Topic 4 Section Analysis

ANALYSIS OF BEAM SECTION UNDER FLEXURE

Flexural members are those subject to bending, e.g. beams and

slabs. EC2 also mentioned on other types of beam namely:

Deep Beams - Deep beams having a clear span of less than three times their overall depth require special consideration and are outside the scope of this module. EC2 gives design guidance.

Slender Beams- Slender beams, where the breadth of the compression face b is small compared with the depth and length, have a tendency to fail by lateral torsional buckling. If this buckling is prevented by, for example, a slab attached to the compression face of the beam, then the beam is considered as laterally restrained and it will be satisfactory if the depth h is not more than 2.5 times the breadth b. If the compression face is unrestrained, then the unrestrained length should not be more than 35 b .In practice, slender beams are rare.

Stress- Strain Relations for the Design of Cross-sections

Downward loads on a simply supported beam induce sagging

bending moments which cause compressive stress in the material

fibres above the neutral axis and tensile stresses in those below.

HJ ROSLAN BIN KOLOPPUSAT PENGAJIAN DIPLOMA PPD

Concrete is good at resisting compressive stresses but its resistance

to tension is so poor that it is ignored. Instead, steel reinforcing

bars are introduced to resist the tension.

A

A

Some of the parameters used to describe the cross-section of a

singly reinforced beam are shown below:

Cross-section A-A of rectangular singly reinforced beam

A singly reinforced beam will be adequate in bending if the ultimate resistance moment Mu of a concrete beam must be greater than or equal to the ultimate design bending moment Md i.e

Mu ≥ Md


The ultimate design bending moment Md is calculated under

ultimate design loads and will be greatest at or near the centre of

the beam.(For simply supported beam with uniformly distributed

load Md = wdL2/8)

To determine the Mu section analysis can be used as shown below:


Figure (b) above shows the strains in the beam when it is subject

to a sagging bending moment equal to M . The neutral axis is at

distance x from the top of the beam.

EC2 in Clause 3.1.7 allows the use of simplified stress-strain

relationship as in Fig (d) which can be used for the design at

ultimate limit state.

In the UK it is considered good practice that x should not be more

than d /2. This will ensure that the beam is under-reinforced and

will fail in a ductile manner by yielding of the reinforcement

giving a robust structural element with some warning of the

failure. If x is much greater than d /2 then the beam is over-

reinforced and may fail by crushing of the concrete, possibly

leading to a sudden collapse. Limiting K to 0.167, as described

later in this section, ensures that the beam is under-reinforced.

During steel yielding, the formation of plastic hinges allows

redistribution of maximum moment, resulting in a safer and

economical structure.

To ensure sufficient rotation of the plastic hinge and also to allow

for others factors, EC2 limits the depth of neutral axis to:

x ≤ 0.45 d for concrete class ≤ C50/60


Figure (c) shows the design values of the stresses at the ULS.

● The concrete cylinder strength f ck is multiplied by 0.85, a factor

which converts compressive strength into bending strength, and

divided by the partial safety factor γ m = 1.5.

● The reinforcement strength fyk is divided by the partial safety

factor γ m = 1.15.

Note that the stress in the concrete below the neutral axis is zero as

it is assumed to be cracked.

Figure (d) is a simplified version of Figure (c) with the curved

stress block replaced by a rectangular stress block of depth 0.8 x

for ease of calculation.

Figure (e) shows the stresses from Figure(d) converted to forces.

● Concrete force F cc = stress multiplied by area

= (0.567 fck )(0.8 x )( b ) = 0.453 xbfck

● Steel force F st = stress multiplied by area

= 0.87 fyk As


Analysis of a singly reinforced beam

Based on Cl 3.1.7 EN 1992-1-1

Design equations can be derived as follows:

Bending will induce a resultant tensile force Fst in the steel bar and a resultant compressive force in the concrete Fcc.

