Shear, Torsion,

UNIT 2 SHEAR, TORSION, CONCRETE Concrete Reinforcement

and Detailing
REINFORCEMENT AND DETAILING
Structure
2.1 Introduction
Objectives
2.2 Shear
2.2.1 Derivation of Shear Force and Shear Stress
2.2.2 Design Shear Strength of Concrete
2.2.3 Design of Shear Reinforcement
2.3 Torsion
2.4 Concrete Reinforcement and Detailing
2.4.1 Development of Stress in Reinforcement
2.4.2 Curtailment of Tension Reinforcement in Flexural Members
2.4.3 Spacing of Reinforcement
2.4.4 Reinforcement Requirement in Members
2.5 Summary
2.6 Answers to SAQs

2.1 INTRODUCTION
A beam, loaded in its own plane, with transverse loads will have applied bending
moment and shear force. Shear force develops internal shear stresses both in
horizontal and vertical planes causing tension and compression in two diagonal
planes (Figure 2.1). The shear reinforcement is provided to resist tensile force
while the compressive force is resisted by concrete itself.
In case, if the transverse loading on a beam is not in the central plane of a beam of
symmetrical cross-section, the beam, in effect, will have applied bending, shear as
well as torsion. Torsional moment is resisted by development of internal shear
force and bending moment and, therefore, additional reinforcement for shear as
well as for bending are provided to resist torsional moment.
Detailing of reinforcement in a member means provision of reinforcement at
proper locations and in appropriate quantities with proper distribution, so that it
can behave as a composite member fulfilling all design criteria – such as
durability, fire resistance, stability, strength, serviceability, etc.
This unit is, therefore, devoted to achieve the following objectives :

Objectives
After studying this unit, you should be able to
• design members for shear,
• design members for torsion, and
• give the detailing of reinforcement in members.
33
Theory of Structures-II
2.2 SHEAR
2.2.1 Derivation of Shear Force and Shear Stress
According to principles of Mechanics of Solid, rate of change of bending moment
⎛ dM ⎞
w.r.t. distance along the span of a beam ⎜ i.e. ⎟ at a section at x is equal to
⎝ dx ⎠
shear force (V) at that section. For beam shown in Figure 2.1, the bending
moment at x and x + δx are M and M + δM, respectively.

Figure 2.1 : Development of Horizontal Shear Force

The horizontal forces due to bending on the portion CGHD of the beam, are C 1
and C2 on sections CE and DF, respectively. If C2 > C1 then a horizontal force
(C2 – C1) is acting on CGHD which is equilibrated by resisting horizontal force,
called horizontal shear force, at surface GH. Similarly, vertical shear force (V) is
acting on section x (Figure 2.2).

Figure 2.2 : Development of Vertical Shear Force on Section at x

Then according to Principles of Mechanics of solid
V Ay
τv = . . . (2.1)
Ib

where τv = Shear stress at section x,

A = Area of concrete (Figure 2.1(b)),
y = Distance of its CG from n.a. of area A,
34
 I = Moment of Inertia about n.a. of the whole cross-section, and Shear, Torsion,
Concrete Reinforcement
b = Breadth, GG′. and Detailing

But the above derivation is for solid section for which the shear stress diagram for
cross section at x is as shown in Figure 2.1(c).
Though the section looks solid, in case of reinforced concrete, the concrete below
the n.a. is not taken into consideration as it has cracked in tension. Only steel
cross section below n.a. exists resulting in the shear stress diagram as shown in
Figure 2.3.

