Previous in Forum: Gas Compressor Arrangement   Next in Forum: AIRFOILED BLADES
Close
Close
Close
7 comments
Rate Comments: Nested
Associate

Join Date: Apr 2008
Posts: 27

Formula for calculating force to bend a metal beam?

11/03/2008 5:04 PM

How do I calculate how much force is required to deflect a tiny beam of spring steel .020" when force is applied to the center of the beam. The beam is .550" long and .100" wide and about .007" thick.

Unfortunately I don't know a lot more about the spring steel characteristics.

Online references would be great.

Thanks,

Dave

Register to Reply
Interested in this topic? By joining CR4 you can "subscribe" to
this discussion and receive notification when new comments are added.

Good Answers:

These comments received enough positive votes to make them "good answers".
4
Guru

Join Date: Nov 2007
Location: Sherwood Park, Alberta, Canada
Posts: 1212
Good Answers: 74
#1

Re: Formula for calculating force to bend a metal beam?

11/03/2008 6:26 PM

Δ = deflection

L = span

E = Modulus of Elasticity (I assumed 30,000,000 psi)

I = Moment of Inertia

b = width

d = depth

Δ = PL3/(48EI)

I = bd3/12 for a rectangular cross section

I think you will need about 1/2 pound to deflect the beam 0.02". This should produce a maximum flexural stress of about 840 psi at midspan (not very much).

__________________
Bruce
Register to Reply Good Answer (Score 4)
Associate

Join Date: Apr 2008
Posts: 27
#2
In reply to #1

Re: Formula for calculating force to bend a metal beam?

11/03/2008 7:29 PM

Thank you Bruce. That's about the force I was looking for. Where did you come up with 30,000,000 psi? Any chance you can list the units used in this equations?

- Dave

Register to Reply
6
Guru

Join Date: Nov 2007
Location: Sherwood Park, Alberta, Canada
Posts: 1212
Good Answers: 74
#3
In reply to #2

Re: Formula for calculating force to bend a metal beam?

11/03/2008 8:37 PM

The modulus of elasticity doesn't vary much among steels. I looked up the properties of spring steel and it listed E as 210,000 MPa (or about 30,450,000 psi).

The units fall into place naturally. They can be metric or Imperial. Since you provided the width and depth of the beam in inches, I worked in Imperial.

So b = 0.1", d = 0.07", L = 0.55", I = 2.858e-9 in4, E = 30,000,000 psi (pounds per square inch). Deflection Δ is in inches and P, the force applied at midspan is in pounds.

If all of these terms had been entered in metric units, i.e. Newtons and millimeters or combinations thereof, P would have been given in Newtons.

The expression Δ=PL3/(48EI) may be rewritten as P=Δ*(48EI)/L3 to solve for P where P is the force required to cause a deflection of Δ.

So P=0.02(48*30e6*2.858e-9)/0.553 = 0.495# (say 1/2 pound).

The units of P must be pounds because all the terms had consistent units and the unit of force was pound.

It is always wise to do a dimensional analysis of an expression to make sure that the units work out to what you expect. Using the letters F and L for force and length, the expression for P is dimensionally L*(F/L2*L4)/L3 which is F*L5/L5 which is F. So P will be in force units which is what we expect.

The expression for Moment of Inertia, bd3/12, dimensionally is L*L3 or L4. So "I" will be in in4 (or mm4 if you work in metric units).

Hope this helps in your understanding.

__________________
Bruce
Register to Reply Good Answer (Score 6)
Associate

Join Date: Apr 2008
Posts: 27
#4
In reply to #3

Re: Formula for calculating force to bend a metal beam?

11/04/2008 10:21 AM

Excellent Bruce.

In case anyone cares, there was one typo (that didn't make it into the equations): d = .007" (not .07").

Thanks again,

Dave

Register to Reply
Guru

Join Date: Nov 2007
Location: Sherwood Park, Alberta, Canada
Posts: 1212
Good Answers: 74
#5
In reply to #4

Re: Formula for calculating force to bend a metal beam?

11/04/2008 1:43 PM

Actually, it did make it into the equations because the moment of inertia, I = bd3/12 includes both b and d. And the expression for P includes I. When you think about it, the beam dimensions must have something to do with deflection.

As it turns out, I used a depth of beam of 0.007" when checking deflections but I used a depth of 0.07" when calculating bending stress and incorrectly reported a maximum stress of 840 psi (my bad). The maximum stress turns out to be 84,100 psi which is a substantial stress.

To convert to psi, multiply by 145. The yield point of this steel is very high (1100 MPa or about 160,000 psi) so a stress of 84,000 psi represents approximately half of its yield value.

__________________
Bruce
Register to Reply
Power-User

Join Date: Apr 2007
Posts: 270
Good Answers: 19
#6

Re: Formula for calculating force to bend a metal beam?

11/05/2008 5:42 AM

This is not a linear calculation and the beam theory that you are applying is inadequate.

Deflection is 20/550=1/27.

Depth to span is 7/550= 1/79

Both of these would indicate that beam theory is an approximation.

Further questions would be: what are the support conditions? Are they fixed and does catenary action play a part (at 1/27 deflection this is the case). Does the way the spring is attached give rise to spring support conditions which increases the difficulty. How accurate does the deflection have to be?

__________________
omw7
Register to Reply
Guru

Join Date: Nov 2007
Location: Sherwood Park, Alberta, Canada
Posts: 1212
Good Answers: 74
#7
In reply to #6

Re: Formula for calculating force to bend a metal beam?

11/05/2008 11:58 AM

You are quite right. The deflection is approximate as most structural calculations are. It is tantamount to finding that a simple beam having a span of 27'-6" deflects a foot at midspan. That is a bit much for small deflection theory.

Boundary conditions are assumed to be hinge at one end and roller at the other, i.e. a simple beam. If the ends are restrained, either horizontally or rotationally, the deflection will be considerably less. Alternatively, it will take a larger force to produce a stipulated deflection.

__________________
Bruce
Register to Reply
Register to Reply 7 comments

Good Answers:

These comments received enough positive votes to make them "good answers".
Copy to Clipboard

Users who posted comments:

ba/ael (4); Non-ME (2); omw7 (1)

Previous in Forum: Gas Compressor Arrangement   Next in Forum: AIRFOILED BLADES

Advertisement