University of Florida/Eas4200c.f08.radsam.d

HOMEWORK 1

1.2 - Basic Structural Elements in Aircraft Structure Aircraft structures are composed of several structural elements, which are designed to withstand various types of loads. It is the combination of these elements that make the entire structure of an aircraft capable of resisting applied loads. For better comprehension of structural mechanics, we must introduce a few important definitions: Stiffness: It is an extensive material property that determines the resistance of an elastic body to deflection. In Engineering, the modulus on elasticity or Young's Modulus (E) is an extremely important characteristic of materials, which shows how much a given material will be able deflect before a permanent deformation occurs. The Young's Modulus or is the radio of applied stress to strain: E = $\displaystyle \sigma$/$\displaystyle \epsilon$ Strength: The strength of a material is the ability to resist an applied force. In engineering, the Yield strength ($\displaystyle \sigma$y) is defined as "the stress at which a material begins to deform plastically." Elastic or elastoplastic materials are able to withstand certain loads and go back to their original shape when the given load is removed. This effect will occur as long as the applied stress is less than the Yield stress (strength). However, if the applied load exceeds the yield stress, the material will deform permanently even when the load is removed. If the strain or deformation keeps increasing, the material may reach its ultimate rupture stress ($\displaystyle \sigma$u) point where it will break. Yield behavior for non-ferrous alloys. Typical yield behavior for non-ferrous alloys. 1: True elastic limit 2: Proportionality limit 3: Elastic limit 4: Offset yield strength True elastic limit This is the stress point where dislocations begin to move. Since these effects occur at microscopic level and are not easy to detect, this definition is not widely used. Proportionality limit Up to this stress point, the stress/strain relationship is linear and directly proportional to each other. The slope of this straight line is the previously mentioned Modulus of Elasticity (E) that corresponds to the stiffness of the material. At this point, the material still behaves elastically, which means the material will return to its original state after an applied load is lifted. Elastic limit (yield strength) Point 3 corresponds to the Elastic limit. If the applied stress is greater than the one corresponding to this point, the material will deform permanently. Offset yield point (proof stress) An offset point is sometimes used instead of the elastic limit since some materials do not have a perfectly linear elastic region. As shown in the figure to the right, a straight line is drawn between point 4 and an offset of .2% to be able to obtain the slope of the line (Young's Modulus) for the material. Point two percent is mostly used for yield stress in metals, but other values may be used for other materials and applications. Contribution by: radsam.d Toughness: The difference between toughness and fracture toughness is that the former is the ability of a material to resist fracture, while the latter is the ability of the material to resist fracture when a crack is already present. If a material has a high value of fracture toughness, it will most likely undergo ductile fracture. This means that it will have the ability to bend (passed the yield strength) before it fractures. On the other hand, if a material has a low value of fracture toughness, such as ceramics, it will undergo brittle fracture. 1.2.1 - Axial Member Axial Member The length of axial members is significantly larger than their cross-sections and are meant to hold compressive and extensional loads applied along its length (axial direction). Deriving from the slope of the stress vs. strain curve, we may obtain a relation for the applied stress: $\displaystyle \sigma$ = E$\displaystyle \epsilon$ The applied stress $\displaystyle \sigma$ is equal to the applied force per the cross-sectional area of the axial member, so substituting in the above equation we obtain: F = EA$\displaystyle \epsilon$ EA is the axial stiffness, which is only a function of the Young's Modulus and the cross-sectional area. This tells us that the stiffness of a member does not vary by changing the shape of a cross-section unless the actual total area increases or decreases. For instance, if a member with a circular cross-section has the same area as a member with a square cross-section (or any other shape), the axial stiffness will be the same for both members. Axial members are good for tensile stresses but may cause buckling failure when compressed. Two ways of improving the buckling strength are to increase the bending stiffness or to shorten the buckle mode by adding supporting members at the joints. In aircraft structures, aside of the longitudinal axial members, supporting members must be added at the wings and the fuselage to be able to withstand stresses. The two most important parts in a wing structure are the spars (2, refer to figure) and ribs (3). Spars extend lengthwise to the wing covering its entire length (from the fuselage to the wing tip). The main purpose of these members is to provide support to the wing and carry all loads back to the fuselage. The ribs on the other hand, give the aerodynamic shape to the wing while providing support to the spars. Contribution by: radsam.d