Figure 2.3 : Shear Stress Distribution Over a Cross-section

Therefore, it can be derived that
vu
τ v, max = . . . (2.2)
b jd
where (jd) is lever arm in flexure and j is nearly equal to 1.
Since exact value of shear stress cannot be derived for reinforced concrete section
due to cracking of concrete in tension, τv, max is represented as τv and called
‘Nominal Shear Stress’. Its value is taken as
Vu
. . . (2.3)
bd
where V = Vu in Limit State Method.
2.2.2 Design Shear Strength of Concrete
For Beams
The design shear strength (τc) of a beam is a function of concrete strength
⎛ As ⎞
(fck) and percentage of tensile reinforcement ⎜ ⎟ . This is concisely given
⎝ bd ⎠
in Table 2.1.
Table 2.1 : Design Shear Strength of Concrete, τ c , (N/mm2)
As
100 Concrete Grade
bd
M 15 M 20 M 25 M 30 M 35 M 40 and above
≤ 0.15 0.28 0.28 0.29 0.29 0.29 0.30
0.25 0.35 0.36 0.36 0.37 0.37 0.38
0.50 0.46 0.48 0.49 0.50 0.50 0.51
0.54 0.56 0.57 0.59 0.59 0.60
1.00 0.60 0.62 0.64 0.66 0.67 0.68
1.25 0.64 0.67 0.70 0.71 0.73 0.74
1.50 0.68 0.72 0.74 0.76 0.78 0.79
1.75 0.71 0.75 0.78 0.80 0.82 0.84
2.00 0.71 0.79 0.82 0.84 0.86 0.88
2.25 0.71 0.81 0.85 0.88 0.90 0.92
2.50 0.71 0.82 0.88 0.91 0.93 0.95
2.75 0.71 0.82 0.90 0.94 0.96 0.98
3.00 and above 0.71 0.82 0.92 0.96 0.99 1.01
Note : The term As is the area of longitudinal tension reinforcement. 35
Theory of Structures-II For Slabs
The design shear strength for slabs is kτc where k is multiplying factor given
in Table 2.2.
Table 2.2 : Multiplying Factor (k) for Design Shear Strength for Slabs
Overall Depth 300 or more 275 250 225 200 175 150 or less
of Slab, (mm)
K 1.00 1.05 1.10 1.15 1.20 1.25 1.30

For Members under Axial Compression

The design shear strength for member under compressive force is δτc
⎛ 3Pu ⎞⎟
where δ = ⎜1 + < 1.5 , . . . (2.4)
⎜ Ag f ck ⎟⎠
⎝
Ag = Gross area of the concrete section, and
fck = Characteristic compressive strength of concrete.
Limiting Shear Stress in Beams
The nominal shear stress (τv) in beams shall not exceed τc, max given in
Table 2.3.
Table 2.3 : Design Shear Strength of Concrete, τc, max (N/mm2)
Concrete Grade M 15 M 20 M 25 M 30 M 35 M 40 and above
τ c ,max 2.5 2.8 3.1 3.5 3.7 4.0
(N/mm2)

For solid slabs the nominal shear stress shall not exceed half the value of
τc, max given in Table 2.3 for beams.

SAQ 1

(a) Discuss the following formula :

V Ay
(i) τv = , and
Ib

Vu
(ii) τ v , max =
b jd
(b) Mention the design parameters on which design shear strength of
concrete depends.
(c) How will you decide the design shear strength of a slab?
(d) Write down in tabular form the Maximum Shear Stress (τc, max) values
for different grades of concrete.
2.2.3 Design of Shear Reinforcement
(a) When the shear stress (τv) is less than design shear strength (τc),
minimum shear reinforcement in the form of stirrups shall be
provided such that
Asv 0.4
≥ . . . (2.5)
b sv 0.87 f y
36
 where Asv = Total cross sectional area of stirrup legs effective in Shear, Torsion,
Concrete Reinforcement
shear, and Detailing
sv = Stirrup spacing along the length of the member,
b = Breadth of rectangular beam or breadth of web of flanged
beam, and
fy = Characteristic or yield strength of shear reinforcement.
Where the value of nominal shear stress (τv) is less than half the shear
⎛ τc ⎞
strength ⎜ i.e. ⎟ and in case of members of minor importance the
⎝ 2 ⎠
above mentioned requirement of minimum shear reinforcement may
not be complied with.
(b) If the shear stress (τv) is more than the design shear strength, the shear
reinforcement shall be provided for the shear force,
Vus =Vu − τc bd . . . (2.6)
(c) If the shear stress (τv) is greater than the maximum shear stress τv, max
(Table 2.3), the section should be redesigned.
Provision of Shear Reinforcement
Vertical Stirrups
The shear reinforcements are normally provided in the form of
vertical stirrups. In addition to resisting shear, this form of stirrups,
very effectively, keeps main reinforcement in position and bind
concrete to prevent it from bursting.
Figure 2.4 below shows different forms of this type of stirrups.

Figure 2.4 : Types of Stirrups with Respect to Number of Vertical Legs

For design
0.87 f y Asv d
Vus = . . . (2.7)
sv
where Asv = Area of cross section of vertical legs, and
37
Theory of Structures-II sv = Spacing of vertical stirrups along the span.

