The member is subject to a shear force of V = 2 kN . Determine the shear flow at points A, B, and C. The thickness of each thin-walled segment in 15 mm.

Question

Accepted Answer

Section Properties:

[latex]\bar{y} =\frac{\Sigma \bar{y} A}{\Sigma A} [/latex]

[latex]=\frac{0.0075(0.2)(0.015)+0.0575(0.115)(0.03)+0.165(0.3)(0.015)}{0.2(0.015)+0.115(0.03)+0.3(0.015)}[/latex]

[latex]=0.08798 \mathrm{m} [/latex]

[latex]I_{N A}=\frac{1}{12}(0.2)\left(0.015^{3}\right)+0.2(0.015)(0.08798-0.0075)^{2} [/latex]

[latex]+\frac{1}{12}(0.03)\left(0.115^{3}\right)+0.03(0.115)(0.08798-0.0575)^{2}[/latex]

[latex]+\frac{1}{12}(0.015)\left(0.3^{3}\right)+0.015(0.3)(0.165-0.08798)^{2}[/latex]

[latex]=86.93913\left(10^{-6}\right) \mathrm{m}^{4}[/latex]

[latex]Q_{A} =0[/latex]

[latex]Q_{B} =\bar{y}_{1}^{\prime} A^{\prime}=0.03048(0.115)(0.015)=52.57705\left(10^{-6}\right) \mathrm{m}^{-3} [/latex]

[latex]Q_{C} =\Sigma \bar{y_{1}}^{\prime} A^{\prime} [/latex]

[latex]=0.03048(0.115)(0.015)+0.08048(0.0925)(0.015) [/latex]

[latex]=0.16424\left(10^{-3}\right) \mathrm{m}^{3}[/latex]

Shear Flow:

[latex]q_{A}=\frac{V Q_{A}}{I}=0 [/latex]

[latex]q_{B}=\frac{V Q_{B}}{I}=\frac{2\left(10^{3}\right)(52.57705)\left(10^{-6}\right)}{86.93913\left(10^{-6}\right)}=1.21 \mathrm{kN/m}[/latex]

[latex]q_{C}=\frac{V Q_{C}}{I}=\frac{2\left(10^{3}\right)(0.16424)\left(10^{-3}\right)}{86.93913\left(10^{-6}\right)}=3.78 \mathrm{kN/m}[/latex]

Question 7.73: The member is subject to a shear force of V = 2 kN . Determ...

Related Answered Questions