- Alternatively, design the member with the right-hand beam
overhanging the support
Compare the two designs for economy. The latter design is summarized in Fig. 21 g. The
total beam weight associated with each scheme is as shown in the following table.
Design 1 Design 2
55(50.2) = 2,761 Ib (12,280.9 N) 62(35.4) = 2,195 Ib (9,763.4 N)
36(25.8) = 929 Ib (4,132.2 N) 45(40.6) = 1,827 Ib (8,126.5 N)
Total 3,690 Ib (16,413.1 N) 4,022 Ib (17,889.9N)For completeness, the column sizes associated with the two schemes should also be
compared.
MECHANISM-METHOD ANALYSIS OF
A RECTANGULAR PORTAL FRAME
Calculate the plastic moment and the reactions at the supports at ultimate load of the
prismatic frame in Fig. 28a. Use a load factor of 1.85, and apply the mechanism
method.
Calculation Procedure:- Compute the ultimate loads to be resisted
There are three potential modes of failure to consider:
a. Failure of the beam BD through the formation of plastic hinges at B, C, and D (Fig.
286)
b. Failure by sidesway through the formation of plastic hinges at B and D (Fig. 28c)
c. A composite of the foregoing modes of failure, characterized by the formation of plas-
tic hinges at C and D
Since the true mode of failure is not readily discernible, it is necessary to analyze
each of the foregoing. The true mode of failure is the one that yields the highest value
OfM^.
Although the work quantities are positive, it is advantageous to supply each angular
displacement with an algebraic sign. A rotation is considered positive if the angle on the
interior side of the frame increases. The algebraic sum of the angular displacements must
equal zero.
Computing the ultimate loads to be resisted yields P 11 = 1.85(40) = 74 kips (329.2 kN);
Qu = 1.85(12) = 22.2 kips (98.75 kN). - Assume the mode of failure in Fig. 2Bb and compute Mp
Thus, A 1 - 10ft WE = 74(100) = 7400. Then indicate in a tabulation, such as that
shown here, where the plastic moment occurs. Include all significant sections for com-
pleteness.