aci concrete
DESCRIPTION
ACI ConcreteTRANSCRIPT
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Lecture 9 - Flexure
June 20, 2003CVEN 444
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Lecture Goals
Load EnvelopesResistance Factors and LoadsDesign of Singly Reinforced Rectangular Beam Unknown section dimensions Known section dimensions
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MomentEnvelopes
Fig. 10-10; MacGregor (1997)
The moment envelope curve defines the extreme boundary values of bending moment along the beam due to critical placements of design live loading.
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MomentEnvelopes ExampleGiven following beam with a dead load of 1 k/ft and live load 2 k/ft obtain the shear and bending moment envelopes
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MomentEnvelopes ExampleUse a series of shear and bending moment diagrams
Wu = 1.2wD + 1.6wL
0
1
2
3
4
5
0 5 10 15 20 25 30 35 40
(ft)
kips
-80
-60
-40
-20
0
20
40
60
80
0 5 10 15 20 25 30 35 40
ft
kips
-250
-200
-150
-100
-50
0
50
100
150
0 5 10 15 20 25 30 35 40
ft
k-ft
Shear Diagram Moment Diagram
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MomentEnvelopes ExampleUse a series of shear and bending moment diagrams
Wu = 1.2wD + 1.6wL
Shear Diagram Moment Diagram
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 5 10 15 20 25 30 35 40
ft
k/ft
-20
-15
-10
-5
0
5
10
15
20
0 5 10 15 20 25 30 35 40
ft
kips
-80
-60
-40
-20
0
20
40
0 5 10 15 20 25 30 35 40
ft
k-ft
(Dead Load Only)
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MomentEnvelopes ExampleUse a series of shear and bending moment diagrams
Wu = 1.2wD + 1.6wL
Shear Diagram Moment Diagram
00.5
11.5
22.5
33.5
44.5
5
0 5 10 15 20 25 30 35 40
ft
k/ft
-60-50-40-30-20-10
01020304050
0 5 10 15 20 25 30 35 40
ft
kips
-200
-150
-100
-50
0
50
100
150
200
0 5 10 15 20 25 30 35 40
ft
k-ft
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MomentEnvelopes ExampleThe shear envelope
Shear Envelope
-80-60-40-20
020406080
0 10 20 30 40
ft
kips
Minimum ShearMaximum Shear
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MomentEnvelopes ExampleThe moment envelope
Moment Envelope
-300
-200
-100
0
100
200
0 5 10 15 20 25 30 35 40
ft
k-ft
Minimum Moment Maximum Moment
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Flexural Design of Reinforced Concrete Beams and Slab Sections
Analysis Versus Design:Analysis: Given a cross-section, fc , reinforcement
sizes, location, fy compute resistance or capacity
Design: Given factored load effect (such as Mu) select suitable section(dimensions, fc, fy, reinforcement, etc.)
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Flexural Design of Reinforced Concrete Beams and Slab Sections
ACI Code Requirements for Strength Design
Basic Equation: factored resistance factored load effect
Ex.
un M M
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ACI Code Requirements for Strength Design
un M M Mu = Moment due to factored loads (required
ultimate moment)
Mn = Nominal moment capacity of the cross-section using nominal dimensions and specified material strengths.
= Strength reduction factor (Accounts for variability in dimensions, material strengths, approximations in strength equations.
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Flexural Design of Reinforced Concrete Beams and Slab Sections
Required Strength (ACI 318, sec 9.2)
U = Required Strength to resist factored loadsD = Dead LoadsL = Live loadsW = Wind Loads E = Earthquake Loads
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Flexural Design of Reinforced Concrete Beams and Slab Sections
Required Strength (ACI 318, sec 9.2)H = Pressure or Weight Loads due to soil,ground
water,etc. F = Pressure or weight Loads due to fluids with
well defined densities and controllable maximum heights.
T = Effect of temperature, creep, shrinkage, differential settlement, shrinkage compensating.
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Factored Load Combinations
U = 1.2 D +1.6 L Always check even if other load types are present.
U = 1.2(D + F + T) + 1.6(L + H) + 0.5 (Lr or S or R)U = 1.2D + 1.6 (Lr or S or R) + (L or 0.8W)U = 1.2D + 1.6 W + 1.0L + 0.5(Lr or S or R) U = 0.9 D + 1.6W +1.6HU = 0.9 D + 1.0E +1.6H
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Resistance Factors, ACI Sec 9.3.2 Strength Reduction Factors
[1] Flexure w/ or w/o axial tension
The strength reduction factor, , will come into the calculation of the strength of the beam.
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Resistance Factors, ACI Sec 9.3.2 Strength Reduction Factors[2] Axial Tension = 0.90
[3] Axial Compression w or w/o flexure(a) Member w/ spiral reinforcement = 0.70(b) Other reinforcement members = 0.65
*(may increase for very small axial loads)
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Resistance Factors, ACI Sec 9.3.2 Strength Reduction Factors
[4] Shear and Torsion = 0.75
[5] Bearing on Concrete = 0.65
ACI Sec 9.3.4 factors for regions of high seismic risk
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Background Information for Designing Beam Sections
1. Location of Reinforcementlocate reinforcement where cracking occurs (tension region) Tensile stresses may be due to :
a ) Flexureb ) Axial Loadsc ) Shrinkage effects
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Background Information for Designing Beam Sections
2. Construction
formwork is expensive - try to reuse at several floors
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Background Information for Designing Beam Sections
3. Beam Depths
• ACI 318 - Table 9.5(a) min. h based on l (span) (slab & beams)
• Rule of thumb: hb (in) l (ft)
• Design for max. moment over a support to set depth of a continuous beam.
