1. Structural Member Properties
Moment of Inertia
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.
©iStockphoto.com
©iStockphoto.com

2. 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

3. Calculating Moment of Inertia
Principles of EngineeringTM
Unit 4 – Lesson Statics
Calculating Moment of Inertia
Calculate beam A Moment of Inertia

4. Moment of Inertia – Composite Shapes
Principles of EngineeringTM
Unit 4 – Lesson Statics
Moment of Inertia – Composite Shapes
Why are composite shapes used in structural design?

5. Beam Deflection Measurement of deformation Importance of stiffness
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Beam Deflection
Measurement of deformation
Importance of stiffness
Change in vertical position
Scalar value
Deflection formulas
**Deflection is the measure of the deformation in a structure and its structural members. Most structures are designed with stresses as the primary design consideration, but deflection of the structure is also important. In fact, deflection is as important a consideration as the strength of buildings materials, especially when designing long support spans such as beams. Stiffness is, therefore, ** an important component of deflection. Deflection is referred to as either a ** change or movement of a structure in a vertical direction when placed under a load or ** as a scalar value that describes the distance that a material bends when placed under a load. ** Formulas are used to calculate the deflection of common structural elements.
Project Lead The Way, Inc.
Copyright 2007

6. Beam Structure Examples
Deflection
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Beam Structure Examples
Beam-like structures are a part of products that humans use every day. These include, but are not limited to, ** diving boards and other cantilever type structures, ** roof supports and trusses, ** axels, ** bones and medical equipment, ** wings of airplanes, and bridges. Before such structures are built, bending analysis calculations must be performed.
Project Lead The Way, Inc.
Copyright 2007

7. What Causes Deflection?
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
What Causes Deflection?
Snow
Live Load
Roof Materials, Structure
Dead Load
Walls, Floors,
Materials, Structure
Occupants, Movable
Fixtures, Furniture
Deflection is caused by more than ** just the live and dead loads it supports. How much a structure will deflect also depends on the ** geometry of the structure, ** the material that is used to build the structure, ** and the physical properties such as environmental extremes and structure support.
Project Lead The Way, Inc.
Copyright 2007

8. Loading Snow Live Load Roof Materials, Structure Dead Load
Deflection
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Loading
Snow
Live Load
Roof Materials, Structure
Dead Load
Occupants, Movable
Fixtures, Furniture
Live Load
Walls, Floors,
Materials, Structure
Dead Load
The load is the amount of weight that the structure must carry. Weight might be the ** snow on a rooftop or the ** individual standing at the end of a springboard. The load is represented by a vector and possesses units of force, either force per length or force per area.
Project Lead The Way, Inc.
Copyright 2007

9. Deflection
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Types of Loads
There are two general types of loading that affect the bending of a beam: a point load and a distributed load. ** The point load is a mass that can be represented by a single vector at one spot along a beam. ** A good example is a person on a balance beam where the point load is established at the person’s center of mass. ** A distributed load can be any shape. A distributed load is comprised of a weight occurring over a span of a beam. The weight can be equally distributed, or it can be unevenly distributed, ** as the case of a pile of rocks in the bed of a truck.
Project Lead The Way, Inc.
Copyright 2007

10. Factors that Affect Bending
Deflection
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Factors that Affect Bending
Material Property
Physical Property
Supports
When we talk about bending, three factors must be considered. They are the ** material property used to build the structure, the ** physical property or the geometry of the structure, and ** the beam or structure supports.
Project Lead The Way, Inc.
Copyright 2007

11. Physical Property - Geometry
Deflection
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Physical Property - Geometry
The geometry of a support structure is the two dimensional image of the structure or beam. ** These are two different support structures made from the same material. One is cut to a different size. The supporting loads, which are on different sides, will deflect differently under the same load.
Project Lead The Way, Inc.
Copyright 2007

