Y

Assume that lateral bracing is adequately provided. The dead load moment MD = 30 in. kip and the live load moment ML = 150 in. kip.

Solution

ASD method

1. Location of neutral axis. For R = 16 in. and t = 0.135 in., the sectional properties of the corner element are as follows:

Ix = Iy = 0.0003889 in.4 A = 0.05407 in.2 x = y = 0.1564 in.

For the unstiffened compression flange w = 1.6775 in., w/t = 12.426 Using k = 0.43 and f= Fy = 50 ksi,

1 = yf /Fcr = (1.052^v/k)(w/t) Vf/E = 0.821 > 0.673 b = [(1 - 0.22/0.821 )/0.821]( 1.6775) = 1.496 in.

Assuming that the web is fully effective, the neutral axis is located at ycg = 4.063 in. Since ycg > d/2, initial yield occurs in the compression flange. Therefore, f= Fy.

2. Check the web for full effectiveness as follows (Figure 6.20):

f = 46.03 ksi (compression) f = 44.48 ksi (tension) C = I/2 /f I =0.966

Using Equation 6.10, k = 4 + 2(1 + C)3 + 2(1 + C) = 23.13 h = 7.355 in. h/f = 54.48

1 = yf/Fcr = (l.052^v/k) (54.48^46.03/29,500 = 0.471 < 0.673

0.3225 in.

FIGURE 6.20 Stress distribution in webs (courtesy of Yu, W.W. 1991).

0.3225 in.

FIGURE 6.20 Stress distribution in webs (courtesy of Yu, W.W. 1991).

Since ho = 8.00 in., bo = 2.00 in., ho/bo = 4. Using Equations 6.12 and 6.13a to compute b1 and b2 as follows:

b1 = be/(3 + C) = 1-855 in. b2 = be/2 = 3.6775 in. Since b1 + b2 = 5.5325 in. > 3.7405 in., the web is fully effective.

3. The moment of inertia Ix is

The section modulus for the top fiber is

4. Based on initiation of yielding, the nominal moment for section strength is

5. The allowable design moment is

Ma = Mn/O = 312.35/1.67 = 187.04 in. kip Based on the given data, the required allowable moment is

M = Md + Ml = 30 + 150 = 180 in. kip Since M < Ma, the I-section is adequate for the ASD method. LRFD method

1. Based on the nominal moment Mn computed above, the design moment is fbMn = 0.90(312.35) = 281.12 in. kip

2. According to the ASCE Standard 7 [38] the required moment for combined dead and live moments is

Since f bMn > Mu, the I-section is adequate for bending strength according to the LRFD approach.

0.2925 in.

0.2925 in.

0.2925 in.

0.2925 in.

1.0475 in.

FIGURE 6.21 Example 6.3 (courtesy of Yu, W.W. 1991).

0.2925 in.

Z (top filter)

1.0475 in. I, .1 I. 0.24 in. FIGURE 6.22 Line elements (courtesy of Yu, W.W. 1991).

b/2

0 0

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