The boundary conditions are now all of the force type, requiring that bending moment and shear force both vanish at x=0 and x=L, i.e.
(8.73)
which, upon substitution into eq (8.58) give
Equating the determinant of the 4×4 matrix to zero yields the frequency equation
(8.74) which is the same as for the clamped-clamped case (eq (8.69)), so that the roots given by eq (8.70) are the values which lead to the first lower frequencies corresponding to the first lower elastic modes of the free-free beam. The elastic eigenfunctions can be obtained by following a similar procedure as in the previous cases. They are
(8.75)
where is the same as in eq (8.72). In this case, however, the system is unrestrained and we expect rigid-body modes occurring at zero frequency, i.e. when On physical grounds, we are considering only lateral deflections and hence we expect two such modes: a rigid translation perpendicular to the beam’s axis and a rotation about its centre of mass.
This is, in fact, the case. Substitution of in eq (8.57) leads to
(8.76a) where A, B, C are constants. Imposing the boundary conditions (8.73) to the solution (8.76a) yields
(8.76b) which is a linear combination of the two functions
(8.77)
where we omitted the constants because they are irrelevant for our purposes.
It is not difficult to interpret the functions (8.77) on a mathematical and on a physical basis: mathematically they are two eigenfunctions belonging to the eigenvalue zero and, physically, they represent the two rigid-body modes considered above.
We leave to the reader the case of a beam which is clamped at one end and simply supported at the other end. The frequency equation for this case is
(8.78) and its first roots are
(8.79)
Also, note that we can approximate
Finally, one more point is worthy of notice. For almost all of the configurations above, the first frequencies are irregularly spaced; however, as the mode number increases, the difference between the two frequency parameters and γnL approaches the value π for all cases. This result is general and indicates, for higher frequencies, an insensitivity to the boundary conditions.
8.5.1 Axial force effects on bending vibrations
Let us consider now a beam which is subjected to a constant tensile force T parallel to its axis. This model can represent, for example, either a stiff string or a prestressed beam.
On physical grounds we may expect that the model of the beam with no axial force should be recovered when the beam stiffness is the dominant restoring force and the string model should be recovered when tension is by far the dominant restoring force. This is, in fact, the case. The governing equation of motion for the free motion of the system that we are considering now is
(8.80) Equation (8.80) can be obtained, for example, by writing the two equilibrium equations (vertical forces and moments) in the free-body diagram of Fig. 8.4 and noting that, from elementary beam theory,
Alternatively, we can write the Lagrangian density
and arrive at eq (8.80) by performing the appropriate derivatives prescribed in eq (3.109). The usual procedure of separation of variables leads to a solution with a harmonically varying temporal part and to the ordinary differential equation
(8.81)
Fig. 8.4 Beam element with tensile axial force (schematic free-body diagram).
where, as in the previous cases, we called u(x) the spatial part of the solution.
If now we let eq (8.81) yields
where is positive and is negative. It follows that we have the four roots ±η and where we defined
(8.82)
The solution of eq (8.81) can then be written as
(8.83) which is formally similar to eq (8.58) but it must be noted that the hyperbolic functions and the trigonometric functions have different arguments. We can now consider different types of boundary conditions in order to determine how the axial force affects the natural frequencies. The simplest case is when both ends are simply supported; enforcing the boundary conditions (8.59) leads to and to the frequency equation
(8.84) which results in because for any nonzero value of η. The allowed frequencies are obtained from this means
(8.85a)
which can be rewritten as
(8.85b)
where it is more evident that for small values of the nondimensional ratio (i.e. when ) the tension is the most important restoring force and the beam behaves like a string. At the other extreme—when R is large—the stiffness is the most important restoring force and in the limit of
we return to the case of the beam with no axial force.
In addition to the observations above, note also that:
• In the intermediate range of values of R, higher modes are controlled by stiffness because of the n2 factor under the square root in eq (8.85b).
• The eigenfunctions are given by (enforcement of the boundary conditions leads also to C2=0)
(8.86) which have the same sinusoidal shape of the eigenfunctions of the beam with no tension (although here the sine function has a different argument).
The conclusion is that an axial force has little effect on the mode shapes but can significantly affect the natural frequencies of a beam by increasing their value in the case of a tensile force or by decreasing their value in the case of a compressive force. In fact, the effect of a compressive force is obtained by just reversing the sign of T. In this circumstance the natural frequencies can be conveniently written as
(8.87)
where we recognize as the critical Euler buckling load. When the lowest frequency goes to zero and we obtain transverse buckling.
In the case of other types of boundary conditions the calculations are, in general, more involved. For example, we can consider the clamped-clamped configuration and observe that placing the origin x=0 halfway between the supports divides the eigenfunctions into even functions, which come from the combination
and odd functions, which come from the combination
In either case, if we fit the boundary conditions at x=L/2, they will also fit at x=–L/2. For the even functions the boundary conditions
lead to the equation
(8.88a) and for the odd functions we obtain
(8.88b) Both eqs (8.88a and b) must be solved numerically: from eq (8.88a) we obtain the natural frequencies while the natural frequencies
are obtained from eq (8.88b).
