Fmkjbd 2

Mrm '

(ft&lb)

(eq 5-15)

2(12)

flexural design equations for T-sections are illustrated in figure 5-4.

(1) *Design coefficients. *In the design of reinforced T-sections, as in the design of rectangular sections,

the first step is to locate the neutral axis. As with rectangular sections, this can be accomplished by

determining the coefficient kT. kT is derived by assuming that the compressive force in the flange, C, is equal

to the tension force in the reinforcement. The contribution of the portion of the web in compression is small

and can be neglected, therefore if;

T.C

Then,

2kTd & ts

pbdfs ' fm

bts

(eq 5-16)

2kTd

Where:

ts = The thickness of the face shell of the unit, inches.

From the strain compatibility relationship it can be determined that;

n

kT '

n % (fs/fm)

Rearranging the equation and solving for fm yields:

kT

fm ' (fs)

n(1 & kT)

Substituting this equation for fm into equation 5-16 yields;

2

np % 1/2(ts/d)

kT '

(eq 5-17)

np % (ts/d)

The coefficient, jT, can be determined by the relationship,

jTd **= **d - z

Where:

z **= **The distance from the extreme compressive fiber to the center of compression (or the center

of gravity of the trapezoid shown in figure 5-4) and is determined as follows:

3kTd & 2ts

ts

z'

2kTd & ts

3

From the above equations jT can be determined by:

6 & 6(ts/d) % 2(ts/d)2 % (ts/d)3[1/(2pn)]

jT '

(eq 5-18)

6 & 3(ts/d)

The resisting moments of the steel and masonry are equal to the product of the moment arm, jTd, and the

tension or compression force, respectively, therefore:

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