Ex 7-1 Rigid Rotation of a Beam
7.6 Kinematically Linear Straight Timoshenko Beams
We shall here make the same assumptions as in Section 7.5 regarding the degree of nonlinearity, i.e. that the strains—here the axial, shear, and cur- vature strains—as well as the rotations are small.
As mentioned in Section 7.4, Timoshenko beam theories are applicable to relatively short beams where the shear strains may not be neglected. Recall that as a rule of thumb, for an elastic beam with a rectangular cross-section the effect of shear strains becomes important when the length of the beam
Kinematically Linear Timoshenko Beams
is less than about four to five times its depth. If the beam is made of, say an orthotropic material with the stiff direction along the beam axis, then shear strains may always play an important role. A similar observation holds for sandwich beams.
The derivations below follow the same pattern as that of Section 7.4 and of Section 7.5. As mentioned in Section 7.4 we did not visualize the shear strain there, but chose to defer it to the present case. In part the reason for this is that the theory developed in Section 7.4 was more complicated than I like—and probably not completely satisfactory. As before, the first consequence of considering shear strains is that we need three generalized displacements to describe deformations of the beam, namely the axial dis- placement component u, the transverse displacement component w and, in addition, to these the rotation ω of the beam “cross-section.”7.10
7.10The reason for the quotation marks is that we are dealing with a one-dimensional beam theory, i.e. the beam theory does not, by itself, entail a thickness, but the meaning should be clear from Fig. 7.6.
Kinematically Linear Timoshenko Beams 111
7.6.1
Kinematics
Here, we must define the three generalized strains, which are strains sketched Timoshenko beam: Shear strain ϕ 6≡ 0
in Fig. 7.6, namely the axial strain ε, the shear strain ϕ, and the curvature strain κ, to the same degree of nonlinearity as in Section 7.3.1, namely
|u′| ≪ 1 , |w′| ≪ 1 and |ω| ≪ 1 (7.64)
7.6.1.1 Axial Strain
Once more (7.57) defines the axial strain
Axial strain ε
ε = u′ (7.65)
7.6.1.2 Shear Strain
We take the shear strain ϕ to be the same as for the moderately nonlinear version of the Timoshenko beam theory, see (7.35).
Shear strain ϕ = w′
− ω
ϕ = w′− ω (7.66)
7.6.1.3 Curvature Strain
Because we have abandoned the assumption of kinematic coupling between the transverse displacement component w and the rotation ω the curvature strain κ must not be defined in accordance with (7.57) but as in (7.9)
Curvature strain κ 6= w′′
κ ≡dωdx= ω′ (7.67) which, by the way, holds regardless of the degree of nonlinearity.
7.6.1.4 Interpretation of Shear Strain
While the physical interpretation of the two strains ε and κ probably is clear, both from Fig 7.6 and from our intuition, the shear strain ϕ deserves special attention. Consider a two-dimensional element of the beam, see Fig. 7.6,7.11 and subject it to the two simple displacement fields shown.
But, first we need to define two directions. One is the direction of the beam axis, which is indicated by the dash-dot line, and the other is the normal to the cross-section of the beam and is given by the arrow. Note that, here, the term cross-section should be taken literally because we are dealing with a two-dimensional beam.
7.11The deformation patterns shown in the figure are not possible for an entire beam cross-section because some ingredient of S-shape is unavoidable. So, the sketches must be taken with a grain of salt.
You may, however, ask if such an S-shape does not violate the idea associated with extension and pure bending that the cross-sections remain plane. The answer is that these different kinds of displacements do not interact. A similar conclusion holds for torsion of beams where most cross-sections experience warping.
Case (a)
Recall that all displacements are infinitesimal. Then, the displacements may be determined as
u(a)1 = 0 and u(a)2 = w′x
1 (7.68) w′ x2 x1 (a) ω x2 x1 (b)
Fig. 7.6: Infinitesimal element of Timoshenko beam. Two fundamental cases of shear strain. In order to avoid clog- ging of the figure, the x1-axis is not coinciding with the
beam axis.
