SINGLY
REINFORCED
BEAM
CE 368 PRINCIPLES OF REINFORCED/PRESTRESSED
CONCRETE
Two Methods of Design:
1. WSD - Working Stress Design
2.
USD - Ultimate Strength Design
Concrete:
Ec
.
W1.54,2_8
w = unit weight of concrete,
,/f'C
kN/m?3
E. = 4700,/f'c
E. = modulus of elasticity of concrete,
MPa
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CE 368 PRINCIPLES
S
NSNS
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SN
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SN
A
OF REINFORCED/PRESTRESSED
f'c = cylinder strength of concrete,
NN
NS
SN
CONCRETE
MPa
= specified compressive strength of concrete, MPa
fc = allowable compressive stress of concrete, MPa
NN
NSNS
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CE 368 PRINCIPLES OF REINFORCED/PRESTRESSED
CONCRETE
Steel:
Es = 200,000 MPa (SI)
=29 x 103 ksi (English)
fs = allowable tensile stress of steel
fy, Fy = yield strength of steel
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CE 368 PRINCIPLES
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OF REINFORCED/PRESTRESSED
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Problem
1. Design — determining the size given the load
2. Investigation — determining the load that a given
section could carry
Two modes of failure:
1. Crushing of concrete — 0.003m/m
2. Yielding of steel — fs > fy
or mm/mm
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CE 368 PRINCIPLES
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NSNS
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SN
A
OF REINFORCED/PRESTRESSED
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State of Reinforcement
1. Under-reinforced — failure is initiated by yielding of steel.
Ductile mode of failure. Beam can undergo large deflections
before final collapse. Ideal design.
2. Over-reinforced — failure is initiated by crushing of concrete.
Failure is sudden without warning. Not advisable.
3. Balanced design - failure is initiated by simultaneously
crushing of concrete and yielding of steel.
D,;Q\)‘E OF 7-6(:%
HW
&
°/v«‘
=
o
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S
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CE 368 PRINCIPLES
Bar
N
S
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OF REINFORCED/PRESTRESSED
AN
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SN
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CONCRETE
Diameter
Philippine Standard
omm
10mm
12mm
l6mm
20mm
25mm
28mm
32mm
36mm
ASTM Designation
# 2
#3
# 4
#5
# 6
#8
#9
# 10
#11
2/8” or 4”)
3/87)
4/87)
5/8%)
6/8” or %)
8/8” or 17)
9/8”0r11/87
10/8” or 1 ¥4”)
11/8”or
1 3/87)
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S
S
S
S
CE 368 PRINCIPLES
N
S
SN
NSNS
OF REINFORCED/PRESTRESSED
Consider the beam
AN
NS
CONCRETE
shown
w
A
R1
R2
SN
NN
NSNS
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NN
CE 368 PRINCIPLES
OF REINFORCED/PRESTRESSED
CONCRETE
Consider the left section a-a
v
Applied moment, My :
X
My = Ri(x) — w(x) (E)
External
Resisting moment, My, :
Internal
To be safe and economical , Mp = My
CE 368 PRINCIPLES
FIGURE
OF REINFORCED/PRESTRESSED
34
CONCRETE
strain variation at
slrese variation at
ultimate load condition
ultimate load condition
Monlincar strcss distribution at ultimate conditions,
CE 368 PRINCIPLES
OF REINFORCED/PRESTRESSED
CONCRETE
0.85/"
concrete
compressive
e
¢
=a
C= 085ahb
stress
-3
1,
{a) Beam
FIGURE
211
{b) Actual compression
stress variation
u
T=Af,
(c) Assumed compression
stress variation
Compression and tension couple at nominal moment.
CE 368 PRINCIPLES
ACI
OF REINFORCED/PRESTRESSED
CONCRETE
Code
In SI units, 8, is to be taken equal to 0.85 for concrete strengths up to and including
30 MPa. For strengths above 30 MPa, B, is to be reduced continuously at a rate of 0.05
for each 7 MPa of strength in excess of 30 MPa but shall not be taken less than 0.65.
For concretes with f = 30 MPa, 8, can be determined with the following expression:
B, = 0.85 — 0.008 (f/ — 30 MPa) > 0.65
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CE 368 PRINCIPLES
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OF REINFORCED/PRESTRESSED
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NSCP 2015 Code Table 422.2.2.4.3
Values of f;for Equivalent Rectangular Concrete Stress Distribution
for 17MPa < f'. < 28MPa; P, = 0.85
for 28MPa < f'.
< 55MPa;
for f'.> 55MPa;
B; = 0.65
p; =0.85—
0.05(f", — 28)
7
NN
CE 368 PRINCIPLES
OF REINFORCED/PRESTRESSED
CONCRETE
0.85f
a2
€ =085 ab
FIGURE
3.3
Beam
internal forces at ultimate conditions.
