Dynamics of axially loaded bar

svishal03

Hi,

I have a simple question and shall be grateful if helped.

See the attached word document.I have been reading the derivation of the consistent mass matrix- for the axially loaded bar.My question is:

1)While formng the equation 10.51 (referring figure 10.7), the author considers the equilibrium of the differential element of length dx.

But the forces on each end ofthe differential element can be different only if there is some load between the two ends.

IS this some kind of error that the author David Hutton of the concerned text on Fundamentals of Finite Element analysis has not considered the force(s) between the ends?

Please help.

Vishal

the_emi_guy

The author is assuming some axial loading initial condition. It is this loading that causes the non-zero net force on the differential elements. A common example would be striking the end of the bar with a hammer. The speed with which the stress wave propagates down the rod is determined using the one dimensional wave equation (10.52) and equals [itex] \sqrt{E/ρ} [/itex].

AlephZero

The bar can move. The equation is just Newton's second law applied to the small element. Force = mass times acceleration.

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the_emi_guy	The author is assuming some axial loading initial condition. It is this loading that causes the non-zero net force on the differential elements. A common example would be striking the end of the bar with a hammer. The speed with which the stress wave propagates down the rod is determined using the one dimensional wave equation (10.52) and equals [itex] \sqrt{E/ρ} [/itex].

AlephZero $Math 2012$ Recognitions: Science Advisor	The bar can move. The equation is just Newton's second law applied to the small element. Force = mass times acceleration.