# Torsional Shearing Stress

## 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).

- Read more about Solution to Problem 350 | Helical Springs
- Log in or register to post comments
- 29288 reads

## Solution to Problem 349 | Helical Springs

- Read more about Solution to Problem 349 | Helical Springs
- Log in or register to post comments
- 20095 reads

## 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.

- Read more about Solution to Problem 347 | Helical Springs
- Log in or register to post comments
- 24578 reads

## Solution to Problem 346 | Helical Springs

**Problem 346**

Compute the maximum shearing stress developed in a phosphor bronze spring having mean diameter of 200 mm and consisting of 24 turns of 20-mm diameter wire when the spring is stretched 100 mm. Use Eq. (3-10) and G = 42 GPa.

- Read more about Solution to Problem 346 | Helical Springs
- Log in or register to post comments
- 18600 reads

## Solution to Problem 345 | Helical Springs

**Problem 345**

A 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 × 10^{6} psi.

- Read more about Solution to Problem 345 | Helical Springs
- Log in or register to post comments
- 21505 reads

## Solution to Problem 343 | 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.

- Read more about Solution to Problem 343 | Helical Springs
- Log in or register to post comments
- 35090 reads

## Solution to Problem 341 | Torsion of thin-walled tube

**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.

- Read more about Solution to Problem 341 | Torsion of thin-walled tube
- Log in or register to post comments
- 10196 reads

## Solution to Problem 340 | Torsion of thin-walled tube

- Read more about Solution to Problem 340 | Torsion of thin-walled tube
- Log in or register to post comments
- 14284 reads

## Solution to Problem 339 | Torsion of thin-walled tube

**Problem 339**

A torque of 450 lb·ft is applied to the square section shown in Fig. P-339. Determine the smallest permissible dimension a if the shearing stress is limited to 6000 psi.

- Read more about Solution to Problem 339 | Torsion of thin-walled tube
- Log in or register to post comments
- 12771 reads

## Solution to Problem 338 | Torsion of thin-walled tube

**Problem 338**

A tube 0.10 in. thick has an elliptical shape shown in Fig. P-338. What torque will cause a shearing stress of 8000 psi?

- Read more about Solution to Problem 338 | Torsion of thin-walled tube
- Log in or register to post comments
- 14484 reads