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Transcript of week2-rectangularandsectionanalysisinbendingandshear-111203094640-phpapp01
Rectangular and section analysis in bending and shear
ARUAN EFENDY BIN MOHD GHAZALI FAKULTI KEJURUTERAAN AWAM
Learning out-comesAt the end of this lecture, the student should be
able to:
a) Identify singly and doubly reinforced section (CO1)
b) Design a rectangular section beams in bending, shear and deflection (CO1, CO2)
c) Details the beams (CO1).
INTRODUCTION TO SIMPLE RC. BEAM DESIGN Here are some examples of Reinforced Concrete beams that you may find in practice
Any of the above arrangements can be employed in
conditions where the beam is simply supported or where
it is continuous over the supports.
Beam design in Bending for Singly-Reinforced Section-Considering the case of a simply supported singly reinforced rectangular beam
The load causes the beam to deflect downwards resulting in tension in the bottom of the beam and
compression in the top. Neutral Axis - The level at which there is
neither tension nor compression OVER-REINFORCED - If a large amount of
reinforcement is present then the concrete will fail first in a brittle
UNDER-REINFORCED - if the section fails due to yielding of the steel reinforcement first and the failure mode is far more ductile resulting in large deformations, cracking and spalling of concrete on the tension face. As this is highly visible it is a much safer mode of failure, and also is more economical
In an under-reinforced section, since the steel has yielded we can estimate the ultimate tensile force in the steel:
Where: fy is the yield stressAs is the area of reinforcementγms is the partial safety factor for the reinforcement (=1.05)
The applied loading on the beam gives rise to an Ultimate Design moment (M) on the beam in this case at mid-span. The resulting curvature produces a compression force in the concrete Fcc and a tensile
force Fst in the steel. For equilibrium of horizontal forces:
The two forces are separated by the lever arm, z which enables the section to resist the applied moment and gives the section it’s Ultimate moment of resistance (Mu). For stability:
which;
Due to concrete (in general z=(d-(0.5X0.9x)) hence;
0.156=M/bd²fcu
Due to steel reinforcement
Example 1. Design of Bending Reinforcement for a Singly
Reinforced Beam
A simply supported rectangular beam of 6m span carries a characteristic dead (gk ) load (inc. Self wt. of beam ), and imposed (qk ) loads of 8 kN/m and 6 kN/m respectively
The beam dimensions are breadth b, 225mm and effective depth d, 425mm. Assuming fcu =30N/mm2 and fy =460N/mm2 calculate the area of reinforcement required
Since Mu>M we can design as a singly reinforced beam orcheck for K
Since K<0.156, design as simply-reinforced beam
Provide 2T20 ( As prov. = 628mm2 )
Home work:
By using appropriate text books, show that:
Mu = 0.156bd²fcu Based on Concrete strength
and
Mu = 0.95fyAsz Based on Steel
Beam design in Bending for Doubly-Reinforced Section
-Considering the case of a doubly reinforced rectangular
beam
The area of compression reinforcement is thus calculated from:
where d’ is the depth to the compression steel from the top surface.
We must now increase the area of tensile reinforcement to maintain compatibility by an equal amount
where ;
and K’ = 0.156
Example 2. Design of Bending Reinforcement for a
Doubly Reinforced Beam
A simply supported rectangular beam of 9m span carries
a characteristic dead (gk ) load (inc. Self wt. of beam ),
and imposed (qk ) loads of 6 kN/m and 8 kN/m respectively
The beam dimensions are breadth b, 225mm and beam
height h, 400mm. Assuming fcu =30N/mm2 and fy =460N/mm2 calculate the area of reinforcement
required
Since we can NOT design as a singly reinforced
beam, we must design as doubly reinforced or check for
k.
Compression Reinforcement-Assuming the compression steel to be 20mm
diameter Bars
PROVIDE 3T20 ( A’s prov. = 943mm2 )
Tension Reinforcement,
Hence, PROVIDE 4T25 (As prov = 1960 mm2 )
Beam design in Shear
-As we have already seen from the examples of failure
modes for RC beams we must consider the capacity of
the beam with respect to shear. Shear failure may be one
of two types
The first of these, diagonal tension, can be prevented by
the provision of shear links; while the second, diagonal
compression, can be avoided by limiting the maximum
shear stress to 5N/mm2 or whichever is the lesser
The design shear stress v, at any cross section is given by:
where V, is the design shear force on the section
Design formulae for link provision:
-If we consider a beam under applied shear force V, the
resulting failure will give rise to a crack which cutsacross
any links as shown:
The failure plane is assumed to lie at an angle of 45° as
shown. The number of links which are therefore intersected by this failure plane is equal to which
allows us to calculate the shear capacity of the links alone as:
The shear resistance of the concrete alone is:
*see Table 3.8 for vc
To avoid failure due to shear the design shear force due to ultimate loads must therefore be less than the sum of these two components i.e.
Although we have arrived at only two possibilities for the calculation of shear reinforcement provision is should be noted that the BS 8110 (Cl. 3.4.5.5) gives us three alternatives to consider:
A further limitation placed on the spacing of links by the BS is that the maximum spacing should be less than 0.75d, which is obviously necessary to avoid a failure plane forming which misses the links altogether
Example 3. Design of Shear Reinforcement for a Singly
Reinforced Beam
A simply supported rectangular beam of 6m span carries
a characteristic dead load, Gk (inc. Self wt. of beam ) and
imposed loads (Qk ) of 10 kN/m respectively.
The beam dimensions are breadth b, 300mm and effective depth d, 547mm. Assuming fcu
=30N/mm2 and fy =460N/mm2 fyv = 250N/mm2 . Calculate the
shear reinforcement provision required for each of the situations given
DESIGN CONCRETE SHEAR STRESS, vc
vc from the table 3.8 = 0.70N/mm2
Design shear stress (v)
Total Ultimate Load, W, is
As beam is symmetrically loaded,
Ultimate shear force V=90kN and design shear stress, v,
is
Diameter and spacing of linksBy inspection,
As described earlier this ratio allows determination of either spacing or area of links. An alternative means of evaluation is presented below:
Remembering also that the maximum spacing of links should not exceed 0.75d which in this case equals
PROVIDE R10 links at 275 mm(<410mm) centres for the whole length of the beam
(Asv/sv = 0.571.)
Deflection (Cl. 3.4.6. BS8110)
BS 8110 details how deflections and the accompanying crack widths may be calculated.
But for rectangular beams some simplified procedures
may be used to satisfy the requirements without too
much effort. This approximate method for
rectangular beams is based on permissible ratios of span/effective
depth.
BS8110 Deflection Criteria:1. Total deflection < span/2502. For spans up to 10 m, deflection after
partitions and finishes < span/500 or 20 mm, whichever lesser.
These criteria are deemed to be satisfied in the following cases:
These BASIC ratios may be enhanced by provision AND
over provision of both tension and compression reinforcement as shown by the following tables
For tension
For compression,
Deflection limit;
1.Singly reinforced beam
Basic span-depth ration x Table 3.10 value
2. Doubly reinforced beam
Basic span-depth ration x Table 3.10 value x Table 3.11 value