2. 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.
3. Design Criteria
• Serviceability
– Crack width limits
– Deflection limits
• Strength – must provide adequate
strength for all possible loads
4. 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
5. Strain and Stress in Concrete Beams
Strain
εs
T
cracked concrete
jd
cracked concrete
d
C
εc
εc
Stress
fs
T
εc=0.003
fs=fy
fs
cracked concrete
C
cracked concrete
c
fc
M = Tjd = Cjd
εs> εy
εs
fc
fc=f’c
where j is some fraction of the ‘effective depth’, d
T = Asfs at failure, T = AsFy
C = T = force in As’ and concrete
6. 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
A f = 0.85f’ ab + A’ f’
7. 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)
8. With No Compression Steel…
Asfy = 0.85f’cabw
a=
As f y
0.85 f 'c bw
a
jd = d −
2
a
j = 1−
2d
For most beams, 5/6 ≤ j ≤ 19/20
9. 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
11. Design Equations
As ≥
Mu
0.85 f y d
Mu
As ≥
0.75 f y d
As ≥
Mu
0.80 f y d
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.
12. 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.
f 'c
ρ b = 0.319β 1
fy
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 beam
bw + 16 hf
13. Balanced Reinforcement Ratio
ρb 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.
14. 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
simply
one end
both ends cantilever
beam
supported continuous continuous
construction
beams or
joists
l /16
l /18.5
l /21
l /8
one way
slabs
l /20
l /24
l /28
l /10
15. Preliminary Design Values
ρ ≤ 5/3 ρb
practical maximum reinforcement ratio
For typical d/bw ratios:
2.5M u
d ≥
f y ρb
3
17. Compression Reinforcement
If ρ > ρb must add compression reinforcement to prevent
failure due to crushing of concrete
A's ≥ ( As − bw dρ b )
8d '
f 'sb = 871 −
3d
fy
f 'sb
18. Crack Control
For 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.
19. 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 ft
24 ft
24 ft
24 ft
30 ft
30 ft
30 ft
