Download Lecture 6: Moderately Large Deflection Theory of Beams and more Study notes Statistical mechanics in PDF only on Docsity!
Lecture 6: Moderately Large Deflection Theory of
Beams
6.1 General Formulation
Compare to the classical theory of beams with infinitesimal deformation, the moderately large deflection theory introduces changes into the strain-displacement relation and vertical equilibrium, but leaves the constitutive equation and horizontal equilibrium unchanged. The kinematical relation, Eq. (2.68) acquires now a new term due to finite rotations of beam element.
◦^ = d dux + (^12)
dw dx
− new term (6.1)
The definition of curvature has also a nonlinear rotation term
κ = −
d^2 w [^ dx^2 1 +
dw dx
) 2 ] 3 / 2 (6.2)
The square of the slope can be large, as compared with the term du dx
and must be retained
in Eq. (6.1). At the same time the square of the slope (beam rotation) are small compared to unity. Why? This is explained in Fig. (6.1), where the square of the slope is plotted against the slope.
Figure 6.1: The significance of the square of the slope term.
At θ = 1 rad = 57 degrees the two terms in the denominator of Eq. (6.2) are equal. However, the theory of moderately large deflections are valid up to θ = 10◦^ ≈ 0 .175 rad. The term θ^2 amounts to 0.03, which is negligible compared to unity. Therefore the curvature is defined in the same way as in the theory of small deflections
κ = − d
(^2) w dx^2
It was shown in Lecture 3 that the equation of equilibrium in the horizontal direction is not affected by the finite rotation. Therefore, we infer from Eq. (3.77) that the axial force is either constant or zero N = constant (6.4)
The vertical equilibrium, given by Eq. (3.79) has a new nonlinear term
d^2 M^ −new term dx^2
(^2) w dx^2
+q = 0 (6.5)
Finally, the elasticity law is unaffected by finite rotation
N = EA◦^ (6.6a) M = EIκ (6.6b)
The solution to the coupled problem depends on the boundary conditions in the hori- zontal direction. Referring to Fig. 5.1, two cases must be considered:
- Case 1, beam free to slide, N = 0, u 6 = 0.
- Case 2, beam fixed, u = 0, N 6 = 0.
6.2 Solution for a Beam on Roller Support
Consider first case 1. From the constitutive equation, zero axial force beams that there is no extension of the beam axis, ◦^ = 0. Then, from Eq. (6.1)
du dx
dw dx
At the same time, the nonlinear term in the vertical equilibrium vanishes and the beam response is governed by the linear differential equation
EI
d^4 w dx^4 =^ q(x)^ (6.8)
which is identical to the one derived for the infinitesimal deflections. As an example, consider the pin-pin supported beam under mid-span point load. From Eqs. (5.54) and (5.55), the deflection profile is
w(x) = wo
[
x l −^4
( (^) x l
) 3 ]
and the slope is dw dx =^
wo l
[
( (^) x l
) 2 ]
6.3 Solution for a Beam with Fixed Axial Displacements
The problem is solved under the assumption of simply-supported end condition, and the line load is distributed accordingly to the cosine function.The beam is restrained in the axial direction. There is a considerable strengthening effect of the beam response due to finite rotations of beam elements. The axial force N (non-zero this time) is calculated from Eq. (6.6) with Eq. (6.1)
N = EA
[
du dx
+^1
dw dx
) 2 ]
From Eq. (6.4) we know that N is constant but unknown. In order to make use of the kinematic boundary conditions, let us integrate both sides of Eq. (6.15) with respect to x
N l EA
= u(l) − u(0) +
∫ (^) l
0
dw dx
dx (6.16)
Using the boundary conditions for u, the axial force is related to the square of the slope by
N l EA =
∫ (^) l
0
dw dx
dx (6.17)
In order to determine the deflected shape of the beam, consider the equilibrium in the vertical direction given by Eq. (6.5)
−EI d
(^4) w dx^4
(^2) w dx^2
Dividing both sides by (−EI) one gets
d^4 w dx^4 −^ λ
2 d^2 w dx^2 =^
qo EI cos^
πx l (6.19)
where λ^2 = (^) EIN (6.20)
The roots of the characteristic equations are 0, 0, ±λ. Therefore the general solution of the homogeneous equation is
