This site uses cookies to deliver our services and to show you relevant ads and presentations. By clicking on "Accept", you acknowledge that you have read and understand our Cookie Policy , Privacy Policy , and our Terms of Use.

Download AIRCRAFT STRUCTURES PowerPoint Presentation

Login   OR  Register

Iframe embed code :

Presentation url :


Description :

AIRCRAFT STRUCTURES Presentation : Download AIRCRAFT STRUCTURES ppt presentation and make your presenation attractive

Tags :


Home / Travel & Tourism / Travel & Tourism Presentations / AIRCRAFT STRUCTURES PowerPoint Presentation

AIRCRAFT STRUCTURES PowerPoint Presentation

Ppt Presentation Embed Code   Zoom Ppt Presentation

About This Presentation

Description : AIRCRAFT STRUCTURES Presentation : Download AIRCRAFT STRUCTURES ppt presentation and make your prese... Read More


Published on : Dec 09, 2014
Views : 576 | Downloads : 1

Download Now

Share on Social Media


PowerPoint is the world's most popular presentation software which can let you create professional AIRCRAFT STRUCTURES powerpoint presentation easily and in no time. This helps you give your presentation on AIRCRAFT STRUCTURES in a conference, a school lecture, a business proposal, in a webinar and business and professional representations.

The uploader spent his/her valuable time to create this AIRCRAFT STRUCTURES powerpoint presentation slides, to share his/her useful content with the world. This ppt presentation uploaded by worldwideweb in this Travel & Tourism category is available for free download,and can be used according to your industries like finance, marketing, education, health and many more. provides a platform to marketers, presenters and educationists along with being the preferred search engine for professional PowerPoint presentations on the Internet to upload their AIRCRAFT STRUCTURES ppt presentation slides to help them BUILD THEIR CROWD!!

