User:Eas4200c.f08.vqcrew.d/hw3

Curved Panels
Figure 1 shows a curved panel with uniform thickness $$t$$ and constant shear stress $$\tau $$. This diagram will be analyzed in order to determine the forces acting on the panel in both the y and z directions. The shear flow $$q$$ is defined as: $$q=t \tau $$

In order to find the total forces acting in the y and z direction, the panel must be divided up into small sections with differential lengths and force vectors. A side view of the panel with dimensions and the differential components is shown in Figure 2. The differential force element, $$d\bar{F}$$, is equal to the shear flow multiplied by the differential length of the element, or: $$d\bar{F}=qd\bar{L}$$



A larger view of the differential length and force element are shown in Figure 3. Figure 3 also shows the components of the differential length element. The differential force vector and length vector can be broken into y and z components and simplified as follows:

$$d\bar{F}=qd\bar{L}$$ $$d\bar{F}=q(dL_{y} \hat{j} + dL_{z} \hat{k})$$ $$d\bar{F}=q(dL cos\theta \hat{j} + dL sin \theta  \hat{k})$$ $$d\bar{F}=q(dy \hat{j} + dz  \hat{k})$$

By integrating along the path from A to B, we can find the total force acting on the curved panel.

$$\bar{F}=\int_{A}^{B}{d \bar{F}}=q\left(\int_{A}^{B}{dy \hat{j}} + \int_{A}^{B}{dz \hat{k}}\right)$$ $$\bar{F}=q\left(a \hat{j} + b\hat{k}\right)$$

The force vector can be broken into y and z components and equated to the above solution to produce equations (1.4) and (1.5): $$\bar{F}=F_{y} \hat{j} + F_{z} \hat{k}$$ $$\Rightarrow F_{y} = qa = t \tau a$$ $$\Rightarrow F_{z} = qb = t \tau b$$

The relationship between the force components in a curved panel can be expressed as: $$\frac{F_y}{F_z}=\frac{a}{b}$$

The magnitude of the force, $$\left|\left| \bar{F} \right| \right|$$, can be expressed as follows: $$\left|\left| \bar{F} \right| \right|=\sqrt{(F_y)^2+(F_z)^2} = q\sqrt{a^2+ b^2}$$

However, $$\sqrt{a^2+ b^2}$$ is equal to the straight line distance, $$d$$, from point A to B. Therefore the force vector can also be expressed as: $$\bar{F}=qd$$

Torsion in a Closed, Thin-walled Beam
Figure XX depicts a generic, closed, thin-walled cross-section for a beam under torsion. The coordinate system may be defined from any arbitrary point in the plane of the section. The torque vector points in the positive x direction, or: $$\bar{T}=T \hat{i}$$ The differential unit of torque can be described as the cross product of the distance from the origin to the differential length segment with the differential force vector, or: $$d \bar{T}= \bar{r} \times d \bar{F}$$ The magnitude of this element is: $$dT=\rho dF$$ where $$\rho$$ is the perpendicular distance from the origin to the force vector and $$dF=qdL$$. To find the total torque, the differential torque element must be integrated about the closed path of the cross-section.

$$T=\oint{dT} = \oint{\rho dF}=q\oint{\rho dL}$$ However, $$\rho dL$$ is also equal to twice the area of the triangle swept out by the $$r$$ vector and $$dL$$. We can evaluate the integral over the average area, $$\bar{A}$$, to find the total torque as follows:

$$T=2q\int _{\bar{A}}{dA}=2q\bar{A}$$