# A Cantilever Beam Problem with Small Deflections and Perturbed Boundary Data

Authors: Ammar Khanfer; Lazhar Bougoffa
Journal: Journal of Function Spaces
 (2021)

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## Abstract

The boundary value problem of a fourth-order beam equationu4=λfx,u,u′,u″,u′′′,0≤x≤1 is investigated. We formulate a nonclassical cantilever beam problem with perturbed ends. By determining appropriate values of λ and estimates for perturbation measurements on the boundary data, we establish an existence theorem for the problem under integral boundary conditions u0=u′0=∫01pxuxdx,u″1=u′′′1=∫01qxu″xdx, where p,q∈L10,1, and f is continuous on 0,1×0,∞×0,∞×−∞,0×−∞,0.


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## Body

## 1. Introduction and Preliminaries

Beams are one of the main structural elements in construction engineering. One of the objectives of the beam theory is to study the behavior of beams to analyze deformations under loads. The deformation of beams occurs when the beam is under a load, which causes the beam to develop bending moment and shear force. The deformation of the beam is modeled by the fourth-order Euler–Bernoulli equation(1)u4=fx,where u represents the deflection of the beam, u′ represents the slope, u″ is the bending moment (torque), u′′′ is the shear force, and u4 is the load density stiffness. The function f is a load on the beam; it is uniformly distributed if the loading is only the weight of the beam, without any further concentrated mass at the free end. The boundary conditions are governed by the particular type of beams under study and the way in which the beam is supported. The most important type of beams that have many useful applications in industry is the cantilever beam, where the beam is fixed (clamped, anchored to a support, or built into a wall) at one end (say at x=0) and free at the other end (say x=1). The fixed end must have zero transitional and zero rotational motions, whereas the free end must have zero bending and zero shearing force. The bending moment is equal to the applied force multiplied by the distance to the point of application; so, it becomes zero at the free end because there is no stress at this end. So, we must have the following boundary conditions
(2)u0=u′0=0,u″1=u′′′1=0.The cantilever beams are widely used in construction engineering and can be found in many structures such as buildings and bridges. Some of the common examples of cantilever beams are aircraft wings, cranes, suspended bridges, balconies, diving board, electronic spring connectors, shelves, basketball backboard, road signs, and many other examples. Because of its importance, the cantilever beam problem received a wide and considerable attention from researchers, and a large number of research about cantilever can be found from a quick literature search [see and the references therein]. In particular, in [1–3], the authors investigated the existence of deflections (solutions) for the case
(3)u4=fx,ux,0≤x≤1,under conditions (2). In [4–6], the case
(4)u4=fx,u,u′,0≤x≤1was investigated under conditions (2). In [7–11], the fully fourth-order nonlinear boundary value problem
(5)u4=fx,u,u′,u′′,u′′′,0≤x≤1was investigated under conditions (2). The authors in [8, 9] assumed a Nagumo-type condition on f, and [10] assumed a L1 Caratheodory condition on f.In all the aforementioned research, the cantilever problem was investigated under the classical condition (2). Some special types of relaxed, restrained, or propped cantilever beams violate these conditions. If the fixed end is loose, the support (or the anchor) is relaxed thenu0≠0 or/and u′0≠0. If the free end is rolled or pinned support, then u″1≠0,u′′′1≠0. If a concentrated force, or translational elastic spring is attached at the free end, then u″1=0 and u′′′1≠0, while if a concentrated moment, or rotational