Lecture 19 Oneway Slab Design

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Lecture 19 - One-way Slab Design

• July 21, 2003
• CVEN 444

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Lecture Goals

• One-way slab design
• Joist
• Slab

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Example One-way Slab Design
Design a one way slab with a clear span of 10 ft.
Use 12 in. wide segment bw. fy 40 ksi, fc 3
ksi and a live load of 325 psf.
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Example One-way Slab Design
Determine the preliminary slab depth for an
internal section
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Example One-way Slab Design
The correction factor is CF 0.4
(fy/100) Self-weight of the element for a 1
ft. segment
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Example One-way Slab Design
The factored load
Factored moments demanded
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Example One-way Slab Design
The factored moment The depth of the
reinforcement (assume 6 bar) is
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Example One-way Slab Design
Obtain a set of the equations to find (c/d)
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Example One-way Slab Design
Find (c/d) for the given slab Therefore, the
slab is in the compression control region.
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Example One-way Slab Design
Need the steel to be in the tension controlled
region. So chose an arbitrary k, k0.3 and
k0.255
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Example One-way Slab Design
Determine the depth of the slab with the new
d. with h 5 in., then d 3.875 in. Go back
and check (c/d) ratio
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Example One-way Slab Design
The new factored load is The maximum moment
is
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Example One-way Slab Design
Determine the new value for k and
(c/d) Therefore, the slab is in the tension
controlled region.
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Example One-way Slab Design
Determine the area of the steel
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Example One-way Slab Design
The area of steel is 0.509 in2 per foot.
Calculate the maximum allowable spacing Select
bars and spacing. Use 6 (0.44 in2) per 9 in.
governs
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Example One-way Slab Design
Transverse reinforcement is due to temperature
and shrinkage so Amin is Select bars and
spacing. Use 3 (0.11 in2) per 11 in.
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Example 2 One-way Slab Design
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Example 2 One-way Slab Design
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Example 2 One-way Slab Design
Compute the factored moments at the faces of the
supports and determine the depth The tributary
area is 3 ft wide on the joists
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Example 2 One-way Slab Design
Compute the moments for all of the sections the
external length is External section
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Example 2 One-way Slab Design
Compute the moments for the sections the
internal length is Internal section
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Example 2 One-way Slab Design
Determine the maximum d for the joist if we use
the maximum moment for (c/d) 0.3 and k 0.255,
thickness b 6 in. and a maximum moment 57.9
k-ft
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Example 2 One-way Slab Design
The additional distance, h d 0.75 in (cover)
db h 13.05 in 0.75 in 0.5 in.13.05 in.
1.25 in. h 14.3 in. From 9.5a
(ACI Table 9.5) Use h 19.5 in. Depth of
joist 19.5in 3.5 in. 16 in.
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Example 2 One-way Slab Design
Find (c/d) for external column, Mu 24.1 k-ft
(289.2 k-in) d 19.5 in. 1.25 in. 18.25
in. The area can be computed using
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Example 2 One-way Slab Design
The area of the steel The
minimum area by definition is
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Example 2 One-way Slab Design
Calculate the maximum allowable spacing
governs
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Example 2 One-way Slab Design
Use 0.37 in2 distributed over 3 ft section of
the joist Use 3 _at_ 10 in spacing
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Example 2 One-way Slab Design
Find (c/d) for middle of beam, Mu 41.3 k-ft
(495.6 k-in) The area can be computed using
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Example 2 One-way Slab Design
Calculate the a value of the beam for the
positive moment Use 2 5 bars (As 0.62 in2
)

