CHAPTER 2 DESIGN OF BEAMS FOR FLEXURE USING WORKING STRESS DESIGN (WSD) METHODS 2.1 Basic Assumption: 1. A section which is plane before bending remains plane after bending. This implies strains across section are linearly varying. This is true for most section of flexural member except deep beam where shear deformation is significant. 2. Beam section behaves elastically when subjected to service load moment. This implies stress in the concrete varies linearly from zero at neutral axis to a maximum at the extreme fiber. 3. Tensile strength of concrete is ignored. The reinforcement assumed to takes all the tension due to flexure. 4. Perfect bond exist between steel bars and concrete such that no slip occurs. This is possible if adequate development length of bars and concrete cover are provided. 5. The modular ratio, n = Es Ec , may be taken as the nearest whole number (but not less than 6 or more than 15). In doubly reinforced sections, to consider creep of concrete in compression zone an effective modular ratio of 2 Es Ec shall be used to transform compression reinforcement for stress computation. 2.2 Design Equations for Singly Reinforced Rectangular Section Consider a singly reinforced rectangular section subjected to a service load moment, M as shown below.
εc
fc
x
M D
fs n
εs fs
b
X-section
fc .b. x 2 x j . d = (d − ) 3 Cc =
d AS
x 3
Strain
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Ts = As . f s Stress
a) From the strain diagram, similarity of triangles gives
εc x = εs d−x
(1)
In elastic range, applying Hooke’s law, the maximum strain in concrete & strain in steel,
fc Ec
εc =
&
εs =
fs Es
And, the ratio of these strains
εc f .E = c s εs f s . Ec By definition,
(2)
Es Ec is the modular-ratio, n
Equating Eq(1) and Eq(2), and substituting n = Es Ec
⇒
x n . fc = =k d f s + n . fc
⇒
x = k . d , thus k is an indicator of the neutral axis position.
(3)
b) Considering equilibrium of a section
i)
For horizontal equilibrium
⇒
[∑ F
Cc = Ts
H
=0
]
Substituting Cc and Ts ,
fc . b . x = As . f s 2
⇔
As b.d
Let
ρ=
Then,
As = ρ .b . d
(4)
--is known as geometric steel
ratio
Substituting it into Eq.(4 )
⇒ With, x = k . d ,
fc . b . x = ρ . b. d . fs 2
⇒
fc . b . k . d = ρ . b. d . fs 2
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2ρ . f s fc
simplifying,
⇒
k=
From Eq.(3 ),
k=
n . fc f s + n . fc
ρ=
or
k . fc 2 fs
(5)
Ratio of stresses in steel to concrete, rearranging the above equation
⇒ but,
f s n . (1 − k ) = fc k
(6)
⇒
fs k = fc 2ρ
⇒
k n . (1 − k ) = 2ρ k
from Eq.(5), and equating with Eq.(6),
Rearranging the following second degree equation in terms of ‘k’ is obtained.
k 2 + ( 2 ρ . n) . k − ( 2 ρ . n) = 0 Solving for k,
k = − ( ρ . n) + ii)
( ρ . n) 2 + (2 ρ . n)
(7)
The internal couple resulting from internal forces Cc and Ts must equal to the external applied service load moment. The convenient moment center is taken usually the line of action of the internal forces.
--Taking moment of internal forces about line of action of Ts ,
M = Cc . ( d − Substituting Cc =
x ) 3
fc . b. x 2
&
x = k . d , and simplifying then equation of service load
moment resistance of section is obtained as,
⇒
M =
fc k . k . b . d 2 . (1 − ) 2 3
(8a)
k 3
Letting j = (1 − ) be lever-arm ratio for internal forces of section of beam, then service load moment resistance of section may be written as,
⇒
M =
fc . k . j . b . d2 2
(8b)
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Letting R =
fc . k . j be relative bending moment of section of beam, then service load 2
moment resistance of section may be written as,
⇒
M =R . b . d2
(8c)
Rearranging Eq.(8c), the effective depth of section required by singly reinforced beam obtained as,
d=
M R.b
--In similar manner, taking moment of internal forces about line of action of Cc ,
M = Ts . ( d −
x ) 3
Substituting Ts = As . f s
x = k . d , simplifying equation of service load moment
&
resistance of section is obtained as,
⇒
M = As . f s . d . (1 −
k ) 3
(9a)
k 3
Letting j = (1 − ) be lever arm ratio for internal forces of section of beam, then service load moment resistance of section may be written as,
⇒
M = As . f s . j . d
(9b)
Rearranging Eq.(9b), the area of tension steel required by beam section is obtained as,
As =
M fs . j . d
2.3 Type of Singly Reinforced Beam Sections-Based on Modes of Stresses
Depending on the amount of steel used by section, singly reinforced sections are divided into three: Balanced section, Over-reinforced section and Under-reinforced section.
