singly reinforced beam

1. Reinforced Concrete Design Module 1
Subject: CE 74B- Reinforced Concrete Design
1. Title of the Module
Introduction and Singly Reinforced Beam
2. Introduction
In this module, the students will be given a clear background on Working Stress
Design, the specifications regarding the same, types of beams, design of rectangular
beams reinforced with tension bars only and investigation of beams both rectangular
and beams with irregular sections.
3. Learning Outcomes
At the end of the course, the students should be able to:
a. be familiar with WSD specifications and apply the same in solving problems with
speed and accuracy
b. be able to design and investigate beams reinforced with tension bars only
c.. investigate beams with irregular cross sections
4. Learning Content
It contains readings, discussions, questions and sets of activities that students can work
on individually or by group.
Topic 1. Introduction with WSD Specifications
Topic 2. Diagonal Tension and Bond Stress
Topic 3. Design and Investigation of Rectangular Beams
Topic 4. Investigation of Beams with Irregular Section

2. Working Stress Design (ACI 318-63)
PHILIPPINE STANDARD REINFORCING BARS
Philippine Near Nominal Unit
Standard ASTM Sectional Weight
Designation Designation Area (mm2
) (kg/m)
6 mm # 2 ( 1/4'') 28.27 0.222
10 mm # 3 ( 3/8'') 78.54 0.616
12 mm # 4 ( 1/2'') 113.10 0.888
16 mm # 5 ( 5/8'') 201.10 1.579
20 mm # 6 1 3/4'') 314.20 2.466
25 mm # 8 (1 ) 491.90 3.854
28 mm # 9 (1 1/8..) 615.75 4.833
32 mm # 10 (1 1/4'') 804.25 6.313
36 mm #11 (1 3/8'') 1017.90 7.991
Philippine Philippine Standard
Standard Compared with
Designation ASTM Standard
6 mm 10.7% smaller than ASTM No.2 or 1/4" diameter
10 mm 10.22% larger than ASTM No.3 or 3/8" diameter
12 mm 10.7% smaller than ASTM No.4 or 1/2" diameter
16 mm 1.6% larger than ASTM No.5 or 5/8" diameter
20 mm 10.22% larger than ASTM No. 6 or 3/4" diameter
25 mm 2.9% smaller than ASTM No. 8 or 1" diameter
28 mm 4.49% smaller than ASTM No.9 or 1 1/8" diameter
32 mm 1.6% smaller than ASTM No. 10 or 1 1/4" diameter

3. 36 mm 0.97% larger than ASTM No. 11 or 1 3/8" diameter
Bar Size
Nominal
Diameter
mm
Nominal
Area
Nominal Mass kg/m
# 10 11.3 100 0.785
# 15 16.0 200 1.570
# 20 19.5 300 2.355
# 25 25.2 500 3.925
# 30 29.9 700 6.495
# 35 35.7 1000 7.850
# 45 43.7 1500 11.775
# 55 66.4 2500 19.625
Minimum Cover for Cast-in Place Concrete Recommended by ACI Code
(ACI 318M-83)
A) If concrete is not exposed to weather:
For beams and columns………………………………………………….. 40mm
For slabs,walls, joists, with No. 35 bars or smaller……………………. 20mm
For slabs,walls, joists, with No. 45 bars or smaller……………………. 40mm
For shells and folded plate members with No. 20 or larger bars…….. 20mm
For shells and folded plate members with No. 10 or No. 15 bars……. 15mm

4. B) If concrete is exposed to the weather:
If reinforcing bars are No. 20 or larger………………………………… 50mm
If reinforcing bars are No. 10 or No. 15 ………………………………. 40mm
C) In all concrete cast directly on the earth……………………………….. 70mm
a) Slabs with beams or drop panels………………………………. 120mm
b) Slabs without beams or drop panels…………………………... 100mm
c) Slabs with beams on all four edges……………………………. 90mm
Temp. bars for slabs. (Area requirements)
As = 0.002 bt for Grade 300 bars
As = 0.0018 bt for Grade 400 bars
Clear distance between bars
a) Not less than the bar diameter nor 25mm for beams
b) Not less than 1.5 bar diameter nor 40mm for columns
c) When bars are placed in several layers, the clear distance between layers
must not be less than 25mm.
Maximum spacing of main reinforcing bars in slabs and walls is equal to 3 times the wall or
slabs thickness or 500mm whichever is smaller.
Spacing of tie wires for columns:
a) Not more than 16 bar diameter
b) Not more than 48 tie diameter
c) Not more than the least dimension of the cross section
Modulus of Elasticity of Concrete:

5. Ec = wc
1.5
0.043√fc’ Mpa for values of wc between 1500-2500 kg/m3
Ec = 4700√fc’ for normal density concrete
fc = 0.45fc’
fc = 165 MPa for Grade 400
fc = 140 MPa for Grade 300
The ratio 0.45 gives a factor of safety of
1
0.45
= 2.22
Es = 2,040,000 kg/cm2
Es = 200,000 MPa for all types of steel
ALLOWABLE STRESSES IN CONCRETE (S.I. VERSION)
(ACI 318-63)
Description Symbol fc’=17.78 MPa fc’=20.73 MPa fc’=27.65MPa