Example 2.1

Determine the shear reinforcement in the form of vertical stirrups of φ 6 of

a rectangular cross section of b × d = 250 × 450 reinforced with 4 φ 20 to
resist 100 kN shear force. Use M 25 concrete, Fe 415 for main
reinforcement and Fe 250 for transverse reinforcement.
Solution
The beam section is shown in Figure 2.5.

Vu 100 × 103
τv = = = 0.889 N/mm 2 < τ c , max (= 3.1 N/mm 2 ) (Table 2.3)
bd 250 × 450

π
× 2024×
For M 25 concrete and pt % = 4 × 100 = 1.12%
250 × 450

Figure 2.5 : Beam Section

τ c = 0.67 N/mm 2 from Table 2.1

τc < τv, hence, vertical shear stirrups are provided as follows :

0.67 × 250 × 450
Vus = Vu − τc bd = 100 − N . . . (2.8)
103

= 24.63 kN
From Eq. (2.8)
0.87 f y Asv d
Vus =
sv

0.87 × 250 × 2 × 28.27 × 450

or, Sv = = 224.68 mm c/c
24.63 × 10 3

sv, max as per Code = 0.75 d = 0.75 × 450 = 337.5 mm c/c > 224.68 mm c/c
Hence, provided φ 6 two legged stirrups @ 220 mm c/c.
A Series of Main Bent up Bars or Inclined Stirrups at Different
Cross-sections

38
 Bent-up main bars may be provided for shear where these are no longer Shear, Torsion,
Concrete Reinforcement
required for flexure (Figure 2.6). Inclined stirrups may also be provided in a and Detailing
similar manner as bent-up bars.

Figure 2.6 : Inclined Reinforcement to Resist a Part of Shear Force

The design formula for such shear reinforcement is

0.87 f y Asv d
Vus = (sin α + cos α) . . . (2.9)
sv

where α = inclination of bent-up bars or inclined stirrups.

Where bent-up bars are provided as shear reinforcement, their contribution
towards shear resistance shall not be more than 50% of the total shear
reinforcement.
When a Single Bar or a Group of Bars are Bent-up at the same Cross-section
Such cases (Figures 2.7(a) and (b)) are economical since these bent-up bars
cover the whole distance (d + d cot α) for which shear reinforcement is
needed.
The design formula in this case is
Vus = 0.87 fy As sin α . . . (2.10)

(a) Single or Group of Parallel Inclined Bars at the same Cross-section

39
Theory of Structures-II

(b) Maximum Distance between Bent-up Bars or Inclined Stirrups

Figure 2.7

Example 2.2

Design shear reinforcement for the beam, shown in Figures 2.7(a) and (b),
for the following data :
No. of bent-up bars = 2 φ 16 at α = 45º
Vus = 180 kN
Grade of shear reinforcement = Fe 415 for vertical stirrups.
Grade of main bars = Fe 250.
Solution
Vus1 = 0.87 fy Asv1 sin α . . . (2.11)
π 1
= 0.87 × 250 × 2 × 162 ×
4 2

180
= 61.83 kN < (i.e. 50% of total SF)
2
= 61.83 kN
Vus2 = (180 – 61.83) kN = 118.17 kN
0.87 f y Asv 2 d
Vus2 = 118.17 × 103 =
sv2

Adopting φ 8 bars for vertical stirrups

π
0.87 × 415 × 2 × × 82 × 450
or, sv 2 = 4 = 137.49 mm
118.17 × 103

Provided φ 8 two legged stirrups @ 135 mm c/c in addition to

2 φ 16 bent-up bars.

SAQ 2

(a) Determine spacing of two legged vertical stirrups of φ 8 for a

rectangular beam b × d = 250 × 500 if τv < τc and fy = 415 N/mm2.
(b) Describe the procedure for design of vertical stirrups for shear in
reinforced concrete beams.
(c) Explain the formulae with figures for the design of shear
reinforcements in the following cases :
(i) Series of bent-up bars or inclined stirrups at different sections.

