Engineering
CIV3284
Design of Concrete & Masonry
Structures
Dr. Ye Lu
Lecture #3 (week 3)
Previously on this topic…
 Strength Check
Rd >= Ed (Design capacity >= Design action effects)
Rd =  Ru
Ed =1.2 G + 1.5 Q
 Serviceability Check
Long term load, Ed = G + ψl Q
Short term load, Ed = G + ψs Q
 Simplified Method for Continuous Beams
Moment = Coef. * Fd Ln2
Today’s Lecture (Chapters 3)
Serviceability Check for RC Beams
• Deflections
• Crack Control
Some Notations for Beam Cross-section
bef
Flange
d
tf
Web
D
Ast
bw
bef : Effective width of flange
bw : Width of web
Ast : Total area of tensile reinforcement
b =bef = bw
tf : Thickness of flange
D: Overall depth
d: Effective depth (from extreme compression fibre to resultant tensile force)
Review of Beam in Bending


y
N.A.


Stress and strain distribution
For elastic range:  = E
Deformation (strain) compatibility
 
My
Ig
y: Distance to neutral axis (N.A.)
Review of Beam in Bending
Which one is correct for S = ?
b
S: First Moment of Area
A.
b*h*d
B.
b*h*d/2
C.
b*h*d/4
h
h/2
d
N.A.
Review of Beam in Bending
b
Which one is correct for I = ?
I: Second Moment of Area
A.
b*h3/8
B.
b*h3/12
C.
b*h3/6
N.A.
h/2
h
What if N.A. is not at the middle of the section?
b
I = b*h3/12+b*h*d2
h
h/2
d
N.A.
Serviceability Check
• Deflections
Short term deflection – instant deflection by loads
Long term deflection – additional sustained-load deflection by creep
UDL: w
EI
∆=?
Simply Supported Beam
Mid-Span Deflection:
Different materials with different elastic moduli (Es, Ec)?
Cracked and Uncracked Sections
a)
uncracked
b)
cracked
N.A.
Different sections with different I?
Section Transformation (Uncracked)
, the modular ratio (steel to concrete)
Section Transformation (Fully Cracked)
b
b
dn
N.A.
D
d
Ast
n Ast
, the modular ratio (steel to concrete)
Uncracked Elastic Section (Ig)
c
c
b
dn
dn
N.A.
d
D
d-dn
st
st
Ast
n Ast
(n-1) Ast
ct
ct
Determine the depth of N.A. (dn)
b
Sabove N.A.=Sbelow N.A.
bd n 
dn 
D  d n   n  1A  d  d 
dn
 b D  d n  
st
n
2
2
D
 ( n  1) A st d
2
bD  ( n  1) A st
bD 
h
h/2
d
N.A.
S=area*d=b*h*d
Uncracked Elastic Section (Ig)
c
c
b
dn
dn
N.A.
d
D
d-dn
st
st
Ast
n Ast
(n-1) Ast
ct
ct
b
Calculate Ig
h
h/2
2
Ig 
1
D

bD 3  bD    d n   ( n  1) Ast  ( d  d n ) 2
12
2

d
N.A.
I = b*h3/12+b*h*d2
c
c
b
dn
dn
N.A.
d
D
d-dn
st
st
Ast
n Ast
(n-1) Ast
ct
ct
Calculate stress
Stress in concrete:
Stress in steel:
c 
Mdn
Ig
 ct 
M ( D  dn )
Ig
 c .at depth d 
M (d  d n )
Ig
st d  dn st Es d  dn
d  dn
d  dn M(D  dn )

 
 st  n
ct st  n
E
D
d
D
d
D  dn
Ig



ct D  dn
n
ct
c
n
 st  n
M (d  d n )
 n  c . at depth d
Ig
Fully Cracked Elastic Section (Icr)
b
b
c
c
kd/3
C
dn=kd
N.A.
d
D
jd=d-kd/3
st
st
Ast
T = Ast st
n Ast
b
Determine the depth of N.A. (dn)
Sabove N.A.=Sbelow N.A.
b dn 
1
d n  n Ast  ( d  d n )
2
Define: dn = kd ; p = Ast/bd

k
np 2  2 np
 np
h
h/2
d
N.A.
S=area*d=b*h*d
Fully Cracked Elastic Section (Icr)
b
b
c
c
kd/3
C
dn=kd
N.A.
d
D
jd=d-kd/3
st
st
Ast
T = Ast st
n Ast
b
Calculate Icr
1
d
3
bd n  bd n ( n ) 2  n Ast  ( d  d n ) 2
12
2
1
3
 bd n  n Ast  ( d  d n ) 2
3
I cr 
h
h/2
d
N.A.
I = b*h3/12+b*h*d2
Fully Cracked Elastic Section (Icr)
b
b
c
c
kd/3
C
dn=kd
N.A.
d
D
jd=d-kd/3
st
Ast
n Ast
st
T = Ast st
Calculate stress
Stress in concrete:
Stress in steel:
c 
M dn
I cr
E d  dn
d  dn
M (d  d n )
 st d  d n


