The pressure under the plate - 2790/[34(3I)] = 2.65 kips/in^2 (18.3 MPa). For a 1-in (25.4-
mm) width of plate, M= V2(2.65)/(l .59)^2 - 76.33 irrkips (8.6 kN-m); S = MIf= 76.33/27
= 2.827 in^3 (46.33 cm^3 ; t = (65)°^5 = 4.12 in (104.6 mm).
Plate thicknesses within this range vary by %-in (3.2-mm) increments, as stated in the
AISC Manual. However, a section of the AISC Specification requires that plates over 4 in
(102 mm) thick be planed at all bearing surfaces. Set t = 4 l/2 in (114.3 mm) to allow for
the planing.
- Design the beams at the lower tier
Try seven beams. Thus, P = 2790/7 = 398.6 kips (1772.9 kN); M = 398.6(62 - 31)/8 =
1545 in-kips (174.6 kN-m); S 3 = 1545/24 = 64.4 in^3 (1055.3 cm^3 ).
Try S15 x 50. Then S = 64.2 in^3 (1052.1 cm^3 ); tw = 0.550 in (14.0 mm); k = 1.25 in
(31.8 mm); b = 5.64 in (143.3 mm). The space between flanges is [60 - 7 x 5.641/6 = 3.42
in (86.9 mm). This result is satisfactory. Then/ 6 = 398.6/[0.550(31 + 2 x 1.25)] = 21.6 <
27 kips/in^2 (186.1 MPa), which is satisfactory; V= 398.6(62 - 31)/[2(62)] = 99.7 kips
(443.5 kN); v = 99.7/[15(0.55O)] - 12.1 < 14.5, which is satisfactory. - Summarize the design
Thus: A = 60 in (1524 mm); B = 62 in (1574.8 mm); base plate is 31 x 34 x 41/2 in (787.4
x 863.6 x 114.3 mm), upper-tier steel, four beams Sl8 x 70; lower-tier steel, seven beams
15150.0.
WIND-STRESS ANALYSIS
BY PORTAL METHOD
The bent in Fig. 15 resists the indicated wind loads. Applying the portal method of analy-
sis, calculate all shears, end moments, and axial forces.Calculation Procedure:- Compute the shear factor for each column
The portal method is an approximate and relatively simple method of wind-stress analysis
that is frequently applied to regular bents of moderate height. It considers the bent to be
composed of a group of individual portals and makes the following assumptions. (1) The
wind load is distributed among the aisles of the bent in direct proportion to their relative
widths. (2) The point of contraflexure in each member lies at its center.
Because of the first assumption, the shear in a given column is directly proportional to
the average width of the adjacent aisles. (An alternative form of the portal method as-
sumes that the wind load is distributed uniformly among the aisles, irrespective of their
relative widths.)
In this analysis, we consider the end moments of a member, i.e., the moments exerted
at the ends of the member by the joints. The sign conventions used are as follows. An end
moment is positive if it is clockwise. The shear is positive if the lateral forces exerted on
the member by the joints constitute a couple having a counterclockwise moment. An axi-
al force is positive if it is tensile.
Figure 160 and b represents a beam and column, respectively, having positive end mo-
ments and positive shear. By applying the second assumption, Ma = Mb = M, Eq. a; V =
2MIL, OTM= VLII, Eq. b;H = 2MIL, or M = HL/2, Eq. c. In Fig. 15, the calculated data
for each member are recorded in the order indicated.