12. Deflection
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Beam Supports
The method in which a beam is supported is a definite factor in how much a beam will deflect. For example, compare a ** person standing in the middle of a diving board and ** a person standing in the middle of a bridge made of the same material. We will assume that the ** springboard and the ** bridge both have the same geometry and therefore the same moment of inertia. Moment of inertia is the rotational analog or measurement of mass. Since both are made from the same material, they have the same modulus of elasticity. ** Modulus of elasticity is the mathematical description of an object or substance's tendency to be deformed elastically (i.e., non-permanently) when a force is applied. The resulting deflections will be much different because of the nature of the beam supports.
Project Lead The Way, Inc.
Copyright 2007

13. Spring Board Deflection
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Beam Deflections
Bridge Deflection
Spring Board Deflection
In reality the resultant bending would be as illustrated. Supports are important because deflection equations are generated from four integrals from the loading points (yes, for you calculus fans, that is integrating an equation 4 times in order to derive the deflection equation). The supports are used as boundary conditions to solve for some of the constants of integration.
Project Lead The Way, Inc.
Copyright 2007

14. Calculating Deflection on a Spring Diving Board
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Calculating Deflection on a Spring Diving Board
Pine Diving Board Dimensions:
Base (B) = 12 in.
Height (H) = 2 in.
72 in. P
 Max ?
250 lb
Now let’s calculate the maximum deflection on a spring diving board. ** We know that the diving board is twelve inches by two inches and 72 inches long. ** Also, the diving board is made out of pine, which has a modulus of elasticity of 1.76x106 psi. ** The person jumping off of the diving board weighs 250 lbs, which is the applied load on the diving board.
Known:
Pine (E) = 1.76 x 106 psi
Applied Load (P)= 250 lb
Project Lead The Way, Inc.
Copyright 2007

15. Deflection of Cantilever Beam with Concentrated Load
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Deflection of Cantilever Beam with Concentrated Load P L
 max
 max = P x L3
3 x E x I
Where:  max is the maximum deflection
P is the applied load
L is the length
E is the elastic modulus
I is the cross section moment of inertia
To calculate the ** maximum deflection on the cantilever beam diving board, we will multiply ** the applied load (P) times the cube of ** the length (L) of the board. Then we will take the answer we just found divided ** by three times modulus of elasticity (E) times ** the cross-sectional moment of inertia (I).
Project Lead The Way, Inc.
Copyright 2007

16. Moment of Inertia (MOI)
Deflection
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Moment of Inertia (I) is a mathematical property of a cross section (measured in inches4) that is concerned with a surface area and how that area is distributed about a centroidal axis.
Formally, the moment of inertia is defined as a mathematics property of the cross section of a shape. It quantifies the stiffness of the beam due to its geometry and is calculated about the center of mass or centroid. The higher the inertia, the more resistant's a shape is from bending.
Let’s solve our problem. We will begin by identifying the moment of inertia.
Project Lead The Way, Inc.
Copyright 2007

17. Calculating Moment of Inertia (I)
Deflection
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Calculating Moment of Inertia (I)
I = (12 in.)(2 in.)3 12
I = (12 in.)(8 in.3) 12
To calculate the moment of inertia, take the area of the cross section, or base times height, and multiply it by the height squared and divide by twelve (the constant 12 comes from the deviation of the equation using calculus). In our problem, ** the first step is find the cube of 2 inches. ** Next we take 12 inches times 8 inches cubed. ** Finally, we take 96 inches to the fourth divided by 12. ** The moment of inertia is 8 inches to the fourth.
I = 96 in.4 12
I = 8 in.4
Project Lead The Way, Inc.
Copyright 2007

18. Cantilever Beam Load Example
Deflection
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Cantilever Beam Load Example
72 in. P
 Max
250 lb
Known:
Pine (E) = 1.76 x 106 psi
Applied Load (P) = 250 lb
 max = P x L3
3 x E x I
 max = (250 lb) (72 in.)3
(3) (1.76 x 106 psi) (8 in.4)
 max = (250 lb) ( in.3)
Now that we know our moment of inertia, we can plug all of the information into our maximum deflection formula. We must find the maximum deflection of a 12 in. x 2 in. pine diving board that is 6 feet long and carries a load of 250 pounds per foot. ** First list the formula. ** Plug in all known information. ** Determine the cube of 72 inches.
Project Lead The Way, Inc.
Copyright 2007

