4487

Since the above computed stress is close to the assumed value, it is alright. 3. Check the effectiveness of the web. Use the AISI Specification to check the effectiveness of the web element. From Figure 6.23, f1 = 50(4.1945/5.513) = 38.04 ksi (compression) f2 = 50(5.2205/5.513) = 47.35 ksi (tension) C = l/2/f1| = 1.245.

Using Equation 6.10, k = 4 + 2(1 + C)3 + 2(1 + C) = 4 + 2(2.245)3 + 2(2.245) = 31.12 h/t = 9.415/0.105 = 89.67 < 200 OK.

FCT ~ v/3Ll2V"9.67^\/297500 be = h = 9.415 in.

Since ho = 10.0 in., bo = 15.0 in., ho/bo = 0.67 < 4, use Equation 6.12 to compute b1:

,f 1.052 , , / 38.04 1 = J— = , (89.6^W-= 0.607 < 0.673

Since C > 0.236, b2 = be/2 = 4.7075 in. b1 + b2 = 6.9255 in.

Because the computed value of (b1 + b2) is greater than the compression portion of the web (4.1945 in.), the web element is fully effective.

FIGURE 6.23 Effective lengths and stress distribution using fully effective webs (courtesy of Yu, W.W. 1991).

4. Moment of inertia and section modulus. The moment of inertia based on line elements is

I'Z = 213 + ^(Iy2) = 890.4493 in.3 (X^1) (yCg)2 = 27.3662(4.487)2 = 550.9683 in.3

The actual moment of inertia is

The section modulus relative to the extreme tension fiber is

Nominai moment. The nominal moment for section strength is

Once the nominal moment is computed, the design moments for the ASD and LRFD methods can be determined as illustrated in Example 6.2.

According to procedure II of the AISI Specification, the nominal moment, Mn, is the maximum bending capacity of the beam by considering the inelastic reserve strength through partial plastification of the cross-section as shown in Figure 6.24. The inelastic stress distribution in the cross-section depends on the maximum strain in the compression flange, which is limited by the Specification for the given width to thickness ratio of the compression flange. On the basis of the maximum compression strain allowed in the Specification, the neutral axis can be located by Equation 6.45 and the nominal moment, Mn, can be determined by using Equation 6.46:

where s is the stress in the cross-section. For additional information, see Yu [1].

ecu = Cysy

Neutral axis bij 2 ^

0 0

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