Derive Equation for Flexural Strain 2

[BN76]

Hi!
I am reading standards for setting 3pt bending test. I read ISO 178 and ASTM D790-10. I would like to know how the flexural strain equation is derived? The equation is given as:

Flexural-Strain, e = 6sh/L^2
where
s: measured deflection
h: rectangular bar thickness
L: length between two support.

Thank you.

 

[dhengr]

BN76:
That formula is correct for a rectangular section and a point load at the center of the beam, but it is also a beautiful example of obfuscation and algebraic manipulation, in that it completely clouds any understanding of the whole problem, and how you get there. It is really a fairly simple engineering problem. I would suggest you dig out of your old Statics/Engineering Mechanics and Strength of Materials text books, and arrive at that result yourself so you have a vague idea what you are doing in this testing program. Check out the bending moment, bending stress, deflection, and Hooke’s law, and do the algebra. You might also want to talk with your boss or a senior engineer in your company, so the company knows what you know or don’t know, as you set this test up. They can help you through this problem, they have a vested interest in seeing you get it right and being successful.

 

[Cockroach]

It is derived from classical beam theory. A simply supported beam has a central load imparting deflection. The cross sectional geometry of the beam is rectangular. The strain energy is simply that, deformation over the original length.

By asking for a derivation, you mean you want it laid out in detail?

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[Cockroach]

Attached is the proof regarding the issue at hand, derivation for the Flexural Stiffness Equation. This is for the specifications ISO 178 or ASTM D790-10 only. Reason for that is how the Fracture Toughness specimen is prepared, namely the beam depth to length is one and a half.

I've chosen singularity equations in order to model the deflection of a mid span point load on a simply supported beam. This is the understanding of the mentioned specifications, the supports must be placed four times the beam depth apart.

What is found is the general equations governing a simply supported beam with load at midspan. From there I have added the constraint of length to depth ratio dictated by the specification. The rest is just mathematics.

Hope this helps you out. I've noticed you posted this same question on "Engineers Edge", really hope it is not some sort of homework problem.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[navaneekrnk]

The standard also has a different equation for maximum stress for large deflection (of the order of thickness):
max stress = 3PL/2bd^2 (1+6(D/L)-4(d/L)(D/L)),
Where,
P: Load Applied, L: Support span, b: width, d: thickness, D: Beam deflection.
How the right part of the equation in paranthesis was derived???

Thank you,
Regards,
Navanee

 

[BN76]

Dear Kenny,
I really appreciate that you spent time showing me how the flexural strain is derived in your neatly written sheet. It clears out some of the confusion I had when I was working that by myself.
One thing that is not very clear to me is that both the standards specified the "span-to-depth" ratio of 16 (SS 7.2.1 of D790-10), why did the standards uses h/L = 3/2?
Regarding your question, I did post this problem on another forum as this is my 1st use of professional forum. I am not sure which forum will get professional responses. I am trying to understand how flexural strain is derived so that I can use it to plot a stres-strain curve of a "one end fixed, one load applied at the opposite end" cantilevered beam. I conducted a series of testing in that setup and want to estimate the force that my specimen (small and short Ti rod)reaches its' yield strength.

Nevertheless, I really appreciate your help.

 

[Cockroach]

I think, and I could be wrong, that h/L = 1.5 is spelled out only to keep the specimen uniform between testing labs. You would want to have identical specimens and a well defined test procedure in order to get consistant and comparable results between testing centres.

This is typical of tensile load specimens with API which spell out exactly where the material is to be obtained from the forging and what exact dimensions it should have. This one is more close to the Fracture Toughness procedure, I noted that even the notch was well defined by the literature.

Afterall, having different specimen dimensions would greatly influence the results. Then how could specification uniformity be correlated?

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[zekeman]

The basic equation for strain of a beam at any point is

y/r

which is derived in any strength of materials text

where

y= distance to neutral axis
1/r= curvature = M/EI from beam theory

So
e=y/r=yM/EI

Now for a simply supported 3 point beam with point force at center,the deflection is

Delta=PL^3/48EI

Dividing the 2 equations

e/delta=48yM/PL^3

But the moment M for this beam at the center is
PL/4
After substituting
e=12*delta*y/L^2

Now if you choose the outer edge for the strain

y=h/2, and you get your formula,

y=6*delta*h/L^2

Note that there is no assumption of a rectangular beam to get this result or any of the ISO or ASTM papers.

 

[Cockroach]

If I have a load P at midspan, L/2, then the moment at the support is PL/2. You (Zekeman) has mentioned PL/4.

The literature calls for "average force" which is half the load. So I believe that Zekeman implies "average moment" for the beam with centre load, then yeah, PL/4. The specifications are somewhat worded differently but get to the same result.

I found another similar specification which dicated distance between supports to be four times the beam width. This again leads to the same Flexural Strain equation, albeit slightly altered mathematical model.

But I prefer Zekeman's solution, it is quite a bit cleaner and falls out of the classical beam deflection theory.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[rb1957]

sorry kenneth but ?

point load P at mid-span, L/2.
reaction P/2
for SS, maximum moment = PL/4, no?

for fixed, moment at the support is PL/8, no?

 

[hydtools]

Max moment is PL/4 for simply supported beam. Reaction at support is P/2. Moment arm is L/2. P/2 x L/2 = PL/4.

There is no moment at the support of a simply supported beam.

Ted

 

[Cockroach]

Just the way it was worded. Yes, the left and right support reaction s are identical at P/2. I have P directed down at P at the mid point of the beam. So the maximum bending moment of the system is PL/2.

Look at the shear and bending moment diagrams one pg 1 of my original attachment. The shear is order zero and therefore +P/2 from the left support thru to mid span. Then it jumps down by P crossing mid span to give me the reaction at right hand support, -P/2. So correct, the end supports are P/2.

This means the moment diagram is first order or linear. The moment is zero at left support, ramps up to PL/2 at mid span and then linear decreasing to zero at right support. So we have the tee-pee with maximum moment of PL/2.

Check this by moment area method using the shear diagram. Damn it......found it.......(P/2))(L/2)=PL/4. Arithmetic was never my strong suit.

Correct, PL/4. I got the wrong number on the bending moment diagram. Should be PL/4, not PL/2. My apologies!

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[SNORGY]

No worries, you just forgot to factor in the fourth law of thermodynamics.
It was the heat of the moment.

Regards,

SNORGY.

 

[Cockroach]

LOL....

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada