Solved numerical examples | Design of Singly reinforced sections

Guide to design of Singly reinforced Sections | Building Construction

For the “design of Singly reinforced sections” article series, we have covered the following:

 

Now we will move on with our next solved numerical example in which we will make use of the formulas that we have derived in our earlier articles.

Numerical Problem

Determine the following:

a)      The position of the neutral axis

b)      Lever arm

c)       Moment of resistance

d)      Percentage of steel

(For a rectangular beam section of width b mm and effective depth d mm. Take σcbc = 5 N/mm2, σst = 140 N/mm2, m = 18.66

To find Neutral Axis (NA)

σcbc /(σst/m) = xc/(d – xc)

5/(140/18.66) = xc/(d – xc)

Therefore, xc  = 0.399d mm= 0.4dmm

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Moment of Resistance | Design of Singly reinforced Sections

Moment of Resistance | Guide to design of Singly reinforced Sections

For the design of Singly reinforced Sections article series, we have covered the following:

 

Now we will move on with our discussion on “Moment of resistance” and derive the formula for Moment of resistance for balanced section, under-reinforced section and over reinforced section.

The moment of resistance of the concrete section is the moment of couple formed by the total tensile force (T) in the steel acting at the centre of gravity of reinforcement and the total compressive force (C) in the concrete acting at the centre of gravity (c.g.) of the compressive stress diagram. The moment of resistance is denoted by M.

The distance between the resultant compressive force (C) and tensile force (T) is called the lever arm, and is denoted by z.

Moment of resistance | Singly reinforced Section
Moment of resistance | Singly reinforced Section

From the diagram above, it is clear that the intensity of compressive stress varies from maximum at the top to zero at the neutral axis.

Therefore, centre of gravity of the compressive force is at a distance x/3 from the top edge of the section.

Therefore, z = d-x/3

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