Direct Design Method for Two-way Slab Analysis

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

1. Introduction:The direct design method consists of a set of rules for distributing moments to slab and beam sections in a two-way slab system. 2. Limitations on use of Direct Design method (ACI 13.6.1):(i) Minimum of 3 continuous spans in each direction (3 × 3 panel). (ii) Rectangular panels with long span/short span ≤ 2. (iii)Successive span in each direction shall not differ by more than 1/3 the longer span. (iv) Columns may be offset from the basic rectangular grid of the building by up to 0.1 times the span parallel to the offset (figure 01). . l 0.1l

Figure 01: Column offset at a distance of 0.1l from the basic rectangular grid. (v) All loads must be due to gravity only (N/A to un-braced laterally loaded frames, mats or pre-stressed slabs). (vi) Service (unfactored) live load ≤ 2 (service dead load). (vii)For panels with beams between supports on all sides, relative stiffness of the beams in the two perpendicular directions:

α1l22 α 2l12 Shall not be less than 0.2 nor greater than 5.0.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Where α is the ratio of flexural stiffness of beam section to flexural stiffness of width of slab bounded laterally by centerlines of adjacent panels (if any) on each side of the beam.

α=

flexural stiffness of beam flexural stiffness of slab

α=

4E cb I b / l

=

4E cb I b

4E cs I s / l

4E cs I s

E cb = Modulus of elasticity of beam concrete E sb = Modulus of elasticity of slab concrete I b = Moment of inertia of uncracked beam I s = Moment of inertia of uncracked slab

The width of slab is bounded laterally by centerline of adjacent panels on each side of the beam (figure 02).

l2 /2 hw < 4hf

l2 /2

l 2 /2 bw + 2hw < bw + 8hf

hf

hw

hf

hw bw

(a) Section for I b (Edge beam)

(c) Section for I b (Interior beam)

(b) Section for Is (Edge beam)

(d) Section for I s (Interior beam)

Figure 02: Ib and Is in case of interior and exterior beams. 3. Definitions related to Direct Design Method:(i) Frames: Slab is considered to be a series of frames in two directions. (ii) Panel (ACI 13.2.3): A panel is bounded by column, beam, or wall centerlines on all sides. A panel includes all flexural elements between column centerlines.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

l1

Direct Design Method

l1

l1

Pa ne l

Exterior Frame EW1 l

Interior Frame EW2

2

N

ln

Half middle strip

0.25l1 or 0.25l 2 (Whichever is less)

Column strip Middle strip

0.25l1 or 0.25l 2 (Whichever is less)

Column strip

l2

r rio EW3 e t In me Fra l2

r erio W4 t x E eE fram

(a)

l1

N ln

Exterior frame NS1

l1

Interior frame NS2

l2

Interior frame NS3 l2

Exterior frame NS4

l1

l2

(b) Figure 03: Slab system divided into EW and NS frames.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

(iii)Column Strip (ACI 13.2.1): Column strip is a design strip with a width on each side of a column centerline equal to 0.25l2 or 0.25l1, whichever is less. Column strip includes beams, if any. (iv) l1: l1 is the length of span in direction that moments are being determined, measured center-to-center of supports. (v) l2: l2 is the length of span transverse to l1, measured center-to-center of supports. (vi) Middle strips (ACI 13.2.2): Middle strip is a design strip bounded by two column strips. 4. Distribution of Moments:(i) Total static Moment, Mo (ACI 13.6.2): The total static moment for a span length ln and width l2 of a given frame is given by ACI equation 13-3 as:

M0 =

wu l2ln2

(ACI 13 - 3)

8

Where, wu = Factored load per unit area. ln = length of clear span in direction that moments are being determined, measured face-to-face of supports. l2 = As defined above. However, two exceptional cases as defined by ACI are given below: ACI 13.6.2.3 — where the transverse span of panels on either side of the centerline of supports varies, l2 in Eq. (13-3) shall be taken as the average of adjacent transverse spans. ACI 13.6.2.4 — when the span adjacent and parallel to an edge is being considered, the distance from edge to panel centerline shall be substituted for l2 in Eq. (13-3). ln is defined by ACI as given below: ACI 13.6.2.5 — clear span ln shall extend from face to face of columns, capitals, brackets, or walls. Value of ln used in Eq. (13-3) shall not be less than 0.65l1. Circular or regular polygon shaped supports shall be treated as square supports with the same area.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

ACI R13.6.2.5 â€” if a supporting member does not have a rectangular cross section or if the sides of the rectangle are not parallel to the spans, it is to be treated as a square support having the same area, as illustrated in Fig. R13.6.2.5.

Figure 04: Equivalent square section for supporting members, ACI fig 13.6.2.5. (ii) Longitudinal Distribution of Static Moment (ACI 13.6.3): For a typical interior panel, the total static moment is divided into positive moment 0.35Mo and negative moment of 0.65Mo. For an exterior panel, the total static moment is dependent on the type of restraint at the outside edge. ACI table 13.6.3.3 (table 13.3 Nilson 13th Ed) as shown in figure 05 of this document can be used for longitudinal distribution. Alternatively, figure 06 of this document can also be used.

Figure 05: Table for longitudinal distribution of static moment in exterior panel.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

First Interior

Exterior

Support

Interior

Support

0.75 1 No restraint

Direct Design Method

0.65

0.65

0 0.35 0.65

0.63 0.65

0.65

0.65

2 Full restraint 0.35

0.35

3 Slab with beam between supports

0.70

0.16

0.65

0.35 0.57

4 Edge beam only (no other beam)

0.65

0.30

0.70

0.65

0.65

0.35 0.50 0.70

0.26

0.65

0.65

5 No beams 0.35 0.52

Figure 06: Longitudinal moment distribution. Note that the longitudinal moment values mentioned are for the entire width of the equivalent building frame i.e., the width of two half column strips and two half-middle strips of adjacent panels.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

(iii)Transverse or Lateral distribution of Longitudinal Moments:Tables 13.6.4.1, 13.6.4.2 and 13.6.4.4 of the ACI as given in figure 08 are used to assign moments to column strip. The remaining moments are assigned to middle strip in accordance with ACI 13.6.3. Beams between supports shall be proportioned to resist 85 percent of column strip moments if α1l2/l1 {Where l2 shall be taken as full span length irrespective of frame location (exterior or interior)} is equal to or greater than 1.0 (ACI 13.6.5.1). As an alternative the ACI tables mentioned above for the assignment of moments to column strips, figure 09 (Graph A4, Nilson 13th Ed) can also be used. Moreover, figure 10 of this document illustrates the summary of lateral distribution of moments for slab system without beams. Transverse distribution of the longitudinal moments to middle and column strips is a function of the ratio of length (l2/l1), α1, and βt. Where, βt = the ratio of torsional stiffness of edge beam section to flexural stiffness of a width of slab equal to span length of beam, center-to-center of supports. If there is no edge beam, βt is taken equal to zero. If there is edge beam, βt is calculated as follows.

βt =

Ecb C 2Ecs Is

Where, C is the torsional constant of the edge beam. This is roughly equal to the polar moment of inertia of edge beam and is given as: C =∑[1–0.63{x/y}×x3y/3] = [1 – 0.63{x1/y1}×x13y1/3] + [1 – 0.63{x2/y2}×x23y2/3] Where, “x” is the shorter side of the rectangle and “y” is the longer one. Slab

y2

x2

y1 x1

Beam Figure 07: Cross-section of torsional member (edge beam) for calculation of βt.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Figure 08: Lateral distribution.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Figure 09: Graph A4, (Nilson 13th Ed) for transverse distribution of longitudinal moments.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Center line of panel 1 2

middle strip l2 /2

l2 /2 A l1/4 or l 2 /4 (whichever is smaller)

1 2

column strip B

1 2

column strip

C

l2 /2

l2 /2

1 2

middle strip

Exterior panel A

M p= 0.52Mo

Panel moment M ne = 0.26Mo A Column strip (-0.26Mo ) 100% A

0

B

M p= 0.35Mo M ni = 0.65Mo

B

B

C

C

(-0.49M o ) 75%

(+0.14M o ) 40%

(-0.175M o ) (-0.16M o ) 25% 25%

Interior panel

M ni = 0.65Mo

(+0.21M o ) 60%

(-0.525M o ) (-0.49M o ) 75% 75% (+0.208Mo ) 40%

Middle strip

Interior panel

M ni = 0.70Mo (+0.312Mo ) 60%

Center line of panel

C

(-0.16M o ) 25%

Figure 10: Summary of longitudinal & lateral distribution for slabs without beams.

