- Compute the shear in each member by analyzing the member
as a free body
The shear is positive if the transverse forces exert a counterclockwise moment. Thus Hab
= (Mab + Mba)/2Q = -2.43 kips (-10.8 kN); Hcd = -3.08 kips (-13.7 kN); Hce = 4.49 kips
(19.9 kN); Vbc = 0.88 kip (3.9 kN). - Compute the axial force in AB and BC
Thus Pab = 0.88 kip (3.91 kN); Pbc = -7.57 kips (-33.7 kN). The axial forces in EC and
CD are found by equating the elongation of one to the contraction of the other. - Check the bent for equilibrium
The forces and moments acting on the structure are shown in Fig. 186. The three equa-
tions of equilibrium are satisfied.
WIND DRIFT OF A BUILDING
Figure 2Qa is the partial elevation of the steel framing of a skyscraper. The wind shear di-
rectly above line 11 is 40 kips (177.9 kN), and the wind force applied at lines 11 and 12 is
4 kips (17.8 kN) each. The members represented by solid lines have the moments of iner-
tia shown in Table 1, and the structure is to be analyzed for wind stress by the portal
method. Compute the wind drift for the bent bounded by lines 11 and 12; that is, find the
horizontal displacement of the joints on line 11 relative to those on line 12 as a result of
wind.
FIGURE 20