User:Eas4200c.f08.aeris.guan/919lecture

Torque, Shear Flow, and Shear Panels
Recall that from the problem 1.1 that the torque of the thin-walled cross section equaled to twice the shear stress multiplied by the product of the dimensions given by the equation $$ \tau=\frac{T}{2abt}$$. This expression for the torque's relationship with the cross sectional dimensions is only for a particular case where the cross section of the thin-walled beam or member being analyzed is rectangular. For a more general case of relating the torque applied on an object, it is necessary to introduce the concept of shear flow. Shear flow is the product of the shear stress and the thickness of the cross section. Usually, this thickness is very small compared to the contour length. For more general case of relating torque with dimensions and shear flow, a formula is introduced in section 3.5 of "Mechanics of Aircraft Structures" by C.T. Sun. $$\displaystyle T= \oint{\rho qds}=\int\int_{A_{bar}}{2qdA}=2qA_{bar}$$ This formula describes the torque of a thin-walled bar with an arbitrary cross section of thickness t. Here, q is the shear flow, t is the thickness, A bar is the average area of the cross section, and T is the torque. In summary, for a nonuniform thin-walled cross section,$$T=2qA_{bar}:$$  and for a thin-walled rectangular cross section$$T=2 \tau ab$$:  Deriving the torque for the thin-walled rectangular cross section was an ad hoc method because of the assumptions used to determine the equation. However, it is logical. The derivation of the non-uniform cross section torque is a method based on elasticity. Upon scrutinization of the stringers of an aircraft that is part of a monocoque structure, the open thin-walled box structure was previously determined to be more practical that the closed structure because of ease of riveting and installation. However, as the stringers were examined more carefully, it is realized that the vertical sections are slightly slanted. Why is this? That is because this structure is practical for stockpiling and storage and at the same time, not losing much moment of inertia due to the slanting of the vertical sections.  Starting chapter two, the structure of a shear panel is introduced. It is a thin sheet of material, usually metal aluminum, that can withstand shear load. There is a relationship that relates the shear modulus of a material and the shear strain with the shear stress that is applied the structure. $$\displaystyle \tau =G \gamma$$, where G is the shear modulus, and gamma is the shear strain. Gamma describes the change in a right angle between two fictitious lines in the material due to deformation. It can be expressed by displacements in an (x,y) coordinate system relative to the structure. $$\displaystyle \gamma = \frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=\frac{\partial u_{x}}{\partial y}+\frac{\partial u_y}{\partial y}$$ where $$u=u_x$$= the displacement along the x direction and similar for v. This shear strain is called the engineering shear strain and not the tensorial shear strain. These two quantities are different in the fact that $$\displaystyle \epsilon_{xy}=$$ (tensorial shear strain)$$\displaystyle = .5 \gamma_{xy}$$. The following picture describes the deformation of an infinitesimally small element (i.e. part of a shear panel) subject to shear forces. The blue shape is the deformed infinitesimally small rectangle. 