Advanced Methods of Structural Analysis

Problems 445

For beam failureQlimD12M
yvert
l. This case is shown on the Fig.12.10e by line
B, which is parallel to horizontal axis because the beam mechanism is realized only
by loadQ.

For sidesway failurePlim D
8Myvert
l. This case is shown in the Fig.12.10eby
lineS, which is parallel to vertical axis because the sidesway mechanism is realized
only by loadP.
For combined failure relationships between limit loads is given formula (12.4).
Corresponding line is shown in Fig.12.10e by lineBCS.SinceQD2P,then

PlimD

16Myvert

3l

D

5:33Myvert

l

: (12.5)

Formulae (12.2), (12.3), and (12.5) present limit loadP, which corresponds to dif-
ferent mechanisms of failure. Failure is governed by minimum load

PlimD

5:33Myvert

l

;

which corresponds to combined mechanism of failure; therefore, this type of failure
will happens.
Note that increasing of all geometrical dimensions of the frame.l; h/byn-times
leads to decreasing of the limiting loads byn-times.

Problems.......................................................................

12.1.A straight rod with cross sectional areaAis located between two rigid sup-
portsMandNand subjected to axial loadP(Fig.P12.1). The yield stress of
material isy. Define the limit loadP. Solve this problem by direct and static
methods.

ab

MN

P

Fig. P12.1

Ans.PlimD2yA.

12.2.A straight uniform rod with cross sectional areaAis attached to a rigid sup-
port at endM, while there is a gap ofbetween the right end of rod and the rigid
support atN; the rod is subjected to axial loadP(Fig.P12.2). The yield stress of
material isy. Calculate the limit loadP. Compare result with problem 12.1 and
explain your answer.