19. Cantilever Beam Load Example
Deflection
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Cantilever Beam Load Example
max = ( x 107 lb)(in.3)
(5.28 x 106 psi)(8 in.4)
(4.224 x 107 psi)(in.4)
max = ( x 107)
(4.224 x 107 in.)
max = 2.21 inches
** Now multiply the top numbers together. Then take 3 times E (1.76x106 psi), which is the elastic modulus. ** Now take the inertia of 8, times the psi number and ** simplify units. ** Divide the top number by the bottom number ** to find the maximum deflection of the beam, which is 2.21 inches.
Project Lead The Way, Inc.
Copyright 2007

20. Calculating Deflection on a Pine Beam in a Structure
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Calculating Deflection on a Pine Beam in a Structure
Beam Dimensions:
Base (B) = 4 in.
Height (H) = 6 in.
Length (L) = 96 in. P L
 max
Now let’s calculate the maximum deflection on a pine beam in a structure. We know that the beam is supported by a pin at one end and a roller at the other end. ** The beam is four inches by six inches and is 96 inches long. ** Also, the beam board is made out of pine, which has a modulus of elasticity of 1.76x106 psi. ** The 200 lb applied load is placed at the center of the beam.
Known:
Pine (E) = 1.76x106 psi
Applied Load (P)= 200 lb
Project Lead The Way, Inc.
Copyright 2007

21. Deflection of Simply Supported Beam with Concentrated Load
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Deflection of Simply Supported Beam with Concentrated Load P L
 max
 max = P x L3
48 x E x I
Note that the simply supported beam is pinned at one end. A roller support is provided at the other end.
Where:  max is the maximum deflection
P is the applied load
L is the length
E is the elastic modulus
I is the cross section moment of inertia
To calculate the ** maximum deflection on a cantilever beam supported by a pin at one end and a roller at the other end, we will multiply ** the applied load (P) times the cube of ** the length (L) of the board. Next we will divide ** by 48 times the modulus of elasticity (E) times ** the cross section moment of inertia (I).
Project Lead The Way, Inc.
Copyright 2007

22. Calculating Moment of Inertia (I)
Deflection
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Calculating Moment of Inertia (I)
I = (4 in.)(6 in.)3 12
I = (4 in.)(216 in.3) 12
To solve our problem we first must calculate the moment of inertia. Take the area of the cross section, which is base times height, multiply it by height squared, and divide by twelve (the constant 12 comes from the deviation of the equation using calculus). In our problem, ** the first step is find the cube of 6 inches. ** Take 4 inches times 216 inches cubed. ** Finally, divide 864 inches to the fourth by 12. ** The moment of inertia is 72 inches to the fourth.
I = 864 in.4 12
I = 72 in.4
Project Lead The Way, Inc.
Copyright 2007

23. Simply Supported Beam Example
Deflection
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Simply Supported Beam Example
96 in.
Known:
Pine (E) = 1.76x106 psi
Applied Load (P) = 200 lb P
 max
 max = P x L3
48 x E x I
 max = (200 lb)(96 in.)3
(48)(1.76x106 psi)(72 in.4)
 max = (200 lb)( in.3)
Now that we know our moment of inertia, we can plug all of the information into our maximum deflection formula. We must find the maximum deflection of a 4 in. by 6 in. pine beam that is 96 inches long and carries a load of 200 pounds per foot in the center of the beam. ** First list the formula. ** Plug in all known information. ** Determine the cube of 96 inches.
Project Lead The Way, Inc.
Copyright 2007

24. Simply Supported Beam Example
Deflection
Principles of EngineeringTM
Lesson 5.2 – Strength of Materials
Simply Supported Beam Example
max = ( x 108 lb)(in.3)
(8.448 x 107 psi)(72 in.4)
( x 109 psi)(in.4)
max = ( x 108)
( x 109 in.)
max = inches
** Now multiply the top numbers together. Take 48 times E (1.76x106 psi), which is the elastic modulus. ** Now take the inertia of 72, multiply by the psi number, and ** simplify units. ** Divide the top number by the bottom number ** to find the maximum deflection of the beam, which is inches.
Project Lead The Way, Inc.
Copyright 2007