40
 (ii) When a single bar or a group of bars are bent-up at the same Shear, Torsion,
Concrete Reinforcement
cross-section. and Detailing

2.3 TORSION
Torsion or twisting moment is a moment about the axis of a member. If the cross
section of the member is circular*, shear stress only will develop on it. * Refer formula
T q Gθ
But if the cross section of a member is other than circular, shearing as well as = =
J r l
bending stresses are produced due to warping of the surface. The phenomenon is which is valid only for
complicated, hence, only codal provisions are explained here. cylindrical member,
where symbols have
2.3.1 Provision for Torsion their usual meanings.
Provision for torsion is not made separately but its effect are taken care of along
with shear and bending. Mathematically, at a section on which shear, bending and
torsion are acting simultaneously,
Tu
Equivalent shear, Ve = Vu + 1.6 . . . (2.12)
b
Equivalent bending moment
⎛ D⎞
⎜1 + ⎟
M e1 = M u + M t = M u + Tu ⎜ b ⎟ . . . (2.13)
⎜ 1 .7 ⎟
⎜ ⎟
⎝ ⎠

⎛ D⎞
⎜1 + ⎟
where M t = Tu ⎜ b ⎟
⎜ 1.7 ⎟
⎜ ⎟
⎝ ⎠
Ve
Now, τve =
bd
(a) If τve < τc only minimum shear reinforcement is provided.
(b) If τc < τve < τc, max and Mt < Mu the shear reinforcement for
Vus = Ve – τc bd as well as longitudinal tensile reinforcement at tension
face for Mel are provided.
(c) But if τc < τve < τc, max and Mt > Mu the shear as well as bending
reinforcement are provided as in (b) and, in addition, longitudinal
tensile reinforcement at compression face is provided for a bending
moment equal to (Mt – Mu).
Shear Reinforcement
Shear reinforcement is provided in the form of closed hoop enclosing all the
corners having an area of cross section (Asv) given by
⎛ Tu sv V u sv ⎞
Asv = ⎜ + ⎟ . . . (2.14)
⎜ b1 d1 (0.87 f y ) 2.5 d1 (0.87 f y ) ⎟
⎝ ⎠

41
Theory of Structures-II ( τ ve − τ c ) b s v
or, Asv = . . . (2.15)
0.87 f y

whichever is more.
where b1 = Centre to centre distance between corner bars along width,
and
d1 = Centre to centre distance between corner bars along depth.

Example 2.3

Design longitudinal as well as transverse reinforcement for a rectangular

beam shown in Figure 2.8 for the following data :
Vu = 100 kN; Mu = 100 kN-m; Tu = 9 kN-m; fck = 25 N/mm2;
fy = 415 N/mm2.

Figure 2.8 : Beam Section

Solution
Tu 9
Ve = Vu + 1.6 = 100 + 1.6 × = 148 kN
0.3 0.3
Ve 148 × 103
τve = = = 1.1 N/mm 2
bd 300 × 450

⎛ Ast f y ⎞
M u = 0.87 f y Ast d ⎜⎜1 − ⎟
⎟
⎝ b d f ck ⎠
⎛ Ast × 415 ⎞
or, 100 × 10 6 = 0.87 × 415 × Ast × 450 × ⎜⎜1 − ⎟
⎝ 300 × 450 × 25 ⎟⎠

or, 100 × 106 = 162472.5 Ast – 19.98 Ast2

or, Ast2 – 8131.76 Ast + 5005005 = 0
8131.76 ± (8131.76 2 − 4 × 5005005)
or, Ast =
2
2
= 670.83 mm
Ast 670.83
∴ pt % = × 100 = × 100 = 0.5%
bd 300 × 450
∴ For M 25 concrete and pt % = 0.5%
42
 Shear, Torsion,
τ c = 0.49 N/mm 2 < τ ve (= 1.1 N/mm 2 ) Concrete Reinforcement
and Detailing
Hence, both longitudinal as well as transverse reinforcements shall be
designed including torsional effects.
⎛ D⎞
⎜1 + ⎟
M e1 = M u + M t = M u + Tu ⎜ b ⎟
⎜ 1.7 ⎟
⎜ ⎟
⎝ ⎠
⎛⎛ 500 ⎞ ⎞
⎜ ⎜1 + ⎟⎟
⎜ ⎝ 300 ⎠ ⎟
= 100 + 9 ⎜ ⎟ = 100 = 14.12
1.7
⎜⎜ ⎟⎟
⎝ ⎠
= 114.12 kN-m
Here Mt < Mu, hence, no tensile reinforcement on compression force will be
required.
⎛ Ast f y ⎞
M e1 = 0.87 f y Ast d ⎜⎜1 − ⎟
⎟
⎝ b d f ck ⎠