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Page 10 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

5. Minimum Slab Thickness for two-way construction (ACI 9.5.3):The definitions of the terms are:h = Minimum slab thickness without interior beams. ln = length of clear span in direction that moments are being determined, measured face-to-face of supports. β = ratio of clear spans in long to short direction of two-way slabs. αm = average value of α for all beams on edges of a panel. a. For 0.2 ≤ αm ≤ 2:

fy ln 0.8 + 200,000 h= 36 + 5β (α m − 0.2 ) But not less than 5 in. fy in psi. b. For αm > 2:

fy ln 0.8 + 200,000 h= 36 + 9β

But not less than 3.5 in. fy in psi. c. For αm < 0.2, use the ACI table 9.5 (c), reproduced in figure 11 of this document.

Figure 11: Table for minimum thickness of slabs without interior beams.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Additionally, slab systems with αm < 0.2 shall also fulfill the following requirements: •

For slabs without drop panels meeting ACI 13.3.7.1 and 13.3.7.2, hmin = 5 in

•

For slabs with drop panels meeting ACI 13.3.7.1 and 13.3.7.2, hmin = 4 in

ACI 13.3.7.1 — Drop panel shall extend in each direction from centerline of support a distance not less than one-sixth the span length measured from center-to-center of supports in that direction. ACI 13.3.7.2 — Projection of drop panel below the slab shall be at least one-quarter the slab thickness beyond the drop. 6. Max Spacing and Min Reinforcement requirement of the ACI code: •

One way slab (ACI 7.6.5 & 7.12.2.2): ¾ Main Reinforcement = 3 hf, or 18 in whichever is less (hf = slab thickness) ¾ Temp reinforcement = 5 hf or 18 in whichever is less.

•

Two way slab (ACI 13.3.2): ¾ 2 hf in each direction.

•

Min Reinforcement in all cases (ACI 7.12.2.1): ¾ 0.0018 b hf for grade 60. ¾ 0.002 b hf for grade 40 and 50.

7. Detailing of flexural reinforcement for column supported two-way slabs: •

For protection of the steel against damage from fire or corrosion, at least 3/4 in. concrete cover must be maintained.

•

Because of the stacking that results when bars are placed in perpendicular layers, the inner steel will have an effective depth 1 bar diameter less than the outer steel.

•

In case of two way slabs supported on sufficiently stiff beams with α l2/l1 greater than 1.0, curvatures and moments in the short direction are greater than in the long direction of a rectangular panel, therefore short-direction bars are normally placed closer to the top or bottom surface of the slab, with the larger effective depth d, and long-direction bars are placed inside these, with the smaller d.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

•

Direct Design Method

In the case of flat plates/slabs, it is clear that the middle-strip positive moments (for example) are larger in the long direction than the short direction, exactly the opposite of the situation for the slab with stiff beams. In the column strips, positive and negative moments are larger in the long than in the short direction. On this basis, the designer is led to place the long-direction negative and positive bars, in both middle and column strips, closer to the top or bottom surface of the slab, respectively, with the larger effective depth.

•

If column-line beams are added and if their stiffness is progressively increased for comparative purposes, it will be found that the short-direction slab moments gradually become dominant, although the long-direction beams carry larger moments than the short-direction beams.

•

The best guide in specifying steel placement order in areas where stacking occurs is the relative magnitudes of design moments obtained from analysis for a particular case, with maximum d provided for the bars resisting the largest moment. No firm rules can be given. For square slab panels, many designers calculate the required steel area based on the average effective depth, thus obtaining the same bar size and spacing in each direction.

•

In case of DDM standard bar cut off points from figure 13 of this document are used as recommended in ACI code, figure 13.3.8.

•

ACI 13.3.8.5 requires that all bottom bars within the column strip in each direction be continuous or spliced with Class A splices (1.0 ld, For development length see ACI 12.2.3 or Nelson 13th Ed, page 172 chapter 5) or mechanical or welded splices. At least two of the column strip bars in each direction must pass within the column core and must be anchored at exterior supports (ACI 13.3.8.5).

8. Reinforcement at exterior corners: •

Reinforcement should be provided at exterior corners in both the bottom and top of the slab, for a distance in each direction from the corner equal to one-fifth the longer span of the corner panel as shown in figure 12, below.

•

The reinforcement at the top of the slab should be parallel to the diagonal from the corner, while that at the bottom should be perpendicular to the diagonal.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

â€˘

Direct Design Method

Alternatively, either layer of steel may be placed in two bands parallel to the sides of the slab. The positive and negative reinforcement, in any case, should be of a size and spacing equivalent to that required for the maximum positive moment in the panel, according to ACI 13.3.6.

Figure 12: Reinforcement at exterior corners of slab

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Figure 13: ACI fig 13.3.8.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Design Pb.1: Design the slab shown below (Follow the Direct Design Method for the slab analysis). Data Given: A 75′ × 60′ building, divided into nine (9) panels using beams supported at their ends on columns. Each panel is 20′ × 25′. fc′ = 4 ksi fy = 60 ksi Height of building = 10′ Column dimensions = 14″ × 14″ Live load = 144 psf 7"

l1 = 25'-0"

l1 = 25'-0"

5'

Half column strip

5'

Half middle strip

5'

Half middle strip

5'

Half column strip

5'

Half column strip

5'

Half middle strip

l2 = 10'-7"

l1 = 25'-0"

A

20'-0"

l2 = 20'

All beams 14" x 20"

20'-0"

All columns 14" square

20'-0"

Figure 14: Given slab beam system.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Solution: Check if the slab system satisfies all the limitations for Direct Design Method. 1) There must be a minimum of three continuous spans in each direction. 2) The panels must be rectangular, with the ratio of the longer to shorter spans within a panel not greater than 2. 3) The successive span lengths in each direction must not differ by more than one third of the longer span. 4) Loads must be due to gravity only and the service live load must not exceed 2 times the service dead load. 5) If beams are used on the column lines, the relative stiffness of the beam in the two perpendicular directions, given by the ratio αl2/l1, must be in between 0.2 and 5.0. 6) Columns may be offset a maximum of 10 percent of the span in the direction of the offset from either axis between centerlines of the successive columns. Step No 1: Sizes for beams, slab and column. •

Beams: Let assume all beam sections equal to 14″ × 20″.

•

Column: Let the column dimensions = 14″ × 14″.

•

Slab Thickness: To find the minimum slab thickness, ACI equations (section 13.8, Nelson 13th Ed) will be used which utilizes αm (average value of α for all beams on edges of a panel). For this purpose, the relevant calculation is given in table 01. Let assume slab thickness (hf) equal to 7″. Then,

•

Effective width for beam: We can now calculate the effective width (beff) for interior and edge beams according to ACI R13.2.4: Effective flange projection = minimum of 4hf and hw 4hf = 4 × 7 =28″ hw = h – hf = 20 – 7 =13″ Therefore, effective flange projection = 13″ beff = bw + 2(Effective flange projection) = 14 + 2 × 13 = 40″ And, for edge beams: beff = bw + (Effective flange projection) = 14 + 13 = 27″

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

hw < 4hf = 13" hf = 7"

Direct Design Method

b w+ 2h w< b w+ 8h f = 40" hf = 7"

hw = 13"

h w= 13" bw = 14"

bw = 14"

Figure 15: Interior and edge beams sections. Table 01: Moment of inertia of beams, slab and corresponding values of α. 1 2 3 4 5 6 7 8 9 Moment of Inertia of Moment of Inertia of α Slab slabs beams Panel Panel height Length Width Value (hf) Value Value (ft) (b) (ft) Notation Notation Notation (in4) 4 4 (in) (in ) (in )

