Structural Member Properties Moment of Inertia 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

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

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

Moment of Inertia – Composite Shapes Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Moment of Inertia – Composite Shapes Why are composite shapes used in structural design?

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

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

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

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

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

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

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

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

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

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

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

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

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

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) (373248 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

Cantilever Beam Load Example Deflection Principles of EngineeringTM Lesson 5.2 – Strength of Materials Cantilever Beam Load Example max = (9.3312 x 107 lb)(in.3) (5.28 x 106 psi)(8 in.4) (4.224 x 107 psi)(in.4) max = (9.3312 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

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

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

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

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)(884736 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

Simply Supported Beam Example Deflection Principles of EngineeringTM Lesson 5.2 – Strength of Materials Simply Supported Beam Example max = (1.769472 x 108 lb)(in.3) (8.448 x 107 psi)(72 in.4) (6.08256 x 109 psi)(in.4) max = (1.769472 x 108) (6.08256 x 109 in.) max = 0.029 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 0.029 inches. Project Lead The Way, Inc. Copyright 2007