FE Cheat Sheet: Engineering Concepts for FE Exam Prep

FE CHEAT SHEET In-depth review of fundamental engineering concepts to gain the knowledge you need to prepare for the FE Exam. www.engenieer.com HOW TO USE THIS FE CHEAT SHEET Knowing the foundational topics of engineering in essential to preparing for the FE Exam. And you’ll find that when you really understand the concepts in these pages, you will begin to see progress in your FE preparations. You will get faster at solving problems. As you gain the knowledge in each section, you will find studying for Dynamics, Mechanics of Materials, Fluid Mechanics, and Structure easier. And you will feel confident taking your FE Exam. Use this Cheat Sheet every time you study for the FE. Take your time to understand the methods I outline and practice using these methods on actual problems. If you are using our course, you will get in-depth lectures on each topic, lots of problems to solve, and quizzes after each section. If you don’t, that’s ok! I have a full playlist of FE problems on my YouTube channel that you can begin with too. Don’t let engineering concepts hold you back from passing the FE Exam. Instead, learn the fundamentals, get faster at solving problems, and build recall so you can take your exam! Kenza FE COURSES BUILT TO HELP YOU PASS FASTER The typical FE journey leaves engineers unprepared to take the exam. Do you struggle with: • Knowing where to start • Complicated study material • Countless hours studying with little results • Missing our on career opportunites • Wasting time googling • Never feeling ready to take the exam • Stuck and not progressing • Endless cycle of testing & failing It shouldn’t be this way. I know every engineer deserves to pass the FE Exam! Simplify the FE Journey with enGENIEer’s Pass Faster Method. With one-stop studying, easy to understand material, and self-paced course you will see results faster. Students go from lost and failing to confident and passing in three months. ACCELERATOR MORNING + AFTERNOON BUNDLE EVERYTHING YOU NEED TO PASS THE FE CIVIL! It will be virtually impossible to find any better content than enGENIEer’s Courses. I finally passed on my 4th attempt. My first attempt I used the School of PE and the PPI Lindeburg Review and they both were ineffective and costly. Kenza has designed a Successful, Coherent, and Masterful System that works for all FUTURE ENGINEERS! Shawn, enGENIEer Student and EIT Math www.engenieer.com 1. Straight Line There are 3 forms of equation: General: 𝑨𝒙 𝑩𝒚 𝑪 𝟎 Standard: 𝒚 𝒎𝒙 𝒃 Point-slope: 𝒚 𝒚𝟏 𝒎 𝒙 𝒙𝟏 m is the slope, b is the y-intercept, (x,y) is the point the line passes through. Two lines are parallel: 𝒎𝟏 𝒎𝟐 𝟏⁄𝒎𝟐 Two lines are perpendicular: 𝒎𝟏 2. Circle Given the equation representing the circle: 𝑨𝒙𝟐 𝑩𝒚𝟐 𝑪𝒙 𝑫𝒚 𝑬 𝟎 To find the center and the radius of circle: Rearrange the equation: 𝑨𝒙𝟐 Complete the square: 𝑨𝒙𝟐 𝑩𝒚𝟐 𝑪𝒙 𝑪𝒙 Find the equation of circle: 𝒙 𝑪 𝟐 ⎯ 𝟐 𝒉 𝟐 𝒚 𝑫𝒚 𝑬 𝑩𝒚𝟐 𝑫𝒚 𝒌 𝟐 𝒓𝟐 𝑫 𝟐 ⎯⎯ 𝟐 𝑬 𝑪 𝟐 ⎯ 𝟐 Center of the circle: 𝒉, 𝒌 ; radius is r 3. Polygons To find the interior angles of polygon: 𝜽 Here are sides of polygons: Triangle: 3 sides Quadrilateral: 4 sides Pentagon: 5 sides Hexagon: 6 sides Heptagon or Septagon: 7 sides Octagon: 8 sides Nonagon: 9 sides Decagon: 10 sides 𝝅 𝒏 𝟐 ⎯⎯⎯⎯⎯, where n is number of sides 𝒏 𝑫 𝟐 ⎯⎯ 𝟐 Math www.engenieer.com 4. Matrix If matrix A 𝒂 𝒃 is multiplied by matrix B 𝒄 𝒅 , the final matrix is 𝒃 𝒄 5. Derivatives Derivative of functions: see page 48 on the reference manual L'Hospital's Rule 𝐟 𝐱 𝐠 𝐱 𝐥𝐢𝐦𝐱→𝐚 ⎯⎯⎯, Plug-in a in f(x) and g(x) If f(a)/g(a) is 0/0 or ∞/∞, then it's indeterminate form, to solve for the limit, take the derivative of f(x) and g(x), and then plug-in a 𝐟 𝐱 𝐥𝐢𝐦 ⎯⎯⎯⎯ 𝐱→𝐚 𝐠 𝐱 𝐟′ 𝐱 𝐥𝐢𝐦 ⎯⎯⎯⎯⎯ 𝐱→𝐚 𝐠′ 𝐱 Maximum and minimum value of a function: 1. Take the derivative of the function, equal it to zero and solve for x 2. Take the second derivative and plug-in x i. If x<0, we have a maximum value ii. If x>0, we have a minimum value 3. To find the maximum or minimum value, plug-in x to the original function 6. Integrals Integral of functions: see page 49 on the reference manual If given 2 functions, where 𝒚𝟏 is the upper function and 𝒚𝟐 is the lower function, then the area of the bounded region is: 𝒃 𝑨= 𝒂 𝒚𝟏 − 𝒚𝟐 Where a and b are points of intersection (use the calculator to solve for the integral) To find the point of intersection, equal the functions 𝒚𝟏 𝒚𝟐 and solve for x. Math www.engenieer.com 7. Cross and Dot Products Unit vector in the direction of vector: 𝑼 If the length is given, 𝑼 𝒂 ⎯⎯ |𝒂| 𝒂 ⎯⎯∗ 𝑳𝒆𝒏𝒈𝒕𝒉 |𝒂| Dot product measures how much the vectors move in the same direction 𝐚 𝐛 𝐚 𝐛 To find the angle between two vectors: cos θ = ⎯⎯⎯⎯⎯ | || | Cross product measures how much the vectors are perpendicular: 𝑨𝒙𝑩 |𝑨 𝒙 𝑩| To find the unit vector that is perpendicular to vectors A and B: ⎯⎯⎯⎯⎯ To check if two vectors are: Parallel: a x b =0 Perpendicular: a ∙ b =0 8. Polar Coordinates A polar coordinate: Indicates the location of a point, in terms of the radius and the angles. Sometimes, it is more convenient to use polar system than Cartesian system. r - Distance from O to P θ - Angle between polar axis and OP Sign Convention for r and θ • θ is positive: CCW from the polar axis • θ is negative: CW from the polar axis • r is positive: As it is given • r is negative: Opposite of the given r Statistics www.engenieer.com 1. Mean, Mode, Median, Standard Deviation, Sample Variance Use your calculator, it's faster and don't forget to include frequency if it's given 2. Confidence Interval Z-distribution: If the sample size (n) is above 30 and standard deviation is known 𝑿 𝝈 𝒁𝜶 ⎯⎯⎯ 𝝁 𝟐 √⎯⎯ 𝒏 𝑿 𝝈 𝒁𝜶 ⎯⎯⎯ 𝟐 √⎯⎯ 𝒏 Where 𝒁𝜶⁄𝟐 value is on table on page 75 reference manual T-distribution: If the sample size (n) is below 30 and standard deviation is unknown 𝑿 𝒔 𝒕𝜶 ⎯⎯⎯ 𝝁 𝟐 √⎯⎯ 𝒏 𝑿 𝒔 𝒕𝜶 ⎯⎯⎯ 𝟐 √⎯⎯ 𝒏 Where 𝒕𝜶⁄𝟐 value is on page 77 and depends on: 𝒗 𝜶 𝒏 𝟐 𝟏 𝟏 𝑪𝑳 ⎯⎯⎯⎯⎯⎯ 𝟐 3. Probability Distribution Find the standard units using the following equation: 𝒁 𝑿 𝝁 ⎯⎯⎯ 𝝈 To find the probability, use the table on page 76 and the notations that are on page 67 on the reference manual 4. Linear Regression To find the slope of the line, the y-intercept, the correlation coefficient, the coefficient of determination use the calculator (under 2nd data if using TI 36X Pro) 5. Curve Fitting If we have a positive slope, R = 1 If we have a negative slope, R = 1 If data is scattered, R = 0 Ethics Review the following sections on the reference manual 1. Rules of Professional Conduct (page 4-5) Licensee's