Advanced Methods of Structural Analysis

506 13 Stability of Elastic Systems

k

j 0

P

l

A

Fig. P13.4

Ans. tanlD

l

^2 l^2 ̨C 1

;D

r

P

EI

, ̨D

EI

kl

13.5.Design diagram of the column is presented in Fig. P13.5. The total length
of the column isl, while the length of the bottom part is ̨l. The column is sub-
jected to forcesN 1 andN 2. Relationship between these forces remains constant,
i.e.,N 2 DˇN 1. Parameters ̨andˇare the given fixed numbers. Stiffness rigid-
ity of the portions areEI 1 andEI 2. Derive the stability equation. Consider three
possible buckling forms. Apply double integration method.

N 1 =N

N 2 =bN

al

l

EI 1

EI 2

Fig. P13.5

Ans. (1) cos.1 ̨/ n 1 lD 0 I(2) cosn 2 ̨lD 0 I(3) For smallest critical force

tan.1 ̨/ n 1 ltann 2 ̨l

n 1

n 2

.1Cˇ/D0; n 1 D

r

N 1

EI 1

;n 2 D

r

.1Cˇ/ N

EI 2

13.6.Derive the stability equation for uniform columns subjected to axial compres-
sive force. Consider the following supports: (a) clamped-free; (b) clamped-pinned;
(c) pinned-pinned.Apply the initial parameter method. Compare the results with
those are presented in Table13.1.

13.7.The clamped beam with elastic support is subjected to axial forceN
(Fig. P13.7). Derive the stability equation. (Hint: Reaction at the right support
isky 1 ,wherey 1 is vertical displacement at the right support. The moment at the
left support should be calculated taking into account the lateral displacementy 1 of
axial forceN, i.e.,M 0 Dky 1 lNy 1 ).