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Lecture 9 - Flexure
June 20, 2003CVEN 444
Lecture GoalsLecture Goals
Load EnvelopesResistance Factors and LoadsDesign of Singly Reinforced Rectangular Beam Unknown section dimensions Known section dimensions
MomentMomentEnvelopesEnvelopes
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.
MomentMomentEnvelopes ExampleEnvelopes ExampleGiven following beam with a dead load of 1 k/ft and live load 2 k/ft obtain the shear and bending moment envelopes
MomentMomentEnvelopes ExampleEnvelopes 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)
kip
s
-80
-60
-40
-20
0
20
40
60
80
0 5 10 15 20 25 30 35 40
ft
kip
s
-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
MomentMomentEnvelopes ExampleEnvelopes 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
kip
s
-80
-60
-40
-20
0
20
40
0 5 10 15 20 25 30 35 40
ft
k-ft
(Dead Load Only)
MomentMomentEnvelopes ExampleEnvelopes 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
0102030
4050
0 5 10 15 20 25 30 35 40
ft
kip
s
-200
-150
-100
-50
0
50
100
150
200
0 5 10 15 20 25 30 35 40
ft
k-ft
MomentMomentEnvelopes ExampleEnvelopes ExampleThe shear envelope
MomentMomentEnvelopes ExampleEnvelopes 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
Flexural Design of Reinforced Flexural Design of Reinforced Concrete Beams and Slab Concrete Beams and Slab SectionsSections
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.)
Flexural Design of Reinforced Flexural Design of Reinforced Concrete Beams and Slab Concrete Beams and Slab SectionsSections
ACI Code Requirements for Strength Design
Basic Equation: factored resistance factored load effect
Ex.
un M M
ACI Code Requirements for ACI Code Requirements for Strength DesignStrength 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.
Flexural Design of Reinforced Flexural Design of Reinforced Concrete Beams and Slab Concrete Beams and Slab SectionsSections
Required Strength (ACI 318, sec 9.2)
U = Required Strength to resist factored loads
D = Dead Loads
L = Live loads
W = Wind Loads E = Earthquake Loads
Flexural Design of Reinforced Flexural Design of Reinforced Concrete Beams and Slab Concrete Beams and Slab SectionsSections
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.
Factored Load Factored Load CombinationsCombinations
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 +1.6H
U = 0.9 D + 1.0E +1.6H
Resistance Factors, Resistance Factors, ACI ACI Sec 9.3.2 Strength Reduction Sec 9.3.2 Strength Reduction FactorsFactors
[1] Flexure w/ or w/o axial tension
The strength reduction factor, , will come into the calculation of the strength of the beam.
Resistance Factors, Resistance Factors, ACI ACI Sec 9.3.2 Strength Reduction Sec 9.3.2 Strength Reduction FactorsFactors
[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)
Resistance Factors, Resistance Factors, ACI ACI Sec 9.3.2 Strength Reduction Sec 9.3.2 Strength Reduction FactorsFactors
[4] Shear and Torsion = 0.75
[5] Bearing on Concrete = 0.65
ACI Sec 9.3.4 factors for regions of high seismic risk
Background Information for Background Information for Designing Beam SectionsDesigning Beam Sections
1. Location of Reinforcement locate reinforcement where cracking occurs (tension region) Tensile stresses may be due to :
a ) Flexureb ) Axial Loadsc ) Shrinkage effects
Background Information for Background Information for Designing Beam SectionsDesigning Beam Sections
2. Construction
formwork is expensive - try to reuse at several floors
Background Information for Background Information for Designing Beam SectionsDesigning 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.
Background Information for Background Information for Designing Beam SectionsDesigning Beam Sections
4. Concrete Cover
Cover = Dimension between the surface of the slab or beam and the reinforcement
Background Information for Background Information for Designing Beam SectionsDesigning Beam Sections
4. 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.
Background Information for Background Information for Designing Beam SectionsDesigning 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
Background Information for Background Information for Designing Beam SectionsDesigning 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
Background Information for Background Information for Designing Beam SectionsDesigning 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
Minimum Cover DimensionMinimum Cover Dimension
Interior beam.
Minimum Cover DimensionMinimum Cover Dimension
Reinforcement bar arrangement for two layers.
Minimum Cover DimensionMinimum Cover Dimension
ACI 3.3.3
Nominal maximum aggregate size.
- 3/4 clear space - 1/3 slab depth - 1/5 narrowest dim.
Example - Singly Example - Singly Reinforced BeamReinforced Beam
Design 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.
Example - Singly Example - Singly Reinforced BeamReinforced BeamFrom the calculation of Mn
u
n
c c
2c 1
2c
size
R
2
10.85 0.85 1
2 2
10.85 1 where,
2
10.85 1
2
aM C d
a a af ba d f bd d
d d
a cf bd k k k
d d
f k k bd
Example - Singly Example - Singly Reinforced BeamReinforced 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 1
2
0.5264 ksi
ck
d
kR f k
Example - Singly Example - Singly Reinforced BeamReinforced BeamCalculate the bd 2
U
2 N
u u
3
12 in225 k-ft
ft0.9
5699 in0.5264 ksi
M
Mbd
R R
Example - Singly Example - Singly Reinforced BeamReinforced Beam
Calculate d, if b = 12 in.
32 25699 in
440.67 in 21.79 in.12 in
d d
Use d =22.5 in., so that h = 25 in.
0.275 0.275 22.5 in 6.1875 in.c d
Example - Singly Example - Singly Reinforced BeamReinforced 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 cA
f
Example - Singly Example - Singly Reinforced BeamReinforced BeamChose one layer of 4 #9 bars
Compute
2 2s 4 1.0 in 4.00 inA
2
s 4.00 in
12.0 in 22.5 in
0.014815
A
bd
Example - Singly Example - Singly Reinforced BeamReinforced Beam
Calculate min for the beam
y
min min
c
y
200 2000.005
400000.005
3 3 30000.00411
40000
f
f
f
0.014815 0.005 The beam is OK for the minimum
Example - Singly Example - Singly Reinforced BeamReinforced 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.
Example - Singly Example - Singly Reinforced BeamReinforced 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
Example - Singly Example - Singly Reinforced BeamReinforced BeamFind a
Find c
2s y
c
4.0 in 40 ksi
0.85 0.85 3 ksi 12.0 in.
5.23 in.
A fa
f b
1
5.23 in.
0.85
6.15 in.
ac
Example - Singly Example - Singly Reinforced BeamReinforced BeamCheck the strain in the steel
Therefore, is 0.9
t cu
22.5 in. 6.15 in.0.003
c 6.15 in.
0.00797 0.005
6.15 in.0.2733
22.5 in.
d c
c
d
Example - Singly Example - Singly Reinforced BeamReinforced BeamCompute the Mn for the beam
Calculate Mu
N s y
2
2
5.23 in.4.0 in 40 ksi 22.5 in.
2
3186.6 k-in
aM A f d
U N
0.9 3186.6 k-in 2863.4 k-in
M M
Example - Singly Example - Singly Reinforced BeamReinforced 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.
c
d Use a smaller c/d
ratio