MODULE II : Analysis and Design of Singly Reinforced Section

(WSD and USD Method)

Course No. : CE - 514
Course Title : Reinforced Concrete Design

Course Description : This course is concerned with the design, applications and code
specifications used in structural reinforced concrete members subjected to flexure
(beams, girders, joists, lintels, girts, etc.), tension and compression members
(columns), combined stressed members (beams-columns), beam column connections
using the Elastic Limit Method, also known as the Alternate Stress Design (ASD) or
Working Stress Design (WSD) and the Plastic Limit Method (PLM) or the Ultimate
Strength Design (USD). Applications and specifications as applied to buildings,
bridges, and other reinforced concrete structures are also given emphasis. A thorough
knowledge and proficiency in Structural Theory is imperative.

Total Learning Time: 4 Units (3 hours lecture, 3 hours computational laboratory per week)
Pre-requisites : Theory I and Theory II
Overview:
Reinforced concrete beam design consists primarily of producing member details
which will adequately resist the ultimate bending moments, shear forces and torsional
moments. At the same time serviceability requirements must be considered to ensure that the
member will behave satisfactorily under working loads.
Design and detailing of the bending reinforcement must allow for factors such as
anchorage bond between the steel and concrete. The area of the tensile bending reinforcement
also affects the subsequent design of the shear and torsion reinforcement. Arrangement of
reinforcement is constrained both by the requirements of the codes of practice for concrete
structures and by practical considerations such as construction tolerances.

Learning Outcomes:
After completing this module, the students should be able to:
1. Analyze the impact of the external loads to the rectangular beam section
using WSD Method.
2. Design the Singly-reinforced section using WSD Method.
3. Analyze the impact of the external loads to the rectangular beam section
using USD Method.
4. Design the Singly-reinforced section using USD Method.

Indicative Content: Analysis of Singly Reinforced Section (WSD or ASD), Design of

Singly Reinforced Section (WSD or ASD), Analysis of Singly Reinforced Section
(USD or PLM), Design of Singly Reinforced Section (USD or PLM)

Pre-Assessment: Answer the following Questions:

1. What is a cracking moment?
2. Discuss the difference between WSD and USD.
3. What do you think is the most efficient and the most economical design?
4. Compute the bending stresses in the beam shown. Use the transformed area
method, n = 8 and M = 110 ft-k.
 Figure:

Discussion:

A. Working Stress Design (WSD) or Alternate Stress Design (ASD)

A.1. Analysis of Singly Reinforced Section (WSD)

In working-stress design a margin of safety is provided by permitting calculated
flexure stress to reach only a certain percentage of the ultimate strength of the concrete or of
the yield strength of the reinforcing. The percentage are sufficiently small so that an
approximately linear relation exists between the stress and strain in the concrete as well as in
the reinforcing. The code permits the design of reinforced concrete members using service
loads and the working-stress design method, except that the method is now referred to as the
Alternate Design Method.

Elastic Stresses:
The cracking moment of a beam is normally quite small compared to the service load
moment. Thus when the service loads are applied, the bottom of the beams cracks. The
cracking of the beam does not necessarily mean that the beam is going to fail. The
reinforcing bars of the concrete side begin to pick up the tension caused by the applied
moment.
The strain in the concrete and in the steel will be equal at equal distance from the
neutral axis. But if the strains in the two materials at a particular point are the same, their
stresses cannot be the same. Since they have different moduli of elasticity. The ratio of the
steel modulus to the concrete modulus is called modular ratio (n).

Thus: n = _Es_
Ec
 If the modular ratio for a particular beam is 10, the stress in the steel will be 10 times
the stress in the concrete at the same distance from the neutral axis or one square inch of steel
will carry the same total force as 10in2 of concrete.
Steel bars are replaced with an equivalent area of fictitious concrete (nAs) which
supposedly can resist tension. This area is referred to as the transformed area.

fc

NA

nAs
transformed area
fs/n

Example: Calculate the bending stresses in the beam by using transformed area method.
n = 9 and M = 70 ft.-k.

