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CEEN 3304 Concrete Design
Flexural Analysis and Design of Beams
Francisco AguíñigaAssistant Professor
Civil and Architectural Engineering ProgramTexas A&M University – Kingsville
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Stress distributions
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Composite beamsdF = σdA = (E1ε)dzdydF’ = σ’dA’ = (E2ε)ndzdyEquating dF and dF’n = E1/E2
The force in material 1 isdF = σdA = σ’dA’σdzdy = σ’ndzdyσ = n σ’
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Uncracked concrete beams
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Stresses on cracked beam
1. Find neutral axis2. Find Icr3. Find stresses
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Stresses on cracked beam
( ) ( ) 02
2
=−− kddnAkdb s
bkdf
C c
2=
ss fAT =
jdfATjdM ss==
jdAMfs
s =
2
22kjbd
fbkdjd
fCjdM cc ===
2
2kjbd
Mfc =
bdAs=ρ
( ) nnnk ρρρ −+= 22
3kddjd −=
31 kj −=
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Ultimate flexural strength
Steel fails when fs = fy
Concrete fails when εc = εu = 0.003Knowing c need to know: C and β
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Ultimate flexural strength
0.72
0.425
0.325
0.56
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Example 3.3
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Ultimate flexural strength
Whitney’s equivalent rectangular stress block
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Example 3.4
Solve the same beam using the rectangular stress block
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Limiting reinforcement ratios
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ACI strength reduction factors
ACI (9.3.2)Tension-controlled failure
When εc = 0.003, εs > 0.005, So φ = 0.9
Compression-controlled failureWhen εc = 0.003, εs < εy = 0.002, So φ = 0.65
Transition-controlled failureWhen εc = 0.003, εy < εs < 0.005So φ = A1 + B1εt
y
yAε
ε−
−=
005.09.000325.0
1y
Bε−
=005.0
25.01
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ACI strength reduction factors
Tension-controlled failureWhen εc = 0.003, εs > 0.005So φ = 0.9
Compression-controlled failure
When εc = 0.003, εs < εy
So φ = 0.65Transition-controlled failure
When εc = 0.003We have εy < εs < 0.005So φ = A1 + B1εt
ρmax
ρtc
ρbal
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ACI transition-controlled failure
Reason for limiting εt > 0.004 in tension-controlled failure (ACI 10.3.5)
In 2002 ACI codeLimit ρ to ρmax to 0.75ρbal
Results in εt = 0.00376, so limit εt to 0.004End up with ρmax < 0.75ρbal
bdAs=ρ
ty
cbal f
fε
βρ
+=
003.0003.085.0 '
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ACI tension-controlled failure
From
For εt = 0.004, c/dt = 3/7And cmax = 3/7 dt
So amax = β1cmax = 3/7β1dt
So ρmax = 0.364 β1 f’c/fy (dt/d)
ttdc
ε+=
003.0003.0
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ACI tension-controlled failure
From
For εt = 0.005, c/dt = 3/8And cmax = 3/8 dt
So amax = β1cmax = 3/8β1dt
So ρmax = 0.319 β1 f’c/fy (dt/d)
ttdc
ε+=
003.0003.0
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ACI minimum reinforcement
ACI 10.5.1 requires that
dbf
dbff
A wy
wy
cs
2003 '
min, ≥=
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Design of R/C beams
ACI 9.5.2.1 minimum span/depth ratios for beams
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Design of R/C slabs
ACI 9.5.2.1 minimum span/depth ratios for one-way slabs
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Design of R/C beams
Selection of widthACI 7.6.1, 7.6.2, and 3.3.2Minimum space for single layer bars
smin = largest of (db or 1 in. or ¾ max aggregate size)
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Design of R/C beams
Minimum space for multiple layers of bars
Bars in upper layer placed directly above bottom layerClear distance between layers > 1 in.Also satisfy single layer requirements
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Design of R/C beamsMinimum cover
Cast in place concrete – protection¾ in. for slabs and 1 ½ in. for beams and columns
Exposed to weather or in contact with soilCover > 2 in.Concrete cast directly on ground - cover >= 3 in.
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Design of R/C beams
Minimum widthUsually #3 or #4 are used for stirrupsMinimum cover for bars in beam is 1.5 in.bmin = 2 x 1.5 in. + 2 x 1/2in. + 4 x 1 in. + 3 x 1 in. = 11 in.
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T-beams
Effective flange width – beams with slab on both sides (ACI 8.10)
beff < ¼ span of beamEff. overhang width on each side < 8 hf and ½ clear distance to next web
Effective flange width – beams with slab on one side only (ACI 8.10)
Effective overhang width less than1/12 span of beam6 hf and ½ clear distance to next web
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T-beams
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