Page 150 - Handbook of Civil Engineering Calculations, Second Edition
P. 150

STRUCTURAL STEEL DESIGN               1.133


                                Since A is the center of rotation of ABD, DD
 is normal to AD and HD; since G is the
                              center of rotation of GF, FF
 is normal to GF and HF. Therefore, H is the instantaneous
                              center of rotation of DF.
                              2. Record the pertinent dimensions and rotations
                              Record the dimensions a, b, and c in Fig. 33, and express 	 2 and 	 3 in terms of 	 1 . Thus,
                              	 2 /	 1   HD/AD;   	 2   	 1 . Also, 	 3 /	 1   HF/GF   49/25;   	 3   1.96	 1 .
                              3. Determine the angular displacement, and evaluate the
                              internal work
                              Determine the angular displacement (in absolute value) at D and F, and evaluate the inter-
                              nal work in terms of 	 1 . Thus, 	 D   	 1 + 	 2   2	 1 ; 	 F   	 1 + 	 3   2.96	 1 . Then W I   M p
                              (	 D + 	 F )   4.96M p 	 1 .
                              4. Apply the theorem of virtual displacements to determine the
                              displacement of each applied load
                              Determine the displacement of each applied load in the direction of the load. Multiply the
                              displacement by the load to obtain the external work. Record the results as shown:

                                                      Displacement in direction
                                        Load                of load               External work
                              Section  kips  kN         ft            m        ft·kips   kN·m
                              B       4    17.8      h   25	 2   25	 1  7.6	 1  100	 1   135.6	 1
                              C      34    151.2     v   10	 2   10	 1  3.0	 1  340	 1   461.0	 1
                              D      25    111.2          v   20	 1   6.1	 1    500	 1   678.0	 1
                              E      22    97.9           v   10	 1   3.0	 1    220	 1   298.3	 1
                                                                               ________  ___________
                              Total                                            1160	 1  1572.9	 1


                              5. Equate the external and internal work to find M p
                              Thus, 4.96M p 	 1   1160	 1 ; M p   234 ft·kips (317.3 kN·m).
                                Other modes of failure may be assumed and the corresponding value of M p computed
                              in the same manner. The failure mechanism analyzed in this procedure (plastic hinges at
                              D and F) yields the highest value of M p and is therefore the true mechanism.



                              REDUCTION IN PLASTIC-MOMENT
                              CAPACITY CAUSED BY AXIAL FORCE

                              A W10   45 beam-column is subjected to an axial force of 84 kips (373.6 kN) at ultimate
                              load. (a) Applying the exact method, calculate the plastic moment this section can devel-
                              op with respect to the major axis. (b) Construct the interaction diagram for this section,
                              and then calculate the plastic moment by assuming a linear interaction relationship that
                              approximates the true relationship.

                              Calculation Procedure:
                              1. Record the relevant properties of the member
                              Let P   applied axial force, kips (kN); P y   axial force that would induce plastification if
                              acting alone, kips (kN)   Af y ; M p 
  plastic-moment capacity of the section in combina-
                              tion with P, ft·kips (kN·m).
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