6. Modulus of n =
𝐸𝑠
𝐸𝑐
10
Elasticity
Flexure:
Compression fc’ = 0.45fc’ 7.78
Tension in
Plain Conc. fc= 0.13√fc’ 0.54
Footing &
Walls
Shear:
Diagonal
tension at
distance “d” v = 0.09√fc’ 0.37
from face of
support
Joist with
no web v = 0.10√fc’ 0.42
reinforcement
Punching shear
at distance “d”/2 vp = 0.17√fc’ 0.71

7. Bearing:
On full area fc = 0.25 fc’ 4.32
On one third
area or less fc = 0.375 fc’ 6.48
ALLOWABLE STRESSES IN CONCRETE (S.I. VERSION)
(ACI 318-63)
Description Symbol fc’=176 kg/cm2
fc’=211 kg/cm2
fc’=281 kg/cm2
Modulus of n =
𝐸𝑠
𝐸𝑐
10 9 8
Elasticity
Flexure:
Compression fc’ = 0.45fc’ 79 94.8 128.5
Tension in
Plain Conc. fc= 0.424√fc’ 5.63 6.16 7.11
Footing &
Walls
Shear:

8. Diagonal
tension at
distance “d” v = 0.292√fc’ 3.87 4.24 4.89
from face of
support
Punching shear
for slabs and vp = 0.53√fc’ 7.03 7.73 8.84
footings
Bearing:
On full area fc = 0.25 fc’ 44 52.7 70.3
On one third
area or less fc = 0.375 fc’ 66 79.1 105.5
Allowable Stresses (Working Stress Design)
A) Modulus of Elasticity of Concrete:
Ec = 4730√fc’ MPa (S.I. System)
Ec = 15100√fc’ kg/cm2 (MKS System)

9. B) Modulus of Elasticity of Steel:
Es = 200,000 MPa (S.I. System)
Es = 2,040,000 kg/cm2
(MKS System)
C) Allowable Shearing Stress:
Va = 0.09√fc’ MPa (S.I. System)
Va = 0.29√fc’ kg/cm2
(MKS System)
D) Allowable Punching Stress:
Vp = 0.17√fc’ MPa (S.I. System)
Vp = 0.53√fc’ kg/cm2
(MKS System)
E) Allowable Bond Stress:
1) For top bars in tension:
Ua =
7.18√𝑓𝑐’
𝐷
MPa (S.I. System)
Ua =
2.29√𝑓𝑐’
𝐷
kg/cm2 (MKS System)
2) Other than top bars in tension:
Ua =
10.14√𝑓𝑐’
𝐷
MPa (S.I. System)
Ua =
3.23√𝑓𝑐’
𝐷
kg/cm2 (MKS System)

10. 3) For all compression bars:
Ua = 0.54√𝑓𝑐’ MPa (S.I. System)
Ua = 1.70√𝑓𝑐’ kg/cm2
(MKS System)
ALLOWABLE BOND STRESS
(ACI 318 - 63)
1. For tension bars with sizes and deformations conforming to ASTM A305 top bars
(Horizontal bars placed that more than 300 mm of concrete is cast in the
member below the bar).
Top bars….u =
7.18√𝑓𝑐’
𝐷
nor 2.42 MPa …u =
2.29√𝑓𝑐’
𝐷
nor 24.6 kg/cm2
Bars other than top bars ….. u =
10.14√𝑓𝑐’
𝐷
nor 3.46 MPa ………..
...u =
3.23√𝑓𝑐’
𝐷
nor 35.2 kg/cm2
For plain bars, use one half of these values but not more than 1.11 MPa or 11.25
kg/cm2
2. For tension bars with sizes and deformations conforming to ASTM A408 top bars
….. u = 0.18√𝑓𝑐’MPa ….u = 0.556√𝑓𝑐’kg/cm2
Bars other than top bars ... . u = 0.25√𝑓𝑐’MPa ……. u = 0.795√𝑓𝑐’kg/cm2

11. 3. For all deformed compression bars:
u = 0.54√𝑓𝑐’nor 2.76 MPa …. u = 1.72 √𝑓𝑐’nor 28.1 kg/cm2
WORKING STRESS DESIGN
THE STRAIGHT-LINE OR ELASTIC-LOAD THEORY
This is based from the assumption that plate sections remain plane
and normal to the longitudinal fiber stress before and after bending and
that both materials are elastic. The stress of concrete varies from fc in
the extreme fiber to zero at the neutral axis.

12. fc = compressive unit stress on the concrete at the surface most
remote from the neutral axis.
fs = tensile unit stress in the longitudinal reinforcement.
b = width of beam.
d = depth from the center of the steel bars to the extreme fiber of
concrete.
kd = distance from the neutral axis to the extreme fiber of concrete.
jd = distance between the compressive force C and the tensile force
T.
As = cross-sectional area of steel bars.
p = ratio of the area of steel to that of the effective area of concrete.
Mc = resisting moment of concrete.
Ms = resisting moment of steel.
Es = modulus of elasticity of concrete.
Δc = deformation per unit length of concrete.
Δs = deformation per unit length of steel.
n = ratio of modulus of elasticity of steel to that of concrete.

