Multiple constraints: Connecting rods for high performance engines

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The discussion focuses on combining formulas to determine the material index for connecting rods in high-performance engines. The user has successfully transposed equations for mass and stress but struggles with isolating length (L) for further calculations. Clarification is sought on whether L should be isolated before substituting into the next formula. The conversation includes specific equations and relationships between variables like area (A) and moment of inertia (I). Overall, the thread emphasizes the mathematical approach needed to solve the problem effectively.
PCarson85
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Summary:: How to combine two formulas to find the material index

Attached is the problem I am having trouble understanding.

I have been able to do the first two combinations by transposing for A in the mass equation then subsituting into the stress equation. The next combinations (in red box) are harder to see. Is L isolated and then inserted into the next formula? How is this broken down?

Thanks for any help on this.
 

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Hello @PCarson85, :welcome: !

(10.5) ##\Rightarrow I = L^2 F / (\pi^2 E) ##
##I = b^3 w/12 ## & ##b = \alpha w \ \Rightarrow I = \displaystyle {\alpha^3w^4\over 12} ##
##A = bw ## & ##b = \alpha w \ \Rightarrow A = \alpha w^2 \Rightarrow I = \alpha A^2/12 \Rightarrow A = \sqrt {\displaystyle {12 I\over \alpha}} = \sqrt{\displaystyle {12 L^2F\over \alpha \pi^2E}}##
Rearrange and substitute in (10.1) to get (10.6) :cool:
 
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Of course... b x w is area... much appreciated!
 
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