Transcript Lecture 4 - Fundamentals

Lecture 9 - Flexure
June 20, 2003
CVEN 444
Lecture Goals
Load Envelopes
Resistance Factors and Loads
Design of Singly Reinforced Rectangular
Beam


Unknown section dimensions
Known section dimensions
Moment
Envelopes
The moment envelope
curve defines the extreme
boundary values of
bending moment along the
beam due to critical
placements of design live
loading.
Fig. 10-10; MacGregor (1997)
Moment
Envelopes Example
Given following beam with a dead load of 1 k/ft and
live load 2 k/ft obtain the shear and bending moment
envelopes
Moment
Envelopes Example
Use a series of shear and bending moment diagrams
5
Wu = 1.2wD + 1.6wL
kips
4
3
2
1
0
0
5
10
15
20
25
30
35
40
(ft)
80
150
60
100
40
50
0
-20 0
5
10
15
20
25
30
35
40
k-ft
kips
20
0
-50 0
10
15
20
25
30
-100
-40
-150
-60
-200
-80
5
-250
ft
Shear Diagram
ft
Moment Diagram
35
40
Moment
Envelopes Example
Use a series of shear and bending moment diagrams
1.4
Wu = 1.2wD + 1.6wL
1.2
k/ft
1
(Dead Load Only)
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
30
35
40
ft
20
40
15
20
10
0
0
k-ft
kips
5
0
-5 0
5
10
15
20
25
30
35
5
10
15
20
25
30
-20
40
-40
-10
-60
-15
-20
-80
ft
Shear Diagram
ft
Moment Diagram
35
40
Moment
Envelopes Example
Use a series of shear and bending moment diagrams
k/ft
Wu = 1.2wD + 1.6wL
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
5
10
15
20
25
30
35
40
25
30
35
40
50
40
30
20
10
0
-10 0
-20
-30
-40
-50
-60
200
150
100
50
5
10
15
20
25
30
35
40
k-ft
kips
ft
0
-50 0
5
10
15
20
-100
-150
-200
ft
Shear Diagram
ft
Moment Diagram
Moment
Envelopes Example
The shear envelope
Shear Envelope
80
Minimum Shear
60
Maximum Shear
40
kips
20
0
-20 0
10
20
-40
-60
-80
ft
30
40
Moment
Envelopes Example
The moment envelope
Moment Envelope
200
k-ft
100
0
-100
0
5
10
15
20
25
30
-200
-300
ft
Minimum Moment
Maximum Moment
35
40
Flexural Design of Reinforced
Concrete Beams and Slab Sections
Analysis:
Design:
Analysis Versus Design:
Given a cross-section, fc , reinforcement
sizes, location, fy
compute
resistance or capacity
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
Ex.
 factored load
effect
 Mn  Mu
ACI Code Requirements for Strength
Design
 Mn  Mu
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 loads
D = Dead Loads
L = Live loads
W = 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
U
U
U
U
= 1.2(D + F + T) + 1.6(L + H) + 0.5 (Lr or S or R)
= 1.2D + 1.6 (Lr or S or R) + (L or 0.8W)
= 1.2D + 1.6 W + 1.0L + 0.5(Lr or S or R)
= 0.9 D + 1.6W +1.6H
= 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
(b) Other reinforcement members
 = 0.70
 = 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 Reinforcement
locate reinforcement where cracking occurs
(tension region) Tensile stresses may be due to :
a ) Flexure
b ) Axial Loads
c ) 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 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
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 in
No. 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
 3t
Max. space = smaller of 
18 in.
Minimum Cover Dimension
Interior beam.
Minimum Cover Dimension
Reinforcement 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
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 Reinforced
Beam
From the calculation of Mn
a

Mn  C  d - 
2

a

 a   1  a 
 0.85 f cba  d -   0.85 f cbd   d 1 -   
2

 d   2  d 
1 
a
c


2
 0.85 f cbd k  1 - k   where, k    1  
d
 2 
d
1  2

 0.85 f c k  1 - k   bd
 2  size
Ru
Example - Singly Reinforced
Beam
Select c/d =0.275 so that  =0.9. Compute k’ and
determine Ru
c

k   1    0.85  0.275
d
 0.23375
k 

Ru  0.85 f c k  1 - 
2

 0.23375 
 0.85  3 ksi  0.23375 1 
2


 0.5264 ksi
Example - Singly Reinforced
Beam
Calculate the bd 2
 MU 
  
MN 

2
bd 

Ru
Ru

 12 in  
 225 k-ft  ft  



0.9




  5699 in 3

0.5264 ksi
Example - Singly Reinforced
Beam
Calculate d, if b = 12 in.
3
5699 in
d 
 440.67 in 2  d  21.79 in.
12 in
2
Use d =22.5 in., so that h = 25 in.
c  0.275d  0.275 22.5 in   6.1875 in.
Example - Singly Reinforced
Beam
Calculate As for the beam
0.85 f cb1c
As 
fy
0.85  3 ksi 12 in. 0.85 6.1875 in.

40 ksi
 4.02 in 2
Example - Singly Reinforced
Beam
Chose one layer of 4 #9 bars
As  4 1.0 in 2   4.00 in 2
Compute r
As
4.00 in 2
r

bd 12.0 in  22.5 in 
 0.014815
Example - Singly Reinforced
Beam
Calculate rmin for the beam
r min
200
 200

 0.005

fy
40000


 r min  0.005
 3 f c  3 3000  0.00411
 fy
40000

0.014815  0.005
The beam is OK for
the minimum r
Example - Singly Reinforced
Beam
Check whether or not the bars will fit into the beam.
The diameter of the #9 = 1.128 in.
b  4d b  3s  2 cover  dstirrup 
 4 1.128 in.  3 1.128 in.  2 1.5 in.  0.375 in.
 11.65 in
So b =12 in. works.
Example - Singly Reinforced
Beam
Check the height of the beam.
db
h  d   cover  dstirrup 
2
1.128 in.

 22.5 in. 
 1.5 in.  0.375 in.
2
 24.94 in
Use h = 25 in.
Example - Singly Reinforced
Beam
Find a
a
As f y
0.85f cb

2
4.0
in

  40 ksi 
0.85  3 ksi 12.0 in.
 5.23 in.
Find c
a
5.23 in.
c

1
0.85
 6.15 in.
Example - Singly Reinforced
Beam
Check the strain in the steel
 d -c
 22.5 in. - 6.15 in. 
t  
  cu  
  0.003
6.15 in.
 c 


 0.00797  0.005
c 6.15 in.

 0.2733
d 22.5 in.
Therefore,  is 0.9
Example - Singly Reinforced
Beam
Compute the Mn for the beam
a

M N  As f y  d - 
2

5.23 in. 

  4.0 in   40 ksi   22.5 in. 
2 

 3186.6 k-in
Calculate Mu
2
MU  MN
 0.9  3186.6 k-in   2863.4 k-in
Example - Singly Reinforced
Beam
Check the beam Mu = 225 k-ft*12 in/ft =2700 k-in
2863.4 - 2700
*100%  6.05%
2700
Over-designed the beam by 6%
c
6.15 in.

 0.2733
d  22.5 in.
Use a smaller c/d
ratio