This document discusses the analysis of singly and doubly reinforced concrete beam sections. It begins by defining singly reinforced sections as having tension reinforcement only, while doubly reinforced sections have reinforcement in both tension and compression zones. Design steps are provided for both section types, including calculating loads, moments, reinforcement areas, and shear reinforcement. Formulas and assumptions used in the design process are also outlined. The goal is for students to learn to properly design reinforced concrete beam sections based on given structural loads and material properties.

1. Design of Structures
Chapter2
ANALYSIS OF SINGLY & DOUBLY REINFORCED SECTIONS

2. ANALYSIS OF SINGLY & DOUBLY REINFORCED SECTIONS
This chapter emphasises on selection of beam sections, comparison of singly reinforced and
doubly reinforced sections. Design formulas and problems on design of sections.
At the end of the chapter, students are expected to satisfactorily meet the following course
objectives and learning outcomes:
OBJECTIVES LEARNING OUTCOMES (LO)
1. Establish design loads,
2. Understand the design code which is used to design
structure
3. Discuss the use of basic approaches and more unique
method to analyzestructure by hand,
4. Identify the responsibility of the engineer to be ethical
in dealing with others and in the presentation of result
from analysis and design.
1. Use the design codes in order to producethe design of structures
2. Apply iteration methods to the initialdesign to converge on an efficient final
3. Maintain ethics within the framework ofprofessional conduct
4. Design and analyze of basic structural elements of reinforced concrete including:
a] Singly and doubly reinforced beams b]One-way slabs c]Columns and footings
5. Calculate the reinforcement details which include:
a.Maximum and minimum reinforcementarea
b.Spacing of reinforcement
c.Curtailment and anchorage of d lapping of reinforcement
6. Present the design details to show reinforcement and size requirements
for basic members

4. SINGLY REINFORCED SECTION:
• The beam that is longitudinally reinforced only in tension zone,
• Tension – reinforcement
• Compression is carried by the concrete.
• To tie the stirrups -two rebars are utilized in the compression zone

5. DOUBLY REINFORCED SECTION:
• The beam that is reinforced with steel both in tension and compression zone
• This type of beam is mainly provided when the depth of the beam is restricted. Depth is
reinforced on the tension side only it might not have sufficient resistance to oppose the
bending moment.
• The moment of resistance cannot be increased by increasing the amount of steel in tension
zone.
• It can be increased by making the beam over reinforced but not more than 25% on the
strained side.
• Thus a doubly reinforced beam is provided to increase the moment of resistance of a beam
having limited dimensions.

6. FLANGED SECTION:
Under certain conditions, T and L beam are more economical than the rectangular
beam.
T-beam (or tee beam)
• The top of the t-shaped cross section serves as a flange - compressive stresses.
• The web (vertical section) of the beam below the compression flange serves to
resist shear stress and coupled forces of bending.
• The T-beam has a big disadvantage compared to an I-beam because it has no
bottom flange with which to deal with tensile forces.
• One way to make a T-beam more efficient structurally is to use an inverted T-
beam with a floor slab or bridge deck joining the tops of the beams.
• Done properly, the slab acts as the compression flange.

9. ASSUMPTIONS
1. Plane section normal to the axis remains plane after bending.
2. The strain diagram is triangular. The maximum strain in concrete at the outermost compression fiber is taken as
0.35% in bending regardless of the strength of concrete.
3. The stress distribution in the concrete is parabolic. The maximum compressive stress is equal to 0.67fcu/ m.
4. The tensile strength of the concrete is ignored.

10. Modes of Failure of Sections in Bending or Flexure
BALANCED SECTION:
• If the ratio of steel to concrete in a beam is such that the maximum strain in the two materials
reaches simultaneously, a sudden failure will occur with less alarming deflection.
b

11. UNDER REINFORCED SECTION:
When the amount of steel is kept less than the balanced condition, the neutral axis shifts upward to
satisfy the equilibrium condition, and the structure fails due to yielding in steel with alarming
deflection.

12. OVER REINFORCED SECTION:
• When the amount of steel is kept more than the balanced condition, the neutral axis shifts
downward to satisfy the equilibrium condition, and the structure fails due to Failure in concrete.

