝒇𝒄 𝑴𝑷𝒂 Linear interpolation is permitted between the values of 0.85 and 1.0 as well as from 0.75 to 0.85 when partial replacement with lightweight material is used.  1. Discuss the use and properties of concrete Steel as Reinforcement to Concrete and steel in constructing structures Assessment (item 4) What are the grades of reinforcing steel according to ASTM  1. Discuss the use and properties of concrete Steel as Reinforcement to Concrete and steel in constructing structures Reinforcing Steel Grades ASTM A615: S (for type of steel) yield strength levels: 40,000 psi (280 MPa); 60,000 psi (420 MPa); 75,000 psi (520 MPa); and 80,000 psi (550 MPa) ASTM A706: Low-alloy deformed and plain bars, W (for type of steel), 60,000 psi (420 MPa) and 80,000 psi (550 MPa), ASTM A996: Deformed rail steel or axle steel bars. R (for type of steel)  1. Discuss the use and properties of concrete Steel as Reinforcement to Concrete and steel in constructing structures Reinforcing Steel Bar Sizes Commercially available diameters (in mm): 8, 10, 12, 16, 20, 25, 32, 36 1. Discuss the use and properties of concrete and steel in constructing structures Engineering Lectures Scan QR  THANK YOU Time for open forum and evaluation Scan QR FLEXURE ANALYSIS OF CONCRETE BEAM Principles of Reinforced Concrete JRPLUCENA by Robert Lucena LEARNING OUTCOMES 1. Investigate the cracking moment of a concrete beam using principles of concrete strength 2. Investigate the allowable moment capacity of a cracked reinforced concrete beam using the principle of elastic stress What we will do 1. Discussion I will discuss concepts, theories, information, etc. related to the topic to attain objectives 1 to 2 2. Open Forum and Student Evaluation Let’s see how much you learn and what are your ideas to share with others 3. Problem Solving Let’s apply what we learn! What we will do 2. Open Forum and Student Evaluation Let’s see how much you learn and what are your ideas to share with others 3. Problem Solving Let’s apply what we learn! 4. Lecturer Evaluation Answer the evaluation form so I can improve my lecture  Inside our toolbox Calculators Notes and pen References  1. Investigate the cracking moment of a concrete Behavior of Concrete Under Increasing Load beam using principles of concrete strength Uncracked Concrete Stage Tensile stress developed is less than the modulus of rupture  1. Investigate the cracking moment of a concrete Behavior of Concrete Under Increasing Load beam using principles of concrete strength Cracked Concrete Stage Tensile stress developed is equal to the modulus of rupture. The amount of moment that caused the start of cracking is the cracking moment, 𝑴𝒄𝒓  1. Investigate the cracking moment of a concrete Behavior of Concrete Under Increasing Load beam using principles of concrete strength Cracked Concrete Stage – Elastic Stresses Stage Concrete at the tension zone cracks, reinforcing bar starts to carry tension. This will continue as long as the compression stress in the top fibers is less than about one-half of the concrete’s compression strength, 𝑓𝑐 ′ , and as long as the steel stress is less than its yield stress  1. Investigate the cracking moment of a concrete Behavior of Concrete Under Increasing Load beam using principles of concrete strength Ultimate Strength Stage (Beam Failure) As the load is increased further so that the compressive stresses are greater than 0.50 𝑓𝑐 ′ , the tensile cracks move farther upward, as does the neutral axis, and the concrete compression stresses begin to change appreciably from a straight line.  1. Investigate the cracking moment of a concrete Behavior of Concrete Under Increasing Load beam using principles of concrete strength Ultimate Strength Stage (Beam Failure) As the load is increased further so that the compressive stresses are greater than 0.50 𝑓𝑐 ′ , the tensile cracks move farther upward, as does the neutral axis, and the concrete compression stresses begin to change appreciably from a straight line.  1. Investigate the cracking moment of a concrete Cracking Moment beam using principles of concrete strength Cracking Moment, 𝑴𝒄𝒓 𝑀𝑐 𝑓𝑟 𝐼𝑔 𝑓= 𝑀𝑐𝑟 = 𝐼𝑔 𝑦𝑡 𝑓𝑟 = modulus of rupture 𝐼𝑔 = gross moment of inertia of the beam* 𝑦𝑡 = fiber distance from neutral axis *The area of reinforcing as a percentage of the total cross-sectional area of a beam is quite small (usually 2% or less), and its effect on the beam properties is almost negligible as long as the beam is uncracked. LEARNING EXERCISES 1 The stress on the extreme tension fiber must be checked to see if the concrete will crack. To check, Assuming the concrete is uncracked, compute the bending stresses in the 𝑀𝑐 20 106 227.5 𝑓𝑡 = = extreme fibers of the beam for a bending 𝐼𝑔 305 4553 moment of 20 kN-m. The normal-weight 12 concrete has an 𝑓𝑐 ′ of 28 MPa 𝒇𝒕 = 𝟏. 