Simplified Design for Beams and One-Way Slabs
3.8 EXAMPLES: SIMPLIFIED DESIGN FOR BEAMS AND ONE-WAY SLABS
3.8.3 Example: Design of the Support Beams for the Standard Pan Joist Floor along a Typical N-S Interior Column Line (Building #1)
(1) Data: › = 4000 psi (normal weight concrete, carbonate aggregate) fy= 60,000 psi
Floors: LL = 60 psf
DL = 130 psf (assumed total for joists and beams + partitions + ceiling & misc.) Required fire resistance rating = 1 hour (2 hours for Alternate (2)).
Preliminary member sizes
Columns interior = 18 18 in.
exterior = 16 16 in.
width of interior beams = 36 in. 2 depth – 2 19.5 = 39.0 in.
The most economical solution for a pan joist floor is making the depth of the supporting beams equal to the depth of the joists. In other words, the soffits of the beams and joists should be on a common plane.
This reduces formwork costs sufficiently to override the savings in materials that may be accomplished by using a deeper beam. See Chapter 9 for a discussion on design considerations for economical formwork. The beams are often made about twice as wide as they are deep. Overall joist floor depth = 16 in. + 3.5 in. = 19.5 in. Check deflection control for the 19.5 in. beam depth. From Table 3-1:
h = 19.5 in. > /18.5 = (28.58 12)/18.5 = 18.5 in. O.K.
where ˜n(end span) = 30 – 0.67 – 0.75 = 28.58 ft (governs)
˜n(interior span) = 30 – 1.50 = 28.50 ft.
(2) Determine factored shears and moments from the gravity loads using the approximate coefficients (see Figs. 2-3, 2-4, and 2-7).
Check live load reduction. For interior beams:
KLLAT= 2(30 30) = 1800 sq ft > 400 sq ft
L = 60(0.25 + 15/ ) = 60(0.604)* = 36.2 psf > 50% Lo
DL = 130 30 = 3.9 klf
LL = 36.2 30 = 1.09 klf
wu= [1.20(130) + 1.6(36.2) = 214 psf] 30 ft = 6.4 klf 1
1000 1 1000 1800
Chapter 3 • Simplified Design for Beams and Slabs
Note: All shear and negative moment values are at face of supporting beams.
Vu@ exterior columns = 6.4(28.58)/2 = 91.5 kips Vu@ first interior columns = 1.15(91.5) = 105.3 kips Vu@ interior columns = 6.4(28.5)/2 = 91.2 kips -Mu@ exterior columns = 6.4(28.58)2/16 = 326.7 ft-kips +Mu@ end spans = 6.4(28.58)2/14 = 373.4 ft-kips -Mu@ first interior columns = 6.4(28.58)2/10 = 522.8 ft-kips +Mu@ interior span = 6.4(28.50)2/16 = 324.9 ft-kips
(3) Design of the column line beams also includes consideration of moments and shears due to wind.
The wind load analysis for Building #1 is summarized in Fig. 2-13.
Note: The reduced load factor (0.5) permitted for load combinations including the wind effect (ACI 9-3 and 9-4) is in most cases, sufficient to accommodate the wind forces and moments without an increase in the required beam size or reinforcement (i.e., the load combination for gravity load only will usually govern for proportioning the beam).
(4) Check beam size for moment strength Preliminary beam size = 19.5 in. 36 in.
For negative moment section:
where d = 19.5 – 2.5 = 17.0 in. = 1.42 ft For positive moment section:
bw= 20 (373.4)/172= 25.8 in. < 36 in.
Check minimum size permitted with ρ = 0.0206:
bw= 14.6(522.8)/172= 26.4 in. < 36 in. O.K.
Use 36 in. wide beam and provide slightly higher percentage of reinforcement (ρ > 0.5 ρmax) at interior columns.
Check for fire resistance: from Table 10-4, required cover for fire resistance rating of 4 hours or less
=3⁄4in. < provided cover. O.K.
bw = 20Mu
d2 = 20(522.8)
172 = 36.2 in. > 36 in.
(5) Determine flexural reinforcement for the beams at the 1st floor level (a) Top bars at exterior columns
Check governing load combination:
• gravity loads
Mu = 326.7 ft-kips ACI Eq. (9-2)
• gravity + wind load
Mu= 1.2(3.9)(28.58)2/16 + 0.8(99.56) = 317.9 ft-kips ACI Eq. (9-3) or
Mu= 1.2(3.9)(28.58)2/16 + 0.5(1.09)(28.58)2/16
+ 1.6(99.56) = 426.1 ft-kips ACI Eq. (9-4)
• also check for possible moment reversal due to wind moment:
Mu= 0.9(3.9)(28.58)2/16 ± 1.6(99.56) = 338.5 klf, 19.9 ft-kips ACI Eq. (9-6)
From Table 3-5: Use 8-No. 8 bars (As= 6.32 in.2)
minimum n = 36[1.5 + 0.5 + (1.0/2)]2/57.4 = 4 bars < 8 O.K.
Check ρ = As/bd = 6.32/(36 17) = 0.0103 > ρmin= 0.0033 O.K.