Solution:
12”

x = 6.78
17” 20” NA

3#9 nAs 17-x = 10.22

3”
12”

As of 3 # 9 = 3 in2
Take moment at the NA: 12x (x/2) = 9 (3in2) (17 - x)
x = 6.78 inches

Moment of Inertia (I) = 1/3 bd3 + Ad2 (at the bottom of the rectangle)
I = 1/3 (12) (6.78)3 + 27 (10.22)2 = 4067 in4
Bending Stresses:
fc = My/I = 12 (70,000) (6.78)/4067 = 1400 psi

fs = nMy/I = 9 (12) (70,000) (10.22)/4067 = 18,998 psi

Note:
a) M – Applied Moment (External Moment) = wL2/8 for simply supported or
moment can be solved using share and moment diagram.
b) M – Internal Moment or Flexural Moment = If/y, fc and fs – are bending stresses
 A.2. Design of Singly Reinforced Section (WSD)
Straight-Line or Elastic Load Theory:
Derivation of Design Equations

b Δc fc

kd/3
kd C
d
NA jd
h
d - kd
As
T
Δs fs/n

cross-section strain diagram stress diagram

where:
C = ½ fc b kd
T = Asfs
Mc = ½ fc b k jd2 : resisting moment of concrete
Ms = As fs jd : resisting moment of steel
k = ranging its ideal value from 0.30 to 0.45
j = ranging its ideal value from 0.85 to 0.90

Notations:
fc = compressive unit stress of concrete at the surface most remote from the NA.
fs = tensile unit stress in the longitudinal reinforcement.
b = width of the beam
d = effective depth from the center of the steel bars to the extreme fiber of concrete.
kd = distance from the NA to the extreme fiber of concrete.
jd = distance between the compressive force C and tensile force T.
As = cross-sectional area of steel bars.
ρ = rho – ratio of the area of steel to that of the effective area of concrete. ρ = As/bd
Es = modulus of elasticity of steel
Ec = 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 the concrete.

Thus: Modulus of Elasticity = unit stress

unit deformation

Es = fs/Δs Δs = fs/Es Ec = fc/Δc Δc = fc/Ec

 Δc = fc/Ec : Δc = fc Es -------------------- (1)
Δs fs/Es Δs fs Ec

From the figure:

Δc = __Δs__ : Δc = __kd___------------------- (2)

kd d – kd Δs d - kd

Equating equation (1) and (2): considering n = Es/Ec

The derive equation for k:

k= n this k is use when designing the beam section.

fs/fc + n

Take Moment at T: Mc = C jd where: C = ½ fc b kd

Mc = ½ fc b k jd2 ---------------------------- (1)

Take Moment at C: Ms = Tjd where: T = As fs

Ms = As fs jd considering ρ = As/bd : As = ρbd
2
Ms = ρb fs jd -------------------------------- (2)

For balanced design: Ms = Mc: simplifying -------_fc_ = _2ρ_ -------------- (3)

fs k
From the equation: _nfc_ = __k__ or __fc__ = ___k___ ----------------------(4)
fs 1–k fs n (1 – k)

Equating equations (3) and (4) and simplify and completing the squares (ρn)2
The derive equation for k:

k = √ 2ρn + (ρn)2 – ρn this k is use when investigating the beam section

Additional Equations:

d = ____Mc_____ : R = ½ fc k j : jd = d – kd/3 : j = 1 – k/3

√ ½ fc k j b

Steps in Designing Singly Reinforced Section

1. Compute the live load moment;

M = _wL2 ; for simply supported beam
8
M = _wL2 ; for continuous beam
12
2. Compute the design constant: k, n, j, R, fc and etc.
3. Solve for the effective depth “d”.
4. Add 50mm to 100mm to the computed “d” to take care of the dead load.
 5. Add the dead load and the live load
6. Re-compute for the total moment
7. Re-check the computed d = √M/Rb
Note: if d is less than the computed “d” in step 3, then it is safe.

8. Compute the steel requirements: As = M/fsjd

8.1. Determine the number of bars needed

9. Check for shearing stress v =Vc/bd

9.1. Vc = critical shear at distance d from the face of the support
= wL/2 – wd

10. If v > va; use 10mmØ stirrups

va = 0.09 √fc’ in MPa

11. Compute the spacing of stirrups; S = Avfv/v’b

Max spacing = d/2

12. Check the bond stress: Uo = V/Σojd

When Uo < Ua then it is safe
V = maximum shear
Ua = 10.14 √fc’
Ø

13. If Ua < Uo then add more steel bars.

14. Check spacing of steel reinforcement:
S should not be less than 25mmØ
S should not be less than the diameter of main bars.
40mm minimum clear cover
60mm covering if there are no stirrups.
65mm covering with stirrups
70mm covering when exposed to earth surface.

Problem: Design a section of a concrete beam reinforced for tension only. The beam is
simply supported on a span of 6.0m and carries a load of 18kN/m. fc’ = 17.2 Mpa, fs = 124
Mpa, n = 12. Assume b = 250mm. Check the beam.

