User:Eml5526.s11.team4/HW1

= Problem 1.1 - Dynamic Response =

Problem Statement


1.1 Derive the following dynamic response of an elastic bar with variable cross-section such as the one shown in the figure

subject to the boundary conditions

and initial conditions

1.2 Repeat for a rectangular cross section

Solution
Starting with D'Alembert's principle,

where




 * $$\mathbf {F}_i$$ || are the applied forces,
 * $$\delta \mathbf r_i$$ || is the virtual displacement of the system, consistent with the constraints,
 * $$ m_i \scriptstyle$$ || are the masses of the particles in the system,
 * $$\mathbf a_i$$ || are the accelerations of the particles in the system,
 * $$m_i \mathbf a_i$$ || together as products represent the time derivatives of the system momenta, and
 * $$i$$ || is an integer used to indicate (via subscript) a variable corresponding to a particular particle.
 * }
 * $$\mathbf a_i$$ || are the accelerations of the particles in the system,
 * $$m_i \mathbf a_i$$ || together as products represent the time derivatives of the system momenta, and
 * $$i$$ || is an integer used to indicate (via subscript) a variable corresponding to a particular particle.
 * }
 * $$i$$ || is an integer used to indicate (via subscript) a variable corresponding to a particular particle.
 * }

The three force contributions are $$N(x+dx)$$, $$N(x)$$, and $$f(x)dx$$. There is one body, the differential body with a mass $$ A(x) \rho(x) dx$$. Note that the differential body mass is approximated and has an error before the limit is taken.

dividing by $$dx$$

Taking the limit

Noting that $$dx$$ changes from the finite difference as shown in the figure to a differential difference.

Dividing the normal force, $$N(x)$$, by the area, $$A(x)$$, yields the stress $$\sigma(x)$$.

In one dimension the strain is

Bringing in Hooke's law

From (1.7) and (1.9)

Substituting into (1.6)

where $$A(x)\rho(x)=m(x)$$ and $$ m(x) $$ is the linear mass density.

Equation (1.1) has been derived.

Problem Statement


Discuss the specific case where the elastic bar from homework problem 1.1 has a rectangular cross section. The cross section is defined as shown in the figure below, where the height $$h(x)$$ varies with axial position. The base $$b$$ is constant for the entire bar.

Solution
Starting with Equation 1.6 which is in a general form,

the area term $$A(x)$$ can be modified for the more specific case of a rectangular cross section. The cross sectional area is given by:

Using the equation for the area of a trapezoid, the mass per unit length $$m(x)$$ can be written as:

where $$\rho$$ is the density. The mid-point of the differential element is used to approximate the density.

Given that:

It should be noted that the normal force terms $$N(x)$$ and $$N(x+dx)$$ contain an area dependence as shown previously in Equations 1.7-1.9. With this in mind, Equation 1.9 and Equation 2.2 can be substituted together into Equation 1.6 to produce:

which is the force balance for an elastic bar with a rectangular cross section.

Again taking the limit:

produces the following Equation 2.6. As noted previously in problem 1, $$dx$$ changes from the finite difference to a differential difference.

Using Equation 1.7 and Equation 1.12, Equation 2.7 is produced.

A simple back substitution of Equation 2.1 into Equation 2.7 yields:

Where $$ A(x)\rho(x)=m(x)$$, Equation 1.1 has again been derived.