obligation to the public Licensee's obligation to employer and clients Licensee's obligation to other licensees 2. General Requirements of Licensure (page 7-8) 3. Grounds for Disciplinary Action (page 8-9) Licensees and interns Unlicensed individuals 4. Intellectual Property (page 11-12) Patents Trademarks Copyrights Trade secrets 5. Societal Considerations (page 12) www.engenieer.com Engineering Economics www.engenieer.com 1. Cash Flow diagram Draw the cash flow diagram when trying to determine the present, annual or future worth Example: Given: First cost (P), maintenance cost (A) and salvage value (F) after 4 years Find: Present value with a 6% interest rate 1- Draw CFD 2- Convert all values (annual and future) to the present worth (see table on page 230) and set up the equation 𝑷 𝑷 𝑨 𝑷⁄𝑨 , 𝟔%, 𝟒 𝑭 𝑷⁄𝑭 , 𝟔%, 𝟒 3- Grab the values of 𝐏⁄𝐀 , 𝟔%, 𝟒 , 𝐏⁄𝐅 , 𝟔%, 𝟒 from the table on page 235 and solve for P Note: The present and annual values are negative because they're a cost 2. Depreciation 1- Straight-line method Depreciation at year 1: 𝑫𝟏 𝑪 𝑺 , where n is the year of the salvage value ⎯⎯⎯ 𝒏 Depreciation at year j: 𝑫𝒋 𝒋 𝑫𝟏 2- MACRS Depreciation at year j: 𝑫𝒋 ∑ 𝒇𝒂𝒄𝒕𝒐𝒓 𝑪, C is the initial cost Factor value is on page 232 and depends on the year of book value and the recovery period Engineering Economics www.engenieer.com 3. Book Value 𝑩𝑽 𝒊𝒏𝒊𝒕𝒊𝒂𝒍 𝒄𝒐𝒔𝒕 𝑫𝒋 To find ∑ 𝑫𝒋 use either the straight-line method or MACRS, depends on the question 4. Capitalized Cost Use capitalized cost when given annual cost that goes forever (perpetual) To find CC, convert everything to annual cost and then use the following equation: 𝑨 ⎯ 𝒊 Capitalized cost 5. Effective Interest Rate 𝒊𝒆 𝟏 𝒓 𝒎 ⎯⎯ 𝒎 𝟏 𝐢𝐞 is the effective rate specified per time period r is the nominal interest rate for same time period as the effective rate m is the number of times interest is compounded per stated time period 6. Benefit Cost Analysis 𝑩 𝐁𝐞𝐧𝐞𝐟𝐢𝐭𝐬 𝐃𝐢𝐬𝐛𝐞𝐧𝐞𝐟𝐢𝐭𝐬 ⎯⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝑪 𝐂𝐨𝐬𝐭𝐬 𝐒𝐚𝐥𝐯𝐚𝐠𝐞 If B/C 1: Project is economically acceptable. If B/C < 1: Project is NOT economically acceptable. 𝑩 𝑪 𝑩 𝑫 𝑪 If 𝑩 If 𝑩 𝑪 0: Project is economically acceptable. 𝑪 < 0: Project is NOT economically acceptable. 7. Breakeven Analysis 𝐅𝐢𝐱𝐞𝐝 𝐂𝐨𝐬𝐭𝐬 𝐅𝐂 BP = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯ 𝐏𝐫𝐢𝐜𝐞 𝐨𝐟 𝐏𝐫𝐨𝐝𝐮𝐜𝐭 𝐩𝐞𝐫 𝐮𝐧𝐢𝐭 𝐕𝐚𝐫𝐢𝐚𝐛𝐥𝐞 𝐂𝐨𝐬𝐭 𝐩𝐞𝐫 𝐮𝐧𝐢𝐭 𝐏 𝐕𝐂 Statics www engenieer o Vector Operations (See Part A) Addition:  Use parallelogram law or triangle rule  Parallelogram law: Tail to tail A A A R B B B R=A+B  Triangle rule: Head to tail A B A R R B A B R=A+B R=B+A Subtraction:  Use parallelogram law or triangle rule A A R B B R=A-B Parallelogram Law B R A R=A-B Triangle Rule Statics www engenieer o Resultant Force (See Part A) i) Graphical Method:  Use the parallelogram law or triangle rule (see part B, ii): Important equations to remember: Law of Sine and Cosine A B A R R B R=A+B R=A+B - Note: Use this method to resolve the force or vector into two components. For resultant force, use the next method because it is faster. ii) Algebraic Method: ⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝐅𝐱 𝟐 𝐅𝐲 𝟐 , 𝛉  Use Algebraic sum of the x and y axis: 𝐅 𝐅𝐲 𝐭𝐚𝐧 𝟏 ⎯⎯ 𝐅 𝐱 Important equations to remember: Trigonometry Functions F F 5 3 4 F = F