- Evaluate the internal work associated with each mechanism
Equate the external and internal work to find Mp. Thus, Mpl = 800/4 = 200; Mp2 =
300/2.25 - 133.3; Mp3 = 300/4 = 75; Mp4 = 1100/4.25 = 258.8; Mp5 = 600/4.25 = 141.2.
Equate the external and internal work to find Mp. - Select the highest value as the correct result
Thus, Mp = 258.8 ft-kips (350.93 kN-m). The frame fails through the formation of plastic
hinges at C and D. - Determine the reactions at ultimate load
To verify the foregoing solution, ascertain that the bending moment does not exceed Mp
in absolute value anywhere in the frame. Refer to Fig. 3Oa.
Thus, MD = -2QHE = -258.8; therefore, HE = 12.94 kips (57.557 KN); Mc = MD + \WE
= 258.8; therefore, VE = 51.76 kips (230.23 KN); then HA = 7.06 kips (31.403 KN); VA =
28.24 kips (125.612 KN).
Check the moments. Thus ^ME = 2OF 4 + 5HA + 20(10) - 80(10) = O; this is correct.
Also, MF = 15HA = 105.9 ft-kips (143.60 kN-m) < Mp. This is correct. Last, MB = 25HA -
20(10) - -23.5 ft-kips (-31.87 kN-m)> -Mp. This is correct.
ANALYSIS OF GABLE FRAME
BY STATIC METHOD
The prismatic frame in Fig. 3 Ia carries the ultimate loads shown. Determine the plastic
moment by applying the static method.
Calculation Procedure:- Compute the vertical shear VA and the bending moment
at every significant section, assuming HA = O
Thus, VA = 41 kips (182.4 KN). Then MB = O; Mc = 386; MD = 432; ME = 276;
MF = -100.
Note that failure of the frame will result from the formation of two plastic hinges. It is
helpful, therefore, to construct a "projected" bending-moment diagram as an aid in locat-
ing these hinges. The computed bending moments are used in plotting the projected bend-
ing-moment diagram. - Construct a projected bending-moment diagram
To construct this diagram, consider the rafter BD to be projected onto the plane of column
AB and the rafter FD to be projected onto the plane of column GF. Juxtapose the two
halves, as shown in Fig. 3 Ib. Plot the values calculated in step 1 to obtain the bending-
moment diagram corresponding to the assumed condition of HA = O.
The bending moments caused solely by a specific value of HA are represented by an
isosceles triangle with its vertex at Z)'. The true bending moments are obtained by super-
position. It is evident by inspection of the diagram that plastic hinges form at D and F and
that HA is directed to the right. - Evaluate the plastic moment
Apply the true moments at D and F. Thus, MD = Mp and MF = — Mp\ therefore, 432 — 37HA
= -{-100 - 25HA); HA = 5.35 kips (23.797 kN) and Mp = 234 ft-kips (317 kN-m).