 st  s
  st  n
c n
 c Ec d n
c
dn
dn
I cr
Doubly Reinforced Fully Cracked Section
c 
(n-1) Asc
b
 sc  n
M (d n  d sc )
I cr
 st  n  c  n
M (d  d n )
I cr
dsc
Asc
M dn
I cr
dn=kd
d
D
Ast
n Ast
Determine the depth of N.A. (dn)
Sabove N.A.=Sbelow N.A. b d  1 d  ( n  1) A  ( d  d )  n A  ( d  d )  d
n
n
sc
n
sc
st
n
n
2
Calculate Icr
I cr 
d
1
3
bd n  bd n  ( n ) 2  n Ast  ( d  d n ) 2  ( n  1) Asc  ( d n  d sc ) 2
12
2
Stiffness (EI) Change with Bending Moment
EIg
EIcr
At service loads (points C1 and C2), the
effective EIef values are between EIg and EIcr
Icr<Ief<Ig
Effective Ief (AS3600 P129)
g
Ig,
(not consider in this unit)
Ig,
0.6Ig,
Short term load: w = G + ψs Q
Mcr.t: Cracking bending moment
b
b
dn=kd
P/Ag, Pe: Prestress contribution (not consider in this unit), we have:
D
d
Mcr.t = Z(f ’ct,f - cs)
Ast
=Ig / (D-dn)
f cf'  0.6 f c '
Es: Young’s modulus of steel reinforcement
Ast / bw d
Clause 3.1.7
n Ast
Short Term Deflection (Mid-Span)
Short term load, w = G + ψs Q
Simply supported beam:
Span in continuous beam:
Continuous beam
Mm
AS3600 p128 or Textbook p108
For simplicity, in practice class question:
Ief = Ief, at mid-span
Long Term Deflection (AS3600 P130)
• ∆l = kcs ∆s.sus
• ∆l : Long-term deflection
• ∆s.sus: immediate deflection due to sustained
load w=(G + ψl Q)
Long term deflection factor
• ∆tot = ∆l + ∆s
• ∆tot ≤ ∆allow (allowable deflection)
• ∆tot/Lef ≤ limit
Table 2.3.2
Deemed to Comply Span-to-Depth Ratios
for Deflection Control (AS3600 P130)
bef
≤
d
tf
D
Ast
bw
Ec: the concrete elastic modulus, bef: the effective width of the flange (Textbook P141)
≤
Fd,ef = effective design load per unit length, taken as:
= (1.0 + kcs) G + (ψs + kcs ψl) Q for total deflection
or
= kcs G + (ψs + kcs ψl) Q for additional deflection, which
occurs after the addition or attachment of partitions.
kcs = Long term deflection factor
∆ / Lef = deflection limit
Crack Control of Beams (Clause 8.6)
1) Check reinforcement ratio
OR
A
D
p  st  p min   b  
bw d
d 
2
f ct' , f
f sy
bef
Ds=tf
bef: the effective width of the flange
D
fsy: the characteristic yield strength of steel reinforcement
f’ct,f: the characteristic tensile strength of concrete
Ds: depth of a slab
f cf'  0.6 f c '
bw
Crack Control of Beams
2) Check concrete cover and bar spacing
c
c
bw
s
c <=100 mm
s <=300 mm
Crack Control of Beams
3) Check tension stress in bars
C
jd
Ms = T j d = Ast fst jd
Ms is determined by short term load (G + 0.7 Q )
Assume j  0.9
fst = Ms / (Ast jd) < the larger stress from tables 8.6.2.2 (a) and (b)
3) Check tension stress in bars
w’max: Characteristic maximum crack width (Clause 8.6.1)
Serviceability Design - Summary
•
•
•
•
Step 1: Locate depth of neutral axis (dn)
Step 2: Define Icr and Ig
Step 3: Define Ief with Mcr
Step 4: Determine serviceability load combinations
(short term and long term)
• Step 5: Check serviceability
Serviceability Design - Summary
 Deflection < Limit deflection
1. calculate ∆tot = ∆l + ∆s ≤ ∆allow
2. check span-to-depth ratios Lef/d
 Crack control
check 1) reinforcement ratio
2) concrete cover and bar spacing
3) steel stresses