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robin November 5, 2011 at 10:38 am

What is the distribution of the mean of a sample of size 49?is it going to be normal?find the parameters of the normal distribution.you need p(mean>1.25)standardise and solve.

Kenny McCormick™ November 5, 2011 at 10:56 pm

I was going to do that at first but the whole scrolling thing ruined my mood.

Vi November 6, 2011 at 10:29 am

The 5 things that I have found are indispensable as a designer are: 1. my dress form 2. my fabric cutting scissors (Mine are Gingher; I got them in college and they were well worth the investment)3. my Reader’s Digest Complete Guide to Sewing (got it in college; you can get it at a used bookstore for $2 – $5)4. my fishing tiered fishing tackle box for carrying my supplies (same one from college)5. my C-thru rulerBONUS6. my french curves7. wide with muslin (for draping on the dress form and for making muslin samples before actually cutting into the good fabric)8. A SEAM RIPPER – the tool you hate to use but love to have9. my mechanical pencils10. a strong magnet (for picking up pins easily)11. my L-square12. my hip curve13. my Draping book from college (teaches techniques for draping on the dress form)14. my cutting mat15. my rotary cutter16. wide width paper on a roll (white or brown)I know I went over, but if you now have a selection. To be honest, I find all of these items critical for me so they aren’t in any particular order.

Gunner0812 November 6, 2011 at 11:18 pm

Sheet Metal Sheers are large scissors used to cut sheet metal. A sheet metal cutting table (or mechanical sheer) is a large mechanically operated blade that slices sheet metal. It can be operated using a manual foot pedal, pneumatics, or hydraulics. The most important safety tip is to keep your fingers clear of the blade! Anytime you are working with cutting or forming metal you should be wearing safety glasses as well

Dave B November 7, 2011 at 11:17 am

You take the area of the cylindrical surface of diameter 22mm and 12mm high. so: A=(PI)xDxH (the “shear” area).Pi = 3.14approxD = 22mmH = 12mm A=828.96mm^2 (Say 829mm^2)So minimum required force is going to be the force that will give you a shear stress of 360MPa. Now, MPa can also be written as N/mm^2,so to get the force in Newtons, we multiply 829mm^2 by 360.Approx 298440N, or 298.44kN. If you convert that to Kg (for the press) you get a required press load of around 30.5 tonnes.BUT,You want the compressive stress in the punch. To get that you use the 298.44kN in the equation stress=force/area, but use the area of the punch = PI()x22^2/4=379.94mm^2, (say 380mm^2)So punch compressive stress = 298.44kN/380mm^2 = 785.4N/mm^2 which can be written as 785.4MPa.I hope that helps. What you are doing is determining the punch load from the shear requirement, then using that load to determine the punch compressive stress.

Grillparzer November 7, 2011 at 11:14 pm

A press or brake bends or stamps metal, a sheer cuts, no idea what a slip/roll does.

Bentley November 8, 2011 at 10:41 am

Any repair of this nature depends on the material and the position on the block. Some things require an unusual amount of disassembly before one can weld in certain areas. However, if you have merely stripped the threads in a piece of metal, this is a relatively straight-forward thing for a competent mechanic.Your main problem, as I see it, is finding someone who doesn’t want to just replace things, which clearly in this case would be costly. Shop around carefully and find someone who is willing to consider repairs that do not involve simply replacing parts.

Z November 8, 2011 at 10:41 pm

I own an Acer 3624WXMi for about 1 yr now and have drop kicked it across an iced over parking lot into a slush puddle. Spilled a number of “fluids” all over it. Smothered it with a blanket and slept on it. Warped/burned a couple keys from cig cherries. not to mention the ol’ trip over the power cable and watch it fly about like it had wings.I have never in my life owned a more durable product. Purchased brandy new for 400 and would never think twice about buying another.I hear ya with the precautions one takes with “some” puters, but for some reason, I guess, this one has something to prove.Thanks for the post

midnite.scribe November 9, 2011 at 10:58 am

It is a “scissor” action, an angled cutting action.