𝟗𝟎 𝑴𝑷𝒂 The modulus of rupture can be obtained: 380 455 ′ 𝑓𝑟 = 0.7𝜆 𝑓𝑐 = 0.7(1.00) 28 75 𝒇𝒓 = 𝟑. 𝟕 𝑴𝑷𝒂 305 LEARNING EXERCISES 1 Since, Assuming the concrete is uncracked, 𝑓𝑡 < 𝑓𝑟 compute the bending stresses in the extreme fibers of the beam for a bending It can be said that the concrete on the tension zone will not moment of 20 kN-m. The normal-weight crack concrete has an 𝑓𝑐 ′ of 28 MPa 380 455 75 305 LEARNING EXERCISES 2 The cracking moment is obtained using the equation: Solve for the cracking moment of the 305 4553 beam cross-section. The normal-weight 𝑓𝑟 𝐼𝑔 3.7 12 concrete has an 𝑓𝑐 ′ of 28 MPa 𝑀𝑐𝑟 = = 𝑦𝑡 227.5 𝑴𝒄𝒓 = 𝟑𝟖. 𝟗𝟒 𝒌𝑵 − 𝒎 380 455 75 305 1. Investigate the cracking moment of a concrete beam using principles of concrete strength  2. Investigate the allowable moment capacity of a Elastic Stress Analysis of Cracked Concrete Beam cracked reinforced concrete beam using the principle of elastic stress Cracked Concrete Stage – Elastic Stresses Stage On the tensile side of the beam, an assumption of perfect bond is made between the reinforcing bars and the concrete. strain in the concrete and in the steel will be equal at equal distances from the neutral axis.  2. Investigate the allowable moment capacity of a Elastic Stress Analysis of Cracked Concrete Beam cracked reinforced concrete beam using the principle of elastic stress Modular Ratio, n ratio of modulus of elasticity of two materials The stress in steel cannot be the same with the stress in the concrete because they have different modulus of elasticity. That is why, the area of steel is converted to equivalent concrete area by using the modular ratio. 𝐸𝑠 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙 𝑛= = 𝐸𝑐 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 *if n=10 it means that steel is 10 times stronger than concrete LEARNING EXERCISES 3 Check if the concrete will crack Compute the maximum bending stress in 𝑀𝑐 70 106 227.5 𝑓𝑡 = = the beam for a bending moment of 50 kN- 𝐼𝑔 305 4553 m. The normal-weight concrete has an 𝑓𝑐 ′ 12 of 28 MPa. Take n = 9. 𝒇𝒕 = 𝟒. 𝟕𝟓 𝑴𝑷𝒂 The concrete will crack since it has a modulus of rupture of 𝑓𝑟 = 3.7 𝑀𝑃𝑎. But the beam will be in elastic stage since 380 the stress is note more than 50% of 𝑓𝑐 ′ 455 3 – 25 mm dia. 75 305 LEARNING EXERCISES 3 From the discussion, since the concrete at the tension zone will not carry tension, the reinforcing steel will carry the Compute the maximum bending stress in stress. The principle of modular ratio will be utilized to the beam for a bending moment of 50 kN- develop a transformed section (an imaginary conversion of m. The normal-weight concrete has an 𝑓𝑐 ′ steel to concrete). Take note that the distance of the tensile of 28 MPa. Take n = 9. force as a resultant from the tensile stress must be the same with the distance of the compressive force which is the resultant of the compressive stress. 305 380 455 𝐹𝑐 3 – 25 mm dia. 455 75 455 - x 𝑛𝐴𝑠 305 𝐹𝑠 LEARNING EXERCISES 3 Locate the neutral axis of the transformed section Compute the maximum bending stress in the beam for a bending moment of 50 kN- m. The normal-weight concrete has an 𝑓𝑐 ′ of 28 MPa. Take n = 9. 380 455 𝑥 305 𝑥 = 9 1,472 455 − 𝑥 3 – 25 mm 2 dia. 75 𝑥 = 160.07 305 LEARNING EXERCISES 3 The moment of inertia of the transformed section is: Compute the maximum bending stress in the beam for a bending moment of 50 kN- m. The normal-weight concrete has an 𝑓𝑐 ′ of 28 MPa. Take n = 9. 380 455 305 160.073 𝐼= + 9 1,472 455 − 160.07 2 3 – 25 mm 3 dia. 75 𝐼 = 1,569 × 106 𝑚𝑚4 305 LEARNING EXERCISES 3 The maximum bending stress for compression is Compute the maximum bending stress in 𝑀𝑦𝑐 50 106 160.07 𝑓𝑐 = = the beam for a bending moment of 50 kN- 𝐼𝑔 1,569 × 106 m. The normal-weight concrete has an 𝑓𝑐 ′ of 28 MPa. Take n = 9. 