September $22^{\mbox{th}}$ - Curved Panels

Curved panel diagram as drawn in class - Figure 1 Figure 1 shows a thin curved panel being exposed to a shear stress. With the help of this diagram, we may derive the following two equations of the book: $V_{y}=\tau \,ta\,$ (1.4) $V_{z}=\tau \,tb\,$ (1.5) where: $V_{y}\,$ is the shear stress in the y direction. $V_{z}\,$ is the shear stress in the z direction. Curved panel diagram from the side - Figure 2 A side view of the curved panel is shown in figure 2. In this diagram we may see how the force(dF) acts on a small section(dl) at and angle theta. This force may be decomposed in two components along Y and Z. Also, knowing that the force is equal to the shear flow(q) times the length, we obtain the following relation: $q=\tau \,t$ $dF=qdl=q(dl_{y}+dl_{z})\,$ Following the geometry on figure 2, we know: $dl_{y}=dlcos\theta \,$ $dl_{z}=dlsin\theta \,$ Therefore: $dF=q(dlcos\theta \,+dlsin\theta \,)$ $dF=q(dy+dz)\,$ Detailed curved panel dl section - Figure 3 A detailed drawing of the shear force being applied on the small element 'dl' is shown in figure 3. We may clearly see how dF is out an an angle and how it may be decomposed into components. To obtain the total value of F, we integrate dF from point A to B. The resultant force is computed as follows: $F=\iint _{A}^{B}dF\,=q(\iint _{A}^{B}dy\,+\iint _{A}^{B}dz\,)\,$ If we look at figure 2 gain, one may see that teh horizontal distance from A to B is equal to 'a', and that the vertical distance between A and B is equal to 'b'. Therefore whe may solve the two integrals above and obtain an expression for F. $F=q(a{\hat {j}}+b{\hat {k}})\,$ This last equation corresponds to equations 1.4 and 1.5, because we know that: $q=\tau \,t$ The expression for F shown above is still decomposed for 'y' and 'z' coordinates, but one may compute the total resultant force in the actual direction of stress: $||F||=(F_{y}^{2}+F_{z}^{2})^{(1/2)}\,$ $||F||=(q(a^{2}+b^{2})^{(1/2)})\,$ Where: $d=(a^{2}+b^{2})^{(1/2)}\,$ -- Length of a straight line. So we conclude: $F=qd\,$ -- Which is essentially Eq. 3.49a

September $22^{\mbox{nd}}$ - Proof of equation 3.48

Figure 1 Figure 1 shows a constant flow on a closed thin-walled section. Equation 3.48 implies that the torque 'T' is equal to 2 times the shear flow and the enclosed area. It is written as follows: $T=2q{\bar {A}}\,$ Where ${\bar {A}}\,$ is the enclosed area and $q\,$ is the shear flow. We know that the torque at a point on the thin wall is the radius times the force being applied. Also, the force is equal to the shear flow times the length of the element. $dT={\vec {r}}\ xd{\vec {F}}\,$ $dT=\rho \,dF\,=\rho \,(qdl)\,$ Figure 2 Figure 2 shows a closer view of the element $dl\,$ and the force $dF$. Due to the irregular shape of the cross-section, the radius may not always be perpendicular to the wall, so we draw a vector $\rho \,$ perpendicular to $dF\,$. This forms the triangle shown in Figure 2, so the area of the triangle $dA$ may de computed: $dA={\frac {1}{2}}\rho \,dl\,$ The total torque applied to the thin-walled is the integral of $dT\,$

HW: Proof of Moment of Inertia for a solid circular cross-section

First, we derive the polar moment of inertia equation from the following relation: $J=\iint _{A}z^{2}~dA\,$ where $z$ is the distance from the edge to the axis about which the torsion occurs. Transforming the expression to polar coordinates, we obtain: $J=\int _{o}^{2\pi \,}\int _{o}^{r}\rho \,^{3}d\rho \,d\theta \,$ where $\rho \,$ is the distance from the origin to the wall. Now, solving the integral, we obtain the expected expression in terms of $r\,$: $J={\frac {1}{2}}\pi \,r^{4}$ Contribution by radsam.d