elastic spring is attached, then u″1≠0. If a concentrated mass is placed at the free end, it will develop a shear force of the form u′′′1=−mg, and when m=0, this reduces the condition to the classical one.The present paper deals with the fully fourth-order nonlinear boundary value problem(6)u4=λfx,u,u′,u″,u′′′,0≤x≤1,where f is continuous on 0,1×0,∞×0,∞×−∞,0×−∞,0, under a perturbed homogeneous conditions, where the two ends are perturbed (the fixed end is slightly relaxed, and the free end is slightly supported). This can be formulated using integral conditions of the form
(7)u0=u′0=∫01pxuxdx,u″1=u′′′1=∫01qxu″xdx,where p,q∈L10,1. Here, f satisfies a growth condition with the variable parameters:
(8)fx,u,v,w,z≤axu+bxv+cxw+dxz+ex,where a,b,c,d, and e are positive continuous functions on 0,1. Moreover, letting
(9)α=∫01p2xdx,β=∫01q2xdx,then we have the following assumptions
(10)0≤α≤δ1<14,0≤β≤δ2<14.Letting(11)λ0=1/2−δ2/1−2δ2M2C1/1−2δ1+2C1+2/1−2δ2,where M=maxa,b,c,d,e, and
(12)C1=2/1−2β1/2−α/1−2α2.Then, we impose the following condition(13)λ≤λ0.Condition (13) provides small deflections. The parameter λ represents the reciprocal of the flexural rigidity which measures the resistance to bend; so, smaller values of λ indicate large flexural rigidity for the material of the beam, and this causes small deflections when load is applied on the beam. Hook’s law is valid as long as the deflection is small. Problems with large deflections cannot be solved in terms of the linear beam theory—in which Euler-Bernoulli Equation applies, since Hook’s law is no longer valid. When the beam is assumed to be homogeneous, made of high rigid material, and behaves in a linear elastic manner, its deflection under bending is usually small. When thin flexible beams are used, large deflections are expected to occur. Many researchers investigated cantilever beams with large deflections [12–26], and others investigated small deflections [27, 28]. Most of the beams used in industry and constructions (buildings, bridges, aircrafts, etc.) undergo small deflections [29]. This is of utmost importance since large deflections can cause cracks in the beams, and this may eventually lead to disastarous damages. So, best efforts are made to limit deflections in the ceilings and walls and in the design of aircraft, see [29] for more details on beam theory. Small deflections usually occur when either the loaded force f is small (hence, M is small) or the material of the beam has high flexural rigidity, which implies that λ is small.Condition (10) provides small values for p,q to produce the required slight changes to the default boundary settings of a cantilever beam. The functions p,q represent perturbation measurements related to bearings, rollers, springs, or any other mechanical settings that will perturb the boundary data.Condition (8) is a growth condition imposed on the load function f. This condition generalizes the boundedness condition and the growth condition with constant parameters
(14)fx,u,v,w,z≤au+bv+cw+dz+efor some positive constants a,b,c,d, and e.The integral boundary conditions (7) results from perturbing conditions (2) by imposing small measurements p,q related. Note that if p=q=0, the conditions in (7) reduce to (2). The integral boundary conditions have been studied and applied extensively in beam theory by many authors, for example, see [30–41].The purpose of this paper is to prove the existence of solutions (small deflections) of the cantilever beam boundary value problem (6) under the conditions (7)-(13). This interesting problem has not been studied in research. The result gives an affirmative answer to the question of existence of solutions of (6)-(13), i.e., the existence of small deflections of general types of perturbed cantilever beams under conditions (7)-(13), including the simple cantilever beam problem when p=q=0.


