General Results - Boundary Value Problem: Formulation - Civil Engineering (CE) PDF Download

General results 

In this section, we record two results that are useful while solving boundary value problems. One result tells when one can expect an unique solution to a given boundary value problem. The other allows us to construct solutions to complex boundary conditions from simple cases.

Uniqueness of solution

Now, we are interested in showing that there could at most be one solution that could satisfy the prescribed displacement or mixed boundary conditions, in a given body made of a material that obeys Hooke’s law and in static equilibrium with no body forces acting on it. If traction boundary condition is specified, we shall see that the stress is uniquely determined but the displacement is not for the bodies made of a material that obeys Hooke’s law and in static equilibrium with no body forces acting on it. Towards this, we consider a general setting with B being some region occupied by the body and ∂B the boundary of the body. Let us also assume that displacement is prescribed over some part of the boundary ∂B_u and traction specified on the remaining part of the boundary, ∂B_σ. Let (u₁, σ₁) and (u₂, σ₂) be the two distinct solutions to a given boundary value problem. Let us define

u = u₁ − u₂, σ = σ₁ − σ₂.                 (7.108)

By virtue of (u₁,σ₁) and (u₂, σ₂) satisfying the specified boundary conditions,

u(x) = o, for x ∈ ∂B_u,                    (7.109)

t(_n)(x) = σ_n = o, for x ∈ ∂B_σ.           (7.110)

First, we examine the term,
 

W = σ · grad(u) = σ ·€,               (7.111)

here to obtain the last equality we have used (2.104) and the fact that the Cauchy stress σ is a symmetric tensor. Since, we are interested in materials that obey isotropic Hooke’s law, we substitute for the stress from⁵

in (7.111) to obtain

As recorded in table 6.1 G > 0 and K > 0. Further recognizing that irrespective of the sign of the Cartesian components of the strain,

on assuming that € 6= 0. Thus, since each term in equation (7.113) is positive we infer that

              W > 0,                               (7.115)

as long as ∈ ≠ 0.
Next, we want to express

in terms of the boundary conditions alone. Towards this, using the result in equation (2.219) we write

Since the body is in static equilibrium and has no body forces acting on it, from equation (7.6) div(σ) = o. Using div(σ) = o in equation (7.117) and substituting the result in (7.116) we get

where to obtain the last equality we have used the result (2.267). Substituting the conditions (7.109) and (7.110) on the displacement and stress, in (7.118) that

Since, from equation (7.115) the integrand in the equation (7.119) is positive and W = 0 only if € = 0, it follows that  = 0 everywhere in the body.Then, it is straight forward to see that σ = 0, everywhere in the body. It can then be shown thet, we do this next for a special case , the conditions € = 0 everywhere in the body and (7.109) imply u = o everywhere in the body.
Integrating the first order differential equations on the Cartesian components of the displacement field when  = 0, we obtain

u = (K₁y+K₂z+K₄)e_x+(−K₁x+K₃z+K₅)e_y−(K2x+K₃y−K₆)e_z,                          (7.120)

where K_i ’s are constants. Knowing the displacement of 3 points, say (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃), that are not collinear to be zero, the constants K_i ’s could be uniquely determined by solving the following system of equations:

As long as the three points are not collinear, it can be seen that the above set of equations are independent and hence, K_i = 0. However, if the displacement is not known at 3 points, as in the case of traction boundary condition being specified, the displacement field cannot be uniquely determined.
Hence, we conclude that the solution to the boundary value problem involving a body in static equilibrium, under the absence of body forces and made of a material that obeys isotropic Hooke’s law is unique except in cases where only traction boundary condition is specified. As a consequence of this theorem, if a solution has been found for a given boundary conditions it is the solution for a body in static equilibrium, under the absence of body forces and made of a material that obeys isotropic Hooke’s law.

Principle of superposition 

This principle states that For a given body made up of a material that obeys isotropic Hooke’s law, in static equilibrium and whose magnitude of displacement is small, if {u (¹) ,σ (¹)} is a solution to the prescribed body forces, b (¹) and boundary conditions,  and {u (²) ,σ (²)} is a solution to the prescribed body forces, b (²) and boundary conditions,  then

{u (¹) + u (²) ,σ (¹) + σ (²)} will be a solution to the problem with body forces, b (¹) + b (²) and boundary conditions,  The proof of the above statement follows immediately from the fact that equation (7.23) is linear. That is,

where we have used the linearity property of the grad, div and ∆ operators (see section 2.8). This is one of the most often used principles to solve problems in engineering.