13. CROSS SECTION STRESS DIAGRAM STRAIN DIAGRAM
Modulus of Elasticity =
𝑈𝑛𝑖𝑡 𝑜𝑓 𝑠𝑡𝑟𝑒𝑠𝑠
𝑈𝑛𝑖𝑡 𝑜𝑓 𝑑𝑒𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛
𝐸𝑠 =
𝑓𝑠
𝛥𝑠
𝛥𝑠 =
𝑓𝑠
𝐸𝑠
𝐸𝑐 =
𝑓𝑐
𝛥𝑐
𝛥𝑐 =
𝑓𝑐
𝐸𝑐

14. 𝛥𝑐
𝛥𝑠
=
𝑓𝑐
𝐸𝑐
𝑓𝑠
𝐸𝑠
𝛥𝑐
𝑘𝑑
=
𝛥𝑠
𝑑−𝑘𝑑
𝛥𝑐
𝛥𝑠
=
𝑘𝑑
𝑑−𝑘𝑑
𝑓𝑐
𝐸𝑐
×
𝐸𝑠
𝑓𝑠
=
𝑘𝑑
𝑑−𝑘𝑑
𝑓𝑐
𝑓𝑠
×
𝐸𝑠
𝐸𝑐
=
𝑘𝑑
𝑑−𝑘𝑑
𝐸𝑠
𝐸𝑐
= 𝑛
𝑓𝑐 𝑛
𝑓𝑠
=
𝑘𝑑
𝑑−𝑘𝑑
𝑓𝑐
𝑓𝑠
=
𝑘
𝑛(1−𝑘)
𝑓𝑐𝑛 − 𝑓𝑐𝑛𝑘 = 𝑘𝑓𝑠
𝑘(𝑓𝑠 + 𝑓𝑐𝑛) = 𝑓𝑐𝑛
𝑘 =
𝑓𝑐𝑛
𝑓𝑠+𝑓𝑐𝑛
(1) 𝑘 =
𝑛
𝑛+
𝑓𝑠
𝑓𝑐
(for design only)
k = ranges from 0.3 to 0.45
j = ranges from 0.9 to 0.85

15. 𝑀𝑐 = 𝐶𝑗𝑑
𝐶 =
1
2
𝑓𝑐(𝑏)𝑘𝑑
(2)𝑀𝑐 =
1
2
𝑓𝑐 𝑘𝑗 𝑏𝑑2
(Resisting moment of concrete)
𝑀𝑠 = 𝑇𝑗𝑑
(3) 𝑀𝑠 = 𝐴𝑠 𝑓𝑠 𝑗𝑑
(Resisting moment of steel)
𝑝 =
𝐴𝑠
𝑏𝑑
𝐴𝑠 = 𝑝𝑏𝑑
𝑀𝑠 = 𝑝𝑏𝑑 𝑓𝑠 𝑗𝑑
𝑀𝑠 = 𝑝 𝑓𝑠 𝑗𝑏𝑑2
𝑀𝑠 = 𝑀𝑐 (for balanced design):
𝑝 𝑓𝑠 𝑗𝑏𝑑2 =
1
2
𝑓𝑐 𝑘𝑏𝑑2
2 𝑝 𝑓𝑠 = 𝑓𝑐𝑘
𝑓𝑐
𝑓𝑠
=
2𝑝
𝑘
But
𝑓𝑐
𝑓𝑠
=
𝑘𝑑
𝑛(𝑑−𝑘𝑑)

16. 𝑓𝑐
𝑓𝑐
=
𝑘
𝑛(1−𝑘)
𝑘
𝑛(1−𝑘)
=
2𝑝
𝑘
𝑘2
= 2𝑝𝑛 − 2𝑝𝑛𝑘
𝑘2
+ 2𝑝𝑛𝑘 − 2𝑝𝑛 = 0
By completing square and solving for k.
𝑘2
+ 2𝑝𝑛𝑘 + (𝑝𝑛)2
= 2𝑝𝑛 + (𝑝𝑛)2
(𝑘 + 𝑝𝑛)2
= 2𝑝𝑛 + 𝑝2
𝑛2
𝑘 + 𝑝𝑛 = √2𝑝𝑛 + 𝑝2𝑛2
(4) 𝑘 = √2𝑝𝑛 + 𝑝2𝑛2 − 𝑝𝑛 -(for investigation only)
(5) 𝑑 = √
𝑀
1
2
𝑓𝑐𝑗𝑘𝑏
𝑅 =
1
2
𝑓𝑐 𝑘𝑗
(6) 𝑑 = √
𝑀
𝑅𝑏
𝑗𝑑 = 𝑑 −
𝑘𝑑
3
(7) 𝑗 = 1 −
𝑘
3
Steps in Designing Beam Reinforced for Tension