a) Balanced Section: The most economical section in terms of material usage. In this section, the maximum stresses in both the reinforcement and the concrete reach simultaneously the respective permissible value. i.e
f s = f s , allow
f c = f c , allow
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From Eq.(3), neutral axis depth ratio of singly reinforced section,
k=
n . fc n = n. fc + f s n + f s fc
For balanced section, ratio of allowable stresses of steel to concrete is denoted by ‘r’ as,
r=
f s , allow f c , allow
Substituting ‘r’ into above equation, the balanced neutral axis depth ratio is obtained as
kb =
n n+r
(10)
From Eq.(5), steel ratio of singly reinforced section,
ρ=
k . fc k = 2 fs 2 f s fc
Substituting ‘r’, the balanced steel ratio is obtained as
ρb =
kb 2r
(11a)
where r—is ratio of allowable stresses of steel to concrete Substituting equation of kb from Eq.(10) in to Eq.(11a), the balanced steel ratio is rewritten as
ρb =
n 2r . ( n + r )
(11b)
This equation would gives the balanced steel ratio of singly reinforced section in such away that the maximum stresses developed in steel and concrete when section subjected to service load moment will reach simultaneously the respective allowable stresses. The corresponding lever-arm ratio and relative bending moment of balanced singly reinforced section are obtained by
jb = (1 −
Rb =
kb ) 3
f c , allow 2
. kb . jb
b) Over-reinforced Section if ρ > ρb: Over-reinforced sections are those that contain more reinforcement than the balanced one. Hence, as the applied moment is increased, the maximum stress in concrete reaches its permissible value first; and by the time the stress in
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reinforcement reaches its permissible stress, the concrete is over stressed. Therefore, the stresses in concrete and steel for such section are as follow:
f s < f s , allow
--determined from stress diagram using similarity of triangles
f c = f c , allow The maximum moment of resistance of over-reinforced section is obtained by the following equations in terms of allowable stress of concrete as,
M =
f c , allow 2
. k . b . d 2 . (1 −
or
M = As . f s . j . d
where
fs =
k ) 3
--used to determine neutral axis depth ratio --used to determine area of tension steel
n . (1 − k ) . f c , allow < f s , allow k
Here, an increase of load produces over stress in concrete earlier than the reinforcement; as a result the concrete crushes in compression. Such failure is sudden and occurs without warning. For this reasons, over-reinforced section is not recommended in design. c) Under-reinforced Section if ρ < ρb: Under reinforced sections are those that contain less reinforcement than the balanced one. In such sections, the tensile reinforcement is insufficient to develop the full strength of the concrete in compression, so that when the reinforcement is fully stressed, the concrete is under-stressed. Therefore, the stresses in concrete and steel for such section are as follow:
f s = f s , allow f c < f c , allow
--determined from stress diagram using similarity of triangles
The maximum moment of resistance of under-reinforced section is obtained by the following equations in terms of allowable stress of steel as,
M =
fc k . k . b . d 2 . (1 − ) 2 3
or
M = As . f s , allow . j . d
where
f c = f s , allow .
--used to determine neutral axis depth ratio --used to determine area of tension steel
k < f c , allow n . (1 − k )
Here, failure is more gradual than over-reinforced section. As when steel is over-stressed, the steel yields but is still able to support the yield stress since steel is a ductile material. Therefore, from
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both safety and economic point of view, it is recommended to design section of flexural member as under-reinforced section. 2.4 Control of Deflection
The deflection of structure or part of structure shall not adversely affect the appearance or efficiency of structure or finishes or partitions. For beams and slabs, the vertical deflection limits may generally be assumed to be satisfied provided that the minimum depth required by deflection specified by code is maintained. ACI code provide minimum depth required by beams and one-way slabs in terms of span length as given in table below can be used as a crude estimate of initial depth to control deflection.
