## What Is Stress -strain Block

Design of a flanged section with the depth of the stress block below the flange fhe T-scction beam shown in figure 4.14 is required to resist an ultimate design moment of 180kNm. The characteristic material strengths are /yk = 500N/mm2 and •. = 25 N/mnr. Calculate the area of reinforcement required.

neutral axis

Figure 4.14

Design example of a T-section, s > hi

Section

Stress Block

In figure 4.14

Fci is the force developed in the flange

Fcv, is (he force developed in the area of web in compression

Moment of resistance, Air. of the flange is

= 0.567 x 25 x 400 x 100(350 - 100/2) x 10"6 = 170 kN m < 180 kNm. the design moment

Therefore, the stress block must extend below the flange.

Ii is now necessary to determine the depth. jw, of the web in compression, where

For equilibrium: Applied momeni

Fcw x Zl = 170 + 0.567/ckfcwsw X Z2 = 170 + 0.567 x 25 x 200iw(250 - iw/2) x 10 6 = 170 + 2835,?w(250 - jw/2) x 10"fi This equation can be rearranged into sw2 - 500iw + 7.05 x lO3 = 0 Solving (his quadratic equation jw = 15 mm so that the depth of neutral axis .v- (/if + jw)/0.8 = (100 + 15)/0.8 = 144 mm = 0.41</

As x < 0.45(/ compression reinforcement is not required. For the equilibrium of the section

0.87 x 5(M) x As = 0.567 x 25(400 x 100 + 200 x 15) = 610 x 10*

Therefore

Analysis of a flanged section

Detcnnine the ultimate momeni of resistance of the T-beam section shown in figure 4.15 given fyk = 500 N/mm2 and yck = 25 N/mnr. The compressive force in the flange is

Fc, = 0.567fAbtht

Tlien tensile force in the reinforcing steel, assuming it has yielded, is F*i = 0.87/ykA,

neutral axis

Figure 4.15

Analysis example of a T-section, s > hf

Section

Stress Block

- .-reforc /•',, > Fc( so that s > hf and the force in the web is F„ = 0.567/ckMi - hf)

■ equilibrium

4.25(5- 150) = 1128 - 957 Hence s = 190 mm a = 5/0.8 = 238 mm = 0.43 d

■■■ ;h this depth of neutral axis the reinforcement has yielded, as assumed, and

If Fet > F,„ the the stress block would not extend beyond the tlange and the section "aid be analysed as in example 4.2 for a rectangular section of dimensions h, x d.) ' .iking moments about the ccntroid of the reinforcement

= [957(550 - 150/2) + 170(550 - 190/2 - 150/2)] x 10"3 = 519kNm

Design of a flanged section with depth of neutral axis x 0 45d

\ sale but conservative design for a Hanged section with s > hf can be achieved by siting the depth of neutral axis to x = 0.45d, the maximum depth allowed in the code. Design equations can be derived for this condition as follows.

Depth of stress block, s = 0.8a- = 0.8 x 0.45J = 0.36d

Figure 4.16

Flanged section with depth of neutral axis x = 0.45d

0.567fck

0.567fck

Figure 4.16

Flanged section with depth of neutral axis x = 0.45d

Section

Stress Block

Section

Stress Block

+1 0

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### Responses

• Columbus
What is stress strain block?
7 years ago
• roma
Why s is 0.8 of depth of neutral axis?
6 years ago