![Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for flexure:
ΦMn ≥ Mu (ΦMn is Mdesign or Mcapacity)
For ΦMn = Mu
ΦAsfy(d – a/2) = Mu
As = Mu/ {Φfy (d – a/2)}
Calculate “As” by trial and success method.
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
43
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for flexure:
First Trial:
Assume a = 4″
As = 2370.24 / [0.9 × 40 × {17.5 – (4/2)}] = 4.25 in2
a = Asfy/ (0.85fc′bw)
Prof. Dr. Qaisar Ali
= 4.25 × 40/ (0.85 × 3 × 14) = 4.76 inches
Reinforced Concrete Design – II
44
22](https://image.slidesharecdn.com/l-01introduction-131121114659-phpapp02/85/Lecture-01-Introduction-to-Reinforced-Concrete-Design-22-320.jpg)
![Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for flexure:
Second Trial:
• As = 2370.24 / [0.9 × 40 × {17.5 – (4.76/2)}] = 4.35 in2
• a = 4.35 × 40/ (0.85 × 3 × 14) = 4.88 inches
• As = 2370.24 / [0.9 × 40 × {17.5 – (4.88/2)}] = 4.37 in2
Third Trial:
• a = 4.37 × 40/ (0.85 × 3 × 14) = 4.90 inches
“Close enough to the previous value of “a” so that As = 4.37 in2 O.K
Prof. Dr. Qaisar Ali
Reinforced Concrete Design – II
45
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1.6
Flexural and Shear Design of Beam as per ACI:
Solution:
Step No. 04: Design.
Design for flexure:
ρmin = 3
3 × 3000 /40000 = 0.004
200/40000 = 0.005
Therefore, ρmin = 0.005
Prof. Dr. Qaisar Ali
Check for maximum and minimum reinforcement allowed by ACI:
Asmin = ρminbwd = 0.005 × 14 × 17.5 = 1.225 in2
f′ /fy ≥ 200/fy (ACI 10.5.1)
Reinforced Concrete Design – II
46
23](https://image.slidesharecdn.com/l-01introduction-131121114659-phpapp02/85/Lecture-01-Introduction-to-Reinforced-Concrete-Design-23-320.jpg)
Lecture - 01: Introduction to Reinforced Concrete Design
- 1. Department of Civil Engineering, University of Engineering and Technology Peshawar Reinforced Concrete Design-II By: Prof Dr. Qaisar Ali Civil Engineering Department UET Peshawar www.drqaisarali.com Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 1 Department of Civil Engineering, University of Engineering and Technology Peshawar Course Content Mid Term Introduction One-Way Slab System Design ACI Coefficient Method for Analysis of One-Way Slabs Two Way Slab System Design Prof. Dr. Qaisar Ali ACI Analysis Method for Slabs Supported on Stiff Beams or Walls ACI Direct Design Method for Slabs with or without Beams Reinforced Concrete Design – II 2 1
- 2. Department of Civil Engineering, University of Engineering and Technology Peshawar Course Content Final Term Introduction to Earthquake Resistant Design of RC Structures Introduction to Pre-stressed Concrete Introduction to various Types of Retaining Walls and Design of Cantilever RW Introduction to Bridge Engineering Design of Stairs Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 3 Department of Civil Engineering, University of Engineering and Technology Peshawar Grading Policy Midterm = 25 % Final Term = 50 % Session Performance = 25 % Assignments = 10 % (6 Assignments ) Quizzes = 15 % (6 Quizzes) Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 4 2
- 3. Department of Civil Engineering, University of Engineering and Technology Peshawar Lectures Availability All lectures and related material will be available on the website: www.drqaisarali.com Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 5 Department of Civil Engineering, University of Engineering and Technology Peshawar Lecture-01 Introduction By: Prof Dr. Qaisar Ali Civil Engineering Department UET Peshawar drqaisarali@nwfpuet.edu.pk Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 6 3
- 4. Department of Civil Engineering, University of Engineering and Technology Peshawar Topics Concept of Capacity and Demand Flexure Design of Beams using ACI Recommendations Shear Design of Beams using ACI Recommendations Example Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 7 Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Demand Demand on a structure refers to all external actions. Gravity, wind, earthquake, snow are external actions. These actions when act on the structure will induce internal disturbance(s) in the structure in the form of stresses (such as compression, tension, bending, shear, and torsion). The internal stresses are also called load effects. Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 8 4
