LA INDUSTRIA AGROALIMENTARIA LEONESA: UN REPLANTEAMIENTO ESTRATÉGICO
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In accordance with ACI Eq. (11-2), the total shear strength is the sum of two components: shear strength provided by concrete (φVc) and shear strength provided by shear reinforcement (φVs). Thus, at any section of the member, Vu≤ φVc+ φVs. Using the simplest of the code equations for shear strength of concrete, Vc, can be calculated as follows:
Vc= φ2λ bwd ACI Eq. (11−3)
where λ is a modification factor to account for the reduced mechanical properties of lightweight concrete.
λ = 1.0 for normal weigh concrete, λ = 0.85 for sand-lightweight concrete and 0.75 for all-lightweight concrete. Figure 3-6 shows the values for nominal shear strength provided by concrete (Vc) for a rectangular section, for normal and light weight concrete and for different concrete strengths. Figure 3-7 and Table 3-7 summarize ACI 318 provisions for shear reinforcement.
ʹf
c
0.08 0.1 0.12 0.14 0.16 0.18
3 4 5 6 7 8
Concrete Shear Strength vc (ksi)
f'c (ksi) Vc=φ vcbwd
Normal weight
Sand lightweight
All lightweight
Figure 3-6 Nominal Shear Strength Provided by Concrete Vc(Kips)
Table 3-7 ACI Provisions for Shear Design
* Members subjected to shear and flexure only; φVc=φ2 bwd, φ = 0.75 (ACI 11.2.1.1)
** Av = 2 Abfor U stirrups; fy≤ 60 ksi (ACI 11.4.2)
*** Maximum spacing based on minimum shear reinforcement (= Af/50b ) must also be considered (ACI 11.4.6.3).
* Vs ≤ 8 bwd
** See ACI Section 11.4.6.1 for exceptions Vu max
φVc
0.5φVc
d/2
Zone 1 Zone 2 Zone 3
φVc ≥ Vu > 0.5 φVc
Vu > Vc Vu≤ 0.5 φVc
Provide calculated shear reinforcement (φVs = Vu – φVc)*
Provide minimum shear reinforcement**
No shear reinforcement
required
φ
Figure 3-7 Shear Reinforcement Requirements
ʹf
c
Chapter 3 • Simplified Design for Beams and Slabs
The design values in table 3-8 are valid for › = 4000 psi:
The selection and spacing of stirrups can be simplified if the spacing is expressed as a function of the effective depth d (see Reference 3.3). According to ACI 11.4.5.1 and ACI 11.4.5.3, the practical limits of stirrup spacing vary from s = d/2 to s = d/4, since spacing closer than d/4 is not economical. With one intermediate spacing at d/3, the calculation and selection of stirrup spacing is greatly simplified. Using the three standard stirrup spacings noted above (d/2, d/3, and d/4), a specific value of φVscan be derived for each stirrup size and spacing as follows:
For vertical stirrups:
ACI Eq. (11.15) By substituting d/n for s (where n = 2, 3, or 4), the above equation can be rewritten as:
Thus, for No.3 U-stirrups @ s = d/2 with fy= 60,000 psi and φ = 0.75 φVs= 0.75(0.22)60 2 = 19.8 kips, say 20 kips
The values φVsgiven in Table 3-9 may be used to select shear reinforcement with Grade 60 rebars.
Table 3-9 Values of φVs(fy= 60 ksi)
It should be noted that these values of φVsare not dependent on the member size nor on the concrete strength.
s #3 U-stirrups #4 U-stirrups #5 U-stirrups d/2 20 kips 36 kips 56 kips
d/3 30 kips 54 kips 84 kips d/4 40 kips 72 kips 112 kips
*Valid for stirrups with 2 legs (double the tabulated values for 4 legs, etc.)