𝑓𝑐 = 5.10 𝑀𝑃𝑎 The maximum bending stress for tension can be obtained in a similar manner, but remember that the steel was 160.07 transformed to concrete, to solve for the stress on the actual steel, the stress must be multiplied by the modular ratio. 294.93 𝑀𝑦𝑡 50 106 294.93 𝑓𝑠 = 𝑛 =9 = 84.59 𝑀𝑃𝑎 𝐼𝑔 1,569 × 106 LEARNING EXERCISES 4 The maximum bending stress considering compression of concrete is Compute the maximum moment at elastic stage of the beam from the previous 0.5𝑓𝑐 ′ 𝐼𝑔 0.5(28)(1,569 × 106 ) problem. The normal-weight concrete has 𝑀𝑐 = = 𝑦𝑐 160.07 an 𝑓𝑐 ′ of 28 MPa. Take n = 9. 𝑀𝑐 = 137.23 𝑘𝑁 − 𝑚 The maximum bending stress considering the elastic limit of steel (for example, 𝑓𝑦 = 280 𝑀𝑃𝑎) 101.25 160.07 𝑓𝑦 𝐼𝑔 280(1,569 × 106 ) 𝑀𝑡 = = 353.75 294.93 𝑛𝑦𝑡 9(294.93) 𝑀𝑡 = 165.51 𝑘𝑁 − 𝑚 Choose the lower value, 𝑴𝒄 = 𝟏𝟑𝟕. 𝟐𝟑 𝒌𝑵 − 𝒎 LEARNING EXERCISES 5 The transformed section diagram Compute the bending stresses in the beam shown. n = 10 and M = 118 ft-k Locating the centroid 𝑥 14 𝑥 + (10 − 1)(2)(𝑥 − 2.5) = 10 4 17.5 − 𝑥 2 𝑥 = 6.45 𝑖𝑛. LEARNING EXERCISES 5 The moment of inertia of the transformed section: Compute the bending stresses in the beam shown. n = 10 and M = 118 ft-k 14 6.453 𝐼= + 20 − 1 2 3.95 2 + 10 4 11.05 2 3 𝐼 = 6,729 𝑖𝑛4 LEARNING EXERCISES 5 The bending stresses are: Compute the bending stresses in the beam 𝑀𝑦𝑐 12(118,000)(6.45) 𝑓𝑐 = = = 𝟏, 𝟑𝟓𝟕 𝒑𝒔𝒊 shown. n = 10 and M = 118 ft-k 𝐼𝑔 6,729 ′ 𝑀𝑦𝑠 12(118,000)(3.95) 𝑓𝑠 = 2𝑛 = 2(10) = 𝟏𝟔, 𝟔𝟐𝟒 𝒑𝒔𝒊 𝐼𝑔 6,729 𝑀𝑦𝑠 12(118,000)(11.05) 𝑓𝑠 = 𝑛 = 10 = 𝟐𝟑, 𝟐𝟓𝟑 𝒑𝒔𝒊 𝐼𝑔 6,729 2. Investigate the allowable moment capacity of a cracked reinforced concrete beam using the principle of elastic stress References: J. McCormac and R. Brown, Design of Reinforced Concrete ACI 318-11 Code Edition, 9th ed. John Wiley & Sons, Inc., 2007. “National Structural Code of the Philippines,” 2015. Engineering Lectures Scan QR  THANK YOU Time for open forum and evaluation Scan QR  INSVESTIGATION OF SINGLY REINFORCED BEAM USING NSCP 2015 Principles of Reinforced Concrete JRPLUCENA by Robert Lucena LEARNING OUTCOMES 1. Investigate the nominal moment capacity of singly reinforced beam 2. Investigate the ultimate moment capacity of singly reinforced beam using strength reduction according to NSCP 2015 What we will do 1. Discussion I will discuss concepts, theories, information, etc. related to the topic to attain objectives 1 to 2 2. Open Forum and Student Evaluation Let’s see how much you learn and what are your ideas to share with others 3. Problem Solving Let’s apply what we learn! What we will do 2. Open Forum and Student Evaluation Let’s see how much you learn and what are your ideas to share with others 3. Problem Solving Let’s apply what we learn! 4. Lecturer Evaluation Answer the evaluation form so I can improve my lecture  Inside our toolbox Calculators Notes and pen References  1. Investigate the nominal moment capacity of Non-linear Stress of Concrete singly reinforced beam 𝒇𝒄 < 𝟎. 𝟓𝟎𝒇𝒄 ′ 𝒇𝒄 > 𝟎. 𝟓𝟎𝒇𝒄 ′ Compressive stress distribution is linear Compressive stress distribution is non- linear (parabolic)  1. Investigate the nominal moment capacity of Non-linear Stress of Concrete singly reinforced beam For ease of calculation, the parabolic stress block is converted to rectangular stress block  1. Investigate the nominal moment capacity of Nominal moment capacity singly reinforced beam The nominal moment capacity of the beam is defined by the couple: 𝑎 𝑀𝑛 = 𝐶 𝑜𝑟 𝑇 𝑑− 2  1. Investigate the nominal moment capacity of Procedure for investigation of nominal moment singly reinforced beam 1. Satisfy equilibrium by establishing: 𝐶=𝑇 0.85𝑓𝑐 ′ 𝑎𝑏 = 𝐴𝑠 𝑓𝑦 2. Solve for 𝑎 and obtain the distance between the line of action of 𝐶 and 𝑇. 3. Calculate the nominal moment using the equation: 𝑎 𝑀𝑛 = 𝐶 𝑜𝑟 𝑇 𝑑− 2 LEARNING EXERCISES 1 Draw the stress diagram of the beam. Determine the nominal moment strength of the beam shown in the figure. Use 𝑓𝑐 ′ = 28 𝑀𝑃𝑎 and 𝑓𝑦 = 450 𝑀𝑃𝑎 Establish the equation 𝐶=𝑇 0.85𝑓𝑐 ′ 𝑎𝑏 = 𝐴𝑠 𝑓𝑦 LEARNING EXERCISES 1 Establish the equation Determine the nominal moment strength 𝐶=𝑇 of the beam shown in the figure. Use 𝑓𝑐 ′ = 28 𝑀𝑃𝑎 and 𝑓𝑦 = 450 𝑀𝑃𝑎 0.85𝑓𝑐 ′ 𝑎𝑏 = 𝐴𝑠 𝑓𝑦 Substituting the values 𝜋252 0.85 28 𝑎(300) = 3 450 4 𝑎 = 92.81 𝑚𝑚 LEARNING EXERCISES 1 The nominal moment capacity of the beam is: Determine the nominal moment strength 𝑎 𝑀𝑛 = 𝐶 𝑜𝑟 𝑇 𝑑− of the beam shown in the figure. Use 𝑓𝑐 ′ = 2 28 𝑀𝑃𝑎 and 𝑓𝑦 = 450 𝑀𝑃𝑎 Using 𝐶 ′ 𝑎 𝑀𝑛 = 0.85𝑓𝑐 𝑎𝑏 𝑑 − 2 92.81 𝑀𝑛 = 0.85(28)(92.81)(300) 430 − 2 𝑴𝒏 = 𝟐𝟓𝟒. 