Solution:
18kN/M

6.0m

Moment: M = wL2/8 = 18 (6.0)2/8 = 81.0 x 106 N – mm

Design Constant: n = 12 fc = 0.45fc’
fc = 0.45 (17.20) = 7.74Mpa
 k = _____12_____ = 12/28.02 = 0.43
124/7.74 + 12
j = 1 – 0.43/3 = 0.857
R = ½ (7.74) (0.43) (0.857) = 1.42 Mpa

Solve for d: d = __81 x 106__ = 477.67mm

√ 1.42 (250)

Add: 50mm d = 477.67mm + 50mm = 527.67 say 525mm

h = 525mm + 65mm –--------- covering
h = 590mm

Dead Load = (0.59m) (0.25m) (23.54kN/m3) = 3.466kN/m

Total Load = 3.466 + 18 = 21.466kN/m or 21,466N/m
Re-compute for Moment: M = 21.466 kN/m (6)2/8 = 96.598kN-m = 96.598 x 106 N-mm

Re-check for effective depth:

d = __96.598 x 106__ = 521.64mm < 525mm: safe
√ 1.42 (250)
Compute for Steel Reinforcements:
As = __M__ = __96.598 x 106___ = 1731.44mm2
fs jd 124 (0.857) (525)

Try: 28mm Ø bars: As1-28 = Π (28)2/4 = 615.75mm2

Number of bars needed: N = 1731.44/615.75 = 2.81pcs say 3 pcs.
Therefore, Use: 3 – 28mm Ø bars.
Check for shearing stress: v = Vc/bd Vc = wL/2 – wd
Vc = 21,466 (6)/2 – 21,466 (0.525)
Vc = 53,128.35N
v = 53,128.35 = 0.405N/mm2 = 0.405MPa
250(525)
Allowable shearing stress: va = 0.09 √ 17.2 = 0.37 MPa
Therefore, v > va: stirrups are needed. Use 10mm Ø bar for stirrups
Av = 2Π (10)2/4 = 157.08mm2
v’= v – va = 0.405 – 0.37 = 0.035 MPa

Spacing of stirrups: S = Av fv S = 157.08 (124) = 2,226.48mm

v’b 0.035 (250)

Maximum S = d/2 = 525/2 = 262.50mm say 260mm. Therefore, Use 260mm o.c.
Check for bond stress: Uo = ___V___ V = wL/2 V = 21,466 (6)/2 = 64,398N
∑ojd

Sum of perimeters of steel bars: ∑o = 3 Π (28) = 264mm

Actual bond stress: Uo = ____64,398 N____ = 0.54MPa
264 (0.857) (525)
Allowable Bond Stress: Ua = 10.14 √ fc’ = 10.14 √ 17.2 = 1.50 MPa
D 28

Note: Uo < Ua, therefore, it is safe.

Check Spacing of bars:
250mm

Note: Use 40mm minimum clear cover

a) 2S + 3 (28) + 80 = 250mm
S = 43mm > 25mm
b) S = 43mm > 28mm

40 S S 40

Therefore: Adopt 250mm x 590mm x 6.0m

Reinforced with 3 – 28mmØ bars

B. Ultimate Strength Design (USD) or Plastic Limit Method (PLM)

In the strength design method (formerly called ultimate strength method), the service
loads are increased by factors to obtain the load at which failure is considered to be
"imminent". This load is called the factored load or factored service load. The structure or
structural element is then proportioned such that the strength is reached when the factored
load is acting. The computation of this strength takes into account the nonlinear stress-strain
behavior of concrete. The strength design method may be expressed by the following:
Strength provided ≥ [strength required to carry factored loads] Where the "strength provided"
(such as moment strength) is computed in accordance with the provisions of a building code,
and the "strength required" is that obtained by performing a structural analysis using factored
loads.
Among the several advantages of the strength design method as compared to the
alternate design method are the following:
1. The derivation of the strength design expressions takes into account the non-linear
shape of the stress-strain diagram. When the resulting equations are applied,
decidedly better estimates of load-carrying ability are obtained.
2. With strength design a more consistent theory is used throughout the designs of
reinforced concrete structures. For instance, with the alternate design method the
transformed area or straight-line method is used for beam design and a strength
design procedure is used for columns.
3. A more realistic factor of safety is used in strength design. With working-stress
design the same safety factor is used for dead and live loads, whereas such is not
the case for strength design. The designer can certainly estimate the magnitudes of
 the dead loads that a structure will have to support more accurately than the live
loads. For this reason, the use of different safety factors in strength design for the
two types of loads is a definite improvement.
4. A structure designed by the strength method will have a more uniform safety
factor against collapse throughout. The strength method takes full advantage of
high-strength steels, whereas working-stress design partly does so.
5. The strength methods permit more flexible designs than does the alternate design
method. For instance, the percentage of steel may be varied quite a bit. As a result,
large sections may be used with small percentages of steel or small sections with
large percentages of steel. Such variations are not the case in the relatively fixed
alternate design method. If the same amount of steel is used in strength design for
a particular beam as would be required by the alternate design method, a smaller
section will result. If the same size section is used as required by working stress
design, a smaller amount of steel will be required.