⎛ Ast × 415 ⎞
or, 114.12 × 10 6 = 0.87 × 415 Ast × 450 × ⎜⎜1 − ⎟
⎝ 300 × 450 × 25 ⎟⎠

or, 114.12 × 106 = 162472.5 Ast – 19.98 Ast2

or, Ast2 – 8131.76 Ast + 5711711.71

8131.76 ± (8131.76 2 − 4 × 5711711.71)

or, Ast = = 776.55 mm 2
2
Hence, provided 4 φ 16 as tensile reinforcement
804 × 100
pt% = % = 0.6%
300 × 450

(0.57 − 0.49)
τ c = 0.49 + × (0.6 − 0.5)
(0.75 − 0.5)

= 0.52 N/mm2 < τve (= 1.1)

Hence, transverse reinforcement will be provided as follows :
Design of Transverse Reinforcement
Tu s v Vu s v
Asv = +
b1 d1 (0.87 f y ) 2.5 d1 (0.87 f y )

Assuming 2-legged φ 8 closed hoops to be provided, then

Asv = 2 × 50 = 100 mm2
Now b1 = 300 – 2 × 50 = 200 mm
d1 = 500 – 2 × 50 = 400 mm
Substituting these values in the above equation,
9 × 10 6 × s v 100 × 10 3 × s v
100 = +
200 × 400 × 0.87 × 415 2.5 × 400 × 0.87 × 415
43
Theory of Structures-II = 0.31 sv + 0.28 sv = 0.59 sv
100
or, sv = = 169.49 mm c/c
0.59
(τve − τc ) bsv
Again Asv </
0.87 f y

0.87 × 415 × 100

or, sv >/ = 207.5 mm c/c
(1.1 − 0.52) × 300

Asv 0.4
≥
b sv 0.87 f y

100 0.4
or, ≥
300 × 169.49 0.87 × 415
or, 0.00197 ≥ 0.0011 (Hence, O.K.)

Figure 2.9 : Designed Section

x1 = 200 + 8 + 10 = 218 mm
8 16 8 10
y1 = 400 + + + + = 421 mm
2 2 2 2
( x1 + y1 )
169.49 > (= 159.75) < x1 (= 218) < 300 mm
4
Hence provided closed hoops of 2-legged φ 8 @ 155 mm c/c.
As depth of beam is more than 450 mm, additional side longitudinal bars shall be
provided as follows :
Side Reinforcement
0.1
Longitudinal Ast, min = × 300 × 450 =135 mm2
100
135
∴ On each face Ast, min = = 67.5 mm2
2
Hence, provided 2 φ 8 as shown on each face (Figure 2.9).

Example 2.4

44
 Design longitudinal as well as transverse reinforcements for a rectangular Shear, Torsion,
Concrete Reinforcement
beam, shown in Figure 2.10, for the following data : and Detailing
Vu = 50 kN; Mu = 75 kN-m; Tu = 50 kN-m
fck = 20 N/mm2; fy = 415 N/mm2.
Solution
Tu 50
Ve = Vu + 1.6 = 50 + 1.6 ×
b 0.3
= 316.67 kN

Figure 2.10 : Beam Section

Ve 316.67 × 103
τve = = = 2.346 N/mm 2
bd 300 × 450

⎛ Ast f y ⎞
M u = 0.87 f y Ast d ⎜⎜1 − ⎟
b d f ck ⎟
⎝ ⎠

⎛ A × 415 ⎞
or, 75 × 106 = 0.87 × 415 × Ast × 450 ⎜⎜1 − st
⎟
⎝ 300 × 450 × 20 ⎟⎠

or, 75 × 106 = 162472.5 Ast – 24.973 Ast2

or, Ast2 – 6506.024 Ast + 3003243.5 = 0

6506.024 ± 6506.024 2 − 4 × 3003243.5

or, Ast =
2
2
= 500.042 mm
Ast 500 × 100
pt% = × 100 = = 0.37%
bd 300 × 450

0.12
∴ τc = 0.36 + × 0.12 = 0.42 N/mm2 < τve (= 2.346 N/mm2)
0.25
Hence, both longitudinal and transverse reinforcements shall be provided.
⎛ ⎛ 500 ⎞ ⎞
⎜ 1 + ⎜ 300 ⎟ ⎟
Mel = Mu + Mt = 75 + 50 × ⎜ ⎝ ⎠⎟ . . . (2.16)
⎜ 1.7 ⎟
⎜ ⎟
⎝ ⎠
= 75 + 78.431 = 153.431 kNm 45
Theory of Structures-II Here Mt > Mu
Hence, longitudinal reinforcement shall be provided on the flexural
compression face, such that the beam can also withstand an equivalent
Me2 = (Mt – Mu), the Me2 being taken opposite to Mu.
Me2 = 78.431 – 75 = 3.431 kN-m
⎛ Aste 2 × 415 ⎞
or, Me2 = 0.87 × 415 × Aste2 × 450 × ⎜⎜1 − ⎟
⎝ 300 × 450 × 20 ⎟⎠

