elongation

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 × 106 psi.
 

Solution 345

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 × 106 psi.
 

Solution 344

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.
 

Problem 343

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 × 106 psi. For the bronze bar, the area is 1.5 in.2 and E = 12 × 106 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.
 

Rigid bar supported by three rods

 

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 × 106 psi.
 

Solution 207

Axial Deformation

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

$\sigma = E \varepsilon$

 

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

$\delta = \dfrac{PL}{AE} = \dfrac{\sigma L}{E}$

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.
 

 
 
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