Department of Civil Engineering, N-W.F.P UET, Peshawar
Concepts of Capacity & Demand
Demand: It is an action on the structure which can cause internal disturbance(s) in the structure in the form of stresses. A demand on the structure may be in the form of Gravity, Wind, Earthquake, Snow or other loads. Capacity: The overall ability of a structure to carry an imposed demand. Internal stresses due to an imposed demand (Load effects): Compression, Tension, Bending, Shear, and Torsion. Failure: It occurs when capacity is less than demand. To avoid failure, capacity to demand ratio should be 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? It depends. Example 1.1: Calculate demand in the form of stresses or load effects on the given concrete pad of size 12″ × 12″. Solution: Demand = 50 tons Load (50 tons) Concrete pad
12
"
12"
Figure 1.1: Concrete pad. 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. In the given example, load of 50 tons is demand. Its effect on the pad will be a compressive stress equal to load divided by the area of the pad. Or: (50 × 2204)/ (12 × 12) = 765.27 psi Therefore compressive stress of 765.27 psi is demand in the form of load effect on the pad. Capacity of the pad in the form of resistance should be able to carry a stress of 765.27 psi. In other words, the compressive strength of concrete pad (capacity) should be more than 765.27 psi (demand). Prof Dr. Qaisar Ali (http://www.eec.edu.pk)
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Department of Civil Engineering, N-W.F.P UET, Peshawar
Concepts of Capacity & Demand
Example 1.2: Determine capacity to demand ratio for the pad given in figure 01 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? Solution: (i) Capacity/ Demand = 500 / 765.27 = 0.653 (ii) 765.27/ 765.27 = 1.0 (iii)1000/ 765.27 = 1.3 (iv) 2000/ 765.27 = 2.6 Comments on the above results: (i) Failure, (ii) Capacity just equal to Demand, (iii)Capacity is 1.3 times greater than Demand, (iv) Capacity is 2.6 times greater than Demand. In (iii) and (iv), there is some margin of safety normally called as factor of safety. 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. 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 (i) Take an increased demand instead of actual demand (load), e.g. 70 ton instead of 50 ton in the previous example, (ii) Take a reduced capacity instead of actual capacity such as 1500 psi for concrete whose actual strength is 3000 psi (iii) Doing both. 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. 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 Prof Dr. Qaisar Ali (http://www.eec.edu.pk)
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Department of Civil Engineering, N-W.F.P UET, Peshawar
Concepts of Capacity & Demand
based research studies. 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. 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. Example 1.3: Design a 12″ × 12″ pad to carry a load of 200 tons. The area of the pad cannot be increased for some reasons. Concrete strength = 3 ksi Allowable strength = 1.5 ksi (for Working Stress method) Solution: Demand = 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: (i) Increase area of the pad (geometry); it cannot be done as required in the example. (ii) Increase the strength by using some other material; using high strength concrete, steel or other material; economical is to use concrete and steel combinely. How? Let’s see. 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: Capacity of pad = Acfc′/2 + Asfy/2 (Force units) {P = Rc + Rs (Demand=Capacity)} Demand (in kips) = (200 × 2204/1000) = 440.8 k Therefore, 440.8 = (144 × 3/2) + (As × 40/2) Prof Dr. Qaisar Ali (http://www.eec.edu.pk)
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Department of Civil Engineering, N-W.F.P UET, Peshawar
Concepts of Capacity & Demand
As = 11.24 in2 Think on how to provide this much area of steel. This is how compression members are designed against axial loading. Example 1.4: Check the capacity of the concrete beam given in figure 1.2 against flexural stresses within the linear elastic range. Concrete compressive strength (fc′) = 3 ksi Solution: M = wl2/8 = {2.0 × (20)2/8} × 12 = 1200 in-k 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-k M (total) = 1200 + 145 = 1345 in-k
w = 2.0 k/ft
l = 20'-0"
20"
12"
Figure 1.2: 20′-0″ long beam. In the linear elastic range, flexural stress in concrete beam can be calculated as: ƒ = My/I (linear elastic range) Therefore, M = ƒI/y y = (20/2) = 10″ I = 12 × 203/12 = 8000 in4 ƒ =?
Prof Dr. Qaisar Ali (http://www.eec.edu.pk)
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Department of Civil Engineering, N-W.F.P UET, Peshawar
Concepts of Capacity & Demand
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√ (fc′), (ACI 9.5.2.3). Therefore, ƒtension = 7.5√ (fc′) = 7.5 × √ (3000) = 411 psi Hence M = Capacity of concrete in bending = 411 × 8000/ (10 × 1000) = 328.8 in-k Therefore, Demand = 1345 in-k, and Capacity = 328.8 in-k Example 1.5: Check the shear capacity of the same beam. Solution: -
2.0 k/ft 20'-0" Vu = 17.1 k
Vmax = 20 k d = 17.5"
Figure 1.3: Shear Force Diagram. Demand (Vu) = (20/10) × {10 – (17.5/12)} = 17.1 k Capacity (Vc) = 2√ (fc′)bh (ACI 11.3.1.1) = 2√ (3000) × 12 × 20/1000 = 26.29 k > 17.1 k
Prof Dr. Qaisar Ali (http://www.eec.edu.pk)
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Department of Civil Engineering, N-W.F.P UET, Peshawar
Concepts of Capacity & Demand
Example 1.6: Flexural and shear design of beam according to ACI 318-02.