φVs = φAvfyd s
φVs = φAvfyn
Equation Design Value ACI Section
Vc = 2 bwd 0.095bwd ACI 11.2.1.1 0.50 Vc = bwd 0.048bwd ACI 11.4.6.1 Maximum ( Vc + Vs) = 10 bwd 0.48bwd ACI 11.4.7.9 Joists defined by ACI 8.13
Vc = 2.2 bwd
0.104bwd ACI 8.13.8
* bw and d are in inches and the resulting shear in kips ʹf
c
ʹf
c
ʹf
c
ʹf
c
Table 3-8 Concrete Shear Strength Design Values for › = 4000 psi
The design charts in Figs. 3-8 through 3-11 offer another simplified method for shear design. By entering the charts with values of d and φVs= Vu- φVcfor the member at the section under consideration, the required stirrup spacing can be obtained by locating the first line above the point of intersection of d and φVs. Values for spacing not shown can be interpolated from the charts if desired. Also given in the charts the values for the minimum practical beam widths bwthat correspond to the maximum allowable for each given spacing s; any member which has at least this minimum bwwill be adequate to carry the maximum applied Vu. Fig. 3-11 can also be used to quickly determine if the dimensions of a given section are adequate: any member with an applied Vuwhich is less than the applicable Vu(max)can carry this shear without having to increase the values of bwand/or d. Once the adequacy of the cross-section has been verified, the stirrup spacing can be established by using Figs. 3-8 through 3-11. This spacing must then be checked for compliance with all maximum spacing criteria.
φVs = φ8 ʹfcbwd
Figure 3-8 Design Chart for Stirrup Spacing, #3-U Stirrups
Chapter 3 • Simplified Design for Beams and Slabs
10 20 30 40 50 60 70 80
10 15 20 25 30 35 40
d, in
18"
16"
14"
12"
10"
8"
6"
spacing, s = 4"
Vu− φVc =φAvfyd s
φVs=Vu−φVc,kips
Figure 3-9 Design Chart for Stirrup Spacing, #4-U Stirrups 30
40 50 60 70 80 90
10 15 20 25 30 35 40
d, in φVs=Vu−φVc,kips
spacing, s = 4"
Vu− φVc = φAvfyd s
18"
(—) (—) (—) (—) (—) (—)
(8") (12")**
16"
14"
12"
10"
8"
6"
* Horizontal line indicates for s = d/2.
** Values in ( ) indicate minimum practical bwcorresponding to bwd for given s.
φVs
φVs= φ8 fcʹ φV = φ8 ʹ
Figure 3-10 Design Chart for Stirrup Spacing, #5-U Stirrups
Chapter 3 • Simplified Design for Beams and Slabs
50 60 70 80 90 100 110 120
10 15 20 25 30 35 40
d, in spacing,
s = 4"
Vu− φVc = φAvfyd s
(—) (—) (—)
(—) (20")**
6" (14")
8"
(10")
10"
(8") 12"
14"
16"
18"
φVs=Vu−φVc,kips
Figure 3-11 - Design Chart for Stirrups Spacing for Different Stirrup Sizes
#5
#4
#3
0.25 0.3 0.35 0.4 0.45 0.5 s/d
φVs = Vu - φVc’ kips 130
120
110
100
90
80
70
60
50
40
30
20
10
0
Chapter 3 • Simplified Design for Beams and Slabs
Figure 3-12 Design Chart for Maximum Shear Force—›= 4 ksi
40 80 120 160 200 240 280 320 360 400
10 15 20 25 30 35 40
d, in.
bw=36"
30"
24"
22"
20"
18"
16"
14"
12"
10"
3.6.1 Example: Design for Shear Reinforcement
The example shown in Fig. 3-13 illustrates the simple procedure for selecting stirrups using design values for Vcand Vs.
(1) Design data: › = 4000 psi, fy= 60,000 psi, wu= 7 kips/ft.
(2) Calculations: Vu@ column centerline: wu˜/2 = 7 24/2 = 84.0 kips Vu@ face of support: 84 – 1.17(7) = 75.8 kips Vu@ d from support face (critical section): 75.8 – 2(7) = 61.8 kips
(φVc+ φVs)max: 0.48 bwd = 0.48(12)(24) = 138.2 kips
φVc: 0.095 bwd= 0.095(12)(24) = 27.4 kips
φVc/2: 0.048 bwd = 0.048(12)(24) = 13.80 kips
(3) Beam size is adequate for shear strength, since 138.2 kips > 61.8 kips (also see Fig. 3-12). φVs(required)
= 61.8 – 27.4 = 34.4 kips. From Table 3-9, No.4 @ d/2 = 12 in. is adequate for full length where stirrups are required since φVs = 36 kips > 34.4 kips. Length over which stirrups are required is (75.8 – 13.8)/7
= 8.86 ft from support face.
