Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 68642 by Maclaurin Stickker last updated on 14/Sep/19

Young′s modulus of a material measures its resistance caused by external stresses. On a vertical wall is a solid mass of specific mass ρ and Young ε modulus in a straight parallelepiped shape, the dimensions of a which are shown in the figure. Based on the correlations between physical quantities, determine the the expression that best represents the deflection suffered by the solid by the action of its own weight.

$$\mathrm{Young}'\mathrm{s}\:\mathrm{modulus}\:\mathrm{of}\:\mathrm{a}\:\mathrm{material}\:\mathrm{measures} \\ $$$$\mathrm{its}\:\mathrm{resistance}\:\mathrm{caused}\:\mathrm{by}\:\mathrm{external}\:\mathrm{stresses}. \\ $$$$\mathrm{On}\:\mathrm{a}\:\mathrm{vertical}\:\mathrm{wall}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solid}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{specific} \\ $$$$\mathrm{mass}\:\rho\:\mathrm{and}\:\mathrm{Young}\:\varepsilon\:\mathrm{modulus}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{parallelepiped}\:\mathrm{shape},\:\mathrm{the}\:\mathrm{dimensions} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{which}\:\mathrm{are}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{figure}.\: \\ $$$$\mathrm{Based}\:\mathrm{on}\:\mathrm{the}\:\mathrm{correlations}\:\mathrm{between}\:\mathrm{physical} \\ $$$$\mathrm{quantities},\:\mathrm{determine}\:\mathrm{the}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{that} \\ $$$$\mathrm{best}\:\mathrm{represents}\:\mathrm{the}\:\mathrm{deflection}\:\mathrm{suffered} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{solid}\:\mathrm{by}\:\mathrm{the}\:\mathrm{action}\:\mathrm{of}\:\mathrm{its}\:\mathrm{own}\:\mathrm{weight}. \\ $$

Commented by Maclaurin Stickker last updated on 14/Sep/19

Commented by mr W last updated on 14/Sep/19

Commented by mr W last updated on 14/Sep/19

δ=((wL^4 )/(8EI)) L=a E=ε I=((bh^3 )/(12)) w=bhρg ⇒deflection δ=((bhρga^4 )/(8ε((bh^3 )/(12))))=((3ρga^4 )/(2εh^2 ))

$$\delta=\frac{{wL}^{\mathrm{4}} }{\mathrm{8}{EI}} \\ $$$${L}={a} \\ $$$${E}=\epsilon \\ $$$${I}=\frac{{bh}^{\mathrm{3}} }{\mathrm{12}} \\ $$$${w}={bh}\rho{g} \\ $$$$\Rightarrow{deflection}\:\delta=\frac{{bh}\rho{ga}^{\mathrm{4}} }{\mathrm{8}\epsilon\frac{{bh}^{\mathrm{3}} }{\mathrm{12}}}=\frac{\mathrm{3}\rho{ga}^{\mathrm{4}} }{\mathrm{2}\epsilon{h}^{\mathrm{2}} } \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com