Handbook of Civil Engineering Calculations

ANALYSIS OFA TWO-SPAN BEAM

WITH CONCENTRATED LOADS

The continuous Wl8 x 45 beam in Fig. 24 carries two equal concentrated loads having
the locations indicated. Disregarding the weight of the beam, compute the ultimate value
of these loads, using both the static and the mechanism method.

Calculation Procedure:

1. Construct the force and
   banding-moment diagrams
   The continuous beam becomes unsta-
   ble when a plastic hinge forms at C and
   at another section. The bending-mo- FIGURE 24
   ment diagram has vertices at B and Z),
   but it is not readily apparent at which of these sections the second hinge will form. The
   answer is found by assuming a plastic hinge at B and at D, in turn, computing the corre-
   sponding value of Pw and selecting the lesser value as the correct result. Part a will use
   the static method; part 6, the mechanism method.
   Assume, for part a, a plastic hinge at B and C. In Fig. 25, construct the force diagram and
   bending-moment diagram for span A C. The moment diagram may be drawn in the manner
   shown in Fig. 25b or c, whichever is preferred. In Fig. 25c, A CH represents the moments
   that would exist in the absence of restraint at C, and A CJ represents, in absolute value, the
   moments induced by this restraint. Compute the load P 11 associated with the assumed hinge
   location. From previous calculation procedures, Mp = 268.8 ft-kips (364.49 kN-m); then MB
   = 14 x 16/V30 - 14M/30 = Mp; Pu = 44(268.8)7224 = 52.8 kips (234.85 kN).


2. Assume another hinge location and compute the ultimate load
   associated with this location
   Now assume a plastic hinge at C and D. In Fig. 25, construct the force diagram and bend-
   ing-moment diagram for CE. Computing the load P 11 associated with this assumed location,
   we find MD = 12 x 24PM/36 - 24M/36 = Mp\ Pn = 60(268.8)7288 = 56.0 kips (249.09 kN).

(a) Force diogrom (c) Moment diagram by parts (d) Force diagram

(b) Moment diagram

FIGURE 25

(e) Moment diagram