Check maximum stirrup spacing:
Since φVs= 36 kips < 54.6 kips, the maximum spacing is the least of the following:
Use 10-No.4 U-stirrups at 12 in. at each end of beam.
The problem may also be solved graphically as shown in Fig. 3-13. φVsfor No.3 stirrups at d/2, d/3, and d/4 are scaled vertically from φVc. The horizontal intersection of the φVsvalues (20 kips, 30 kips, and 40 kips) with the shear diagram automatically sets the distances where the No.3 stirrups should be spaced at d/2, d/3, and d/4.
The exact numerical values for these horizontal distances are calculated as follows (although scaling from the sketch is close enough for practical design):
No.3 @ d/4 = 6 in.: (75.8 – 57.4)/7 = 2.63 ft (31.5 in.) use 6 @ 6 in.
@ d/3 = 8 in.: (57.4 – 47.4)/7 = 1.43 ft (17.0 in.) use 2 @ 8 in.
@ d/2 = 12 in.: (47.4 – 13.8)/7 = 4.8 ft (57.6 in.) use 5 @ 12 in.
A more practical solution may be to eliminate the 2 @ 8 in. and use 9 @ 6 in. and 5 @ 12 in.
sm ax =
d / 2= 12in.(governs) 24in.
Avfy / 50bw = 40in.
⎧
⎨⎪⎪
⎩⎪
⎪
φ4 ʹfcbwd= 0.19bwd= 54.6 kips
Chapter 3 • Simplified Design for Beams and Slabs
As an alternative, determine the required spacing of the No.3 U-stirrups at the critical section using Fig. 3-8.
Enter the chart with d = 24 in. and φVs= 38.9 kips. The point representing this combination is shown in the design chart. The line immediately above this point corresponds to a spacing of s = 6 in. which is exactly what was obtained using the previous simplified method.
CL Span
5 @ 12"
2 @ 8"
6 @ 6"
2"
14"
12'-0"
wu = 7 kips/ft CL Column
bw = 12"
d = 24"
h = 27"
#3 @ d/2 d/3 d/4 d from support face
Face of column
84 kips
75.8 kips
slope = 7 kips/ft 61.8 kips
1.17' 2.0'
57.4 kips (27.4 + 30)
47.4 kips ( 27.4 + 20)
27.4 kips = φVc
20 kips
30 kips40 kips
φVc/2 13.8 kips
Stirrups Required No
stirrups 1.17'
φVc + φVs = 27.4 + 40 = 67.4 > 61.8 kips
φVs
3.6.2 Selection of Stirrups for Economy
Selection of stirrup size and spacing for overall cost savings requires consideration of both design time and fabrication and placing costs. An exact solution with an intricate stirrup layout closely following the variation in the shear diagram is not a cost-effective solution. Design time is more cost-effective when a quick, more conservative analysis is utilized. Small stirrup sizes at close spacings require disproportionately high costs in labor for fabrication and placement. Minimum cost solutions for simple placing should be limited to three spacings: the first stirrup located at 2 in. from the face of the support (as a minimum clearance), an intermediate spacing, and finally, a maximum spacing at the code limit of d/2. Larger size stirrups at wider spacings are more cost-effective (e.g., using No.4 for No.3 at double spacing, and No.5 and No.4 at 1.5 spacing) if it is possible to use them within the spacing limitations of d/2 and d/4.
In order to adequately develop the stirrups, the following requirements must all be satisfied (ACI 12.13):
(1) stirrups shall be carried as close to the compression and tension surfaces of the member as cover requirements permit, (2) for No.5 stirrups and smaller, a standard stirrup hook (as defined in ACI 7.1.3) shall be provided around longitudinal reinforcement, and (3) each bend in the continuous portion of the stirrup must enclose a longitudinal bar. To allow for bend radii at corners of U stirrups, the minimum beam widths given in Table 3-10 should be provided.
Table 3-10 Minimum Beam Widths for Stirrups
Note that either the No.3 or the No.4 stirrup in the example of Fig. 3-8 can be placed in the 12 in. wide beam.