17. 1. Compute live load moment, depending on type of load and
support
𝑀 =
𝑤𝐿2
8
(simply supported)
𝑀 =
𝑤𝐿2
12
(continuous beam)
2. Compute the design constant
𝐸𝑠
𝐸𝑐
= 𝑛
𝑓𝑐 = 0.45𝑓𝑐′
𝑘 =
𝑛
𝑛+
𝑓𝑠
𝑓𝑐
𝑗 = 1 −
𝑘
3
𝑅 =
1
2
𝑓𝑐 𝑘𝑗
3. Solve for the depth “d” by assuming a value of “b” ranging from
250 mm to 300 mm.
𝑑 = √
𝑀
𝑅𝑏
4. Add 50 mm to 100 mm to this computed “d” to take care of the
dead load. Use the relation
𝑏 =
𝑑
2
𝑜𝑟 𝑏 =
𝑑
1.75
5. Add the dead load to the live load to obtain the total load.
6. Recompute the total moment
7.Recheck the computed 𝑑 = √
𝑀
𝑅𝑏
(The result would be the same as
that values which you have added 50 mm or 100 mm)
8. Check for shear 𝑣 =
𝑉
𝑏𝑑

18. 9. If v > va use 10 mm stirrups
10. Spacing of stirrups
𝑆 =
𝐴𝑣𝑓𝑣
𝑣′𝑏
𝑣′ = 𝑣 − 𝑣𝑎
𝑣𝑎 = .09√𝑓𝑐′𝑀𝑃𝐴
𝑆 𝑖𝑠 𝑛𝑜𝑡 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛
𝑑
2
(max. spacing of stirrups)
11. Check for Bond Stress
𝑈𝑜 =
𝑉
∑ 𝑗𝑑
𝑂
𝑉 = max 𝑠ℎ𝑒𝑎𝑟
∑ = 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑏𝑎𝑟𝑠
𝑜
12. If 𝑢 > 𝑈𝑜, 𝑡ℎ𝑒𝑛 𝑎𝑑𝑑 𝑚𝑜𝑟𝑒 𝑠𝑡𝑒𝑒𝑙 𝑏𝑎𝑟𝑠

20. MOMENT COEFFICIENT FOR CONTINUOUS BEAMS AND SLABS
Limitations to this coefficients:
1. Adjacent clear spans should not differ by more than 20% of the shorter pan.
2. Ratio of live load to deadload should not exceed 3.
3. The loads should be uniformly distributed only.
For Positive Moment.
End spans:
If discontinuous end is unrestrained- - - - - - - - - - - - - -
𝑤𝐿2
11
If discontinuous end is integral with the support- - - - -
𝑤𝐿2
14
Interior spans- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
𝑤𝐿2
16
For Negative Moment:
Negative moment at exterior face of first interior support:
Two spans- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
𝑤𝐿′2
9
More than two spans- - - - - - - - - - - - - - - - - - - - - - - - -
𝑤𝐿′2
10
Negative moment at other faces interior supports- - - -
𝑤𝐿′2
11
Negative moment at face of all supports for:
a) Slabs with spans less than 3 m.- - - - - - - - - - - - - - - - -
𝑤𝐿′2
12
b) Beams and girders where the ratio of the sum of column stiffness to beam stiffness exceeds 8 at each
end of span- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
𝑤𝐿′2
12
Negative moment at interior faces of exterior supports for members built integrally with their supports.
a) When the support is a spandrel beam of girder- - - - - -
𝑤𝐿2
24
b) When the support is a column- - - - - - - - - - - - - - - - - - - -
𝑤𝐿2
16
Shear Forces:
a) Shear in end members at first interior support- - - - - - - - 1.15
𝑤𝐿
2

21. b) Shea0r at all other supports- - - - - - - - - - - - - - - - - - - - - - -
𝑤𝐿
11
The figure below shows a free body diagram of a segment of a concrete beam dx in length. The vertical
shear V.
∑𝑀𝐴=0
T(jd) + V(dx) = (T + 𝛥𝑇)jd (1) 𝛥𝑇 =
𝑉(𝑑𝑥)
𝑗𝑑
T(jd) + V(dx) – T(jd) = 𝛥𝑇(jd) ∑𝐹ℎ= 0
V(dx) = 𝛥𝑇(jd) H + T = T + 𝛥𝑇
H = vb(dx) (shearing force along the horizontal plane)
bv(dx) + T = T + 𝛥𝑇
(2) 𝛥𝑇 = vb(dx)
(1) & (2)
𝑉(𝑑𝑥)
𝑗𝑑
= vb(dx)
v =
𝑉
𝑏𝑗𝑑
(1956 ACI CODE)
v =
𝑉
𝑏𝑑
(1963 ACI CODE)
V = maximum shear at a distance “d” from support
v = shearing stress
d = effective depth
b = width of beam
Allowable shearing stress is
va = 0.09√𝑓𝑐′MPa
va = 0.09√𝑓𝑐′kg/cm²