𝟏𝟗 𝒌𝑵 − 𝒎 1. Investigate the nominal moment capacity of singly reinforced beam  2. Investigate the ultimate moment capacity of singly Advantages of Strength Design reinforced beam using strength reduction according to NSCP 2015 1. Better estimates of load-carrying ability are obtained. 2. a more consistent theory is used throughout the designs 3. A more realistic factor of safety 4. The strength method takes considerable advantage of higher strength steels, whereas working-stress design did so only partly. The result is better economy for strength design. 5. The strength method permits more flexible designs than did the working- stress method.  2. Investigate the ultimate moment capacity of singly Concept of Ultimate Strength reinforced beam using strength reduction according to NSCP 2015 Nominal strength is reduced by a reduction factor to consider uncertainties in material strengths, dimensions, and workmanship. The result is the ultimate strength. In flexure analysis of beam: ∅𝑴𝒏 ≥ 𝑴𝒖 ∅ is the strength reduction factor  2. Investigate the ultimate moment capacity of singly Strength Reduction reinforced beam using strength reduction according to NSCP 2015 ∅, Strength Reduction Factor  2. Investigate the ultimate moment capacity of singly Strength Reduction reinforced beam using strength reduction according to NSCP 2015 ∅, Strength Reduction Factor  2. Investigate the ultimate moment capacity of singly Strength Reduction reinforced beam using strength reduction according to NSCP 2015 ∅, Strength Reduction Factor  2. Investigate the ultimate moment capacity of singly Beam Expressions reinforced beam using strength reduction according to NSCP 2015 Conversion of stress diagram  2. Investigate the ultimate moment capacity of singly Beam Expressions reinforced beam using strength reduction according to NSCP 2015 Conversion of stress diagram 𝜷𝟏 = 𝟎. 𝟖𝟓 − 𝟎. 𝟎𝟎𝟖 𝒇𝒄 ′ − 𝟑𝟎𝑴𝑷𝒂 ≥ 𝟎. 𝟔𝟓  2. Investigate the ultimate moment capacity of singly Strains in Flexural Member reinforced beam using strength reduction according to NSCP 2015 1. Maximum usable strain in concrete is 0.003. 2. Grade 60 steel has a strain at yielding of 0.002 𝒅−𝒄 𝝐𝒕 = 𝟎. 𝟎𝟎𝟑 𝒄 𝒅 is the effective depth of the beam  2. Investigate the ultimate moment capacity of singly Minimum Steel Reinforcement reinforced beam using strength reduction according to NSCP 2015 Steel Ratio, 𝜌 𝑨𝒔 𝝆= 𝒃𝒅  2. Investigate the ultimate moment capacity of singly Balanced Steel Ratio reinforced beam using strength reduction according to NSCP 2015 Balanced Steel Ratio, 𝜌𝑏𝑎𝑙 At ultimate load for such a beam, the concrete will theoretically fail (at a strain of 0.00300), and the steel will simultaneously yield 𝑐 0.003 = 𝑑 0,003 + 𝑓𝑦 /𝐸𝑠  2. Investigate the ultimate moment capacity of singly Balanced Steel Ratio reinforced beam using strength reduction according to NSCP 2015 𝑐 0.003 Balanced Steel Ratio, 𝜌𝑏𝑎𝑙 = 𝑑 0,003 + 𝑓𝑦 /𝐸𝑠 Considering: 0.85𝑓𝑐 ′ 𝑎𝑏 = 𝐴𝑠 𝑓𝑦 𝜌𝑓𝑦 𝑑 𝑎= 0.85𝑓𝑐 ′ 𝑎 𝜌𝑓𝑦 𝑑 𝑐= = 𝛽1 0.85𝛽1 𝑓𝑐 ′  2. Investigate the ultimate moment capacity of singly Balanced Steel Ratio reinforced beam using strength reduction according to NSCP 2015 𝑐 0.003 Balanced Steel Ratio, 𝜌𝑏𝑎𝑙 = 1 𝑑 0,003 + 𝑓𝑦 /𝐸𝑠 𝑎 𝜌𝑓𝑦 𝑑 𝑐= = 2 𝛽1 0.85𝛽1 𝑓𝑐 ′ Solving equations 1 and 2 simultaneously and and taking E equal to 200,000 MPa: 𝟎. 𝟖𝟓𝜷𝟏 𝒇𝒄 ′ 𝟔𝟎𝟎 𝝆𝒃𝒂𝒍 = 𝒇𝒚 𝟔𝟎𝟎 + 𝒇𝒚  2. Investigate the ultimate moment capacity of singly Procedure for Investigation reinforced beam using strength reduction according to NSCP 2015 Procedure for investigation of ultimate moment capacity of singly reinforced beam 1. Check for steel ratio and compare it with maximum and minimum steel ratio 2. Solve for 𝑎 (depth of equivalent rectangular compressive stress block). This will allow the calculation of distance between the couple forces from the equation 𝐶=𝑇 3. Solve for c (depth of the parabolic compressive stress block). This will allow the determination of strain on steel at tension zone. 