A tension spring is to be used as shown in the figure below. The spring must be pretensioned to exert a 50 N force at the "bottom" of the stroke, and should have a spring rate of 3800 N/m. The peak-to-peak deflection for this spring is 40 mm. The spring has Sut = 1400 MPa, and a spring index of c = 10, use factor of safety of n = 1.8 .Design the spring; namely determine the wire diameter, the mean coil diameter and the number of active coils. Neglect the effect of end hooks on the spring rate. Note: Among standard wire diameters are 1.4mm, 2 mm, 3.5 mm, 4.5 mm, 5.5 mm, Use Ssy = 0.45 Sut and G = 80 GPa.
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A helical compression spring, made of circular wire, is subjected to an axial force, which varies from 4 kN to 5.5 Kn. Over this range of force, the deflection of the spring should be approximately 6 mm. The spring index can be taken as 4. The spring has square and ground ends. The spring is made of patented and cold-drawn steel wire with ultimate tensile strength of 1180 N/mm2 and modulus of rigidity of 82140 N/mm2 . The permissible shear stress for the spring wire should be taken as 40% of the ultimate tensile strength. Design the spring and calculate (i) wire diameter; (ii) mean coil diameter; (iii) number of active coils; (iv) total number of coils; (v) solid length of the spring; (vi) free length of the spring; (vii) required spring rate; and (viii) actual spring rate
Adi S.
A vertical wire $5.0 \mathrm{~m}$ long and of $0.0088 \mathrm{~cm}^{2}$ cross-sectional area has a modulus $Y=200 \mathrm{GPa}$. A $2.0-\mathrm{kg}$ object is fastened to its end and stretches the wire elastically. If the object is now pulled down a little and released, the object undergoes vertical SHM. Find the period of its vibration. The force constant of the wire acting as a vertical spring is given by $k=F / \Delta L$, where $\Delta L$ is the deformation produced by the force (weight) $F$. But, from $F / A=Y\left(\Delta L / L_{0}\right)$, $$ k=\frac{F}{\Delta L}=\frac{A Y}{L_{0}}=\frac{\left(8.8 \times 10^{-7} \mathrm{~m}^{2}\right)\left(2.00 \times 10^{11} \mathrm{~Pa}\right)}{5.0 \mathrm{~m}}=35 \mathrm{kN} / \mathrm{m} $$ Then for the period we have $$ T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{2.0 \mathrm{~kg}}{35 \times 10^{3} \mathrm{~N} / \mathrm{m}}}=0.047 \mathrm{~s} $$
(a) Referring to the metal rod in Figure $9.2 \mathrm{a}$ (under tensile stress), show that Eq. 9.4 can be rewritten to resemble a Hooke's law type of spring relationship for the rod. That is, show that it can be written as $F=k \Delta L,$ where $k$ is the "effective" spring constant for the rod. Express $k$ symbolically in terms of the rod's cross-sectional area $A$, its Young's modulus $Y,$ and its unstressed length $L_{\mathrm{o}}$ and show that it has the proper SI units. (b) Now consider a thin rod of iron that is subjected to a tensile force of $2.00 \times 10^{3} \mathrm{~N}$. If it has a cross-section of radius $1.00 \mathrm{~cm}$ and an unstressed length of $25.0 \mathrm{~cm},$ determine its effective spring constant. (c) By how much does this rod stretch when this force is applied? (d) How much work is done by this stretching force?
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