Handbook of Civil Engineering Calculations

3, the member behaves as a mechanism (i.e., a constrained chain of pin-connected rigid
bodies, or links).
In Fig. 23, indicate, in hyperbolic manner, the virtual displacement of the member
from its initial position ABC to a subsequent position AB'C. Use dots to represent plastic
hinges. (The initial position may be represented by a straight line for simplicity because
the analysis is concerned solely with the deformation that occurs during phase 3.)

2. Express the linear
   displacement under the load
   and the angular displacement
   at every plastic hinge
   Use a convenient unit to express these dis-
   placements. Thus, A = a0A = bSc; therefore,

ec = aBA/b = (^26) A; O 8 = BA + (^8) C = (^30) A

3. Evaluate the external and
   internal work associated with
   the virtual displacement
   FIGURE 23 The work performed by a constant force
   equals the product of the force and its dis-
   placement parallel to its action line. Also,
   the work performed by a constant moment
   equals the product of the moment and its angular displacement. Work is a positive quanti-
   ty when the displacement occurs in the direction of the force or moment. Thus, the exter-
   nal work WE = PMA = P^BA = 20PUBA. And the internal work W 1 = Mp(BB + Oc) = 5MpBA.


4. Equate the external and internal work to evaluate
   the ultimate load

Thus, 2QP 11 OA = 5MP (^0) A; Pu = (5/20)(268.8) = 67.20 kips (298.906 kN).
The solution method used here is also termed the virtual-work, or kinematic, method.

ANALYSIS OF A FIXED-END BEAM UNDER

CONCENTRATED LOAD

If the beam in the two previous calculation procedures is fixed at A as well as at C, what is

the ultimate load that may be applied at Bl

Calculation Procedure:

1. Determine when failure impends
   When hinges form at A 9 B, and C, failure impends. Repeat steps 3 and 4 of the previous
   calculation procedure, modifying the calculations to reflect the revised conditions. Thus

WE = 2QPU (^0) A; W 1 = MP(8a +O 8 + Oc) = 6MpOA; 20PU (^6) A = 6MP (^6) A; Pn = (6/20)(268.8) =
80.64 kips (358.687 kN).

2. Analyze the phases through which the member passes
   This member passes through three phases until the ultimate load is reached. Initially, it
   behaves as a beam fixed at both ends, then as a beam fixed at the left end only, and final-
   ly as a simply supported beam. However, as already discussed, these considerations are
   extraneous in plastic design.