Col7/Col5

25 20 25 20

20 25 10.6 13.1

7 7 7 7

IIS25=bhf3/12 IIS20=bhf3/12 IES25=bhf3/12 IES20=bhf3/12

6860 8575 3636 4493

IBINT = 2bwh3/12

18667

IBEXT = 1.5bwh3/12

14000

α INT25 α INT20 α EXT25 α EXT20

2.7 2.2 3.9 3.1

Note: IIS25 = Moment of inertia (MOI) of 25′ long interior slab. IES25 = Moment of inertia (MOI) of 25′ long exterior slab. IIS20 = Moment of inertia (MOI) of 20′ long interior slab. IES20 = Moment of inertia (MOI) of 20′ long exterior slab. IBINT = Moment of inertia of interior beam. IBEXT = Moment of inertia of exterior beam. α INT25 = Ratio of MOI of interior beam to MOI of 25′ long interior slab. α EXT25 = Ratio of MOI of exterior beam to MOI of 25′ long exterior slab. α INT20 = Ratio of MOI of interior beam to MOI of 20′ long interior slab. α EXT20 = Ratio of MOI of exterior beam to MOI of 20′ long exterior slab αm = (α INT25 + 2 × α INT20+ α EXT25)/4

{for panel A as shown in fig. 14}

= (2.7 + 2.2 + 2.2 + 3.9)/4 = 2.75 β = larger clear span / smaller clear span = [{25 – (2 × (14/2)/12)}]/ [{20 – (2 × (14/2)/12)}] = 23.8 / 18.8 = 1.27 Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

•

Direct Design Method

According to given conditions, Since αm > 2, following equation applies: h = ln {0.8+ (fy/200000)}/ (36 + 9 β) [ACI 9.5.3] h = (23.8 × 12) × {0.8+ (60000/200000)}/ (36 + 9 × 1.27) = 6.63″ < 7″ Therefore hf = 7″ is O.K. If not then revise assumed thickness.

Step No 2: Load on slab. Service Dead Load (D.L) = γslabhf = 0.15 × (7/12) = 0.0875 ksf Service Live Load (L.L) = 144 psf or 0.144 ksf Factored Load (wu) = 1.2D.L + 1.6L.L = 1.2 × 0.0875 + 1.6 × 0.144 = 0.336 ksf Step No 3: Analysis. Though four frames are required to be analyzed for this specific slab system, only two of the frames will be analyzed and designed for demonstration purpose. The details are given in appendix A. I. Analysis of E-W Interior Frame: Step (A): Frame Data. •

Design Span of frame (c/c) = l1 = 25′

•

Design Length of frame = ln = 25 – (2 × 14/2)/12 = 23.8′

•

Width of frame = l2 = 20′

•

Column strip width = (Shorter span)/ 4 = 20/4 = 5′

Step (B): Total static moment. Mo = wul2ln2/8 (for Mo, l2 is the width of frame) = 0.336 × 20 × 23.82/8 = 476 ft-k Step (C): Longitudinal distribution of Total static moment (Mo). Table 02: Longitudinal Distribution of Total Static Moment. Exterior span Interior span 25' 476

Static Moment Mo (ft-k) Section Distribution Factor, ACI 13.6.3 (D.F) Longitudinal Moment (L.M) = Mo x D.F

25' 476

Exterior Interior Positive Negative Positive Negative Negative Negative 0.16

0.57

0.7

0.65

0.35

0.65

77

272

334

310

167

310

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

272 ft-k

Direct Design Method

272 ft-k

167 ft-k

77 ft-k 77 ft-k 334 ft-k

310ft-k

310 ft-k

334 ft-k

Figure 16: Longitudinal Distribution of Total Static Moment (Mo) Step (D): Lateral Distribution of Longitudinal moment (L.M). βt calculation is as follows: C = ∑ [(1- (0.63X/Y)X3Y/3] C = {(1- (0.63 × 14/20)) x 143 × 20/3 + (1- (0.63 × 7/13)) × 73 × 13/3}= 11210 βt = C/ (2IIS25) = 11210/ (2 × 6860) = 0.81 Minimum of hw or 4h f =13"

hf =7"

20"

14"

C = 11210 Figure 17: βt calculation. Other terms required are: α INT25 =2.7 l2/l1 = 20/25 = 0.8{l2 shall be taken as full span length irrespective of frame location (exterior or interior)} α INT25l2/l1 = 2.2 The values of column strip and middle strip moments obtained from lateral distribution of longitudinal moments are given in table 03.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Span Interior

Exterior

Section

Direct Design Method

Table 03: Lateral Distribution of Longitudinal moment. Column Strip Column Strip Moment Beam Moment Slab Moment (C.S.M) ft-k (M.D.S) ft-k (C.S.S.M) ft-k

Middle Strip Moments (M.S.M) ft-k

Negative

0.81 x 310 = 252

0.85 x 252 = 214

0.15 x 252 = 38

0.19 x 310 = 58

Positive Interior negative Positive Exterior negative

0.81 x 167 =135

0.85 x 135 = 115

0.15 x 135 = 21

0.19 x 167 = 32

0.81 x 334 =271

0.85 x 271 = 230

0.15 x 271 = 41

0.19 x 334 = 63

0.81 x 272 =220

0.85 x 220= 188

0.15 x 220 = 33

0.19 x 272 = 52

0.93 x 77 =72

0.85 x 72 =61

0.15 x 72 = 11

0.07 x 77 = 5

Note: Coefficients for lateral distribution have been taken from graph A.4 (pg 755) Nelson, using αl2/l1, βt, and l2/l1.

Step (E): Moment transferred to beam B1. Self weight of beam = γbeambwhw = 0.15 × (14/12) × (13/12) = 0.20 k/ft Factored load (wu) = 1.2 × 0.20 = 0.24 k/ft Moment due to self weight of beam B1 (M) = wuln2/8 = 0.24 × 23.82/8 = 17 ft-k Table 04: Moment transferred to beam B1 Exterior span

Interior span

25' 17

25' 17

Static Moment M of beam (ft-k) Section Distribution Factor (D.F) Moment due to self weight (MDSB) = M x D.F (ft-k) Beam moment from slab (MDS) (ft-k) Total Moment (ft-k)

Exterior Interior Positive Negative Positive Negative Negative Negative 0.16 0.57 0.7 0.65 0.35 0.65 3

10

12

12

6

12

61

188

230

214

115

214

64

198

242

226

121

226

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

B4 B2

B1

B3

Direct Design Method

B3

B4

C1

C2

C2

C1

C3

C4

C4

C3

C3

C4

C4

C3

C1

C2

C2

C1

B1

B2

B4

B3

B3

B2

B1

B1

B2

B4

Figure 18: Beam and Column plan. Step (F): Moment transferred to columns (ACI 13.6 9). Exterior column (C3) moment = MDSB + LM…………………………..…… (A) Where, MDSB = moment due to self weight of beam as given in table 04. LM = Longitudinal moment as given in table 02. Exterior column (C3) moment = 3 + 77 = 80 ft-k Interior column (C4) moment = MDSB(unbalanced)+(0.65/8){0.5wuLLl2ln2}…(B) = (12–12)+(0.65/8) × {0.5 × (1.6 × 0.144) × 20 × 23.82} = 106.3 ft-k Step (G): Shear in beam B1 (ACI 13.6.8). Tributary area (A) = 2 × (10 × 20/2) +5 × 20 =300 ft2 wub = wuslabA/l1 + Self weight of beam = 0.336 × 300/ 25 + 0.24 = 4.27 k/ft Vext = 46.17 k Vcr, ext = 40.1 k Vint = 60.59 k Vcr, int = 54.52 k

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Page 22 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

64 k-ft

242 k-ft

4.27 k/ft

Direct Design Method

226 k-ft

25' RDL

53.3 k

53.3 k

RDM

7.12 k

7.12 k

Total

46.17 k

113.89 k

53.3 k

Vcr, ext =40.1 k Vext = 46.17 k

14.19' 1.42'

Vint = 60.59 k

10.81' Vcr, int =54.52 k

45° 20'

B1

B1

10'

5'

10'