SAFETY PROVISIONS
Structures and structural members must always be designed to carry some reserve
load above what is expected under normal use. Such reserve capacity is provided to account
for a variety of factors, which may be grouped in two general categories:
 Factors relating to overload.
 Factors relating to understrength (that is, less strength than computed by acceptable
calculating procedures).
Overloads may arise from changing the use for which the structure was designed,
from underestimation of the effects of loads by oversimplification in calculation procedures,
and from effects of construction sequence and methods. Understrength may result from
adverse variations in material strength, workmanship, dimensions, control, and degree of
supervision, even though individually these items are within required tolerances. In the
strength design method, the member is designed to resist factored loads, which are obtained
by multiplying the service loads by load factors. Different factors are used for different
loadings. Because dead loads can be estimated quite accurately, their load factors are smaller
than those of live loads, which have a high degree of uncertainty. Several load combinations
must be considered in the design to compute the maximum and minimum 2 design forces.
Reduction factors are used for some combinations of loads to reflect the low
probability of their simultaneous occurrences. The ACI Code presents specific values of load
factors to be used in the design of concrete structures. In addition to load factors, the ACI
Code specifies another factor to allow an additional reserve in the capacity of the structural
member. The nominal strength is generally calculated using accepted analytical procedure
based on statistics and equilibrium; however, in order to account for the degree of accuracy
within which the nominal strength can be calculated, and for adverse variations in materials
and dimensions, a strength reduction factor, Ø, should be used in the strength design method.
To summarize the above discussion, the ACI Code has separated the safety provision
into an overload or load factor and to an under capacity (or strength reduction) factor, Ø.
A safe design is achieved when the structure's strength, obtained by multiplying the nominal
strength by the reduction factor, Ø, exceeds or equals the strength needed to withstand the
factored loadings (service loads times their load factors).
The requirement for strength design may be expressed:
𝐷𝑒𝑠𝑖𝑔𝑛 𝑆𝑡𝑟𝑒𝑛𝑔ℎ𝑡 ≥ 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑 (𝑖. 𝑒., 𝑟𝑒𝑞𝑢𝑖𝑟𝑑 𝑆𝑡𝑟𝑒𝑛𝑔ℎ𝑡)
Ø𝑃𝑛 ≥ 𝑃𝑢
Ø𝑀𝑛 ≥ 𝑀𝑢
Ø𝑉𝑛 ≥ 𝑉𝑢

Where 𝑃𝑛, 𝑀𝑛 𝑎𝑛𝑑 𝑉𝑛 are "nominal" strengths in axial compression, bending

moment, and shear, respectively, using the subscript n.

𝑃𝑢, 𝑀𝑢 𝑎𝑛𝑑 𝑉𝑢 are the factored load effects in axial compression, bending moment,
and shear, respectively, using the subscript u.

Given a load factor of 1.2 for dead load and a load factor of 1.6 for live load, the
overall safety factor for a structure loaded be a dead load, 𝐷L, and a live load, 𝐿L, is:
𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 = 1.2𝐷L + 1.6𝐿L (1)
𝐷L + 𝐿L (∅ )

LOAD FACTORS AND STRENGTH REDUCTION FACTORS

Overload Factors 𝑼, the factors 𝑼 for overload as given by ACI-9.2 are:
𝑈 = 1.4(𝐷 + 𝐹)
𝑈 = 1.2(𝐷 + 𝐹 + 𝑇) + 1.6(𝐿 + 𝐻) + 0.5(𝐿𝑟 𝑜𝑟 𝑆 𝑜𝑟 𝑅)
𝑈 = 1.2𝐷 + 1.6(𝐿𝑟 𝑜𝑟 𝑆 𝑜𝑟 𝑅) + (1.0𝐿 𝑜𝑟 0.8𝑊)
𝑈 = 1.2𝐷 + 1.6𝑊 + 1.0𝐿 + 0.5(𝐿𝑟 𝑜𝑟 𝑆 𝑜𝑟 𝑅)
𝑈 = 1.2𝐷 + 1.0𝐸 + 1.0𝐿 + 0.2𝑆
𝑈 = 0.9𝐷 + 1.6𝑊 + 1.6𝐻 𝑈 = 0.9𝐷 + 1.0𝐸 + 1.6𝐻