24. DIAGONAL TENSION
When a simple beam is subjected to bending the fibers above the neutral axis are either in
compression or tension while tensile stresses occur in the fibers below or above the neutral axis. In
addition to these stresses, there are also inclined stresses of diagonal stresses which tend to produce
cracks. These cracks appear to be vertical at the center of span and become more inclined as they
approach the supports in which it makes an angle of about 45o
with the horizontal. The stresses that
cause these cracks are called diagonal tension. To resist these diagonal tension, reinforcing bars called
stirrups are used. The vertical stirrup is assumed to carry the vertical component of the diagonal tension
while the longitudinal bars carries the horizontal component of the diagonal tension.
Consider a strip having a length of S(mm).
Assume the diagonal tension as Td acting at an angle of 45o
with the horizontal. T is the component of Td,
both in horizontal and vertical. Since we could not place reinforcement in inclined position, it is then
placed at a vertical position to carry a load of T. But the horizontal force T is the only force that could
resist shearing stress as shown.
T = v’ bs
𝐴𝑣𝑓
𝑣 = v’ bs
S =
𝐴𝑣𝑓𝑣
v’ b
Spacing of stirrups
S = Spacing of stirrups
Av = cross- sectional area of stirrups (10mm)

25. (min. 3/8 dia. with an area of 0.11 in2
but
we would be using U-stirrups therefore
Av = 0.11(2)
Av = 0.22(157mm2
)
𝑓
𝑦= allowable tensile stress of stirrups if not given, it is equal to fs.
v’ = v - 𝑣𝑎
v =
𝑉
𝑏𝑑
(actual shearing stress)
𝑉
𝑎 = allowable shearing stress
𝑉
𝑎 = 0.29 √𝑓𝑐′ kg/cm2
= 0.09 √𝑓𝑐′ MPa
Max. spacing of stirrups =
𝑑
2
HOW TO DETERMINE THE PORTION OF THE
BEAM REQUIRING WEB REINFORCEMENT
(STIRRUPS)
v’ = v – 𝑣𝑐 (excess shearing stress)
𝑣𝑐 = allowable shearing stress
L = span of beam
X = length of beam from support in which
stirrups are required.

26. For Simply Supported Beams
𝑣′
𝑥
=
𝑣
𝐿/2
x =
𝐿 𝑣′
2𝑣
For Cantilever Beam
𝑣′
𝑥
=
𝑣
𝐿
x =
𝐿 𝑣′
𝑣
BOND STRESS
∑ MA = 0
V(dx) + Tjd = (T + ∆T)jd
V(dx) + Tjd - Tjd = ∆Tjd
∆T =
𝑉(𝑑𝑥)
𝑗𝑑
𝜋d = ∑0 = perimeter of steel bar

27. F = u𝜋d(dx)
F = u∑0(dx)
∆T = F
∆T = u∑0(dx)
𝑉(𝑑𝑥)
𝑗𝑑
= u∑0(dx)
u =
𝑉
∑0𝑗𝑑
(bond stress)
Allowable Bond Stresses:
Tension Bars: (Deformed Bars)
A) ASTM A 305 U =
7.18√𝑓𝑐′
𝐷
MPa
Top Bars
Bar other than top bars:
B) ASTM A 408 U =
10.14√𝑓𝑐′
𝐷
MPa
Top Bars U = 0.18 √𝑓𝑐′ Mpa
Bars other than top bars: U = 0.25 √𝑓𝑐′ Mpa
Compression bars U = 0.54 √𝑓𝑐′ Mpa
Note: For plain bars, use one half of ASTM A 305 but not more than 1.14 MPa.

28. ANCHORAGE BOND
T = 𝐴𝑠𝑓
𝑠
F = ∑0 UL
T = F
𝐴𝑠𝑓
𝑠 = ∑0 UL
𝜋𝑑2
4
= 𝜋dUL
L =
𝑑𝑓𝑠
4𝑢
Where L = length of embedment 𝑓
𝑠= tensile stress of steel bars
d = diameter of steel bars u = bond stress of concrete
ELASTIC THEORY IS BASED ON THE FOLLOWING ASSUMPTIONS
1. Plane sections remain plane before and after bending occurs.
2. Concrete is elastic; that is the stress of concrete varies from zero at the neutral axis to a maximum at
the extreme fibers.
3. Concrete is not the food in carrying tension and only the steel bars are carrying all the stresses due to
bending which is tension.
4. There must be no slippage between concrete and steel bars.

29. Balanced design = a design so proportional such that the maximum stresses in the concrete and
steel are reach simultaneously so they fail together.
Under-reinforced design = the steel reinforcement is lesser than what is needed for a balanced
design. In this particular type of design, the steel fails first while the concrete has not yet reached its
allowable values but the failure is gradual with the steel yielding.
Over-reinforced design = the steel reinforcement is more than what is needed for a balanced
design. This type of design is not advisable cause concrete fails suddenly in compression.
CE BOARD – AUGUST 1965
Design a section of a concrete beam reinforced for tension only. The beam is simply supported
on a span of 6 m and carries a load of 18 KN/m. What stirrups will be required? Use ACI Specs, with fc’ =
17.2 MPa, fs = 124 MPa, n = 12. Assume b = d/2
Solution:
fc = 0.45 fc’
fc = 0.45 (17.2)
fc = 7.74 MPa
k =
𝑛
𝑛+
𝑓𝑠
𝑓𝑐
k =
12
12+
124
7.74
k = 0.428 M =
𝑤𝐿2
8
j = 1 -
𝑘
3
M =
21473(6)2(1000)
8
j = 1 -
0.428
3
M = 96.63 x 106
N.mm
j = 0.857 d = √𝑀
𝑅𝑏

