Determine the shear-stress variation over the cross section of the thin-walled tube as a function of elevation y and show that τ max = 2V/A , where A = 2π rt . Hint: Choose a differential area element dA = Rt dθ . Using dQ = y dA ,

Question

Determine the shear-stress variation over the cross section of the thin-walled tube as a function of elevation y and show that [latex] au _{\max} = 2 V/A[/latex] , where [latex]A=2\pi rt[/latex] . Hint: Choose a differential area element dA = Rt dθ . Using [latex] dQ = y dA [/latex], formulate Q for a circular section from θ to (π - θ ) and show that [latex]Q=2R^{2}t\cos heta [/latex] , where [latex]\cos heta =\sqrt{R^{2}-y^{2}}/R [/latex] .

Accepted Answer

[latex]d A = R t d heta[/latex]

[latex]d Q=y d A = y R t d heta[/latex]

Here [latex] y=R \sin heta[/latex]

Therefore [latex]d Q=R^{2} t \sin heta d heta[/latex]

[latex]Q=\int_{ heta}^{\pi- heta} R^{2} t \sin heta d heta=\left.R^{2} t(-\cos heta) ight|_{ heta} ^{\pi- heta}[/latex]

[latex]\quad=R^{2} t[-\cos (\pi- heta)-(-\cos heta)]=2 R^{2} t \cos heta [/latex]

[latex]d I=y^{2} d A=y^{2} R t d heta=R^{3} t \sin ^{2} heta d heta[/latex]

[latex]I=\int_{0}^{2 \pi} R^{3} t \sin ^{2} heta d heta=R^{3} t \int_{0}^{2 \pi} \frac{(1-\cos 2 heta)}{2} d heta[/latex]

[latex]\quad=\left.\frac{R^{3} t}{2}\left[ heta-\frac{\sin 2 heta}{2} ight] ight|_{0} ^{2 \pi}=\frac{R^{3} t}{2}[2 \pi-0]=\pi R^{3} t[/latex]

[latex] au=\frac{V Q}{I t}=\frac{V\left(2 R^{2} t \cos heta ight)}{\pi R^{3} t(2 t)}=\frac{V \cos heta}{\pi R t} [/latex]

Here [latex]\cos heta=\frac{\sqrt{R^{2}-y^{2}}}{R}[/latex]

[latex] au=\frac{V}{\pi R^{2} t} \sqrt{R^{2}-y^{2}}[/latex]

[latex] au_{\max } ext { occurs at } y=0 ;[/latex] therefore

[latex] au_{\max }=\frac{V}{\pi R t}[/latex]

[latex]A=2 \pi R t ;[/latex] therefore

[latex] au_{\max }=\frac{2 V}{A} \quad \mathbf{Q E D}[/latex]

Question 7.62: Determine the shear-stress variation over the cross section ...

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