Where: 𝐷 = dead load; L = live load; 𝐿𝑟 = roof live load; S = snow load;
F = load due to weights and pressures of fluids;
R = rain load; W = wind load;
E = earthquake load; −defined densities and controllable maximum heights;
H = load due to weight and pressure of soil, water in soil or other materials;
T = the cumulative effect of temperature, creep, shrinkage and differential settlement.

Strength Reduction Factors ∅

The factors ∅ for understrength are called strength reduction factors according to ACI-9.3
and are as follows:
Strength Condition ∅ 𝑭𝒂𝒄𝒕𝒐𝒓𝒔
Flexure (with or without axial force)
 Tension-controlled sections …………………………….. 0.90
 Compression-controlled sections
Spirally reinforced …………….………………….. 0.75
Others ………………………………..……………. 0.65
Shear and torsion …..……………………………….……..……..... 0.75
Bearing on concrete …….…………………………….….………... 0.65
Post-tensioned anchorage zones ………….……………………….. 0.85
Struts, ties, nodal zones, and bearing areas in strut-and-tie models...0.75
 FLEXURE IN BEAMS
Reinforced concrete beams are nonhomogeneous in that they are made of two entirely
different materials. The methods used in the analysis of reinforced concrete beams are
therefore different from those used in the design or investigation of beams composed entirely
of steel, wood, or any other structural material.

Two different types of problems arise in the study of reinforced concrete:

1. Analysis. Given a cross section, concrete strength, reinforcement size and location,
and yield strength, compute the resistance or strength. In analysis there should be one unique
answer.
2. Design. Given a factored design moment, normally designated as 𝑀𝑢 select a
suitable cross section, including dimensions, concrete strength, reinforcement, and so on. In
design there are many possible solutions.
The Strength Design Method requires the conditions of static equilibrium and strain
compatibility across the depth of the section to be satisfied.

The following are the assumptions for Strength Design Method:

1. Strains in reinforcement and concrete are directly proportional to the distance from
neutral axis. This implies that the variation of strains across the section is linear, and
unknown values can be computed from the known values of strain through a linear
relationship.
2. Concrete sections are considered to have reached their flexural capacities when
they develop 0.003 strain in the extreme compression fiber.
3. Stress in reinforcement varies linearly with strain up to the specified yield strength.
The stress remains constant beyond this point as strains continue increasing. This implies that
the strain hardening of steel is ignored.
4. Tensile strength of concrete is neglected.
5. Compressive stress distribution of concrete can be represented by the
corresponding stress-strain relationship of concrete. This stress distribution may be simplified
by a rectangular stress distribution as described later.

B.1. Analysis of Singly Reinforced Section (USD)

A. Material Properties:
A.1. Concrete
a) Compressive Strength (fc’)
fc’= is the cylindrical strength or ultimate strength. It is taken from the
compression test of a 6” Ø x 12” cylinder @ 28days of curing period at
a uniform loading applied.
b) Stress-Strain Diagram
fc’ reached a strain from 0.002 to 0.003. In calculating fc’ the strain is
assumed to be reached @ Єcu = 0.003
Thus, Єcu = fc’/Ec
Where: Єcu = ultimate concrete strain, fc’ = compressive strength of
concrete, and Ec = modulus of elasticity of concrete.
 c) Modulus of Elasticity
Ec = 4,700√ fc’ in MPa for Normal Weight Concrete
For other weight of concrete:
Ec = wc1.5 (0.043) √ fc’ : fc’ in MPa
d) Tensile Strength of Concrete
tensile strength of concrete ranges from 10% to 15% of fc’

A.2. Reinforcing Steel

a) Yield Strength (fy)
- measured at a strain of 0.005
- stress at which there is a large increase in strain or deformation with
small increase in load or stress.