- 5. Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Capacity The overall ability of a structure to carry an imposed demand. Beam will resist the Applied Load (Demand) applied load up to its capacity and will fail when demand exceeds capacity Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 9 Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Failure Occurs when Capacity is less than Demand. To avoid failure, capacity to demand ratio should be kept greater than one, or at least equal to one. It is, however, intuitive to have some margin of safety i.e., to have capacity to demand ratio more than one. How much? Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 10 5
- 6. Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Failure Failure (Capacity < Demand) Reinforced Beam Test Video Prof. Dr. Qaisar Ali 11 Reinforced Concrete Design – II Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Example 1.1 Calculate demand in the form of stresses or load effects on the given concrete pad of size 12″ × 12″. 50 Tons Concrete pad 12″ 12″ Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 12 6
- 7. Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Example 1.1 Solution: Based on convenience either the loads or the load effects as demand are compared to the load carrying capacity of the structure in the relevant units. 50 Tons Demand in the form of load: Capacity of the pad in the form Load = 50 Tons of resistance should be able to carry a stress of 765.27 psi. Demand in the form of Load effects: In other words, the compressive The effect of load on the pad will be 12″ a compressive stress equal to load strength of concrete pad (capacity) should be more than divided by the area of the pad. 12″ Load Effect=(50 × 2204)/ (12 × 12) 765.27 psi (demand). = 765.27 psi Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 13 Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Example 1.2 Determine capacity to demand ratio for the pad of example 1.1 for the following capacities given in the form of compressive strength of concrete (i) 500 psi (ii) 765.27 psi (iii) 1000 psi (iv) 2000 psi. Comment on the results? 50 Tons 12″ 12″ Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 14 7
- 8. Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Example 1.2 Solution: As calculated in example 1.1, demand = 765.27 psi. Therefore capacity to demand ratios are as under: i. ii. 765.27/ 765.27 = 1.0 (Capacity just equal to Demand) iii. 1000/ 765.27 = 1.3 (Capacity is 1.3 times greater than Demand) iv. Capacity/ Demand = 500 / 765.27 = 0.653 (Failure) 2000/ 765.27 = 2.6 (Capacity is 2.6 times greater than Demand) In (iii) and (iv), there is some margin of safety normally called as factor of safety. Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 15 Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Safety Factor It is always better to have a factor of safety in our designs. It can be achieved easily if we fix the ratio of capacity to demand greater than 1.0, say 1.5, 2.0 or so, as shown in example 1.2. Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 16 8
- 9. Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Safety Factor For certain reasons, however, let say we insist on a factor of safety such that capacity to demand ratio still remains 1.0. Then there are three ways of doing this: Take an increased demand instead of actual demand (load), e.g. 70 ton instead of 50 ton in the previous example, Take a reduced capacity instead of actual capacity such as 1500 psi for concrete whose actual strength is 3000 psi Doing both. How are these three situations achieved? Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 17 Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Working Stress Method In the Working Stress or Allowable Stress Design method, the material strength is knowingly taken less than the actual e.g. half of the actual to provide a factor of safety equal to 2.0. Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 18 9
