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

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

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

4. Centroid Location
Centroids
Principles of EngineeringTM
Unit 4 – Lesson 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.

5. Moment of Inertia Principles
Principles of EngineeringTM
Unit 4 – Lesson 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.
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©iStockphoto.com

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

7. Moment of Inertia Principles
Principles of EngineeringTM
Unit 4 – Lesson 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

8. Moment of Inertia Principles
Principles of EngineeringTM
Unit 4 – Lesson 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?

9. Calculating Moment of Inertia – Rectangles
Moment of Inertia Principles
Moment of Inertia
Principles of EngineeringTM
Unit 4 – Lesson 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

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

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

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

13. Simple Shape vs. Flange Beams
Moment of Inertia
Principles of EngineeringTM
Unit 4 – Lesson 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 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 = in.4
I = 6.08 in.4
Area = 8.00 in.2
Area = 2.75 in.2

14. Moment of Inertia – Composites
Principles of EngineeringTM
Unit 4 – Lesson 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)

15. Structural Member Properties
Moment of Inertia
Principles of EngineeringTM
Unit 4 – Lesson 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.

16. 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

17. 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

18. 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

19. 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.

20. Modulus of Elasticity Principles
Moment of Inertia
Principles of EngineeringTM
Unit 4 – Lesson 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

21. Modulus of Elasticity Principles
Moment of Inertia
Principles of EngineeringTM
Unit 4 – Lesson 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?

22. Modulus of Elasticity Principles
Moment of Inertia
Principles of EngineeringTM
Unit 4 – Lesson 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

23. Calculating Beam Deflection
Moment of Inertia
Principles of EngineeringTM
Unit 4 – Lesson 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

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

25. Calculating Beam Deflection
Moment of Inertia
Principles of EngineeringTM
Unit 4 – Lesson Statics
ΔMax = F L3
48 E I
Calculate beam deflection for beam B
ΔMax = 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

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

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

28. Equilibrium Static equilibrium: Translational equilibrium:
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson 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
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©iStockphoto.com
Balanced
Unbalanced

29. Equilibrium Rotational equilibrium:
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson 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
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30. Truss Analysis
Primary truss loads – loads calculated with ideal assumptions Used in welded steel-tube fuselages, piston-engine motor mounts, ribs, and landing gear

31. 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

32. 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

33. 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

34. Summary
Equilibrium
Translational
Rotational

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