𝑑−𝑐 𝜖𝑡 = 0.003 𝑐  2. Investigate the ultimate moment capacity of singly Procedure for Investigation reinforced beam using strength reduction according to NSCP 2015 Procedure for investigation of ultimate moment capacity of singly reinforced beam 4. Based on the calculated strain on steel, determine the strength reduction factor, ∅ 5. Obtain the nominal moment capacity of the beam 𝑎 𝑀𝑛 = 𝐶 𝑜𝑟 𝑇 𝑑− 2 6. Solve for the ultimate moment capacity ∅𝑀𝑛 ≥ 𝑀𝑢 LEARNING EXERCISES 2 Check for the percentage of steel: Calculate the ultimate bending moment 𝜋 282 capacity of the beam in accordance with 𝐴𝑠 4 × 4 𝜌= = = 0.0104 ′ NSCP 2015. Take 𝑓𝑐 = 28 𝑀𝑃𝑎 and 𝑏𝑑 380 625 𝑓𝑦 = 414 𝑀𝑃𝑎 The minimum steel ratio is 0.25 𝑓𝑐 ′ 0.25 28 𝜌𝑚𝑖𝑛 = = = 0.0032 𝑓𝑦 414 1.4 1.4 𝜌𝑚𝑖𝑛 = = = 0.0034 𝑓𝑦 414 The maximum steel ratio can also be checked LEARNING EXERCISES 2 Solve for 𝑎 and 𝑐 using the equilibrium equation: Calculate the ultimate bending moment 𝐶=𝑇 capacity of the beam in accordance with ′ NSCP 2015. Take 𝑓𝑐 = 28 𝑀𝑃𝑎 and 0.85𝑓𝑐 ′ 𝑎𝑏 = 𝐴𝑠 𝑓𝑦 𝑓𝑦 = 414 𝑀𝑃𝑎 𝜋 282 𝐴𝑠 𝑓𝑦 4× (414) 𝑎= 4 ′ = 0.85𝑓𝑐 𝑏 0.85(28)(380) 𝑎 = 143 𝑚𝑚 𝑎 143 𝑐= = = 168.24 𝑚𝑚 𝛽 0.85 LEARNING EXERCISES 2 Check for strain Calculate the ultimate bending moment 𝑑−𝑐 625 − 168.24 𝜖𝑡 = 0.003 = 0.003 capacity of the beam in accordance with 𝑐 168.24 ′ NSCP 2015. Take 𝑓𝑐 = 28 𝑀𝑃𝑎 and 𝑓𝑦 = 414 𝑀𝑃𝑎 𝜖𝑡 = 0.008 Since the strain is greater than 0.005, the beam is tension- controlled. The value of strength reduction factor to be used is 0.90. (∅ = 0.90). Solving for the nominal moment capacity: 𝑎 𝑀𝑛 = 𝐶 𝑜𝑟 𝑇 𝑑− 2 𝑎 143 𝑀𝑛 = 𝐴𝑠 𝑓𝑦 𝑑 − = (87.96)(414) 625 − 2 2 LEARNING EXERCISES 2 𝑀𝑛 = 𝐴𝑠 𝑓𝑦 𝑑 − 𝑎 2 = (87.96)(414) 625 − 143 2 Calculate the ultimate bending moment capacity of the beam in accordance with 𝑀𝑛 = 20.16 𝑘𝑁 − 𝑚 ′ NSCP 2015. Take 𝑓𝑐 = 28 𝑀𝑃𝑎 and 𝑓𝑦 = 414 𝑀𝑃𝑎 The ultimate moment capacity of the beam is: 𝑀𝑢 = ∅𝑀𝑛 = 0.90(20.16) 𝑴𝒖 = 𝟏𝟖. 𝟏𝟒 𝒌𝑵 − 𝒎 LEARNING EXERCISES 3 Check for the percentage of steel: Calculate the ultimate bending moment 𝜋 322 capacity of the beam in accordance with 𝐴𝑠 10 × 4 𝜌= = = 0.0339 ′ NSCP 2015. Take 𝑓𝑐 = 28 𝑀𝑃𝑎 and 𝑏𝑑 380 625 𝑓𝑦 = 414 𝑀𝑃𝑎 The minimum steel ratio is 0.25 𝑓𝑐 ′ 0.25 28 𝜌𝑚𝑖𝑛 = = = 0.0032 𝑓𝑦 414 10 – 32 mm 1.4 1.4 dia. 𝜌𝑚𝑖𝑛 = = = 0.0034 𝑓𝑦 414 The maximum steel ratio can also be checked LEARNING EXERCISES 3 Solve for 𝑎 and 𝑐 using the equilibrium equation: Calculate the ultimate bending moment 𝐶=𝑇 capacity of the beam in accordance with ′ NSCP 2015. Take 𝑓𝑐 = 28 𝑀𝑃𝑎 and 0.85𝑓𝑐 ′ 𝑎𝑏 = 𝐴𝑠 𝑓𝑦 𝑓𝑦 = 414 𝑀𝑃𝑎 𝜋 322 𝐴𝑠 𝑓𝑦 10 × (414) 𝑎= 4 ′ = 0.85𝑓𝑐 𝑏 0.85(28)(380) 𝑎 = 368.15 𝑚𝑚 – 32 810 mm – 32 mm dia. dia. 𝑎 368.15 𝑐= = = 433.12 𝑚𝑚 𝛽 0.85 LEARNING EXERCISES 3 Check for strain Calculate the ultimate bending moment 𝑑−𝑐 625 − 433.12 𝜖𝑡 = 0.003 = 0.003 capacity of the beam in accordance with 𝑐 433.12 ′ NSCP 2015. Take 𝑓𝑐 = 28 𝑀𝑃𝑎 and 𝑓𝑦 = 414 𝑀𝑃𝑎 𝜖𝑡 = 0.001 Since the strain is less than 0.002, the beam is compression- controlled. This type of beam is not recommended for use. The value of strength reduction factor to be used is 0.65 (∅ = 0.65), but since the steel did not yield, 𝑓𝑠 must be – 32 810 mm used. – 32 mm dia. dia. 0.85𝑓𝑐 ′ 𝑎𝑏 = 𝐴𝑠 𝑓𝑠 0.85𝑓𝑐 ′ 𝑎𝑏 = 𝐴𝑠 𝐸𝜖𝑡  0.85𝑓𝑐 ′ 𝑎𝑏 = 𝐴𝑠 𝑓𝑠 LEARNING EXERCISES 3 ′ 𝑑−𝑐 Calculate the ultimate bending moment 0.85𝑓𝑐 𝛽𝑐𝑏 = 𝐴𝑠 (200,000) 0.003 capacity of the beam in accordance with 𝑐 ′ NSCP 2015. Take 𝑓𝑐 = 28 𝑀𝑃𝑎 and 625 − 𝑐 𝑓𝑦 = 414 𝑀𝑃𝑎 0.85(28)(0.85)𝑐(380) = 8,042(200,000) 0.003 𝑐 𝑐 = 386.73 𝑎 = 𝛽𝑐 = 0.85 386.73 = 328.72 The nominal moment capacity of the beam is: – 32 810 mm – 32 mm dia. dia. 𝑎 𝑀𝑛 = 0.85𝑓𝑐 ′ 𝑎𝑏 𝑑 − 2 328.72 𝑀𝑛 = 0.85(28)(328.78)(380) 625 − 2 LEARNING EXERCISES 3 𝑀𝑛 = 0.85(28)(328.78)(380) 625 − 328.72 Calculate the ultimate bending moment 2 capacity of the beam in accordance with ′ NSCP 2015. Take 𝑓𝑐 = 28 𝑀𝑃𝑎 and 𝑀𝑛 = 1,370 𝑘𝑁 − 𝑚 𝑓𝑦 = 414 𝑀𝑃𝑎 The ultimate moment capacity is: 𝑀𝑢 = ∅𝑀𝑛 = 0.65(1,370) 𝑴𝒖 = 𝟖𝟗𝟎. 