• question_answer Two helical springs of the same material and of equal circular cross-section and length and number of turns, but having radii 90 mm and 40 nun, kept concentrically (smaller radius spring within the larger radius spring), arc compressed between two parallel planes with a load P. The inner spring will carry a load equal to: A) $\frac{P}{2}$                          B) $\frac{P}{3}$C) $\frac{P}{1.088}$                    D) $\frac{P}{2.088}$

${{G}_{1}}={{G}_{2}},$ ${{d}_{1}}={{d}_{2}}$ ${{D}_{wl}}=180\,mm,\,D$ $\frac{8{{W}_{1}}D_{m2}^{3}{{n}_{1}}}{{{G}_{1}}d_{1}^{4}}=\frac{8{{W}_{2}}D_{m2}^{3}{{n}_{2}}}{{{G}_{2}}d_{2}^{4}}$ $\frac{{{W}_{1}}}{{{W}_{2}}}={{\left( \frac{D{{m}_{2}}}{D{{m}_{1}}} \right)}^{3}}={{\left( \frac{80}{180} \right)}^{3}}=\frac{1}{11.39}$ ${{W}_{1}}+{{W}_{2}}=P$ ${{W}_{1}}(1+11.39)=P$ ${{W}_{1}}=\frac{1139}{1.088}\times P=\frac{P}{1.088}$