Handbook of Civil Engineering Calculations

*«/„ - *W, = No, - °-^9 °X2^X36kSi - 73.4 kip-ft (99.5 kNm)

which is also the tabulated value for (J)^Mp f°r a W8 x 28 in the Beam Selection Table in
Part 3 of the AISC LRFD Manual.
Since

*L_MMa_a37>0>2

QtPn 267 kips

the first of two interaction formulas applies.

^_<t> =8/J^_ + J^\

tpn^9 \<M4c <M4W

8 / 50 kip-ft \

037 + ± „ A* ^ +01 = 0.37 + 0.61 = 0.98 < 1.0 o.k.

9 \ 73.4 kip-ft I

2. Analyze the second orientation being considered
   For orientation (b) in Fig. 41

Pw = 1OO kips (444.8 kN), M^ = O, MM>; = 50 kip-ft (67.8 kNm)

Again, try a W8 x 28. For all Lb. the design flexural strength for^-axis bending

<Nony=<Nop=<hZyFy

0.9Ox 10.1in^3 x36ksi

= r-r = 27.2 kip-ft (36.9 kNm)

12 in/ft

Because Muy = 50 kip-ft > (f>bMny =27.2 kip-ft, a W8 x 28 is inadequate. Try a W8 x 48:
Ag = 14.1 in
2
(90.0 cm
2
), Zy = 22.9 in
3
(375.3 cm
3
)

*"„- °-

90X2

^;

X36kSi

= 61.8^(83.SkNn 1 )

<t>tPn = QfyAg = 0.90 x 36 ^j- x 14.1 in^2 = 457 kips (2032.7 kN)

Because (PJQfn) = (100 kips/457 kips) = 0.22 > 0.2, interaction formula (Hl-Id) again
applies.

_P^+8/JC_ + J^_\

Q 1 Pn 9\QbMnx <No„)

0.22 + - (O +^5 ° ^"^ } = 0.22 + 0.72 = 0.94 < 1.0 o.k.

9 \ 61.8 kip-ft/

3. Find the section for a load eccentric with respect to both
   principal axes
   For orientation (c) in Fig. 41, assume that the load is eccentric with respect to both princi-
   pal axes. Referring to Fig. 41 c