Thin Plates

PAGE 137

Bending of Plates Under the Combined Action of Lateral Loads and Forces

in the Middle Plane of the PlatePositive directions of Stresses

In Plane Stress Resultants

Equilibrium conditions in the middle plane

Eq.(156) are independent of the bending of the plate

In the present consideration, we assume that all in plane stress resultants are known and their displacements are not affected by the bending of the plate and eq.(156) is independent of the equilibrium equations for bending and can be treated separately.

However in considering, say buckling problems, the influence of Nx , Ny and Nxy on ( Fz = 0 must be considered.

Thus,

The contribution of Nx to ( Fz is therefore

Neglecting the higher order term it becomes

Similarly the contribution of Ny to ( Fz is

The contribution of Nyx to ( Fz is

Similarly, The contribution of Nxy to ( Fz is

The total contribution to ( Fz of the in plane stress resultants are

The contribution to ( Fz of the transverse forces is

Finally, ( Fz = 0 yields

Since

The last equation becomes

That is

Physically, is Nx times rate of change of slope if slope equals a constant, there will be no contribution of Nx.

Discussion

Unknowns ( dependent variables )Equations8 stress resultants5Equilibrium equations

Mx , My , Mxy

( Fx = 0 ( Mx = 0

Qx , Qy

( Fy = 0 ( My = 0

Nx , Ny , Nxy

( Fz = 0

6strain resultants6Stress Strain relations

(x , (y , (xy

Mx , My , Mxy ~ (x , (y , (xy

(x , (y , (xy

Nx , Ny , Nxy ~ (x , (y , (xy3displacements6Straindisplcement relations

u , v , w

(x , (y , (xy ~ w

(x , (y , (xy ~ u , v

(17 unknowns(17 equations

Uni Directionally Compressed Simply Supported Rectangular Plates

In this case, Ny = Nxy = 0and q = 0 Nx = ( p ( compression )

In which Eq.(157) becomes

which is homogeneous. The question can be asked at what value of p does (158) yields non trivial solution (buckling).

The b.c. are satisfied if we take

Subst. (159) in (158) leads to

For w to be non zero , the expression in [ ] must vanish, from which we obtain

The smallest p occurs when n = 1 for which (160) becomes

or

where

For given a & b , the critical p is obtained by choosing m to yield pminBi Directionally Compressed Simply Supported Rectangular Plates

In this case, (157) becomes

Take

Subst. (162) in (161) leads to

Thus the buckling criteria is

or

Note that the RHS is non negative , hence buckling can occur only when either px or py is positive i.e., in compression

Example if a = b and px = py = p

Then (163) becomes

pmin occur when m = n = 1 for which

Simply Supported Rectangular Plates Under Combined Action of Uniform Transverse Load and Uniform Compression

In this case, (157) becomes

Take (considering double symmetry)

and expand qo in the double series

subst. (165) and (166) in (164) leads to

for which (165) becomes

if p were in tension, then p in (167) should be replaced by ( p and (167) becomes

Comparing eq.(168) with solution eq.(57) , it can be concluded from the presence of the term in the brackets of the denominator that the deflection of the plate is somewhat diminished by the action of the tensile forces Nx = p . This is as would be expected.

For buckling to occur, observe that w ( ( when the quantity in [ ] in the denominator of eq.(167) vanishes i.e.,

for which

which is identical to (160).

Bending of Anisotropic Plates

In the previous discussion we have assumed that the elastic properties of the material of the plate are the same in all directions. There are, however, cases in which an anisotropic material must be assumed if we wish to bring the theory of plates into agreement with experiments.

Let us assume that the material of the plate has three planes of symmetry with respect to its elastic properties ( such plates are called orthotropic ) and thus we have,

Orthotropic Plates

Stress Strain Law

Strain Displacement Relations

Note : Ex( , Ey( , E(( , and G are needed to characterize the elastic properties of a material in the case of plane stress.

Equilibrium Conditions

Subst. (170) in (169) leads to

The stress resultants are then given by

In which,

Subst. (173) in (171 a,b) gives

where

Governing Differential Equation

Subst. (173) in (171 c) yields the differential equation for the bending of orthotropic plates,

For isotropic plates,

for which

subst. the latter expressions in (177) leads to

which is identical to our previous plate equation

Some Special Cases of Orthotropic Plates

Reinforced Concrete Slabs

LetEsbe Youngs modulus of steel

Ecfor concrete

(cPoissons ratio of concrete

n= Es/Ec

in terms of the elastic constants given before we have approximately

For a slab with two way reinforcement in x and y directions

where Icx is the moment of inertia of the slab material, Isx that of the reinforcement taken about the neutral axis in the section x = constant , and Icy and Isy are the respective values for section y = constant.

In the special case where

Eq.(177) becomes

Introducing a new variable

Eq.(178) becomes

or

where

Eq.(180) can be treated in the same manner as an isotropic plate with flexural rigidity DxBending of S.S. Orthotropic Plates Under Uniformly Distributed Load

Expand qo in the double series

and take

subst. (182) and (183) in (177) leads to

for which (183) becomes

Note that (182) satisfies the b.c. and for isotropic case, Dx = H = Dy = D and (184) becomes (57)

S.S. Infinitely Long Orthtropic Plates Under Line Load

In this case, q(x,y) = 0 and (177) becomes

Take the solution in the form

which satisfies the b.c. along x = 0 and x = a

subst. (186) in (185) leads to

The roots of the corresponding characteristic equations are

Introducing the notation,

There are 3 cases to be considered:

Case 1,(>1orH2>DxDyCase 2,(=1orH2=DxDyCase 3,(