13. Design Steps – Singly Reinforced Simply Supported
• Step1: Ultimate / Factored Load Calculation
• Dead Load = B x D x 1m x Density of concrete
▪ Step 02 : Moment Calculation
Design Moment = 𝑀 =
𝑊 𝑥 𝑙2
8
ULTIMATE MOMENT OF RESISTANCE, MU = 0.156 * fcu * b * d2

14. Design Steps - Continued
• Step No 03 – Check for section (Singly Reinforced / Doubly
Reinforced)
• If Mu > M Singly Reinforced - no compression reinforcement is required
• If Mu < M Doubly Reinforced - compression reinforcement is required – See Design
steps for Doubly Reinforced
• Step No 04 MAIN STEEL, As
• As =
𝑀
0.87 ∗𝑓𝑦 ∗𝑍
• 𝑍 = 𝑑 0.5 + 0.25 − 𝐾/0.9
• K =
𝑀
𝑓𝑐𝑢 ∗𝑏 ∗ 𝑑2
• Calculate nos. pf bar from As

15. Design Steps - Continued
• Step 05 -SHEAR REINFORCEMENT - The spacing of links should not exceed 0.75d
• Shear stress,  =
𝑉
𝑏 ∗𝑑
; V = (W * l )/2
• DESIGN CONCRETE SHEAR STRESS, c from table
• Values of design concrete shear stress, υc (N/mm2) for fcu = 25
N/mm2 concrete (Table 3.8, BS 8110)
For other values of cube
strength up to a maximum of
40 Nmm−2, the design
shear stresses can be
determined by multiplying
the values in the table by the
factor ( fcu/25) 1/3

16. Design Steps - Continued
• Form and area of links in beams (Table 3.7, BS 8110)
υc /2 < υ < (υc + 0.4) then
Where υ > (υ c + 0.4)
The Part which is more need to be designed

17. Design Steps - Continued
• Step 06 – Check for EFFECTIVE SPAN
• The effective span is the lesser of
• center-to-center distance between support
• Clear distance between supports plus the effective depth
• Step 07 Check for DEFLECTION – Actual < Permissible(Span/Depth* Modification Factor)
• Span/effective depth

18. Design of a simply supported concrete beam
(BS 8110)
A reinforced concrete beam which is 300 mm wide and 600 mm
deep is required to span 6.0 m between the centres of
supporting piers 300 mm wide .The beam carries dead and
imposed loads of 25 kNm−1 and 19 kNm−1 respectively.
Assuming fcu = 30 Nmm−2, fy = fyv = 500 Nmm−2 and the
exposure class is XC1, design the beam.

19. Step1: Ultimate / Factored Load Calculation
Dead
Self weight of beam = 0.6 × 0.3 × 24 = 4.32 kNm−1
Total dead load (gk) = 25 + 4.32 = 29.32 kNm−1
Imposed
Total imposed load (qk) = 19 kNm−1
Ultimate load
Total ultimate load (W) = (1.4gk + 1.6qk)
= (1.4 × 29.32 + 1.6 × 19)= 71.45 kN/m
Step 02 : Moment Calculation
Design Moment = 𝑀 =
𝑊 𝑥 𝑙2
8
=
71.45 𝑥 62
8
= 321.5 kNm
Effective depth, d
Assume diameter of main bars (Φ) = 25 mm , Assume diameter of links (Φ′) = 8 mm From Table 3.6, cover for
exposure class XC1 = 15 + Δc = 25 mm.
d = h − c − Φ′ − Φ/2 ; = 600 − 25 − 8 − 25/2 = 554 mm

20. ULTIMATE MOMENT OF RESISTANCE, MU
Mu = 0.156 * fcu * b * d2 = 0.156 × 30 × 300 × 5542
= 430.9 × 106 Nmm = 430.9 kNm
Step No 03 – Check for section (Singly Reinforced / Doubly
Reinforced)
Since Mu > M no compression reinforcement is required.
• Step No 04 MAIN STEEL, As
• As =
𝑀
0.87 ∗𝑓𝑦 ∗𝑍
• 𝑍 = 𝑑 0.5 + 0.25 − 𝐾/0.9 𝑍 = 554ቄ
ቅ
0.5 +
0.25 − 0.116/0.9 =470mm
• K =
𝑀
𝑓𝑐𝑢 ∗𝑏 ∗ 𝑑2 K =
321.5 ∗ 106
30 ∗300 ∗ 5542 = 0.116
• As =
321.5 ∗ 106
0.87 ∗500 ∗470
= 1573 mm2 provide 4H25 (As = 1960 mm2).