30. R =
1
2
fc kj d = √
96.63 x 106
1.42(250)
= 522 < 525
R =
7.74(.428)(.857)
2
Steel requirements:
R = 1.42 As =
𝑀
𝑓𝑠𝑗𝑑
M =
𝑤𝐿2
8
As =
96.63 x 106
124(.857)(525)
M =
18000(6)2(1000)
8
As = 1732 𝑚𝑚2
M = 81 x 106
N.mm Try 28 mmϴ;
d = √𝑀
𝑅𝑏
A =
𝜋
4
(282
) = 615.75 𝑚𝑚2
Try b = 250 mm 615.75 N = 1732
d = √
81000000
1.42(250)
N = 2.81 say 3 bars
d = 477.7 mm
Try d = 525 mm
Total depth = 525 + 65 = 590mm
Dead Load = 0.25(.59)(2400)(9.81)
Dead Load = 3473 N/m
Total Load = 18000 + 3473 = 21473 N/m
Check for Shear;
V =
1
2
(21473) (6) – 21473 (.525)
V = 53146
v =
𝑉
𝑏𝑑
v =
53146
250(525)
v = 0.40 MPa
va = 0.09 √fc′
va = 0.09 √17.2 = 0.37 MPa

31. Use 10mm ϴ;
Av = 2 (
𝜋
4
)(102
) = 157 𝑚𝑚2
, fv = 124 MPa
S =
𝐴𝑣 𝑓𝑣
𝑣′𝑏
v’ = 0.40 – 0.37 = 0.03
S =
157(124)
0.03(250)
=2596
Max S. =
𝑑
2
=
525
2
= 262.5mm say 260 mm
Check for bond stress;
Va =
10.14√fc′
𝐷
Va =
10.14√17.2
28
= 1.5 MPa
Ʃo = 𝜋(28)(3) = 264 mm
U =
𝑉
Ʃo jd
U =
64419
264(.857)(525)
U = 0.54 < 1.5 MPa (safe)

32. CE BOARD – FEBRUARY 1975
A concrete beam reinforced for tension only is required to carry a uniformly distributed load of
27000 N/m (including its own weight) on a simple span of 9 m. The overall depth ‘’D’’ is twice the width
‘’b’’ and the center of the steel reinforcement is to be located at a distance 1/10 D from the underside of
the beam. Compute the dimension of the beam and the area of reinforcement required. fs = 124 MPa, fc
= 5.53 MPa, n = 15.
Solution:
k =
𝑛
𝑛+
𝑓𝑠
𝑓𝑐
k =
15
15+
124
5.53
k = 0.401
j = 1 -
𝑘
3
j = 1 -
.401
3
j = .866
R =
1
2
fc kj
R =
5.53(.401)(.866)
2
R =0.96 Recheck;
M = R b 𝑑2
d = √𝑀
𝑅𝑏
M =
𝑤𝐿2
8
d = √
273375000
0.96(450)
M =
27000(9)2(1000)
8
d = 795mm < 810 mm (safe)
M = 273375 x 103
N.mm As =
𝑀
𝑓𝑠𝑗𝑑

33. 273375 x 103
= 0.96 b𝑑2
As =
273375000
124(.866)(810)
b𝑑2
= 284765625 As = 3143 𝑚𝑚2
(
𝐷
2
)(0.90D)2
= 284765625 Use 450 mm x 900 mm
D = 889 say 900 mm with As = 3143 𝒎𝒎𝟐
b = 450 mm
d = 0.90(900) = 810 mm
Problem 3.
A reinforced concrete beam rectangular in cross-section, is loaded to produce a
maximum moment of 7100 N-m. if 𝑏 = 200 𝑚𝑚, 𝑑 = 250 𝑚𝑚, 𝑓𝑐′
= 4.8 𝑀𝑃𝑎 and 𝑓𝑠 =
124 𝑀𝑃𝑎, determine As for balanced design.
Solution
𝐶 =
1
2
𝑓𝑐′
(𝑘𝑑)(200)
𝐶 =
4.8𝑘(250)(200)
2
𝐶 = 120000 𝑘
𝑀 = 𝐶𝑗𝑑
𝑀 = 120000(1 −
𝑘
3
)(250𝑘)
7100000 = 120000(250𝑘)(1 −
𝑘
3
)
3𝑘−𝑘2
3
= 0.24

34. 3𝑘 − 𝑘2
= 0.72
𝑘2
− 3𝑘 + 0.72 = 0
𝑘 =
3±2.40
2
𝑘 = 0.26
𝑗 = 1 −
𝑘
3
𝑗 = 1 −
0.26
3
𝑗 = 0.912
𝑀 = 𝐴𝑠 𝑓𝑠 𝑗𝑑
7100000 = 𝐴𝑠 (124)(.912)(250)
𝐴𝑠 = 251.13 𝑚𝑚2
PROBLEM:
A rectangular reinforced concrete beam shown, carries a total moment of 119.2 KN.m. If fc = 5.2
Mpa, fs = 138 MPa and n = 12, determine the steel area required for a balanced design.