At equilibrium , ultimate design moment M = moment of resistance of the section

M = Fccz =Fstz ………… (1)HJ ROSLAN BIN KOLOP

PUSAT PENGAJIAN DIPLOMA PPD

s/2

εst

s =0.8xNeutralAxis

b0.0035

Fst

z=lad

Fcc

0.85fck/γc = 0.567fck

As

x

Section Strain Stress Block

d

Where z is the lever arm.

Fcc = stress x area of action = 0.567fck x b x s …………..(2)

and z = d – s/2

substitute in (2)

M = Fccz = 0.567fck x b x s x z

Replacing s,

M = 1.134fckb(d-z)z

Rearranging and substitute K = M/bd2fck

(z/d)2 - (z/d) + K/1.134 = 0

Solving the quadratic equation:

z = d {0.5 + √(0.25 – K/1.134)}

In equation (1) ,

Fst = (fyk/γs) x As γs - safety factor for steel = 1.15 = 0.87fykAs

Therefore,

As = M 0.87fykz

This equation can be used to design the tension steel area for singly reinforced beam section to resist an ultimate design moment M.


Balance Section

For a singly reinforced beam, EC2 allows the maximum value for x (depth to neutral axis)= 0.45d (for C50/60 concrete) in order to provide ductile section and a gradual tension type failure. This is often referred as the balance section because at ultimate limit state the concrete and tension in steel reach their ultimate strain at the same time.

x bal = 0.45 d

The depth of the stress block,

s = 0.8xbal = 0.8 x 0.45d = 0.36d

The force on the concrete stress block is Fccbal = 0.567fck x b x s = 0.204fckbd

For equilibrium the force in concrete = force in steel

Fstbal = 0.87fykAsbal = Fccbal = 0.204fckbd

Therefore,Asbal = 0.234fckbd/fyk

So that100Asbal = 23.4 fck percent bd fyk

which is the steel percentage for a balanced section which should not be exceeded for a ductile singly reinforced section.


Example: If fck = 25 N/mm2 and fyk=500 N/mm2

100Asbal = 23.4 x 25 = 1.17 percent bd 500

The ultimate moment of resistance of the balanced section is

Mbal = Fccbalzbal wherezbal = d – s/2 = 0.82d

Substitue Fccbal and z

Mbal = 0.167fckbd2

and

Md = 0.167 = Kbal

fckbd2

If

Md > Kbal = 0.167 than the section can no longer be singly fckbd2 reinforced and compression steel is needed in the Compression zone of the section

If the value of K for a singly reinforced beam is found to exceed K lim > 0.167 then the beam is likely to fail by crushing of the concrete before the reinforcement yields.(Over-reinforced)

This should be avoided by either:● using stronger concrete and so reducing K● increasing the size of the beam and so reducing K● providing compression reinforcement above the neutral axis to form a doubly reinforced beam.


Moment of Resistance of a Section

Given a cross-section of a reinforced concrete beam, the moment of resistance of the section can be calculated.

For equilibrium of the compressive force in the concrete and the tensile force in the steel as shown below:

Fcc = Fst

therefore

0.567fckb x s = 0.87 fyk As

Therefore depth of stress block is

s = 0.87fykAs/0.567fckb

And x = s/0.8


Therefore moment of resistance of the section is

M = Fst x z

This equation assume the tension reinforcement has yielded, which will be the case if x< 0.617d

Example:

Determine the moment of resistance of the cross-section shown below. Given fyk=500 N/mm2; fck = 25 N/mm2.

Fcc = Fst

Therefore0.567fckb x s = 0.87 fyk As

0.567 x 25 x 300 x s = 0.87 x 500 x 1470


b=300mm

d =520mm

As=1470 mm2

s = 150 mmandx = s/0.8 = 150/0.8 = 188 mm

This value is less than 0.617. Therefore the steel yielded and fst = 0.87fyk as assumed.

