Question Number 226572 by mr W last updated on 05/Dec/25
Answered by mr W last updated on 07/Dec/25
Commented by mr W last updated on 07/Dec/25
![ρ=(m/h) I=((mh^2 )/3) Iα=((mgh sin θ)/2) ((mh^2 α)/3)=((mgh sin θ)/2) ⇒α=((3g sin θ)/(2h)) ((Iω^2 )/2)=((mgh(1−cos θ))/2) ((mh^2 ω^2 )/6)=((mgh(1−cos θ))/2) ⇒ω^2 =((3g(1−cos θ))/h) bending moment in cross−section at x: dM=ρ(x+ξ)αξdξ−ρgξ sin θdξ dM=((ρg sin θ)/(2h))[(3x−2h)ξ+3ξ^2 ]dξ M(x)=((ρg sin θ)/(2h))∫_0 ^(h−x) [(3x−2h)ξ+3ξ^2 ]dξ M(x)=((ρg sin θ)/(2h))[(((3x−2h)(h−x)^2 )/2)+(h−x)^3 ] M(x)=((ρg sin θ x(h−x)^2 )/(4h)) with λ=(x/h) M(x)=((mgh sin θ)/4)(1−λ)^2 λ axial force in this cross−section: dN=ρω^2 (x+ξ)dξ−ρg cos θ dξ dN=((3(1−cos θ)ρg)/h)[x−((h cos θ)/(3(1−cos θ)))+ξ]dξ N(x)=((3(1−cos θ)ρg)/h)∫_0 ^(h−x) [x−((h cos θ)/(3(1−cos θ)))+ξ]dξ N(x)=((3(1−cos θ)ρg)/h)[(x−((h cos θ)/(3(1−cos θ))))(h−x)+(((h−x)^2 )/2)] N(x)=((ρg)/(2h))[(3−5 cos θ)h^2 +2h cos θ x−3(1−cos θ)x^2 ] ⇒N(x)=((mg)/2)[3−5 cos θ+2 cos θ λ−3(1−cos θ)λ^2 ] if we neglect the stress caused through N(x), then the cross section with the largest bending moment M(x) is most likely to break. ((dM(x))/dλ)=0 3λ^2 −4λ+1=0 (3λ−1)(λ−1)=0 ⇒λ=(x/h)=(1/3) ✓ generally the largest tension stress in the cross−section is σ=(M/W)+(N/A) σ=((mgh sin θ)/(4W))(1−λ)^2 λ+((mg)/(2A))[3−5 cos θ+2 cos θ λ−3(1−cos θ)λ^2 ] with μ=(h/(2k))=((Ah)/(2W)) with k=(W/A) ((2Aα)/(mg))=μ sin θ (1−λ)^2 λ+3−5 cos θ+2 cos θ λ−3(1−cos θ)λ^2 ((2Aα)/(mg))=μ sin θ λ^3 −(3−3 cos θ+2μ sin θ)λ^2 +(2 cos θ+μ sin θ)λ+3−5 cos θ for σ_(max) : 3μ sin θ λ^2 −(3−3 cos θ+2μ sin θ)λ+2 cos θ+μ sin θ=0 ⇒λ=((3(1−cos θ)+2μ sin θ−(√(18(1−cos θ)−(9+8μ^2 ) sin^2 θ+12μ sin θ (1−3 cos θ))))/(6μ sin θ)) example: chimney is thin−walled cylinder with radius R and wall thickness t. A=2πRt W=πR^2 t k=((πR^2 t)/(2πRt))=(R/2) ⇒μ=(h/(2k))=(h/R) example: chimney is solid cylinder with radius R. A=πR^2 W=((πR^3 )/4) k=(R/4) ⇒μ=(h/(2k))=((2h)/R)](https://www.tinkutara.com/question/Q226617.png)
$$\rho=\frac{{m}}{{h}} \\ $$$${I}=\frac{{mh}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${I}\alpha=\frac{{mgh}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\frac{{mh}^{\mathrm{2}} \alpha}{\mathrm{3}}=\frac{{mgh}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\frac{\mathrm{3}{g}\:\mathrm{sin}\:\theta}{\mathrm{2}{h}} \\ $$$$\frac{{I}\omega^{\mathrm{2}} }{\mathrm{2}}=\frac{{mgh}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{2}} \\ $$$$\frac{{mh}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{6}}=\frac{{mgh}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{2}} \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{3}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{h}} \\ $$$${bending}\:{moment}\:{in}\:{cross}−{section} \\ $$$${at}\:{x}: \\ $$$${dM}=\rho\left({x}+\xi\right)\alpha\xi{d}\xi−\rho{g}\xi\:\mathrm{sin}\:\theta{d}\xi \\ $$$${dM}=\frac{\rho{g}\:\mathrm{sin}\:\theta}{\mathrm{2}{h}}\left[\left(\mathrm{3}{x}−\mathrm{2}{h}\right)\xi+\mathrm{3}\xi^{\mathrm{2}} \right]{d}\xi \\ $$$${M}\left({x}\right)=\frac{\rho{g}\:\mathrm{sin}\:\theta}{\mathrm{2}{h}}\int_{\mathrm{0}} ^{{h}−{x}} \left[\left(\mathrm{3}{x}−\mathrm{2}{h}\right)\xi+\mathrm{3}\xi^{\mathrm{2}} \right]{d}\xi \\ $$$${M}\left({x}\right)=\frac{\rho{g}\:\mathrm{sin}\:\theta}{\mathrm{2}{h}}\left[\frac{\left(\mathrm{3}{x}−\mathrm{2}{h}\right)\left({h}−{x}\right)^{\mathrm{2}} }{\mathrm{2}}+\left({h}−{x}\right)^{\mathrm{3}} \right] \\ $$$${M}\left({x}\right)=\frac{\rho{g}\:\mathrm{sin}\:\theta\:{x}\left({h}−{x}\right)^{\mathrm{2}} }{\mathrm{4}{h}} \\ $$$${with}\:\lambda=\frac{{x}}{{h}} \\ $$$${M}\left({x}\right)=\frac{{mgh}\:\mathrm{sin}\:\theta}{\mathrm{4}}\left(\mathrm{1}−\lambda\right)^{\mathrm{2}} \lambda \\ $$$${axial}\:{force}\:{in}\:{this}\:{cross}−{section}: \\ $$$${dN}=\rho\omega^{\mathrm{2}} \left({x}+\xi\right){d}\xi−\rho{g}\:\mathrm{cos}\:\theta\:{d}\xi \\ $$$${dN}=\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\rho{g}}{{h}}\left[{x}−\frac{{h}\:\mathrm{cos}\:\theta}{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}+\xi\right]{d}\xi \\ $$$${N}\left({x}\right)=\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\rho{g}}{{h}}\int_{\mathrm{0}} ^{{h}−{x}} \left[{x}−\frac{{h}\:\mathrm{cos}\:\theta}{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}+\xi\right]{d}\xi \\ $$$${N}\left({x}\right)=\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\rho{g}}{{h}}\left[\left({x}−\frac{{h}\:\mathrm{cos}\:\theta}{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}\right)\left({h}−{x}\right)+\frac{\left({h}−{x}\right)^{\mathrm{2}} }{\mathrm{2}}\right] \\ $$$${N}\left({x}\right)=\frac{\rho{g}}{\mathrm{2}{h}}\left[\left(\mathrm{3}−\mathrm{5}\:\mathrm{cos}\:\theta\right){h}^{\mathrm{2}} +\mathrm{2}{h}\:\mathrm{cos}\:\theta\:{x}−\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:\theta\right){x}^{\mathrm{2}} \right] \\ $$$$\Rightarrow{N}\left({x}\right)=\frac{{mg}}{\mathrm{2}}\left[\mathrm{3}−\mathrm{5}\:\mathrm{cos}\:\theta+\mathrm{2}\:\mathrm{cos}\:\theta\:\lambda−\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\lambda^{\mathrm{2}} \right] \\ $$$${if}\:{we}\:{neglect}\:{the}\:{stress}\:{caused} \\ $$$${through}\:{N}\left({x}\right),\:{then}\:{the}\:{cross}\:{section}\: \\ $$$${with}\:{the}\:{largest}\:{bending}\:{moment}\: \\ $$$${M}\left({x}\right)\:{is}\:{most}\:{likely}\:{to}\:{break}. \\ $$$$\frac{{dM}\left({x}\right)}{{d}\lambda}=\mathrm{0} \\ $$$$\mathrm{3}\lambda^{\mathrm{2}} −\mathrm{4}\lambda+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{3}\lambda−\mathrm{1}\right)\left(\lambda−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda=\frac{{x}}{{h}}=\frac{\mathrm{1}}{\mathrm{3}}\:\checkmark \\ $$$${generally}\:{the}\:{largest}\:{tension}\:{stress} \\ $$$${in}\:{the}\:{cross}−{section}\:{is} \\ $$$$\sigma=\frac{{M}}{{W}}+\frac{{N}}{{A}} \\ $$$$\sigma=\frac{{mgh}\:\mathrm{sin}\:\theta}{\mathrm{4}{W}}\left(\mathrm{1}−\lambda\right)^{\mathrm{2}} \lambda+\frac{{mg}}{\mathrm{2}{A}}\left[\mathrm{3}−\mathrm{5}\:\mathrm{cos}\:\theta+\mathrm{2}\:\mathrm{cos}\:\theta\:\lambda−\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\lambda^{\mathrm{2}} \right] \\ $$$${with}\:\mu=\frac{{h}}{\mathrm{2}{k}}=\frac{{Ah}}{\mathrm{2}{W}}\:{with}\:{k}=\frac{{W}}{{A}} \\ $$$$\frac{\mathrm{2}{A}\alpha}{{mg}}=\mu\:\mathrm{sin}\:\theta\:\left(\mathrm{1}−\lambda\right)^{\mathrm{2}} \lambda+\mathrm{3}−\mathrm{5}\:\mathrm{cos}\:\theta+\mathrm{2}\:\mathrm{cos}\:\theta\:\lambda−\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\lambda^{\mathrm{2}} \\ $$$$\frac{\mathrm{2}{A}\alpha}{{mg}}=\mu\:\mathrm{sin}\:\theta\:\lambda^{\mathrm{3}} −\left(\mathrm{3}−\mathrm{3}\:\mathrm{cos}\:\theta+\mathrm{2}\mu\:\mathrm{sin}\:\theta\right)\lambda^{\mathrm{2}} +\left(\mathrm{2}\:\mathrm{cos}\:\theta+\mu\:\mathrm{sin}\:\theta\right)\lambda+\mathrm{3}−\mathrm{5}\:\mathrm{cos}\:\theta \\ $$$${for}\:\sigma_{{max}} : \\ $$$$\mathrm{3}\mu\:\mathrm{sin}\:\theta\:\lambda^{\mathrm{2}} −\left(\mathrm{3}−\mathrm{3}\:\mathrm{cos}\:\theta+\mathrm{2}\mu\:\mathrm{sin}\:\theta\right)\lambda+\mathrm{2}\:\mathrm{cos}\:\theta+\mu\:\mathrm{sin}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)+\mathrm{2}\mu\:\mathrm{sin}\:\theta−\sqrt{\mathrm{18}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)−\left(\mathrm{9}+\mathrm{8}\mu^{\mathrm{2}} \right)\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{12}\mu\:\mathrm{sin}\:\theta\:\left(\mathrm{1}−\mathrm{3}\:\mathrm{cos}\:\theta\right)}}{\mathrm{6}\mu\:\mathrm{sin}\:\theta} \\ $$$${example}: \\ $$$${chimney}\:\:{is}\:\:{thin}−{walled}\:{cylinder} \\ $$$${with}\:{radius}\:{R}\:{and}\:{wall}\:{thickness}\:{t}. \\ $$$${A}=\mathrm{2}\pi{Rt} \\ $$$${W}=\pi{R}^{\mathrm{2}} {t} \\ $$$${k}=\frac{\pi{R}^{\mathrm{2}} {t}}{\mathrm{2}\pi{Rt}}=\frac{{R}}{\mathrm{2}}\:\Rightarrow\mu=\frac{{h}}{\mathrm{2}{k}}=\frac{{h}}{{R}} \\ $$$${example}: \\ $$$${chimney}\:\:{is}\:\:{solid}\:{cylinder}\:{with} \\ $$$${radius}\:{R}. \\ $$$${A}=\pi{R}^{\mathrm{2}} \\ $$$${W}=\frac{\pi{R}^{\mathrm{3}} }{\mathrm{4}} \\ $$$${k}=\frac{{R}}{\mathrm{4}}\:\Rightarrow\mu=\frac{{h}}{\mathrm{2}{k}}=\frac{\mathrm{2}{h}}{{R}} \\ $$