Material Properties and Forces Aerospace Engineering © 2011 Project Lead The Way, Inc.

Centroid Principles Centroids Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Object’s center of gravity or center of mass. Graphically labeled as

Centroid Principles Centroids Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics One can determine a centroid location by utilizing the cross-sectional view of a three-dimensional object.

Centroid Location Centroids Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics 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.

Moment of Inertia Principles Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Moment of Inertia (I) is a mathematical property of a cross section (measured in inches4) that gives important information about how that cross-sectional area is distributed about a centroidal axis. Stiffness of an object related to its shape. In general, a higher moment of inertia produces a greater resistance to deformation. ©iStockphoto.com ©iStockphoto.com

Watch the following clip: https://www.youtube.com/watch?v=uyU25DdONjo

Moment of Inertia Principles Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Two beams of equal cross-sectional area Difference is the orientation of the load Beam Material Length Width Height Area A Douglas Fir 8 ft 1 ½ in. 5 ½ in. 8 ¼ in.2 B

Moment of Inertia Principles Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics What distinguishes beam A from beam B? Because of the orientation, beam A with the long dimension oriented parallel to the load has a greater moment of inertia. This orientation is 13.5 times as stiff as beam B oriented with the short dimension aligned parallel to the load. 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?

Calculating Moment of Inertia – Rectangles Moment of Inertia Principles Moment of Inertia Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Why did beam B have greater deformation than beam A? Difference in moment of inertia due to the orientation of the beam Calculating Moment of Inertia – Rectangles Significant influence b = Width of sample h = Thickness of sample

Calculating Moment of Inertia Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Calculate beam A moment of inertia

Calculating Moment of Inertia Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Calculate beam B moment of inertia

Moment of Inertia 13.5 times stiffer Beam A Beam B Moment of Inertia Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics 13.5 times stiffer Beam A Beam B

Simple Shape vs. Flange Beams Moment of Inertia Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Doing more with less Calculating the moment of inertia for a composite shape, such as an I-beam, is beyond the scope of this presentation. The values of moment of inertia and cross-sectional area were given for purposes of comparison. Lead students to the observation that the beam’s stiffness is most influenced by the major cross-sectional area’s distance from the center of gravity. Further analysis: Both of these shapes are 2 in. wide x 4 in. tall, and both beams are comprised of the same material. The I-beam’s flanges and web are 0.38 in. thick. The moment of inertia for the rectangular beam is 10.67 in.4. Its area is 8 in.2. The moment of inertia for the I-Beam is 6.08 in.4. Its area is 2.75 in.2. The I-beam may be 43% less stiff than the rectangular beam, BUT it uses 66% less material. Increasing the height of the I-beam by about 1 in. will make the moment of inertia for both of the shapes equal, but the I-beam will still use less material (61% less). Complex Shapes Use This Power I = 10.67 in.4 I = 6.08 in.4 Area = 8.00 in.2 Area = 2.75 in.2

Moment of Inertia – Composites Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Moment of Inertia – Composites Why are composite materials used in structural design? + = Fiberglass Sandwich Composite shapes allow a weak material such as Styrofoam to act as a support for the stronger fiberglass material which is located farther away from the beam’s center of gravity. This design creates strong material that is lightweight . (Weak) Styrofoam Styrofoam + Fiberglass (Weak) (Strong)

Structural Member Properties Moment of Inertia Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics 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.

σ = 𝐹 𝐴 Tension Stress A body being stretched Applied load divided by cross-sectional area σ = 𝐹 𝐴 The shape of the cross section is not important Appropriate cross section is the smallest area in the loaded part

Compression σ = 𝐹 𝐴 A body being squeezed Load divided by area σ = 𝐹 𝐴 Only for parts that are very short compared to cross sectional dimensions or parts that are laterally constrained

Tensile Test – Stress-Strain Curve 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 σ = 𝐹 𝐴 stress = load or E = σ ϵ Area This slide is a review of POE content. Refer to the POE curriculum lesson 2.3 Material Testing if your students need more background information. ϵ = 𝛿 𝐿 0 strain = amount of stretch or original length

Tensile Test – Stress-Strain Curve Plastic Deformation Unrecoverable elongation beyond the elastic limit When the load is removed, only the elastic deformation will be recovered This slide is a review of POE content. Refer to the POE curriculum lesson 2.3 Material Testing if your students need more background information.

Modulus of Elasticity Principles Moment of Inertia Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Beam Material Length Width Height Area I A Douglas Fir 8 ft 1 ½ in. 5 ½ in. 8 ¼ in.2 20.8 in.4 B ABS plastic

Modulus of Elasticity Principles Moment of Inertia Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics 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?

Modulus of Elasticity Principles Moment of Inertia Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Why did beam B have greater deformation than beam A? Difference in material modulus of elasticity The ability of a material to deform and return to its original shape Characteristics of objects that impact deflection (ΔMAX) Applied force or load Length of span between supports Modulus of elasticity Moment of inertia

Calculating Beam Deflection Moment of Inertia Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics ΔMax = F L3 48 E I Beam Material Length (L) Moment of Inertia (I) Modulus of Elasticity (E) Force (F) A Douglas Fir 8 ft 20.8 in.4 1,800,000 psi 250 lbf B ABS Plastic 419,000 psi

Calculating Beam Deflection Moment of Inertia Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics ΔMax = F L3 48 E I Calculate beam deflection for beam A ΔMax = 250 lbf (96 in.)3 48 (1,800,000 psi) (20.8 in.4) ΔMax = 0.123 in. Beam Material Length I E Load A Douglas Fir 8 ft 20.8 in.4 1,800,000 psi 250 lbf

Calculating Beam Deflection Moment of Inertia Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics ΔMax = F L3 48 E I Calculate beam deflection for beam B ΔMax = 250 lbf (96 in.)3 48 (419,000 psi) (20.8 in.4) ΔMax = 0.53 in. Beam Material Length I E Load B ABS Plastic 8 ft 20.8 in.4 419,000 psi 250 lbf

Douglas Fir vs. ABS Plastic Moment of Inertia Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics 4.24 Times less deflection ΔMax B = 0.53 in. ΔMax A = 0.123 in.

Statics Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics The study of forces and their effects on a system in a state of rest or uniform motion ©iStockphoto.com ©iStockphoto.com ©iStockphoto.com

Equilibrium Static equilibrium: Translational equilibrium: Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics 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 Translational equilibrium: The state in which there are no unbalanced forces acting on a body ©iStockphoto.com ©iStockphoto.com Balanced Unbalanced

Equilibrium Rotational equilibrium: Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Rotational equilibrium: The state in which the sum of all clockwise moments equals the sum of all counterclockwise moments about a pivot point Remember Moment = F x D ©iStockphoto.com

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

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

Significant influence 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

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 E = σ ϵ σ = 𝐹 𝐴 ϵ = 𝛿 𝐿 0 ΔMax = F L3 48 E I

Summary Equilibrium Translational Rotational

Write a one page paper (handwritten) with pictures of what Moment of Inertia is.