Ved November 9, 2011 at 10:58 pm

The data is incomplete.Does the checking involve replacement or without replacement.if replacement is done,a.10C5*(1/10)^5*(9/10)^5b.10C5*(1/10)^5*(9/10)^5+10C6*(1/10)^6*(9/10)^4+10C7*(1/10)^7*(9/10)^3+10C8*(1/10)^8*(9/10)^2 + 10C9*(1/10)^9*(9/10)+10C10*(1/10)^10

Master Tiki November 10, 2011 at 10:24 am

Sounds like it could either be a binder or hole punch.

micky_b_good November 10, 2011 at 10:42 pm

In general there are 3 methods of bolting, snug tight and fully tensioned and fully tensioned – friction grip. If snug tight and fully tensioned, the plates that have the shear force will slip slightly and the holes in the plate will come into contact with the bolts causing the bolts to take the force out in shear.If you are designing a friction grip joint (especially usefull in vibrating machinery) the plates surfaces have a roughned finish and when the bolts are tightened, they must be tensioned to such a level that the shear force in the plates are resisted in friction by the faying surfaces. The bolts must be tensioned to the required level to ensure that the faying surfaces can resist the shear force. The bolts are generally tightened to just above the elastic limit with the tension in the bolts estimated by a torque wrench or load indicating washers, or by ultrasonics measuring the elongation of the bolts.I hope this helps.

blahb31 November 11, 2011 at 10:44 am

First you would find the probability of getting a shear strength of more than 1000 pounds. That means that you have to integrate the pdf over values above 1.int(1 to 2) x/2dx = (x^2/4)(1 to 2)= (4 – 1)/4= 3/4.The number of screws that have shear strengths exceeding 1000 pounds out of the three that were sampled would then have a binomial distribution with number of trials n = 3 and probability of success 3/4.P(2 successes) = {3!/(2!1!)} * (3/4)^2 * (1/4)^1= 0.421875So the probability that exactly 2 of the 3 screws have shear strengths exceeding 1000 pounds is 0.421875.

retropink November 11, 2011 at 10:44 pm

The idea, with shear curtains, is not to block out the light. If you don’t like the “shearness” of the shears, you could layer them by putting more shears onto the pole and bunching them together more. Or you could put another layer of shears over or under your current shears. If you like the light, but want a little more privacy, you could install mini blinds or a roller blind that you would pull down only when you needed it. One last suggestion is to keep the shears but layer a valance and drapes over them. This way the center of your window lets in light but you have a better feeling of privacy with the drapes & valance.

Whippet keeper November 12, 2011 at 11:12 am

Sorry, I’ve never sharpened my own. We had a service come to the shop and later to the vet office I worked at.You can try asking a sharpener and perhaps they’ll be kind enough to tell you.There is a great booth here in CA at our local dog shows, they do the best job. Sorry don’t remember their name.If the sharpener knows his stuff a pair of scissors can be sharpened many times and will keep a good edge for a while (if they are good scissors). I’d say start with a light hand and go slow. I’ve had scissors ruined by new sharpeners that I was desperate and tried them. They didn’t know how to deal with curved scissors and rather than tell me they just ruined several pair.Good luck.

Kelvin November 12, 2011 at 11:01 pm

Let X be the number of defective pieces out of 10 pieces.X ~ Binomial(10, 0.1)P(Exactly 5 defective) =P(X = 5) = 10C5 * (0.1)^5 * (0.9)^5 = 0.001488P(5 or more defective) = P(X ≥ 5)= P(X = 5) + P(X = 6) + P(X = 7) + P(X = + P(X = 9) + P(X = 10)= 10C5 * (0.1)^5 * (0.9)^5 + 10C6 * (0.1)^6 * (0.9)^4 + … + (0.1)^10= 0.001635

Sarita Bonana November 13, 2011 at 10:51 am

call the person or company you purchased the machine from. Tell them to come look at the machine.Good luck.

rgfmss November 13, 2011 at 10:27 pm

a dimension change act as sensing element ;this change help to change of resistance when it is conected to an electrical ckt with sup[ply voltage &change of resistance creat change in voltage this change in voltage is futher processed & and finally at display screen give results; it also can be done by LVDT in which again displacement is converted to displace ment of a core( solid metalicplate type str)movable in atranse former in between primary & secondry with equalnomber of windings;with the position of core out putvoltage changing;