or, 3.431 × 106 = 162472.5 Aste2 – 24.973 Aste

2
2

2
or, Aste 2 − 6506.024 Aste2 + 137388.38 = 0

(6506.024 ± (6506.024 2 − 4 × 1 × 137388.38) )

or, Aste 2 =
2
2
= 21.186 mm
Hence, provided 2 φ 10.
Ast for Mel
⎛ Ast f y ⎞
Mel = 0.87 fy Ast d ⎜⎜1 − ⎟
⎝ b d f ck ⎟⎠

⎛ Ast × 415 ⎞
or, 153.431 × 106 = 0.87 × 415 Ast × 450 × ⎜⎜1 − ⎟
⎝ 300 × 450 × 20 ⎟⎠

or, 153.431 × 106 = 162472.5 Ast – 24.973 Ast2 2

or, Ast2 – 6506.024 Ast + 6143875.38 = 0

(6506.024 ± (6506.024 2 − 4 × 6143875.38 ) )

or, Ast =
2
= 1146.31 mm2
Hence, provided 4 φ 20.
4 × 314
pt% = × 100% = 0.93%
300 × 450
(0.62 − 0.56)
∴ τc = 0.56 + × (0.93 – 0.75) = 0.60 N/mm2 < τve
(1.0 − 0.75)
(= 1.93 N/mm2).
Hence, transverse reinforcement will be provided.
Design of Transverse Reinforcement
⎛ Tu s v Vu s v ⎞
Asv = ⎜ + ⎟
⎜ b1 d1 (0.87 f y ) 2.5 d1 (0.87 f y ) ⎟
⎝ ⎠

Assuming 2-legged φ 8 closed hoops to be provided, then

Asv = 2 × 50 = 100 mm2
b1 = 300 – 2 × 50 = 200 mm
d1 = 500 – 2 × 50 = 400 mm
46
 Substituting these values in above equation, Shear, Torsion,
Concrete Reinforcement
and Detailing
50 × 10 6 × s v 50 × 10 3 s v
100 = +
200 × 400 × 0.87 × 415 2.5 × 400 × 0.87 × 415

= 1.731 sv + 0.138 sv = 1.869 sv

100
or, sv = = 53.51 mm c/c < 300 < 0.75 d (= 337.5)
1.869
(τ ve − τ c ) b s v
Asv </
0.87 f y

0.87 × 415 × 100

or, sv >/ = 68.93 mm
(2.346 − 0.6) × 300

Asv 0.4
Again, ≥
b sv 0.87 f y

100 0.4
≥
300 × 53.51 0.87 × 415
0.006 > 0.0011 (Hence, O.K.)
x1 = 200 + 8 + 10 = 218 mm
8 20 8 10
y1 = 400 + + + + = 423 mm
2 2 2 2
x1 + y1
53.51 < (= 160.25) < x1 (= 218) < 300 mm
4
Hence, provided closed hoops of 2-legged φ 8 @ 50 mm c/c.
As depth of beam is more than 450 mm, additional side longitudinal
bars shall be provided as follows :
Side Reinforcement
0.1
Longitudinal, Ast, min = × 300 × 450 = 135 mm2
100
135
∴ On each face, Ast, min = = 67.5 mm2
2
Hence, provided 2 φ 8 as shown on each face (Figure 2.11).