Table: ACI-code minimum depth of beams and one-way slab to control deflection Simply
End
Types of Member
supported
spans
Spans
spans
-Beams or one–way S-400MPa
l 16
l 18.5
l 21
l 8
S-300MPa
l 20
l 23
l 26
l 10
S-400MPa
l 20
l 24
l 28
l 10
S-300MPa
l 25
l 30
l 35
l 12.5
ribbed slab -One-way solid slab
Interior Cantilever
Note: For other grades of steel, the value given for S − 400 MPa is modified by multiplying factor of ( 0.4 +
fy 690
).
EBCS-2 provide minimum effective depth, ‘d’ to be used to control deflection is given as,
d = ( 0.4 + 0.6 where
fy
).
le
400 β a
f y --characteristic yield strength of steel in MPa
le --effective span length; and for two-way slabs, the shorter span length
β a --constant as given in table below; and for slabs carrying partition walls likely to crack, shall be taken as β a ≤ 150 l0
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l0 --distance in meter between points of zero moment (for continuous beam, may be taken approximately as 0.7 times length of span), and, for a cantilever span, twice the length to the
face of the support
Table: Values of β a Simply Member
End
supported spans
-Beams
Interior Cantilever spans
Spans
20
24
28
10
a) span ratio, ll ls = 2 (includes one-way slabs)
25
30
35
12
b) span ratio, ll ls = 1
35
40
45
10
-Slabs
Note: For slabs with intermediate span ratio interpolate linearly.
2.5 Doubly-Reinforced Rectangular Beam Section
If the section of RC beam is limited in dimension (usually depth), it can not develop the compressive force required to resist the applied bending moment as singly reinforced section. That is, the applied moment is greater than the balanced moment capacity of singly reinforced section. For small increase of moment over the balanced one, over-reinforced section can be used, which is not recommended in design.
A more economical and safe way of designing section in such case is to provide reinforcement in compression zone of RC section. This section termed as doubly reinforced. Doubly reinforced section can also be used if the required depth of section of beam as singly reinforced is unacceptable. The purpose of reinforcement in compression zone of RC section is to assist the concrete in resisting compressive force and to keep the neutral axis at the ideal position ensuring balanced type failure.
In doubly reinforced beam section, concrete and steel act together to take compression. If both steel and concrete behave elastically, the stress in compression steel is ‘ n = Es Ec ’ (modular ratio) times the concrete stress at the same level. However, concrete under sustained compressive stress deforms continuously with time due to creep effect and concrete is also subjected to
30
shrinkage over a period of time. Whereas these time dependent effects do not occur in steel. As RC beam deforms, even at low loading, there is a continuous transfer of stress from concrete to compression steel. Therefore, the actual stress in compression steel is larger than that computed on the basis of elastic behavior of materials. i.e f s1 > n . f c1 where f c1 is concrete stress at the level of compression steel.
To approximate the effect of creep of concrete, ESCP-2/83 code species that an effective modular ratio of ‘ 2n = 2 Es Ec ’ is to be used to transform compression reinforcement for stress computation with the stress in compression reinforcement not to exceed the allowable stress of steel, f s , allow .
2.6 Design Equation for Doubly-Reinforced Rectangular Section
Consider a doubly reinforced rectangular section subjected to a service load moment, M as shown below.
