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and Every cross-section that initially is plane and also normal to the longitudinal axis, remains plane and and normal to the deflected axis too. A sandwich beam of length ‘L’, load ‘P’ and width ‘W’ in three-point bending with simply-supported end conditions has the following properties: NOTE: I have defined stresses in terms of moment and shear force per unit width because that is typically how beam solutions are presented.

In the following table, the formulas describing the static response of the simple beam under a trapezoidal load distribution, as depicted in the schematic above, are presented.

At any case, the moment application area should spread to a small length of the beam, so that it can be successfully idealized as a concentrated moment to a point. The following rules are adopted here: These rules, though not mandatory, are rather universal.

The following table presents the formulas describing the static response of the fixed beam, with both ends fixed, under a partially distributed trapezoidal load.

w , where Optional properties, required only for deflection/slope results: Simply supported beam with uniform distributed load, Simply supported beam with point force in the middle, Simply supported beam with point force at a random position, Simply supported beam with triangular load, Simply supported beam with trapezoidal load, Simply supported beam with slab-type trapezoidal load distribution, Simply supported beam with partially distributed uniform load, Simply supported beam with partially distributed trapezoidal load, The material is homogeneous and isotropic (in other words its characteristics are the same in ever point and towards any direction), The loads are applied in a static manner (they do not change with time), The cross section is the same throughout the beam length.
W Therefore axial forces can be commonly neglected.

w Calculating beam deflections should typically be done using Timoshenko, or Shear Beam, theory.

, L

In this case the load is distributed uniformly to only a part of the beam span, having a constant magnitude

are force per length. the lengths at the left and right side of the beam respectively, where the load distribution is varying (triangular). The static analysis of any load carrying structure involves the estimation of its internal forces and moments, as well as its deflections. Typically, for a plane structure, with in plane loading, the internal actions of interest are the axial force The bending moment is positive when it causes tension to the lower fiber of the beam and compression to the top fiber. In this case, a moment is imposed in a single point of the beam, anywhere across the beam span. The geometry of a symmetric sandwich layup is shown below. the span length and

. In order to consider the force as concentrated, though, the dimensions of the application area should be substantially smaller than the beam span length.

a In the following table, the formulas describing the static response of the simple beam under a uniform distributed load

, where In some cases the total force

By a similar approach we can derive the equation for mid-span deflection of a beam in 3-point bending (simply using in the general solution equation from Zenkert): The deflection ‘‘ for this case is illustrated in the following figure.

The total amount of force applied to the beam is The following table presents the formulas describing the static response of a fixed beam, with both ends fixed, under a concentrated point force and P L In this case, the load is distributed over the entire beam span, with linearly varying magnitude, starting from , where

\theta_A=-\frac{w(15L^4 - 20L^2a^2 - 10L^2b^2 + 15La^3 - 3a^4 + 3b^4)}{360EIL}, \theta_B=\frac{w (15L^4 - 10L^2a^2 - 20L^2b^2 + 15Lb^3 + 3a^4 - 3b^4)}{360E I L}, s_1(x)=xa^3+2ax^3-2a^2x^2-x^4-{a^4\over5}. Some basic terms for stiffness are defined here.

In the following table, the formulas describing the static response of the simple beam under a concentrated point force L
W={L\over2}(w_1+w_2)

Calculate the moment of inertia of various beam cross-sections, using our dedicated calculators. w_2 W=w\left(L-a-b\right)

W={L-a-b\over2}(w_1+w_2)

It is pretty hard work to follow, so I couldn't recommend it in general. b Deflections and slopes of simply supported beam.

the span length. The formulas for partially distributed uniform and triangular loads can be derived by appropriately setting the values of beam as presented in figure 1: were t is the thickness of the skins; c is the thickness of the core, overall thickness of the sandwich is d = c + 2t, and b and l are the width and length of the specimen. w_2

w_2 Every cross-section that initially is plane and also normal to the longitudinal axis, remains plane and and normal to the deflected axis too.

, at the right fixed end.

w

The shape of the distributed load is trapezoidal, as illustrated in the following figure.

from the left end. To the contrary, a structure that features more supports than required to restrict its free movements is called redundant or indeterminate structure.

w b

The solution for this case case is taken directly from (Zenkert) as follows: Now we will use this general solution for a single point load to calculate a specific case for 4-point-bending with quarter-point loading, .

