Author Topic: 11/17/2014 Dynamics  (Read 1611 times)

FastTrack Guru

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11/17/2014 Dynamics
« on: November 17, 2014, 08:05:45 AM »
Question:

An applied force of 150 pounds acts on a helical spring with mean spring diameter of .25 inches.  Find the shear stress (lb/in2) if the wire diameter is .08 inch. 

(A) 1879
(B) 2100
(C) 3812
(D) 4103

levi2144

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Re: 11/17/2014 Dynamics
« Reply #1 on: November 23, 2014, 03:59:05 PM »
answer: c

FastTrack Guru

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Re: 11/17/2014 Dynamics
« Reply #2 on: December 02, 2014, 08:22:55 AM »
Refer to page 224 of the FE Handbook: Helical Springs

The equation of shear stress is:

T = Ks (8FD)/ πd3

You just have to calculate Ks:

Ks= (2C + 1)/(2C)
C = D/d
C= .25 in/ .08 in = 3.124
Ks= (2(3.124) + 1)/(2(3.124))
= (7.248)/6.248
= .022

T = Ks (8FD)/ πd3
T = .022(8)150 lb (.25in) / π(.08 in )3
T = 6.6 in lb / .00160845 in3
T = 4103 lb / in2   

Answer is (D) 4103 lb / in2   

cwl

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Re: 11/17/2014 Dynamics
« Reply #3 on: December 10, 2014, 03:19:12 PM »
Is Ks actually .022?

andrei

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Re: 11/17/2014 Dynamics
« Reply #4 on: February 24, 2015, 08:40:14 AM »
It appears that Ks should be 1.16:

Ks= (2C + 1)/(2C)
C = D/d
C= .25 in/ .08 in = 3.124
Ks= (2(3.124) + 1)/(2(3.124))
= (7.248)/6.248
= .022 (!?)
= 1.16

cwl

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Re: 11/17/2014 Dynamics
« Reply #5 on: February 26, 2015, 09:09:14 AM »
Using Ks = 1.16, I come up with T = 216351 lb/in2