d1
εc 1 s
A
M
f c , allow
xb = kb . d
εs
AS
f s, allow
≡
a) Doubly reinforced section
As1. f s1 As 2
T1 = As1. f s , allow b) Balanced singly reinforced section
C2 =
f s1
+
n
As1
b
Strain
As1
C1
d
f c1
c) Comp. steel plus Excess tens. steel
f s , allow T2 = As 2 . f s , allow
Two couples method is used to determine the required areas of tension and compression reinforcement by treating doubly reinforced section into two parts. The total resisting moment is equal to the sum of two resisting couples: one of which is provided by given cross-section of beam without compression reinforcement with a partial tension steel area, As1 that balance concrete in compression; and the other by compression steel, As1 and the remainder of tension steel area, As 2 . Thus, the section with compression steel is designed as balanced reinforced
31
section in such away that compression
steel and extra tension steel are proportioned by
maintaining the balanced neutral axis depth. Let
M 1 --balanced moment capacity of a section if singly reinforced M 2 --excess moment produced by compression steel plus excess tension steel
Then, total moment capacity of doubly reinforced section is,
M = M1 + M 2 Balanced moment capacity of a section if singly reinforced and the corresponding area of tension steel balancing the section are obtained by
M 1 = M b = Rb . b . d 2 &
As1 =
M1 f s , allow . jb . d
Excess moment resisted by compression steel plus excess tension steel from couple produced by internal forces developed in the section,
M 2 = ( M − M 1 ) = C2 . ( d − d 1 ) = T2 . ( d − d 1 ) Rearranging the above equation, internal forces developed in compression steel and excess tension steel are obtained as
C2 = T2 =
M2 ( d − d1 )
Then, area of excess tension steel is obtained as,
As 2 =
T2 f s , allow
=
M2 f s , allow . ( d − d 1 )
Therefore, total area of tension steel required by doubly reinforced section is obtained as,
As = As1 + As 2 From similarity of triangles shown in fig.(b) above, the stress in the concrete at the level of compression reinforcement, f c1 ,
f c1 = f c , allow .
( kb . d − d 1 ) kb . d
Therefore, the stress in compression reinforcement,
f s1 = ( 2n . f c1 ) ≤ f s , allow Due to the presence of reinforcement in compression zone, there is a loss of concrete area of magnitude, As1 . And, this will cause a corresponding loss in compression force of ( As1 . f c1 ) .
32
1 Therefore, if f s = ( 2n . f c1 ) < f s , allow
C2 = f s1 . As1 − f c1 . As1 = 2n . f c1 . As1 − f c1 . As1 ⇔
C2 = As1 . f c1 . ( 2n − 1)
⇒
As1 =
C2 ( 2n − 1) . f c1
1 If f s = ( 2n . f c1 ) ≥ f s , allow , then f s , allow
C2 = f s , allow . As1 − f c1 . As1 = As1 . ( f s , allow − f c1 ) ⇒
As1 =
C2 ( f s , allow − f c1 )
2.7 Flanged Section (T- or L-section) under Flexure
In construction of building structures, the slab is usually supported by a system of beams. If the connection between the beam and the slab adequately transmit longitudinal shear force, then the beam and slab together may act as a homogeneous section of T- or L-forms. For loaded beamslab if subjected to a moment which produces compression at the top surface, the slab therefore becomes parts of the compression flange of the beam, resulting in a greater zone of compression and giving a more economical section. For a reinforced concrete beam-slab section, adequate connection between the beam and the slab is easily provided by casting the section as monolithic, and by extending beam-stirrups and bent bars up into the slab.
It is known that the compressive stress caused by flexure in the upper flange decreases as the distance from the web increases. This is because the shear deformation of flange relieves some of compressive stresses as the element becomes more remote to the web. Therefore, this makes exact analysis of flanged section of infinite wide-flange complex. In order to simplify the design of flanged section of infinite wide-flange, it is usual to assume a uniform stress over a reduced width of flange. This reduced width is known as effective width. Effective width of flange is determined equating forces on compressive flange due to actual compressive stress on infinite wide-flange with equivalent uniform compressive stress on reduced width of flange.
The effective width has been found to depend primarily on the type loading, span length, spacing of beams, width of the web, and the relative thickness of the slab with respect to the total beam
33
depth. For practical design of flanged section, effective width of flange recommended by codes may be used.