𝟓𝟎 𝒌𝑵 − 𝒎 – 32 810 mm – 32 mm dia. dia. References: J. McCormac and R. Brown, Design of Reinforced Concrete ACI 318-11 Code Edition, 9th ed. John Wiley & Sons, Inc., 2007. “National Structural Code of the Philippines,” 2015. Engineering Lectures Scan QR  THANK YOU Time for open forum and evaluation Scan QR  DESIGN OF SINGLY REINFORCED BEAM USING NSCP 2015 Principles of Reinforced Concrete JRPLUCENA by Robert Lucena LEARNING OUTCOMES 1. Design a singly reinforced rectangular beam in accordance with NSCP 2015. What we will do 1. Discussion I will discuss concepts, theories, information, etc. related to the topic to attain objectives 1 to 2 2. Open Forum and Student Evaluation Let’s see how much you learn and what are your ideas to share with others 3. Problem Solving Let’s apply what we learn! What we will do 2. Open Forum and Student Evaluation Let’s see how much you learn and what are your ideas to share with others 3. Problem Solving Let’s apply what we learn! 4. Lecturer Evaluation Answer the evaluation form so I can improve my lecture  Inside our toolbox Calculators Notes and pen References  1. Design a singly reinforced rectangular beam Beam Proportions in accordance with NSCP 2015.  1. Design a singly reinforced rectangular beam Beam Proportions in accordance with NSCP 2015. ratio of d to b is in the range of 1.5 to 2.  1. Design a singly reinforced rectangular beam Selection of Reinforcing Bars in accordance with NSCP 2015.  1. Design a singly reinforced rectangular beam Spacing of Reinforcing Bars in accordance with NSCP 2015.  1. Design a singly reinforced rectangular beam Concrete Cover in accordance with NSCP 2015.  1. Design a singly reinforced rectangular beam Design Procedure in accordance with NSCP 2015. Procedure for design of singly reinforced rectangular beam 1. Establish dimension of the beam 2. Compute for the factored load and the ultimate moment, 𝑀𝑢 3. Assuming a tension-controlled beam, ∅ = 0.90, solve for the steel ratio using the equation: 0.85𝑓𝑐 ′ 2𝑅𝑛 𝜌= 1− 1− 𝑓𝑦 0.85𝑓𝑐 ′ Where: 𝑀𝑢 𝑅𝑛 = ∅𝑏𝑑 2 Supplemental reading: Section 3.4 J. McCormac and R. Brown, Design of Reinforced Concrete ACI 318-11 Code Edition, 9th ed. John Wiley & Sons, Inc., 2007.  1. Design a singly reinforced rectangular beam Design Procedure in accordance with NSCP 2015. Procedure for design of singly reinforced rectangular beam 4. Solve for the area of steel and compare with minimum steel ratio. Redesign for beam dimensions when needed. 5. Draw the cross section of the beam including required concrete cover and rebar spacing 6. Investigate for the ultimate moment capacity of the beam. It should be greater than the initially determined 𝑀𝑢 LEARNING EXERCISES 1 The minimum depth can be obtained by following: A 6-m simply supported beam is to carry the following: DL = 25 kN/m and LL = 29 kN/m. Design a singly reinforced rectangular beam using 𝑓𝑐 ′ = 28 MPa and 𝑓𝑦 = 420 MPa 𝑙 6,000 ℎ= = = 375 16 16 Note that this is only the minimum depth, architectural requirement must also be checked. In this problem we assume that the beam h is required to be 600 mm LEARNING EXERCISES 1 For an overall depth of 600 mm an ideal width should be obtained by following the ratio: of 1.5 to 2 (d to b). Say, b = A 6-m simply supported beam is to carry 350 mm. the following: DL = 25 kN/m and LL = 29 kN/m. Design a singly reinforced Determine the loads to be carried by the beam: rectangular beam using 𝑓𝑐 ′ = 28 MPa and 𝑓𝑦 = 420 MPa LL = 29 kN/m DL = 25 kN/m + 4.96 kN/m = 29.96 kN/m The factored load is 𝑤𝑢 = 1.2 29.96 + 1.6 29 = 𝟖𝟐. 𝟑𝟓𝟐 𝒌𝑵/𝒎 LEARNING EXERCISES 1 𝑤𝑢 = 1.2 29.96 + 1.6 29 = 𝟖𝟐. 𝟑𝟓𝟐 𝒌𝑵/𝒎 A 6-m simply supported beam is to carry Solving for 𝑀𝑢 the following: DL = 25 kN/m and LL = 29 kN/m. Design a singly reinforced 𝑤𝑢 𝐿2 82.352 62 𝑀𝑢 = = rectangular beam using 𝑓𝑐 ′ = 28 MPa and 8 8 𝑓𝑦 = 420 MPa 𝑀𝑢 = 370.58 𝑘𝑁 − 𝑚 For a tension-controlled beam the steel ratio required can be calculated using the equation: 0.85𝑓𝑐 ′ 2𝑅𝑛 𝜌= 1− 1− 𝑓𝑦 0.85𝑓𝑐 ′ LEARNING EXERCISES 1 Solving for 𝑅𝑛 A 6-m simply supported beam is to carry 𝑀𝑢 𝑅𝑛 = the following: DL = 25 kN/m and LL = 29 ∅𝑏𝑑 2 kN/m. Design a singly reinforced Saying 𝑑 = 600 rectangular beam using 𝑓𝑐 ′ = 28 MPa and 𝑓𝑦 = 420 MPa 370.58 106 𝑅𝑛 = 2 = 3.27 0.90 350 600 LEARNING EXERCISES 1 Obtaining the steel ratio: A 6-m simply supported beam is to carry 0.85𝑓𝑐 ′ 2𝑅𝑛 the following: DL = 25 kN/m and LL = 29 𝜌= 1− 1− kN/m. Design a singly reinforced 𝑓𝑦 0.85𝑓𝑐 ′ rectangular beam using 𝑓𝑐 ′ = 28 MPa and 𝑓𝑦 = 420 MPa 0.85 28 2 3.27 𝜌= 1− 1− 420 0.85 28 𝜌 = 0.0084 LEARNING EXERCISES 1 𝜌 = 0.0084 A 6-m simply