considers that the bent behaves as a vertical cantilever. Consequently, the direct stress in a
column is directly proportional to the distance from the column to the centroid of the
combined column area. As in the portal method, the assumption is made that the point of
contraflexure in each member lies at its center. Refer to the previous calculation proce-
dure for the sign convention.
Computing the shear and moment on the bent at midheight, we have the following.
Upper row: H= 3900 Ib (17.3 kN); M= 3900(7.5) = 29,250 ft-lb (39,663.0 N-m). Center
row: H= 3900 + 7500 = 11,400 Ib (50.7 kN); M= 3900(22.5) + 7500(7.5) 144,000 ft-lb
(195.3 kN-m). Lower row: H= 11,400 + 8250 = 19,650 Ib (87.5 kN); M= 3900(39) +
7500(24) + 8250(9) = 406,400 ft-lb (551.1 kN-m), or M = 144,000 + 11,400(16.5) +
8250(9) = 406,400 ft-lb (551.1 kN-m), as before.
- Locate the centroidal axis of the combined column area,
and compute the moment of inertia of the area with respect
to this axis
Take the area of one column as a unit. Then x = (30 + 54 + 75)/4 = 39.75 ft (12.12 m); / =
39.75^2 + 9.75^2 + 14.25^2 + 35.25^2 = 3121 ft^2 (289.95 m^2 ). - Compute the axial force in each column
Use the equation/= MyIL The yll values are
ABCD
y 39.75 9.75 -14.25 -35.25
yll 0.01274 0.00312 -0.00457 -0.01129Then column A-2-3, P = 29,250(0.01274) = 373 kips (1659 kN); column 5-0-1, P =
406,400(0.00312) = 1268 kips (5640 kN).- Compute the shear in each beam by analyzing each joint
as a free body
Thus, beam A-3-B, V= 373 I b (165 9 N) ; beam 5-3-C, V= 37 3 + 91 = 464 I b (2. 1 kN) ;
beam C-3-D, V= 46 4 - 134 = 330 I b (146 8 N) ; bea m A-2-B, V= 1835 - 373 = 1462 I b
(6.5 kN); bea m B-I-C, V= 1462 + 449 - 91 = 1820 I b (8. 1 kN). - Compute the end moments of each beam
Apply Eq. b of the previous calculation procedure. Or for beam A-3-B, M= !4(373)(30) =
5595 ft-lb (7586.8 N-m). - Compute the end moments of each column
Do this by equating the algebraic sum of the end moments at each joint to zero. - Compute the shear in each column
Apply Eq. c of the previous calculation procedure. The sum of the shears in each horizon-
tal row of columns should equal the wind load above that plane. For instance, for the cen-
ter row, ^H = -(2178 + 4348 + 3522 + 1352) - -11,400 Ib (-50.7 kN), which is correct. - Compute the axial force in each beam by analyzing each joint
as a free body
Thus, beam A-3-B, P = -3900 + 746 - -3154 Ib (-14.0 kN); beam 5-3-C, P = -3154
- 1488 = -1666 Ib (-7.4 kN).