Figure 19: Load on Beam (B1). II. Analysis of E-W Exterior Frame: Step (A): Frame Data. •

Design Span of frame (c/c) = l1 = 25′

•

Design Length of frame = ln = 25 – (2 × 14/2)/12 = 23.8′

•

Width of frame = l2 = (20/2) + 14/(2 × 12) = 10.6′

•

Column strip width = (shorter span)/ 4 = 20/4 = 5′

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 23 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Step (B): Total static moment. Mo = wul2ln2/8 (for Mo, l2 is the width of frame) = 0.336 × 10.6 × 23.82/8 = 252 ft-k Step (C): Longitudinal distribution of Total static moment (Mo). Table 05: Longitudinal Distribution of Total Static Moment. Exterior span

Interior span

25'

25'

252

252

Static Moment Mo (ft-k)

Exterior Interior Positive Negative Positive Negative Negative Negative

Section Distribution Factor(D.F)

0.16

0.57

0.7

0.65

0.35

0.65

Longitudinal Moment (L.M) = Mo x D.F

40

144

176

164

88

164

144 ft-k

144 ft-k

88 ft-k

40 ft-k 40 ft-k 176 ft-k

164 ft-k

164ft-k

176 ft-k

Figure 20: Longitudinal Distribution of Total Static Moment (Mo) Step (D): Lateral Distribution of Longitudinal moment (L.M). α EXT25 = 3.9 l2/l1 = 20/25 = 0.8 {l2 shall be taken as full span length irrespective of frame location (exterior or interior)} α EXT25l2/l1 = 3.12

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Page 24 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Table 06: Lateral Distribution of Longitudinal moment. Span

Interior

Exterior

Section

Column Strip Moment (C.S.M) ft-k

Beam Moment (M.D.S) ft-k

Column Strip Slab Moment (C.S.S.M) ft-k

Middle Strip Moments (M.S.M) ft-k

Negative

0.81 x 164 = 133

0.85 x 133= 113

0.15 x 133 = 20

0.19 x 164 = 31

Positive Interior negative Positive Exterior negative

0.81 x 88 =71

0.85 x 71 = 61

0.15 x 71 = 11

0.19 x 88 = 17

0.81 x 176=143

0.85 x 144= 121

0.15 x 144 = 22

0.19 x 176 = 33

0.81 x 144 =117

0.85 x 117= 99

0.15 x 117 = 18

0.19 x 144= 27

0.88 x 40 =35

0.85 x 35 =30

0.15 x 35 = 5

0.12 x 40 = 5

Note: Coefficients for lateral distribution have been taken from graph A.4 (pg 755) Nelson, using αl2/l1, βt, and l2/l1.

Note: ACI 13.6.5.1 states that “Beams between supports shall be proportioned to resist 85 percent of column strip moments if α1l2/l1 is equal to or greater than 1.0”. Where, βt calculation for 10.06′ width of slab strip is given below: C = ∑ [{1- (0.63X/Y)}X3Y/3] C = {(1- (0.63× 14/20)) × 143 × 20/3 + (1- (0.63 × 7/13)) × 73 × 13/3}= 11210 βt = C/ (2IES25) = 11210/ (2 × 3636) = 1.54 Minimum of hw or 4h f =13"

hf =7"

20"

14"

C = 11210 Figure 21: βt calculation. Step (E): Moment transferred to beam. Self weight of beam = γbeambwhw = 0.15 × (14/12) × (13/12) = 0.20 k/ft Factored load (wu) = 1.2 × 0.20 = 0.24 k/ft Moment due to self weight of beam (M) = wuln2/8 = 0.24 × 23.82/8 = 17 ft-k Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 25 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Table 07: Moment transferred to beam (B2) Exterior span

Interior span

25' 17

25' 17

Static Moment M of beam (ft-k) Section Distribution Factor (D.F) Moment due to self weight (MDSB) = M x D.F Beam moment from slab (MDS) Total T.M

Exterior Interior Positive Negative Positive Negative Negative Negative 0.16 0.57 0.7 0.65 0.35 0.65 3

10

12

12

6

12

30

99

121

113

61

113

33

109

132

125

67

125

Step (F): Moment transferred to columns Exterior column (C1) moment = MDSB + L.M ……………………………… (A) =3 + 40 = 43 ft-k Interior column (C2) moment = MDSB(unbalanced)+(0.65/8){0.5wuLLl2ln2}…(B) = (12 – 12)+ (0.65/8) × {0.5 × (1.6 × 0.144) ×10.6 × 23.82} = 56.29 ft-k Step (G): Shear in beam (B2). Tributary area (A) = 2 × (10 × 10/2) + 5 × 10 =150 ft2 wub = wuslabA/l1 + Self weight of beam = 0.336 × 150/25 + 0.24 = 2.256 k/ft Vext = 24.24 k Vcr, ext = 21 k Vint = 32.16 k Vcr, int = 28.9 k

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 26 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar 33 k-ft

2.256 k/ft

132 k-ft

Direct Design Method 125 k-ft

25' RDL

28.2 k

28.2 k

RDM

3.96 k

3.96 k

Total

24.24 k

60.36 k

28.2 k

Vcr, ext =21.0 k Vext =24.24 k

14.26' 1.42'

Vint =32.16 k

10.74' Vcr, int =28.9k

B2

B2 10'

10'

5'

10'

Figure 22: Load on Beam (B2).

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 27 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Step No 4: Design. (1) Design of slab strips. A. E-W Interior slab strip: dl = 5.5"

hf = 7"

Direction along 25' length.

Figure 23: Section of slab at E-W interior strip. ds = hf – 1 = 7 – 1 = 6″ dl = ds – bar dia = 6 – (4/8) = 5.5″ (for # 4 bar) Asmin = 0.0018bhf (for fy = 60 ksi) = 0.0018 × 12 × 7 = 0.151 in2 (ρmin = 0.0023, in terms of actual effective depth) Now, Equation used to calculate (ρ) in (table 1.4) is as follows: Mu = Φfyρbdl2{1-0.59ρfy/fc′) = 0.9 × 60 × ρ × 12 × 5.52 × {1-0.59 x ρ x 60/4} After solving the above equation for ρ, we get: ρ = [19602 ± √{(196022) - (4 x 173477.7 x Mu′ x 12)}]/2(173477.7)…….(A) Table 08: Slab Design. Frame Span

Section

Column Strip Moment Interior Middle Strip Moment

negative

106

5.5

38

4.30

0.0027

1.57

#4

13.48

9

Positive

106

5.5

21

2.38

0.0023

1.34

#4

15.81

12

negative

120

5.5

58

5.90

0.0037

2.47

#4

9.73

9

Positive

120

5.5

32

3.20

0.0023

1.52

#4

15.81

12

106

5.5

41

4.64

0.0029

1.70

#4

12.47

9

106

5.5

33

3.73

0.0023

1.36

#4

15.60

12

106

5.5

11

1.24

0.0023

1.34

#4

15.81

12

120

5.5

63

6.30

0.0040

2.64

#4

9.09

9

120

5.5

52

5.20

0.0033

2.16

#4

11.09

9

120

5.5

5

0.50

0.0023

1.52

#4

15.81

12

Column Strip Moment Exterior Middle Strip Moment

Interior negative Positive Exterior negative Interior negative Positive Exterior negative

b (in) d (in)

Mu (ft-k) Mu′ = ρ (eqn As = ρ x b x d Bar Spacing = (Table Recommended Mu x 12/b (in2) A) taken b x Ab/As 1.3)

Location

Note: “b” is the width of strip.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

B. E-W Exterior slab strip: dl = 5.5"

hf = 7"

Direction along 25' length.