Grades of Steel
40 ------------ fy = 40 ksi = 276 MPa
50 ------------ fy = 50 ksi = 345 MPa
60 ------------ fy = 60 ksi = 414 MPa

b) Modulus of Elasticity (Es)

Es = 200,000 MPa
Es = 29 x 106 psi

Stress-Development in Singly Reinforced Rectangular Beams

Figure: P a
W
_ _ _ _ _ _ _ _ _ __ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _

b fc fc’ fc’ 0.85fc’

a c C a C

c
d NA (d–a/2)

As
As fs fs fs<fy fs<fy T

Section uncracked stage cracked stage actual stress rectangular

diagram stress block
C = Compressive Force
c = from actual stress diagram, c is the distance from the NA to the outer most fiber of the
compression side.
The actual parabolic stress is converted to rectangular stress block
C = (average stress x ab)
C = 0.85fc’ x ab
The stress at any section a-a will vary depending on the magnitude of the applied load.
1. Uncracked Stage:
- At low loads both concrete and steel will resist tensile.
- Stresses are elastic
- Stress is proportional to strain
2. Cracked Stage:
- At moderate loads section will be cracked with only steel resisting tensile stresses
- Stresses are elastic
- Stress is proportional to strain
3. Ultimate Stage:
- At ultimate loads cracked becomes larger
- Stresses are elastic
- Stress is no longer proportional to strain

Notations:
b = width of the section
d = effective depth from the compression side of the beam to the center of the
reinforcing bars.
h = total depth of the section
As = total steel area of reinforcement
Mn = theoretical (nominal) ultimate moment
Mu = ØMn = usable ultimate moment
Ø = capacity reduction factor

b Єcu = 0.003 fc’ 0.85fc’

a C a C
c
d d
h NA (d-a/2)
(d-c)
As

Єs T = Asfs T = Asfs

Note:
a = β1c
β1 = 0.85 when fc’< 30 MPa
β1 = 0.85 – 0.008 (fc’- 30) when fc’> 30 MPa
C = 0.85 fc’ab
T = Asfs
By Static Equilibrium: ∑Fh = 0: C=T
0.85fc’ab = Asfs
a = Asfs/0.85fc’b

∑M@As = 0
Mn = C (d - a/2)
Mn = 0.85 fc’ab (d – a/2) or Mn = Asfs (d – a/2)
Mu = ØMn

The value of “a” depends on fs – whether the steel will yield or not.
fs = Єs Es ------- for Єs < Єsy
fs = fy ----------- for Єs > Єsy

From Strain Compatibility:

___Єs___ = __0.003__
d–c c

Єs = 0.003 (d – c)
c
Conditions or Modes of Failure:
Beam fails at:
1. Primary Compression Failure:
- Over reinforced section
- Єs < Єsy ------------------ fs = Єs Es
- Concrete Fail First
- Єcu = 0.003
2. Balanced Failure:
- Balanced Section
- Єs = Єsy ------------------ fs = fy
- Simultaneous Failure
- Єcu = 0.003
3. Primary Tension Failure:
- Under Reinforced Section
- Єs > Єsy ------------------ fs = fy
- Steel Fail First
- Єcu = 0.003
Єcu = ultimate concrete strain
4. Load Factors: DL = 1.40 and LL = 1.70 -------- ACI 2001

At balanced Condition: ρb = 0.85 β1 fc’ __600__

fy 600 + fy

ρb = balance steel ratio (rho balance)

 Derivation: At balanced condition, ρ = ρb
Єs = Єsy ----------fs = fy
By Static Equilibrium: C = T
0.85fc’ab = Asfy
a = Asfy/0.85fc’ab : ρ = As/bd or ρb = As/bd or As = ρb bd

a = ρb bd fy or a = ρb d fy but a = β1c
0.85fc’b 0.85 fc’

c = __ρb d fy__ --------------- (1)

0.85 β1 fc’

From Strain Diagram: by ratio and proportion

___Єsy___ = __0.003__
d–c c

Єsy = __0.003 (d - c)_ but: Єsy = fy/Es

c Es = 200,000 MPa
_fy_ = _0.003 (d – c)_
Es c
Simplify:
c = _0.003 Es d__ -------------- (2)
fy + 0.003Es

Equate eq. (1) and eq. (2) and simplify, the derive equation for rho balanced is:

ρb = 0.85 β1 fc’ __600__

fy 600 + fy

Problem: A rectangular beam has a width of 250mm and an effective depth of 500mm.
the allowable stresses are fc’= 28.0 MPa and fy = 410.0 MPa. Calculate the usable ultimate
moment when: a) As = 1,638mm2
b) As = 4,914mm2
Given:
fc’= 28 MPa b = 250mm a) As = 1,638mm2
fy = 410 MPa d = 500mm b) As = 4,914mm2

