Convert loading velocity rate to strain rate on 3-point bending tests

[Efstia]

Hi,

I would like to ask if anyone knows how to convert 100 cm/sec (loading velocity) to strain rate which units are strain per second.

Thanks a lot.

 

[Ron]

The load rate you are noting is absurdly high. This would constitute shock loading.

For most flexure tests the load application is given in terms of the extreme fiber stress rate, usually less than about 200 psi/min.

 

[Efstia]

You are right. The loading rate is for impact loading actually. I am reading a paper based on impact properties of concrete.

They claim that the loading velocity of 100 cm/sec corresponds to 1.2 strain per second. I would like to know how they did work out the strain rate in an impact event.

Thanks a lot.

 

[BAretired]

I just noticed that this thread is included twice in this forum. You should remove the other one.

What is meant by loading velocity? The load falls on the beam with the stated velocity at impact, is that correct? As the beam absorbs the impact by deflecting, the velocity of load decreases to zero, then reverses as the beam recovers (assuming it remains in the elastic range).



BA

 

[rb1957]

maybe the idea is that the pointing point is moving (in the direction of the load) at 0.1m/s

then as the beam deflection increases by 0.001m there'll be an increase in strain in the beam (max bending stress > strain) and this should be linear.

Quando Omni Flunkus Moritati

 

[BAretired]

The pointing point?

BA

 

[rb1957]

where the impact is happening

Quando Omni Flunkus Moritati

 

[BAretired]

I confess I don't understand the original question. Something is moving at a rate of 100 cm/sec. In so doing, it causes a change in the strain in the fibers of the beam. The unit strain is greater in the outer fibers than those closer to the neutral axis but the rate of change of unit strain is the same for all fibers.

It is not clear to me how 1.2 radians per second is derived from a loading rate of 100 cm/sec.



BA

 

[Efstia]

The loading velocities were: 4.23 x 10^-3, 0.846, 70 and 100 cm/sec. Corresponding strain rates were 0.5 x 10^-5, 0.01, 0.8 and 1.2 strain/sec.

The paper provides beam dimensions of 12.5 mm (thick), 75 mm (wide) and 300 mm (long).

The tests with the smaller velocities were used normally under flexural test. The other two higher velocities were performed using a drop-weight tower.

I don't know how did they work out the strain rates. What I have in mind is that they used strain gauges attached on the beams to record the strains during the impact event and work out the strain rate from the strain versus time plot.

What do you think?

 

[rb1957]

bending strain is proportional to load point deflection;
strain rate is proportional to load point deflection rate.

for an assumed deflection, you can calculate the bending strain.

together with a known loading rate, you can calculate the strain rate.

note, very high loading rates (impacts) have different internal strains compared to slow loading rates, so that's the tricky bit to calc.

i had expected the slow rate loadings to have a proportional strain rate, but they seem to be out by a factor of 10 ?
4.23E-3 5E-6 ... 4.23/5E-3 = 846
0.846 0.01 ... 0.846/0.01 = 84.6 ?
70 0.8 ... 70/0.8 = 87.5 (round-off?, 70/0.827 = 84.6)

Quando Omni Flunkus Moritati

 

[BAretired]

Well, I thought I found the answer but now I'm not so sure. Here is what I did:

The load point is moving down at a speed of 10 cm/sec under an instantaneous load P. P is not the actual load applied...it is the effective load at time t (the instantaneous load).

At time t, the bending moment is PL/4 if the load is applied at midspan. The instantaneous stress at any distance y from the neutral axis is PL.y/4I and the unit strain e_y at the 'y' fiber is PL.y/4EI.

The instantaneous deflection of the beam is PL³/48EI.

But e_y = PL.y/4EI = Δ.y/12L²

If dΔ/dt = 10 cm/sec
then de_y/dt = 10y/12L²
L = beam span and remains constant.
For any given fiber, y is constant.
so if dΔ/dt = 10 cm/sec then de_y/dt = 10/12 radians/sec = 1/1.2 rad/sec which is the inverse of what the OP stated.

Help!

BA

 

[BAretired]

Try again:
M = PL/4
f_y = My/I (stress at fiber y from N.A.)
e_y = My/EI = PL.y/4EI

Δ = PL³/48EI
e_y = Δ*12y/L²

dΔ/dt = 100 cm/sec = 1000 mm/sec
y = 12.5/2 = 6.25 mm
L = 300 mm
de/dt = 1000*12*6.25/300² = 0.8333 rad/sec

Still have a problem relating to the correct answer.

BA

 

[rb1957]

BA ... i wish you'd use "strain" as the dimensionless unit for, well, strain. i find your use of the dimensionless angle "radian" confusing.

Quando Omni Flunkus Moritati

 

[BAretired]

rb...Sorry for the confusion. I could be wrong, but so far as I know the word "strain" is not a recognized unit. Unit strain has no units, that is it is dimensionless, i.e. mm/mm or inch per inch.

I guess I could say that the strain rate de_y/dt = 0.8333 sec-1 but I chose to use radians/sec which is the same thing.

BA

 

[rb1957]

i think strain and microstrain are well understood engineering units. sec^-1 might be mathematically correct but it lacks something in meaning, no?

Quando Omni Flunkus Moritati

 

[BAretired]

Now that you mention it rb, I have heard the term microstrain. Okay, I concede the point. Read strain/sec for my radians/sec in my earlier post.

BA

 

[Efstia]

You can record microstrains using strain gauges and convert them to strains. By plotting strain versus time you have the strain rate. But, can you work out the strain rate from deflection versus time data?

 

[BAretired]

Deflection and strain are related as I have indicated above. If you know the deflection rate at time t under dynamic loading, then you should be able to calculate the strain rate at the same time (or vice versa).

BA

 

[rb1957]

i agree, BA's calc was what i had in mind. strain and deflection are proportional to one another.

Quando Omni Flunkus Moritati

 

[Efstia]

There is also a paper that they worked out the strain rate by applying the Hooke's law. So they got the peak load and by applying an E value, they calculated the strain rate; you need time to calculate the strain rate.