3
BEAM DESIGN FOR FLEXURE
Beams can be categorized into shallow beams and deep beams. Shallow beams are those beams having a length to depth ratio (Ln/d) greater than 5 and deep beams less than 5. Shallow beams may also act as a deep beam when the loading is concentrated near the support or M/V is less than 2. Shallow beams re one of th most common elements in reinforced concrete design . Since the flexural strength normally controls the dimensions of beams, the beam is initially designed for moment and only later checked for shear and reinforced additionally if required. The check for shear occurs at a later stage and is independent of the design for moment. In this chapter we shall focus on shallow beams designed for flexural strength using two design methods: working stress method (WSD) and ultimate strength design method (USD). 3.1 Beam cross sections b
d’
b hf
N.A.
N.A.
d
d
bw Rectangular
T-Section
U-Section
3.2 Bending Failures in beams 3.2.1 Brittle failure-failure is initiated by the crushing of the concrete followed by the sudden disintegration of the compression zone.When the beam is over reinforced the stress in the relatively large area of steel has not reached its yield point. When the beam is under reinforced, the steel will snap and total rupture of the section will occur suddenly To prevent brittle failure, the reinforcement must yield while the strain in the concrete is less than the failure strain of 0.003. 3.2.2 Ductile failure- beam has moderate percentage of steel and failure is initiated vy a yielding of steel while strain of concrete are relatively low. Such beams can continue to carry load and are able gto undergo large deflections before final collapse.
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3.3 Working Stress Design method Assumptions 1. Plane section before bending remain plane after bending 2. Stress varies linearly from the linear axis 3. No slip occurs between the surrounding steel and the concrete. 3.3 Cracking moment (Mcr) If the moment is small, tensile bending stresses are low or less than the tensile strength of concrete and no cracking occurs. For this condition the entire cross section carries flexural stresses and steel reinforcement is neglected. The analysis for bending stresses in the uncracked beam can be based on the properties of the gross cross sectional area using elastic equations.
where: M=the bending moment of the cross section y=distance from the centroid to a point which f can be evaluated Ig =moment of inertia of the gross cross section. f =stress at distance y from the centroid
h/2
M
h
Centroidal axis
fr
Equation 3.1 is valid when external moments will not exceed the tensile strength limit of the concrete. If a moment is applied that causes to reach the modulus of rupture, the cross section will be on the verge of cracking. The moment that produces a tensile stress just equal to the modulus of rupture is called the cracking moment
where: Mcr – cracking moment Ig- gross moment of inertia fr – modulus of rapture 2
3.4 Investigation of Singly Reinforced Concrete Beam If the moment is bigger than Mcr tensile stress at the outermost surface exceeds the modulus of rupture and therefore the tensile strength of concrete is no longer considered as shown on figure below. b
fc C
y
M
d
arm
N.A .
d -y fs
T
Two equilibrium conditions apply to a section subjected to bending only
The resultant internal compressive force must be equal to the resultant internal tensile force
The moment of the internal couple composed of the resultant compressive and tensile force must be equal to the applied bending moment.
3.4.1 Neutral axis of rectangular compression zones In this case the neutral axis becomes the reference axis and not the centroidal axis. To determine the location of the N.A. two methods may be used: Moment-Area Method. By equating the product of the area of concrete compression zone and moment arm to the product of the transformed area of steel and moment arm we get ( ) or
A quadratic solution would give the value for y. 3.3 Method of Transformed Section In the method of transformed section, the cross section containing steel and concrete is transformed into a homogenous section of one material all having the modulus of elasticity of concrete. 3
In the transformation, two conditions must be satisfied:
Equilibrium condition requires that the total tensile force be the same, or
Compatibility of deformation condition requires that the unit elongation be the same, or
Utilizing the modular ratio n=Es/Ec
y
C
Ac
d
arm nAs b Cross section
N.A.
T
Transformed section
Maximum allowable extreme fiber in compression is 0.45 fc’ and tensile stress in reinforcement fs shall not exceed 140 Mpa for Grade 275 and 170 Mpa for Grade 415 steel. 3.3.1Neutral axis at ideal location The neutral axis is at the ideal location if the allowable stresses for both steel and concrete are reached simultaneously.
If the neutral axis is nearer to the compression face than the ideal location, then the allowable steel stress is reached first, one may say steel controls. If the neutral axis is farther from the compressive than the ideal location, only the allowable concrete stress can be reached. In this case concrete controls.