- 10. Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Strength Design Method In the Strength Design method, the increased loads and the reduced strength of the material are considered, but both based on scientific rationale. For example, it is quite possible that during the life span of a structure, dead and live loads increase. The factors of 1.2 and 1.6 used by ACI 318-02 (Building code requirements for structural concrete, American Concrete Institute committee 318) as load amplification factors for dead load and live load respectively are based on probability based research studies. Prof. Dr. Qaisar Ali Note: We shall be following ACI 318-02 throughout this course Reinforced Concrete Design – II 19 Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Strength Design Method Similarly, the strength is not reduced arbitrarily but considering the fact that variation in strength is possible due to imperfections, age factor etc. Strength reduction factors are used for this purpose. Factor of safety in Strength Design method is thus the combined effect of increased load and reduced strength, both modified based on a valid rationale. Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 20 10
- 11. Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand About Ton 1 metric ton = 1000 kg or 2204 pound 1 long ton: In the U.S., a long ton = 2240 pound 1 short ton: In the U.S., a short ton = 2000 pound In Pakistan, the use of metric ton is very common; therefore we will refer to Metric Ton in our discussion. Prof. Dr. Qaisar Ali 21 Reinforced Concrete Design – II Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Example 1.3 Design the 12″ × 12″ pad to carry a load of 200 tons. The area of the pad cannot be increased for some reasons. Concrete strength (fc′) = 3 ksi, therefore Allowable strength = fc′/2 = 1.5 ksi (for Working Stress method) 200 Tons Concrete pad 12″ 12″ Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 22 11
- 12. Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand 200 Tons Concrete pad Example 1.3 12″ Solution: 12″ Demand in the form of load (P) = 200 Tons = 200 × 2204/1000 = 440.8 kips Demand in the form of load effects (Stress) = (200 × 2204)/ (12 × 12) = 3061.11 psi = 3.0611 ksi Capacity in the form of strength = 1.5 ksi (less than the demand of 3.0611 ksi). There are two possibilities to solve this problem: Increase area of the pad (geometry); it cannot be done as required in the example. Increase the strength by using some other material; using high strength concrete, steel or other material; economical is to use concrete and steel combine. Prof. Dr. Qaisar Ali 23 Reinforced Concrete Design – II Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand 200 Tons Concrete pad Example 1.3 12″ Solution: 12″ Let us assume that we want to use steel bar reinforcement of yield strength fy = 40 ksi. Then capacity to be provided combinely by both materials should be at least equal to the demand. And let us follow the Working Stress approach, then: {P = Rc + Rs (Demand=Capacity)} (Force units) Capacity of pad = Acfc′/2 + Asfy/2 (Force units) Therefore, 440.8 = (144 × 3/2) + (As × 40/2) As = 11.24 in2 (Think on how to provide this much area of steel? This is how compression members are designed against axial loading). Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 24 12
- 13. Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Example 1.4 Check the capacity of the concrete beam given in figure below against flexural stresses within the linear elastic range. Concrete compressive strength (fc′) = 3 ksi 2.0 kip/ft 20″ 20′-0″ 12″ Beam section Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 25 Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Example 1.4 Solution: M = wl2/8 = {2.0 × (20)2/8} × 12 = 1200 in-kips Self-weight of beam (w/ft) = (12 × 20 × 0.145/144) = 0.24167 k/ft Msw (moment due to self-weight of beam) = (0.24167×202×12/8) = 145 in-kips M (total) = 1200 + 145 = 1345 in-kips In the linear elastic range, flexural stress in concrete beam can be calculated as: Prof. Dr. Qaisar Ali ƒ = My/I (linear elastic range) Therefore, M = ƒI/y Reinforced Concrete Design – II 26 13
