Appunti corso Advanced structural design

Thin plates: introduction and load types

We design our objects by preventing some critical situations, which we name limit states. To introduce simplified mechanical models (in this case for plates), we need some definitions and conditions to limit the application of our theory. The design of the beam can be done by knowing the ratio L/h.

Type of loads

Kinematic model: Plate: structural member for which 2 dimensions are much greater than the third (the thickness). We have two contributions to deformation: bending and warping. To take into account the additional rotation due to shear, we add the Reissner-Mindlin theory. We can see that the theory of plates is developed from the Saint-Venant beam theory but is more complex. So, we have to study two new effects:

• Plates are internally statically indeterminate (not valid for beams);
• In plates, lateral contraction is not permitted (due to the presence of the Poisson coefficient).

With these, we can define our theory for plates: we start with the kinematic model and introduce the hypothesis.

Kinematic model and strains

We start from the theory of Reissner-Mindlin, which takes into account the deformability due to shear (but not the total one) because, looking at the picture on top, the total deformability of shear in warping of the C-S of our element (and in terms of our theory, is impossible to take into account the warping). By considering the angle of rotation γ, we can neglect warping. The displacements xzu, v, w depend linearly on the z-axis, perpendicular to the mid-surface of the C-S, and from the generalized rotations βx and βy. They are called "generalized" because some are displacements, others are rotations.

We then define strains: strains are unidimensional! To define these, we assume two fundamental hypotheses:

• No strains in the z direction;
• When we consider a bending rotation in the plane, this also means a torsion rotation in the perpendicular plane. γxz and γyz come from two different contributions. A rotation and a derivative of the deflection. This highlights that we can define the two rotations βx, βy as the sum of two contributions, one coming from shear strain and one from variation of the deflection. We have a minus sign because when we look at the variation of w w.r.t x, the minus sign is associated with the sign convention.

General approach to thick and thin plates

This is the most general approach applicable to thick and thin plates. As the thickness reduces, the contribution of the shear component reduces, in our case so much that it is negligible. Since this neglection gives us a simpler theory, this is developed as a special formulation, according to the Kirchhoff-Love hypothesis (this can be applied in cases where shear strain is small, so only for thin plates):

In this case, since u, v are derived directly from w, the only unknown is w, so the problem is much simpler. Looking again at the generalized strains for this case (second derivative of the displacements is a curvature if we take into account the hypothesis of small displacements). Also, strains in x and y directions are linearly dependent on the curvature. In the same way, we can define the bending curvatures (by using the properties of the mixed derivatives, we get the torsional curvatures): These are the compatibility equations.

Stress resultants and equilibrium

Now we develop equilibrium:

We assume stresses = 0 (in reality no, but since they are associated with the intensity of the applied load in the vertical direction which is much smaller than the stresses that arise in the plane, they can be considered negligible). We can neglect these because they are of the same order as the surface applying loading. We have only five independent stress components of the stress tensor. We can define the internal generalized stress in terms of bending moments, torsional moments by integration of the stress distribution along the x-axis, and by vertical equilibrium, we get shear in x and y directions.

Having γxz and yz =0 does not mean that the corresponding stresses are =0!

Stress distribution and shear components

If we want to see the distribution over the thickness of the stress components, we consider linear elastic material, so the strains linearly vary over the thickness, and since we have homogeneous material, we have a linear variation of the stresses. The classical distribution that we use in Saint Venant theory of the beam is reported in the figure. We know that sigma can be defined as a constant x z. We need to find this constant and relate it to the moment point x by point: The value of "a" that we get is exactly the inertia moment of a rectangular section (for a beam). We can do the same for sigma, y, and sigma, z.

Shear components and equilibrium

What about shear components?

When we approach the shear problem, we don’t deal with compatibility conditions, but we use only equilibrium because compatibility can’t be used for shear. In this picture, we see the typical Jourawsky approach for a beam. (Jourawsky predicts the parabolic distribution of the shear tension). Shear stresses along thickness vary according to a parabolic profile because when regarding the distribution of the shear stresses, some BCs (shear stresses at top and bottom must be =0. Also, we will not have a non-symmetric distribution because if it is symmetric, the derivative of the stress over the thickness over the mid-span is =0 and so we will have three conditions).

Average shear stress = shear force / thickness. This procedure can be also obtained by direct equilibrium.

Equilibrium equations

At this point, we write equilibrium equations in terms of stress resultant and not in terms of point stresses (this is why we do all the preliminary passages):

We have to use two hypotheses for the equilibrium equations:

• We don’t have volume forces in x and y direction;
• Sigma is equal to zero, so the derivative is equal to zero;

In the initial passage, what we have done is replace in the derivatives the functions in terms of internal stress resultants. By doing this, we obtain equations in terms of stress resultants (bending moment and shear) on the right. For the three equations ((*) Is the Jourawsky contribution), we have the z coordinate involved, and we want to change this, and what we can do is:

We take this and we integrate over the thickness. Doing this, we divide the integral into two parts. The second term is a volume force component, force x unit area (=p= out of plane loading). Concerning the other integral, Qx and Qy can be taken outside of the integral since they don’t depend on the thickness.