HW: Proof of Moment of Inertia for a hollow circular cross-section

Assumtion: $t<<r\,$ Since the moment of inertia for a circle was derived in the previous section, we may use this equation for the outer and inner radius: $J_{o,out}={\frac {1}{2}}\ \pi \,r_{o}^{4}$ $J_{o,in}={\frac {1}{2}}\ \pi \,r_{i}^{4}$ Therefore, we subtract $J_{out}-J_{in}\,$ to compute for the moment of inertia of the thin wall: $J_{o}={\frac {1}{2}}\ \pi \,(r_{o}^{4}-r_{i}^{4})$ One may also assume that $r_{o}=r_{i}+t\,$. And with this assumption, one obtains an reasonable average radius ${\bar {r}}\,$: ${\bar {r}}\,={\frac {r_{o}+r_{i}}{2}}={\frac {2r_{i}+t}{2}}={\frac {2r_{o}-t}{2}}\,$ ${\bar {r}}\,=r_{i}+{\frac {t}{2}}=r_{o}-{\frac {t}{2}}\,$ But, assuming $t<<r\,$: ${\bar {r}}\,=r_{i}=r_{o}\,$ Going back to $J_{o}\,$: $J_{o}={\frac {1}{2}}\ \pi \,(r_{o}-r_{i})(r_{o}+r_{i})(r_{o}^{2}+r_{i}^{2})\,$ $J_{o}={\frac {1}{2}}\pi \,t~2{\bar {r}}\,~2{\bar {r}}\,^{2}\,$ $J_{o}=2\pi \,t{\bar {r}}\,^{3}\,$ Contribution by radsam.d

HW 4

HW 4: 10/8/08

Introduction to Multicell Sections Multicell Thin-Walled Sections (Refer to Section 3.6 on book) Wing sections are many times composed of multicell construction. The skin of the airfoil is supported by thin vertical webs and these are supported by stiffeners. In order to calculate the torsion of a single cell, the Prandtl stress function must be constant along each one of the boundaries. For instance, for a two cell section, there are three boundary contours: $\phi \ (S_{0})=C_{0}\,$ $\phi \ (S_{1})=C_{1}\,$ $\phi \ (S_{2})=C_{2}\,$ Where $C_{0},C_{1},C_{2}\,$ are constants. The shear flow is considered to be positive in the counterclockwise direction, and is the difference between the inside contour and the outside contour. In addition, the total shear flow on the system is the superposition of the two obtained shear flows(one from each cell), and with this, the total torque may be computed with the following relation[1]: $T=2{\bar {A}}_{1}\ q_{1}+2{\bar {A}}_{2}\ q_{2}\,$ 1. Mechanics of Aircraft Structures Contribution by radsam.d Given: $t_{1}=0.3cm\,$ $t_{2}=0.5cm\,$ $t_{12}=0.4cm\,$ $a=30cm\,$ $b=60cm\,$ $c=40cm\,$ Find $\theta \,$ as a function of torque $T\,$ and torsional constant $J\,$ Due to superposition of shear flows, we know the total torque is the summation of $T_{1}\,$ and $T_{2}\,$ $T=T_{1}+T_{2}=2{\bar {A}}_{1}\ q_{1}+2{\bar {A}}_{2}\ q_{2}\,$ $Eqn.1\,$ Computing the areas for each one of the cells: ${\bar {A}}_{1}\ =ac\,$  ; ${\bar {A}}_{2}\ =bc$ The expression for the twist angle $\theta \,$ was previously computed. Refer to Eqn. 3.56 in "Mechanics of Aircraft Structures": $\theta _{1}\ ={\frac {1}{2G{\bar {A}}_{1}}}\oint {\frac {q_{1}}{t_{1}}}ds\,$ The computed area is substituted into twist angle equation: $\theta _{1}\ ={\frac {1}{2G{\bar {A}}_{1}}}\left[{\frac {2q_{1}a}{t_{1}}}+{\frac {q_{1}c}{t_{1}}}{\frac {(q_{1}-q_{2})c}{t_{12}}}\right]$ $Eqn.2\,$ The same procedure followed above is used to obtain the twist angle for cell 2: $\theta _{2}\ ={\frac {1}{2G{\bar {A}}_{2}}}\oint {\frac {q_{2}}{t_{2}}}ds\,$ $\theta _{2}\ ={\frac {1}{2G{\bar {A}}_{2}}}\left[{\frac {2q_{2}b}{t_{2}}}+{\frac {q_{2}c}{t_{2}}}{\frac {(q_{2}-q_{1})c}{t_{12}}}\right]$ $Eqn.3\,$ $c_{1}\,$ and $c_{2}\,$ have the same rate of twist angle, therefore: $\theta _{1}\ =\theta _{2}\,$ $Eqn.4\,$ With the four equations above, expressions for $q_{1}\,$ and $q_{2}\,$ in terms of $T\,$ may be obtained. For instance: $q_{1}=\beta _{1}\ T\,$ $q_{2}=\beta _{2}\ T\,$ Where: $\beta _{1}\ =0.01\,$ and $\beta _{2}=0.016\,$ in N/m. "Computation" In addition, a general expression for twist angle $\theta \,$ may be obtained: $\theta \ =\theta _{1}\ =\theta _{2}\ ={\frac {T}{2GJ}}$