47
Theory of Structures-II

Figure 2.11 : Designed Section

2.4 CONCRETE REINFORCEMENT AND

DETAILING
Reinforced concrete is a composite material, hence a proper bond between the
two materials – concrete and reinforcement – is the first requirement. A minimum
length of reinforcing bar is needed to develop the full bond strength between
concrete and steel. This length is expressed in terms of development length.
‘Development length’ on each side of any section is the length over which the
force in the reinforcement at that section will be developed without bond failure
between these two materials.
Sometimes a reinforcing bar is extended and/or bent at its ends to satisfy
development length requirement. Such extension and/or bending of a bar at its
ends is called anchorage. A bent bar provides a greater safety against bond failure
as during a pull-out the whole concrete is to be crushed.
Reinforcing bars have limited lengths for ease of handling and transporting;
hence in case of a continuous member or a member of large span it is necessary
for continuity to join two bars by overlapping the ends at the joint. The
overlapping portion are joined together either by concrete itself by providing
proper development length or by welding in case of limitation of length for
overlapping. Such jointing of two bars for continuity of reinforcing bar at any
section is called splicing.
Reinforcements are round bars. They are provided as straight or shaped
appropriately to suit the requirements. These reinforcements are placed at certain
spacing to meet the design requirements. But these spacing must be within a
range of the minimum spacing and the maximum spacing for ease of casting,
compaction, control of cracking, etc. as laid down in the Standard Code of
Practice.
All types of reinforcements must have sufficient concrete cover to protect them
from environmental exposure conditions and also against fire. Such cover is
called Nominal Cover.
Other requirements of detailing are the minimum and the maximum amount of
reinforcement, side face reinforcement, distribution of reinforcement, etc.
All detailing requirements defined above have been discussed in the following
sections and explained in the examples given in appropriate Units.
2.4.1 Development of Stress in Reinforcement
The calculated compression or tension in any reinforcing bar must be developed
on each side of a section by providing appropriate development length or
anchorage or a combination thereof.
The development length,
φ σs
Ld =
4τ bd

where, φ = Nominal diameter of the bar,

σs = Stress in bar at the section considered at design load, and

48 τbd = Design bond stress defined below.

Bond Stress is longitudinal shear stress at the interface between concrete and Shear, Torsion,
Concrete Reinforcement
reinforcing bar. The Design Bond Stress in Limit State Method for plain bars in and Detailing
tension are given in Table 2.4.
Table 2.4 : Design Bond Stress for Plain Bars in tension
Grade of Concrete M20 M25 M30 M35 M40 and above

Design Bond Stress τbd, (N/mm2) 1.2 1.4 1.5 1.7 1.9

It is evident from Table 2.4 that the design bond stress increases with increase in
concrete strength.
Design bond stress for deformed bars of Grades Fe 415 and Fe 500 in tension
shall be greater by 60% of those given in Table 2.4. Similarly design bond stress
for above mentioned bars in compression shall be greater by 25% of those given
in Table 2.4.
Where sufficient development length in tension is not available such as at ends of
a beam, anchorage is provided by bends or hooks. The anchorage value of a bend
shall be taken as 4 times the diameter of the bar for each 45° bend subject to a
maximum of 16 times the diameter of the bar. The U-type standard hook shall
have anchorage value of 16 φ (Figure 2.12).

Figure 2.12 : Standard Hooks and Standard 90o Bend

In compression, only projected length of the bar along with its bend, hooks or
straight length beyond bend shall be considered for development length
(Figure 2.13).

49
Theory of Structures-II

Figure 2.13 : Projected Lengths only to be Considered for Ld for Bar in Compression
Splices are provided to maintain the continuity of the bar. Lap length including
anchorage, if any, for flexural tension shall be Ld or 30 φ whichever is greater.
For bar carrying compression, the lap length shall be Ld or 24 φ whichever is
greater (Figure 2.14).