sin F = F (4/5) F = F cos 𝐩𝐩 𝐇𝐲𝐩 𝐀𝐝 𝐇𝐲𝐩 𝐩𝐩 𝐀𝐝 Sin θ = ⎯⎯⎯, Cos θ = ⎯⎯⎯, Tan θ = ⎯⎯⎯ θ = 𝐭𝐚𝐧 𝟏 𝐩𝐩 ⎯⎯⎯ 𝐀𝐝 F = F (3/5) ⎯⎯ Similar Triangles: ⎯⎯ 𝟒 ⎯ 𝟓 θ = 𝐭𝐚𝐧 ⎯⎯ 𝟑 𝟏 𝐩𝐩 𝐀𝐝 ⎯⎯⎯ ⎯⎯ Statics www engenieer o Cables and Pulleys (See Part B) A. Steps For Analysis: i) Draw the Free Body Diagram (FBD) ii) Identify all forces iii) Equilibrium equations: ∑ 𝐅𝐱 = 0, ∑ 𝐅𝐲 = 0. Moment of Force (See Part C) M=rF F is the force and r is the moment arm, which is the perpendicular distance to the line of action of the force. Examples: M r F M=rF r r sin M r r cos F M = (r sin ) F M F r M = r F = (0) F = 0 Same line of action! Statics Types of Supports (See Part C) www engenieer o a. Roller: ◊ Reaction prevents translation in the vertical direction. ◊ Reaction force acts perpendicular to the surface. R ɽ R ɽ b. Pin or hinge: ◊ Reaction prevents translation in any direction. ◊ The resultant reaction is broken into y and x components. R୶ R୷ c. Fixed: ◊ Reaction prevents translation and rotation. ◊ Reaction has a force (in the x and y direction) and a moment. R୶ M R୷ Statics www engenieer o Trusses (See Part D) (I) Identifying Zero Force Members members are connected to a joint that is not subjected to any external i) If two loads or reactions, then both members are zero. ii) If three members, two of which are collinear and are connected to a joint that is not subjected to any external loads or reactions, then the force member that is not collinear is zero. (II) Fast Method Of Joint (FMJ) Four Rules: 1. Equilibrium Equations (Mentally) 2. Break diagonal vectors into 2 components 3. Draw vectors head-head or tail-tail 4. Every action, there is an equal and opposite reaction F y F y ⎯⎯F F x ⎯⎯F x F Forces on members: ◊ If member elongates Tension (Force pulling the joint) ◊ If member shrinks Compression (Force pushing into the joint) Method of Section Cut the truss where we want to determine the forces. Draw Free Body Diagram. Take moment where two unknown forces intersect, so we have only one unknown in the moment equation. Statics www engenieer o Centroids (See Part E) Procedure For Analysis: Given A Function 1. Draw a small vertical or a horizontal rectangle 2. Find the area of the small rectangle, A = base x height 3. Find 𝒙 and 𝒚 𝒙 𝐝𝐀 , 𝐲𝐜 ⎯⎯⎯⎯ 𝐀 4. Plug in to the equation: 𝐱 𝐜 𝒚 𝐝𝐀 ⎯⎯⎯⎯ 𝐀 Procedure For Analysis: Given Geometric Shapes 1. Break the drawing into geometric parts. 2. Find the centroid, 𝐱 𝐧 or 𝐲𝐧 for each geomteric part & always with respect to the datum (0,0)! 3. Find the area of the geometric shapes. 4. Plug in to the equations: 𝐱 𝐜 ∑ 𝐱 𝐧 𝐚𝐧 𝐲𝐜 ⎯⎯⎯⎯⎯, 𝐀 ∑ 𝐲𝐧 𝐚𝐧 ⎯⎯⎯⎯⎯ 𝐀 Moment of Inertia (See Part F) Procedure For Analysis: 1. Break the drawing into geometric parts. 2. Find the moment of inertia for each geometric shape. 3. Find 𝐝𝐱 and 𝐝𝐲 , if necessary! 4. Plug in to the equations: 𝐈 𝐱 𝐈𝐱𝐜 𝐝𝟐 𝐲 𝐀 , 𝐈 𝐲 𝐈𝐲𝐜 𝐝𝟐 𝐱 𝐀 𝐈𝐱 or 𝐈𝐲 𝐈𝐱𝐜 or 𝐈𝐲𝐜 Use These Equations Not Applicable 𝐈𝐱 or 𝐈𝐲 Not Applicable Use These Equations 𝐈𝐱𝐜 or 𝐈𝐲𝐜 Zero – Shape is on the axis, there is NO distance! Not Applicable Shape is NOT symmetric Not Applicable (find centroid x and y Use These Equations Use These Equations Zero – Centroid of the small shape is at the same point as the centroid of the whole shape! Distance from centroid of the small shape to the centroid of the whole shape! (Also, use fast method, e.g. 