Force calculation to deflect music wire

The equation is:

d = F L³/(48EI)

In this case, I = pi*D⁴/64

So d = FL³*64/(48E*pi*D⁴)

or D = [64L³F/(48E*pi*d)]^0.25

For L=0.55, F=0.4, d=0.25.......D=0.0139"

Maximum Stress

M = 0.4*0.55/4 = 0.055"#

S = pi*D³/32 = 2.64e-7

f_max = M/S = 208,600 psi

This greatly exceeds the yield point of the steel, so it is not an acceptable solution.

What figure do you use for the yield point of ASTM A228 music wire? The website

http://www.matweb.com/search/datasheet.aspx?matguid=4bcaab41d4eb43b3824d9de31c2c6849&ckck=1

indicates a "Tensile Strength, Yield" of "231,000 to 399,000". Am I looking at the wrong property?

To GUEST question:

The wire is unattached to anything. It is freely supported by 2 'ledges' that are .55" apart. It is not under tension.

Thank you,

Dave

The wire must be attached to something.

Or else it wouldn't hold any force?!?!?!

You say it is "free supported" by two ledges - are they welded to these ledges, clamped to the ledges?

And how flexible are these ledges?

Can you attach a picture or sketch?

Please see below a chart(?) of the tensile strength of A228 music wire - sorry about the formatting, I didn't want to spend the time to go through it.

Anyways, note that each diameter size of wire has a different specified strenght.

The "chart" is: [diameter "] [min ksi] [max ksi] [diameter "] [min ksi] [max ksi]

dia min max dia min max

0.004 439 485 0.055 300 331
0.005 426 471 0.059 296 327
0.006 415 459 0.063 293 324
0.007 407 449 0.067 290 321
0.008 399 441 0.072 287 317
0.009 393 434 0.076 284 314
0.010 387 428 0.080 282 312
0.011 382 422 0.085 279 308
0.012 377 417 0.090 276 305
0.013 373 412 0.095 274 303
0.014 369 408 0.100 271 300
0.015 365 404 0.102 270 299
0.016 362 400 0.107 268 296
0.018 356 393 0.110 267 295
0.020 350 387 0.112 266 294
0.022 345 382 0.121 263 290
0.024 341 377 0.125 261 288
0.026 337 373 0.130 259 286
0.028 333 368 0.135 258 285
0.030 330 365 0.140 256 283
0.032 327 361 0.145 254 281
0.034 324 358 0.150 253 279
0.036 321 355 0.156 251 277
0.038 318 352 0.162 249 275
0.040 315 349 0.177 245 270
0.042 313 346 0.192 241 267
0.045 309 342 0.207 238 264
0.048 306 339 0.225 235 260
0.051 303 335

Wire lies in a 'slot' in ABS plastic on 2 ledges as shown in picture below.

If I'm not mistaken, your chart shows .014" wire strength of 369,000 - 408,000 psi.

This is well beyond the stress calculated above so there shouldn't be undue stress on the wire and it should not deform under the given force.

Alright - that clears things up for me - I was envisioning something else.

And yes, you read that "chart" correctly - a 0.014" diameter wire with have a tensile strength between 369,000 and 408,000 psi.

I am trying to think about this problem a little more -

the solution that Ba/el provided is for elastic beam theory, and I just want to confirm that it is appropriate for this application (that is a fairly large deflection for a short beam)

****Lastly - you suggested that there will not be any deformation. Yes, there will be deformation - there is a difference between deflection and stress. And note that deflection is not based on strength, rather on modulus of elasticity*****

Hello Non-ME,

I have not been able to find the yield stress of this material. For most steels encountered in structural engineering, the two are not the same, the yield being only about one half of the ultimate tensile strength.

In the case of music wire, however, that will not be the case as the wire has been drawn through a die and hardened. So you won't exceed the yield point, but there is also a safety factor to consider. What are the consequences of failure? If they are not too significant, you may want to try the 0.014 wire to confirm that it works. You should calibrate it to confirm the load/deflection ratio found by theory.

A deflection of 0.025" in a span of 0.55" does not exactly fit with small deflection theory, so the calculation is really an approximation.

Is it possible to increase the span to arrive at a better span/deflection ratio? This would require a larger beam, but that would not be a significant cost as it is so small anyway...just a thought.

Another thought - is this wire cyclically loaded? Do you have to consider fatigue?

The wire is not cyclical loaded. It is designed to apply constant force against a cylindrical metal object that is against a printed circuit board gold electrical contact pad.

Some more critical info is needed -

How are the ends of the wire supported? Is there any 'give' or flexibility?

Is the wire tensioned at all?