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Young-s-modulus-of-a-material-measures-its-resistance-caused-by-external-stresses-On-a-vertical-wall-is-a-solid-mass-of-specific-mass-and-Young-modulus-in-a-straight-parallelepiped-shape-the-dim




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

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