User:Eml5526.s11.team4.cook/HW1

= Dynamic Response =

Problem Statement


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

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 stresses, $$N(x)$$, 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 $$A(x)\rho(x)=m(x)$$

Equation (1.1) has been derived.