## 2. Existence and Uniqueness Theorems

The problem (1.6)-(7) can be converted into the following system:
(15)u″=v,u0=u′0=∫01pxuxdx,v″=λfx,u,u′,v,v′,v1=v′1=∫01qxvxdx.We need the following lemma.Lemma 1.
Ifh∈C10,1 with h0=∫01rxhxdx or h1=∫01rxhxdx,x∈0,1, where r∈C0,1, then
(16)∫01h2xdx≤21−2∫01r2xdx∫01h′2xdx
provided1−2∫01r2xdx>0.Proof.
Forh0=∫01rxhxdx, we note that
(17)hx=∫0xh′ξdξ+h0=∫0xh′ξdξ+∫01rxhxdx.
Sincex≤1, we have
(18)hx≤∫01h′xdx+∫01rxhxdx.
Hence,(19)h2x≤2∫01h′xdx2+∫01rxhxdx2.
Using the Cauchy-Schwarz inequality, we obtain(20)h2x≤2∫01h′2xdx+∫01r2xdx∫01h2xdx.
Consequently,(21)∫01h2xdx≤2∫01h′2xdx+∫01r2xdx∫01h2xdx.
The proof is complete. Similarly, forh1=∫01rxhxdx, we also note that
(22)−hx=∫x1h′ξdξ−h1=∫x1h′ξdξ−∫01rxhxdx.
Then, the argument is similar to the proof of the above.☐Proposition 2.
If (8)-(13) hold, then there exists a constant L>0 such that for any x∈0,1 and any solution u to Eq.(6), we have
(23)max0≤x≤1ux+max0≤x≤1u″x≤L.Proof.
Multiplying both sides of the first equation of (15) by φxu′, where φx=1−x and integrating the resulting equation from 0 to 1, then employing integration by parts with φ0=1,φ1=0 and φ′x=−1, we obtain
(24)12∫01u′x2dx−12u′02=∫011−xvxu′xdx.
Taking into accountu′0=∫01pxuxdx, we have
(25)12∫01u′x2dx=12∫01pxuxdx2+∫011−xvxu′xdx.
The integrals∫01pxuxdx and ∫011−xvxu′xdx can be estimated by means of the Cauchy-Schwarz inequality
(26)∫01pxuxdx2≤∫01p2xdx∫01u2xdx,∫011−xvxu′xdx≤∫01v2xdx1/2∫01u′x2dx1/2.
Thus,(27)12∫01u′x2dx≤12∫01p2xdx∫01u2xdx+∫01v2xdx1/2∫01u′x2dx1/2.
Applying now Lemma1 to the functions u and v appearing in the right-hand side of this inequality with u′0=∫01pxuxdx and v1=∫01qxvxdx, respectively, we obtain
(28)12∫01u′x2dx≤∫01p2xdx1−2∫01p2xdx∫01u′x2dx+21−2∫01q2xdx1/2∫01v′x2dx1/2∫01u′x2dx1/2.
It follows that(29)12−∫01p2xdx1−2∫01p2xdx∫01u′x2dx1/2≤21−2∫01q2xdx1/2∫01v′x2dx1/2.
Consequently,(30)12−∫01p2xdx1−2∫01p2xdx2∫01u′x2dx≤21−2∫01q2xdx∫01v′x2dxprovided 1/2−∫01p2xdx/1−2∫01p2xdx>0; that is, 1−4∫01p2xdx>0.
It follows that(31)∫01u′x2dx≤C1∫01v′x2dx,where
(32)C1=2/1−2∫01q2xdx1/2−∫01p2xdx/1−2∫01p2xdx2=2/1−2β1/2−α/1−2α2.
Proceeding as before, multiplying both sides of the second equation of (15) by ψxv′, where ψx=−x and integrating the resulting equation from 0 to 1, then employing integration by parts, taking into account ψ0=0,ψ1=−1,ψ′x=−1 and the nonlocal boundary condition v1=∫01qxvxdx, we obtain
(33)12∫01v′x2dx=12∫01qxvxdx2−λ∫01xfx,u,u′,v,v′v′xdx.
Applying the growth condition (8) to fx,u,u′,v,v′ by assuming that ax≤a,bx≤b,cx≤c,dx≤d,ex≤e,∀x∈0,1 with a,b,c,d,e>0, we obtain
(34)12∫01v′x2dx≤12∫01q2xdx∫01v2xdx+aλ∫01uxv′xdx+bλ∫01u′xv′xdx+cλ∫01vxv′xdx+dλ∫01v′x2dx+λ∫01exv′xdx.
The integrals appearing in the right-hand side of this inequality can be estimated by means of theε−inequality:
(35)∫01AxBxdx≤1ε∫01A2xdx+ε∫01B2xdx,ε>0.
Thus,(36)12∫01v′x2dx≤12∫01q2xdx∫01v2xdx+aλε1∫01u2xdx+aλε1∫01v′x2dx+bλε2∫01u′x2dx+bλε2∫01v′x2dx+cλε3∫01v2xdx+cλε3∫01v′x2dx+dλ∫01v′x2dx+λε4∫01v′x2dx+e2λε4,εi>0,i=1,⋯,4.
