# elongation

## Solution to Problem 347 | Helical Springs

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

## 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

## Solution to Problem 344 | Helical Springs

**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 × 10^{6} psi.

- Read more about Solution to Problem 344 | Helical Springs
- Log in or register to post comments

## 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

## Solution to Problem 238 Statically Indeterminate

**Problem 238**

The lower ends of the three bars in Fig. P-238 are at the same level before the uniform rigid block weighing 40 kips is attached. Each steel bar has a length of 3 ft, and area of 1.0 in.^{2}, and E = 29 × 10^{6} psi. For the bronze bar, the area is 1.5 in.^{2} and E = 12 × 10^{6} psi. Determine (a) the length of the bronze bar so that the load on each steel bar is twice the load on the bronze bar, and (b) the length of the bronze that will make the steel stress twice the bronze stress.

- Read more about Solution to Problem 238 Statically Indeterminate
- Log in or register to post comments

## Solution to Problem 219 Axial Deformation

**Problem 219**

A round bar of length *L*, which tapers uniformly from a diameter *D* at one end to a smaller diameter d at the other, is suspended vertically from the large end. If *w* is the weight per unit volume, find the elongation of ω the rod caused by its own weight. Use this result to determine the elongation of a cone suspended from its base.

- Read more about Solution to Problem 219 Axial Deformation
- Log in or register to post comments

## Solution to Problem 218 Axial Deformation

**Problem 218**

A uniform slender rod of length L and cross sectional area A is rotating in a horizontal plane about a vertical axis through one end. If the unit mass of the rod is ρ, and it is rotating at a constant angular velocity of ω rad/sec, show that the total elongation of the rod is ρω^{2} L^{3}/3E.

- Read more about Solution to Problem 218 Axial Deformation
- Log in or register to post comments

## Solution to Problem 207 Axial Deformation

**Problem 207**

A steel wire 30 ft long, hanging vertically, supports a load of 500 lb. Neglecting the weight of the wire, determine the required diameter if the stress is not to exceed 20 ksi and the total elongation is not to exceed 0.20 in. Assume E = 29 × 10^{6} psi.

- Read more about Solution to Problem 207 Axial Deformation
- Log in or register to post comments

## Axial Deformation

In the linear portion of the stress-strain diagram, the tress is proportional to strain and is given by

since $\sigma = P / A$ and $\varepsilon = \delta / L$, then $\dfrac{P}{A} = E \dfrac{\delta}{L}$

To use this formula, the load must be axial, the bar must have a uniform cross-sectional area, and the stress must not exceed the proportional limit.

- Read more about Axial Deformation
- 1 comment
- Log in or register to post comments