Helical SpringsWhen 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 stress1= P/A and the torsional shearing stress2= 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.M. Wahl formula a more precise, it is given by:

where m is called thespring indexand (4m - 1)/(4m - 4) is theWahl Factor.The elongation of the bar is

Notice that the deformationis directly proportional to the applied load P. The ratio of P tois called thespring constantk and is equal to

Springs in SeriesFor 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 ParallelFor two or more springs in parallel, the resulting spring constant is

Problem 343Determine 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.Problem 343HideClick here to show or hide the solution Equation (3-10)Where:P = 1.5 kN = 1500 N; R = 90 mmd = 20 mm; n = 20 turnsm = 2R/d = 2(90)/20 = 9

answer

answer

Problem 344Determine 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 106psi.Solution 344HideClick here to show or hide the solution Equation (3-10)WhereP = 500 lb; R = 4 ind = 1 in; n = 20 turnsm = 2R/d = 2(4)/1 = 8

answer

answer

Problem 344Determine 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 106psi.Solution 344HideClick here to show or hide the solution Equation (3-10)WhereP = 500 lb; R = 4 ind = 1 in; n = 20 turnsm = 2R/d = 2(4)/1 = 8

answer

answerProblem 345A helical spring is fabricated by wrapping wire 3/4 in. in diameter around a forming cylinder 8 in. in diameter. Compute the number of turns required to permit an elongation of 4 in. without exceeding a shearing stress of 18 ksi. Use Eq. (3-9) and G = 12 106psi.Solution 345HideClick here to show or hide the solution Equation (3-9)

answer

Problem 346Compute the maximum shearing stress developed in a phosphor bronze spring having mean diameter of 200 mm and consisting of 24 turns of 200-mm diameter wire when the spring is stretched 100 mm. Use Eq. (3-10) and G = 42 GPa.Solution 346HideClick here to show or hide the solution

Where= 100 mm; R = 100 mmd = 20 mm; n = 24 turnsG = 42 000 MPa

Equation (3-10)Wherem = 2R/d = 2(100)/20 = 10

answer

Problem 347Two steel springs arranged in series as shown inFig. P-347supports a load P. The upper spring has 12 turns of 25-mm-diameter wire on a mean radius of 100 mm. The lower spring consists of 10 turns of 20-mm diameter wire on a mean radius of 75 mm. If the maximum shearing stress in either spring must not exceed 200 MPa, compute the maximum value of P and the total elongation of the assembly. Use Eq. (3-10) and G = 83 GPa. Compute the equivalent spring constant by dividing the load by the total elongation.

Solution 347HideClick here to show or hide the solution Equation (3-10)For Spring (1)

For Spring (2)

Use answerTotal elongation:

answerEquivalent spring constant, kequivalent:

answer

Problem 348A rigid bar, pinned at O, is supported by two identical springs as shown inFig. P-348. Each spring consists of 20 turns of 3/4-in-diameter wire having a mean diameter of 6 in. Determine the maximum load W that may be supported if the shearing stress in the springs is limited to 20 ksi. Use Eq. (3-9).

Solution 348HideClick here to show or hide the solution Equation (3-9)

For this problem, the critical spring is the one subjected to tension. Use P2= 519.75 lb.

answer

Problem 349A rigid bar, hinged at one end, is supported by two identical springs as shown inFig. P-349. Each spring consists of 20 turns of 10-mm wire having a mean diameter of 150 mm. Compute the maximum shearing stress in the springs, using Eq. (3-9). Neglect the mass of the rigid bar.

Solution 349HideClick here to show or hide the solution

Equation (3-9)For spring at left:

answerFor spring at right:

answer

Problem 350As shown inFig. P-350, a homogeneous 50-kg rigid block is suspended by the three springs whose lower ends were originally at the same level. Each steel spring has 24 turns of 10-mm-diameter on a mean diameter of 100 mm, and G = 83 GPa. The bronze spring has 48 turns of 20-mm-diameter wire on a mean diameter of 150 mm, and G = 42 GPa. Compute the maximum shearing stress in each spring using Eq. (3-9).

Solution 350HideClick here to show or hide the solution

Equation (1)

Equation (2)

Equation (3)From Equation (1)

Substitute P1to Equation (3)

Equation (4)From Equation (2)

Substitute P2to Equation (4)

For steel at left: answerFor steel at right: answerFor phosphor bronze: answer