22. • Step 05 -SHEAR REINFORCEMENT - The spacing of links should not exceed 0.75d
• Shear stress,  =
𝑉
𝑏 ∗𝑑
; V = (W * l )/2 = (71.45 * 6)/2 =214.4kN
 =
214.4 𝑥 103
300 ∗554
= 1.29 Nmm−2 < permissible = 0.8 30 = 4.38 Nmm−2
Design concrete shear stress, υc
100 ∗𝐴𝑠
𝑏 ∗𝑑
= 1.18
1.18 = 0.63 +
0.72−0.63
1.5−1.0
∗ (1.18 − 1)
=0.66
υc = (30/25)1/3 × 0.66 = 0.70 Nmm−2
1.00 0.63
1.5 0.72

23. Diameter and spacing of links
Where υ < (υc + 0.4) = 0.7 + 0.4 = 1.1 Nmm−2, nominal links are required according to
Hence from Table 3.13, provide H8 links at 300 mm centres where υ < 1.10 Nmm−2, i.e. 2.558 m
either side of the mid-span of beam.

24. • Where υ > (υc + 0.4) = 1.10 Nmm−2 design links required according to
• Maximum spacing of links is 0.75d = 0.75 × 554 = 416 mm. Hence from
Table provide 8 mm diameter links at 225 mm centres (Asv/sv = 0.447)
where v > 1.10 Nmm−2, i.e. 0.442 m in from both supports.

25. • Step 06 – Check for EFFECTIVE SPAN
The effective span is the lesser of
(1) centre-to-centre distance between support, i.e. 6 m, and
(2) clear distance between supports plus the effective depth, i.e. 5700 + 554 =
6254 mm. Therefore assumed span length of 6 m is correct.

26. Design of bending reinforcement for a doubly
reinforced beam (BS 8110)

27. Design Steps – Doubly Reinforced Simply Supported
• Step1: Ultimate / Factored Load Calculation
• Dead Load = B x D x 1m x Density of concrete
▪ Step 02 : Moment Calculation
Design Moment = 𝑀 =
𝑊 𝑥 𝑙2
8
ULTIMATE MOMENT OF RESISTANCE, MU = 0.156 * fcu * b * d2

28. Design Steps - Continued
• Step No 03 – Check for section (Singly Reinforced / Doubly Reinforced)
• If Mu > M Singly Reinforced - no compression reinforcement is required - See Design steps for Singly Reinforced
• If Mu < M Doubly Reinforced - compression reinforcement is required –
• Step No 04 MAIN STEEL, A
• Compression Reinforcement
• 𝑍 = 𝑑 0.5 + 0.25 − 𝐾′/0.9 d’ Cover Provided in compression zone; K’ =
𝑴𝒖
𝒇𝒄𝒌 ∗𝒃 ∗𝒅𝟐
• x =
𝑑−𝑧
0.45
• If d’/z , 0.37 then As’ =
𝑴−𝑴𝒖
𝟎.𝟖𝟕 ∗𝒇𝒚 ∗(𝒅−𝒅′)
• If d′/x > 0.37, the compression steel will not have yielded and, therefore, the compressive stress will be less than 0.87fy. In such
cases, the design stress can be obtained
• z = d − 0.9x/2

29. Design Steps - Continued
• TENSION REINFORCEMENT
• As =
𝑀𝑢
0.87 ∗𝑓𝑦 ∗𝑍
+ As’
• 𝑍 = 𝑑 0.5 + 0.25 − 𝐾/0.9
• K =
𝑀
𝑓𝑐𝑢 ∗𝑏 ∗ 𝑑2
• Calculate nos. of bar from As