supported beam is to carry Check for the minimum steel ratio the following: DL = 25 kN/m and LL = 29 kN/m. Design a singly reinforced 0.25 𝑓𝑐 ′ 0.25 28 rectangular beam using 𝑓𝑐 ′ = 28 MPa and 𝜌𝑚𝑖𝑛 = = = 0.0032 𝑓𝑦 420 𝑓𝑦 = 420 MPa 1.4 1.4 𝜌𝑚𝑖𝑛 = = = 0.0033 𝑓𝑦 420 The calculated steel ratio is more than the minimum steel ratio 𝐴𝑠 = 𝜌𝑏𝑑 = 0.0084 350 600 𝐴𝑠 = 1,764 𝑚𝑚2  𝐴𝑠 = 1, 764 𝑚𝑚2 LEARNING EXERCISES 1 A 6-m simply supported beam is to carry Designing the beam cross section the following: DL = 25 kN/m and LL = 29 kN/m. Design a singly reinforced Using 25 mm dia. Steel rectangular beam using 𝑓𝑐 ′ = 28 MPa and 𝑓𝑦 = 420 MPa 𝐴𝑠 1,764 𝑛= = = 3.59 𝑠𝑎𝑦 𝟒 𝒑𝒄𝒔. 𝐴𝑏𝑎𝑟 𝜋 252 4 LEARNING EXERCISES 1 Investigate the ultimate moment capacity of the beam A 6-m simply supported beam is to carry Solve for 𝑎 and 𝑐 using the equilibrium equation: the following: DL = 25 kN/m and LL = 29 kN/m. Design a singly reinforced 𝐶=𝑇 rectangular beam using 𝑓𝑐 ′ = 28 MPa and 𝑓𝑦 = 420 MPa 0.85𝑓𝑐 ′ 𝑎𝑏 = 𝐴𝑠 𝑓𝑦 𝜋 252 𝐴𝑠 𝑓𝑦 5× (420) 𝑎= 4 ′ = 0.85𝑓𝑐 𝑏 0.85(28)(350) 𝑎 = 99.00 𝑚𝑚 𝑎 99.00 𝑐= = = 116.47 𝑚𝑚 𝛽 0.85 LEARNING EXERCISES 1 Check for strain A 6-m simply supported beam is to carry 𝑑−𝑐 600 − 116.47 𝜖𝑡 = 0.003 = 0.003 the following: DL = 25 kN/m and LL = 29 𝑐 116.47 kN/m. Design a singly reinforced rectangular beam using 𝑓𝑐 ′ = 28 MPa and 𝜖𝑡 = 0.010 𝑓𝑦 = 420 MPa LEARNING EXERCISES 1 Investigate for the ultimate moment capacity A 6-m simply supported beam is to carry 𝑎 𝑀𝑛 = 𝐶 𝑜𝑟 𝑇 𝑑− the following: DL = 25 kN/m and LL = 29 2 kN/m. Design a singly reinforced rectangular beam using 𝑓𝑐 ′ = 28 MPa and 99.00 𝑀𝑛 = 1,963(420) 600 − 𝑓𝑦 = 420 MPa 2 𝑀𝑛 = 453.98 𝑘𝑁 − 𝑚 The ultimate moment capacity of the beam is: 𝑀𝑢 = ∅𝑀𝑛 = 0.90(453.98) 𝑴𝒖 = 𝟒𝟎𝟖. 𝟓𝟖 𝒌𝑵 − 𝒎 Design a singly reinforced rectangular beam in accordance with NSCP 2015. Engineering Lectures Scan QR  THANK YOU Time for open forum and evaluation Scan QR  INVESTIGATION OF DOUBLY REINFORCED BEAM USING NSCP 2015 Principles of Reinforced Concrete JRPLUCENA by Robert Lucena LEARNING OUTCOMES 1. Investigate the ultimate moment capacity of a doubly reinforced rectangular beam in accordance with NSCP 2015. What we will do 1. Discussion I will discuss concepts, theories, information, etc. related to the topic to attain objectives 1 to 2 2. Open Forum and Student Evaluation Let’s see how much you learn and what are your ideas to share with others 3. Problem Solving Let’s apply what we learn! What we will do 2. Open Forum and Student Evaluation Let’s see how much you learn and what are your ideas to share with others 3. Problem Solving Let’s apply what we learn! 4. Lecturer Evaluation Answer the evaluation form so I can improve my lecture  Inside our toolbox Calculators Notes and pen References  1. Investigate the ultimate moment capacity of a Compression Steel doubly reinforced rectangular beam in accordance with NSCP 2015. compression steel Used to increase the ultimate moment capacity of a beam beyond that of a beam with a maximum percentage of steel. (without increasing the size of the beam). Compression steel reinforce the concrete by carrying compressive stress  1. Investigate the ultimate moment capacity of a Beam Proportions doubly reinforced rectangular beam in accordance with NSCP 2015. 𝑎 𝑀𝑛 = 𝐴𝑠1 𝑓𝑦 𝑑 − + 𝐴𝑠2 𝑓𝑦 𝑑 − 𝑑′ 2  1. Investigate the ultimate moment capacity of a Relationship of Compression and Tensile Steel doubly reinforced rectangular beam in accordance with NSCP 2015. 𝐴𝑠2 𝑓𝑦 = 𝐴′𝑠 𝑓′𝑠  1. Investigate the ultimate moment capacity of a Strain on Compression Steel doubly reinforced rectangular beam in accordance with NSCP 2015. 𝑐 − 𝑑′ 𝜖′𝑠 = 0.003 𝑐  1. Investigate the ultimate moment capacity of a Investigation Procedure doubly reinforced rectangular beam in accordance with NSCP 2015. Procedure for design of singly reinforced rectangular beam 1. Solve for the depth of the parabolic stress block by taking equilibrium analysis along horizontal axis 𝐶𝑐 + 𝐶𝑠 = 𝑇𝑠 𝐴′𝑠 𝑓𝑦 + 0.85𝑓𝑐 ′ 𝛽1 𝑐𝑏 = 𝐴𝑠 𝑓𝑦 2. Compute the strain in compression and tension steel 𝑐−𝑑′ 𝑑−𝑐 𝜖′𝑠 = 0.003 and 𝜖𝑡 = 0.003 𝑐 𝑐 If steel did not yield, recompute for equation along horizontal axis but considering 𝑓𝑠 or 𝑓′𝑠 or both.  1. Investigate the ultimate moment capacity of a Investigation Procedure doubly reinforced rectangular beam in accordance with NSCP 2015. Procedure for design of singly reinforced rectangular beam 3. Solve for 𝐴𝑠2 using: 𝐴𝑠2 𝑓𝑦 = 𝐴′𝑠 𝑓′𝑠 4. Solve for 𝐴𝑠1 using: 𝐴𝑠1 = 𝐴𝑠 − 𝐴𝑠2 5. Calculate the ultimate moment strength 𝑎 𝑀𝑛 = 𝐴𝑠1 𝑓𝑦 𝑑 − + 𝐴𝑠2 𝑓𝑦 𝑑 − 𝑑′ 2 LEARNING EXERCISES 1 Solving for the depth of the parabolic stress block Determine the design moment capacity of 𝐴′𝑠 𝑓𝑦 + 0.85𝑓𝑐 ′ 𝛽1 𝑐𝑏 = 𝐴𝑠 𝑓𝑦 the beam shown in the figure. 