Figure 2.14 : Transverse Reinforcement at a Splice

2.4.2 Curtailment of Tension Reinforcement in Flexural

Members
To economise the design of a flexural member, the tensile bars are curtailed at the
section beyond which it is no longer required to resist flexure. Such curtailed bars
are extended for a distance equal to 12 φ or effective depth, whichever is greater,
except at simple support or free end of cantilever.
At simple support, positive moment tension reinforcement shall be limited to a
diameter such that Ld computed for fd does not exceed
M1
+ L0
V
where M1 = Moment of resistance of the section assuming all reinforcement at
the section to be stressed to fd,
fd = 0.87 fy,
V = Shear force at the section due to design loads,
L0 = Sum of anchorage beyond the centre of the support and the
equivalent anchorage value of hook, etc., and
φ = Diameter of bar.
M1
The value of in the above expression may be increased by 30% when the
V
ends of reinforcement are confined by a compressive reaction. At least one-third
of positive moment reinforcement in simple member shall extend along the same
Ld
face of the member into the support, to a length equal to .
3
50
Flexural reinforcement shall not be terminated in tension zone unless one of the Shear, Torsion,
Concrete Reinforcement
following conditions are fulfilled : and Detailing
(a) The shear at the cutoff point does not exceed two-thirds of that
permitted.
(b) Stirrup area in excess of that required for shear and torsion is
provided along each terminated bar over a distance from cutoff point
equal to three-fourth the effective depth of the member. The excess
stirrup area shall be not less than 0.4 bs/fy, where b is the breadth of
the beam and s is the spacing. The resulting spacing shall not exceed
d/8 βb where βb is the ratio of the area of bars cutoff to the total area
of bars at the section.
(c) For 36 mm and smaller bars, the continuing bars provide double the
area required for flexure at the cutoff point and the shear does not
exceed three-fourths of that permitted.
2.4.3 Spacing of Reinforcement
Minimum Spacing between Bars in Tension
The minimum horizontal spacing between two parallel main bars shall be
diameter of larger bar or maximum size of coarse aggregate plus 5 mm.
However, where compaction is done by needle vibrator, the spacing may be
further reduced to two-third of the nominal maximum size of the coarse
aggregate.
The minimum vertical distance between two main bars shall be
(a) 15 mm,
(b) two-third of the nominal size of coarse aggregate, or
(c) maximum size of the bar or whichever is greater.
Maximum Spacing between Bars in Tension
Normally these spacing will be as mentioned below :
(a) For beams, these distances are 300 mm, 180 mm and 150 mm
for grades of main reinforcement of Fe 250, Fe 415 and Fe 500,
respectively.
(b) For slabs
(i) the maximum spacing between two parallel main
reinforcing bars shall be 3d or 300 mm or whichever is
less, and
(ii) the maximum spacing between two secondary parallel
bars shall be 5d or 450 mm or whichever is less.
2.4.4 Reinforcement Requirement in Members
Beams* * For Flanged
Beams b = bw.
Ast 0.85
(a) Minimum tensile steel is given by the ratio, = .
bd fy

(b) Maximum Tensile Reinforcement in Beams shall not exceed 0.04 bD.
51
Theory of Structures-II (c) Maximum area of compression reinforcement shall not exceed
0.04 bD.
(d) Beam having depth exceeding 750 mm, side face reinforcement of
0.1% of web area shall be provided. This reinforcement shall be
equally distributed on two faces at a spacing not exceeding 300 or
web thickness or whichever is less.
Detailing for the other types of members have been explained in appropriate
Units.

2.5 SUMMARY
Design of a section for shear is carried on in the following steps :
Vu
(a) Determine nominal shear stress, τ v = .
bd
(b) Determine design shear strength of concrete, τc (Table 2.1).
(c) Provide shear reinforcements as per requirement and convenience.
Similarly, for design of a section for torsion one may proceed as given below :
Tu
(a) Determine the equivalent shear force, Ve = Vu + 1.6 .
b
(b) Calculate nominal shear stress, τv and if τc > τv, only nominal shear
reinforcement is to be provided. But if τc < τv, both longitudinal and
shear reinforcements are required.
(c) Provide longitudinal reinforcement on tensile side for equivalent
bending moment, Me1 = Mu + Mt
⎛ D⎞
⎜1 + ⎟
where M t = Tu ⎜ b ⎟
⎜ 1.7 ⎟
⎜ ⎟
⎝ ⎠
If Mt > Mu, provide longitudinal reinforcement on flexural
compression face such that the beam can also withstand an equivalent
Me2 = (Mt – Mu), the moment Me2 being taken as acting in the opposite
sense to the moment Mu.
(d) Provide two-legged closed hoops enclosing corner longitudinal bars
of area Asv, given by
Tu s v Vu s v
Asv = +
b1 d1 (0.87 f y ) 2.5 d1 (0.87 f y )

But the total transverse reinforcement shall not be less than

( τ ve − τ c ) b s v
0.87 f y

For composite and integral behaviour of a reinforced concrete beam,

the reinforcements are detailed in such a way that the stresses at every
section are appropriately developed with the provision of appropriate
amount and type of reinforcements. Proper cover to all reinforcements
are also provided for durability and fire resistance.
52
 Shear, Torsion,
2.6 ANSWERS TO SAQs Concrete Reinforcement
and Detailing
SAQ 1
(a) Refer Section 2.2.1.
(b) Refer Table 2.1.
(c) Refer Table 2.2.
(d) Refer Table 2.3.
SAQ 2
(a) Refer Section 2.2.3.
(b) Refer Section 2.2.3.
(c) Refer Section 2.2.3.

53