30. Design Steps - Continued
• Step 05 -SHEAR REINFORCEMENT - The spacing of links should not exceed 0.75d
• Shear stress,  =
𝑉
𝑏 ∗𝑑
; V = (W * l )/2
• DESIGN CONCRETE SHEAR STRESS, c from table
• Values of design concrete shear stress, υc (N/mm2) for fcu = 25
N/mm2 concrete (Table 3.8, BS 8110)
For other values of cube
strength up to a maximum of
40 Nmm−2, the design
shear stresses can be
determined by multiplying
the values in the table by the
factor ( fcu/25) 1/3

31. Design Steps - Continued
• Form and area of links in beams (Table 3.7, BS 8110)
υc /2 < υ < (υc + 0.4) then
Where υ > (υ c + 0.4)
The Part which is more need to be designed

32. Design Steps - Continued
• Step 06 – Check for EFFECTIVE SPAN
• The effective span is the lesser of
• center-to-center distance between support
• Clear distance between supports plus the effective depth
• Step 07 Check for DEFLECTION – Actual < Permissible(Span/Depth* Modification Factor)
• Span/effective depth

33. Design of bending reinforcement for a doubly
reinforced beam (BS 8110)
• The reinforced concrete beam shown in Fig. has an effective span
of 9 m and carries uniformly distributed dead (including self
weight of beam) and imposed loads of 4 and 5 kNm−1 respectively.
Design the bending reinforcement assuming the following: Cover to
main steel = 40 mm
• fcu = 30 Nmm−2
• fy = 500 Nmm−2

34. Step1: Ultimate / Factored Load Calculation
Ultimate load
Total ultimate load (W) = (1.4gk + 1.6qk)
= (1.4 × 4 + 1.6 × 5)= 13.60 kN/m
Step 02 : Moment Calculation
Design Moment = 𝑀 =
𝑊 𝑥 𝑙2
8
=
13.6 𝑥 92
8
= 137.7 kNm
Effective depth, d
Assume diameter of main bars (Φ) = 25 mm
d = h − c − Φ′ − Φ/2 ; = 370 − 25/2-40 = 317 mm

35. ULTIMATE MOMENT OF RESISTANCE, MU
Mu = 0.156 * fcu * b * d2 = 0.156 × 30 × 230 × 3172
= 108.2 × 106 Nmm = 108.2 kNm
Step No 03 – Check for section (Singly Reinforced / Doubly Reinforced)
Since Mu < M compression reinforcement is required.
• Step No 04 MAIN STEEL, As
• As’ =
𝑀−𝑀𝑢
0.87 ∗𝑓𝑦 ∗(𝑑−𝑑′)
Assume diameter of compression bars (φ) = 16 mm. Hence d′ = cover + φ/2 = 40 + 16/2 = 48 mm
• 𝑍 = 𝑑 0.5 + 0.25 − 𝐾′/0.9 𝑍 = 317 0.5 + 0.25 − 0.156/0.9 =246mm
• K’ =
𝑀𝑢
𝑓𝑐𝑢 ∗𝑏 ∗ 𝑑2 K’ =
108.2
30 ∗230 ∗3172 = 0.116
• x =
𝑑−𝑧
0.45
=
317−246
0.45
=158mm d’/z = 48/158 = 0.3<0.37
• As’ =
𝑴−𝑴𝒖
𝟎.𝟖𝟕 ∗𝒇𝒚 ∗(𝒅−𝒅′)
=
𝟏𝟑𝟕.𝟕−𝟏𝟎𝟖.𝟐 𝒙 𝟏𝟎𝟔
𝟎.𝟖𝟕 ∗𝟓𝟎𝟎 ∗(𝟑𝟏𝟕−𝟒𝟖)
= 𝟐𝟓𝟐 𝒎𝒎𝟐
• Hence from Table 3.10, provide 2H16 (As′ = 402 mm2)

36. Design Steps - Continued
• TENSION REINFORCEMENT
• As =
𝑀𝑢
0.87 ∗𝑓𝑦 ∗𝑍
+ As’ = 108.2 𝑥 106
0.87 ∗500 ∗246
+ 252 = 1263 mm2
• Hence provide 3H25 (As = 1470 mm2).