Fundamentals of Engineering Exam Review Statics by WenWhai

Fundamentals of Engineering Exam Review Statics by Wen-Whai Li, Ph. D. , P. E. Department of Civil Engineering the University of Texas at El Paso Feb. 21, 2004 This document can be found in http: //utminers. utep. edu/wli/classes. htm User ID: civil Password: engineering Force Vector

Things you need to know Before the FE Exam • Know the reference materials available to you in advance Go to http: //www. ncees. org/exams/study_materials/fe_handbook/#single and review the handbook before you go to the exam. Familiarize yourself with the notations used in the handbook. Page 4 Page 6 Page 14 Pages 21 -22 Pages 27 -30 Law of Sines and Law of Cosines Vector Properties Derivatives and Integrals Topics and formula you need to know Geometric properties • Know your strength, focus on the subjects you are confident, and solve as many practice problems as you can. FE-Review - 2

1. Forces and Moments 1. 1 Resolution of a Vector • Resolution of a Vector – A vector may be resolved into two “components” having known lines of action by using the parallelogram law. • Law of Sines and Law of Cosines FE-Review - 3

1. 2 Vector Operations • Dot Product • Cross Product C = A B = A B sin uc C = A B sin FE-Review - 4

• Cartesian Vector Representation A = Ax i + Ay j + Az k • Magnitude of a Cartesian Vector • Unit Vector – A unit vector in the direction of a vector A is : u. A = A/A or FE-Review - 5

1. 3 Direction Cosines – To determine the coordinate direction angles, consider the projections of the Vector onto x, y and z axes as shown in the Figure. The cosines of the angles are called direction cosines. FE-Review - 6

1. 3 Direction Cosines – An easy way to obtain the direction cosines is by the following: – Substituting the direction cosines into the above equation, we have – Because, this is a unit vector, it follows that FE-Review - 7

• Example 1. 1 Express the force shown in the figure as a Cartesian vector FE-Review - 8

1. 4 Vector Expression of a Force using a Position Vector • Determine a position vector r. AB using the Cartesian coordinates A(x. A, y. A , z. A)and B(x. B, y. B , z. B), • Determine a unit vector from the position vector • Express the force/moment vector as the product of the magnitude of the force and the unit vector FE-Review - 9

• Example 1. 2 The cables exert forces FAB = 100 N and FAC = 120 N at A. Determine the magnitude of the resultant force acting at A. A 4 m 4 m B 2 m C FE-Review - 10

1. 5 Moment of a Force – Vector Formulation The moment of a force F about a point O is Mo = r F r is a position vector from point O to any point on the line of action of F The magnitude of the moment Mo is M = r F sin o is the angle formed between r and F Mathematically, i j k Mo = r F = rx ry rz Fx Fy Fz FE-Review - 11

Resultant Moment of a System of Forces The resultant moment of a system of forces about the same point O can be determined by adding the individual moment vectors resulting from r F MRO = r F FE-Review - 12

• Example 1. 3 A rectangular plate is supported by brackets at A and B and by a wire CD. Knowing that the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire on point C. FE-Review - 13

1. 8 Moment of a Couple Scalar Formulation The magnitude of the moment produced by two equal, opposite, and noncollinear forces is M=Fd F is the magnitude of the force, N or lb d is the distance perpendicular to the forces, m or ft FE-Review - 14

• Example 1. 4 Determine the moment of the couple acting on the member shown below. FE-Review - 15

• Example 1. 5 A square foundation mat supports the four columns shown. Determine the magnitude and point of application of the resultant of the four loads. FE-Review - 16

Review Problems Set I: Forces and Moments FE-Review - 17

2 Equilibrium 2. 1 State of Equilibrium • State of Equilibrium When a system is in equilibrium, the sum of all forces is zero and the sum of the moments about any point is zero. i. e. S F = 0 S Mo = 0 These conditions come from Newton’s First Law. • Equilibrium of a Particle For a particle, the moment equation is automatically satisfied since it is assumed to occupy a point in space. F = 0 The line of action of all forces are assumed to pass through the same point (Concurrent System) FE-Review - 18

2. 2 Equations of Equilibrium • General Equilibrium Conditions: F=0 • Many point = 0 Two-dimensional equilibrium conditions: Three equations for three variables, i. e. 1) S Fx = 0; S Fy = 0; S Mo = 0 or 2) S Fa = 0; S MA = 0; S MB = 0 or 3) S MA = 0; S MB = 0; S MC = 0 FE-Review - 19

