1. Material Properties and Forces

2. Centroid Principles Object’s center of gravity or center of mass. Graphically labeled as

3. Centroid Principles One can determine a centroid location by utilizing the cross-sectional view of a three-dimensional object.

4. Centroid Location Symmetrical Objects Centroid location is determined by an object’s line of symmetry. Centroid is located on the line of symmetry. When an object has multiple lines of symmetry, its centroid is located at the intersection of the lines of symmetry.

5. Moment of Inertia Principles Moment of Inertia (I) is a mathematical property of a cross section (measured in inches 4 ) that gives important information about how that cross-sectional area is distributed about a centroidal axis. In general, a higher moment of inertia produces a greater resistance to deformation. Stiffness of an object related to its shape. ©iStockphoto.com

6. BeamMaterialLengthWidthHeightArea ADouglas Fir8 ft1 ½ in.5 ½ in.8 ¼ in. 2 BDouglas Fir8 ft5 ½ in.1 ½ in.8 ¼ in. 2 Moment of Inertia Principles Two beams of equal cross-sectional area Difference is the orientation of the load

7. Will beam A or beam B have a greater resistance to bending, resulting in the least amount of deformation, if an identical load is applied to both beams at the same location? What distinguishes beam A from beam B? Moment of Inertia Principles

8. Calculating Moment of Inertia – Rectangles Why did beam B have greater deformation than beam A? Moment of Inertia Principles Difference in moment of inertia due to the orientation of the beam Significant influence b = Width of sample h = Thickness of sample

9. Calculating Moment of Inertia Calculate beam A moment of inertia

10. Calculating Moment of Inertia Calculate beam B moment of inertia

11. Moment of Inertia 13.5 times stiffer Beam A Beam B

12. Simple Shape vs. Flange Beams Doing more with less I = 10.67 in. 4 Area = 8.00 in. 2 I = 6.08 in. 4 Area = 2.75 in. 2 Complex Shapes Use This Power

13. Moment of Inertia – Composites Why are composite materials used in structural design? Styrofoam (Weak) Fiberglass Sandwich (Weak) Styrofoam + Fiberglass (Strong) + =

14. Modulus of Elasticity (E) The ratio of the increment of some specified form of stress to the increment of some specified form of strain. Also known as Young’s Modulus. In general, a higher modulus of elasticity produces a greater resistance to deformation. Structural Member Properties

15. Tension Stress A body being stretched

16. Compression A body being squeezed

17. Modulus of Elasticity (E) The proportional constant (ratio of stress and strain) A measure of stiffness – The ability of a material to resist stretching when loaded Tensile Test – Stress-Strain Curve stress = load Area strain = amount of stretch original length or

18. Plastic Deformation Unrecoverable elongation beyond the elastic limit When the load is removed, only the elastic deformation will be recovered Tensile Test – Stress-Strain Curve

19. Modulus of Elasticity Principles BeamMaterialLengthWidthHeightAreaI ADouglas Fir8 ft1 ½ in.5 ½ in.8 ¼ in. 2 20.8 in. 4 BABS plastic8 ft1 ½ in.5 ½ in.8 ¼ in. 2 20.8 in. 4

20. Modulus of Elasticity Principles What distinguishes beam A from beam B? Will beam A or beam B have a greater resistance to bending, resulting in the least amount of deformation, if an identical load is applied to both beams at the same location?

21. Why did beam B have greater deformation than beam A? Modulus of Elasticity Principles Difference in material modulus of elasticity The ability of a material to deform and return to its original shape Applied force or load Length of span between supports Modulus of elasticity Moment of inertia Characteristics of objects that impact deflection (Δ MAX )

22. Calculating Beam Deflection BeamMaterialLength (L) Moment of Inertia (I) Modulus of Elasticity (E) Force (F) ADouglas Fir8 ft20.8 in. 4 1,800,000 psi 250 lbf BABS Plastic8 ft20.8 in. 4 419,000 psi 250 lbf Δ Max = F L 3 48 E I

23. Calculating Beam Deflection BeamMaterialLengthIELoad ADouglas Fir8 ft20.8 in. 4 1,800,000 psi 250 lbf Calculate beam deflection for beam A Δ Max = F L 3 48 E I Δ Max = 250 lb f (96 in.) 3 48 (1,800,000 psi) (20.8 in. 4 ) Δ Max = 0.123 in.

24. Calculating Beam Deflection BeamMaterialLengthIELoad BABS Plastic8 ft20.8 in. 4 419,000 psi 250 lbf Δ Max = F L 3 48 E I Calculate beam deflection for beam B Δ Max = 250 lb f (96 in.) 3 48 (419,000 psi) (20.8 in. 4 ) Δ Max = 0.53 in.

25. Douglas Fir vs. ABS Plastic 4.24 Times less deflection Δ Max A = 0.123 in. Δ Max B = 0.53 in.

26. Statics The study of forces and their effects on a system in a state of rest or uniform motion ©iStockphoto.com

27. Equilibrium Translational equilibrium: The state in which there are no unbalanced forces acting on a body ©iStockphoto.com Balanced Unbalanced Static equilibrium: A condition where there are no net external forces acting upon a particle or rigid body and the body remains at rest or continues at a constant velocity

28. Equilibrium Rotational equilibrium: The state in which the sum of all clockwise moments equals the sum of all counterclockwise moments about a pivot point ©iStockphoto.com Remember Moment = F x D

29. Truss Analysis Primary truss loads – loads calculated with ideal assumptions Used in welded steel-tube fuselages, piston- engine motor mounts, ribs, and landing gear

30. Truss Analysis – Engine Mount Example Line of force is from the center of gravity of the engine Rigid connection from the fuselage and engine to the truss 3,200lbf

31. Summary Centroid is object’s center of gravity or center of mass Moment of Inertia (I) is a mathematical property of a cross section Significant influence

32. Summary Composite shapes used in structural design to create lightweight, strong material Modulus of Elasticity (E) The ratio of the increment of some specified form of stress to the increment of some specified form of strain Deflection calculated using modulus of elasticity Δ Max = F L 3 48 E I

33. Summary Equilibrium –Translational –Rotational