HW10 Hints & Tips


HW10.1
The spring force when it just opens and when it's fully open is just the corresponding pressure times the piston area, or 23.1 and 69.3 N.  Between those forces, the piston travels 4mm compressing the spring, so the spring rate must be 11.54 N/mm. With that, we now have everything in Eqn. 17.18 except Na, so we can solve for the number of active coils, 19.2.  (19.6 if you skimped and used Eqn. 17.15 without the C^2 term.)  Using Table 17.3, we can calculate the solid length as (Na + 2) x dwire = 52.9mm.  The problem statement says that the spring is at the solid length when the valve is just fully open, so there is no added margin.  And the spring expands by 69.3 N divided by the spring rate as it goes to its free condition with no load.  The solid length plus that expansion equals 58.9mm, the free length.  We go back to Table 17.3 to figure out the pitch.  Finally, with a Kwahl of 1.325 for dynamic loading, the max shear stress is 179.6 MPa.


HW10.2
Spring rate is 7.13 N/mm; solid length is 24mm.  The Shear Yield strength for 2mm Music wire is 0.4 x Sut = 784.5 MPa.  Setting this as the shear stress, we solve Eqn. 17.8 to get a max force of 176 N, using the static shear factor, Kd.  If that force makes the spring solid, then it expands 176N ÷ 7.13 N/mm = 24.67mm going from solid to free. Calculating MaxDeflection/FreeLength = 0.506 and Lfree/Dmean = 3.74 and going to Fig. 17.8, we see that this parallel-ends spring is stable regarding buckling.


HW10.3
The spring index is 13.33.  The stiffness is 0.348 N/mm.  The Shear Yield Strength for this 1.5mm Chromium Vanadium wire is 747.6MPa, so the 1.5 FOS limits the shear stress to 498.4MPa.  Using Dmean because this is static loading, we calculate the max force as 31.8N.  At that load, the deflection P/K = 91.6mm.  Free length is 121.6mm.

Hw10.4
Sut for this 2mm Hard Drawn wire is 1532 MPa and the tensile Yield Strength is 0.6 times that, or 919MPa.  Remember, helical torsional springs bend the wire so the stress is bending, not shear.  A FOS of 2 limits the stress to 919/2 = 459.6MPa.  With Ki = 1.0635, the maximum moment on the spring is 339.4 Nmm.  With a moment arm of 40mm, this means P cannot exceed 8.5N.  The number of active turns from Eqn. 17.44 is 7.84.  The max moment results in an angular deflection of 0.204 turns, bringing the total turns to 7.84 + 0.204 = 8.044.  Eqn. 17.45 gives the loaded ID as 22.4mm, down from its relaxed 23mm.