wg = Co + C 1 x + C 2 cosh λx + C 3 sinh λx (6.21)
As the particular solution of the inhomogeneous equation we can try
wp(x) = C cos πxl (6.22a) d^2 wp dx^2 =^ −^
π^2 l^2 C^ cos^
πx l (6.22b) d^4 wp dx^4 =^
π^4 l^4 C^ cos^
πx l (6.22c)
Substituting the above solution to the governing equation (6.18) one gets [ π^4 l^4
C − λ^2 π
2 l^2
C − Po EI
]
cos πx l
The above solution satisfy the differentia equation if the amplitude C is related to input parameters and the unknown tension N
C =
qo EI π^2 l^2
λ^2 + π
2 l^2
) = qo EI
( (^) π l
+ N
( (^) π l
The general solution of Eq. (6.18) is a sum of the particular solution of the inhomogeneous equation wp and general solution of the homogeneous equation, wg
w(x) = wg + wp (6.25)
There are five unknowns, Co, C 1 , C 2 , C 3 and N and five equations. Four boundary conditions for the transverse deflections
w = 0, d
(^2) w dx^2 = 0^ at^ x^ =^ ±^
l 2 (6.26)
and equation (6.17) relating the horizontal and vertical response. The determination of the integration constants is straightforward. Note that the problem is symmetric. Therefore the solution should be an even function^1 of x. The terms C 1 x and C 3 sinh λx are odd functions. Therefore the respective coefficients should vanish
C 1 = C 3 = 0 (6.27)
w(x) = Co + C 2 cosh λx + C cos
πx l (6.28) The remaining two coefficients are determined only from the boundary conditions at one side of the beam w(x = l 2
d^2 w dx^2
x= 2 l
Co + C 2 cosh λl 2 = 0
C 2 λ^2 cosh λl 2
The solution of the above system is
Co = C 2 = 0 (6.30)
The slope of the deflection curve, calculated from Eq. (6.28) is
dw dx =^ −C
π l sin^
πx l (6.31) (^1) The function is even when f (−A) = f (A). The function is odd when f (−A) = −f (A).
(i) The bending solution is obtained by dropping the cubic term in Eqs. (6.36) or (6.38)
C =
qol^4 EI
π^4 (6.39) where the coefficient π^4 = 97.4. this result for the sinusoidal distribution of the line load should be compared with the uniform line load (coefficient 77) and point load (coefficient 48).
(ii) The membrane solution is recovered by neglecting the first linear term
C^3 =
qol^4 EI
π^4 (6.40) The plot of the full bending/membrane solution and two limiting cases is shown in Fig. (6.3).
Figure 6.3: Comparison of a bending and membrane solution for a beam.
The question is at which magnitude of the central deflection relative to the beam thick- ness the non-dimensional load qo Eh
is the same. This is the intersection of the straight line
with the third order parabola. By eliminating the load parameter between Eqs. (6.39) and (6.40) one gets
C^2 =
4 I
A =
4 ρ^2 A A = 4ρ
where ρ is the radius of gyration of the cross-section. For a square cross-section
C = 2ρ = 2
I
ρ
h^4 12 h^2
= √h 3
∼= 0. 58 h (6.42)
It is concluded that the transition from bending to membrane action occurs quite early in the beam response. As a rule of thumb, the bending solution in the beam restrained from axial motion is restricted to deflections equal to half of the beam thickness. If deflections are larger, the membrane response dominate. For example, if beam deflection reaches three
thicknesses, the contribution of bending and membrane action is 3:81. In the upper limit of the applicability of the theory of moderately large deflection of beams C ∼= 10h, the contribution of bending resistance is only 0.3% of the membrane strength. The rapid transition from bending to membrane action is only present for axially re- strained beams. If the beam is free to slide in the axial direction, no membrane resistance is developed and load is always linearly related to deflections. The above elegant closed-form solution was obtained for the particular sinusoidal distri- bution of line load, which coincide with the deflected shape of the beam. For an arbitrary loading, only approximate solutions could be derived. One of such solution method, appli- cable to the broad class of non linear problems for plates and shells is called the Galerkin method.