User Presentation
Related Presentation
Free PowerPoint Templates
Slide 2 - Course Objective The purpose of the course is to teach the principles of solid and structural mechanics that can be used to design and analyze aerospace structures, in particular aircraft structures.
Slide 3 - ppt slide no 3 content not found
Slide 4 - Airframe
Slide 5 - Function of Aircraft Structures General The structures of most flight vehicles are thin walled structures (shells) Resists applied loads (Aerodynamic loads acting on the wing structure) Provides the aerodynamic shape Protects the contents from the environment
Slide 6 - Definitions Primary structure: A critical load-bearing structure on an aircraft. If this structure is severely damaged, the aircraft cannot fly. Secondary structure: Structural elements mainly to provide enhanced aerodynamics. Fairings, for instance, are found where the wing meets the body or at various locations on the leading or trailing edge of the wing.
Slide 7 - Definitions… Monocoque structures: Unstiffened shells. must be relatively thick to resist bending, compressive, and torsional loads.
Slide 8 - Definitions… Semi-monocoque Structures: Constructions with stiffening members that may also be required to diffuse concentrated loads into the cover. More efficient type of construction that permits much thinner covering shell.
Slide 9 - ppt slide no 9 content not found
Slide 10 - ppt slide no 10 content not found
Slide 11 - Function of Aircraft Structures: Part specific Skin reacts the applied torsion and shear forces transmits aerodynamic forces to the longitudinal and transverse supporting members acts with the longitudinal members in resisting the applied bending and axial loads acts with the transverse members in reacting the hoop, or circumferential, load when the structure is pressurized.
Slide 12 - Function of Aircraft Structures: Part specific Ribs and Frames Structural integration of the wing and fuselage Keep the wing in its aerodynamic profile
Slide 13 - Function of Aircraft Structures: Part specific Spar resist bending and axial loads form the wing box for stable torsion resistance
Slide 14 - Function of Aircraft Structures: Part specific Stiffener or Stringers resist bending and axial loads along with the skin divide the skin into small panels and thereby increase its buckling and failing stresses act with the skin in resisting axial loads caused by pressurization.
Slide 15 - Simplifications The behavior of these structural elements is often idealized to simplify the analysis of the assembled component Several longitudinal may be lumped into a single effective longitudinal to shorten computations. The webs (skin and spar webs) carry only shearing stresses. The longitudinal elements carry only axial stress. The transverse frames and ribs are rigid within their own planes, so that the cross section is maintained unchanged during loading.
Slide 16 - UNIT-I Unsymmetric Bending of Beams The learning objectives of this chapter are: •Understand the theory, its limitations, and its application in design and analysis of unsymmetric bending of beam.
Slide 17 - UNIT-I UNSYMMETRICAL BENDING The general bending stress equation for elastic, homogeneous beams is given as where Mx and My are the bending moments about the x and y centroidal axes, respectively. Ix and Iy are the second moments of area (also known as moments of inertia) about the x and y axes, respectively, and Ixy is the product of inertia. Using this equation it would be possible to calculate the bending stress at any point on the beam cross section regardless of moment orientation or cross-sectional shape. Note that Mx, My, Ix, Iy, and Ixy are all unique for a given section along the length of the beam. In other words, they will not change from one point to another on the cross section. However, the x and y variables shown in the equation correspond to the coordinates of a point on the cross section at which the stress is to be determined.                                                                         (II.1)
Slide 18 - Neutral Axis: When a homogeneous beam is subjected to elastic bending, the neutral axis (NA) will pass through the centroid of its cross section, but the orientation of the NA depends on the orientation of the moment vector and the cross sectional shape of the beam. When the loading is unsymmetrical (at an angle) as seen in the figure below, the NA will also be at some angle - NOT necessarily the same angle as the bending moment. Realizing that at any point on the neutral axis, the bending strain and stress are zero, we can use the general bending stress equation to find its orientation. Setting the stress to zero and solving for the slope y/x gives                                           (
Slide 19 - UNIT-II SHEAR FLOW AND SHEAR CEN Restrictions: Shear stress at every point in the beam must be less than the elastic limit of the material in shear. Normal stress at every point in the beam must be less than the elastic limit of the material in tension and in compression. Beam's cross section must contain at least one axis of symmetry. The applied transverse (or lateral) force(s) at every point on the beam must pass through the elastic axis of the beam. Recall that elastic axis is a line connecting cross-sectional shear centers of the beam. Since shear center always falls on the cross-sectional axis of symmetry, to assure the previous statement is satisfied, at every point the transverse force is applied along the cross-sectional axis of symmetry. The length of the beam must be much longer than its cross sectional dimensions. The beam's cross section must be uniform along its length.
Slide 20 - Shear Center If the line of action of the force passes through the Shear Center of the beam section, then the beam will only bend without any twist. Otherwise, twist will accompany bending. The shear center is in fact the centroid of the internal shear force system. Depending on the beam's cross-sectional shape along its length, the location of shear center may vary from section to section. A line connecting all the shear centers is called the elastic axis of the beam. When a beam is under the action of a more general lateral load system, then to prevent the beam from twisting, the load must be centered along the elastic axis of the beam.
