Calculate the shear flow distribution in the channel section shown in Fig. 20.7 produced by a vertical shear load of 4.8 kN acting through its shear centre. Assume that the walls of the section are only effective in resisting shear stresses while the booms, each of area 300 mm²,

Question

Calculate the shear flow distribution in the channel section shown in Fig. 20.7 produced by a vertical shear load of 4.8 kN acting through its shear centre. Assume that the walls of the section are only effective in resisting shear stresses while the booms, each of area 300 mm², carry all the direct stresses.

Accepted Answer

The effective direct stress carrying thickness [latex]t_D[/latex] of the walls of the section is zero so that the centroid of area and the section properties refer to the boom areas only. Since Cx (and Cy as far as the boom areas are concerned) is an axis of symmetry [latex]I_{xy}[/latex] = 0; also [latex]S_x[/latex] = 0 and Eq. (20.6) thereby 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)\end{aligned}[/latex] (20.6)

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

in which [latex]I_{x x}=4 imes 300 imes 200^2=48 imes 10^6 \mathrm{~mm}^4[/latex]. Substituting the values of [latex]S_y ext { and } I_{x x}[/latex] in Eq. (i) gives

[latex]q_s=-\frac{4.8 imes 10^3}{48 imes 10^6} \sum_{r=1}^n B_r y_r=-10^{-4} \sum_{r=1}^n B_r y_r[/latex] (ii)

At the outside of boom 1,[latex]q_s[/latex] = 0. As boom 1 is crossed the shear flow changes by an amount given by

[latex]\Delta q_1=-10^{-4} imes 300 imes 200=-6 \mathrm{~N} / \mathrm{mm}[/latex]

Hence [latex]q_{12}[/latex] = −6 N/mm since, from Eq. (i), it can be seen that no further changes in shear flow occur until the next boom (2) is crossed. Hence

[latex]q_{23}=-6-10^{-4} imes 300 imes 200=-12 \mathrm{~N} / \mathrm{mm}[/latex]

Similarly

[latex]q_{34}=-12-10^{-4} imes 300 imes(-200)=-6 \mathrm{~N} / \mathrm{mm}[/latex]

while, finally, at the outside of boom 4 the shear flow is
[latex]-6-10^{-4} imes 300 imes(-200)=0[/latex]
as expected. The complete shear flow distribution is shown in Fig. 20.8. It can be seen from Eq. (i) in Example 20.3 that the analysis of a beam section which has been idealized into a combination of direct stress carrying booms and shear stress only carrying skin gives constant values of the shear flow in the skin between the booms; the actual distribution of shear flows is therefore lost. What remains is in fact the average of the shear flow, as can be seen by referring to Example 20.3. Analysis of the unidealized channel section would result in a parabolic distribution of shear flow in the web 23 whose resultant is statically equivalent to the externally applied shear load of 4.8 kN. In Fig. 20.8 the resultant of the constant shear flow in the web 23 is 12 × 400 = 4800 N = 4.8 kN. It follows that this constant value of shear flow is the average of the parabolically distributed shear flows in the unidealized section.
The result, from the idealization of a beam section, of a constant shear flow between booms may be used to advantage in parts of the analysis. Suppose that the curved web 12 in Fig. 20.9 has booms at its extremities and that the shear flow [latex]q_{12}[/latex] in the web is constant. The shear force on an element [latex]\delta s[/latex] of the web is [latex]q_{12} \delta s[/latex], whose components horizontally and vertically are [latex]q_{12} \delta s[/latex] cos Φ and [latex]q 12 \delta s \sin \phi[/latex]. The resultant, parallel to the x axis, [latex]S_x[/latex], of [latex]q_{12}[/latex] is therefore given by

[latex]S_x=\int_1^2 q_{12} \cos \phi \mathrm{d} s[/latex]

or

[latex]S_x=q_{12} \int_1^2 \cos \phi \mathrm{d} s[/latex]

which, from Fig. 20.9, may be written

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

Similarly the resultant of [latex]q_{12}[/latex] parallel to the y axis is

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

Thus the resultant, in a given direction, of a constant shear flow acting on a web is the value of the shear flow multiplied by the projection on that direction of the web.
The resultant shear force S on the web of Fig. 20.9 is

[latex]S=\sqrt{S_x^2+S_y^2}=q_{12} \sqrt{\left(x_2-x_1 ight)^2+\left(y_2-y_1 ight)^2}[/latex]

i.e.
[latex]S=q_{12} L_{12}[/latex] (20.9)
Therefore, the resultant shear force acting on the web is the product of the shear flow and the length of the straight line joining the ends of the web; clearly the direction of the resultant is parallel to this line.

The moment [latex]M_q[/latex] produced by the shear flow [latex]q_{12}[/latex] about any point O in the plane of the web is, from Fig. 20.10

[latex]M_q=\int_1^2 q_{12} p \mathrm{~d} s=q_{12} \int_1^2 2 \mathrm{~d} A[/latex]

or
[latex]M_q=2 A q_{12}[/latex] (20.10)
in which A is the area enclosed by the web and the lines joining the ends of the web to the point O. This result may be used to determine the distance of the line of action of the resultant shear force from any point. From Fig. 20.10

[latex]\mathrm{Se}=2 A q_{12}[/latex]

from which

[latex]e=\frac{2 A}{S} q_{12}[/latex]

Substituting for [latex]q_{12}[/latex] from Eq. (20.9) gives

[latex]e=\frac{2 A}{L_{12}}[/latex]

Question 20.3: Calculate the shear flow distribution in the channel section...

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