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fc < 0.45fc’
0.45fc’
0.45fc’
y N.A.
N.A. (Ideal)
N.A.
d
d
d-y
fs/n
N.A. at ideal location
d
fs/n
ft < fs/n
Steel controls
Concrete controls
To locate the ideal location of the neutral axis,
⁄
Allowable moment can then be computed either by
Internal couple method,
Flexure formula
where,
3.4 Design of Singly Reinforced Concrete Beam In design problems, the bending moment, the modular ratio, and the allowable working stress are given. The designer is to select the values of b, d and As. The procedure for determining the theoretical values of b, d and As, the ideal location of the neutral axis is described below.
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⁄ k=ratio of the ideal y to the effective depth d.
or
j = ratio of the ideal moment arm jd to the effective depth d. Since
and letting
ρ= reinforcement ratio to place the NA at the ideal location. Equating bending moment Mw to Cjd
In which
Equating Mw to Tjd, Mw = As(allow fs) jd, from which
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b
allow fc
C Ideal y=kd
d
ideal moment arm= jd
As T allow fs /n
The design constants k,j,ρ, and R is aimed to place the ideal location of the NA. Their values depend on the allowable working stresses fc, fs and n. The procedure for determining the theoretical values of b,d, and As is summarized as follows: 1. Find the required value of bd2 from Mw/R. 2. Assume a value of b and determine d ( select b and h and check weight) 3. Determine As from ρbd and check its value from Mw/[(allow fs)(jd)] and check its value from
3.4.1 Bar Placement and concrete cover Minimum clear spacing between parallel bars in a layer shall be equal to the bar diameter but not less than 25 mm. (NSCP 407.7.1) Clear distance between bars with two or more layers shall not be less than 25 mm. (NSCP 407.7.2) Minimum cover for concrete for beams and columns permanently exposed to earth ……….. 75 mm exposed to earth or weather 20 mm to 36 mm diameters ……….50 mm 16 mmm or smaller………………... 40 mm Not exposed to earth or weather Main reinforcement, ties, stiirrups, spirals …..40 mm
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A=concrete cover B= stirrup or ties C=for 20 mmφ or smaller-20 mm; for 25 mmφ or bigger -0.5 x bar φ D/E= clear spacing
B
A
C
D
E
3.4 Design of Doubly reinforced concrete beam Steel located in the compression zone, or compression steel, is denoted by A’s. Beams reinforced with both tension and compression steel are doubly reinforced. The compression steel: increases flexural strength reduce long term deflections due to creep increase ductility
d’
As’
C1
d
d
=
kd
C2
2nAs’
kd
+
d-d’
d-y nAs T1
As
T2
b Cracked transformed section
Doubly reinforced cross section
The modular ratio n is multiplied by 2 in the compression steel. Doubling the modular ratio accounts for the increase in stress that occurs with time as the concrete in the compression zone creeps. Creep produces additional strain and gradually raises stress approximately twice that of the initial value. The “two couple method” is commonly used to determine the required values of As and As’ in a doubly reinforced section. In this method, the N.A. is first placed at the ideal location, then the bending moment is taken as the sum of the two resisting couples. (
)
(
)
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One of the resisting couples is provided by the concrete stress solid along with a portion of As1 of the tension steel and the remainder As2 of the tension steel. The procedure for determining As and As’ is as follows: 1. Compute Mw1 = Rbd2 2. Compute As1 = ρbd and check As1 = Mw1/(fsjd) 3. Compute Mw2 = Mw – Mw1 4. Compute C2 = T2 = Mw2/(d-d’) 5. Compute As2 = T2/fs 6. Compute As = As1+ As2 7. Compute
8. Compare 2nfc1 with fs if 2nfc1 ≤ fs;
if 2nfc1 > fs;
3.5 Investigation of Doubly Reinforced Concrete Beam To locate the neutral axis, equate moments of compression area and transformed compression steel to the moment of the transformed tension steel to determine y.
The moment strength can then be calculated, Mw = Mw1 + Mw2 where (d
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)
3.6 Investigation of Beams with Non-Rectangular Compression Zones The basic concept of locating the position of the neutral axis by summing moments of the appropriate areas about the neutral axis is applicable to cross sections of any shape. w t
AC1
Y
N.A.
A c2
d
nAs b v
Transformed section
(
)
(
)
A quadratic solution would give the value of y. The service load moment can then be derived
or, (
)
where
The stresses in the concrete and steel can then be computed as:
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