Required: Mu = usable ultimate moment: fc’< 30MPa, therefore β1 = 0.85

ρb = 0.85 (0.85) __28__ ___600___ = 0.0293

410 600 + 410

a) When As = 1,638mm2

ρ = __As = ___1638___ = 0.0131 < 0.0293 ; under-reinforced section

bd 250 (500) primary tension failure
 use: fs = fy

solve for a: a = __Asfy__ = _1,638 (410)_ = 112.87 mm

0.85fc’b 0.85 (28) 250

Mn = T (d – a/2) = C (d – a/2)
Mn = Asfy (d – a/2)
= 1638 (410) (500 – 112.87/2) = 297, 889, 382. 70 N-mm x 1/106
= 297.89 kN-m

Mu = Ø Mn
= 0.90 (297.89)
= 268.10 kN-m.

b) As = 4,914mm2

ρ = __4914__ = 0.0393 > (ρb) 0.0293 ; over-reinforced section

250 (500) primary compression failure
fs < fy, use fs = Єs Es
fs = (d – c) 0.003
c
fs = (β1d – a) 600
a
Solve for a:
C=T
0.85 fc’a b = As fs
0.85 (28) a (250) = 4914 (β1d – a) 600
a
2
a + 495.53a – 210, 600 = 0
a = 273.76mm

Solve for fs: fs = 0.85 (500 – 273.76) 600 = 331.47 MPa

273.76

Mn = T (d – a/2) = Asfs (d – a/2)

Mn = 4914 (331.47) (500 – 273.76/2) x 1/106 = 591.47 kN-m
Mu = 0.90 (591.47)
Mu = 532. 32 kN-m

B.2. Design of Singly Reinforced Section (USD)

Free Design: bd2 = _______Mu________

ϕρfy (1 – 0.59ρfy/fc’)

Mu = ultimate factored moment

 ρmin < ρ < ρmax
ρmin = 1.4/fy ----minimum steel ratio
ρmax = 0.75ρbal ------maximum steel ratio
For Restricted Design:

ρ = 0.85 fc’ 1- 1 – ____2Mu____

fy √ 0.85ϕfc’bd2

Recommended Steel Ratio:

ρ = 0.18fc’/fy or ρ = ½ ρmax

Problem: Design the beam structure considering the given below:

fc’= 27.58 MPa Assume beam weight = 2.20kN/m
fy = 413.70 MPa Dead Load = 14.60 kN/m
Live Load = 21.90 kN/m

3.60m 1.80m

Total beam weight = 14.60 + 2.20 = 16.80kN/m

Factored Loads: DL = 1.4 (16.80) = 23.52 kN/m
LL = 1.7 (21.90) = 37.23 kN/m

Determine the position of the live load that will give a maximum moment,
considering shear and moment diagram.

a)
LL = 37.23 kN/m
DL = 23.52 kN/m

b)
LL = 37.23 kN/m
DL = 23.52 kN/m

c) LL=37.23kN/m
DL = 23.52 kN/m

The maximum moment computed using shear and moment diagram:

Mu = - 98.31 kN-m
 Mu = + 80.50 kN-m

Design the beam using the largest moment: It occurs at the cantilever section.
Mu = - 98.31 kN-m

bd2 = _______Mu________ use: ρ = 0.18fc’/fy

ϕρfy (1 – 0.59ρfy/fc’) ρ = 0.18 (27.58)/413.70
ρ = 0.012

Compute: ρmin = 1.4/fy = 1.4/413.70 = 0.0034 < ρ ----- ok

ρmax = 0.75 ρbal

ρb = 0.85 β1 fc’ __600__ = 0.85 (0.85) 27.58 (600) = 0.0285

fy 600 + fy 413.70 (600 + 413.70)