Concrete is assumed to crush at a strain of about 0.003 (which is a little
conservative for most concretes) and the steel to yield at fy
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M,=T (d — %)
M,, = nominal or unfactored moment
capacity
a
M,, = Asfy (d — E)
M, = ultimate or factored moment
CE 368 PRINCIPLES
N
S
SN
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AN
OF REINFORCED/PRESTRESSED
NS
SN
NN
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CONCRETE
Resisting moment of steel
Mu
M,
=
qj Mn
¢ = strength reduction factor
a
¢ = 0.90 for bending
= pAsfy (d - E)
¢ = 0.75 for shearing
capacity
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S
CE 368 PRINCIPLES
N
S
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NSNS
OF REINFORCED/PRESTRESSED
Resisting moment
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of concrete
a
My =c(a-3)
M, = 085 f', ab(d—i)
a
Mu
=
NS
¢Mn
M, = ¢ 085 f’, ab(d—g>2
SN
NN
NSNS
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CE 368 PRINCIPLES
OF REINFORCED/PRESTRESSED
CONCRETE
B
=aj:
-—
¢
,
cC=T
f c
a b=
=
pbd
g = Pbafy
ASfS
at yielding of steel ,f; = f,
.
percentage of tensile or steel ratio
bd
As
=0
f,c
.
p=—
.
0.85
0.85h
As
2
T= A_Jy
,
=
a
C=0RSfah
T
d-
ZFH
Asfy
B
~—
0.85h
f’C
_ pdfy
0.85
f,c
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S
S
S
N
CE 368 PRINCIPLES
S
SN
NSNS
OF REINFORCED/PRESTRESSED
Pl
Letw ==~
AN
NS
SN
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CONCRETE
o
=085
fe
M, = ¢ 0.85 f, ab(d—§>
a
wd
My =¢ f'c wb d*(1 - 0.59w)
)
Mu:gbf’cwbdz(l—%)
My = ¢ pfyb d? (1 - 0.59
Ultimate Resisting moment of concrete
Pfj)
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CE 368 PRINCIPLES
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AN
OF REINFORCED/PRESTRESSED
Strain diagram
NS
SN
CONCRETE
ByR&P:
€& T &
&
d
¢
d
e d
c
=
€t g
Hooke’s
f=Ee¢
Law:
f o ¢
Thus;
g, = —
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CE 368 PRINCIPLES
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SN
NSNS
OF REINFORCED/PRESTRESSED
AN
NS
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CONCRETE
At balanced design
fs=1
e. = 0.003 (concrete strain)
ap = Picp
0.003d
7 = 00035, ]
cp =
=5
0.003 +
0.003d
Es
E; =200000MPa
Es
0.003d(200000)
> = 0.003 (200000) + £,
600d
600d
=600+,
W =Pigoory,
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CE 368 PRINCIPLES
_
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OF REINFORCED/PRESTRESSED
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CONCRETE
Asfy
@ = 085h 1,
__Asfy
% = 085b 1,
_A4s
Pb = ba
ppbdfy _
600d
0.85b f'. ~
"1600+ f,
_085f'cf;
Pb
fy
As
=
pbbd
600
f, + 600
balanced ratio
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CE 368 PRINCIPLES
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CONCRETE
To ensure yielding of steel p;,in< P < Pmax
1.4
Pmin
=
E
Pmax= 0.75p
max
pr
Spmin:
_4s
US€Pmin
Ifp > Pinax US€ Prmax
P
g
Pmax
bd
=0.75
0.85f'.f,
600
fy
f, + 600
=
For design use:
p= 0.18% ( to control deflection)
y
p = 50% pnax(for both economy and control deflection)
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CE 368 PRINCIPLES
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OF REINFORCED/PRESTRESSED
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CONCRETE
Minimum depth of beam to control deflection
L
f;
h = 16 (0.4 + 7—8/0)
Simply supported beam
h = L
0.4 + f_y
One-end
18.5
700
h =_ o1L
fy
0.4+ 700
L
h= g <O.4 + %)
Both
continuous
.
-end continuous
Cantilever beam
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CONCRETE
Steel reinforcement if size of sections is given
My, = ¢ f'. wb d?*(1—0.59w)
b d2
¢f,
M,
R.=
M,
=
((1) —
059(1)2)
Let
n—
¢
b d?
Cc
R,
I3
=|\w
w?
2(0.85)
w?
2(0.85)
R
—w+——=0
f
Resistance
reduction factor
NN
CE 368 PRINCIPLES
OF REINFORCED/PRESTRESSED
CONCRETE
Use quadratic formula
a4 =
_ —b++b%—4ac
2(0.85)
2a
*=
1
1
) 7.R a
1+ j( [(-1 =4 2—4( 20085))
w
=
2(
1
2(0.85)
)
b= —1¢=
.
T
=085|1-
|1 —W
Ry
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CE 368 PRINCIPLES
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¢
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CONCRETE
I
f P
<
Brittle failure when
:
<Pmins
tensile
stress exceeds modulus of
rupture
If p>Pimaxs
Failure is initiated by
crushing of concrete
prmin
<
p
<
Pmax»
Failure is initiated by
yielding of steel
CE 368 PRINCIPLES OF REINFORCED/PRESTRESSED CONCRETE
APPLIED ULTIMATE
LOAD
(FACTORED
1. U=12DL+16LL
2. Basic Wind
U=12DL+1.6LL+0.5WL
U=0.9DL + 1.0WL
3. Earthquake
U=12DL+1.0LL
U=0.9DL + 1.0E
+ 1.0E
LOAD)
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CE 368 PRINCIPLES
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ACI/NSCP Specifications
S, = spacing between bars
a. S, >25mm
b. S, > bar diameter
c. S>1 3 (maximum size of aggregate)
If bar diameters will be placed in two layers:
a.vertical spacing between bars > 25mm
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Using 2 layers of bars.
