Compression Reinforcement

What happens if the beam section is fixed and

K

M

bd f

ck

=

2

> allowable limit

Is there any way in which the section can be reinforced so that it

still fails in a ductile manner?

Consider the effect of adding extra steel in the compression

region. Effectively this has the same effect as increasing the area

of the compression block. Thus providing compression

reinforcement provides a means of designing a stronger ductile

section than is possible with tension steel only.

If

K

M

bd f

ck

=

2

> allowable limit then compression steel is

required. The standard procedure adopted by most codes, and

the associated reasoning, is as follows;

First consider the case when

K

is the maximum allowable value

K

M

bd f

M bd

ck

ck

= =

⇒ =

'

.

'.

2

2

0166

0166

f

This

M’

is the largest moment which the single reinforced

section can resist if it is to fail in a ductile manner. The area of

tension steel

A

s

required to resist this moment is easily found

using the standard formulae.

{ }

min

1

maxmin

2

87.0

'

13.125.05.0

'

zf

M

A

dKdz

fbd

M

K

y

s

cu

=⇒

−+=

=

Consider the effect of adding additional tensile steel

A

s2

and at

the same time adding a matching amount of steel

A’

in the

compression region such that;

A f

A

f

s

s y

s

''=

2

γ

In this case the additional forces are self-equilibrating,

equilibrium of axial forces is satisfied without having to re-

calculate

x

. In most simple cases the compression steel is placed

close to the extreme compression fibre, which at failure has a

strain of 0.0035, and hence has a strain greater than yield strain,

so that typically

A’=A

s2

.

The addition of this extra steel increases the ultimate moment

capacity of the beam by

(

)

(

)

∆M

f

A d d A f d d

ult

y

s

s s s

= − =

γ

2

'''

−'

Therefore to design a section for a moment M greater than M’ it

is necessary to provide compression reinforcement

A’

where,

(

)

(

)

A

M M

f d d

s

s

'

'

''

=

−

−

and tension reinforcement

A

s

where,

(

)

(

)

A A A

M

f z

M M

f

d d

s s s

y

y

s

= + = +

−

−

1 2

087

'

.

'

'

min

γ

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