Figure 24: Section of slab at E-W exterior strip. ds = hf – 1 = 7 – 1 =6″ dl = ds – bar dia = 6 – (4/8) = 5.5″ (for # 4 bar) Asmin = 0.0018 × 12 × 7 = 0.151 in2 (ρmin = 0.0023, in terms of actual effective depth) Now, Equation used to calculate (ρ) values is as follows: Mu = Φfyρbdl2{1-0.59ρfy/fc′} = 0.9 × 60 × ρ × 12 × 5.52 × {1-0.59 × ρ × 60/4} After solving the above equation for ρ ρ = [19602 ± √ {(196022) – (4 × 173477.7 × Mu′ × 12)}]/2(173477.7) Table 09: Slab Design. Frame Span

Location

Section

b (in)

d (in)

Mu (ft-k) (Table 1.7)

Mu′ = Mu x 12/b

ρ (eqn A)

As =ρ x b x d (in2)

Bar taken

Spacing = b x Ab/As

Recommended

negative

53

5.5

20

4.52

0.0028

0.83

#4

12.81

9

Interior

Column Strip Moment Middle Strip Moment

Positive

53

5.5

11

2.48

0.0023

0.67

#4

15.81

12

negative

60

5.5

31

6.20

0.0039

1.30

#4

9.25

9

Positive

60

5.5

17

3.40

0.0023

0.76

#4

15.81

12

53

5.5

22

4.98

0.0031

0.91

#4

11.60

9

53

5.5

18

4.07

0.0025

0.74

#4

14.27

12

Exterior negative

53

5.5

5

1.13

0.0023

0.67

#4

15.81

12

Interior negative

60

5.5

33

6.60

0.0042

1.38

#4

8.67

9

Positive

60

5.5

27

5.40

0.0034

1.12

#4

10.67

9

Exterior negative

60

5.5

5

1.00

0.0023

0.76

#4

15.81

12

Column Strip Moment Exterior Middle Strip Moment

Interior negative Positive

Note: “b” is the width of strip.

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Page 29 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Step No 5: Detailing. All the frames may be analyzed and designed by the same procedure as given in steps of analysis and design. However, the complete detail of reinforcement placement in slab is given below.

# 4 @ 9" c/c (T)

# 4 @ 12" c/c (T)

# 4 @ 9" c/c (T)

# 4 @ 12" c/c (B)

# 4 @ 12" c/c (T) # 4 @ 9" c/c (B)

B

# 4 @ 9" c/c (T)

# 4 @ 12" c/c (B)

# 4 @ 12" c/c (B)

# 4 @ 12" c/c (B)

# 4 @ 12" c/c (B)

# 4 @ 9" c/c (T)

B

# 4 @ 9" c/c (B)

# 4 @ 9" c/c (T)

20'

# 4 @ 9" c/c (T)

# 4 @ 9" c/c (B)

# 4 @ 12" c/c (T)

# 4 @ 9" c/c (T) # 4 @ 9" c/c (T)

B1 # 4 @ 12" c/c (T)

# 4 @ 9" c/c (B)

# 4 @ 12" c/c (B)

# 4 @ 12" c/c (B)

# 4 @ 12" c/c (B)

# 4 @ 9" c/c (B)

# 4 @ 9" c/c (T)

# 4 @ 12" c/c (B)

# 4 @ 9" c/c (T)

# 4 @ 9" c/c (T)

# 4 @ 9" c/c (T)

20'

# 4 @ 9" c/c (T)

# 4 @ 9" c/c (T)

# 4 @ 12" c/c (T)

# 4 @ 12" c/c (B)

B1

# 4 @ 12" c/c (B)

# 4 @ 9" c/c (T)

B2

A

PLAN Figure 25: Reinforcement details (plan view)

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

B4

25'

# 4 @ 9" c/c (T)

# 4 @ 9" c/c (B)

# 4 @ 9" c/c (T)

20'

25'

# 4 @ 9" c/c (T)

# 4 @ 12" c/c (T)

# 4 @ 12" c/c (B)

25'

B2

A

B3

B3

B4

Page 30 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

#4 @ 9" c/c

#4 @ 12" c/c

5'-0"

Direct Design Method

#4 @ 9" c/c 6'-6"

#4 @ 12" c/c 6'-6" #4 @ 9" c/c

#4 @ 9" c/c

#4 @ 12" c/c

#4 @ 12" c/c

18.83'

18.83'

SECTION A-A

#4 @ 12" c/c 6'-0"

#4 @ 12" c/c

#4 @ 9" c/c

#4 @ 9" c/c

#4 @ 9" c/c

8'-0"

8'-0"

#4 @ 12" c/c

#4 @ 9" c/c

#4 @ 12" c/c

23.83'

23.83'

SECTION B-B Figure 26: Sectional views of slab.

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

9. Shear Strength of Slab without beams:There are two types of shear that needs to be addressed. (a) One-way shear or beam shear at distance “d” from the column, (b) Two-way or punch out shear which occurs along a truncated cone.

Figure 27: Beam shear and punching shear. Shear design in flat plates & flat slabs. Punching shear: For punching shear ΦVn = ΦVc + ΦVs ΦVc is least of (a), (b), and (c). (a) ΦVc = Φ4√ (fc′)bod (b) ΦVc = (2 + 4/βc) √ (fc′)bod (c) {(αsd/bo +2} √ (fc′)bod βc = longer side of column/shorter side of column αs = 40 for interior column = 30 for edge column = 20 for corner columns

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

d/2

Column

c2

c1 d/2

d/2

d = Depth of member bo = 2{(c1 + d) + (c2 + d)} bo = 2(c1+ c2+ 2d) bo = 2c1 + 2c2+ 4d)

Figure 28: Critical perimeter for punching shear. When ΦVc ≥ Vu (Φ = 0.75) O.K, Nothing required. When ΦVc < Vu, then either increase ΦVc by: (a)

Increasing d ,depth of slab ( Drop Panel)

(b)

Increasing bo, critical shear perimeter (Column capital)

(c)

fc′ (high Strength Concrete)

Or provide shear reinforcement in the form of: (a)

Shear heads

(b)

Bent Bars

(c)

Integral beams

(d)

Shear studs.

Drop Panels and Column Capitals: (1) Drop panel: A drop panel with dimensions conforming to ACI 9.5.3.2 and 13.3.7.1 can be used for: (i) Increasing the area of critical shear perimeter. (ii) Increasing the depth of slab, reducing the amount of negative reinforcement. (iii) Stiffening slab and reducing deflections.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

la (length of panel)

Direct Design Method

l b (length of panel)

h/4

l a /6

l b /6

h

Drop panel

Figure 29: Drop Panel. ACI 13.3.7.1 states that drop panel shall extend in each direction from and of the support a distance not less than 1/6 of the span length measured from centre of supports in that direction. ACI 13.3.7.2 states that Projection of drop panel below the slab shall be at least one-quarter the slab thickness beyond the drop. (2) Column Capitals: Occasionally, the top of the columns will be flared outward, as shown in figure 34. This is done to provide a larger shear perimeter at the column and to reduce the clear span, ln, used in computing moments. ACI 6.4.6 requires that the capital concrete be placed at the same time as the slab concrete. As a result, the floor forming becomes considerably more complicated and expensive. For this reason, other alternatives, such as drop panels or shear reinforcement in the slab, should be considered before capitals are selected. If capital must be used, it is desirable to use the same size throughout the project. The diameter or effective dimension of the capital, “dc” or “c”, is defined in ACI 13.1.2 as that part of the capital lying within the largest circular cone or pyramid with a 90o vertex that can be induced within the outlines of the supporting column. The diameter is measured at the bottom of the slab or drop panel, as illustrated in figure 30.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

"dc" or "c"

Direct Design Method

Slab

45째

Drop Panel Capital

Figure 30: Column Capital.

Figure 31: Circular column with column capital and drop panel.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Shear Reinforcements: (1)

Shear Heads: The shear heads shown in Fig. 32a consist of standard structural steel shapes embedded in the slab and projecting beyond the column. They serve to increase the effective perimeter bo of the critical section for shear. In addition, they may contribute to the negative bending resistance of the slab. It consists of short lengths of I or wide-flange beams, cut and welded at the crossing point so that the arms are continuous through the column. Normal negative slab reinforcement passes over the top of the structural steel, while bottom bars are stopped short of the shear head. Column bars pass vertically at the corners of the column.

(2)

Bent-bar: The bent-bar arrangement in Fig. 32b is suited for use with concrete columns. The bars are usually bent at 45째 across the potential diagonal tension crack, and extend along the bottom of the slab a distance sufficient to develop their strength by bond.

