The thin-walled single cell beam shown in Fig. 20.11 has been idealized into a combination of direct stress carrying booms and shear stress only carrying walls. If the section supports a vertical shear load of 10 kN acting in a vertical plane through booms 3 and 6, calculate the distribution of shear flow around the section.

Boom areas: $B_1=B_8=200 \mathrm{~mm}^2, B_2=B_7=250 \mathrm{~mm}^2, B_3=B_6=400 \mathrm{~mm}^2, B_4=$ $B_5=100 \mathrm{~mm}^2$

Question

The thin-walled single cell beam shown in Fig. 20.11 has been idealized into a combination of direct stress carrying booms and shear stress only carrying walls. If the section supports a vertical shear load of 10 kN acting in a vertical plane through booms 3 and 6, calculate the distribution of shear flow around the section.

Boom areas: $B_1=B_8=200 \mathrm{~mm}^2, B_2=B_7=250 \mathrm{~mm}^2, B_3=B_6=400 \mathrm{~mm}^2, B_4=$ $B_5=100 \mathrm{~mm}^2$

Boom areas: [latex]B_1=B_8=200 \mathrm{~mm}^2, B_2=B_7=250 \mathrm{~mm}^2, B_3=B_6=400 \mathrm{~mm}^2, B_4=[/latex] [latex]B_5=100 \mathrm{~mm}^2[/latex]

Accepted Answer

The centroid of the direct stress carrying area lies on the horizontal axis of symmetry so that [latex]I_{xy}[/latex] = 0. Also, since [latex]t_D [/latex] = 0 and only a vertical shear load is applied, Eq. (20.11) reduces to

[latex]\begin{aligned}q_s=&-\left(\frac{S_x I_{x x}-S_y I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} x \mathrm{~d} s+\sum_{r=1}^n B_r x_r ight) \&-\left(\frac{S_y I_{y y}-S_x I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} y \mathrm{~d} s+\sum_{r=1}^n B_r y_r ight)+q_{s, 0}\end{aligned}[/latex] (20.11)

[latex]q_s=-\frac{S_y}{I_{x x}} \sum_{r=1}^n B_r y_r+q_{s, 0}[/latex] (i)

in which

[latex]I_{x x}=2\left(200 imes 30^2+250 imes 100^2+400 imes 100^2+100 imes 50^2 ight)=13.86 imes 10^6 \mathrm{~mm}^4[/latex]

Equation (i) then becomes

[latex]q_s=-\frac{10 imes 10^3}{13.86 imes 10^6} \sum_{r=1}^n B_r y_r+q_{s, 0}[/latex]

i.e.

[latex]q_s=-7.22 imes 10^{-4} \sum_{r=1}^n B_r y_r+q_{s, 0}[/latex] (ii)

‘Cutting’ the beam section in the wall 23 (any wall may be chosen) and calculating the ‘basic’ shear flow distribution [latex]q_b[/latex] from the first term on the right-hand side of Eq. (ii) we have

[latex]\begin{aligned}&q_{\mathrm{b}, 23}=0 \&q_{\mathrm{b}, 34}=-7.22 imes 10^{-4}(400 imes 100)=-28.9 \mathrm{~N} / \mathrm{mm} \&q_{\mathrm{b}, 45}=-28.9-7.22 imes 10^{-4}(100 imes 50)=-32.5 \mathrm{~N} / \mathrm{mm} \&q_{\mathrm{b}, 56}=q_{\mathrm{b}, 34}=-28.9 \mathrm{~N} / \mathrm{mm} ext { (by symmetry) } \&q_{\mathrm{b}, 67}=q_{\mathrm{b}, 23}=0 ext { (by symmetry) } \&q_{\mathrm{b}, 21}=-7.22 imes 10^{-4}(250 imes 100)=-18.1 \mathrm{~N} / \mathrm{mm} \&q_{\mathrm{b}, 18}=-18.1-7.22 imes 10^{-4}(200 imes 30)=-22.4 \mathrm{~N} / \mathrm{mm} \&q_{\mathrm{b}, 87}=q_{\mathrm{b}, 21}=-18.1 \mathrm{~N} / \mathrm{mm} ext { (by symmetry) }\end{aligned}[/latex]

Taking moments about the intersection of the line of action of the shear load and the horizontal axis of symmetry and referring to the results of Eqs (20.7) and (20.8) we have, from Eq. (17.18)

[latex]S_x=q_{12} \int_1^2 \mathrm{~d} x=q_{12}\left(x_2-x_1 ight)[/latex] (20.7)

[latex]S_y=q_{12}\left(y_2-y_1 ight)[/latex] (20.8)

[latex]0=\oint p q_{\mathrm{b}} \mathrm{d} s+2 A q_{s, 0}[/latex] (17.18)

[latex]\begin{aligned}0=& {\left[q_{\mathrm{b}, 81} imes 60 imes 480+2 q_{\mathrm{b}, 12}(240 imes 100+70 imes 240)+2 q_{\mathrm{b}, 23} imes 240 imes 100 ight.} \&\left.-2 q_{\mathrm{b}, 43} imes 120 imes 100-q_{\mathrm{b}, 54} imes 100 imes 120 ight]+2 imes 97200 q_{s, 0}\end{aligned}[/latex]

Substituting the above values of [latex]q_b[/latex] in this equation gives

[latex]q_{s, 0}=-5.4 \mathrm{~N} / \mathrm{mm}[/latex]

the negative sign indicating that [latex]q_{s, 0}[/latex] acts in a clockwise sense.
In any wall the final shear flow is given by [latex]q_s=q_{\mathrm{b}}+q_{s, 0}[/latex] so that

[latex]\begin{aligned}&q_{21}=-18.1+5.4=-12.7 \mathrm{~N} / \mathrm{mm}=q_{87} \&q_{23}=-5.4 \mathrm{~N} / \mathrm{mm}=q_{67}\end{aligned}[/latex]

[latex]\begin{aligned}&q_{34}=-34.3 \mathrm{~N} / \mathrm{mm}=q_{56} \&q_{45}=-37.9 \mathrm{~N} / \mathrm{mm}\end{aligned}[/latex]

and

[latex]q_{81}=17.0 \mathrm{~N} / \mathrm{mm}[/latex]

giving the shear flow distribution shown in Fig. 20.12.

Question 20.4: The thin-walled single cell beam shown in Fig. 20.11 has bee...

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