(b) Bottom bars in end spans:
As= 373.4/4(17) = 5.49 in.2 Use 8-No. 8 bars (As= 6.32 in.2) (c) Top bars at interior columns:
Check governing load combination:
• gravity load only
Mu = 522.8 ft-kips ACI Eq. (9-2)
• gravity + wind loads:
Mu= 1.2(3.9)(28.54)2/10 + 0.8(99.56) = 460.8 ft-kips ACI Eq. (9-3) or
Mu= 1.2(3.9)(28.54)2/10 + 0.5(1.09)(28.54)2/10
+ 1.6(99.56) = 584.9 ft-kips ACI Eq. (9-4)
As = Mu
4d = 426.1
4(17)= 6.27 in.2
Chapter 3 • Simplified Design for Beams and Slabs
As= 584.9/4(17) = 8.6 in.2 Use 11-No. 8 bars (As= 8.6 in.2) (d) Bottom bars in interior span:
As= 324.9/4(17) = 4.78 in.2 Use 7-No. 8 bars (As= 5.53 in.2)
(6) Reinforcement details shown in Fig. 3-22 are determined directly from Fig. 8-3(a).* Provided 2-No. 5 top bars within the center portion of all spans to account for any variations in required bar lengths due to wind effects.
Since the column line beams are part of the primary wind-force resisting system. ACI 12.11.2 requires at least one-fourth the positive moment reinforcement to be extended into the supporting columns and be anchored to develop full fy at face of support. For the end spans: As/4 = 8/4 = 2 bars. Extend 2-No. 8 center bars anchorage distance into the supports:
• At the exterior columns, provide a 90° standard end-hook (general use). From Table 8-5, for No. 8 bar:
˜n= 14 in. = 16 – 2 = 14 in. O.K.
• At the interior columns, provide a Class B tension splice (ACI 13.2.4). Clear space between No. 8 bars = 3.4 in. = 3.4db. From Table 8-2, length of splice = 1.0 30 = 30 in. (ACI 12.15).
(7) Design of Shear Reinforcement
Design shear reinforcement for the end span at the interior column and use the same stirrup requirements for all three spans.
Check governing load combination:
• gravity load only
Vu= at interior column = 105.5 kips (governs)
• gravity + wind loads:
Vu= 1.2(1.15)(3.9)(28.54)/2 + 0.8(6.64) = 82.1 kips or
Vu= 1.2(1.15)(3.9)(28.54)/2 + 0.5(1.15)(1.09)(28.54)/2 + 1.6(6.64) = 95.7 kips
• wind only at span center
Vu= 1.6(6.64) = 10.6 kips
Vu@ face of column = 105.5 kips
Vuat distance d from column face = 105.5 – 6.4(1.42) = 96.4 kips (φVc+ φVs)max= 0.48 bwd = 0.48(36)17 = 293.8 kips > 96.4 kips O.K.
φVc= 0.095 bwd = 0.095(36)17 = 58.1 kips φVc/2 = 29.1 kips
7'-10"
2"
clear 6"
10'-4" 10'-4"
30'-0" 30'-0"
2-#8 2-#8
2-#5 11-#8 2-#5
5-#8 5-#8 8-#8
Class B tension splice 18"
1'-3" 1'-3"
4'-3" 4'-3"
16"
0"
1.5" clear to stirrups
11-#8 at interior columns 8-#8 at eterior columns
36"
16"3.5"
12
#4 U-stirrups 8-#8 in end spans
7-#8 in interior spans Figure 3-22 Reinforcement Details for Support Beams along N-S Interior Column Line
Chapter 3 • Simplified Design for Beams and Slabs
Length over which stirrups are required: (105.5 – 29.1)/6.4 = 11.9 ft φVs(required) = 96.4 – 58.1 = 38.3 kips
Try No. 4 U-stirrups
From Fig. 3-4, use No.4 @ 8 in. over the entire length where stirrups are required (see Fig. 3-23).
#4 U-stirrups
Face of column
18 @ 8" = 12'-0"
2"
Figure 3-23 Stirrup Spacing Layout
References
3.1 Fling. R.S., “Using ACI 318 the Easy Way,” Concrete International, Vol. 1, No. 1, January 1979.
3.2 Pickett, C., Notes on ACI 318-77, Appendix A – Notes on Simplified Design, 3rd Edition, Portland Cement Association, Skokie, Illinois
3.3 Rogers, P., “Simplified Method of Stirrup Spacing,” Concrete International, Vol. 1, No. 1, January 1979.
3.4 Fanella D. A, “Time-Saving Design Aids for Reinforced Concrete – Part 1”, Structural Engineer, August 2001 3.5 Fanella D. A, “Time-Saving Design Aids for Reinforced Concrete - Part 2 Two-Way Slabs”, Structural
Engineer, October 2001
3.6 Fanella D. A, “Time-Saving Design Aids for Reinforced Concrete - Part 3 Columns & Walls”, Structural Engineer, November 2001
3.7 PSI – Product Services and Information, Concrete Reinforcing Steel Institute, Schaumburg, Illinois.
(a) Bulletin 7901A Selection of Stirrups in Flexural Members for Economy (b) Bulletin 7701A Reinforcing Bars Required – Minimum vs. Maximum (c) Bulletin 7702A Serviceability Requirements with Grade 60 Bars
3.8 Notes on ACI 318-08, Chapter 6, “General Principles of Strength Design,” EB708, Portland Cement Association, Skokie, Illinois, 2008.
3.9 Design Handbook in Accordance with the Strength Design Method of ACI 318-89: Vol. 1–Beams, Slabs, Brackets, Footings, and Pile Caps, SP-17(91), American Concrete Institute, Detroit, 1991.
3.10 CRSI Handbook, Concrete Reinforcing Steel Institute, Schaumburg, Illinois, 10th Edition, 2008.
3.11 ACI Detailing Manual – American Concrete Institute, Detroit, 1994.
Chapter 3 • Simplified Design for Beams and Slabs