S.D.L=1.0 k/ft; L.L = 1.5 k/ft 20'-0" Figure 1.4: 20′-0″ long beam. Solution: Step No 1: Sizes. For 20 ft length, a 20-inch deep beam would be appropriate (assumption). Width of beam cross section (bw) = 14″ (assumption) Self Study 1: Study ACI 9.5.2.1 for recommendations on depth of beams.
20" 17.5"
14" Figure 1.5: Beam section with assumed sizes. Step No 2: Loads. Self weight of beam = γcbwh = 0.15 × (14 × 20/144) = 0.292 k/ft wu = 1.2D.L + 1.6L.L (ACI 9.2) = 1.2 × (1.0 + 0.292) + 1.6 × 1.5 = 3.9504 k/ft Step No 3: Analysis. Flexural Analysis: Mu = wu l2/8 = 3.9504 × (20)2 × 12/8 = 2370.24 in-k Analysis for shear in beam: Vu = 39.5 × {10 – (17.5/12)}/10 = 33.74 k
Prof Dr. Qaisar Ali (http://www.eec.edu.pk)
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Department of Civil Engineering, N-W.F.P UET, Peshawar
Concepts of Capacity & Demand
3.9504 k/ft 20'-0"
Vu = 33.74 k 39.5 k
10'
10'
Figure 1.6: Shear Force Diagram. Step No 4: Design. 1. 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. First Trial: Assume a = 4″ As = 2370.24 / [0.9 × 40 × {17.5 – (4/2)}] = 4.25 in2 a = Asfy/ (0.85fc′bw) = 4.25 × 40/ (0.85 × 3 × 14) = 4.76 in 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 in Third Trial: As = 2370.24 / [0.9 × 40 × {17.5 – (4.88/2)}] = 4.37 in2 a = 4.37 × 40/ (0.85 × 3 × 14) = 4.90 in This is close enough to the previous value of “a” so that As = 4.37 in2 O.K Prof Dr. Qaisar Ali (http://www.eec.edu.pk)
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Department of Civil Engineering, N-W.F.P UET, Peshawar
Concepts of Capacity & Demand
Check for maximum and minimum reinforcement allowed by ACI: ρmin = 3√ (fc′)/fy ≥ 200/fy (ACI 10.5.1) 3 × √ (3000)/40000 = 0.004 200/40000 = 0.005 Therefore, ρmin = 0.005 Asmin = ρminbwd = 0.005 × 14 × 17.5 = 1.225 in2 And, 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 Food for Thought: Why flexural reinforcement is compared with maximum and minimum requirements of ACI? Note that ρmin & ρmax can also be found using table A.4, Nilson 13th Ed. 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, 8 #7 bars (4.80 in2), 6 #8 bars (4.74 in2), or combination of two different size bars. Curtailment of flexural reinforcement: •
Positive steel can be curtailed 50 % at a distance (l/8) from face of the support. Graph A2 and A3 in “Appendix A” of Nilson 13th Ed can be used to find cutoff points in simply supported and continuous beams respectively. Figure 5.15 of chapter 5 in Nilson 13th Ed can be used to find cutoff or bend points for bars in continuous beams with approximately equal spans with uniformly distributed loads.