(3)

Integral Beams: Another type of shear reinforcement is illustrated in Fig. 32c, where vertical stirrups have been used in conjunction with supplementary horizontal bars radiating outward in two perpendicular directions from the support to form what are termed integral beams contained entirely within the slab thickness. These beams act in the same general way as the shear heads shown in Fig. 32a. Adequate anchorage of the stirrups is difficult in slabs thinner than about 10 in. ACI 11.12.3 requires the slab effective depth d to be at least 6 in., but not less than 16 times the diameter of the shear reinforcement. In all cases, closed hoop stirrups should be used with a large diameter horizontal bar at each bend point, and the stirrups must be terminated with a standard hook (Ref. 13.18).

(4)

Shear Stud: A more recent development is the shear stud reinforcement shown in Fig. 32d. This consists of large-head studs welded to steel strips.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

(c)

Direct Design Method

(d) Figure 32: Shear reinforcement for flat plates.

The strips are supported on wire chairs during construction to maintain the required concrete cover to the bottom of the slab below the strip and the usual cover is maintained over the top of the head. Because of the positive anchorage provided by the stud head and the steel strip, these devices are Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 37 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

more effective, according to tests, than either the bent bar or integral beam reinforcement. In addition, they can be placed more easily, with less interference with other reinforcement, than other types of shear steel. Design for punching shear: Design Problem 02 (pg 455, Nelson 13th Ed): A flat plate has thickness h = 7 ½ in and is supported by 18″ square columns spaced 20 ft on centers each way. The floor will carry a total factored load of 300 psf. Check the adequacy of the slab in resisting punching shear at a typical interior column, and provide shear reinforcement, if needed. d = 6 in, fc′ = 4000 psi, fy = 60000 psi. Solution: (a) Vu = 300{(20)2 – (2)2} = 118800 lb = 118.8 k d/2 = 3" d/2 = 3" 18"

24"

18" 24"

24"

20'

24"

20'

Figure 33: Critical perimeter calculation. (b) Shear capacity of concrete in punching shear: ΦVc = 0.75 × 4 √ (fc′) bod bo = 2(c1 + d) +2(c2 + d) bo = 4(18 + 6) = 96″

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

ΦVc = 0.75 × 4 √ (4000) × 96 × 6/1000 = 109 k ΦVc < Vu Options available: (i) Capital: Determine minimum (bo). Equating the applied critical shear to shear capacity. Vu = ΦVc 118.8 = 0.75 × 4 √ (fc′) × bo × 6 bo = 104.5″ Now bo = 4 (c + d) = 4(c + 6) =104.5 c = 20.13 ≈ 21″

1.5"

18"

1.5"

1.5"

C

y

θ

Figure 34: Column capital. According to ACI code, θ < 45o Let θ = 30o, then y = 2.6″ For θ = 20o, then y = 4″ (ii) Drop panel: To determine the minimum “d” required. Vu = ΦVc 118.8 = 0.75 × 4 × √ (4000) × 96 × d/1000 d = 6.5″ and h = 6.5 + 1.5 = 8″ Thickness of drop panel = 2″ (say) {h/4, ACI recommendation} h = 2 + 7.5 = 9.5″

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

6'-8"

7.5"

2″ 20 /6=3'-4"

20 /6=3'-4"

Figure 35: Drop panel. (iii) Increasing column size or slab thickness (overall) would not be economical. (iv) Shear reinforcement: (a) Bent bar reinforcement: When bent bars are to be used, ACI 11.12.3 reduces ΦVc by 2. ΦVc = Φ × 4 × √ (fc′) × bo × d Therefore, ΦVc = 109/2 = 54.5 kip Reinforcement required, Av = (Vu – ΦVc)/ Φfysinα

Vs

Vs sinα

α Figure 36: Av = (118.8 – 54.5)/ (0.75 × 60 × sin45o) = 2.03 in2 Using total 4 bars (two in each direction), providing 8 legs crossing the critical section, the area per bar = 2.03/8 = 0.25, using # 5 bar.

1

2 Figure 37: Bent bar.

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Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

According to ACI, Vs = Vu – ΦVc = Avfysinα is not to exceed 3√ (fc′)bod. 3√ (fc′)bod = 3 × √ (4000) × 96 × 6/1000 = 109 kip Vs = 2.4 × 60 × sin45 = 100 O.K. No 5 bar

5"

15" Development Length

Figure 38: Bent bar reinforcement. (b) Stirrup reinforcement: Using 3/8″ Φ, 2 legged (0.22 in2), 4 (side) = 4 × 0.22 = 0.88 in2 Spacing (s) = ΦAvfyd/ (Vu – ΦVc) s = 0.75 × 0.88 × 60 × 6/ (118.8 – 54.5) = 3.68 ≈ 3.5″ Maximum spacing allowed d/2 = 6/2 = 3″ controls. Four #5 bars are to be provided in each direction to hold the stirrups. The beams such formed should be extended to a length at which bo can provide shear capacity. We know minimum bo = 104.5″ bo = 4R + 4c1 R = √ (x2 + x2) x = (3/4)(lv – c1/2) R = √ (2) x bo = 4√ (2) x + 4c1 bo = 4√ (2){(3/4)(lv – c1/2)} + 4c1 Or bo = 4.24lv – 2.12c1 + 4c1= 4.24lv + 1.88c1 Therefore lv = 16.67 ≈ 17″

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 41 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

x = 43 (l v - c1 /2) 45째 x R

Figure 39:

Figure 40: Stirrups Reinforcement (ACI R11.12.3).

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 42 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Design Pb. 03: Design the slab for the building shown below. (Follow the Direct Design Method). Data Given: A 60′ × 45′ building, divided into nine (9) panels, supported at their ends on columns. Each panel is 20′ × 15′. fc′ = 4 ksi fy = 60 ksi Height of building = 10′ Column dimensions = 14″ × 14″ Live load = 144 psf 7"

l1 = 20'-0"

l1 = 20'-0"

3.75'

Half column strip

3.75'

Half middle strip

3.75'

Half middle strip

3.75'

Half column strip

3.75'

Half column strip

3.75'

Half middle strip

l1 = 20'-0"

l2 = 8.08' 15'-0"

l2 = 15'

15'-0"

All columns 14" square

15'-0"

Figure 41: Given flat plate.

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 43 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Solution: The given slab system satisfies all the necessary limitations for Direct Design Method to be applicable. Step No 1: Sizes for slab and columns. Slab: To find minimum slab thickness (hf), ACI 9.5.3.2 {ACI Table 9.5 (c) or table 13.50, Nelson 13th Ed} will be used.

Table 10: Minimum thickness of slabs without interior beams. Without drop panels With drop panels Yield strength, fy (psi) 40000 60000 75000

Exterior panels Without With edge edge beams beams ln/33 ln/36 ln/30 ln/33 ln/28 ln/31

Interior panels ln/36 ln/33 ln/31

Exterior panels Without With edge edge beams beams ln/36 ln/40 ln/33 ln/36 ln/31 ln/34

Interior panels ln/40 ln/36 ln/34

For our case (Slab without drop panels, interior and edge beams) hf = ln/30 (Exterior panel) hf = ln/33 (Interior panels) Exterior panel governs. Therefore, hf = ln/30 = [{20 – (2 × 14/2)/12}/30] × 12 = 7.53″ (Minimum requirement) Take hf = 8″; as there are no beams; α = 0 Columns: Let the column dimensions = 14″ × 14″. Step No 2: Load on slab. Service Dead Load (D.L) = γslabhf = 0.15 × (8/12) = 0.1 ksf Service Live Load (L.L) = 144 psf or 0.144 ksf Factored Load (wu) = 1.2D.L + 1.6L.L = 1.2 × 0.1 + 1.6 × 0.144 = 0.3504 ksf

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 44 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Step No 3: Analysis. Though four frames are required to be analyzed for this specific slab system, only two of the frames will be analyzed and designed for demonstration purpose. The details are given in appendix A. I. Analysis of E-W Interior Frame. Step (A): Frame Data. •

