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 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

 

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.