concrete flexural design
TRANSCRIPT
Reinforced Concrete Flexural Members
Reinforced Concrete Flexural Members
Concrete is by nature a continuous material
Once concrete reaches its tensile strength ~400 psi, concrete will crack.Stress in steel will be ~ 4000 psi.
Design Criteria
• Serviceability– Crack width limits– Deflection limits
• Strength – must provide adequate strength for all possible loads
As area of steel in tension zone
As’area of steel in compression zone
d distance from center of tension reinforcement to outermost point in compression
d’ distance from center of compression reinforcement to outermost point in compression
Strain and Stress in Concrete Beams
T
C
fs
fc
cracked concrete
fs
fc
fs=fy
cracked concrete
fc=f’c
c
Stress
T
C
εs
εc
cracked concrete
ε s
εc
εs> εy
cracked concrete
εc=0.003
Strain
djd
M = Tjd = Cjd where j is some fraction of the ‘effective depth’, d
T = Asfs at failure, T = AsFy C = T = force in As’ and concrete
Stress in Concrete at Ultimate
ACI 318 approximates the stress distribution in concrete as a rectangle 0.85f’c wide and ‘a’ high, where a = β1c.
Cconcrete = 0.85f’cabw
Csteel = A’s f’s
Asfy = 0.85f’cabw + A’s f’s
Definitions
• β1 shall be taken as 0.85 for concrete strengths f’c up to and including 4000 psi. For strengths above 4000 psi, β1 shall be reduced continuously at a rate of 0.05 for each 1000 psi of strength above 4000 psi, but β1
shall not be taken less than 0.65.
• bw = width of web
• f’s = stress in compression reinforcement (possibly fy)
With No Compression Steel…
Asfy = 0.85f’cabw
wc
ys
bf
fAa
'85.0
d
aj
21
2
adjd
For most beams, 5/6 ≤ j ≤ 19/20
Moment Equation
recall, M = Tjd = Cjd and T = AsFy
φ = 0.9 for flexure
Mu ≤ ΦMn=0.9Tjd = 0.9Asfyjd
substituting 5/6 ≤ j ≤ 19/20
0.75Asfyd ≤ Mu ≤ 0.85Asfyd
Reinforcement Ratio
db
A
w
s
db
A
w
s'' Compression reinforcement ratio
Reinforcement ratio for beams
Design Equations
df
MA
y
us 85.0
df
MA
y
us 75.0
df
MA
y
us 80.0
For positive moment sections of T-shaped beams, and for negative moment sections of beams or slabs where ρ ≤ ⅓ ρb.
For negative moment sections where ρ ≥ ⅔ ρb
and for positive moment sections without a T flange and with ρ ≥ ⅔ ρb.
For intermediate cases where ⅓ ρb < ρ < ⅔ ρb
regardless of the direction of bending.
Balanced Reinforcement Ratio, ρb
To insure that steel tension reinforcement reaches a strain εs ≥ fy/Es before concrete reaches ε = 0.003 (steel yields before concrete crushes) the reinforcement ratio must be less than ρb. Where ρb is the balanced reinforcement ratio or the reinforcement ratio at which the steel will yield and the concrete will crush simultaneously.
y
cb f
f '319.0 1 For rectangular compression zones (negative bending)
For positive bending (T-shaped compression zone) reinforcement ratio is usually very low (b very large)
b = effective flange width, least of:bw + half distance to the adjoining parallel beam on each side of the web ¼ the span length of the beambw + 16 hf
Balanced Reinforcement Ratioρb for rectangular compression zone
Fy, ksi f’c = 3000 psi 4000 5000 6000
40 0.0203 0.0271 0.0319 0.0359
50 0.0163 0.0217 0.0255 0.0287
60 0.0136 0.0181 0.0213 0.0239
Note: if ρ > ρb can add compression reinforcement to prevent failure due to crushing of concrete.
Depth of Beam for Preliminary Design
The ACI code prescribes minimum values of h, height of beam, for which deflection calculations are not required.
Minimum values of h to avoid deflection calculations
Type of beam
construction
simply supported
one end continuous
both ends continuous
cantilever
beams or joists
l /16 l /18.5 l /21 l /8
one way slabs
l /20 l /24 l /28 l /10
Preliminary Design Values
ρ ≤ 5/3 ρb practical maximum reinforcement ratio
For typical d/bw ratios:
by
u
f
Md
5.23
Beam AnalysisACI 318 Approximate Moments and Shears
Compression Reinforcement
If ρ > ρb must add compression reinforcement to prevent
failure due to crushing of concrete
sb
ybwss f
fdbAA
')('
d
df sb 3
'8187'
Crack ControlFor serviceability, crack widths, in tension zones, must be limited.
ACI 318 requires the tension reinforcement in the flanges of T-beams be distributed over an effective flange width, b, or a width equal to 1/10 span, whichever is smaller. If the effective flange width exceeds 1/10 the span, additional reinforcement shall be provided in the outer portions of the flange.
Flexure Design Example p. 21 notes
The partial office building floor plan shown had beams spanning 30 ft and girders spanning 24 ft. Design the slab, beams, and girders to support a live load of 80 psf and a dead weight of 15 psf in addition to the self weight of the structure. Use grade 60 reinforcing steel and 4000 psi concrete.
30 ft30 ft30 ft30 ft
24 ft
24 ft
24 ft
Reinforcing Steel