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