ρmax = 0.75 (0.0285) = 0.0214 > ρ -------- ok

bd2 = ______________98.31 x 106__________________

0.90(0.012) 413.70 [1 – 0.59(0.012) (413.70/27.58)]

bd2 = 24,617,733.70mm3

for economical design: ½ < b/d < 2/3 say use: b/d =2/3 b = 2d/3

2/3 d(d2) = 24, 617, 733.70 mm3

d = 333.0mm
b = 222.0mm
Try:
b = 230mm
d = 340mm
Assume one layer bars, with 40mm clear cover and 10mm diameter stirrups:
Total effective covering = 40mm + 10mm + Ø/2 < 65mm, say total cover = 70mm
Unit weight of concrete = 23.54kN/m3
Total depth h = 340mm + 70mm = 410mm
To check: actual weight < assumed weight
Actual weight = 0.23 (0.41) 23.54kN/m3 = 2.216 ≈ 2.2 say ok.
As = ρbd ------------------using #8 bar; Ø = 25.4mm; Ab = 510mm2
d = 410 – 50 – 25.40/2 : d = 347.30mm > 340mm -------- ok
As = 0.012 (230) (347.30) = 958.55mm2
Number of bars needed: N = As/Ab = 958.55/510 = 1.88 say 2 pcs. ---- top bars
Check for bars spacing: S = 230 – 40(2) – 10(2) – 25.4(2) = 79.20mm > 25.40mm--ok

For Positive Moment: Mu = 80.50 kN-m b = 230mm d = 410mm

Using restricted steel ratio:

ρ = 0.85 fc’ 1- 1 – ____2Mu____

 fy √ 0.85ϕfc’bd2

Using # 7 bar: As = 380mm2 Ø = 22mm

d = 410 – 50 – 22/2 = 349mm

ρ = 0.85 _(27.58)_ 1 - 1 – ______2 (80.50 x 106)______ = 0.00832

413.70 √ 0.85(0.90) 27.58 (230) (349)2

ρmin < ρ = 0.00832 < ρmax ---------- ok

As = 0.00832 (230) (349) = 667.85mm2
Number of bars: N = 667.85/380 = 1.76 say 2 pcs. ---- bottom bars
Check spacing of bars: S = 230 – 40(2) – 10(2) – 22(2) = 86mm > 22mm---ok

Evaluation: Solve the following Problems:

1. Compute the bending stresses in the beam using transformed area method.
M = 100kN-m and n = 10.

100 100 100 mm 200

100mm x 100

x - 100
500mm

3#9 430-x nAs

70mm

2. A concrete beam reinforced for tension only is required to carry a uniformly

distributed load of 27,000N/m including its own weight on a simple span of 9.0m. The
overall depth “h” is twice the width “b” and the center of steel reinforcement is to be located
at a distance (1/10) h 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 and n = 15.

3. A reinforced concrete beam 40cm wide by 60cm deep and with tensile
Reinforcement of 4 – 25mmØ steel bars is simply supported on a clear span of 8.0m.
Calculate the maximum concentrated load it can carry at mid-span if the tensile stress fc and
fs do not exceed 45kg/cm2 and 1,300kg/cm2 respectively. Assume n = 12 and disregard the
weight of the beam. Use effective covering of 10cm.

4. Find the safe ultimate moment that the beam could carry. It has an effective depth
of 500mm and a width of 280mm. The effective covering is 60mm with reinforcement of
4 - 36mmØ. fc’= 25MPa and fy = 400MPa.

5. Given a singly reinforced section with a width of 250mm and an effective depth of
500mm. Assuming fc’= 25 MPa, fy = 230 MPa, and using the rectangular stress block
method, compute the ultimate moment for:
a) As = 3,000mm2 b) As = 9,000mm2 c) ρ = ρb
d) Plot Mu (vertical axis) against ρ (horizontal axis) and comments on the curve.
Use and increment of 1000mm2 with As varying from 3000 to 9000mm2

6. Compute the ultimate moment (Mu) for the section shown considering the spread
of reinforcement. Use fc’= 25MPa and fy = 410MPa.

300mm

L 600mm

12 #7

150mm
150mm

Additional Readings: Analysis and Design of Doubly Reinforced Section for:

a) Working Stress Design Method
b) Ultimate Strength Design Method.

References:
: Association of Structural Engineers of the Philippines; National Structural Code
of the Philippines (NSCP C101, Vol. 1 Buildings and Other Vertical Structures),
7th Edition, 2016
: McCormac, Jack C.; (2005); Design of Reinforced Concrete; 7th Edition; John
Wiley & Sons, New York, USA
: Nilson, Arthur H. and Winter, George et al.; (2000); Design of Concrete
Structures; 13th Edition; McGraw-Hill, Inc.
: Gambhir,M.L.; (2010); Fundamentals of Reinforced Concrete Design
: Meyer, C., (1997); Design of Concrete Structures
: American Concrete Institute; ACI 318 Building Code Requirements for
Reinforced Concrete; Latest Edition

Prepared by:
Engr. Joel C. Villaruz, Ph.D.
Assistant Professor I