Ch 3-a.pdf

REINFORCED CONCRETE (1)
(CE411)
Chapter 3-a
Flexural Analysis and Design of
Beams
Instructor:
Eng. Abdallah Odeibat
Civil Engineer, Structures , M.Sc.
1

TYPICAL SINGLY REINFORCED CONCRETE
RECTANGULAR BEAM
6

FLEXURAL STRESS
12
The beam is a structural member used
to support the internal moments and
shears. It would be called a beam-
column if a compressive force existed.
C = T
M = C*(jd)
= T*(jd)

ASSUMPTIONS
 Plane sections before bending remain plane under load
 Strain distribution is linear in both concrete & steel and is directly
proportional to the distance from N.A.
 Concrete in the tension zone is neglected in the flexural analysis &
design computation
 Maximum concrete strain = 0.003 (in compression)
13
εc=0.003
εs = fy / Es
h d
c
0.85fc’
a a/2
d-a/2
b
C
T

RC BEAM BEHAVIOR
 Assume that a small transverse load is placed on a
concrete beam with tensile reinforcing and that load is
gradually increased in magnitude until the beam
fails. As this happen the beam will go through three
behavior stages before collapse:
1. Uncracked Concrete Stage
2. Concrete Cracked–Elastic Stresses Stage
3. Beam Failure—Ultimate-Strength Stage
14

 For the beam cross section shown, Assuming the concrete is
uncracked, and neglecting tension steel; Calculate
i. the concrete stress on the compression face,
ii. the concrete stress on the tension face,
iii. and the steel stress in tension.
(fc’ = 28 MPa , normal weight concrete,
a) For bending moment of 70 kN.m
b) For the Cracking bending moment.
EXAMPLE
17

RC BEAM BEHAVIOR
21
c
a 1



FLEXURE EQUATIONS
22
bd
As



EXAMPLE
 Determine nominal moment capacity and design
(ultimate) moment capacity of R.C. rectangular
section.
Given: fc’ = 35 MPa
fy= 400 MPa
b = 350 mm
d = 550 mm
As = 4 ø 32; Area = 3217 mm2
24

SOLUTION
 Assume stress in steel reaches yield stress :
T=C
3217 (400) = 0.85 (35)(350)x a
Get a= 123.6 mm a/2 = 61.8 mm
Mn = 1286.8 (550-61.8) x 10-3 = 628.216 kN.m
Check if steel has yielded
β1 = 0.85 – (35-28)(0.05)/7=0.8
c = a/ β1 = 123.6/0.8 =154.5 mm
25

SOLUTION (CONT’)
 From strain distribution,
εs = 0.00768 > εy
Knowing that εy = fy / Es = 400 /200,000 =0.002
So assumption is acceptable;
Mn = 628.216 kN.m
Mu = øMn = 0.9(628.216) kN.m = 565.4 kN.m
26

THE BALANCED SECTION
27
εc=0.003
εs = fy / Es
h d
c
0.85fc’
a a/2
d-a/2
b
C
T
ACI 10.3.2 — Balanced strain conditions exist at a
cross section when tension reinforcement reaches the
strain corresponding to fy just as concrete in
compression reaches its assumed ultimate strain
of 0.003.

THE BALANCED SECTION
28
εc=0.003
εs = fy / Es
h d
c
0.85fc’
a a/2
d-a/2
b
C
T
c
a 1

 bd
As



MAXIMUM RATIO OF STEEL
(FOR ϵT =0.004, Φ = 0.817)
 ACI code limits the maximum ratio of steel
 If we use compression block depth,
29

 If (under-reinforced) (ductile failure)
30

 If (over-reinforced) (brittle failure)
In this case, we have 2 unknowns : fs , a
Solution of this equation will be done by
trial and error; or by quadratic
equation in term of c.
 ACI code recommends to use ρmax .
31

TENSION CONTROLLED SECTION
(FOR ϵT =0.005, Φ = 0.9)
 ACI code limits the maximum ratio of steel
 If we use compression block depth,
32

EXAMPLE
 Determine ultimate Moment Capacity of
R.C. rectangular section.
Given: fc’ = 25 MPa
fy= 280 MPa
b = 200 mm
d = 300 mm
If
a) As = 4 ø 18 mm;
b) As = 4 ø 22 mm(HW)
c) As = 4 ø 25 mm
d) As = 4 ø 28 mm(HW)
33

CONCRETE DIMENSIONS AND REINFORCEMENT
DESIGN
38
d
b
As .
.