Design Span of frame (c/c) = l1 = 20′

•

Design Length of frame = ln = 20 – (2 × 14/2)/12 = 18.83′

•

Width of frame = l2 = 15′

•

Column strip width = (Shorter span)/ 4 = 15/4 = 3.75′ l1 = 20'-0"

l1 = 20'-0"

l1 = 20'-0"

15'-0" 3.75'

Half middle strip

3.75'

Half column strip

3.75'

Half column strip

3.75'

Half middle strip

l2 = 15'

15'-0"

15'-0"

Figure 42: East West Interior Frame. Step (B): Total static moment. Mo = wul2ln2/8 (for Mo, l2 is the width of the frame) = 0.3504 × 15 × 18.832/8 = 233 ft-k

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 45 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Step (C): Longitudinal distribution of Total static moment (Mo). Table 11. Longitudinal distribution of Total static moment (Mo) Exterior span

Interior span

20'

20'

233

233

Static Moment Mo (ft-k) Distribution Factor (D.F) ACI 13.6.3 (D.F) Longitudinal Moment (L.M) = Mo x D.F

0.26

0.52

0.7

0.65

0.35

0.65

61

121

163

151

82

151

Step (D): Lateral Distribution of Longitudinal moment (L.M). α= 0 l2/l1 = 15/20 = 0.75 {l2 shall be taken as full span length irrespective of frame location (exterior or interior)} αl2/l1 = 0 βt = 0 Table 12: Lateral Distribution of Longitudinal moment (L.M). Column Strip Middle Strip Span Section Moment Moments (C.S.M) ft-k (M.S.M) ft-k Negative 0.75 x 151=113 0.25 x 151=38 Interior Positive 0.6 x 82=49 0.4 x 82=33 Interior negative 0.75 x 163=122 0.25 x 163=41 Exterior Positive 0.6 x 121=73 0.4 x 121=48 Exterior negative 1.00 x 61=61 0 x 61=0 Note: Coefficients for lateral distribution have been selected from graph A.4 (pg 755) Nelson, using αl2/l1, βt, and l2/l1.

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 46 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

C1

C2

C2

C1

C3

C4

C4

C3

C3

C4

C4

C3

C1

C2

C2

C1

Figure 43: Column plan. Step (E): Moment transferred to columns. Exterior column (C3) moment = L.M……………………..……….……..……… (A) = 61 ft-k Interior column (C4) moment = (0.65/8)x{0.5xwuLLx l2xln2}…………………….. (B) = (0.65/8) x {0.5x (1.6x0.144) x 15 x 18.832} = 49.78 ft-k

II. Analysis of E-W Exterior Frame. Step (A): Frame Data. •

Design Span of frame (c/c) = l1 = 20′

•

Design Length of frame = ln = 20 – (2 x 14/2)/12 = 18.83′

•

Width of frame = l2 = (15/2) + 14/(2 x 12) = 8.08′

•

Column strip width = (shorter span)/ 4 = 15/4 = 3.75′

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 47 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar 7"

l1 = 20'-0"

Direct Design Method

l1 = 20'-0"

3.75'

Half column strip

3.75'

Half middle strip

l1 = 20'-0"

l2 = 8.08' 15'-0"

15'-0"

All columns 14" square

15'-0"

Figure 44: East west exterior frame. Step (B): Total static moment. Mo = wul2ln2/8 = 0.3504 x 8.08 x 18.832/8 = 126 ft-k Step (C): Longitudinal distribution of Total static moment (Mo). Table 13: Longitudinal Distribution of Total Static Moment (Mo) Exterior span

Interior span

20' 126

20' 126

Static Moment Mo (ft-k) Distribution Factor (D.F) ACI 13.6.3 (D.F)

0.26

0.52

0.7

0.65

0.35

0.65

Longitudinal Moment (L.M) = Mo x D.F

33

66

89

82

45

82

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 48 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Step (D): Lateral Distribution of Longitudinal moment (L.M). α= 0 l2/l1 = 15/20 = 0.75 αl2/l1 = 0 βt = 0 (no edge beam)

Span Interior

Exterior

Table 14: Lateral Distribution of Longitudinal moment (L.M). Column Strip Middle Strip Moments (M.S.M) Section Moment (C.S.M) ft-k ft-k Negative 0.75 x 82 = 62 0.25 x 82 = 21 Positive 0.60 x 45 =27 0.40 x 45 = 18 Interior 0.75 x 89=67 0.25 x 89 = 22 negative Positive 0.60 x 66 =40 0.40 x 66= 26 Exterior 1.00 x 33 =33 0 x 33 = 0 negative

Note: Coefficients for lateral distribution have been selected from graph A.4 (pg 755) Nelson, using αl2/l1, βt, and l2/l1.

Step (E): Moment transferred to columns. Exterior column (C1) moment = L.M ………………………….…………… (A) =33 ft-k Interior column (C2) moment = (0.65/8){0.5wuLLl2ln2}………..………….… (B) = (0.65/8) x {0.5 x (1.6 x 0.144) x 8.08 x 18.832} = 27 ft-k

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 49 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Step No 4: Design. (1) Design of Slab strips: (A) E-W Interior slab strip: ds = hf – ¾ - (4/8)/2 = 8 – 1 =7″ Asmin = 0.0018x 12 x 8 = 0.1728 in2 (ρmin = 0.002, in terms of actual effective depth) Now, Equation used to calculate (ρ) values is as follows: Mu = Φfyρbds2{1-0.59ρfy/fc′) = 0.9 x 60 x ρ x 12 x 72 x {1-0.59 x ρ x 60/4) 281005 ρ2 - 31752 ρ + Mu = 0 After solving the above equation for ρ ρ = [31752 ± √{(317522) -(4 x 281005 x Mu′ x 12)}]/2(281005)…………(A)

d s= 7"

hf = 8"

Direction along 20' length. Figure 45: Slab section at E-W interior strip. Table 15: Slab Design (E-W, Interior Frame). Frame Span

Interior

Location

Section

b (in)

d (in)

Mu (ftk)

Mu′ = Mu x12/b

ρ (equation A)

As = ρ*b*d (in2)

BAR taken

Spacing = b*Ab/As

Recommended

Column Strip Moment

negative

90

7

113

15.07

0.0060

3.79

#4

4.75

3

Positive

90

7

49

6.53

0.0025

1.59

#4

11.31

9

Middle Strip Moment

negative

90

7

38

5.07

0.0020

1.26

#4

14.29

9

Positive

90

7

33

4.40

0.0020

1.26

#4

14.29

9

90

7

122

16.27

0.0065

4.11

#4

4.38

3

90

7

73

9.73

0.0038

2.40

#4

7.51

3

90

7

61

8.13

0.0032

1.99

#4

9.03

9

90

7

41

5.47

0.0021

1.33

#4

13.57

9

90

7

48

6.40

0.0025

1.56

#4

11.55

9

90

7

0

0.00

0.0020

1.26

#4

14.29

9

Interior negative Column Strip Positive Moment Exterior negative Exterior Interior negative Middle Strip Positive Moment

Exterior negative

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 50 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

(B) E-W Exterior slab strip: ds = hf – 1 = 8 – 1 =7″ Asmin = 0.0018x 12 x 8 = 0.1728 in2 (ρmin = 0.002, in terms of actual effective depth) Now, Equation used to calculate (ρ) values is as follows: Mu = Φfyρbds2{1-0.59ρfy/fc′) = 0.9 x 60 x ρ x 12 x 72 x {1-0.59 x ρ x 60/4) 281005 ρ2 - 31752 ρ + Mu = 0 After solving the above equation for ρ ρ = [31752 ± √{(317522) -(4 x 281005 x Mu′ x 12)}]/2(281005)…………(A)

d s= 7"

hf = 8"

Direction along 20' length. Figure 46: Slab section at E-W exterior strip. Table 16: Slab Design (E-W, Exterior Frame). Frame Span

Location

Section

b (in)

d (in)

Mu (ft-k) (Table 1.6)

Mu′ = Mu x 12/b

ρ (eqn A)

As = ρ x bxd (in2)

Bar taken

Spacing = b x Ab/As

Recommended

Column Strip Moment

negative

45

7

62

16.53

0.0066

2.09

#4

4.30

3

Positive

45

7

27

7.20

0.0028

0.88

#4

10.24

9

Middle Strip Moment

negative

45

7

21

5.60

0.0022

0.68

#4

13.24

9

Positive

45

7

18

4.80

0.0020

0.63

#4

14.29

9

Interior negative

45

7

67

17.87

0.0072

2.27

#4

3.96

3

Positive

45

7

40

10.67

0.0042

1.32

#4

6.82

3

Exterior negative

45

7

33

8.80

0.0034

1.08

#4

8.33

9

Interior negative

52

7

22

5.07

0.0020

0.73

#4

14.29

9

Positive

52

7

26

6.00

0.0023

0.84

#4

12.34

9

Exterior negative

52

7

0

0.00

0.0020

0.73

#4

14.29

9

Interior

Column Strip Moment Exterior Middle Strip Moment

Note: “b” is the width of strip.