8.5.2 The effects of shear deformation and rotatory inertia (Timoshenko beam)
It was stated in Section 8.5 that the Euler-Bernoulli theory of beams provides satisfactory results as long as the wavelength is large compared to the lateral dimensions of the beam which, in turn, may be identified by the radius of gyration r of the beam section. Two circumstances may arise when the above condition is no longer valid:
1. The beam is sufficiently slender (say, for example, ) but we are interested in higher modes.
2. The beam is short and deep.
In both cases the kinematics of motion must take into account the effects of shear deformation and rotatory inertia which—from an energy point of view—result in the appearance of a supplementary term (due to shear deformation) in the potential energy expression and in a supplementary term (due to rotatory inertia) in the kinetic energy expression.
Let us consider the effect of shear deflection first. Shear forces t result in an angular deflection θ which must be added to the deflection due to bending alone. Hence, the slope of the elastic axis ∂y/∂x is now written as
(8.89) and the relationship between bending moment and bending deflection (from elementary beam theory) now reads
(8.90)
Moreover, the shear force Q is related to the shear deformation θ by (8.91) where G is the shear modulus, A is the cross-sectional area and is an adjustment coefficient (sometimes called the Timoshenko shear coefficient) whose value must generally be determined by stress analysis considerations and depends on the shape of the cross-section. In essence, this coefficient is introduced in order to satisfy the equivalence
and accounts for the fact that shear is not distributed uniformly across the section. For example, for a rectangular cross-section and other values can be found in Cowper [7].
In the light of these considerations, the potential energy density consists of two terms and is written as
(8.92)
where in the last term we take into account eq (8.89).
On the other hand, the kinetic energy density must now incorporate a term that accounts for the fact that the beam rotates as well as bends. If we call J the beam mass moment of inertia density, the expression for the kinetic energy density is written as
(8.93)
Moreover, J is related to the cross-section moment of inertia I by
(8.94) where is the beam mass density and is the radius of gyration of the cross-section. Taking eqs (8.92) and (8.93) into account, we are now
in the position to write the explicit expression of the Lagrangian density
(8.95)
and perform the appropriate derivatives prescribed in eq (3.109) to arrive at the equation of motion. In this case, however, both y and are independent variables and hence we obtain two equations of motion. As a function of y, the Lagrangian density is a function of the type where, following the notation of Chapter 3, the overdot indicates the derivative with respect to time and the prime indicates the derivative with respect to x.
So, we calculate the two terms
and
to arrive at the first equation of motion
(8.96a)
and to the boundary conditions (eq (3.110))
(8.96b)
which take into account the possibility that either the term in brackets or δy can be zero at x=0 and x=L.
A similar line of reasoning holds for the variable ; the Lagrangian function is of the type and the derivatives we must find are now given by
so that the second equation of motion is
(8.97a)
with the boundary conditions (eq (3.110))
(8.97b)
which take into account the possibility that either EI(∂/∂x) or δ are zero at x=0 and x=L.
Equations (8.96a) and (8.97a) govern the free vibration of a Timoshenko beam; we note that, in the above treatment, there are two ‘modes of deformation’ whose physical coupling translates mathematically into the coupling of the two equations.
If now, for simplicity, we assume that the beam properties are uniform throughout its length we can arrive at a single equation for the variable y.
From eq (8.96) we get
(8.98) and by differentiating eq (8.97) we obtain
(8.99)
Substitution of eq (8.98) into eq (8.99) yields the desired result, i.e.
(8.100)
Now, a closer look at eq (8.100) shows that:
1. The term
arises from shear deformation and vanishes when the beam is very rigid in shear, i.e. when
2. The term
is due to rotatory inertia and vanishes when 3. The term
results from a coupling between shear deformation and rotatory inertia.
Note that this term vanishes when either or
4. When both shear deformation and rotatory inertia can be neglected we recover the Euler-Bernoulli case, which is represented by the first two terms.
With reference to the brief discussion on the velocity of wave propagation of Section 8.5, it may be interesting to note at this point that the unphysical result of infinite velocity as the wavelength is removed by the introduction of the effects of shear deformation and rotatory inertia in the equations of motion. As a matter of fact, the introduction of rotatory inertia alone is sufficient to obtain finite velocities at any wavelength, but the results in the short-wavelength range are not in good agreement with the values of velocities calculated from the exact general elastic equations. A much better agreement is obtained by including the effect of shear deformation (e.g.
Graff [8]).
From eq (8.100), if we assume a solution we arrive at the ordinary differential equation
(8.101)
which can be written in other forms (if the reader can find it convenient) by
taking into account the relationships where
v is Poisson’s ratio. From here, at the price of more cumbersome calculations, we can proceed as in the preceding cases to arrive at the values of the natural frequencies of the Timoshenko beam. We will not do so but we will investigate briefly the effects of shear deflection alone and of rotatory inertia alone on the eigenvalues of a pinned-pinned beam.