Example

CE 331, Fall 2010

Example: Analysis of Reinforced Concrete Floor

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beam web

Beam Spacing

bw h

tslab

A

slab

Beam Spacing

slab span

Beam Span

Beam Spacing

Beam Spacing

Beam Spacing

A

Column Spacing

Column Spacing

Column Spacing

Section A‐A: Slab & Beam Elevation

Beam Spacing

Typical Reinforced Concrete Building Cast‐in‐place reinforced concrete structures have monolithic slab to beam and beam to column connections. Monolithic comes from the Greek words mono (one) and lithos (stone). Consequences of monolithic construction include: • beam‐to‐column connections transfer moment making reinforced concrete frames highly indeterminate, and • the deck slab and the beam are one piece of concrete, making the beams “T‐shaped” A typical reinforced concrete floor system is shown in the sketches below. Exterior Span Interior Span Exterior Span

Plan View of Floor System

Column Spacing

Column Spacing Ln

Ln (clear span)

Column Spacing

Ln

slab beam web

column

Elevation View It’s important to have a clear picture in one’s mind of the concrete structure represented by the 2D sketches above. The orthographic views on the following page of the same floor system are intended to help in this regard.

CE 331, Fall 2010


Isometric View of Reinforced Concrete Floor System

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Cut‐away showing beam dimensions

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Placement of Reinforcement Steel reinforcement is placed in the tension zones of reinforced concrete beams, as indicated by the crack patterns shown below.

-'ve M steel +'ve M steel

Reinforcement in Tensile Zones

Factored Moments due to Dead + Live Loads (Mu) Because concrete frames are highly indeterminate, the moments due to factored dead and live loads are typically calculated with the aid of a computer program. Alternatively, designers often use the American Concrete Institute (ACI) moment coefficients (shown below) which represent the envelope of moments due to dead load plus various live load span load patterns. wu L2n 14 Mu

wu L2n 10

ACI Moment Coefficients

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Example. Check the flexural strength in the office building indicated below at three locations: 1. a one‐foot‐wide section of slab 2. the middle of an exterior span of an interior beam 3. at the interior support of an exterior span of an interior beam

1

4 @ 12 ft

1

2

3

2

3

3 @ 29 ft Col. Size = 24” x 24” t_slab = 5 in h = 16 in b_w = 13 in

f'c = 4000 psi fy = 60,000 psi unit wt of conc. = 150 pcf wSD = 10 psf wLL = 40 psf

Slab Reinforcement: #4 @ 15” Beam Reinforcement: Midspan: 3 #7 bars Support: 8 #6 bars #3 stirrups

1. One‐foot‐wide strip of slab εc = 0.003

12”

0.85f'c a

C

c

d

t_slab

As

εs

As

Strain Distribution in the Concrete

fs Stresses

T Forces

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Calc. Mu: Super‐imposed Dead Load

wSDL = (SDL)(tributary width) = (0.010ksf)(1ft) = 0.010klf

Self‐weight

wsw = (unit_wt) (width)(depth) = (0.150kcf)(1ft)(5in/12in/ft) = 0.0625klf

Live load

wL = (LL)(tributary width) = (0.040ksf)(1ft) = 0.040klf

Factored load

wu = 1.2(wsw + wSDL) + 1.6(wL) = 1.2(0.0625klf + 0.010klf) + 1.6(0.040klf) = 0.151klf

Moment due to factored Mu = (1/Mcoef)(wu)(Ln)2 loads

Mcoef = 10, max. moment for continuous beam, [FE Ref. pg. 144]

Clear span

Ln =beam spacing – width of beam web = 12ft – 13in/12in/ft = 10.92ft

Mu = (1/10)(0.151klf)(10.92ft)2,

Mu = 1.80kft

Calc. φMn: Calc. depth of stress block from ΣFH: C = T Æ 0.85 f’c a b = As fy

(Assume steel yields)

Calculate the area of steel in a one‐foot wide strip of slab: As of one #4 bar = 0.20 in2 [FE Ref. pg 144] 2 in 2 1bar 12 in As in a 1 ft − wide strip = (0.20 in / bar )( in ) = 0 . 160 1 ‐ ft strip 1 ft − wide strip 15 0.85 (4ksi) a (12in) = (0.160 in2)(60ksi),

a = 0.235in

Calc. effective depth, d cover = 0.75in, [ACI Clear Cover Reqt’s: Concrete not exposed to weather, Slabs, No 11 bar and smaller] d = h – cover – φbar/2 = 5in – 0.75in – 0.50in / 2, Calc. strain in steel, 0.003 ε s + 0.003 = a / β1 d

s

β1 = 0.85 for fc' = 4 ksi ε s + 0.003 0.003 0.235 in / 0.85

=

[FE Ref. pg. 145] [FE Ref. pg. 144]

4.00 in

• • •

εs > εy = 0.002, so steel yields as assumed εs > 0.004 = min. for beams [FE Ref. pg. 145] εs = 0.90 since s > 0.005 [FE Ref. pg. 145]

d = 4.00in

εs = 0.0404

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Calc.φMn φMn = φ As fy (d – a/2) = (0.90)(0.160 in2)(60ksi)(4.00in – 0.235in/2) / 12in/ft