𝑓𝑦 = 420 𝑀𝑃𝑎, 𝑓′𝑐 = 21 𝑀𝑃𝑎 𝜋282 𝜋362 2× 420 + 0.85 21 0.85 𝑐 360 = 4 × 420 4 4 65 𝑐 = 218.38 2 – 28 𝑎 = 0.85 218.38 = 185.62 mm dia. 690 Compute strain on compression steel 550 𝑐 − 𝑑′ 4 – 36 𝜖′𝑠 = 0.003 mm dia. 𝑐 75 218.38 − 65 𝜖′𝑠 = 0.003 = 0.0021 360 218.38 LEARNING EXERCISES 1 𝜖′𝑠 = 218.38 − 65 0.003 = 0.0021 Determine the design moment capacity of 218.38 the beam shown in the figure. 𝑓𝑦 = Therefore, the compression steel yields 𝜖′𝑠 > 0.002. Check for 420 𝑀𝑃𝑎, 𝑓′𝑐 = 21 𝑀𝑃𝑎 yielding on the tension steel 65 𝑑−𝑐 𝜖𝑡 = 0.003 𝑐 2 – 28 615 − 218.38 mm dia. 𝜖𝑡 = 0.003 = 0.0054 690 218.38 550 Therefore, the compression steel yields 𝜖𝑠 > 0.002. The beam is 4 – 36 tension-controlled (∅ = 0.90). mm dia. 75 360 LEARNING EXERCISES 1 Divide 𝐴𝑠 into 𝐴𝑠1 and 𝐴𝑠2 using the relationship Determine the design moment capacity of 𝐴𝑠2 𝑓𝑦 = 𝐴′𝑠 𝑓′𝑠 the beam shown in the figure. 𝑓𝑦 = Since the compression steel yields, we use 𝑓′𝑠 = 𝑓𝑦 420 𝑀𝑃𝑎, 𝑓′𝑐 = 21 𝑀𝑃𝑎 65 𝐴′𝑠 𝑓𝑦 1,232(420) 𝐴𝑠2 = = = 1,232 𝑚𝑚2 𝑓𝑦 420 2 – 28 And 𝐴𝑠1 can be solved using mm dia. 690 𝐴𝑠1 = 𝐴𝑠 − 𝐴𝑠2 = 4,072 − 1,232 = 2,840 𝑚𝑚2 550 4 – 36 mm dia. 75 360 LEARNING EXERCISES 1 The ultimate moment capacity of the beam is: 𝑎 Determine the design moment capacity of 𝑀𝑛 = 𝐴𝑠1 𝑓𝑦 𝑑 − + 𝐴𝑠2 𝑓𝑦 𝑑 − 𝑑′ 2 the beam shown in the figure. 𝑓𝑦 = 420 𝑀𝑃𝑎, 𝑓′𝑐 = 21 𝑀𝑃𝑎 185.62 𝑀𝑛 = 2,840 420 615 − + 1,232 420 615 − 65 65 2 𝑀𝑛 = 907.46 𝑘𝑁 − 𝑚 2 – 28 mm dia. Using (∅ = 0.90). 690 𝑀𝑢 = ∅𝑀𝑛 550 4 – 36 𝑀𝑢 = 0.90 907.46 mm dia. 75 𝑴𝒖 = 𝟖𝟏𝟔. 𝟕𝟏 𝒌𝑵 − 𝒎 360 LEARNING EXERCISES 2 Solving for the depth of the parabolic stress block Determine the design moment capacity of 𝐴′𝑠 𝑓𝑦 + 0.85𝑓𝑐 ′ 𝛽1 𝑐𝑏 = 𝐴𝑠 𝑓𝑦 the beam shown in the figure. 𝑓𝑦 = 420 𝑀𝑃𝑎, 𝑓′𝑐 = 28 𝑀𝑃𝑎 𝜋202 𝜋322 2× 420 + 0.85 28 0.85 𝑐 360 = 4 × 420 4 4 65 𝑐 = 149.29 2 – 20 𝑎 = 0.85 149.29 = 126.90 mm dia. 690 Compute strain on compression steel 550 𝑐 − 𝑑′ 4 – 32 𝜖′𝑠 = 0.003 mm dia. 𝑐 75 149.29 − 65 𝜖′𝑠 = 0.003 = 0.0017 360 149.29 LEARNING EXERCISES 2 𝜖′𝑠 = 149.29 − 65 0.003 = 0.0017 Determine the design moment capacity of 149.29 the beam shown in the figure. 𝑓𝑦 = Therefore, the compression steel do not yield 𝜖′𝑠 < 0.002. 420 𝑀𝑃𝑎, 𝑓′𝑐 = 28 𝑀𝑃𝑎 Recomputing for 𝑐 using 𝑓′ 𝑠 65 𝐴′𝑠 𝑓′𝑠 + 0.85𝑓𝑐 ′ 𝛽1 𝑐𝑏 = 𝐴𝑠 𝑓𝑦 2 – 20 𝑐 − 𝑑′ 𝐴′𝑠 0.003 𝐸𝑠 + 0.85𝑓𝑐 ′ 𝛽1 𝑐𝑏 = 𝐴𝑠 𝑓𝑦 mm dia. 𝑐 690 𝑐 − 65 550 628 0.003 200,000 + 0.85 28 0.85 𝑐 360 𝑐 4 – 32 = 3,217 420 mm dia. 75 𝑐 = 155.42 360 𝑎 = 0.85 155.42 = 132.11 LEARNING EXERCISES 2 Compute strain on compression steel Determine the design moment capacity of 𝑐 − 𝑑′ 𝜖′𝑠 = 0.003 the beam shown in the figure. 𝑓𝑦 = 𝑐 420 𝑀𝑃𝑎, 𝑓′𝑐 = 28 𝑀𝑃𝑎 155.42 − 65 𝜖′𝑠 = 0.003 = 0.0017 65 155.42 Check for yielding on the tension steel 2 – 20 mm dia. 𝑑−𝑐 𝜖𝑡 = 0.003 690 𝑐 550 615 − 155.42 𝜖𝑡 = 0.003 = 0.0089 4 – 32 155.42 mm dia. 75 Therefore, the compression steel yields 𝜖𝑠 > 0.002. The beam is tension-controlled (∅ = 0.90). 360 LEARNING EXERCISES 2 Divide 𝐴𝑠 into 𝐴𝑠1 and 𝐴𝑠2 using the relationship Determine the design moment capacity of 𝐴𝑠2 𝑓𝑦 = 𝐴′𝑠 𝑓′𝑠 the beam shown in the figure. 𝑓𝑦 = 𝑓′𝑠 = 0.0017 200,000 = 340 𝑀𝑃𝑎 420 𝑀𝑃𝑎, 𝑓′𝑐 = 28 𝑀𝑃𝑎 65 𝐴′𝑠 𝑓′𝑠 628 340 𝐴𝑠2 = = = 508 𝑚𝑚2 𝑓𝑦 420 2 – 20 And 𝐴𝑠1 can be solved using mm dia. 690 𝐴𝑠1 = 𝐴𝑠 − 𝐴𝑠2 = 3,217 − 508 = 2,709 𝑚𝑚2 550 4 – 32 mm dia. 75 360 LEARNING EXERCISES 2 The ultimate moment capacity of the beam is: 𝑎 Determine the design moment capacity of 𝑀𝑛 = 𝐴𝑠1 𝑓𝑦 𝑑 − + 𝐴𝑠2 𝑓′𝑠 𝑑 − 𝑑′ 2 the beam shown in the figure. 𝑓𝑦 = 420 𝑀𝑃𝑎, 𝑓′𝑐 = 28 𝑀𝑃𝑎 65 132.11 𝑀𝑛 = 3,217 420 615 − + 508 340 615 − 65 2 2 – 20 𝑀𝑛 = 859.05 𝑘𝑁 − 𝑚 mm dia. 690 Using (∅ = 0.90). 550 𝑀𝑢 = ∅𝑀𝑛 4 – 32 mm dia. 𝑀𝑢 = 0.90 859.05 75 𝑴𝒖 = 𝟕𝟕𝟑. 𝟏𝟒 𝒌𝑵 − 𝒎 360 1. Investigate the ultimate moment capacity of a doubly reinforced rectangular beam in accordance with NSCP 2015. Engineering Lectures Scan QR  THANK YOU Time for open forum and evaluation Scan QR