Direct Design Method

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Direct Design Method

• Design of Two way floor system for
• Flat slab

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Given data
Figure-1 shows a flat slab floor with a total
area of 12,500 sq ft. It is divided into 25
panels with a panel size of 25 ft x 20 ft.
Concrete strength is and steel
yield strength is fy40,000 psi. Service live
load is to be taken as 120 psf. Story height is
10 ft. Exterior columns are 16 in. square and
interior columns are 18 in. round. Edge beams
are14 ? 24 in. Overall Thickness of slab is 7.5
in. outside of drop panel and 10.5 in. through
the drop panel. Sizes of column capital and drop
panels are shown in Fig.-1.
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Fig.-1
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(No Transcript)
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Problem-1
Compute the total factored static moment in the
long and short directions of an interior panel in
the flat slab design as shown in Fig.1.
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• Neglecting the weight of the drop panel, the
  service dead load is (150/12)(7.5)94 psf thus
• wu1.2wD 1.6wL
• 1.2(94) 1.6(120)
• 132 204
• 336 psf

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The equivalent square area for the column capital
has its side equal to 4.43 ft then, using Eq.1,
with Ln measured to the face of capital (i.e.,
equivalent square),
So far as flat slabs with column capitals are
concerned, it appears that the larger values of
395 ft-kips and 292 ft-kips should be used
because Eq.8 is specially suitable in
particular, ACI states that the total factored
static moment shall not be less than that given
by Eq.1.
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Given data Problem-2
Review the slab thickness and other nominal
requirements for the dimensions in this flat slab
design described in problem-1.
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• Stiffness of edge beams.
• Before using Table-1, the a values for the edge
  beams are needed. The moment of inertia of the
  edge beam section shown in Fig.2 is 22,9000 in4.
  Thus the a value for the long edge beam is

Fig.-2
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Stiffness of edge beams. The moments of
inertia Ib and Is refer to the gross sections of
the beam and slab within the cross-section. ACI
permits the slab on each side of the beam web to
act as a part of the beam, this slab portion
being limited to a distance equal to the
projection of the beam above or below the slab,
whichever is greater, but not greater than four
times the slab thickness, as shown in Fig.
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In which
where
h overall beam depth t overall slab
thickness bE effective width of flange bw
width of web
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and for the short edge beam, it is
These ? values are entered on Fig.3.
(b) Minimum slab thickness using Table-1. The
long and short clear spans for deflection control
are Ln 25-4.43 20.57 ft Sn
20-4.43 15.57 ft. From which
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Fig.-3
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Table-1 Minimum thickness of slab without
interior beams
WITHOUT DROP PANELS WITHOUT DROP PANELS WITHOUT DROP PANELS WITH DROP PANELS WITH DROP PANELS WITH DROP PANELS
fy EXTERIOR PANELS EXTERIOR PANELS INTERIOR EXTERIOR PANELS EXTERIOR PANELS INTERIOR
(ksi) a 0 a ? 0.8 PANELS a 0 a ? 0.8 PANELS
40


60


75

For fy between 40 and 60 ksi, min. t is to be
obtained by linear interpolation.
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• For fy40 ksi, a flat slab with drop panel, and a
  smaller of 4.34 and 5.42, Table-1 gives
• For both exterior and interior panels.

• Nominal requirement for slab thickness.
• The minimum thickness required is, from part (b),
  6.17 in. The 7.5 in. slab thickness used is more
  than ample 6.5 in. should probably have been
  used.

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(d) Thickness of drop panel. Reinforcement within
the drop panel must be computed on the basis of
the 10.5 in. thickness actually used or 7.5 in.
plus one-fourth of the projection of the drop
beyond the column capital, whichever is smaller.

• In order that the full 3-in. projection of the
  drop below the 7.5 in. slab is usable in
  computing reinforcement, the 6 ft 8 in. side of
  the drop is revised to 7 ft so that one-fourth of
  the distance between the edges of the 5-ft column
  capital and the 7-ft drop is just equal to (10.5
  - 7.5) 3 in.

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Given data Problem-3
For the flat slab design problem-1, Compute the
longitudinal moments in frames A, B, C and D as
shown Fig.-1, 4 5.
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Fig.-4
19
Fig.-5
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Longitudinal moments in the Frame. The
longitudinal moments in frames A, B, C, D are
computed using Case 4 of Fig.23(DDM) for the
exterior span and Fig.19(DDM) for the interior
span. The computations are shown in Table-1 and
the results are summarized in Fig.45.
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• Check the five limitations (the sixth limitation
  does not apply here) for the direct design
  method. These five limitations are all satisfied.
• Total factored static moment M0.
• Referring to the equivalent rigid frames A, B,
  C, and D in Fig. 45, the total factored static
  moment may be taken from the results previously
  found thus

• M0 for A 395 ft-kips
• M0 for B 0.5(395) 198 ft-kips
• M0 for C 292 ft-kips
• M0 for D 0.5(292) 146 ft-kips

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Table-1 Longitudinal Moments (ft-kips) for the
flat slab
FRAME A B C D
M0 395 198 292 146
Mneg at exterior support, 0.30M0 118 59 88 44
Mpos in exterior span, 0.50M0 198 99 146 73
Mneg at first interior support, 0.70M0 276 139 204 102
Mneg at typical interior support, 0.65M0 257 129 190 95
Mpos in typical interior span, 0.35M0 138 69 96 51
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Given data Problem-4
For the flat slab design problem-1, compute the
torsional constant C for the edge beam and the
interior beam in the short and long directions.
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• For the short or long edge beam Fig.6(a), the
  torsional constant C is computed on the basis of
  the cross-section shown in Fig.6(a).