Prof Dr. Qaisar Ali (http://www.eec.edu.pk)
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Department of Civil Engineering, N-W.F.P UET, Peshawar
Concepts of Capacity & Demand
Self Study 2: Study cutoff, bends and development length (chapter 5, Nilson 13th Ed). 2. Design for shear:
The above table 01 shows the summary of web reinforcement requirement in beams. (a) When ΦVc/2 ≥ Vu, no web reinforcement is required. (b) 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. Maximum spacing and minimum reinforcement requirement as permitted by ACI 11.5.4 and 11.5.5.3 shall be minimum of: i) smax = Avfy/(50bw), ii) d/2 iii) 24 inches iv) Avfy/ {0.75√ (fc′)bw} Self Study 3: Study ACI 11.5.5R. (c) When ΦVc < Vu, web reinforcement is required as: Vu = ΦVc + ΦVs ΦVs = Vu – ΦVc Prof Dr. Qaisar Ali (http://www.eec.edu.pk)
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Department of Civil Engineering, N-W.F.P UET, Peshawar
Concepts of Capacity & Demand
ΦAvfyd/s = Vu – ΦVc s = ΦAvfyd/(Vu – ΦVc) Self Study 4: Study Example 4.1, page 133, Nilson 13th Ed. Study example 4.3, page 137 to 139, Nilson 13th Ed. Now, Vu = 33.74 k ΦVc = (Capacity of concrete in shear) = Φ2√ (fc′)bwd ΦVc=0.75×2×√(3000)×14×17.5/1000 = 20.13 k (Φ=0.75, ACI 9.3.2.3) ΦVc < Vu {Shear reinforcement is required} 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″ Maximum spacing and minimum reinforcement requirement as permitted by ACI 11.5.4 and 11.5.5.3 is minimum of: i) smax = Avfy/(50bw) =0.22 × 40000/(50 × 14) = 12.57″ ii) smax = d/2 = 17.5/2 = 8.75″ iii) smax = 24″ iv) Avfy/ 0.75√(fc′)bw = 0.22×40000/ {(0.75×√(3000)×14} =15.30″ Other checks: (a) Check for depth of beam: ΦVs ≤ Φ8√ (fc′)bwd (ACI 11.5.6.9) Φ8√ (fc′)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. (b) Check if “ΦVs ≤ Φ4√ (fc′)bwd” (ACI 11.5.4.3): If “ΦVs ≤ Φ4√(fc′)bwd”, the maximum spacing (smax) is O.K. Otherwise reduce spacing by one half. 13.61 k < 40.26 k O.K. Arrangement of stirrups in the beam: With #3, 2 legged vertical stirrups @ 8.75″ c/c (maximum spacing and minimum reinforcement
Prof Dr. Qaisar Ali (http://www.eec.edu.pk)
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Department of Civil Engineering, N-W.F.P UET, Peshawar
Concepts of Capacity & Demand
requirement as permitted by ACI), the shear capacity (ΦVn) of the beam will be equal to: ΦVn = ΦVc + ΦVs ΦVs = (ΦAvfyd)/ smax ΦVs = (0.75 × 0.22 × 40 × 17.5/8.75) = 13.2 k Therefore ΦVn = 20.13 + 13.2 = 33.33 k < (Vu = 33.74 k) It means that theoretically, from a section at a distance equal to s/2 up to a section where shear is 33.33 k, #3, 2 legged vertical stirrups @ 8.5″ c/c shall be provided. After that, #3, 2 legged vertical stirrups @ 8.75″ of spacing can be provided from section where shear is 33.33 kips up to a section where shear is equal to ΦVc/2. Beyond the value of ΦVc/2, no shear reinforcement is theoretically required. However # 3, 2 legged vertical stirrups @ 12″ c/c are recommended to hold the flexural reinforcement bars. Practically it will be more feasible to provide # 3, 2 legged vertical stirrups @ 8.5″ c/c up to ФVc/2 and #3, 2 legged vertical stirrups @ 12″ c/c beyond ФVc/2 as shown in figure 1.6. Start providing stirrups at a distance s/2 = 8.5/2 = 4.25″, from the face of the support.
Prof Dr. Qaisar Ali (http://www.eec.edu.pk)
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Department of Civil Engineering, N-W.F.P UET, Peshawar
Concepts of Capacity & Demand
V = 38.01 k Vu = 33.74 k 33.33 k φVn Vmax= 39.5 k
φVc = 20.13 k
20.13 k
φVc /2 = 10.065 k
10.065 k
d = 17.5"
2.5' 5.09'
s/2 = 4.25" #3 @ 8.5" c/c 9" Brick Wall
Theoretically no stirrups needed
#3, 2 legged stirrups @ 12" c/c Figure 1.6: Stirrups arrangement.
Prof Dr. Qaisar Ali (http://www.eec.edu.pk)
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Department of Civil Engineering, N-W.F.P UET, Peshawar
Concepts of Capacity & Demand
Step No 5: Drafting. 4 # 4 Bars
A
B 2 # 4 Bars
l/5 = 4'
#3, 2 legged vertical stirrups @ 12" c/c
#3, 2 legged vertical stirrups @ 8.5" c/c
Standard hooks ACI 7.1
s/2 = 4.25"
(5+5) # 6 Bars
A
B
7.5'
2.5'
Seismic hook ACI 21.1 4 # 4 Bars
2 # 4 Bars
# 3 Stirrups @ 8.5" c/c 1" spacer bar @ 3' c/c
20"
10 # 6 Bars
# 3 Stirrups @ 12" c/c 1" spacer bar @ 3' c/c
20"
10 # 6 Bars
14"
14"
Section A-A
Section B-B
Figure 1.7: Reinforcement Details (Long Section & Cross Sections of beam)
eNote 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 (http://www.eec.edu.pk)
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Department of Civil Engineering, N-W.F.P UET, Peshawar
Concepts of Capacity & Demand
References 他 Design of Concrete Structures (13th Ed.) by Nilson, Darwin & Dolan. 他 ACI 318-02 (Building code requirements for structural concrete, American Concrete Institute committee 318).
Prof Dr. Qaisar Ali (http://www.eec.edu.pk)
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