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Background Information for Designing Beam Sections
4. Concrete Cover
Cover = Dimension between the surface of the slab or beam and the reinforcement
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Background Information for Designing Beam Sections4. Concrete Cover
Why is cover needed?[a] Bonds reinforcement to concrete[b] Protect reinforcement against corrosion[c] Protect reinforcement from fire (over
heating causes strength loss)[d] Additional cover used in garages, factories,
etc. to account for abrasion and wear.
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Background Information for Designing Beam Sections
Minimum Cover Dimensions (ACI 318 Sec 7.7)
Sample values for cast in-place concrete
• Concrete cast against & exposed to earth - 3 in.
• Concrete (formed) exposed to earth & weather No. 6 to No. 18 bars - 2 in.No. 5 and smaller - 1.5 in
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Background Information for Designing Beam Sections
Minimum Cover Dimensions (ACI 318 Sec 7.7)
•Concrete not exposed to earth or weather- Slab, walls, joists
No. 14 and No. 18 bars - 1.5 inNo. 11 bar and smaller - 0.75 in
- Beams, Columns - 1.5 in
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Background Information for Designing Beam Sections
5.Bar Spacing Limits (ACI 318 Sec. 7.6)
- Minimum spacing of bars
- Maximum spacing of flexural reinforcement in walls & slabs
Max. space = smaller of
.in 18 t3
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Minimum Cover Dimension
Interior beam.
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Minimum Cover DimensionReinforcement bar arrangement for two layers.
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Minimum Cover Dimension
ACI 3.3.3
Nominal maximum aggregate size.
- 3/4 clear space - 1/3 slab depth - 1/5 narrowest dim.
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Example - Singly Reinforced BeamDesign a singly reinforced beam, which has a moment capacity, Mu = 225 k-ft, fc = 3 ksi, fy = 40 ksi and c/d = 0.275
Use a b = 12 in. and determine whether or not it is sufficient space for the chosen tension steel.
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Example - Singly Reinforced BeamFrom the calculation of Mn
u
n
c c
2c 1
2c
size
R
210.85 0.85 1
2 210.85 1 where, 2
10.85 12
aM C d
a a af ba d f bd dd d
a cf bd k k kd d
f k k bd
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Example - Singly Reinforced BeamSelect c/d =0.275 so that =0.9. Compute k’ and determine Ru
1
u c
0.85 0.275
0.23375
0.85 12
0.233750.85 3 ksi 0.23375 12
0.5264 ksi
ckd
kR f k
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Example - Singly Reinforced BeamCalculate the bd 2
U
2 N
u u
3
12 in225 k-ftft
0.9
5699 in0.5264 ksi
MMbdR R
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Example - Singly Reinforced BeamCalculate d, if b = 12 in.
32 25699 in 440.67 in 21.79 in.
12 ind d
Use d =22.5 in., so that h = 25 in.
0.275 0.275 22.5 in 6.1875 in.c d
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Example - Singly Reinforced BeamCalculate As for the beam
c 1s
y
2
0.85
0.85 3 ksi 12 in. 0.85 6.1875 in.40 ksi
4.02 in
f b cAf
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Example - Singly Reinforced BeamChose one layer of 4 #9 bars
Compute
2 2s 4 1.0 in 4.00 inA
2
s 4.00 in12.0 in 22.5 in
0.014815
Abd
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Example - Singly Reinforced BeamCalculate min for the beam
y
min minc
y
200 200 0.00540000
0.0053 3 3000 0.00411
40000
f
ff
0.014815 0.005 The beam is OK for the minimum
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Example - Singly Reinforced BeamCheck whether or not the bars will fit into the beam. The diameter of the #9 = 1.128 in.
b stirrup4 3 2 cover
4 1.128 in. 3 1.128 in. 2 1.5 in. 0.375 in.11.65 in
b d s d
So b =12 in. works.
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Example - Singly Reinforced BeamCheck the height of the beam.
Use h = 25 in.
bstirrupcover
21.128 in.
22.5 in. 1.5 in. 0.375 in.2
24.94 in
dh d d
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Example - Singly Reinforced BeamFind a
Find c
2s y
c
4.0 in 40 ksi0.85 0.85 3 ksi 12.0 in.5.23 in.
A fa
f b
1
5.23 in.0.85
6.15 in.
ac
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Example - Singly Reinforced BeamCheck the strain in the steel
Therefore, is 0.9
t cu22.5 in. 6.15 in. 0.003
c 6.15 in.0.00797 0.0056.15 in. 0.273322.5 in.
d c
cd
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Example - Singly Reinforced BeamCompute the Mn for the beam
Calculate Mu
N s y
2
25.23 in.4.0 in 40 ksi 22.5 in.
23186.6 k-in
aM A f d
U N
0.9 3186.6 k-in 2863.4 k-inM M
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Example - Singly Reinforced BeamCheck the beam Mu = 225 k-ft*12 in/ft =2700 k-in
Over-designed the beam by 6%
2863.4 2700 *100% 6.05%2700
6.15 in. 0.273322.5 in.
cd Use a smaller c/d
ratio