Procedures for Analysis • Isolate a system • Draw FBD • Identify all forces and moments acting on the system in the FBD • Include all dimensions • Identify all unknowns • Select a coordinate systems • Apply the equations of equilibrium • Solve for unknowns FE-Review - 20

• Two-Force Member In two-force members, the forces at the two ends must be collinear, equal in magnitude and opposite in direction. FE-Review - 21

• Example 2. 1 The 20 -lb create is suspended with three cables as shown. Determine the tension in each rope segment. FE-Review - 22

• Example 2. 2 Determine the horizontal and vertical components of reaction for the beam loaded as shown in the figure. Neglect the weight of the beam in the calculations. FE-Review - 23

Example 2. 3 The plate has a mass of 50 kg/m 2. Determine the force in each cable of it is suspended in the horizontal plane. FE-Review - 24

Review Problems Set II: Equilibrium FE-Review - 25

3 Structural Analysis 3. 1 Method of Joints The method of joints takes advantage that the force system at each joint is concurrent. Thus, the only equations of equilibrium to be satisfy are: S Fx = 0; S Fy = 0; • Example: FE-Review - 26

3. 3 Zero-Force Members • Zero-force members arise from special joints: 1. Joints with no load with two non-collinear members. 2. Joints with no load with three members, two of which are collinear. FE-Review - 27

• Example 3. 1 – Determine the forces in members BC and BD. Indicate whether the force is in tension or compression. FE-Review - 28

3. 2 Method of Sections • The method of sections is helpful to find the forces of few members • A section is identified through which the truss will be imaginarily “cut” in two segments • The members cut are replaced by forces in the direction of the members (use actionreaction). Always assume tension. FE-Review - 29

• Select the FBD of the segment with less number of forces • Use the resulting FBD of one of the segments to solve for the unknowns using: Fx = 0; Fy = 0; M=0 • The most useful equation is M = 0 • Try to take ( M = 0) about a joint (on or off the FBD) at which two of the unknowns cancel • If two of the unknowns are parallel, then use ( F = 0) in the direction perpendicular to the members that are parallel. FE-Review - 30

Example 3. 2 Determine the force in member IJ. FE-Review - 31

3. 3 Frames and Machines • Frames and machines are two common types of structures which are often composed of two or more pin-connected multi-force members (more that two forces). • Frames are stationary and are used to support loads, whereas machines contain moving parts and are design to transmit and alter the effects of forces. • To solve forces in the members of frames and machines, the following procedure is used: FE-Review - 32

• Procedure – Determine the external reactions at the supports of the structure. – Isolate each part of the frame or machine. Show all forces and couple moments acting on the part. – Identify two-force members (if any) and represent their forces with equal and opposite forces sharing the same line of action – Forces common to any two contacting members act with equal magnitudes but opposite in direction (action-reaction) – Use equilibrium to solve for the forces after the FBDs of the individual parts are drawn. FE-Review - 33

• Example 3. 3 Find all the zero-force members. FE-Review - 34

• Example 3. 4 Determine the forces in members of BC, GC and GE using the method of sections. FE-Review - 35

• Example 3. 5 Determine the reactions of the compounded beam shown (frame) FE-Review - 36

Review Problems Set III: Structural Analysis FE-Review - 37

4 Geometric Properties 4. 1 Center of Gravity and Center of Mass for a System of Particles • The location of the center of gravity can be determined using the principle of equilibrium system. The sum of all weights is The location of the center of gravity is FE-Review - 38

4. 2 Center of Gravity and Centroid for a Body Center of Gravity Mathematically, the center of gravity is where d. W is the weight of a particle with an infinitesimal length; and x, y, z describe the location of the integration element in space. x FE-Review - 39

• If represents the specific weight of the body, as weight per volume, the center of gravity can be expressed as: • Express as the product of density and gravitational acceleration g, we can define the mass center if g is constant in = g. FE-Review - 40

• Centroid is the point which defines the geometric center of an object. Centroid for the Volume of an Object ( = constant) FE-Review - 41

Centroid of the Area of an Object ( and t are constants, V=A t) FE-Review - 42

Centroid of a Line ( is a constant and the cross sectional area of a rod is constant) – Centroid does not need to be within the object – Centroid may be partially or completely defined by symmetry FE-Review - 43

4. 3 Composite Bodies • A composite body consists of a series of connected “simple” shape bodies, which may be circular, rectangular, or triangular. • Any object can be treated as a composite body by subdividing it into parts. • Use the knowledge of centroid on known geometry to find out the centroid of a complicate body. • Holes or voids can be treated with negative values FE-Review - 44

