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Lecture 9 - Flexure June 20, 2003 CVEN 444
Lecture 9 - Flexure
June 20, 2003CVEN 444
Lecture Goals
Load EnvelopesResistance Factors and LoadsDesign of Singly Reinforced Rectangular Beam Unknown section dimensions Known section dimensions
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.
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
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
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)
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
MomentEnvelopes ExampleThe shear envelope
Shear Envelope
-80-60-40-20
020406080
0 10 20 30 40
ft
kips
Minimum ShearMaximum Shear
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
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.)
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
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.
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
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.
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
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.
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)
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
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
Background Information for Designing Beam Sections
2. Construction
formwork is expensive - try to reuse at several floors
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.
Background Information for Designing Beam Sections
4. Concrete Cover
Cover = Dimension between the surface of the slab or beam and the reinforcement
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.
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
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
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
Minimum Cover Dimension
Interior beam.
Minimum Cover DimensionReinforcement bar arrangement for two layers.
Minimum Cover Dimension
ACI 3.3.3
Nominal maximum aggregate size.
- 3/4 clear space - 1/3 slab depth - 1/5 narrowest dim.
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.
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
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
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
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
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
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
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
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.
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
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
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
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
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
