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

$\text { Given } P_{\max }=150 N \quad P_{\min .}=50 N \quad d=3 mm$.

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

Step I Mean and amplitude shear stresses

$C=\frac{D}{d}=\frac{18}{3}=6$..

From Eq. (10.7) and (10.5),

$K=\frac{4 C-1}{4 C-4}+\frac{0.615}{C}$              (10.7).

$K_{s}=\left(1+\frac{0.5}{C}\right)$                        (10.5).

$K=\frac{4 C-1}{4 C-4}+\frac{0.615}{C}=\frac{4(6)-1}{4(6)-4}+\frac{0.615}{6}$.

= 1.2525.

$K_{s}=1+\frac{0.5}{C}=1+\frac{0.5}{6}=1.0833$.

$P_{m}=\frac{1}{2}\left(P_{\max .}+P_{\min .}\right)=\frac{1}{2}(150+50)=100 N$.

$P_{a}=\frac{1}{2}\left(P_{\max .}-P_{\min .}\right)=\frac{1}{2}(150-50)=50 N$.

From Eq. (10.18) and (10.19),

$\tau_{m}=K_{s}\left(\frac{8 P_{m} D}{\pi d^{3}}\right)$          (10.18).

$\tau_{a}=K\left(\frac{8 P_{a} D}{\pi d^{3}}\right)$                      (10.19).

$\tau_{m}=K_{s}\left(\frac{8 P_{m} D}{\pi d^{3}}\right)=(1.0833)\left(\frac{8(100)(18)}{\pi(3)^{3}}\right)$.

$=183.91 N / mm ^{2}$.

$\tau_{a}=K\left(\frac{8 P_{a} D}{\pi d^{3}}\right)=(1.2525)\left(\frac{8(50)(18)}{\pi(3)^{3}}\right)$.

$=106.32 N / mm ^{2}$.

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

$S_{s y}=0.45 S_{u t}$                    (10.21).

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

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

From Eq. (10.22),

$\frac{\tau_{a}}{\frac{S_{s y}}{\left(f_{s}\right)}-\tau_{m}}=\frac{\frac{1}{2} S_{s e}^{\prime}}{S_{s y}-\frac{1}{2} S_{s e}^{\prime}}$                    (10.22).

$\frac{\tau_{a}}{\left(\frac{S_{s y}}{f s}\right)-\tau_{m}}=\frac{\frac{1}{2} S_{s e}^{\prime}}{S_{s y}-\frac{1}{2} S_{s e}^{\prime}}$.

$\text { or } \frac{106.32}{\left(\frac{643.5}{f s}\right)-183.91}=\frac{\frac{1}{2}(314.6)}{643.5-\frac{1}{2}(314.6)}$.

$\frac{106.32}{\left(\frac{643.5}{f s}\right)-183.91}=\frac{157.3}{486.2}$.

$\frac{643.5}{(f s)}-183.91=\frac{106.32(486.2)}{157.3}=328.63$.

$\frac{643.5}{(f s)}=183.91+328.63$.

(fs) = 1.26.