Therefore moment of resistance of the section is

M = Fst x z = 0.87fykAs(d – s/2) = 0.87 x 500 x 1470(520 – 150/2) x 10-6

= 284 kNm

Doubly Reinforced Concrete Section

When M > 0.167fckbd2 i.e when the design moment exceeds the moment of resistance of concrete ( Mbal) then compression reinforcement is required.


s/2

εst

s =0.8xNeutralAxis

b0.0035

Fst

z bal

Fcc

0.85fck/γc = 0.567fck

As

x

Section Strains Stress Block

Fsc

d

d’

As’

For this condition the maximum depth of neutral axis allows in EC2 is x <0.45d to ensure ductile failure of the section.

Using equilibrium, Fst = Fcc + Fsc

0.87fykAs = 0.567fckbs +0.87fykA’s

And taking moments about the centroid of the tension steel M = Fcc x zbal + Fsc (d – d’)

It can be proved that:

As’ =

and

As = + As’

Where Kbal = 0.167 and K= If d’/d < 0.171 than the compression yielded

EXAMPLE

The section shown below is to resist an ultimate design moment of 285 kNm. Given fyk=500 N/mm2; fck = 25 N/mm2. Determine the area of reinforcement required.


K= = 285 x 106/260x4402x25 = 0.226 >0.167

Therefore compression steel is required.

Check d’/d = 50/400 = 0.11 < 0.171Compression steel will have yielded

Compression steel:

As’ =

= (0.226 – 0.167) x 25 x 260 x 440 2 0.87 x 500 x (400 – 50) = 438 mm2

Tension steel:

As = + As’

= 0.167 x 25 x 260 x 440 2 + 438 0.87 x 500(0.82 x 400)


b=260 mm

d =440 mm

d’=50 mm

As’

As

= 1339 + 438 = 1777 mm2


Moment of Resistance of Doubly Reinforced Beam

Example

Determine the ultimate moment of resistance of the cross section shown below. Given fyk = 500 N/mm2 and fck = 25 N/mm2

Using equilibrium (tension) (compression)Fst = Fsc + Fcc

0.87fykAs = 0.567fckbs + 0.87 fykAs’

s = 0.87fyk(As – As’)/ 0.567fckb = 0.87 x 500 (2410 – 628)/0.567 x 25 x 280 = 195 mm

x = s/0.8 = 244 mm


s/2

s =0.8xNeutralAxis

b=280

Fst

z bal

Fcc

0.85fck/γc = 0.567fck

As=2410

x

Section Stress Block

Fsc

d=510

d’=50

As’=628

x/d = 244/510 = 0.48 < 0.617

so tension steel yielded.

d’/x = 50/225 = 0.22 < 0.38

so compression steel yielded.

Taking moment about tension steel:

Moment of resistance = Fcc(d – s/2) + Fsc(d – d’) = 0.567fckbs(d-s/2) + 0.87fykAs’(d-d’)={ 0.567x25x280x195(510-195/2) + 0.87x500x620(510-50)}x10-6

= 443 kNm

TUTORIAL


S1 The singly reinforced concrete beam shown below is made of C30/37 concrete. Using stress block as recommended in EC2, check whether the section can resist an ultimate sagging bending moment of 525 kNm.

Given As = 2348 mm2

S2 The singly reinforced concrete beam shown below is made of C35/45 concrete and is required to resist an ultimate sagging bending moment of 150 kNm. The breadth b is 250 mm. Choose a suitable beam depth and determine the area of reinforcement required.

ASSIGNMENT


A simply supported reinforced concrete beam with an effective span of 7.0 m is 500 mm deep overall by 250 mm wide. It supports the following characteristic loads:

Permanent dead loads: 12.0 kN/m plus beam self-weightVariable imposed loads: 11.0 kN/m.

The concrete is grade C40/50, and 25-mm cover is required to all reinforcement. Assuming that the shear links are H10 and the main bars are H25, check whether the beam size isadequate for the bending ULS and determine suitable tension reinforcement.


Topic 4 Section Analysis

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