| Introduction | ||
 Aircraft Fuselage Loading | A new passenger plane is being designed with a composite fuselage skin. The fibers will be placed in four different directions; 1) along the length of the fuselage or longitudinal (long) direction, 2) around the circumferential (circ) direction, 3) 30o from the longitudinal fibers, and 4) -30o from the longitudinal fibers. The skin is made from multiple layers of graphite reinforced epoxy, but the fibers are only placed in the four directions as described. What is known:
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| Question | ||
| What is the normal and shear stress in the direction of the 30o and -30o fibers? | ||
| Approach | ||
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| Combined Stress (or Loads) | |||||
| Stresses at a point (Stress Element) for a Cantilever Beam (Element on Beam is Movable) | In the previous sections, both the bending and shear stresses were presented for beams. Recall, the bending stress will cause a normal stress (either tension or compression, depending on vertical location) and the shear stress will cause a tearing stress. Both of the stresses can act at the same point and should be considered at the same time. Since only linear elastic materials are considered in this eBook, both stresses can be added together using the principle of superposition. This is shown at the left on a square element called a 'stress element'. This element is really just a point, but to see the stress direction, the point is shown as a square element. | ||||
| Stress Element | |||||
 Stresses at a point (Stress Element) for Pressurized Pipe | The beam above is just one possible configuration for multiple stresses acting at a point. Another possibility is a pipe that is pulled, pressurized and twisted at the same time. These three loads on the pipe will cause tension normal stress in both directions (axial and circumferential) and cause a twisting or shear stress. All three loads and their associate stresses can be combined together to give a total stress state at any point. The stress element for this example is shown at the left. This section will examine a stress element to better understand stresses at a point and how they can be analyzed. | ||||
| Sign Convention for Stress Element | |||||
 Sign Convention for Stress Element Positive Directions | The sign convention for stresses at a point is similar to other stresses. Normal tension stress in both the x and y direction are assumed positive. The shear stress is assumed positive as shown in the diagram at the left. Shear stress act on four sides of the stress element, causing a pinching or shear action. All shear stresses on all four sides are the same, thus τxy = τyx | ||||
| Stress Rotation | |||||
 Stress on an Inclined Plane | In the Normal Stress section, stress on an inclined plane was presented. It was noted that stresses are not vector quantities, and are not rotated by just using a single sin or cos function. For the normal and shear stress on an inclined plane, the rotated stresses were found to be
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 Rotated Stress Element | Similarly, the stresses at a point (stress element) can also be rotated to give a new stress state at any particular angle. The rotation angle, θ, is assumed positive using the right hand rule (counter-clockwise in the x-y plane is positive). The new coordinate system is labeled as x' and y'. The new rotated stresses are shown in the diagram at the left. The shear stresses, τx'y' and τy'x, are still equal. | ||||
 Plane Section through Stress Element  Surface Area on Stress Element Plane | The objective is to relate the new stress in x' and y' coordinate system to the original stresses in the x and y coordinate. To do this, the original stress element is sliced at an angle θ, as shown in the diagram at the left. The stresses on the cut plane must be in equilibrium with the stresses on the outside surfaces of the stress element. Remember, nothing is moving, so all stresses and their associated forces must obey static equilibrium equations, ΣF = 0 and ΣM = 0. Before the stresses are actually summed, the area on each surface needs to be defined. The plane section at the angle θ is assumed to have a basic area of dA. The stress element is really just a point, so the area is infinitesimal, or just dA. The other two surfaces are based on dA. The bottom surface will be 'sinθ dA' and the left surface will be 'cosθ dA', which are shown in the diagram at the left. Summing the forces in each direction gives ΣFx = 0 = (σx´ dA) cosθ - (τx´y´ dA) sinθ - σx (cosθ dA) - τxy (sinθ dA) ΣFy = 0 = (σx´ dA) sinθ - (τx´y´ dA) cosθ - σy (sinθ dA) - τxy (cosθ dA) There are two unknowns, σx´ and τx´y´ and two equations, so they can be determined, giving σx´ = σx cos2θ + σy sin2θ + 2 τxy sinθ cosθ τx´y´ = - (σx - σy ) sinθ cosθ +τxy (cos2θ - sin2θ) The y' direction can be developed in the same way, but the section plane is 90o offset. The final equation is σy´ = σx sin2θ + σy cos2θ - 2 τxy sinθ cosθ | ||||
 Rotated Stress Element | Using double angle trigonometry identities, these three equations can be simplified to
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 Stress Element from Plane Fuselage (Fibers oriented in 0o, 90o, 30o, and -30o) | A plane fuselage undergoes both a pressure and twist load. This causes a tension stress in both the longitudinal and circumferential directions and a twisting load or a shear stress. If a stress element is cut from the fuselage, the induced stresses can be shown in a common x-y coordinate system. This longitudinal, or x-direction would be 25 ksi. The circumferential or y-direction would be 50 ksi. The shear stress would be 25 ksi. These stresses are shown in the diagram at the left. The shear arrows in the diagram are in the negative direction, and thus the shear is a negative shear stress. | |
| Stresses in the +30o Fiber Direction | ||
 Initial and 30o Rotated Stress Element | The stresses are requested in both the +30o and -30o. For the +30o, the initial stress element is shown at the left with the positive directions and thus the shear stress is negative. The stress element needs to be rotated 30o in the positive direction. Using the stress transformation equations, the stresses in the new x'-y' coordinate system are   Simplifying gives σx´ = 37.5 - 12.5(0.5) - 25(0.8660) = 9.600 ksi σy´ = 37.5 + 12.5(0.5) + 25(0.8660) = 65.40 ksi τx´y´ = 12.5(0.8660) - 25(0.5) = -1.675 ksi | |
| Stresses in the -30o Fiber Direction | ||
 Initial and -30o Rotated Stress Element | The method to find the stresses in the -30o is the same as for the +30o. Starting with the basic stress transformation equations, gives  Simplifying gives σx´ = 37.5 - 12.5(0.5) - 25(-0.8660) = 52.90 ksi σy´ = 37.5 + 12.5(0.5) + 25(-0.8660) = 22.10 ksi τx´y´ = 12.5(-0.8660) - 25(0.5) = -23.33 ksi | |
 Stress vs. Rotation Angle | ||
| Any Angle | ||
| It is interesting to plot the changing stresses as a function of angle. As expected, the stresses vary in a periodic cycle. Due to the double angle trigonometry terms in all three equations, the period is 180o. | ||
 Stress Element | Example | |
| What is the shear stress on a plane 42o from the horizontal (plane a-a)? | ||
| Solution | ||
 Stress Element Rotated 42o | The stress state at other orientations can be determined using the stress rotation equations. This problems asks for the shear stress on a plane 42o from the horizontal. The basic parameters are σx = -10 ksi σy = -20 ksi τxy = 30 ksi θ = 42o Notice, the two normal stress are negative since the arrows in the original problem diagram are pointing in the negative direction. Using the shear stress rotation equation gives, τx´y´ = - [(σx -σy) sin 2θ] / 2 + τxy cos 2θ = -[(-10 -(-20)) sin84] / 2 + 30 cos84 = -4.973 + 3.136 ksi τx'y' = -1.837 ksi | |
