Solution to Problem 350 | Helical Springs

Problem 350
As shown in Fig. 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).
 

Bar supported by three springs

 

Solution to Problem 347 | Helical Springs

Problem 347
Two steel springs arranged in series as shown in Fig. P-347 supports 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.
 

347-two-springs-in-series.jpg

 

Solution to Problem 344 | Helical Springs Jhun Vert Tue, 04/21/2020 - 02:44 pm

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 to Problem 341 | Torsion of thin-walled tube Jhun Vert Tue, 04/21/2020 - 02:37 pm

Problem 341
Derive the torsion formula τ = Tρ / J for a solid circular section by assuming the section is composed of a series of concentric thin circular tubes. Assume that the shearing stress at any point is proportional to its radial distance.