Question

A compression coil spring is required to provide a variable force on the time to go from a minimum of 100 to a maximum of 300 lbs. in one working deflection of 1 in. It needs to work freely on a 1.25-in-diameter shaft. Use a cold-drawn carbon steel wire with a Sut = 250 ksi. You want an index of resource of 6, with a striking clearance of 15%. The ends are rectified, minus 1.3.

          A compression coil spring is required to provide a variable force on the time to go from a minimum of 100 to a maximum of 300 lbs. in one working deflection of 1 in. It needs to work freely on a 1.25-in-diameter shaft. Use a cold-drawn carbon steel wire with a Sut = 250 ksi. You want an index of resource of 6, with a striking clearance of 15%. The ends are rectified, minus 1.3.
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A compression coil spring is required to provide a variable force on the time to go from a minimum of 100 to a maximum of 300 lbs. in one working deflection of 1 in. It needs to work freely on a 1.25-in-diameter shaft. Use a cold-drawn carbon steel wire with a Sut = 250 ksi. You want an index of resource of 6, with a striking clearance of 15%. The ends are rectified, minus 1.3.
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Transcript

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00:01 In this problem, we'd like to find the wire diameter, the mean coil diameter, and the number of coils.
00:07 Coils.
00:09 So, to first find a question one, or just to start solving for question one, let's start with our equation for stiffness.
00:21 For stiffness, k is equal to w divided by delta, which is equal to g, d to the fourth power, divided by 64 r cubed times h.
00:32 We have that our k is equal to 900 ntons per meter.
00:41 So we can plug in 0 .9 equal to 40 times 10 to the 3 for g times d to the 4th power divided by 64 r3 times h.
00:54 So r3h divided by d fourth is going to be equal.
01:04 Let's just let us solve this.
01:07 So if we multiply this by this, we get 0 .9 times 64, which is equal to 57 .6 r3h equals to 40 times 10 to the 3, d to the fourth power.
01:26 If we divide by d4th, and divide by 57 .6, we get r3h over d4th, equal to a 40 times 10 to the third power divided by 57 .6, which is 694 .4.
01:49 Now this is going to be useful to us in a bit.
01:51 So let's call this equation 1.
01:53 Now from our max shearing stress, t -max is equal to 16 w -r divided by pi d cubed.
02:03 Now we have our max shearing stress to be 120.
02:05 So 120 is equal to 16 times our load.
02:12 Our load is given as 45 newtons times r divided by pi d cubed.
02:19 So now let's simplify this.
02:23 We can set 120 equal to, and let's multiply all our numbers on our left hand side, 16 times 45 divided by pi.
02:34 That's equal to 229 .1, r over d .2.
02:38 Cubed.
02:39 So dividing by 229 .1, r divided by d cubed is equal to 120 divided by 229 .1, which is 0 .523.
02:49 And that's what we got from that sheer stress equation.
02:52 So r is equal to 0 .523d cubed.
02:57 Let's call this equation 2.
03:00 Now, when we have the length of a spring and all the coils touch each other, then n times is going to be equal to that load, which is 45.
03:15 And so n is equal to 45 divided by d.
03:18 Let's call that equation 3.
03:21 Now we can use r from our equation 2 and n from our equation 3 back into our equation 1.
03:28 So we plug in our 0 .523 d cubed cubed, times 45 divided by d times 1 over d 4th, equal to 694 .4.
03:46 So now we can simplify and solve for d.
03:50 So let's simplify everything on our left hand side.
03:54 We have 0 .523 cubed.
03:58 That's equal to 0 .143 d to the 6 times 45 over d times 1 over d 4...
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