𝐝𝐱 𝐱𝐜 𝐱𝐧 , see examples in the course) Centroid about x & y axis Shape on x or y-axis Shape NOT on x or yaxis Centroid about x & y centroidal axis Shape is symmetric 𝐝𝐱 or 𝐝𝐲 Distance from x or y-axis to the centroid of the small shape. 𝐝𝐱 or 𝐝𝐲 Statics www engenieer o Friction Force (See Part G) Free body diagram: W P F=F µ N N Surface or floor exerts a normal force and a frictional force. Frictional Force always tangent to the surface. For equilibrium, normal force acts upward and perpendicular to the surface to balance out the weight. Frictional force acts opposite of the force P to reach equilibrium and to prevent motion to the right. Friction Forces: F, the static frictional force, is when equilibrium is reached. F, the limiting static frictional force, is when the maximum value is reached but still maintaining equilibrium. 𝐅𝐤, the kinetic frictional force, is when an object starts moving and sliding occurs Procedure For Analysis: i. Draw the Free Body Diagram (FBD) ii. Identify all forces iii. Equilibrium and friction force equations: ∑ 𝐅𝐱 = 0, ∑ 𝐅𝐲 = 0, ∑ M = 0, F = µ𝐬 𝐍 - Note: Normal force is NOT Always equal to weight. Belt Friction (See Part G) 𝐅𝟏 𝐅𝟐 𝒆µ𝜽 , where 𝐅𝟏 𝐅𝟐 Dynamics www.engenieer.com 1. Particle Kinematics If given distance, velocity or acceleration as a function of time, use the following equations: Given Find Use 𝒂 𝒕 𝒗 𝒗 𝒕 𝒔 𝒗 𝒕 𝒂 𝒂 𝒅𝒗⁄𝒅𝒕 𝒔 𝒕 𝒗 𝒗 𝒅𝒔⁄𝒅𝒕 𝒗 𝒂 𝒕 𝒅𝒕 𝒔 𝒗 𝒕 𝒅𝒕 When acceleration is constant and numbers are provided for distance, velocity or acceleration, use the following equations: 𝒗 𝒗𝒐 𝒂𝒐 𝒕, use when distance is not known 𝒔 𝒔𝒐 𝒗𝒐 𝒕 𝟏 𝒕𝟐 , use when final velocity is not known ⎯𝒂 𝟐 𝒐 𝒔 𝒔𝒐 𝟏 𝒗𝒐 ⎯ 𝟐 𝒗 𝒕, use when acceleration is not known 𝒗𝟐 𝒗𝒐 𝟐 𝟐𝒂𝒐 𝒔 𝒔𝒐 , use when time is not known When a particle moves along a circular path and angular acceleration is constant, the following formulas are used: 𝛚 𝛚𝒐 𝜶𝒐 𝒕, use when angular distance is not known 𝜽 𝜽𝒐 𝝎𝒐 𝒕 𝛚𝟐 𝛚𝒐 𝟐 𝟏 𝒕𝟐 , use when final angular velocity is not known ⎯𝜶 𝟐 𝒐 𝟐𝜶𝒐 𝛉 𝛉𝒐 , use when time is not known Dynamics www.engenieer.com 2. Normal and Tangential Components When a particle moves in a curve path and the path is known, the velocity and acceleration can be expressed using n and t coordinate axes: 𝒗 𝒂 𝒂 𝒗 𝒕 𝒆𝒕 𝒂 𝒕 𝒆𝒕 𝒗𝒕 𝟐 ⎯⎯⎯ 𝒆𝒏 𝝆 ⎯⎯⎯⎯⎯⎯⎯ 𝒂𝟐𝒕 𝒂𝟐𝒏 3. Newton's Second Law Steps for Analysis: 1- Draw the free body diagram 2- Identify all the forces 3- Break the forces in x and y components 4- Draw the acceleration body diagram 5- Assign the positive sign direction for acceleration 6- Set up the equations of motion: Rectilinear Motion: ∑ 𝐅𝐲 Normal & Tangential: ∑ 𝐅𝐭 𝐦𝐚𝐲 , ∑ 𝐅𝐱 𝐦𝐚𝐭 , ∑ 𝐅𝐧 𝐦𝐚𝐱 , 𝐅𝐟 𝐦𝐚𝐧 𝝁𝒌 𝑵, 𝐅𝐬 𝐤𝐬 Dynamics www.engenieer.com 4. Work and Energy 𝐓𝟏 𝐕𝟏 𝐔𝟏→𝟐 𝐓𝟐 𝐕𝟐 Steps for Analysis: 1- Draw the free body diagram 2- Identify the energies (kinetic, gravitational and spring) at stage 1 3- Identify the energies (kinetic, gravitational and spring) at stage 2 4- Determine work done by nonconservative forces, 𝐔𝟏→𝟐 5- Apply the work-energy equation, 𝐓𝟏 𝐕𝟏 𝐔𝟏→𝟐 𝐓𝟐 F (cos θ) s 𝐕𝟐 5. Principle of Linear Impulse and Momentum Problems that involve force, velocity and time, use the following equation 𝐦 𝐯𝟏 𝐅𝐝𝐭 Where ∑ 𝐅𝐝𝐭 𝐦 𝐯𝟐 𝐅 𝐭𝟐 𝐭𝟏 6. Impact When two objects collide, momentum is