Applying Lemma1 to the functions u and v, we obtain
(37)12∫01v′x2dx≤∫01q2xdx1−2∫01q2xdx∫01v′x2dx2aλε11−2∫01p2xdx∫01u′x2dx+aλε1∫01v′x2dx+bλε2∫01u′x2xdx+bλε2∫01v′x2dx+2cλε31−2∫01q2xdx∫01v′x2dx+cλε3∫01v′x2dx+dλ∫01v′x2dx+λε4∫01v′x2dx+e2λε4,εi>0,i=1,⋯,4..
From (31), the inequality (37) becomes
(38)12∫01v′x2dx≤∫01q2xdx1−2∫01q2xdx∫01v′x2dx+2aλC1ε11−2∫01p2xdx∫01v′x2dx+aλε1∫01v′x2dx+bλC1ε2∫01v′x2dx+bλε2∫01v′x2dx+2cλε31−2∫01q2xdx∫01v′x2dx+cλε3∫01v′x2dx+dλ∫01v′x2dx+λε4∫01v′x2dx+e2λε4.
Choosingεi=1,i=1,⋯,3,ε4=M, and using the hypothesis (10) with M=maxa,b,c,d,e, we obtain
(39)12Δ≤δ21−2δ2Δ+λ2MC11−2δ1+MC1+5M+2M1−2δ2Δ+Mλ,where Δ=∫01v′x2dx. Note that C1≥8. This gives
(40)λ0=1/2−δ2/1−2δ2M2C1/1−2δ1+2C1+2/1−2δ2≤1/2−δ2/1−2δ2M2C1/1−2δ1+C1+5+2/1−2δ2.
Let(41)γ=12−δ21−2δ2−λ2MC11−2δ1+MC1+5M+2M1−2δ2.
We see thatγ>0. Thus,
(42)∫01v′x2dx≤Mλγ=C2and consequently from (31) and (42), we obtain
(43)∫01u′x2dx≤C1C2.
On the other hand, we have(44)ux=∫0xu′ξdξ+u0=∫0xu′ξdξ+∫01pxuxdx.
Thus,(45)ux≤∫01u′x2dx1/2+∫01p2xdx1/2∫01u2xdx1/2.
Hence,(46)ux≤C1C21/2+2∫01p2xdx1/2∫01p2xdx1−2∫01p2xdx1/2C1C21/2=L1.
Similarly,(47)vx≤L2=C21/2+2∫01q2xdx1−2∫01q2xdx1/2C21/2.
These two inequalities imply the required result and complete the proof of the proposition.☐The fundamental theorem used in proving the existence of the solution is Schauder’s fixed theorem. In order to make use of this theorem, it is sufficient to present the following lemmas.Lemma 3.
Letg:0,1⟶ℝ be a continuous function. The unique solution z of the following initial value problem
(48)z″=hxsubject to the nonlocal boundary conditions z0=z′0=∫01rxzxdx, is given by
(49)zx=∫01G1x,yhydy+∫01G2x,yhydy,where G1x;y is the Green function given by
(50)G1x,y=x−y,0≤y≤x≤1,0,0≤x≤y≤1G2x,y=x+11−∫01x+1rxdx∫01rxG1x,ydx.Proof.
Integrating this equation twice, we obtain(51)zx=∫0x∫1yhsdsdy+k1x+k2,where ki,i=1,2 are constants of integration. Integrations by parts of the integral with respect to y in this equation give
(52)zx=−x∫x1hydy−∫0xyhydy+k1x+k2.
We determinek1 and k2 from z0=z′0=∫01rxzxdx. It follows that
(53)zx=∫0xx−yhydy+x+1∫01rxzxdx.
Multiplying both sides of this equation byrxzx and integrating the resulting from 0 to 1, we obtain
(54)∫01rxzxdx=11−∫01x+1rxdx∫01rx∫0xx−yhydydx.
Thus,(55)zx=∫0xx−yhydy+x+11−∫01x+1rxdx∫01rx∫01rx∫0xx−yhydydx,(56)zx=∫01G1x,yhydy+x+11−∫01x+1rxdx∫01rx∫01rx∫01G1x,yhydydx.
The proof is complete.☐We also have the following lemma.Lemma 4.
Letg:0,1⟶ℝ be a continuous function. The unique solution z of the following initial value problem
(57)z″=hxsubjects to the nonlocal boundary conditions z1=z′1=∫01rxzxdx, is given by
(58)zx=∫01G3x,yhydy+∫01G4x,yhydy,where G3x;y is the Green function given by
(59)G3x,y=0,0≤y≤x≤1,y,0≤x≤y≤1,G4x,y=x1−∫01xrxdx∫01rxG3x,ydx.
Thus, problem (15) is equivalent to the following system of integral equations
