arrangement of reinforcement in typical concrete members

8/2/2019 Arrangement of Reinforcement in Typical Concrete Members

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ARRANGEMENT OF REINFORCEMENT IN TYPICAL CONCRETE MEMBERS

Simple Beams and Slabs

Simply supported slab:

Curtailment of tension steel in simply supported slab construction.
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DETAILING OF BEAMS

Beam carries transverse external loads that cause bending moment, shear forces and in

some cases torsion

Concrete is strong in compression and very weak in tension.

Steelreinforcement is used to take up tensile stresses in reinforced concrete beams.

Mild steel bars or Deformed or High yield strength deformed bars (HYSD) are used.

HYSD bars have ribs on the surface and this increases the bond strength at least by 40%

Bar Bending Schedule

Drawings generally include a bar bending schedule

The bar bending schedule describes the length and number, position and the shape of the bar
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Anchorage in steel bars is normally provided in the form of bends and hooks

The anchorage value of bend of bar is taken as 4 times the diameter of bar for every 45 0 bend

subjected to maximum of 16 times the diameter of bar.

Standard hooks


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The beams are classified as:

According to shape: Rectangular, T, L, Circular etc

According to supporting conditions: Simply supported, fixed, continuous and cantilever

beams

According to reinforcement: Singly reinforced and doubly reinforced

Cover in Beam

Minimum cover in beams must be 25 mm or shall not be less than the larger diameter of bar

for all steel reinforcement including links.

Nominal cover specified in Table 16 and 16A of IS456-2000 should be used to satisfy

the durabilitycriteria.

Types of Reinforcement in Beams

Generally a beam consists of following steel reinforcements:

Longitudinal reinforcement at tension and compression face.

Shear reinforcements in the form of vertical stirrups and or bent up longitudinal bars are

provided.
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Side face reinforcement in the web of the beam is provided when the depth of the

web in a beam exceeds 750 mm. (0.1% of the web area and shall be distributed

equally on two faces at a spacing not exceeding 300 mm or web thickness

whichever is less)Specification for the reinforcement in beams is given in clause 8.1 to 8.6 of SP34


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While drawing the details of a beam following convention representation of bars are used

Mild steel bars : f; HYSD bars: # or

Main bars are shown by thick single line.

Hanger bars are shown by medium thick lines.

Different forms of stirrups used in beams

Typical drawing of a simply supported beam


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REINFORCED CONCRETE DESIGN

Assumptions

Concrete has no tensile resistance

The reinforcing resists all the tensile forces.

Maximum tensile stress in the steel:

The characteristic cube stress of the concrete = fcuand the compressive stress in the concrete =

0,4 fcu

The maximum neutral axis depth (dn) is limited to 0.5 d.

The internal moments are in equilibrium with the external moments.

Resistance moment (Mr) of a singly reinforced concrete beam orslab.
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A partial material factor of 1,15 is applied to the steel and 1,5 to the concrete.

The following formula may be derived:

Resistance moment of the concrete in the compression zone:

Take moments about the tension zone, i.e., centroid of the tensile steel.

Assume a practical limit for the maximum: maximum dn = 0,5 d

The resistance moment of the tensile zone:

Take moments about the centroid of the compression zone.

Neutral axis depth and lever arm with known beam section details:

Note that b, d and As are known.


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Horizontal equilibrium of forces:

Fc = Ft

The lever arm z

Determination of the lever arm z = a1xd and tensile area As for a known section and known

dimensions b and d with a moment Mu acting on the section.

Substitute n1 and z in the equation for Mu

Solve for a1:

This can be written in a tabular form:

0,15 0,13 0,10 0,07 0,04

a1 0,75 0,80 0,85 0,90 0,95

With a known value of a1 the lever arm z may be calculated. Once the lever arm is know, the

steel may be calculated.


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arrangement of reinforcement in typical concrete members

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