User:Eas4200c.f08.aero6.villalba/shear1-5

(1.5)
Problem statement: Derive the relations given by equations [1.4] and [1.5] of the textbook.

If we have constant shear $$\tau\!$$ on a curved panel, we can break down the shear force into two components:


 * $$V_x=\tau\,t\,a \qquad \qquad [1.4] \!$$
 * $$V_y=\tau\,t\,b \qquad \qquad \, [1.5] \!$$



From Figure 2, we can find an expression for the differential resulting force $$d\vec V$$:


 * $$d\vec V=q\,\vec {dl}$$

Where $$q\!$$ is the shear flow and $$\vec {dl}$$ > is the differential length that is being studied.

Knowing $$\vec {dl}$$ > has two components (one in $$x\!$$ and another in $$y\!$$), we are able to work out our answer.


 * $$d\vec V=q\,\left( dl_x \, \hat i + dl_y \, \hat j\right)$$
 * $$=q\,\left( dl\, \cos \theta \, \hat i + dl\, \sin \theta \, \hat j\right)$$
 * $$=q\,\left( dx\, \hat i + dy\, \hat j\right)$$

Now, integrating both sides,


 * $$\int {d \vec V}=q\, \left[ \left( \int_{x_a}^{x_b} {dx} \right)\, \hat i + \left( \int_{y_a}^{y_b} {dy} \right)\, \hat j\right]$$

and we also know $$\int_{x_a}^{x_b} {dx}=a$$ and $$\int_{y_a}^{y_b} {dy}=b$$

therefore,


 * $$\vec V=q\, \left( a\, \hat i + b\, \hat j\right)=qa\, \hat i + qb\, \hat j$$

and knowing $$q = \tau \, t \!$$ we now have


 * $$\vec V=\tau t  a\ \hat i + \tau  t  b\ \hat j$$

So this gives us the two components previously stated,


 * $$V_x=\tau\,t\,a \!$$
 * $$V_y=\tau\,t\,b \!$$