always conserved and the impact can be expressed as: 𝐦𝟏 𝐯𝟏 𝐦𝟐 𝐯𝟐 𝐦𝟏 𝐯′𝟏 𝐦𝟐 𝐯′𝟐 The restitution coefficient is also used, 𝐞 𝐯 𝐯 𝟐 𝐧 𝟏 𝐧 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝐯 𝐯 𝟏 𝐧 𝟐 𝐧 Coefficient of Restitution cases: e = 1: Perfectly elastic collision (energy is conserved) e = 0: Plastic collision (energy is not conserved) Mechanics of Materials www.engenieer.com 1. Resultant Forces Normal Force, N: Acts perpendicular to the surface, developed when external force pushes/pulls Shear Force, V: Acts parallel to the surface, developed when external force slides Torque, T: Acts perpendicular to the surface, twists one part of the body with respect to the other end Moment, M: Acts within the plane, bends the body 2. Stress Normal Stress, : 𝐅 𝐀 𝐧  = ⎯⎯ , where force is acting perpendicular (normal) to the area Shear Stress, τ: 𝐅 𝐀 𝐭 τ = ⎯⎯ , where force is acting parallel or tangent to the area 3. Elastic Deformation The deformation of one end with respect to the other end of the member is calculated using; 𝐏𝐋 𝐀𝐄 δ = ⎯⎯ Steps for Analysis: 1- Use method of section 2- Find the internal axial forces using ∑ 𝐅𝐱 𝟎 or ∑ 𝐅𝐲 𝐏𝐋 𝐀𝐄 3- Apply the elastic deformation equation using δ = ⎯⎯ Sign Convention: Tension: +P and +δ Compression: -P and ‐δ 𝟎 Mechanics of Materials www.engenieer.com 4. Thermal Stress The deformation due to temperature can be calculated using: 𝛅𝐓 𝛂∆𝐓𝐋 𝜶 𝑻 𝑻𝒐 5. Strain Normal Strain, ε: Measures change in length 𝐋𝐟 𝐋𝐢 ⎯⎯⎯⎯ 𝐋𝐢 ε Shear Strain, γ: Measures change in angle τ 𝐆 ⎯ 𝜸 6. Torque When an object is subjected to a torsional loading, to determine the shear stress the following equation is used: 𝐓𝐫 ⎯⎯⎯ 𝐉 𝛕 Note: This equation can only be used if the thickness of the shaft is greater than 10% of the shaft radius and elastic deformation 7. Angle of Twist When a shaft is subjected to torque, it rotates or twists. The angle of twist is calculated using: 𝚽 𝐓𝐋 ⎯⎯, where is 𝚽 in radians 𝐉𝐆 Mechanics of Materials www.engenieer.com 8. Shear and Moment Diagram See statics cheat sheet to review the basics of mechanics of materials When given a beam with different type of loading: To draw the shear diagram, take the area under the loading beam To draw the moment diagram, take the area under the shear diagram Note: 1- The shear and moment diagram have to always close 2- The maximum values obtained from the shear and moment diagram represent the maximum shear and maximum moment, respectively. Type of Loading Shear Diagram Moment Diagram Point Load Constant Linear Distributed Load Linear Parabolic Triangle Load Parabolic Cubic 9. Bending Stress When a beam is subjected to bending, the maximum stress is calculated using: 𝑴𝒄 𝝈𝒎𝒂𝒙 − ⎯⎯⎯ 𝑰 M is the maximum bending moment, which can be calculated by drawing the moment diagram or by using the equations on table 140-141 C is the distance from the neutral axis to the outer fibers of the beam 10. Transverse Shear When a beam is subjected to shear, the transverse shear stress is calculated using: 𝛕 𝐕𝐐 ⎯⎯, where 𝑸 𝐈𝐛 𝑨′𝒚′ V is the maximum shear stress which is determined from the shear diagram A' is the area above the point where we want to determine the shear stress 𝒚′ is the distance from the neutral axis to the centroid of the area A' b is thickness of the beam where we