Homework: 10/8/08 - Computation of $\beta \,$

$T=T_{1}+T_{2}=2{\bar {A}}_{1}\ q_{1}+2{\bar {A}}_{2}\ q_{2}\,$ $Eqn.1\,$ $\theta _{1}\ ={\frac {1}{2G{\bar {A}}_{1}}}\left[{\frac {2q_{1}a}{t_{1}}}+{\frac {q_{1}c}{t_{1}}}{\frac {(q_{1}-q_{2})c}{t_{12}}}\right]$ $Eqn.2\,$ $\theta _{2}\ ={\frac {1}{2G{\bar {A}}_{2}}}\left[{\frac {2q_{2}b}{t_{2}}}+{\frac {q_{2}c}{t_{2}}}{\frac {(q_{2}-q_{1})c}{t_{12}}}\right]$ $Eqn.3\,$ $\theta _{1}\ =\theta _{2}\,$ $Eqn.4\,$ Given: $G=27Gpa\,$ ${\bar {A}}_{1}\ =(30)(40)=1200cm^{2}\,$ ${\bar {A}}_{2}\ =(60)(40)=2400cm^{2}\,$ $T=2q_{1}{\bar {A}}_{1}\ +2q_{2}{\bar {A}}_{2}\,$ $T=2400q_{1}+4800q_{2}\,$ $Eqn.5\,$ Substituting the given values for the dimensions of the cells, we obtain expressions for $\theta _{1}\,$ and $\theta _{2}\,$ : $\theta _{1}\ =1.54e^{-12}(200q_{1}+133q_{1}+100q_{1}-100q_{2})\,$ $\theta _{1}\ =6.7e^{-10}q_{1}-1.54e^{-10}q_{2}\,$ $Eqn.6\,$ $\theta _{2}\ =7.7e^{-13}(240q_{2}+80q_{2}+100q_{2}-100q_{1})\,$ $\theta _{2}\ =3.23e^{-10}q_{2}-7.7e^{-11}q_{1}\,$ $Eqn.7\,$ Since $\theta _{1}\ =\theta _{2}\,$, we combine equations 6 and 7: $6.7e^{-10}q_{1}-1.54e^{-10}q_{2}=3.23e^{-10}q_{2}-7.7e^{-11}q_{1}\,$ $7.47e^{-10}q_{1}=4.77e^{-10}q_{2}\,$ $Eqn.8\,$ $q_{1}=0.64q_{2}\,$ || $q_{2}=1.57q_{1}\,$ Substituting the values for $q_{1}\,$ and $q_{2}\,$ into equation 5: $q_{1}=0.010\left({\frac {N}{m}}\right)T$ $q_{2}=0.016\left({\frac {N}{m}}\right)T$ Contribution by radsam.d

HW5

Equations of Equilibrium in a Nonuniform Stress Field

Equations of equilibrium may be written in indicial notation as shown in the last meeting, and it is shown below once again: $\sum _{i=1}^{3}{\frac {\partial \sigma _{ij}}{\partial x_{i}}}=0~~~{\mbox{for}}~~~j=1,~2,~3\,$ Expressing this equation for x,y,z separately, we obtain: ${\frac {\partial \sigma _{xx}}{\partial \ x}}+{\frac {\partial \sigma _{yx}}{\partial \ y}}+{\frac {\partial \sigma _{zx}}{\partial \ z}}=0\,$ ${\frac {\partial \sigma _{xy}}{\partial \ x}}+{\frac {\partial \sigma _{yy}}{\partial \ y}}+{\frac {\partial \sigma _{zy}}{\partial \ z}}=0\,$ ${\frac {\partial \sigma _{xz}}{\partial \ x}}+{\frac {\partial \sigma _{yz}}{\partial \ y}}+{\frac {\partial \sigma _{zz}}{\partial \ z}}=0\,$ Contribution by radsam.d

For simplification purposes, we look at a a 1-dimensional model first. Equating all forces on the x-direction, we obtain:

$\sum \ F_{x}=0=-\sigma _{(x)}\ A+\sigma _{(x+dx)}\ A+f_{(x)}dx\,$

$~~~0=A[\sigma _{(x+dx)}\ -\sigma _{(x)}\ ]+f_{(x)}dx\,$

Applying Taylor series to the term inside the brackets, it becomes: $~~~{\frac {\partial \sigma _{x}}{\partial \ x}}dx+higher~order~temrs\,$

Recall: $F_{(x+dx)}=F_{(x)}+{\frac {\partial \ F_{(x)}}{\partial \ dx}}dx+{\frac {1}{2}}{\frac {\partial ^{2}F_{(x)}}{\partial x^{2}}}(dx)^{2}+.~.~.\,$

Neglecting higher order terms:

${\frac {\partial \sigma }{\partial \ x}}+{\frac {f_{(x)}}{A}}=0~~~~~~~~Applied~load\,$

Now, we may look at a non-uniform 3-D field without applied loads, and focusing on the x-direction.