Observations on compatibility and equilibrium

Observations:

• Compatibility equations are written in terms of generalized strains, bending curvature, and torsional curvature;
• Equilibrium is written in terms of stress resultants (bending moment, torsional moment, and shear forces).

Equilibrium of a plate element

We can consider the equilibrium of a plate element (dx, dy are infinitesimal dimensions, and h is finite). Some terms generate terms of higher order which can be neglected by considering rotation around xz, yy, and the free cardinal equilibrium equations. We get the shear as a function of the bending moment. In this picture, we have defined the torque moment with a double arrow. Now we need to put them together using constitutive laws.

Constitutive laws

Constitutive laws must take into account the nature of the material. We start from:

Isotropic case:

In writing constitutive laws, we use G = shear modulus. It’s more convenient because using G is directly related to tau, gamma, so G is not independent. The next step is replacing the relationship that we found between sigma, x, sigma, y, sigma, xy, and the stress resultants Mx, My, Mxy. Doing this, we can write:

D = bending stiffness of the plate. NB! The stiffness of a plate is generally larger than the stiffness of a beam, and this is the reason why we define E’. This causes the stiffness of a strip of unit length to be E / inertia moment of the C-S. However, for a series of strips, they are cooperating, so we have a higher stiffness. This is put into evidence with the presence in E’ of the Poisson effect.

(We can apply our theory also to orthotropic plates, which means plates obtained by using a net of beams connected together with different inertia in direction x and direction y.)

Now we have compatibility, equilibrium, and for an isotropic material, the constitutive laws. Putting them together, we obtain the Sophie Germain - Lagrange equation:

To solve this equation, we must use a homogeneous + particular solution of the total equation which we are able to find.

Note on the shear stress

This slide is for how we can obtain the parabolic profile of shear through equilibrium condition, with an application of Jourawsky theory. For the beam, to use Jourawsky contribution, we have to put in the equilibrium condition the derivative of the bending moment w.r.t x = 0.

Cylindrical bending

The central part of the plate is not affected by any constraints, this means that on the bottom we have free edges. Cylindrical mean with only one curvature. In a plant in which b > h, we can solve this in a closed form because the solution is polynomial in the x variable at power 4.

Plates in cylindrical bending vs beams

Boundary conditions

When we obtain the solution of our problem, we need to enforce BC’s. In our case, BC’s are applied on the mid-surface only. Also, we can have kinematic conditions (generalized) and static conditions, or again a combination of those two. For our case, two BC’s must be satisfied.

Kinematic BC's

Effective shear force

In a rectangular simply supported plate, on corners we have concentrated forces (so corners lift up). By calculating the concentrated reactions with equilibrium, we can justify this behavior. This is named unbalanced shear force.

Looking at this C-S in the plane x-z, we consider along the side of the plate a series of infinitesimal segments adjacent and we look at torsional moments of each one of these segments. Looking at the torsional moment of the first segment, we have the torsional moment M dx.

Considering the adjacent segment, we have an increment since we are moving along the x-axis. So we have two moments. Let’s represent them as a couple of forces. We have two forces, one upward and one downward. The value of those forces is the moment divided by the length:

Focusing now at the interface between the two segments:

We have a superposition of the two forces. So summing up these components, we have an unbalanced component, which is the first derivative of the moment. This means that looking at the segment centered at the interface, at the center we have a downward force.

So looking at this point, and if we want to know the stress resultant in terms of shear force, we have:

(This behavior is very important in design).

Static BC's

We need to enforce the effective shear force (ESF):

(Kinematic BC’s are directly expressed in terms of w function and we can relate w to ESF). We have a particular situation in which, when Mxy = 0, static BC’s apply directly to the pure shear. This happens in the case of cylindrical bending:

Example of application

Homogeneous BC's

We start by considering:

1. Fixed edge: mean no displacement and no rotation;
2. Simply supported edge (x=a): mean no vertical displacement and no bending moment (by introducing the relationships that we know, we obtain the result in terms of rotation);
3. Free edge: we have shear and bending moment = 0.

Since the quantities are all equal to zero, those are homogeneous BC’s. Then we have, when we have a stiffening beam in which we can’t prescribe for this function a prescribed value, but we must enforce compatibility between the plate and the stiffening beam. This leads to non-homogeneous BC’s.

Let’s consider the beam characterized by a flexural stiffness (EI)b and a torsional stiffness (GJ)b:

For the edge beam, we have two conditions:

• The deflection must be equal for the plate and for the beam (deflection of the beam = wb);
• Then we have the condition βx. This must be a torsional rotation.