6.4 Galerkin Method of Solving Non-linear Differential Equa-
tion
Beris Galerkin, a Russian scientist, mathematician and engineer was active in the first forty ears of the 20th^ century. He is an example of a university professor who applied methods of structural mechanics to solve engineering problems. At that time (World War I), the unsolved problem was moderately large deflections of plates. In 1915, he developed an approximate method of solving the above problem and by doing it made an important and everlasting contribution to mechanics. The theoretical foundation of the Galerkin method goes back to the Principle of Virtual Work. We will illustrate his idea on the example of the moderately large theory of beams. If we go back to Lecture 3 and follow the derivation of the equations of equilibrium from the variational principle, the so called “weak” form of the equilibrium is given by Eq. (3.54). Adding the non-linear term representing the contribution of finite rotations, this equation can be re-written as ∫ (^) l
0
(M ′′^ + N w′′^ + q)δwdx +
∫ (^) l
0
N ′δudx + Boundary terms (6.43)
where
M = −EIw′′^ (6.44a)
N = EA[u′^ +
2 (w
′) (^2) ] (6.44b)
From the weak (global) equilibrium one can derive the strong (local) equilibrium by considering an infinite class of variations. But, what happens if, instead of a “class”, we consider only one specific variation (shape) that satisfies kinematic boundary conditions? The equilibrium will be violated locally, but can be satisfied globally in average ∫ (^) l
0
[
−EIwIV^ + EA
u′^ +^1 2
(w′)^2 w′′^ + q
)]
δw dx = 0 (6.45)
Replacing q 1 by
π qo, the pure membrane solution takes the final form ( (^) wo h
=^6.^4
π^4
( (^) qo Eh
) (^ l h
One can see that not only the dimensionless form of the exact and approximate solutions are identical, but also the coefficient 6.4 in Eq. (6.53) is of the same order as the coefficient 4 in Eq. (6.41).
6.5 Generalization to Arbitrary Non-linear Problems in Plates
and Shells
The previous section felt with the application of the Galerkin method to solve the non-linear ordinary differential equation for the bending/membrane response of beams. Galerkin name is forever attached to the analytical or numerical solution of partial differential equation, such as describing response of plates and shells. In the literature you will often encounter such expression as Galerkin-Bubnov method, Petrov-Galerkin method, the discontinuous Galerkin method or the weighted residual method. The essence of this method is sketched below. Denote by F (w, x) the non-linear operator (the left hand side of the partial differential equation) is defined over a certain fixed domain in the 2-D space S. Now, a distinction is made between the exact solution w∗(x) and the approximate solution w(x). The approxi- mate solution is often referred to as a trial function. The exact solution makes the operator F to vanish F (w∗, x) = 0 (6.54)
The approximate solution does not satisfy exactly the governing equation, so instead of zero, there is a residue on the right hand of the Eq. (6.55)
F (w, x) = R(x) (6.55)
The residue can be positive over part of S and negative elsewhere. If so, we can impose a weaker condition that the residue will become zero “in average” over S, when multiplied by a weighting function w(x) (^) ∫
S
R(x)w(x) dS = 0 (6.56)
Mathematically we say that these two functions are orthogonal. In general, there are also boundary terms in the Galerkin formulation. For example, in the theory of moderately large deflection of plates, Eq. (6.56) takes the form ∫
S
(D∇^4 w − Nαβ w,αβ )w dS = 0 (6.57)
The counterpart of Eq. (6.57) in the theory of moderately large deflection of beams is Eq. (6.47) which was solved in the previous section of the notes. The solution of partial differ- ential equations for both linear and non-linear problems is extensively covered in textbooks on the finite element method and therefore will not be covered here.