35. SOLUTION:
fc
x
=
fs
n
450−x
5.2
x
=
138
12
450−x
5.2(450 – x) = 11.5x
x = 140.12 mm
I =
250x3
3
+ nAs (450 - x)2
I =
250(140.123)
12
+ 12As (450 – 140.12)2
I = 229255170.8 + 1152307.4 As
fc =
Mc∁
I
5.2 =
119.2( 140.12 )(10)6
229255170.8 + 1152307.4 As
1192126888 + 5991998.5 As = 119.2 (140.12) (10)6
As = 2588 mm2

36. PROBLEM:
The beam in the figure shown has a 20 Mpa concrete and is reinforced with 4-25 mm bars. If
it sustains a bending moment of 100 kN.m, calculate the actual bending stress in steel concrete n
= 10.
Solution:
2
2
19635
)
4
(
)
25
(
4
10 mm
nAs 









37.  
 
   
4
6
2
2
2
10
2012
264
19635
3
186
300
186
264
450
0
58905
130
450
19635
150
450
2
300
mm
I
I
mm
x
mm
x
x
x
x
x
x
nAs
x
x




















Stress in concrete:
   
Mpa
fc
fc
Mcx
fc
24
.
9
10
2012
186
1000
100
1
6
2




Stress in steel:
 
   
Mpa
fs
fs
I
x
M
n
fs
131
10
2012
264
1000
100
10
450
6
2






38. PROBLEM:
A rectangular simply supported reinforced concrete beam has a cross-section of 250 mm x
500 mm and a simple span of 6 m. It is reinforced with 4 - 20 mm Փ bars in one row, the center
of bars being 60 mm above the bottom of the beam. Using fs = 124 Mpa, fc = 7 Mpa, n = 12,
determine the additional load P that could be carried by this section at midspan besides a
uniform load of 12kN/m including the weight of beam.
Solution:

39.    
 
 
 
   
   
     
   
   
m
kN
Ms
mm
N
Ms
Ms
d
j
fs
As
Ms
m
kN
mm
N
Mc
Mc
bd
kj
fc
Mc
k
j
k
k
n
n
n
n
k
bd
A
mm
s
.
74
.
60
.
60741760
450
866
.
0
124
1257
.
35
.
61
.
61531194
450
250
866
.
0
40
.
0
7
2
1
2
1
866
.
0
3
1
401
.
0
134
.
0
134
.
0
134
.
0
2
134
.
0
12
011
.
0
2
011
.
0
450
250
1257
1257
4
20
4
A
2
2
2
2
2
2
s
































Safe resisting moment
m
kN
Ms
M .
74
.
60


External moment
   
kN
P
P
WL
PL
M
5
.
4
8
6
12
4
6
74
.
60
8
4
2
2






40. Compute P from shear
 
 
 
18.67kN
P
450
250
1000
6
.
30
2
P
355
.
0
355
.
0
15.56
0.09
allow.v
56
.
15
'
'
45
.
0
7
45
.
0
f
0.09
allow.v
6
.
30
2
P
V
12
45
.
0
36
2
P
V
bd
V
v
'
'





















MPa
MPa
c
f
c
f
c
f
fc
c
kN
Compute P from bond:
  
 
   
241kN
P
450
866
.
0
251
1000
36
2
P
1.6
mm
251
4
20
36
2
P
V
1.6MPa
allow.u
20
15.56
10.14
u
D
c
f'
10.14
allow.u
V
u
0
0




















jd
Safe P = 4.5kN

41. Problem: Page 62

42. The cross-section shown is reinforced with 2-32mm ø and 2-28 mm ø bars
placed in one layer. This beam has single span of 8m and carries the moving loads
shown. Neglecting its own weight, determine whether this beam could carry such
load. f’c= 20 MPa, n= 12, v= 0.4 MPa, u= 1.40 MPa
Solution:
nAs = 12 [
π
4
(32)2
(2) +
π
4
(28)2
(2) ] = 34079.997 mm2
400x (
x
2
) = 34079.997 ( 600 – x )
x = 245.706 mm
600 – x = 354.294 mm
I =
400 ( 245.706 )3
3
+ 34079.997 ( 354.294 )2
= 6255.682 x 106
mm4
Moment capacity of concrete:
f’c =
Mcx
I
0.45( 20 ) =
Mc ( 246 )
6255.682 x 106

43. Mc = 228.866 kN.m
Moment capacity of steel:
fs
n
=
Ms ( 600−x )
I
124
12
=
Ms ( 354.294 )
6255.682 x 106
Ms = 182.453 kN.m
Safe resisting moment M = 182.453 kN.m
For maximum moment, place the moving loads so that the centerline of the
beam is midway between the biggest load and the resultant of the loads.

44. R = 50 + 40 + 10 = 100 kN
100x = 40(3) + 10(4)
x = 1.6 m
ΣMR2 = 0
R1 (8) = 100 (3.2)
R1 = 40 kN
Max. M occurs under the biggest load:
Max. m = 40(3.2) = 128 kN.m ˂ 182.453 kN.m (ok)
Check shear:
For maximum shear, place the biggest load almost above the support R1,
ΣMR2 = 0
8R1 = 100(6.4)
R1 = 80 kN
Max. shear = R1 = 80 kN
v =
V
bd
=
80 ( 1000 )
400 ( 600 )
= 0.333 MPa
allow. v = 0.4 MPa ˃ 0.333 MPa (ok)
Check bond:
u =
V
∑ o jd
Σo = π (32) (2) + π (28) (2) = 377 mm
jd = 600 -
x
3
jd = 600 -
245.706
3
= 518.098 mm
u =
80 (1000)
377 (518.098)
= 0.419 MPa ˂ 1.4 MPa (ok)
Therefore, the beam could carry the moving loads.

