Two steel springs are arranged in series as shown. The upper spring has 24 furns of 50-mm- diameter wire on a mean radius of 186 mm. The lower spring consists of 16 turns of 28-mm diameter wire on a mean radius of 109 mm . If the ultimate shearing stress in either spring must not exceed 596 MPa, (a) compute the maximum value of P and total elongation using factor of safety equal to 3.0. (b) What is the equivalent spring constant? Use G = 83 GPa.

Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter3: Torsion
Section: Chapter Questions
Problem 3.4.8P: Two sections of steel drill pipe, joined by bolted flange plates at Ä are being tested to assess the...
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5. Two steel springs are arranged in series as shown. The upper spring has 24 turns of 50-mm-
diameter wire on a mean radius of 186 mm. The lower spring consists of 16 turns of 28-mm
diameter wire on a mean radius of 109 mm . If the ultimate shearing stress in either spring
must not exceed 596 MPa, (a) compute the maximum value of P and total elongation using
factor of safety equal to 3.0. (b) What is the equivalent spring constant? Use G = 83 GPa.
Transcribed Image Text:5. Two steel springs are arranged in series as shown. The upper spring has 24 turns of 50-mm- diameter wire on a mean radius of 186 mm. The lower spring consists of 16 turns of 28-mm diameter wire on a mean radius of 109 mm . If the ultimate shearing stress in either spring must not exceed 596 MPa, (a) compute the maximum value of P and total elongation using factor of safety equal to 3.0. (b) What is the equivalent spring constant? Use G = 83 GPa.
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