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.

Solution 347

\tau_{max} = \dfrac{16PR}{\pi d^3} \left( \dfrac{4m - 1}{4m - 4} + \dfrac{0.615}{m} \right) \, \to \,\, Equation (3-10)

For Spring (1)
200 = \dfrac{16P(100)}{\pi (25^3)} \left[ \dfrac{4(8) - 1}{4(8) - 4} + \dfrac{0.615}{8} \right]
P = 5182.29 \, \text{N}

For Spring (2)
200 = \dfrac{16P(75)}{\pi (20^3)} \left[ \dfrac{4(7.5) - 1}{4(7.5) - 4} + \dfrac{0.615}{7.5} \right]
P = 3498.28 \, \text{N}

Use P = 3498.28 \, \text{N} \,\, answer

Total elongation:
\delta = \delta_1 + \delta_2
\delta = \left( \dfrac{64PR^3n}{Gd^4} \right)_1 + \left( \dfrac{64PR^3n}{Gd^4} \right)_2
\delta = \dfrac{64(3498.28)(100^3)12}{83\,000(25^4)} + \dfrac{64(3498.28)(75^3)10}{83\,000(20^4)}
\delta = 153.99 \, \text{mm} \,\, answer

Equivalent spring constant, k_equivalent:
k_{equivalent} = \dfrac{P}{\delta} = \dfrac{3498.28}{153.99}
k_{equivalent} = 22.72 \, \text{N/mm} \,\, answer