- 14. Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Example 1.4 Solution: y = (20/2) = 10″ ; I = 12 × 203/12 = 8000 in4 ƒ =? The lower fibers of the given beam will be subjected to tensile stresses. The tensile strength of concrete (Modulus of rupture) is given by ACI code as 7.5 f′ , (ACI 9.5.2.3). f′ = 7.5 × 3000 = 411 psi Therefore, ƒtension = 7.5 Hence M = Capacity of concrete in bending = 411 × 8000/ (10 × 1000) = 328.8 in-kips Therefore, Demand = 1345 in-kips and Capacity = 328.8 in-kips Prof. Dr. Qaisar Ali 27 Reinforced Concrete Design – II Department of Civil Engineering, University of Engineering and Technology Peshawar Concept of Capacity and Demand Example 1.5 Check the shear capacity of the same beam. 2.0 kip/ft Solution: 20′-0″ Shear Demand: Vu = (20/10) × {10 – (17.5/12)} = 17.1 kips 20 kips 17.1 kips Shear Capacity: Vc = 2 f′ bh (ACI 11.3.1.1) = 2 3000 ×12×20/1000 17.5″ = 1.46′ = 26.29 kips > 17.1 kips Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 28 14
- 15. Department of Civil Engineering, University of Engineering and Technology Peshawar Flexural Design of Beams Using ACI Recommendations Load combinations: ACI 318-02, Section 9.2. Load Combinations: ACI 318-02, Section 9.2. U = 1.4(D + F) U = 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R) (9-2) U = 1.2D + 1.6(Lr or S or R) + (1.0L or 0.8W) (9-3) U = 1.2D + 1.6W + 1.0L + 0.5(Lr or S or R) (9-4) U = 1.2D + 1.0E + 1.0L + 0.2S (9-5) U = 0.9D + 1.6W + 1.6H (9-6) U = 0.9D + 1.0E + 1.6H Prof. Dr. Qaisar Ali (9-1) (9-7) Reinforced Concrete Design – II 29 Department of Civil Engineering, University of Engineering and Technology Peshawar Flexural Design of Beams Using ACI Recommendations Strength Reduction Factors: ACI 318-02, Section 9.3. Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 30 15
- 16. Department of Civil Engineering, University of Engineering and Technology Peshawar Flexural Design of Beams Using ACI Recommendations Design: ΦMn ≥ Mu (ΦMn is Mdesign or Mcapacity) For ΦMn = Mu As = Mu/ {Φfy (d – a/2)} Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 31 Department of Civil Engineering, University of Engineering and Technology Peshawar Flexural Design of Beams Using ACI Recommendations Design: ρmin = 3 fc′ /fy ≥ 200/fy (ACI 10.5.1) ρmax = 0.85β1(fc′/fy){εu/(εu + εt)} Where, εu = 0.003 εt = Net tensile strain (ACI 10.3.5). When εt = 0.005, Φ = 0.9 for flexural design. β1= 0.85 (for fc′ ≤ 4000 psi, ACI 10.2.7.3) Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 32 16
- 17. Department of Civil Engineering, University of Engineering and Technology Peshawar Flexural Design of Beams Using ACI Recommendations Design: ρmax and ρmin for various values of fc′ and fy Table 01: Maximum & Minimum Reinforcement Ratios fc′ (psi) 3000 4000 5000 fy (psi) 40000 60000 40000 60000 40000 60000 ρmin 0.005 0.0033 0.005 0.0033 0.0053 0.0035 ρmax 0.0203 0.0135 0.027 0.018 0.0319 0.0213 Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 33 Department of Civil Engineering, University of Engineering and Technology Peshawar Shear Design of Beams using ACI Recommendations When ΦVc/2 ≥ Vu, no web reinforcement is required. When ΦVc ≥ Vu, theoretically no web reinforcement is required. However as long as ΦVc/2 is not greater than Vu, ACI 11.5.5.1 recommends minimum web reinforcement. Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 34 17
- 18. Department of Civil Engineering, University of Engineering and Technology Peshawar Shear Design of Beams using ACI Recommendations Maximum spacing and minimum reinforcement requirement as permitted by ACI 11.5.4 and 11.5.5.3 shall be minimum of: smax = Avfy/(50bw), d/2 24 inches Avfy/ {0.75 f′ bw} Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 35 Department of Civil Engineering, University of Engineering and Technology Peshawar Shear Design of Beams using ACI Recommendations When ΦVc < Vu, web reinforcement is required as: Vu = ΦVc + ΦVs ΦVs = Vu – ΦVc ΦAvfyd/s = Vu – ΦVc s = ΦAvfyd/(Vu – ΦVc) Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 36 18