• Center of Gravity • Center of Volume • Center of Area FE-Review - 45

Example 4. 1 Locate the centroid of the area shown in the figure. FE-Review - 46

Example 4. 2 The gravity wall is made of concrete. Determine the location (x, y) of the gravity G for the wall. FE-Review - 47

Example 4. 3 A beam supporting a variable but linearly distributed load is shown below. Determine the vertical reaction at support B a) 120 b) 180 c) 240 d) 360 N FE-Review - 48

4. 4 Moments of Inertia for Areas In the calculation of the moment of distributed forces about an axis or in the calculation of beam deflections, an integral of the form appears in the calculation. This integral is known as the moment of inertia or the second moment of the area. FE-Review - 49

Moment of Inertia about the x axis Moment of Inertia about the y axis Product of Inertia Polar Moment of Inertia FE-Review - 50

• Radius of Gyration kx : the radius of gyration about the x axis ky : the radius of gyration about the y axis ko : the polar radius of gyration (about O), and ko 2 = kx 2 + ky 2 FE-Review - 51

4. 5 Parallel-Axis Theorem • Parallel-axis Theorem is used to determine the moment of inertia for an area about an axis that is parallel to another axis, where the moment of inertia for an area about that axis is known. The bar notation on the moment of inertia indicates that it is about the centrod of its area FE-Review - 52

• Similarly, FE-Review - 53

4. 6 Moments of Inertia for Composite Areas • Similar to the exercise in determining the centroid for an composite area, we want to take advantage of the knowledge of moments of inertia developed for specific shapes of area with respect to a specific axis and apply to a composite area without going through lengthy integration. • Find the moment of inertia for a simple area from Appendix C, apply the parallel-axis theorem to find the moment of inertia about the desired axis, and take summation of the values calculated for the composite parts. FE-Review - 54

Example 4. 4 FE-Review - 55

Review Problems Set IV: Geometric Properties FE-Review - 56

5 Friction • Dry Friction occurs between the contacting surfaces of bodies in the absence of a lubricating fluid • Friction coefficients - Coefficient of Static Friction, µs - Coefficient of Dynamic Friction, µk • Characteristics of Dry Friction - Tangent to the contacting surface and opposite to the direction of motion Independent of the area of contact The maximum static friction is generally greater than the maximum dynamic friction The maximum friction force is proportional to the normal force, Fs = µs N, when slipping is about to occur When slipping occurs, the kinetic friction force is proportional to the normal force, Fk = µk N FE-Review - 57

5. 1 Coefficient of Static Friction State of Friction Angle of Friction For static friction, For kinetic friction, FE-Review - 58

5. 2 Analysis of Friction Problems • Draw the necessary FBD and identify all forces showing the frictional forces • Determine whether the problem is deterministic by checking the number of equations vs. the number of unknowns • Make sure the friction force is acting in the right direction • If the number of unknowns is greater than the number of equations, look for additional frictional equation at some points of contact • Select a coordinate system • Apply the equations of equilibrium and solve the equation with least unknowns. FE-Review - 59

Example 5. 1 The crate has a mass of 350 kg and is subjected to a towing force P acting at a 20 o angle with the horizontal. Determine the magnitude of P to just start the crate moving downward the plane. FE-Review - 60

Example 5. 2 The 100 lb box is stationary on the inclined surface. a) What is the magnitude of the friction force on the box? b) What is the smallest value of the coefficient of static friction that will permit the box to remain stationary? FE-Review - 61

5. 3 Wedges • A wedge is a bifacial tool with the faces set at a small angle. When a wedge is pushed forward, the faces exert large lateral forces as a result of the small angle between them. FE-Review - 62

Example 5. 3 Determine the force P needed to lift the 100 lb load. Smooth rollers are placed between the wedges. The coefficient of static friction between A and C and between B and D is s=0. 3. Neglect the weight of each wedge. FE-Review - 63

5. 5 Belts • Belt is a device which uses the principle of friction to augment power or reduce loads FE-Review - 64

Example 5. 4 The 50 kg crate is suspended from a rope that passes over two fixed cylinders. The coefficient of static friction is 0. 2 between the rope and the left cylinder and 0. 4 between the rope and the right cylinder. What is the smallest force the woman can exert and support the crate? 0. 38. FE-Review - 65

Review Problems Set V: Friction FE-Review - 66