The consequence of this is that we can express a condition also to βy:

This concerning the static BC’s. Concerning static BC’s, we must have:

Corner reactions

We have some specific forces that may arise in corners due to our physical model. Those forces typically have the same orientation as the load. To understand this, we look at the external edges of the figure and we divide the figure into infinitesimal segments. We replace each moment generated with two forces (one upward and one downward). These take into account positive moments. Taking the positive moments, the two forces associated with the variation of the torsional moment are characterized by the same orientation.

This is important in design. Regarding the direction of the forces, reactions are considered positive if upward. However, if we change the corner, the direction of the forces will change. So for this, we have to look also at the location of the corners. So we can define:

It depends also on the sign of the torsional moment: if the torsional moment is positive, the reaction is upward and vice versa.

Shear forces on the boundary

Suppose to have as in figure a simply supported plate with a force. Moments normal to the side are zero; what we have are the effective shear force. The distribution of those forces will have a maximum value along the mid-span of the beam and a zero value at the corner. We will show that the torsional moment will be negative on the upper left corner and the same on the opposite one. And negative on the other.

Observations

• Corner reactions are self-equilibrated with a portion of the effective shear force (ESF): so if we have a vertical load P supported by ESF, and we have reactions downward, ESF must balance also the corner reactions and not only the forces.
• If the reaction is downward, in order to have this reaction arising, the support must be bilateral: this means that due to force, corners tend to uplift. To avoid this uplift, we must ensure to have a strong connection at corners, using bilateral support. Also, to avoid uplift, we need to use reinforcements. So we must have that: R < A fS y

Structural analysis of rectangular plates

Introduction

We have a possibility to solve the second order differential equation in red. To understand this statement, it is important to introduce integration properties. This approach can be adopted for the solution of closed form solutions, for example, elliptical plate clamped around the perimeter or simply supported cylinder. The development of our theory can be done using derivatives.

Sinusoidal deflection (compatibility)

If we introduce the previous development in the Lagrange equation, we obtain (using also a double series development for the applied load). In the row before the last one, the term in the bracket is a binomial term:

With these results now, we consider the first problem:

This satisfies the BC’s:

Sinusoidal loading (equilibrium)

Now we consider a particular distribution of load, a load with sinusoidal distributions. This means that we can put p11 with the same shape (distributed load has a similar shape of the deflection).

Solution by Fourier series expansions

Series coefficients

At the end, we get:

Convergence and available solutions

Comment on 3rd statement:

On the upper right, we have two equations: one (on the left) is the Lagrange equation and the other is the corresponding homogeneous equation. We consider the homogeneous associated in the form:

We discover that Yn is a solution if it satisfies the homogeneous equation of the 4th order:

The solution is given by:

As an example, considering a rectangular plate with uniform distributed load, we have the cylindrical solution:

We haven’t considered Mxy, Myx, because in the center of the plate, those are zero.

Poisson's ratio influence

By using this solution, considering the ratio between My, max, and Mx, max, we can see the influence of the Poisson’s ratio on our problem. In RC plates, due to cracking, the influence of Poisson's ratio is reduced. For a typical 2-way plates, the important range that we consider is between 0.5 and 1, or is the same between 1 and 2. We use 0.2 to calculate the transversal reinforcements. Usually, the amount of reinforcements is given by: As = Mx, max / 0.85 d fyd. NB! Poisson coefficient is more efficient on actions (bending moments, …) than displacements.

Sinusoidal loading: torsion and shear

Qx and Qy are forces per unit length, and when we derive Qx for ex. w.r.t. x, we obtain a force divided by a surface.

Stress resultants for sinusoidal loading

The last observation means that our theory with the simplifying hypothesis of thin plates has the consequence that forces are concentrated on corners, but with theories such as M-R, they are not exactly concentrated forces, but they are distributed forces, which resultant of the distributed force is a concentrated force.

We see that the distribution of the torque moment is antisymmetric. (Also, the unique value which respects the antisymmetry is zero).

Shear force on a generic face

We have a small element by considering the hypothesis that dx, dy are very small, but the hypotenuse must be = 1 (usually not necessary). By equilibrium in the vertical direction:

Moments on a generic face

The two equations are obtained by rotational equilibrium around n, nt. Mnt is the torque moment on the diagonal surface while Mn is the bending moment on the diagonal surface.

Mohr's circle

Representing the Mohr’s circle by defining the points of coordinates (My, Mxy) and (Mx, -Mxy), we obtain, in particular, the radius Rc by applying Pythagoras’ theorem. On the axis Mn, we represent the bending moments.

Simply supported square plate

We consider the case in which we have a 0.30 value for the Poisson coefficient. We observe this case in which we have the maximum efficiency of the 2-way behavior (because the two sides are the same). In the center of the plate, the bending moment stresses in tension the fiber on the bottom, so this is why the maximum bending moments are positive. Considering a unitary strip for the plate, we can compute the max. bending moment for a beam: So we have a strong reduction, comparing the... (text cuts off)