- 19. Department of Civil Engineering, University of Engineering and Technology Peshawar Shear Design of Beams using ACI Recommendations Check for Depth of Beam: ΦVs ≤ Φ8 f′ bwd (ACI 11.5.6.9) If not satisfied, increase depth of beam. Check for Spacing: ΦVs ≤ Φ4 f′ bwd (ACI 11.5.4.3) If not satisfied, reduce maximum spacing requirement by one half. Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 37 Department of Civil Engineering, University of Engineering and Technology Peshawar Shear Design of Beams using ACI Recommendations Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 38 19
- 20. Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Design the beam shown below as per ACI 318-02. W D.L = 1.0 kip/ft W L.L = 1.5 kip/ft 20′-0″ Take f ′c = 3 ksi & fy = 40 ksi Prof. Dr. Qaisar Ali 39 Reinforced Concrete Design – II Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 01: Sizes. For 20′ length, a 20″ deep beam would be appropriate (assumption). Width of beam cross section (bw) = 14″ (assumption) W D.L = 1.0 kip/ft W L.L = 1.5 kip/ft 20′-0″ 20″ 14″ Beam section Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 40 20
- 21. Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 02: Loads. Self weight of beam = γcbwh = 0.15 × (14 × 20/144) = 0.292 kips/ft W u = 1.2D.L + 1.6L.L (ACI 9.2) = 1.2 × (1.0 + 0.292) + 1.6 × 1.5 = 3.9504 kips/ft Prof. Dr. Qaisar Ali 41 Reinforced Concrete Design – II Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 03: Analysis. Flexural Analysis: Mu = W u l2/8 = 3.9504 × (20)2 × 12/8 = 2370.24 in-kips 3.9504 kip/ft Analysis for Shear in beam: Vu = 39.5 × {10 – (17.5/12)}/10 = 33.74 k 33.74 kips 39.50 SFD 2370.24 BMD Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 42 21
- 22. Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 04: Design. Design for flexure: ΦMn ≥ Mu (ΦMn is Mdesign or Mcapacity) For ΦMn = Mu ΦAsfy(d – a/2) = Mu As = Mu/ {Φfy (d – a/2)} Calculate “As” by trial and success method. Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 43 Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 04: Design. Design for flexure: First Trial: Assume a = 4″ As = 2370.24 / [0.9 × 40 × {17.5 – (4/2)}] = 4.25 in2 a = Asfy/ (0.85fc′bw) Prof. Dr. Qaisar Ali = 4.25 × 40/ (0.85 × 3 × 14) = 4.76 inches Reinforced Concrete Design – II 44 22
- 23. Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 04: Design. Design for flexure: Second Trial: • As = 2370.24 / [0.9 × 40 × {17.5 – (4.76/2)}] = 4.35 in2 • a = 4.35 × 40/ (0.85 × 3 × 14) = 4.88 inches • As = 2370.24 / [0.9 × 40 × {17.5 – (4.88/2)}] = 4.37 in2 Third Trial: • a = 4.37 × 40/ (0.85 × 3 × 14) = 4.90 inches “Close enough to the previous value of “a” so that As = 4.37 in2 O.K Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 45 Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 04: Design. Design for flexure: ρmin = 3 3 × 3000 /40000 = 0.004 200/40000 = 0.005 Therefore, ρmin = 0.005 Prof. Dr. Qaisar Ali Check for maximum and minimum reinforcement allowed by ACI: Asmin = ρminbwd = 0.005 × 14 × 17.5 = 1.225 in2 f′ /fy ≥ 200/fy (ACI 10.5.1) Reinforced Concrete Design – II 46 23
- 24. Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 04: Design. Design for flexure: ρmax = 0.85β1(fc′/fy){εu/(εu + εt)} εt = Net tensile strain (ACI 10.3.5). When εt = 0.005, Φ = 0.9 for flexural design. β1= 0.85 (for fc′ ≤ 4000 psi, ACI 10.2.7.3) ρmax = 0.85 × 0.85 × (3/40) × (0.003/(0.003+0.005) = 0.0204 = 2 % of area of concrete. Asmax = 0.0204 × 14 × 17.5 = 4.998 in2 Asmin (1.225) < As (4.37) < Asmax (4.998) O.K Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 47 Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 04: Design. Design for flexure: Bar Placement: 10 #6 bars will provide 4.40 in2 of steel area which is slightly greater than required. Other options can be explored. For example, 6 #8 bars (4.74 in2), Prof. Dr. Qaisar Ali 8 #7 bars (4.80 in2), or combination of two different size bars. Reinforced Concrete Design – II 48 24