If bars are placed in 2 layers, vertical clear spacing between bars should not
be less than 25 mm.
b =270
h = 630
d =540
<
25 mm clear
4
40 mm
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CE 368 PRINCIPLES
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CONCRETE
Using Bundled bars.
The ACI/NSCP Code permits bundling of up to four bars provided the bundle is
enclosed with stirrups or ties. Bars larger than 32 mmJ should not be bundled
in beams. Spacing limitations and concrete cover of most members are based
on a single bar diameter, thus bundled bars shall be treated as a single bar
with equivalent diameter derived from the equivalent total area.
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CE 368 PRINCIPLES
OF REINFORCED/PRESTRESSED
CONCRETE
Bundled bars arrangements:
o ltdan
oo |
Equivalent diameter D:
&
Example:
‘
3 -28 mm
h =610
e
Equivalent diameter D
-
.
4
4
—D2=3."(28)?
D =48.5 mm
.
!
i
-
oY
cover =70
f
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CE 368 PRINCIPLES
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OF REINFORCED/PRESTRESSED
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CONCRETE
Steps in Designing Singly Reinforced Concrete
Beam
1. Compute for the ultimate applied moment
2. Compute for steel ratio p = 50%pmax
14
= —
Pmin = "y
0.85 f'.B, 600
—0.75
fy
f, +600
Pmax
for steel to yield ppin < P < Pmaxs
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CE 368 PRINCIPLES
N
S
SN
NSNS
AN
OF REINFORCED/PRESTRESSED
3. Using the resisting ultimate moment
then
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of concrete, compute for d, if b =d/2
M, = ¢ f', wb d?(1 — 0.59w)
Ro= of, (1 B 0_59?1’})
M, = ¢ pf,b d? (1 ~0.59 %)
y
B=
b a2
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CE 368 PRINCIPLES
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NSNS
OF REINFORCED/PRESTRESSED
AN
CONCRETE
4. Compute for the steel area A; = pbd
5. Compute for the number of bars n, = A,/ 4,
6. Check spacing between bars
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CE 368 PRINCIPLES
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NSNS
OF REINFORCED/PRESTRESSED
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CONCRETE
Steps in Investigating Singly Reinforced Concrete Beam
1.
Check steel ratio if pjpin < P < Pmax (Steel yields)
1.4
Pmin
=
Pmax=
E
0.75
0.85f'.p;
600
fy
fy + 600
_As
P = bd
2. If steel yields then f; = f, , use ultimate resisting moment
M, = pasfy(d-3)2
of steel
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CE 368 PRINCIPLES
S
SN
NSNS
OF REINFORCED/PRESTRESSED
AN
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CONCRETE
3. Solve for a
SF; =0
cC=T
0.85f'. ab = Asfs
at yielding of steel , f; = f,
B
ASfy
¢ = 0.85b f.
NSNS
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S
S
S
CE 368 PRINCIPLES
N
S
SN
NSNS
OF REINFORCED/PRESTRESSED
AN
NS
SN
NN
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CONCRETE
Design a reinforced concrete beam to carry a service dead load of
10kN/m ( including its own weight) and a concentrated service
live load of 100kN acting at the center of the beam. Simple span
is bm. Use f'_ = 28MPa, f, = 276MPa, b = d/2
NN
CE 368 PRINCIPLES
OF REINFORCED/PRESTRESSED
CONCRETE
Determine the ultimate moment capacity of
the beam section shown from bending using
f'.= 35MPa, f, = 414MPa. If the beam is
simply supported on a span of 6m, find the
total ultimate uniform load and uniform live
420 mm
mim
4-28mm ¢
load that it can carry. yeone = 23.54 kN /m3.
e
o
o
@
le——
300
mm
—=
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S
CE 368 PRINCIPLES
N
S
SN
NSNS
OF REINFORCED/PRESTRESSED
AN
NS
SN
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CONCRETE
Design a reinforced concrete beam to carry a service live load of
20kN/m and
dead load of 12kN/m ( including its own weight)
Use f'_ = 21MPa, f, = 276MPa, b = d/2. The beam is to be
reinforced in tension only.
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S
N
CE 368 PRINCIPLES
S
SN
NSNS
OF REINFORCED/PRESTRESSED
AN
NS
SN
NN
NSNS
NN
NN
CONCRETE
Design a reinforced concrete beam to carry a service live load of
20kN/m and dead load of 15kN/m ( including its own weight) Use f'_ =
21MPa, f, = 345MPa, b = d/2. The beam is to be reinforced in tension
only. Clear cover = 50mm
9
I
[
8m
E=T0GPa
-‘
I
& m
[=1300(10% mm*
]
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