φMn = 2.80k‐ft

φMn = 2.80k‐ft > 1.80k‐ft = Mu, OK

2. Midspan of exterior span: T‐beam

Calc. Mu Self‐weight of T‐beam: The self‐weight of the T‐beam is equal to the unit weight of reinforced concrete (typically 150 pcf) times the cross‐section area of the T‐beam. The flange of the T‐beam extends half‐ way to the nearest beam on either side, which is typically equal to the beam spacing. wsw = unit_wt x AT‐beam AT‐beam = the area of the slab + the area of the beam web that extends below the slab AT‐beam = (beam spacing x t_slab) + (h – t_slab) x bw wsw = (0.150kcf) [ 12ft x (5"/12in/ft) + (16in – 5in) x 13in x (1 ft2/144 in2)]= 0.899 klf wSD = (wSD)(beam spacing) = (0.010ksf)(12') = 0.120 klf wL = (wL)(beam spacing) = (0.040ksf)(12') = 0.480 klf wu = 1.2 (wsw + wSD) + 1.6 wL wu = 1.2 (0.899 klf + 0.120 klf) + 1.6 (0.480 klf) = 1.991 klf Mu = wu Ln2 / 14 (moment coefficient for exterior span, midspan) Ln = clear span = column spacing – width of column = 29ft – 2ft = 27ft Mu =103.7 k‐ft Mu = (1.991 klf) (27 ft)2 / 14

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Calc. φMn: effective flange width, b_f: [FE Reference, pg 146] b_f = minimum of: clear span / 4 16 t_slab + b_w beam spacing 27' (12"/') / 4 = 81" 16(5") + 13" = 93" 12' x 12'/" = 144" b_f = 81" Calc. stress resultants bf

εc = 0.003

t

0.85f'c Cf

aweb c

Cw

d εs

bw

fs

T

Assume the compression block does not extend below the bottom of the flange (typical). ΣFH: C = T 0.85 f’c a b_f = As fy Assume steel yields As = 3(0.60in2) = 1.80in2 0.85 (4 ksi) a (81”) = (1.80 in2) (60 ksi), a = 0.392” t, then ΣFH: C = T 0.85 f’c [(bf – bw) t + bw a] = As fy Assume steel yields Solve equation above for a, and proceed as below, except: when calculating φ Mn add the moment due to the compression in the flange φ Mn = φ [As fy (d – a/2) + Cf (a/2 – t/2)], where Cf = 0.85 f’c t (bf – bw)

a = 0.392”

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Calc. strain in steel cover = 1.5in, [ACI Clear Cover Reqt’s: Concrete not exposed to weather, Beams & columns] 0.003 0.003 + ε s = c d d = 16"−[1.5"+

3" 1 7 + " ] = 13.69" 8 28

a = β1c, β1 = 0.85 for f c' = 4 ksi 0.392" = 0.85 c,

c = 0.461"

0.003 0.003 + ε s = , ε s = 0.0861 0.461" 13.69" ∴ (1) steel has yeilded (ε s > ε y = 0.002) , as assumed (2) ε s > 0.004 ( ACI minfor beams), OK (3) φ = 0.90 (ε s >= 0.005) φ Mn = φ [As fy (d – a/2)] φ Mn = 0.90 [(1.80in2)(60 ksi) (13.69” – 0.392”/2)] (1' / 12"),

φMn = 109 > 104 = Mu, OK k‐ft

k‐ft

3. At Interior Side of Interior Support The bottom of the web is in compression, everything above the neutral axis is cracked. The top steel is spread laterally across the flange (therefore it is always in one layer).

bf

εs

t T

d c bw

.003

a

Cc

0.85f'c

Calc. Mu Mu = wu Ln2 / 10 (moment coefficient for exterior span, interior support) Mu = 145.1 k‐ft = (1.991 klf) (27 ft)2 / 11

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Calc. Mn Calc. depth of stress block

ΣFH: C = T 0.85 f’c a bw = As fy (Assume steel yields) As of one #6 bar = 0.44 in2 As = (8 bars)(0.44 in2 per bar) = 3.52 in2 0.85 (4 ksi) a (13”) = (3.52 in2) (60 ksi), Calc. strain in steel.

a = 4.778”

0.003 0.003 + ε s = c d d = 16"−[1.5"+

3" 1 6 + " ] = 13.75" 8 28

a = β1c, β1 = 0.85 for f c' = 4 ksi 4.778" = 0.85 c,

c = 5.621"

0.003 0.003 + ε s = , ε s = 0.00434 5.621" 13.75" ∴ steel has yeilded (ε s > ε y = 0.002) , as assumed &

φ = 0.48 + 83(0.00434) = 0.840 &

ε s > 0.004 (OK ) Calc. φ Mn

φ Mn = φ [As fy (d – a/2)] φ Mn = 0.840 [(3.52in2)(60 ksi) (13.75” – 4.778”/2)] (1' / 12"),

φMn = 168k‐ft > 145k‐ft = Mu, OK