Rigid Body Stability

Structural Stability Fall 2006
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Prof. Kang,YoungJong

RIGID BODY STABILITY

1.1 Basic Concept of Stability

1.2 The Stability of a rigid block resting on a table

1.3 Stability of a block restrained

by a spring & backlash

1.4 Stability of a loaded ball in a cup

1.5 Compressed Rigid bar Elastically restrained

1.6 Rod & Spring system with 2 - DOF

＊CHAP. 1 Intro (Rigid body Stab.)

§1. Basic Concept of Stability

- Structural members fail due to

a) yielding (strength requirement)

b) buckling (stiffness requirement)

In thin-walled metal structures

- Buckling usually depends on

1) Dimensions of members

2) Support Conditions

3) Material Properties

- Stability of Equilibrium

a) Stable Equilibrium

b) Neutral Equilibrium

c) Unstable Equilibrium

-Neutral Equilibrium is the dividing line for stable  unstable equil. conditions.

So necessary to find Neutral Equilibrium conditions

§2. The Stability of a rigid block resting on a table

(Fig.1)

Consider Static Equilibrium ( Fig.1.a)

(1.1.a)

(1.1.b)

(1.1.c)

Solving equations 1 for gives

(1.2)

From Fig.1.(a), As we increase , the location of reaction of will move forward the point A. When , further increase of will make the block rotate about point A.

The limiting value of in Eq.(1.2) is “a” (without tilting),

substituting into Eq.(1.2) gives

(1.3)

Eq.(1.3) can also be obtained by taking moment about A.

For large rotation (Fig. 1.c) taking moment about A yields

Hence

(1.4)

Plotting Eq.(1.3)  (1.4) (vs. )

(Fig.2)

For

(H.W #.1) A solid Rigid block of has a slight imperfection at the bottom. Compute and Plot vs. for

case 1; is very small

case 2; is large enough to call for large displ. theory

hint ;

Use Approx. Geometry

§3. Stability of a block restrained by a spring  backlash

(Fig.3)

 Backlash has a slack of , so that it does not exert a force until the block has rotated through an angle .

The load – deformation relationship and and are shown with arrows indicating positive direction.

Taking Moment about A,

(1.5)

If the rotation of the block is not large, each force (generalized sense) is a function of its corresponding displ. as shown in Fig.3

(1.6)

The equations of geometrical compatibility

(1.7)

Using Eq. (1.6)  (1.7) in conjunction of Eq. (1.5), We obtain,

(1.8)

(1.9)

From Eq.(1.8) , when

or if takes nonzero value, provided

then , Hence

From Eq.(1.9) in the region , i.e.,

(Fig. 4)

The plot is shown in Fig.4

If , i.e., no rotational spring, AB is flat

To handle this problem by Energy Method, we define the total potential energy functional,

; strain energy functional

; loss of Total Potential Energy due to applied loads

Then

Let where

strain energy stored in spring

Necessary condition for equilibrium

; Stable Equilibrium

; Neutral Equilibrium

; Unstable Equilibrium

Now, of the spring block problem is

(1.10)

(1.10.a)

(1.10.b)

Noting , this is exactly the same

as Eq.(1.9)

implying that the block is in stable equil.

regardless of the value of

§4. The Stability of a loaded ball

in a Cup

(Fig. 5)

Consider a system shown in Fig. 5 consisting of a spherical ball of radius and weight , resting in a cup which is a portion of a sphere of radius . The ball is acted by a vertical force P applied by a plunger guided vertically. The displ. of the ball is measured by as shown in Fig. 5(b)

Then it is evident that the ball can remain in equil. in position for whatever the value . This is indicated in the equil. diagram 5(c) by line OB.

In Fig.5(b), the system is disturbed (disturbing

force D) and is displaced by . The stability analysis of this system is to determine.

Whether it tends to remain to its original position(S.E.), to be displaced further(U.E.) or if it can remain there in equil.(N.E.).

For equil. in displaced position of the system.

We have, provided is small,

※Here is considered as friction force which prevents motion to go to original position

for small value of , (eliminate )

Therefore

(1.13)

Canceling out and rearranging for , yields

(1.14)

Thus, assuming equil. in displaced position. We have arrived at the condition that P must have a FINITE value, which we shall call , At this value of the ball is in equil. in the position and also in neighboring position ( but very small).

In examining the equil (1.13), we have either

ii) in which may

take any small value.

iii) both i) ii) are true

In case that disturbing force D acts on the ball for small displ.

(1.15)

Using the same approximation for small

and

This gives

(1.16)

Eq. (1.16) indicates that when

is undefined. And for small value of , can be quite large  upset the system equil.

If we fix a deformation by setting , then

(1.17)

vs. is shown in Fig. 5 (d)

For (Positive)

(Negative)

indicating that the system must be restrained against displ. (added Energy)

These 3 stages of loading are Stable, Neutral and Unstable equil. Conditions as indicated by Fig. 5 (c)

※If there is disturbing force, bifurcation buckling

will not occur

H.W #. 2) Use Energy Method for this problem of this section

§5. Compressed Rigid Bar

Elastically Restrained

(Fig. 6)

Equil. of displ. system

S.E.

N.E.

U.E.

 Dynamic Disturbance

If the mass moment of inertia of the bar about the hinge is “”, then the equation of motion by d’Alembert’s principle is

(∵inertia force acts in opposite direction)

let

if we consider phase difference, we may write

(1.18)

A  B is can be determined by proper I.C.

, when and

§6. Rod  Spring System with 2-DOF

(Fig. 7)

for bar BC

for the system

(1.19)

(1.20)

For Non-trivial Solution

(Characteristic Polynomial)

say

(Critical Value or Eigenvalues)

for

same

(1st mode)

(2nd mode)

H.W #.3 Redo the problems in §6 using the energy concept and extract the same results as in the class note.