A helical compression spring of a cam-mechanism is subjected to an initial preload of 50 N. The maximum operating force during the load cycle is 150 N. The wire diameter is 3 mm, while the mean coil diameter is 18 mm. The spring is made of oil-hardened and tempered valve spring wire of Grade-VW

Question

A helical compression spring of a cam-mechanism is subjected to an initial preload of 50 N. The maximum operating force during the load cycle is 150 N. The wire diameter is 3 mm, while the mean coil diameter is 18 mm. The spring is made of oil-hardened and tempered valve spring wire of Grade-VW [latex] \left(S_{u t}=1430 N / mm ^{2}\right) [/latex]. Determine the factor of safety used in the design on the basis of fluctuating stresses.

Accepted Answer

[latex] ext { Given } P_{\max }=150 N \quad P_{\min .}=50 N \quad d=3 mm [/latex].

[latex] D=18 mm \quad S_{u t}=1430 N / mm ^{2} [/latex].

Step I Mean and amplitude shear stresses

[latex] C=\frac{D}{d}=\frac{18}{3}=6 [/latex]..

From Eq. (10.7) and (10.5),

[latex] K=\frac{4 C-1}{4 C-4}+\frac{0.615}{C} [/latex] (10.7).

[latex] K_{s}=\left(1+\frac{0.5}{C} ight) [/latex] (10.5).

[latex] K=\frac{4 C-1}{4 C-4}+\frac{0.615}{C}=\frac{4(6)-1}{4(6)-4}+\frac{0.615}{6} [/latex].

= 1.2525.

[latex] K_{s}=1+\frac{0.5}{C}=1+\frac{0.5}{6}=1.0833 [/latex].

[latex] P_{m}=\frac{1}{2}\left(P_{\max .}+P_{\min .} ight)=\frac{1}{2}(150+50)=100 N [/latex].

[latex] P_{a}=\frac{1}{2}\left(P_{\max .}-P_{\min .} ight)=\frac{1}{2}(150-50)=50 N [/latex].

From Eq. (10.18) and (10.19),

[latex] au_{m}=K_{s}\left(\frac{8 P_{m} D}{\pi d^{3}} ight) [/latex] (10.18).

[latex] au_{a}=K\left(\frac{8 P_{a} D}{\pi d^{3}} ight) [/latex] (10.19).

[latex] au_{m}=K_{s}\left(\frac{8 P_{m} D}{\pi d^{3}} ight)=(1.0833)\left(\frac{8(100)(18)}{\pi(3)^{3}} ight) [/latex].

[latex] =183.91 N / mm ^{2} [/latex].

[latex] au_{a}=K\left(\frac{8 P_{a} D}{\pi d^{3}} ight)=(1.2525)\left(\frac{8(50)(18)}{\pi(3)^{3}} ight) [/latex].

[latex] =106.32 N / mm ^{2} [/latex].

Step II Factor of safety
From. Eq. (10.21), the relationships for oil-hardened and tempered steel wires are as follows:

[latex] S_{s y}=0.45 S_{u t} [/latex] (10.21).

[latex] S_{s e}^{\prime}=0.22 S_{u t}=0.22(1430)=314.6 N / mm ^{2} [/latex].

[latex] S_{s y}=0.45 S_{u t}=0.45(1430)=643.5 N / mm ^{2} [/latex].

From Eq. (10.22),

[latex] \frac{ au_{a}}{\frac{S_{s y}}{\left(f_{s} ight)}- au_{m}}=\frac{\frac{1}{2} S_{s e}^{\prime}}{S_{s y}-\frac{1}{2} S_{s e}^{\prime}} [/latex] (10.22).

[latex] \frac{ au_{a}}{\left(\frac{S_{s y}}{f s} ight)- au_{m}}=\frac{\frac{1}{2} S_{s e}^{\prime}}{S_{s y}-\frac{1}{2} S_{s e}^{\prime}} [/latex].

[latex] ext { or } \frac{106.32}{\left(\frac{643.5}{f s} ight)-183.91}=\frac{\frac{1}{2}(314.6)}{643.5-\frac{1}{2}(314.6)} [/latex].

[latex] \frac{106.32}{\left(\frac{643.5}{f s} ight)-183.91}=\frac{157.3}{486.2} [/latex].

[latex] \frac{643.5}{(f s)}-183.91=\frac{106.32(486.2)}{157.3}=328.63 [/latex].

[latex] \frac{643.5}{(f s)}=183.91+328.63 [/latex].

(fs) = 1.26.

Question 10.13: A helical compression spring of a cam-mechanism is subjected...

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