45. Problem: Page 72
What must be the maximum concentrated load at the third points of the
beam shown whose span is 6 m. Assume f’c = 9 MPa, fs = 1224 MPa, n = 12.

46. Solution:
Assume the location of the N.A. as shown.
nAs = 12 (
π
2
) (20)2
(4) = 30159.289 mm2
75x (
x
2
) (2) + 150 (
x−150
2
) (x – 150) = 30159.289 (400-x)
x = 238.716 mm
x – 150 = 88.716 mm
400 – x = 161.284 mm
I =
75 ( 238.716 )3
3
(2) +
150 ( 88.716 )3
3
+ 30159.289 (161.284)3
= 127244.287
x106
mm4
For concrete:
f’c =
Mcx
I
9 =
Mc (238.716)
127244.287 x106
Mc = 4797.326 kN.m
For steel:
fs
n
=
Ms (400−x)
I
124
12
=
Ms ( 161.284 )
127244.287 x106

47. Ms = 8152.437 kN.m
Safe M = 4797.326 kN.m
max. M =
PL
3
+
wL2
8
w = [ 0.3(0.46) – 0.15(.15) ] (9.81) (2.4) = 2.719 kN/m
4797.326 =
P( 6 )
3
+
2.719 ( 6 )2
8
P = 2393.545 kN
Problem:
Find the actual stresses in concrete and steel if the reinforced beam shown
carries a bending moment (due to dead load and live load) equal to 150 kN.m.
Neglect the lack of symmetry n = 10.

48. Solution:
nAs = 10 (
π
4
) (28)2
(4) = 24630.086 mm2
24630.086 (750 – x) = 300x (
x
2
) + 200 (x – 150) (
x − 150
2
)
x = 265.702 mm
750 – x = 484.298 mm
X – 150 = 115.702 mm
I =
200 ( 115.702 )3
3
+
300 ( 265.702 )3
3
+ 24630.086 (484.298)3
= 2799697.166 x106
mm4
Stress in concrete:
f’c =
𝐌𝐱
𝐈
=
𝟏𝟓𝟎 ( 𝟏𝟎𝟎𝟎 )𝟐 ( 𝟐𝟔𝟓.𝟕𝟎𝟐 )
𝟐𝟕𝟗𝟗𝟔𝟗𝟕.𝟏𝟔𝟔 𝐱𝟏𝟎𝟔
= 0.014 MPa

49. Stress in steel:
𝐟𝐬
𝐧
=
𝐌 ( 𝟕𝟓𝟎−𝐱 )
𝐈
𝐟𝐬
𝟏𝟎
=
𝟏𝟓𝟎 ( 𝟏𝟎𝟎𝟎)𝟐 ( 𝟒𝟖𝟒.𝟐𝟗𝟖 )
𝟐𝟕𝟗𝟗𝟔𝟗𝟕.𝟏𝟔𝟔 𝐱𝟏𝟎𝟔
= 0.026 MPa
5. Teaching and Learning Activities
A. Solve the following problems for mastery of the procedures and
specifications:
1. Design a section of a concrete beam reinforced for tension only. The beam is simply
supported on a span of 10m and carries a load of 120 KN/m. What stirrups will be
required? Use ACI Specs, with fc’ = 17.2 MPa, fs = 124 MPa, n = 12. Assume b = d/2
2. Design a section of a concrete beam reinforced for tension only. The beam is a
continuous beam with same spans of 5m and carries a load of 30 KN/m. What stirrups
will be required? Use ACI Specs, with fc’ = 20MPa, fs = 138MPa, n = 10. Assume 1.75b
= d
Note: Additional problems will also be given on investigation of rectangular and beams
of irregular sections.
B. Project: Prepare the floor plans, elevations and perspective drawings of a
building with a specified number of storeys to be assigned by the professor. The
architectural design of the building must be “millennial” or modern. To be submitted to
on Monday of the following week.
6. Recommended learning materials and resources for supplementary reading.

50. Reinforced Concrete Design by Gillesania, Chapter1
Other instructional materials may also be used by students like related textbooks and
videos to enrich their knowledge about the subject.
7. Flexible Teaching Learning Modality (FTLM) adopted
Remote
Asynchronous (modules, exercises, problem sets, etc…)
8. Assessment Task
In this part, students are given Self-assessment Questions (SAQs) and asked to
consider broader aspects of the different topics taken up. Quizzes have been prepared
for this part and can be found in the original module by the author.
9. References
Besavilla, V., Reinforced Concrete Design, 2016
Esplana, Dindo Civil Engineering Review Manual, 2015
Gillesania, DIT Reinforced Concrete Design, 3rd
Edition, 2015
National Structural Code of the Philippines 2015
Nilson, W. Reinforced Concrete Design, 2010