EXAMPLE
 Find the concrete cross section and the
steel area required for a simply supported
rectangular beam with a span of 4.5 m
that to carry a service dead load of
20kN/m and service live load of 31kN/m ,
material strength are fc’ = 28MPa and
fy=420 MPa ?
 Solution:
39

EXAMPLE (CONT’)
 If we want to minimize the concrete section, it is
desirable to use
40

EXAMPLE (CONT’)
41
 Use b = 250 mm then d =342 mm
or
Use b = 200 mm then d = 383mm
Selection of any alternative
is the designer option

EXAMPLE (CONT’)
42
 If we select b = 250 mm and d =342 mm
 Assuming 65 mm concrete cover
 h= d+cover = 342+65=407 mm
 h and b are often rounded up to the
nearest multiple of 50 mm.
 Use h=450 mm
 The actual d = h – cover = 450-65=385 mm

EXAMPLE (CONT’)
 Required As based on the selected dimensions :
43
For Mu = 186.3 *10^6 N.mm, fc’ = 28 MPa , fy = 420 MPa
b= 250 mm , d= 385 mm , Ф =0.9 (to be checked)
ρ = 0.015398 , check
As = ρ b d =1482 mm2

44
Select diameter &
no. of bars to have
As >1482 mm2
Use 4 Ф 22 mm

EXAMPLE (CONT’)
 Use 4 Ф 22 mm ….As provided = 1521 mm2
 Check spacing of bars : it is required that :
…………………………………………………………….
45
107.4 mm
=107.4 mm / 0.85 =126.4 mm

EXAMPLE (CONT’)
46
The Previous
calculations assuming
ø=0.9 are correct

EXAMPLE (CONT’)
47
Mu = 190.48 kN.m > 186.3 kN.m
……...ok

EXAMPLE (CONT’)
 A more economical and practical design can be
obtained using
HW : Redesign the previous
beam using
48

EXAMPLE
 For concrete section of b= 250 mm, d=435 mm
and h=500mm, find the steel area required to
resist Mu = 150 kN.m, , material properties are
fc’ = 28MPa and fy=420 MPa ?
Solution:
.
49

EXAMPLE (CONT’)
50
ρ = 0.00913 , check
As = ρ b d = 992.6 mm2
For Mu = 150 *10^6 N.mm, fc’ = 28 MPa , fy = 420 MPa
b= 250 mm , d= 435 mm , Ф =0.9 (to be checked)

EXAMPLE (CONT’)
51
Select diameter & no. of bars to have As >992.6mm2
Use 4 Ф 18mm …..As provided =1018 mm2
…………………………………………………………….
71.86 mm
= 71.86 mm / 0.85 =84.54 mm

EXAMPLE (CONT’)
52
The Previous
calculations assuming
ø=0.9 are correct
Mu = 153.56 kN.m > 150 kN.m ……...ok

PRACTICAL CONSIDERATION (COVER)
53

PRACTICAL CONSIDERATION (DIMENSIONING)
 h and b are almost rounded up to the nearest
multiple of 25 mm, and often to the next multiple
of 50 mm.
 b and d are economically and practically chosen
with
d = (2 to 3)*b.
54

PRACTICAL CONSIDERATION (BAR SPACING)
55

EXAMPLE
 Determine the nominal moment strength Mn,
if fc’=28MPa and fy = 420 MPa.
56

HW
 Determine the nominal moment strength Mn, if
fc’=30 MPa and fy = 350 MPa.
57

EXAMPLE
 Determine the nominal flexural strength Mn and
the design flexural strength φMn for the shown
beam section. fc‘ =21 MPa fy = 420 MPa.
58

EXAMPLE
 Determine the nominal flexural strength Mn and
the design flexural strength φMn for the shown
beam section. fc‘ =28 Mpa fy = 420 MPa.
59

EXAMPLE
 Design the shown cantilevered beam , fc’=28MPa
and fy = 420 MPa. Use ρ ≈ 0.5 ρmax
60

Ch 3-a.pdf

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