The calculated results in the page are based on the following assumptions: The last two assumptions satisfy the kinematic requirements for the Euler Bernoulli beam theory that is adopted here too.

The fixed beam features more supports than required to be statically sound.

are force per length. w_1

,

Fixed-pinned beam calculator. , where are force per length. In the close vicinity of the force, stress concentrations are expected and as result the response predicted by the classical beam theory maybe inaccurate.

The response resultants and deflections presented in this page are calculated taking into account the following assumptions: The last two assumptions satisfy the kinematic requirements for the Euler-Bernoulli beam theory and are adopted here too.

, the transverse shear force One pinned support and a roller support.

, where As we move away from the force location, the results become valid, by virtue of the Saint-Venant principle. w_2 The roller support also permits the beam to expand or contract axially, though free horizontal movement is prevented by the other support.

w_2

M

a a

\theta_P={P a^2 b^2(L-2a) \over 2 E I L^3}. , where to zero.

, where L In the following table, the formulas describing the static response of the simple beam under a varying distributed load, of trapezoidal form, are presented.

from the left end, are presented. Restraining rotations results in zero slope at the two ends, as illustrated in the following figure. This is only a local phenomenon however. w_2

Read more about us here. This is the case when the cross-section height is quite smaller than the beam length (10 times or more) and also the cross-section is not multi layered (not a sandwich type section). Obviously this is unwanted for a load carrying structure.

The orientation of the triangular load is important for the use of the table!

The total amount of force applied to the beam is P Typically, when performing a static analysis of a load bearing structure, the internal forces and moments, as well as the deflections must be calculated.

The load w is distributed throughout the beam span, having constant magnitude and direction.

w_1

Because there is so little guidance online, I am going to summarise what I think are a few key equations and principles of sandwich structures.

In this case, a moment is imposed in a single point of the beam, anywhere across the beam span. The use of Ansys to calculate sandwich structures Vincent Manet Ecole des Mines de Saint-Etienne, Material and Mechanical Department, 158, cours Fauriel, 42023 Saint-Etienne cedex 2, France fax: (+33) 4-77-42-00-00, email: manet@emse.fr Abstract In this article, we make a comparative study on a simply supported sandwich beam , imposed in the middle, are presented. Read more about us here. ,

The maximum load magnitude is , while the remaining span is unloaded. The total amount of force applied to the beam is For a simply supported beam that carries only transverse loads, the axial force is always zero, therefore it is often neglected.

a

w_1 The following result represents the final mid-span deflection solution for the case shown below. The total amount of force applied to the beam therefore is:

is distributed uniformly, over the entire beam span, having constant magnitude and direction. can be freely assigned. The total amount of force applied to the beam is

The following table presents the formulas describing the static response of the fixed beam, with both ends fixed, under a partially distributed uniform load. To the contrary, a structure without redundancy, would turn to a mechanism, if any of its supports were removed. The structures that offer no redundancy, are called critical or determinant structures.

The sandwich beam consists of: two metal facings, the metal foam core and two binding layers between the faces and the core.

the span length.

P The printed results consist of the following quantities: SIGfmax - The maximum in-plane stress in the laminated face sheet. There are few textbook solutions available (again Zenkert has some simple cases). the span length.

The dimensions of The total amount of force applied to the beam is therefore:

and and

Instead, it is varying linearly, starting from zero at the left fixed end, gradually increasing, up to its peak value

However, as we move away, the predicted results become perfectly valid, as stated by the Saint-Venant principle, provided the loading area remains substantially smaller than the total beam length.

The formulas presented in this section have been prepared for the case of an ascending load (left-to-right), as shown in the schematic.

and

Removing any of the supports or inserting an internal hinge, would render the simply supported beam to a mechanism, that is body the moves without restriction in one or more directions. w_m={w_1+w_2\over2}, s_1 = 10\left[ (L^2+a^2)(L+a) - (a^2+b^2)(a-b) - L b(L+b) -a^3\right], s_2 = L_w \left[L(2L+a+b) - 3(a-b)^2 - 2 a b\right], s_3=120ab (a+L_w) + 10L_w( 6a^2 + 4LL_w - 3L_w^2 ).

at the right end. Therefore, the simply supported beam offers no redundancy in terms of supports. The total amount of force applied to the beam is