Fig.-6
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The torsional constant C equals,
where x shorter dimension of a component
rectangle y longer dimension of a component
rectangle
and the component rectangles should be taken in
such a way that the largest value of C is
obtained.
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Fig.-7
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Fig.-8

• For the short or long interior beam Fig.6(b), a
  weighted slab thickness of 8.5 in. is used, on
  the assumption that one-third of the span has a
  10.5 in. thickness and the remainder has a 7.5
  in. thickness.

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Given data Problem-5
Divide the five critical moments in each of the
equivalent rigid frames A, B, C and D in the flat
slab design problem-1, as shown in Fig.45, into
two parts one for the half column strip (for
frames B and D) or the full column strip (for
frames A and C), and the other for the half
middle strip (for frames B and D) or the two half
middle strips on each side of the column line
(for frames A and C)
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• The percentages of the longitudinal moments
  going into the column strip width are shown in
  lines 10 to 12 of Table-2. The column strip width
  shown in line 2 is one-half of the shorter panel
  dimension for both frames A and C, and one-fourth
  of this value for frames B and D. The sum of the
  values on lines 2 and 3 should be equal to that
  on line 1, for each respective frame.

The moment of inertia of the slab equal in
width to the transverse span of the edge beam is
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• These values are shown in line 5 of Table-2.
• The percentages shown in lines 10 to 12 are
  obtained from Table-2A, by interpolation if
  necessary.

Having these percentages, the separation of each
of the longitudinal moment values shown in
Fig.45 into two parts is a simple matter and
thus is not shown further.
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Table-2 Transverse distribution of longitudinal
moment for flat slab
LINE NUMBER EQUIVALENT RIGID FRAME A B C D
1 Total transverse width (in.) 240 120 300 150
2 Column strip width (in.) 120 60 120 60
3 Half middle strip width (in.) 2_at_60 60 2_at_90 90
4 C(in4) from previous calculations 18,500 18,500 18,500 18,500
5 Is(in.4) in ßt 8,440 8,440 10,600 10,600
6 ßt EcbC/(2EcsIs) 1.10 1.10 0.87 0.87
7 a1 from previous calculations 0 5.42 0 4.34
8 L2/L1 0.80 0.80 1.25 1.25
9 a1 L2/L1 0 4.33 0 5.43
10 Exterior negative moment, percent to column strip 89.0 91.6 91.3 88.7
11 Positive moment, percent to column strip 60.0 81.0 60.0 67.5
12 Interior negative moment, percent to column strip 75.0 81.0 75.0 67.5
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Table-2APercentage of longitudinal moment in
column strip
ASPECT RATIO L2/L1 0.5 1.0 2.0
Negative moment at a1L2/L1 0 ßt0 100 100 100
exterior support ßt?2.5 75 75 75
a1L2/L1 gt 1.0 ßt 0 100 100 100
ßtgt 2.5 90 75 45
Positive moment a1L2/L1 0 60 60 60
a1L2/L1 gt 1.0 90 75 45
Negative moment at a1L2/L1 0 75 75 75
interior support a1L2/L1 gt 1.0 90 75 45
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Given data Problem-6
Design the reinforcement in the exterior and
interior spans of a typical column strip and a
typical middle strip in the short direction of
the flat slab problem-1. As described earlier,
fy40,000 psi.
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• (a) Moments in column and middle strips
• The typical column strip is the column strip of
  equivalent rigid frame C of Fig.5 but the
  typical middle strip is the sum of two half
  middle strips, taken from each of the two
  adjacent equivalent rigid frames C. The factored
  moments in the typical column and middle strips
  are shown in Table-3.