"Picture is worth a thousand words" - Vu-Quoc

The figure shows an infinitesimal element in which the stress is not uniform. However, the element must remain in equilibrium, therefore the six equations of equilibrium must be satisfied. For example, forces along the x-direction are:

$\sum \ F_{x}=0=dydz[-\sigma _{xx}\ (x,y,z)+\sigma _{xx}\ (x+dx,y,z)]~~~\,$ Facets with normal X

$+dzdx[-\sigma _{yx}\ (x,y,z)+\sigma _{yx}\ (x,y+dy,z)]~~~\,$ Facets with normal Y

$+dxdy[-\sigma _{zx}\ (x,y,z)+\sigma _{zx}\ (x,y,z+dz)]~~~\,$ Facets with normal Z

$0=(dxdydz)\left[{\frac {\partial \sigma _{xx}}{\partial \ x}}+{\frac {\partial \sigma _{yx}}{\partial \ y}}+{\frac {\partial \sigma _{zx}}{\partial \ z}}\right]\,$

HW 6

Consider plate of dimensions:

a = dimension along the x-axis
b = dimension along the y-axis

First, we look at a 1-D case:

$[f]={\frac {F}{L}}~~~~~~~~~~~~~~[A]=L^{2}\,\,$

Therefore:

$\left[{\frac {f}{A}}\right]={\frac {F}{L^{3}}}$

$[\sigma ]={\frac {F}{L^{2}}}~~\Rightarrow ~~\left[{\frac {\partial \sigma }{\partial x}}\right]={\frac {\frac {F}{L^{2}}}{L}}\,$

$\epsilon ={\frac {du}{dx}}~~and~~\epsilon ={\frac {\triangle L}{L}}~~\Rightarrow [\epsilon ]={\frac {[du]}{[dx]}}={\frac {L}{L}}=1\,$

$\nu =-{\frac {\epsilon _{yy}}{\epsilon _{xx}}}~~\Rightarrow ~~[\nu ]=1\,$

From book...pg 71

The vector $t$ vanishes because no loads are applied on the lateral surface:
${t}=[\sigma ]{n}$

this way, the stress vector may be evaluated on the lateral surface, knowing that $n_{z}=0$, thus:

${\begin{Bmatrix}t_{x}\\t_{y}\\t_{z}\end{Bmatrix}}={\begin{bmatrix}0&0&\tau _{xz}\\0&0&\tau _{yz}\\\tau _{xz}&\tau _{yz}&0\end{bmatrix}}{\begin{Bmatrix}n_{x}\\n_{y}\\0\end{Bmatrix}}\,$

Therefore, we have: $t_{x}=0~~~~~~~~~~~~t_{y}=0$

$t_{z}=\tau _{xz}n_{x}+\tau _{yz}n_{y}={\frac {\partial \phi }{\partial y}}n_{x}-{\frac {\partial \phi }{\partial y}}n_{y}$

If we look at Figure 2, we may easily derive:

$n_{x}=\sin \eta ={\frac {dy}{ds}}\,$

$n_{y}=\cos \eta ={\frac {dx}{ds}}$

So we may express $t_{z}\,$ as:

$t_{z}={\frac {\partial \phi }{\partial y}}{\frac {dy}{ds}}+{\frac {\partial \phi }{\partial x}}{\frac {dx}{ds}}={\frac {d\phi }{ds}}\,$

So, the free boundary condition $t_{z}=0\,$ is given by: ${\frac {d\phi }{ds}}=0~~or~~\phi =constant$ on the lateral surface.

Note: For solid sections with a single contour boundary, this constant may be approximated to zero.

We are interested in the shear stresses $\tau _{xz}~and~\tau _{yz}\,$ and the resultant torque. Considering a small area $dA=dxdy\,$, the torque is:

$dT=x\tau _{yz}dA-y\tau _{xz}dA\,$

$dT={\begin{pmatrix}-x{\frac {\partial \phi }{\partial x}}-y{\frac {\partial \phi }{\partial y}}\end{pmatrix}}dA\,$