-ACI code prescribes a limit on the effective flange width, ‘ b f ’ as follows. a) For interior T-section, effective flange width shall be the smallest of:
⎧ be = l 4 ⎪ ⎨ be = bw + 16t ⎪ b = center − to − center spacing of the beams ⎩ e Where
l
--is span length of the beam.
bw --is width of the web
t
--is thickness of the slab
b) For exterior T-section (L-forms), effective flange width shall be the smallest of:
⎧ ⎪ be = bw + l 12 ⎪ ⎨ be = bw + 6t ⎪ 1 ⎪ be = bw + ( clear dis tan ce to the next beam ) 2 ⎩ c) For isolated T-sections, effective flange width shall be
be ≤ 4bw and, also ACI code requires that t ≥ bw 2
-EBCS-2/95 also specifies the effective flange width, ‘ b f ’ as follows. a) For symmetrical T-beam, effective flange width shall not exceed the lesser of:
⎧ be = bw + l 5 ⎨ ⎩ be = center − to − center spacing of the beams b) For edge beams (L-section), effective flange width shall not exceed the lesser of:
⎧ be = bw + l 10 ⎪ ⎨ 1 ⎪⎩ be = bw + 2 ( clear dis tan ce to the next beam )
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2.8 Design of Flanged beams for Flexure
Design of flanged beams are made depending on the sign of design moment develop in the section producing either tension or compression on flange side of beam. If the design moment developed in section produces tension on flange side, the section is to be designed as if it were rectangular beam of width equal to the width of the web of the section. For such section, no advantage is gained in using slab as flange of section.
On other hand, if the design moment developed in section produces compression on flange side, the section is to be designed depending on the position the neutral axis. The position of the neutral axis depends up of the proportions of the cross-section, the amount of tension steel, and the strength of the materials. If the neutral axis lies in the flange, the section is to be designed as if it were a rectangular beam of width equal to the effective flange width. When the neutral axis lies in the web, the section is to be designed as T-beam section. 2.9 Design Equations of T-beam Section -Working Stress (Elastic) Method
Consider a flanged section subjected to a service load moment, M as shown below. Assume the neutral axis lies in the web so that the section is designed as T-beam section.
be
ε cε
fc
c
z
t
Cc
k. d
N. A
d
M
fc .
εs
As
Ts = As . f s
bw
T − beam Section
(k . d − t ) k .d
Strain
35
Stress
From geometry of strain diagram and assuming perfect elasticity of both materials, expression for neutral axis depth ratio, k is obtained as:
k=
n
(*)
f n+ s fc
Since the compression area provided by the slab is so large (large be ), the actual maximum compressive stress in concrete, f c will be some unknown fraction of its allowable stress value. Hence, neutral axis depth ratio, k has to be given in terms of the maximum compressive stress in concrete, f c that does not related to the allowable stress value, f c , allow . To simplify the derivation of design equations, the compressive stress in the web above the neutral axis is ignored. Therefore, total compressive force in the flange is equal to:
Cc ≅
f c + f c . (k . d − t ) k . d ( 2k . d − t ) . be .t . be . t = f c . 2k . d 2
and, resultant tensile force in steel,
Ts = As . f s For horizontal equilibrium, Ts = Cc
⇒ From Eq.(*)
As . f s = ρ . be . d . f s = f c . fc =
( 2k . d − t ) . be . t 2k . d
(**)
k . fs n . (1 − k )
Substituting this expression of
f c into Eq.(**) to eliminate unit stresses and then gives
expression of the neutral axis depth ratio of T-beam section as:
k=
1 (t d ) 2 2 n . ρ + (t d )
n. ρ +
The distance to the center of compression (centriod of the trapezoidal area of compressive stress) from the upper face of the beam is:
z=
(3k . d − 2t ) t . ( 2k . d − t ) 3
Then, the lever-arm of the couple formed by the internal tensile and compressive force is:
j . d = (d − z )
36
Substituting z and k , and solving for j , the following expression lever-arm ratio of T-beam section is obtained,
6 − 6(t d ) + 2(t d ) 2 + (t d )3 . (
j =
1 ) 2ρ . n
6 − 3(t d )
The resisting moments of T-beam section are equal to the product of the lever-arm, j. d of the internal force couple and the total tension or compression. Hence,
M = As . f s . j . d or
M = f c . (1 −
--in terms of steel stress
t ) .be . t . j . d 2k . d
--in terms of concrete stress
Approximate equation for resisting moments can be obtained using the limiting values for leverarm between internal forces, ( j . d ) > (d − t 2) and average compressive stress in the flange,
Cc = f c . (1 − t 2k . d ) > f c 2 as, M = As . f s . (d − t 2) M =
--used to determine trail area of tension steel
fc . be . t . (d − t 2) 2
--used to check maximum stress in concrete
37