- 25. Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 04: Design. Design for flexure: Curtailment of flexural reinforcement: Positive steel can be curtailed 50 % at a distance (l/8) from face of the support. For Curtailment and bent up bar details refer to the following figures provided at the end of this lecture: Graph A2 and A3 in “Appendix A” of Nilson 13th Ed. Figure 5.15 of chapter 5 in Nilson 13th Ed. Prof. Dr. Qaisar Ali 49 Reinforced Concrete Design – II Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 04: Design. Design for Shear: Vu = 33.74 kips ΦVc = (Capacity of concrete in shear) = Φ2 f′ bwd = 0.75×2× 3000 ×14×17.5/1000 = 20.13 k (Φ=0.75, ACI 9.3.2.3) Prof. Dr. Qaisar Ali As ΦVc < Vu, Shear reinforcement is required. Reinforced Concrete Design – II 50 25
- 26. Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 04: Design. Design for Shear: Assuming #3, 2 legged (0.22 in2), vertical stirrups. Spacing required (s) = ΦAvfyd/ (Vu – ΦVc) = 0.75×0.22×40×17.5/ (33.74–20.13) ≈ 8.5″ Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 51 Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 04: Design. Design for Shear: Maximum spacing and minimum reinforcement requirement as permitted by ACI 11.5.4 and 11.5.5.3 is minimum of: smax = 24″ Prof. Dr. Qaisar Ali smax = d/2 = 17.5/2 = 8.75″ smax = Avfy/(50bw) =0.22 × 40000/(50 × 14) = 12.57″ Avfy/ 0.75√(fc′)bw = 0.22×40000/ {(0.75×√(3000)×14} =15.30″ Therefore smax = 8.75″ Reinforced Concrete Design – II 52 26
- 27. Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 04: Design. Design for Shear: Other checks: Check for depth of beam: ΦVs ≤ Φ8 Φ8 f′ bwd (ACI 11.5.6.9) f′ bwd = 0.75 × 8 × 3000 × 14 × 17.5/1000 = 80.52 k ΦVs = Vu – ΦVc = 33.74 – 20.13 =13.61 k < 80.52 k, O.K. Therefore depth is O.K. If not, increase depth of beam. Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 53 Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 04: Design. Design for Shear: Other checks: Check if “ΦVs ≤ Φ4 If “ΦVs ≤ Φ4 f′ bwd” (ACI 11.5.4.3): f′ bwd”, the maximum spacing (smax) is O.K. Otherwise reduce spacing by one half. Prof. Dr. Qaisar Ali 13.61 kips < 40.26 kips O.K. Reinforced Concrete Design – II 54 27
- 28. Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 04: Design. Design for Shear: Arrangement of stirrups in the beam: Shear capacity of RC beam is given as: ΦVn = ΦVc + ΦVs ΦVc = 20.13 kips With #3, 2 legged vertical stirrups @ 8.75″ c/c (maximum spacing and minimum reinf. requirement as permitted by ACI), ΦVs = (ΦAvfyd)/ smax ΦVs = (0.75 × 0.22 × 40 × 17.5/8.75) = 13.2 kips Therefore ΦVn = 20.13 + 13.2 = 33.33 k < (Vu = 33.74 k) Prof. Dr. Qaisar Ali 55 Reinforced Concrete Design – II Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 ΦVn #3 @ 8.5″ c/c Prof. Dr. Qaisar Ali #3 @ 8.75″ c/c Reinforced Concrete Design – II Not Required theoretically 56 28
- 29. Department of Civil Engineering, University of Engineering and Technology Peshawar Example 1.6 Flexural and Shear Design of Beam as per ACI: Solution: Step No. 05: Drafting. Note that some nominal negative reinforcement has been provided at the beam ends to care for any incidental negative moment that may develop due to partial restrain as a result of friction etc. between beam ends and walls. In other words, though the beam has been analyzed assuming hinge or roller supports at the ends, however in reality there will always be some partial fixity or restrain at the end. Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 57 Department of Civil Engineering, University of Engineering and Technology Peshawar References ACI 318-02 Design of Concrete Structures (13th Ed.) by Nilson, Darwin and Dolan Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 58 29
- 30. Department of Civil Engineering, University of Engineering and Technology Peshawar Appendix Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 59 Department of Civil Engineering, University of Engineering and Technology Peshawar Appendix Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 60 30
- 31. Department of Civil Engineering, University of Engineering and Technology Peshawar Appendix Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 61 Department of Civil Engineering, University of Engineering and Technology Peshawar The End Prof. Dr. Qaisar Ali Reinforced Concrete Design – II 62 31