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Example 2 One-way Slab Design
Find (c/d) for internal column, Mu 57.9 k-ft
(694.8 k-in) The area can be computed using
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Example 2 One-way Slab Design
Use 0.75 in2 distributed over 3 ft section of
the joist Use 4 _at_ 9 in spacing
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Example 2 One-way Slab Design
The summary
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Example 2 One-way Slab Design
Slab reinforcement normal to the ribs is often
located at mid-depth of the slab to resist both
positive and negative moments. The length
between joist is 36 in. 6 in. 30 in. or 2.5
ft.
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Example 2 One-way Slab Design
The moment is The d 3.5 in. 0.75 in.
0.25 in. 2.5 in. and the k value will be
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Example 2 One-way Slab Design
The area of steel The minimum reinforcement is
governed by temperature and shrinkage from ACI
7.12.2.1
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Example 2 One-way Slab Design
Calculate the maximum allowable spacing (ACI
7.12.2.2) Select bars and spacing. Use 3
(0.11 in2) per 16 in.
governs
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Example 2 One-way Slab Design
Shear at supports must be checked. Since the
joists meet the required in ACI 8.11.8, the
contribution of the concrete to the shear
strength, Vc is permitted to be 10 more than
that specified in 11ACI
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Example 3 One-way Slab Design
Design a slab for an interior span of concrete
joist floor system using 30 in forms. LL 80
psf fc 4 ksi
fy 60 ksi DL
20 psf (excluding self-weight)
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Example 3 One-way Slab Design
Assume an initial thickness of 3 in. For a single
slab without ribs.
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Example 3 One-way Slab Design
Find the minimum height of the joist from Table
9.5a in ACI Code The slab is 3 in. thick so
the joist must extend 15 in. 3 in. 12 in.
below the slab .
Joist Design
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Example 3 One-way Slab Design
Including the ribs in the calculation of the
weight with with a depth of 12 in.
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Example 3 One-way Slab Design
Compute the moments for the sections the
internal length is 24.5 ft. For the rib with a
negative moment The d 15 in. cover db/2
15 in. 0.75 in. 0.25 in. d 14 in.
(Guess a 4 bar with minimum cover)
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Example 3 One-way Slab Design
Find (c/d) for external column, Mu 36.3 k-ft
(435.6 k-in) d 14 in The area can be computed
using the equilibrium equation
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Example 3 One-way Slab Design
The area of the steel The
minimum area required
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Example 3 One-way Slab Design
Use 4 and 2 5 bars (As 0.81 in2 ), to find
the neutral axis

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Example 3 One-way Slab Design
Determine the moment capacity of the
slab Compute the factored nominal moment

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Example 3 One-way Slab Design
Verification for shear strength

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Example 3 One-way Slab Design
Verification for shear strength

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Example 3 One-way Slab Design
Compute nominal V Since fn lt freq we will need
to enlarge the ribs _at_ the ends

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Example 3 One-way Slab Design
The shear width at 13.5 in. using a commercial
form for b 6.66 in. and 3 in. over 36 in. So 36
in - 13.5 in 22.5 in. increase in thickness
22.5/36(3 in.) 1.875 in. if it is on both sides
5in 3.75 in. 8.75 in. gt 6.66 in.
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Example 3 One-way Slab Design
Compute the moments for the sections the
internal length is 24.5 ft. For the rib with a
positive moment
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Example 3 One-way Slab Design
Determine the beff of the joist For the rib
positive
governs
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Example 3 One-way Slab Design
Find (c/d) for external column, Mu 25.0 k-ft
(300 k-in) d 13.5 in.
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Example 3 One-way Slab Design
The minimum k required for the t section
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Example 3 One-way Slab Design
The minimum k required is 0.07843 Use 25
(As 0.62 in2) or 24 13 (As0.51 in2)
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Example 3 One-way Slab Design
Look at the transverse moment along the joist
spacing is 30 in., w 0.197 k/ft2
Using an elastic analysis for concrete for the
positive moment
tension in concrete
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Example 3 One-way Slab Design
The thickness of the slab is checked with an
elastic analysis So a 3 in. slab will work
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Example 3 One-way Slab Design
The minimum reinforcement is governed by
temperature and shrinkage from ACI 7.12.2.1
Determine the spacing between the bars
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Example 3 One-way Slab Design
Calculate the maximum allowable spacing (ACI
7.12.2.2) Use a wire mesh to get the necessary
steel, because 3 bar is 0.11 in2 and with a
maximum spacing of 9 in. the amount of steel will
still be too large. So use a wire mesh.
governs
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Example Interaction Diagrams
The final design is given for joist 12 in deep
and slab thickness of 3 in.