A T-section is to be used for a beam to resist positive moment. The following data are given: beam span = 18 ft [5.49 m], beams are 9 ft [2.74 m] center to center, slab thickness is 4 in. [0.102 m], beam stem dimensions are bw = 15 in. [0.381 m], and d = 22 in. [0559 m], fc′ = 4 ksi [27.6 MPa],

Question

A T-section is to be used for a beam to resist positive moment. The following data are given: beam span = 18 ft [5.49 m], beams are 9 ft [2.74 m] center to center, slab thickness is 4 in. [0.102 m], beam stem dimensions are [latex]b_w[/latex] = 15 in. [0.381 m], and d = 22 in. [0559 m], [latex]f_c^′[/latex] = 4 ksi [27.6 MPa], [latex]f_y[/latex] = 60 ksi [414 MPa], [latex]f_s[/latex] = 24 ksi [165 MPa]. Find the required area of steel and select reinforcing bars for a moment of 200 kipft [272 kN-m].

Accepted Answer

Determine the effective flange width (necessary only for a check on the concrete stress). The maximum value for the flange width is

[latex]b_f= \frac{ ext{beam span}}{4}= \frac{18×12}{4}=[/latex] 54 in. [1.37 m]

or

[latex]b_f[/latex] = center-to-center beam spacing = 9(12) = 108 in. [2.74 m]

or

[latex]b_f[/latex] = beam stem width plus 16 times the slab thickness =

15 + 16(4) = 79 in. [201 m]

The limiting value is therefore 54 in. [1.37 m].
Next, find the required steel area as

[latex]A_s=\frac{M}{f_s[d-(t / 2)]}=\frac{200 imes 12}{24[22-(4 / 2)]}=5.00 \mathrm{in} .^2\left[3364 \mathrm{~mm}^2 ight][/latex]

Select bars using Table 15.5, which incorporates consideration for the adequacy of the stem width. From the table, choose five No. 9 bars, actual [latex]A_s[/latex] = 5.00 in.² . Consideration for the beam width and adequate spacing of the bars is discussed in Example 1 in Section 15.2.
Check the concrete stress:

[latex]C=\frac{M}{jd}=\frac{200×12}{20}= 120[/latex] kips [535 kN]

[latex]f_c= \frac{c}{b_ft}=\frac{120}{54×4}=[/latex] 0.556 ksi [3.83 MPa]

Compare this to the limiting stress of

0.45[latex]f_c^′[/latex] = 0.45(4) = 1.8 ksi [12.4 MPa]

Thus, compressive stress in the flange is clearly not critical. Using the beam stem width of 15 in. and the effective flange width of 54 in., the minimum area of reinforcement is determined as the greater of

[latex]A_s=\frac{6 \sqrt{f_c^{\prime}}}{f_y}\left(b_w d ight)=\frac{6 \sqrt{4000}}{60,000}(15)(22)=2.09 \mathrm{in} .^2\left[1350 \mathrm{~mm}^2 ight][/latex]

or

[latex]A_s=\frac{3 \sqrt{f_c^{\prime}}}{f_y}\left(b_f d ight)=\frac{3 \sqrt{4000}}{60,000}(54)(22)=2.56 \mathrm{in} .^2\left[1650 \mathrm{~mm}^2 ight][/latex]

As both of these are less than the computed area, minimum area is not critical in this case.

The example in this section illustrates a procedure that is reasonably adequate for beams that occur in ordinary beam and slab construction. When special T-sections occur with thin flanges (t less than d/8 or so), these methods may not be valid. In such cases, more accurate investigation should be performed, using the requirements of the ACI Code (Ref. 4).

Question 15.5: A T-section is to be used for a beam to resist positive mome...

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