Slide 21 - Shear Center The two following points facilitate the determination of the shear center location. The shear center always falls on a cross-sectional axis of symmetry. If the cross section contains two axes of symmetry, then the shear center is located at their intersection. Notice that this is the only case where shear center and centroid coincide.
Slide 24 - EXAMPLES For the beam and loading shown, determine: (a) the location and magnitude of the maximum transverse shear force 'Vmax', (b) the shear flow 'q' distribution due the 'Vmax', (c) the 'x' coordinate of the shear center measured from the centroid, (d) the maximun shear stress and its location on the cross section. Stresses induced by the load do not exceed the elastic limits of the material. NOTE:In this problem the applied transverse shear force passes through the centroid of the cross section, and not its shear center. FOR ANSWER REFER
Slide 25 - Shear Flow Analysis for Unsymmetric Beams SHEAR FOR EQUATION FOR UNSUMMETRIC SECTION IS
Slide 26 - SHEAR FLOW DISTRIBUTION For the beam and loading shown, determine: (a) the location and magnitude of the maximum transverse shear force, (b) the shear flow 'q' distribution due to 'Vmax', (c) the 'x' coordinate of the shear center measured from the centroid of the cross section. Stresses induced by the load do not exceed the elastic limits of the material. The transverse shear force is applied through the shear center at every section of the beam. Also, the length of each member is measured to the middle of the adjacent member. ANSWER REFER
Slide 27 - Beams with Constant Shear Flow Webs Assumptions: Calculations of centroid, symmetry, moments of area and moments of inertia are based totally on the areas and distribution of beam stiffeners. A web does not change the shear flow between two adjacent stiffeners and as such would be in the state of constant shear flow. The stiffeners carry the entire bending-induced normal stresses, while the web(s) carry the entire shear flow and corresponding shear stresses.
Slide 28 - Analysis Let's begin with a simplest thin-walled stiffened beam. This means a beam with two stiffeners and a web. Such a beam can only support a transverse force that is parallel to a straight line drawn through the centroids of two stiffeners. Examples of such a beam are shown below. In these three beams, the value of shear flow would be equal although the webs have different shapes. The reason the shear flows are equal is that the distance between two adjacent stiffeners is shown to be 'd' in all cases, and the applied force is shown to be equal to 'R' in all cases. The shear flow along the web can be determined by the following relationship
Slide 29 - Important Features of Two-Stiffener, Single-Web Beams: Shear flow between two adjacent stiffeners is constant. The magnitude of the resultant shear force is only a function of the straight line between the two adjacent stiffeners, and is absolutely independent of the web shape. The direction of the resultant shear force is parallel to the straight line connecting the adjacent stiffeners. The location of the resultant shear force is a function of the enclosed area (between the web, the stringers at each end and the arbitrary point 'O'), and the straight distance between the adjacent stiffeners. This is the only quantity that depends on the shape of the web connecting the stiffeners. The line of action of the resultant force passes through the shear center of the section.
Slide 30 - EXAMPLE For the multi-web, multi-stringer open-section beam shown, determine (a) the shear flow distribution, (b) the location of the shear center Answer
Slide 31 - UNIT-III Torsion of Thin - Wall Closed Sections Derivation Consider a thin-walled member with a closed cross section subjected to pure torsion.
Slide 32 - Examining the equilibrium of a small cutout of the skin reveals that
Slide 33 - ppt slide no 33 content not found
Slide 34 - Angle of Twist By applying strain energy equation due to shear and Castigliano's Theorem the angle of twist for a thin-walled closed section can be shown to be Since T = 2qA, we have If the wall thickness is constant along each segment of the cross section, the integral can be replaced by a simple summation
Slide 35 - Torsion - Shear Flow Relations in Multiple-Cell Thin- Wall Closed Sections The torsional moment in terms of the internal shear flow is simply
Slide 36 - Derivation For equilibrium to be maintained at a exterior-interior wall (or web) junction point (point m in the figure) the shear flows entering should be equal to those leaving the junction Summing the moments about an arbitrary point O, and assuming clockwise direction to be positive, we obtain The moment equation above can be simplified to
Slide 37 - Shear Stress Distribution and Angle of Twist for Two-Cell Thin-Walled Closed Sections The equation relating the shear flow along the exterior wall of each cell to the resultant torque at the section is given as This is a statically indeterminate problem. In order to find the shear flows q1 and q2, the compatibility relation between the angle of twist in cells 1 and 2 must be used. The compatibility requirement can be stated as              where
Slide 38 - The shear stress at a point of interest is found according to the equation To find the angle of twist, we could use either of the two twist formulas given above. It is also possible to express the angle of twist equation similar to that for a circular section
Slide 39 - Shear Stress Distribution and Angle of Twist for Multiple-Cell Thin-Wall Closed Sections In the figure above the area outside of the cross section will be designated as cell (0). Thus to designate the exterior walls of cell (1), we use the notation 1-0. Similarly for cell (2) we use 2-0 and for cell (3) we use 3-0. The interior walls will be designated by the names of adjacent cells. the torque of this multi-cell member can be related to the shear flows in exterior walls as follows
Slide 40 - For elastic continuity, the angles of twist in all cells must be equal The direction of twist chosen to be positive is clockwise.
Slide 42 - EXAMPLE For the thin-walled single-cell rectangular beam and loading shown, determine (a) the shear center location (ex and ey), (b) the resisting shear flow distribution at the root section due to the applied load of 1000 lb, (c) the location and magnitude of the maximum shear stress ANSWER REFER