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 51 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

(2) Shear Design for Flat-plate: (i) At Column C1: hf = 8″ d = 8 – 1 = 7″

d/2 = 3.5"

14"

14"

C1

C2

C2

C1

C3

C4

C4

C3

C3

C4

C4

C3

C1

C2

C2

C1

d/2 =3.5"

Figure 47: Critical perimeter for column (C1). (a) Vu = 0.3504{(7.5 x 10) – (17.5/12)2} = 25.53 k (b) Shear capacity of concrete in punching shear: ΦVc = 0.75 x 4 √ (fc′) bod bo = (c1 + d/2) +(c2 + d/2) bo = 2(14 + 7/2) = 35″ ΦVc = 0.75 x 4 √ (4000) x 35 x 7/1000 = 46.48 k ΦVc > Vu, O.K.

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 52 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

(ii) At Column C2:

d/2 = 3.5"

d/2 = 3.5"

14"

14"

d/2 = 3.5"

C1

C2

C2

C1

C3

C4

C4

C3

C3

C4

C4

C3

C1

C2

C2

C1

Figure 48: Critical perimeter for column (C2). (a) Vu = 0.3504{(7.5 x 20) – (21 x 17.5/144)} = 51.66 k (b) Shear capacity of concrete in punching shear: ΦVc = 0.75 x 4 √ (fc′) bod bo = (c1 + d) +2(c2 + d/2) bo = (14 + 7) +2(14 + 7/2) = 56″ ΦVc = 0.75 x 4 √ (4000) x 56 x 7/1000 = 74.37 k ΦVc > Vu, O.K.

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 53 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

(iii) At Column C3:

d/2 = 3.5"

14"

C1

C2

C2

C1

C3

C4

C4

C3

C3

C4

C4

C3

C1

C2

C2

C1

d/2 = 3.5"

14"

d/2 = 3.5"

Figure 49: Critical perimeter for column (C3). (a) Vu = 0.3504{(10 x 15) – (21 x 17.5/144)} = 51.66 k (b) Shear capacity of concrete in punching shear: ΦVc = 0.75 x 4 √ (fc′) bod bo = 2(c1 + d/2) +(c2 + d) bo = 2(14 + 7/2) + (14 + 7) = 56″ ΦVc = 0.75 x 4 √ (4000) x 56 x 7/1000 = 74.37 k ΦVc > Vu, O.K.

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 54 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

(iv) At Column C4: d/2 = 3.5"

14"

d/2 = 3.5"

14"

d/2 = 3.5"

d/2 = 3.5"

C1

C2

C2

C1

C3

C4

C4

C3

C3

C4

C4

C3

C1

C2

C2

C1

Figure 50: Critical perimeter for column (C4). (a) Vu = 0.3504{(20 x 15) – (21 x 21/144)} = 104 k (b) Shear capacity of concrete in punching shear: ΦVc = 0.75 x 4 √ (fc′) bod bo = 2(c1 + d) +2(c2 + d) bo = 4(14 + 7) = 84″ ΦVc = 0.75 x 4 √ (4000) x 84 x 7/1000 = 111 k ΦVc > Vu, O.K.

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 55 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

Step No 5: Detailing. All the frames may be analyzed and designed by the same procedure as given in steps of analysis and design. However, the complete detail of reinforcement placement in slab is given below.

B

N

#4 @ 9" c/c (T)

#4 @ 9" c/c (T)

#4 @ 3" c/c (T)

C4

#4 @ 5" c/c (B)

#4 @ 5" c/c (B) #4 @ 3" c/c (T)

#4 @ 5" c/c (B)

#4 @ 5" c/c (T) #4 @ 9" c/c (B)

#4 @ 9" c/c (T)

#4 @ 9" c/c (B)

#4 @ 9" c/c (T)

#4 @ 9" c/c (B)

#4 @ 9" c/c (B) #4 @ 9" c/c (B)

7'-6"

#4 @ 9" c/c (T) #4 @ 3" c/c (T)

A

#4 @ 5" c/c (B)

B

#4 @ 3" c/c (T)

#4 @ 5" c/c (B)

C4

PLAN

#4 @ 9" c/c (T)

#4 @ 9" c/c (T)

C3

#4 @ 5" c/c (T)

#4 @ 5" c/c (T)

3.17'

C4

#4 @ 3" c/c (T)

#4 @ 5" c/c (B)

15'-0"

D

#4 @ 9" c/c (B)

#4 @ 9" c/c (B)

C3

#4 @ 9" c/c (T)

7'-6"

#4 @ 5" c/c (T)

#4 @ 5" c/c (T)

#4 @ 3" c/c (T)

#4 @ 5" c/c (T)

#4 @ 5" c/c (T)

#4 @ 9" c/c (B)

E C S

C2

#4 @ 5" c/c (T)

7'-6"

#4 @ 5" c/c (B)

15'-0"

#4 @ 5" c/c (T)

D

#4 @ 3" c/c (T)

#4 @ 9" c/c (B)

#4 @ 9" c/c (T)

#4 @ 5" c/c (B)

W

3.17'

#4 @ 9" c/c (T)

#4 @ 9" c/c (B)

#4 @ 5" c/c (T)

#4 @ 5" c/c (B)

12'-6"

C2

#4 @ 9" c/c (T)

3.17'

7'-6"

#4 @ 5" c/c (B)

C1

#4 @ 9" c/c (T)

C

12'-6"

#4 @ 9" c/c (T)

3.17'

20'-0"

#4 @ 9" c/c (T)

20'-0"

#4 @ 9" c/c (T)

A

C4

According to ACI 13.3.8.5, At least two of the column strip bottom bars or wires in each direction shall pass within the column core and shall be anchored at exterior supports.

Figure 51: Reinforcement details (Plan view)

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 56 of 58

Department of Civil Engineering, N-W.F.P U.E.T, Peshawar

Direct Design Method

#4 @ 5" c/c

#4 @ 9" c/c

0.3Ln= 4'-3"

0.3Ln = 4'-3"

0.3Ln = 4'-3"

#4 @ 5" c/c

Ln = 13.83'

Ln = 13.83'

SECTION A-A (N-S exterior column strip) #4 @ 9" c/c

#4 @ 9" c/c

0.3Ln = 4'-3"

0.3Ln = 4'-3"

0.3Ln = 4'-3"

#4 @ 9" c/c

Ln = 13.83'

Ln = 13.83'

SECTION B-B (N-S middle strip)

#4 @ 5" c/c

#4 @ 3" c/c

0.3Ln = 6'-0"

0.3Ln = 6'-0"

0.3Ln = 6'-0"

#4 @ 5" c/c

Ln = 18.83'

Ln= 18.83'

SECTION C-C (E-W exterior column strip) #4 @ 9" c/c

#4 @ 9" c/c

0.3Ln = 6'-0"

0.3Ln = 6'-0"

0.3L n = 6'-0"

#4 @ 9" c/c

L n = 18.83'

Ln = 18.83'

SECTION D-D (E-W middle strip)

Figure 52: Sectional view.

Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)

Page 57 of 58

lecture-05-direct-design-method

Published on Apr 13, 2012

Direct Design Method for Two-way Slab Analysis Shall not be less than 0.2 nor greater than 5.0. Prof. Dr. Qaisar Ali (http://www.eec.edu.pk)...

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