Helical Springs

Helical Springs

When close-coiled helical spring, composed of a wire of round rod of diameter d wound into a helix of mean radius R with n number of turns, is subjected to an axial load P produces the following stresses and elongation:

The maximum shearing stress is the sum of the direct shearing stress τ1 = P/A and the torsional shearing stress τ2 = Tr/J, with T = PR.

This formula neglects the curvature of the spring. This is used for light spring where the ratio d/4R is small. For heavy springs and considering the curvature of the spring, a more precise formula is given by: (A.M.Wahl Formula)

where m is called the spring index and (4m – 1) / (4m – 4) is the Wahl Factor. The elongation of the bar is

Notice that the deformation δ is directly proportional to the applied load P. The ratio of P to δ is called the spring constant k and is equal to

SPRINGS IN SERIES For two or more springs with spring laid in series, the resulting spring constant k is given by

where k1, k2,… are the spring constants for different springs.

SPRINGS IN PARALLEL

Solved Problems in Helical Springs Problem 343 Determine the maximum shearing stress and elongation in a helical steel spring composed of 20 turns of 20-mm-diameter wire on a mean radius of 90 mm when the spring is supporting a load of 1.5 kN. Use Eq. (3-10) and G = 83 GPa. Solution 343

Strength of Materials

Chapter One

Simple Stresses

Ex:-5- A load (P) is supported by two steel spring arranged in series as shown in figure . The upper spring has (20 turns) of (20mm) diameter wire on mean diameter of (150mm) . The lower spring consists of (15 turns) of (10mm) diameter wire on mean diameter of (130mm) . Determine the maximum shearing stress in each spring if the total deflection is (80mm) and (Gs=83GPa) . Sol:64 * P * R 3 * N G*d 4 64 * P  (0.075) 3 (20) (0.065) 3 *15  0.08     83 *10 9  (0.02) 4 (0.01) 4 

 

 P  233N For upper spring (m 

2 R 2 * 0.075   7.5) d 0.02

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Strength of Materials

Chapter One

Simple Stresses

16 P * R  4m  1 0.615   m   * d 3  4m  4 16 * 233 * 0.075  30  1 0.615    max .  7.5   * (0.02) 3  30  4  max .  12.7 MPa

 max . 

For lower spring (m 

2 R 2 * 0.065   13) d 0.01

 4m  1 0.615   4m  4  m  16 * 233 * 0.065  52  1 0.615   max .   52  4  13   * (0.01) 3  max .  81.MPa

 max . 

16 P * R  *d3

Ex:-6- A load (P) is supported by two concentric steel spring arranged as shown in figure . The inner spring consist of (30 turns) of (20mm) wire diameter on mean diameter of (150mm) , the outer spring has (20 turns) of (30mm) wire diameter on mean diameter of (200mm) . Compute the maximum load that will not exceed a shearing stress of (140 MPa) in either spring . Gs=83 GPa . Sol:-

64 P * R 3 * N 1   2 ,   G*d4 64 P1 * (0.075)3 * 30 64 P2 * (0.1) 3 * 20  G * (0.02) 4 G * (0.03) 4 P1=0.312 P2 2 R 2 * 0.075   7.5) d 0.02 2 R 2 * 0.1   6.67) For outer spring (m  d 0.03

For inner spring (m 

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Strength of Materials

 max . 

Chapter One

16 P * R  4m  1 0.615    * d 3  4m  4 m  

For (P1) :-

140 *106 

16 * P1 * 0.075  30  1 0.615    * (0.02)3  30  4 7.5 

P1=2.45 kN and P2= 7.85kN

 P= P1+ P2=10.3 kN

For (P2) :-

140 *106 

16 * P2 * 0.1  26.7  1 0.615    * (0.03)3  26.8  4 6.67 

P2=6.06 kN and P1= 1.9 kN



 P= P1+ P2=7.96 kN

P= 7.96 kN

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Simple Stresses

Problem 344 Determine the maximum shearing stress and elongation in a bronze helical spring composed of 20 turns of 1.0-in.-diameter wire on a mean radius of 4 in. when the spring is supporting a load of 500 lb. Use Eq. (3-10) and G = 6 × 106 psi. Solution 344