Angular
Section displacement Moment W 1
A -6
B +9 Mp MpQ
C
D -6 Mp MpB
E +6
Total 2MpO
Then 2MpO = 532.80; Mp = 266.4 ft-kips (361.24 kN-m).
- Assume the composite mode of failure and compute Mp
Since this results from superposition of the two preceding modes, the angular displace-
ments and the external work may be obtained by adding the algebraic values previously
found. Thus, WE = 7400 + 532.80 1272.80. Then the tabulation is as shown:
Angular
Section displacement Moment W 1
A -B
B
C +26 Mp 2Mp 6
D -26 Mp 2MP 6
E +6
Total 4MP 6
Then 4M^0= 1272.80; Mp = 318.2 ft-kips (431.48 kN-m).
- Select the highest value of Mp as the correct result
Thus, Mp = 318.2 ft-kips (431.48 kN-m). The structure fails through the formation of plas-
tic hinges at C and Z). That a hinge should ap-
pear at D rather than at B is plausible when it
is considered that the bending moments in-
duced by the two loads are of like sign at D
but of opposite sign at B.
- Compute the reactions
at the supports
Draw a free-body diagram of the frame at ul-
timate load (Fig. 29). Compute the reactions
at the supports by applying the computed val-
ues of Mc and MD. Thus, ^ME = 2OF 4 +
22.2(24) - 74(10) = O; VA = 10.36 kips
(46.081 kN); VE = 74 - 10.36 = 63.64 kips
(283.071 kN); Mc = 1OF 4 + 24HA = 103.6 +
24HA = 318.2; HA = 8.94 kips (39.765 kN);
HE = 22.2 - 8.94 = 13.26 kips (58.980 kN);
M 0 = -24HE = -24(13.26) - -318.2 ft-kips
(-431.48 kN-m). Thus, the results are veri-
FIGURE 29 fied.
