10.8 Shear flow for stringer-web sections

Stinger-Web section @Sun2006

Recall the relationship between shear flow and axial stress: \[{\frac{\partial q}{\partial s}} = -t {\frac{\partial {\sigma_{zz}}}{\partial z}}\]

Which led to: \[\begin{align} q =& - \frac{I_{xx} S_x - I_{xy} S_y}{I_{xx}I_{yy}-I_{xy}^2} \int_0^s x t(s) ds \\ & - \frac{I_{yy} S_y - I_{xy} S_x}{I_{xx}I_{yy}-I_{xy}^2} \int_0^s y t(s) ds \\ & + q (s=0) \\ \end{align}\]

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• The integrations are related to the first moment of the area measured from the centroid. Recall that an integration is a sum of a series of very small pieces.

• When stringers are present, we make the assumption that the stringers carry all the bending load.

  • Thus, \({\sigma_{zz}}\) is zero in web and non-zero in the stringers. Thus, in the web: \[{\frac{\partial q}{\partial s}} = 0\] and we conclude that the shear flow is approximately constant in these web segments. \[q = \mathrm{constant}\]

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The change in shear flow that occurs over a stringer section can be determined by examining the integral. \[\begin{align} \int_0^s x t(s) ds =& \; \displaystyle \sum_{k=1}^i x_k A_k = Q_{xi}\\ \int_0^s y t(s) ds =& \; \displaystyle \sum_{k=1}^i y_k A_k = Q_{yi}\\ \end{align}\] where \(A_k\) is the area of stringer section \(k\), \(x_k\) and \(y_k\) are the distances from the centroid of the area.

Thus, we have “step changes” in the shear flow at each stringer.

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10.8.1 Example

Consider:

Stinger-Web section @Sun2006

Assume:

• \(A_4=A_1\), \(A_3=A_2\)

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The shear flow, \(q_i\) produced by a vertical shear force \(S_y\) is: \[q_i=\frac{-S_y Q_i}{I_z}\] where \[Q_i = \displaystyle \sum_{k=1}^i y_k A_k\]

\[I_{xx} = 2 h^2 (A_1 + A_2)\] and \(y_k\) is the vertical position of the stringer \(A_k\).

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for \(q_1\): \[Q_1 = A_1 h\]

Thus: \[\begin{align} q_1 = - \frac{S_y A_1 h}{2 h^2 (A_1+A_2)} \\ q_1 = - \frac{S_y A_1}{2 h (A_1+A_2)} \end{align}\]

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Similarly: \[\begin{align} q_2 =& \, - \frac{S_y h (A_1 +A_2)}{2 h^2 (A_1+A_2)} \\ =& \, - \frac{S_y}{2 h} \\ q_3 =& \, - \frac{S_y h (A_1 + A_2 - A_2) }{2 h^2 (A_1+A_2)} \\ =& \, - \frac{S_y A_1}{2 h (A_1+A_2)} \\ \end{align}\] Note, the direction of shear flow is opposite that shown in the figure.

We could sanity check the compute shear flows sum to the applied shear force resultants.