want to determine the shear stress Mechanics of Materials www.engenieer.com 11. Pressure Vessels Thin walled pressure vessel 𝑷𝒊 𝒓 ⎯⎯⎯ 𝒕 Tangential or hoop stress: 𝝈𝒕 Axial stress: 𝝈𝒂 𝑷𝒊 𝒓 ⎯⎯⎯ 𝟐𝒕 Thick walled pressure vessel Under internal Pressure 𝒓𝟐 + 𝒓𝟐 𝒐 𝒊 Tangential or hoop stress: 𝝈𝒕 = 𝑷𝒊 ⎯⎯⎯⎯⎯ 𝟐 𝟐 𝒓𝒐 − 𝒓𝒊 𝒓𝟐 𝒓𝒐 𝒓𝒊 Axial stress: 𝝈𝒂 𝒊 𝑷𝒊 ⎯⎯⎯⎯⎯ 𝟐 𝟐 Radial stress: 𝝈𝒓 𝑷𝒊 Under external Pressure 𝒓𝟐 𝒓𝟐 𝒓𝒐 𝒓𝒊 𝒐 𝒊 𝑷𝒐 ⎯⎯⎯⎯⎯ 𝟐 𝟐 Tangential or hoop stress: 𝝈𝒕 Radial stress: 𝝈𝒓 𝑷𝒐 12. Deflection To determine the slope and deflection, use the tables on page 140-141 If the beam has more than one load, use superposition Note: When you plug-in the numbers in the equations, make sure the units cancel!! 13. Mohr's Circle Principal stresses: 𝝈𝒂 , 𝝈𝒃 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝝈𝒙 𝝈𝒚 𝟐 𝝉𝟐𝒙𝒚 ⎯⎯⎯⎯⎯ 𝝈𝒙 𝝈𝒚 ⎯⎯⎯⎯⎯ 𝟐 Maximum in-plane stress: 𝑅 𝟐 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝝈𝒙 𝝈𝒚 𝟐 𝝉𝟐𝒙𝒚 ⎯⎯⎯⎯⎯ 𝟐 Note: 𝝈𝒙 , 𝝈𝒚 , 𝝉𝒙𝒚 can be negative if their direction is opposite from the direction shown in the above figure. Materials www.engenieer.com 1. Stress and Strain Graph A. proportional limit B. elastic limit C. yield stress D. ultimate stress E. fracture stress 2. Failure There are three different type of failures: fracture, fatigue and creep Fracture Ductile Fracture: Crack is stable and it does NOT propagate only if there is an increase in the applied load Brittle Fracture: Crack is unstable and it does propagate even if there is no increase in the applied load Fatigue Occurs to structures that are subjected to dynamic and cyclic stresses Creep Occurs when material is subjected to a static, constant stress Materials www.engenieer.com 3. Tests Brinell and Rockwell test are used to determine material's resistance to localized plastic deformation Charpy test is used to find energy required to fracture and to identify the ductile to brittle transition. Stress-relaxation or creep test is used to determine over time under continuous load 4. Fabrication of Metals Forming Processes: Hot working: Is when a metal is processed through a temperature that is above the recrystallization temperature and goes through plastic deformation Cold working: Is when a metal is processed at room temperature and shaped by applying pressure Thermal Processes: Quenching: Is when a metal is processed and heated to a temperature that is above the recrystallization temperature and it is cooled rapidly Annealing: Is a heat treatment process that is conducted after cold working to remove any residual stresses Tempering: Is a heat treatment process that is conducted after quenching to remove any internal stresses Fluid Mechanics www.engenieer.com 1. Shear Stress The shear stress can be calculated using: 𝒅𝒗 𝛍 ⎯⎯⎯ 𝒅𝒚 𝛕 If velocity is given as a function of distance, take the derivative If velocity is not given as a function of distance, plug-in the numbers in the calculator 2. Bernoulli H1 𝐇𝟐 𝐇𝐋 𝐇𝟏 𝐏𝟏 ⎯⎯ 𝛄 𝐕𝟏𝟐 ⎯⎯⎯ 𝐙𝟏 𝟐𝐠 𝐇𝟐 𝐏𝟐 ⎯⎯ 𝛄 𝐕𝟐𝟐 ⎯⎯⎯ 𝐙𝟐 𝟐𝐠 Note: The velocity sometimes needs to be determined using the continuity equation, Q=VA 𝑯𝑳 can be calculated using Darcy-Weisbach or Hazen-Williams, depending on the question 3. Pressure Drop If the flow is laminar, to determine the pressure drop use the following equation: 𝑸 ∆𝑷𝒇 𝝅𝑫𝟒 ∆𝑷𝒇 ⎯⎯⎯⎯⎯⎯⎯⎯ 𝟏𝟐𝟖𝝁𝑳 𝑸 𝟏𝟐𝟖𝝁𝑳 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝝅𝑫𝟒 If the flow is turbulent, to determine the pressure drop use the following steps: 1. Set up Bernoulli equation 2. For head loss, use either Darcy-Weisbach or Hazen-Williams 3. Rearrange the equation and solve for ∆𝑷𝒇 Fluid Mechanics Major Head Losses Darcy-Weisbach In the problem, the following variables are giving: V- Velocity of the fluid D- Diameter of the pipe L- Length of the pipe µ or v - Dynamic (µ) or kinematic (v) viscosity (see table on page 199) 𝛜 Roughness factor (see table on page 201) To determine the head loss: Step 1: Find Reynolds Number If Re < 2100: Laminar flow, use f=64/Re If Re > 10,000: Turbulent flow, use Moody chart f(Re, 𝛜/D) Step 2: Calculate the head loss using Darcy 𝒉𝒇 𝒇𝒕 𝑳 𝑽𝟐 𝒇 ⎯⎯⎯⎯⎯ 𝑫 𝟐𝒈 f is dimensionless L - m or ft D - m or ft V - m/s or ft/s g - 9.81 m/𝑠 or 32.2 ft/𝑠 Hazen-Williams In the problem, the following variables are giving: V- Velocity of the fluid D- Diameter of the pipe L- Length of the pipe C- Roughness coefficient (see table on page 297) Step 1: Find the flow rate, Q=VA Step 2: calculate head loss using Hazen-Williams equations 𝒉𝒇 𝒇𝒕 𝟒. 𝟕𝟑 𝑳 𝑸𝟏.𝟖𝟓𝟐 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝟏.𝟖𝟓𝟐 𝑪 𝑫𝟒.𝟖𝟕 C is dimensionless L - m or ft D - m or ft Q - 𝑚 /s or 𝑓𝑡 /s www.engenieer.com Fluid Mechanics www.engenieer.com Pump Efficiency or Brake Power To determine the brake power, the following equation is used 𝑾 𝜸𝑯𝑸 ⎯⎯⎯⎯⎯⎯ 𝜼𝒑𝒖𝒎𝒑 H: Head added by pump (Use Bernoulli Equation) Q: Flow rate, VA (Continuity equation) 𝜼𝒑𝒖𝒎𝒑 : Efficiency of the pump 𝝆𝒈 𝒐𝒓 𝜸: 1000 𝑘𝑔⁄𝑚 𝝆𝒈 9.81 𝑚 ⁄𝑠 9810 𝑁⁄𝑚 SI Units 9.81 𝑘𝑁⁄𝑚 10 English Units 𝛒𝐠 𝐨𝐫 𝛄 (water) 𝟗. 𝟖𝟏 𝐤𝐍⁄𝐦𝟑 𝟔𝟐. 𝟒 𝐥𝐛/𝐟𝐭 𝟑 H m ft Q 𝐦𝟑 /𝐬 𝐟𝐭 𝟑 /𝐬 𝛈𝐩𝐮𝐦𝐩 (%) /100 /100 SI Units: 𝐖 𝛄𝐇𝐐 𝐤𝐍⁄𝐦𝟑 𝐦 𝐦𝟑 ⁄𝐬 ⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝐤𝐍 𝛈𝐩𝐮𝐦𝐩 𝛈𝐩𝐮𝐦𝐩 𝐦⁄𝐬 𝐤𝐉⁄𝐬 𝐤𝐖 English Units: 𝐖 𝛄𝐇𝐐 𝐥𝐛⁄𝐟𝐭 𝟑 𝐟𝐭 𝐟𝐭 𝟑 ⁄𝐬 ⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝛈𝐩𝐮𝐦𝐩 𝛈𝐩𝐮𝐦𝐩 𝐥𝐛 𝟏 𝒉𝒑 𝐟𝐭⁄𝐬 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 𝟓𝟓𝟎 𝒍𝒃 𝒇𝒕⁄𝒔 𝐡𝐩 𝜸 SIMPLIFY YOUR FE JOURNEY 1 2 3 ENROLL & START TODAY STUDY SMARTER PASS WITH CONFIDENCE! enGENIEer FE PREP COURSES ACCELERATOR MORNING + AFTERNOON BUNDLE EVERYTHING YOU NEED TO PASS THE FE CIVIL! WE KNOW HOW IMPORTANT PASSING THE FE EXAM IS TO YOU. This is the only course that worked for me. I’ve tried PPI2Pass and it was too difficult. Follow the schedule and trust the process and you will pass. I’ve failed the FE Civil 6 times before passing the 7th time with Kenza’s courses. Rodel, enGENIEer Student and EIT I failed my FE 5 times and I passed the 6th time, what’s the difference? Hands down, Kenza’s course. This is definitely a game changer. Naga, enGENIEer Student and EIT Getting the EIT license is important in the engineering industry and passing the FE exam is stressful. People spend time and energy studying without getting actual results. With enGENIEer, we make it easy to pass your FE exam faster. We simplify engineering concepts and problems and we teach exactly what you need to know for your exam. In 3 months, you can pass your FE exam and excel in your career. Kenza CEO, enGENIEer CONTACT US hello@engenieer.com www.engenieer.com

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