The result is that the strains (4.5) become ε(a)11 = 0 , ε(a)12 = ε(a)21 =1
2w′ and ε (a)
22 = 0 (7.69)
and thus the beam shear strain ϕ(a)is
ϕ(a)= 2ε(a)
12 = w′ (7.70)
Case (b)
In this case the displacements are
u(b)1 = −ωx2 and u(b)2 = 0 (7.71) and ε(b)11 = 0 , ε(b)12 = ε(b)21 = −1 2ω and ε (b) 22 = 0 (7.72) and thus ϕ(b)= 2ε(b) 12 = −ω (7.73)
Combination of Cases (a) and (b)
Any strain situation can be composed of cases (a) and (b) and therefore we may take the beam shear strain ϕ to be
Shear strain ϕ = w′
− ω ϕ = w
′− ω (7.74)
Kinematically Linear Timoshenko Beams 113 7.6.1.5 Interpretation of Operators
Here, we only need to interpret the linear operator l1 which enters the
expression for the generalized strains ε, see Chapter 33. The generalized displacements are u∼ wu ω (7.75)
and the interpretation of the linear operator l1is
l1(u) ∼ u′ (w′− ω) ω′ (7.76)
while the other operators l2, and l11do not enter the description due to its
linearity.
7.6.2
Generalized Stresses
As mentioned above, our choice of strain measures, the generalized strains, dictates our stress measures, the generalized stresses, because these two sets must be each other’s work conjugate in the Principle of Virtual Displace- ments. Since we work with three generalized strains we must have three generalized stresses which, as before, we shall denote N , V and M . They are the work conjugate of ε, ϕ and κ, respectively.
7.6.3
Equilibrium Equations
When the beam end points are located at x = a and x = b the internal virtual work σ · δε is Internal virtual work σ · δε = Zb a (N δε + V δϕ + M δκ)dx (7.77) In order to proceed we need an expression for δε, δϕ and δκ. The strain definitions (7.65), (7.74) and (7.67) provide
Generalized strains ε, ϕ and κ
δε = δu′, δϕ = (δw′− δω) and δκ = δω′ (7.78) and thus (7.77) becomes
Internal virtual work σ · δε = Zb a N δu′+ V (δw′− δω) + Mδω′dx = [N δu]b a+ [V δw]ba+ [M δω]ba − Zb a N′δudx −Zb a V′δwdx −Z b a (M′+ V )δωdx (7.79)
Provided that there are no distributed moment loads, i.e. that the loads are the ones shown in Fig. 7.2, which for convenience is repeated here
¯ pu Pu(a) Pu(b) Pw(a) C(b) Pw(b) C(a) ¯ pw
Fig. 7.7: Undeformed beam with loads and end forces. as Fig. 7.7, the external virtual work T · δu is
External virtual work T · δu = Zb a (¯puδu + ¯pwδw)dx
− Pu(a)δu(a) − Pw(a)δw(a) − C(a)δw′(a)
+ Pu(b)δu(b) + Pw(b)δw(b) + C(b)δw′(b)
(7.80)
which obviously is the same as (7.21).
By equating the internal and external virtual work we can immediately conclude that
First static field
equation M
′+ V = 0 , x ∈]a, b[ (7.81)
which indicates that V is the same as the static quantity V introduced by (7.61). Furthermore, from (7.79) and (7.80) we may also get the other two static field equations. The first is
Second static field
equation V
′+ ¯p
w= 0 , x ∈]a, b[ (7.82)
which agrees with (7.62b), and the second is
Third static field
equation N
′+ ¯p
u= 0 , x ∈]a, b[ (7.83)
which is the same as (7.62a). Thus we recover the equilibrium equations (7.62).
As regards the possible static boundary conditions we get
Possiblestatic boundary conditions N (a) = ¯Pu(a) , N (b) = ¯Pu(b) V (a) = ¯Pw(a) , V (b) = ¯Pw(b) M (a) = ¯C(a) , M (b) = ¯C(b) (7.84) which are the same as (7.63).
We may now realize that equilibrium equations of the linear Bernoulli-
Kinematic linearity : Same static equations for B-E and Timoshenko
Euler theory and the linear Timoshenko theory are the same, although the generalized strains are different,7.12 but recall that in the Bernoulli-Euler
theory the shear force V is not a generalized quantity.
Elastic Bernoulli-Euler Beams 115