- Select the lesser value of the ultimate load
The correct result is the lesser of these alternative values, or Pu = 52.8 kips (234.85 kN).
At this load, plastic hinges exist at B and C but not at D. - For the mechanism method, assume a plastic-hinge location
It will be assumed that plastic hinges are located at B and C (Fig. 26). Evaluate Pu. Thus,
Sc = 14<yi6; SB = 306^/16; A = 14^; WE =
PMA - 14P 11 ^; W 1 = Mp(0B + BC) = 2.75MpOA;
\4PU (^6) A = 2.15MP (^6) A- Pn = 52.8 kips (234.85
kN).
- Assume a plastic hinge
at another location
Select C and D for the new location. Repeat
the above procedure. The result will be identi-
FIGURE 26 cal with that in step 2.
SELECTION OF SIZES FOR
A CONTINUOUS BEAM
Using a load factor of 1.70, design the member to carry the working loads (with beam
weight included) shown in Fig. 21 a. The maximum length that can be transported is 60 ft
(18.3m).
Calculation Procedure:- Determine the ultimate loads to be supported
Since the member must be spliced, it will be economical to adopt the following design:
a. Use the particular beam size required for each portion, considering that the two por-
tions will fail simultaneously at ultimate load. Therefore, three plastic hinges will exist
at failure—one at the interior support and one in the interior of each span.
b. Extend one beam beyond the interior support, splicing the member at the point ofcon-
traflexure in the adjacent span. Since the maximum simple-span moment is greater for
AB than for BC, it is logical to assume that for economy the left beam rather than the
right one should overhang the support.
Multiply the working loads by the load factor to obtain the ultimate loads to be sup-
ported. Thus, w - 1.2 kips/lin ft (17.51 kN/m); wu = 1.70(1.2) - 2.04 kips/lin ft (29.77
kN/m); P=IO kips (44.5 kN); Pu = 1.70(10) = 17 kips (75.6 kN). - Construct the ultimate-load and corresponding
bending-moment diagram for each span
Set the maximum positive moment MD in span AB and the negative moment at B equal to
each other in absolute value. - Evaluate the maximum positive moment in the left span
Thus, RA = 45.9 - M 5 /40; jc = RA/2M; MD = l/2RAx = RA/4No = MB. Substitute the value of
RA and solve. Thus, MD = 342 ft-kips (463.8 kN-m).
An indirect but less cumbersome method consists of assigning a series of trial values
to MB and calculating the corresponding value ofMD, continuing the process until the re-
quired equality is obtained. - Select a section to resist the plastic moment
Thus,Z = Mp/fy = 342(12)136 = 114 in^3 (1868.5 cm^3 ). Referring to the AISC Manual, use
a W21 x 55 with Z= 125.4 in^3 (2055.31 cm^3 ).