7/27/2019 Class Examples Stability

1/4

2L

L

L

P

A

B

C

D

E

PE= !2EI/L2

!= P/PE

AB PE= 0.375!2EI/L2 !1= 2.667P/(!2EI/L2)

BC PE= !2EI/L2 !2= P/(!2EI/L2)

!2= 0.375!1

Using the slope-deflexion equations:

CD MCD= (-4"C+ 2"D)EI/L

MDC= (-2"C+ 4"D)EI/L

Since D is pinned MDC= 0 and so MCD= -3"CEI/L

CB MCB= (s2c2"B-s2"C)EI/L

MBC= (s2"B-s2c2"C)EI/L

BE MBE= 4"BEI/L, E is fixed and so "E= 0.

AB MBA= 0.75s1"BEI/L, A is fixed and so "A= 0.

MB= MBE+ MBA+ MBC and MC= MCD+ MCB we have in matrix form

MC

=

MB

-3 -s2 s2c2

-s2c2 4 + s2+ 0.75s1

"C

"B

Example No. 1

In the plane frame shown below all the members have the same flexural

rigidity, EI, except the lower column, AB, which has a value of 1.5EI. Show

that the elastic critical load is equal to 1.175!2EI/L2. Make use of the EXCEL

spreadsheet to calculate the stability functions.

"B

"A=0

#"C

"D

"E=0

7/27/2019 Class Examples Stability

2/4

(3 + s2) - (s2c2)2= 0(4 + s2+ 0.75s1)

For instability the determinant of the stiffness matrix is zero. Therefore

Using the stability functions we get by direct substitution !2 = 1.175 In

other words Pcr= 1.175!2EI/L2.

TABL OF STABILITY FUNC TIONS

!1 $1 s1 c1 !2 $2 s2 c2

3.1338 2.781 -6.444 -1.287 1.1752 1.703 2.139 1.211

3 + s2 5.139

4 + s2 + 0. 75s1 1.306

s2*c2 2.591

-0.0028

7/27/2019 Class Examples Stability

3/4

Example No.2

In the plane frame shown in Fig. * the members have the same flexural

rigidity, EI. Show that the elastic critical load is equal to 0.61!2EI/L2.

Make use of published tables of stability functions.

Using the slope-deflexion equations:

BC MBC= [4"B#2"C] EI/L

MCB= [2"B#4"C] EI/L

Since C is pinned MCB= 0 and so MBC= 3"BEI/L

AB MAB= [sc"B- s(1 + c) (%/L)] EI/L

MBA= [s"B- s(1 + c) (%/L)] EI/L

MB= MBC+ MBA

= [(s + 3)"B - s(1 + c) (%/L)] EI/L

The shear -FB is given by

MAB+ MBA

L+

P%

L

= [s(1 + c)"B- 2s(1 + c) (%/L)] EI/L2+ !!

2EI/L

2(%/L)

where P = !!2EI/L

2

Taking EI/L = 1, we may write in matrix form

L

P

A

BC

"A=0

%"B, %

L

#"C

7/27/2019 Class Examples Stability

4/4

MB

=

L.FB

s + 3 -s(1 + c)

-s(1 + c) 2s(1 + c) - !!2

"B

%B/L

For instability the determinant of the stiffness matrix is zero. Therefore

(s + 3) - s2(1 + c)

2= 0(2s(1 + c) - !!

2)

Using the stability functions we get

det(!= 0.60) = 0.765 and det (!= 0.64) = -2.002

and so, by interpolation, != 0.61.

In other words

Pcr= 0.61!2EI/L2.