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Table-3 Factored moments in a typical column
strip and middle strip
EXTERIOR SPAN EXTERIOR SPAN EXTERIOR SPAN INTERIOR SPAN INTERIOR SPAN INTERIOR SPAN
Line number Moments at critical section (ft-kips) Negative moment Positive moment Negative moment Negative moment Positive moment Negative moment
1 Total M in column and middle strips (Fig.5) (frame C) -88 146 -204 -190 96 -190
2 Percentage to column strip (Table-2) 91.3 60 75 75 60 75
3 Moment in column strip -80 88 -153 -142 58 -142
4 Moment in middle strip -8 58 -51 -48 38 -48
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• Slab thickness for flexure

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Table-4 Design of reinforcement in column strip
EXTERIOR SPAN EXTERIOR SPAN EXTERIOR SPAN INTERIOR SPAN INTERIOR SPAN INTERIOR SPAN
LINE NUMBER ITEM NEGATIVE MOMENT POSITIVE MOMENT NEGATIVE MOMENT NEGATIVE MOMENT POSITIVE MOMENT NEGATIVE MOMENT
1 Moment, Table-2, line 3 (ft-kips) -80 88 -153 -142 58 -142
2 Width b of drop or strip (in.) 100 120 100 100 120 100
3 Effective depth d (in.) 8.81 6.44 8.81 8.81 6.44 8.81
4 Mu/Ø (ft-kips) -89 98 -170 -158 64 -158
5 Rn(psi) Mu/(Øbd2) 138 236 263 244 154 244
6 ?, Eq. or Table A.5a 0.35 0.62 0.70 0.64 0.39 0.64
7 As ?bd 3.08 4.79 6.17 5.63 3.01 5.63
8 As 0.002bt 2.40 1.80 2.40 2.40 1.80 2.40
9 Nlarger of (7) or(8)/0.31 9.9 15.5 19.9 18.2 9.7 18.2
10 Nwidth of strip/(2t) 5 8 5 5 8 5
11 N required, larger of (9) or (10) 10 16 20 19 10 19
bt100(10.5)20(7.5)1200 in2 for negative
moment region.
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Table-5 Design of reinforcement in Middle Strip
EXTERIOR SPAN EXTERIOR SPAN EXTERIOR SPAN EXTERIOR SPAN EXTERIOR SPAN INTERIOR SPAN INTERIOR SPAN INTERIOR SPAN
LINE NUMBER ITEM NEGATIVE MOMENT POSITIVE MOMENT NEGATIVE MOMENT NEGATIVE MOMENT POSITIVE MOMENT NEGATIVE MOMENT
1 Moment, Table 3, line 4 (ft-kips) -8 58 -51 -48 38 -48
2 Width b of strip (in.) 180 180 180 180 180 180
3 Effective depth d (in.) 6.44 5.81 6.44 6.44 5.81 6.44
4 Mu/Ø (ft-kips) -9 64 -57 -53 42 -53
5 Rn(psi) Mu/(Øbd2) 14 126 92 85 83 85
6 ? 0.04 0.32 0.23 0.22 0.21 0.22
7 As ?bd 0.46 3.30 2.67 2.53 2.20 2.53
8 As 0.002bt 2.70 2.70 2.70 2.70 2.70 2.70
9 Nlarger of (7) or(8)/0.31 8.7 10.6 8.7 8.7 7.1 8.7
10 Nwidth of strip/(2t) 12 12 12 12 12 12
11 N required, larger of (9) or (10) 12 12 12 12 12 12
A mixture of 5 and 4 bars could have been
selected.
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• Design of Reinforcement
• The design of reinforcement for the typical
  column strip is shown in Table-4 for the typical
  middle strip , it is shown in Table-5. Because
  the moments in the long direction are larger than
  those in the short direction, the larger
  effective depth is assigned to the long direction
  wherever the two layers of steel are in contact.
  This contact at crossing occurs in the top steel
  at intersection of middle strips and in the
  bottom steel at the intersection of middle
  strips. Assuming 5 bars and ¾ in. clear cover,
  the effective depths provided at various critical
  sections of the long and short directions are
  shown in Fig.9

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Fig.-9
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Given data Problem-7
Investigate the shear strength in wide-beam and
two-way actions in the flat slab design for an
interior column with no bending moment to be
transferred. We have
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• Wide beam action. Investigation for wide beam
  action is made for sections 1-1 and 2-2 in the
  long direction, as shown in fig.10. The shot
  short direction has a wider critical section and
  short span thus it does not control. For section
  1-1, if the entire width of 20 ft is
  conservatively assumed to have an effective depth
  of 6.12 in.

Fig.-10
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• Wide beam action.
• if however bw is taken as 84 in. and d as 9.12
  in. on the contention that the increased depth d
  is only over a width of 84 in.

This later value is probably unrealistically low.
For section 2-2 the shear resisting section has a
constant d of 6.12 in thus
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It will be rare that wide beam (one way) action
will govern.
(b)Two way action. The critical sections for two
way action are the circular section 1-1 at d/2
4.56 in. from the edge of the column capital and
rectangular section 2-2 at d/2 3.06 in. from the
edge of drop, as shown in fig.11. Since there are
not shearing forces at the centerlines of the
four adjacent panels, the shear forces around the
critical sections 1-1 and 2-2 in fig.11 are
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Fig.-11
46
In the second term, the 0.038 is the weight of
the 3in. drop in ksf.
Compute the shear strength at section 1-1 around
the perimeter of the capital( fig.11)
47
Though both sections 1-1 and 2-2 have ?Vn
significantly greater than Vu, the section around
the drop panel is loaded to slightly higher
percentage of its strength (50 for section 2-2
vs 47 for section 1-1). Shear reinforcement is
not required at this interior location.