Page 146 - Handbook of Civil Engineering Calculations, Second Edition

STRUCTURAL STEEL DESIGN 1.129

5. Evaluate the internal work associated with each mechanism
Equate the external and internal work to find M p . Thus, M p1 800/4 200; M p2
300/2.25 133.3; M p3 300/4 75; M p4 1100/4.25 258.8; M p5 600/4.25
141.2. Equate the external and internal work to find M p .
6. Select the highest value as the correct result
Thus, M p 258.8 ft·kips (350.93 kN·m). The frame fails through the formation of plastic
hinges at C and D.
7. Determine the reactions at ultimate load
To verify the foregoing solution, ascertain that the bending moment does not exceed M p
in absolute value anywhere in the frame. Refer to Fig. 30a.
Thus, M D –20H E –258.8; therefore, H E 12.94 kips (57.557 kN); M C M D +
10V E 258.8; therefore, V E 51.76 kips (230.23 kN); then H A 7.06 kips (31.403 kN);
V A 28.24 kips (125.612 kN).
Check the moments. Thus M E 20V A + 5H A + 20(10) – 80(10) 0; this is correct.
Also, M F 15H A 105.9 ft·kips (143.60 kN·m) < M p . This is correct. Last, M B 25H A
– 20(10) –23.5 ft·kips (–31.87 kN·m) > –M p . This is correct.

ANALYSIS OF GABLE FRAME
BY STATIC METHOD

The prismatic frame in Fig. 31a carries the ultimate loads shown. Determine the plastic
moment by applying the static method.

Calculation Procedure:
1. Compute the vertical shear V A and the bending moment
at every significant section, assuming H A 0
Thus, V A 41 kips (182.4 kN). Then M B 0; M C 386; M D 432; M E 276;
M F –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
bending-moment diagram.
2. 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. 31b. Plot the values calculated in step 1 to obtain the bending-
moment diagram corresponding to the assumed condition of H A 0.
The bending moments caused solely by a specific value of H A are represented by an
isosceles triangle with its vertex at D
. 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 H A is directed to the right.
3. Evaluate the plastic moment
Apply the true moments at D and F. Thus, M D M p and M F – M p ; therefore, 432 –
37H A –(–100 – 25H A ); H A 5.35 kips (23.797 kN) and M p 234 ft·kips (317 kN·m).