(60)u=∫01G1x,yvydy+∫01G2x,yvydy,v=λ∫01G3x,yfx,uy,u′y,vy,v′ydy+λ∫01G4x,yfx,uy,u′y,vy,v′ydy.
Define the Banach spaceX=C0,1×C0,1 with norm u,v∞=u∞+v∞, where u∞=max0≤x≤1ux. Also, define the operator T:X⟶X by Tu,v=T1u,v,T2u,v, where
(61)T1u,v=∫01G1x,yvydy+∫01G2x,yvydy,(62)T2u,v=λ∫01G3x,yfx,uy,u′y,vy,v′ydy+λ∫01G4x,yfx,uy,u′y,vy,v′ydy.Lemma 5.
Under the hypothesis of Proposition2, there exists Ki>0,i=1,2 such that
(63)max0≤x≤1T1u,v≤K1andmax0≤x≤1T2u,v≤K2.Proof.
SinceGix,y≤2,i=1,3,G2x,y≤∫01pxdx/1−∫01x+1pxdx=α∗ and G4x,y≤2∫01qxdx/1−∫01xqxdx=β∗,∀x,y∈0,1, thus,
(64)T1u,v≤∫01G1x,yvydy+∫01G2x,yvydy.
Usingvx≤L2, we get
(65)T1u,v≤2+α∗L2=K1.
ForT2u,v, we have
(66)T2u,v≤λ∫01G3x,yfx,uy,u′y,vy,v′ydy+λ∫01G4x,yfx,uy,u′y,vy,v′ydy,(67)T2u,v≤λ2+β∗∫01fx,uy,u′y,vy,v′ydy.
Applying the growth condition,(68)∫01fx,uy,u′y,vy,v′ydy≤a∫01uydy+b∫01u′ydy+c∫01vydy+d∫01v′y+e,(69)∫01fx,uy,u′y,vy,v′ydy≤a∫01uydy+b∫01u′y21/2+c∫01vydy+d∫01v′y21/2+e,and employing the following inequalities that used in the proof of Proposition 2:
∫01v′x2dx≤C2,∫01u′x2dx≤C1C2,ux≤L1, and vx≤L2, we obtain
(70)∫01fx,uy,u′y,vy,v′ydy≤ML1+C1C212+L2+C212+1.
Hence,(71)T2u,v≤λ2+β∗ML1+C1C21/2+L2+C21/2+1=K2.
Consider now the closed and convex set(72)S=u,v∈X:u,v∞≤K1+K2=K.☐Based on the above results, we haveLemma 6.
For anyu,v∈S,Tu,v is contained in S.Proof.
From Lemma5, we have T1u,v≤K1 and T2u,v≤K2. Since Tu,v=T1u,v,T2u,v, we obtain Tu,v≤K1+K2=K. This shows that Tu,v is contained in S.
To prove thatTu,v is compact, we use the Arzela-Ascoli lemma; that is, TS must be closed, bounded, and equicontinuous. Consequently, Tu,v has a fixed point by the Schauder’s fixed point theorem.☐Thus, we have the following theorem.Theorem 7.
There exists a continuous solutionu,v which satisfies problem (15) with (8)-(13).


## 3. Conclusion

The cantilever beam problem modeled by the nonlinear fourth-order equation (6) is investigated under the integral conditions (7) and assuming the conditions (8)-(13). As illustrated in Section 2, the proposed conditions stand for perturbed conditions at the boundary, which occurs if the cantilever beam is not perfectly cantilevered in the sense that the free end is rolled and/or the fixed end is loose. The integral condition generalizes the standard boundary conditions that are usually proposed in literature for a classical cantilever beam problem. Moreover, these boundary conditions are more practical, in the sense that they represent the actual conditions that may arise in a real-world cantilever beam, which is very useful to researchers in construction engineering. The objective of this research is to determine whether small deflections occur in the cantilever beam under perturbed boundary data. The result shows that the solution to the problem exists, which implies that small deflections continue to exist on the beam whether it is perfectly cantilevered or not.



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