iMechanica - Time dependent vs Independent - Comments

Thank you Biswajit!

HariKrishna — Mon, 16 Jun 2008 09:39:32 -0400

Thank you Biswajit!

re: Strain rate

Biswajit Banerjee — Fri, 13 Jun 2008 01:23:26 -0400

Consider a test where you put a specimen in a loading machine and apply a displacement. The displacement starts at u=0 at t=0 and reaches a value of u=d_1 at t=t_1. If the plot of u vs. t is linear, i.e. u = (d_1/t_1) t, then the rate at which the displacement is applied (du/dt) is constant and equal to d_1/t_1. This is the displacement rate.

Associated with the displacement is a strain e=du/dx (say). For a linear elastic material, at t=0 the strain is e=0 and at t=t_1 the strain is e=d_1/L The strain rate is the quantity de/dt = d_1/(t_1 L).

For plastic deformations we use a similar idea of incremental linearity and can therefore calculate a strain rate associated with small increments of time. Read Hill's book on the theory of plasticity for more details.

-- Biswajit

Re:Time dependent vs Independent

HariKrishna — Thu, 12 Jun 2008 13:02:47 -0400

Can Someone please point me to some source where I can read about this?

Thank you

Time dependent vs Independent

HariKrishna — Wed, 11 Jun 2008 10:33:36 -0400

In Computational mechanics, for rate(time) dependent calculations (eg. Creep, plasticity), when we say time step dt, does it mean physical time.

Can I think this way:

I start at time 0 and calculate stress, strain etc after 1 sec, 2 sec, 3 sec..

Also, what exactly is strain rate? why do some books say ..'when a strain rate is applied'''..I'm confused..dont we just apply displacement or force all the time.

Does it mean that..the expression of strain rate gives how the strain evolves with time...(eg; In creep, under constant load, how the strain evolves with time)

Kindly throw somelight on these..I'm new to computational mechanics..

Harikrishna