### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Engineering Statics ENGR 210

UI

GPA 3.67

### View Full Document

## 22

## 0

## Popular in Course

## Popular in General Engineering

This 53 page Class Notes was uploaded by Mr. Walton Littel on Friday October 23, 2015. The Class Notes belongs to ENGR 210 at University of Idaho taught by Ahmed Abdel-Rahim in Fall. Since its upload, it has received 22 views. For similar materials see /class/227747/engr-210-university-of-idaho in General Engineering at University of Idaho.

## Reviews for Engineering Statics

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/23/15

Engr210 7 Spring 2005 Lesson 34 Moment of Inertia Instructor Ahmed AbdelRahim Page 1 of 2 Today s Obiectives a Apply the parallelaxis theorem b Determine the moment of inertia for a composite area ParallelAxis Theorem for an Area Relates the moment of inertia of an area about an axis passing through the area s centroid to the Moment of Inertia of the area about a corresponding parallel axis ParallelAxis Theorem for an Area 1 1x Adjand I 2 1y Adj Where xv andlyv Moment of Inertia with respect to centroidal X and y axis 1 a dly Moment of Inertia with respect to any X and y axis Parallel to X and y axis 61 and dy perpendicular distances between X and y and X and y Moments of inertia of Composite areas Composite area is made by adding or subtracting a series of simple shaped areas like rectangles triangles and circles The Moments of Inertia ofthese simpler shaped areas about their centroidal axes are known Using these data and the parallelaxis theorem the Moment of Inertia for a composite area can easily be calculated Steps for the analysis 1 Divide the given area into its simpler shaped parts 2 Locate the centroid of each part and indicate the perpendicular distance from each centroid to the desired reference axis Determine the Moments of Inertia of each simpler shaped part about the desired reference axis using the parallelaxis theorem IX IX Amy2 4 The Moments of Inertia ofthe entire area about the reference aXis is determined by performing an algebraic summation of the individual Moments of Inertia obtained in Step 3 Please note that Moments of Inertia ofa hole is subtracted A v Instructor Ahmed AbdelRahim Engr210 7 Spring 2005 Lesson 34 Moment of Inertia Page 2 of 2 Example Example 3in T 3 in Find The moment ofinertia ofthe area about the yaXis and the radius of gyration ky39 Find The moment ofinertia for the area about the X aXis and the radius of gyration kX Engr210 7 Sp ng 2005 Lesson 26 Internal Forces Instructor Ahmed AbdelRahim Page 1 of 1 Today s Ob39ective 1 Identify types of internal forces 2 Determine internal forces in 2D Where we started 1 Activities 2 Load applied 3 Reactions 4 Forces at Joints 5 Internal Forces 6 Properties of materials Resistance 7 Properties of sections 8 Final Design 9 Internal forces in 2D A section divides the structure into two completetly separate pa s Both parts are under equilibrium Internal forces at the section equal and opposite Analyze 2D structures 1 for the entire structure 2 Find support reactions 3 Take a section where you need to find the internal forces imaginary cut 4 Decide which resulting part is easier to analyze 5 Draw a FBD of the piece of the structure you ve decided to analyze Remember to show the N V and M loads at the section 6 Apply equilibrium equations to find N V and N Example 400 N rial Find Internal force at point C l 3 m Engr210 7 Spring 2005 Instructor Ahmed AbdelRahim Lesson 21 Machine Parts Page 1 of 1 Today s Obiectives Problem 694 a Draw the free body diagram of a machine part and its members b Determine the forces acting at the joints and supports of a frame or machine Very similar to frames with moving rather than stationary parts Also similar to frames machine parts are structures that have at least one multiforce member Do you expect loads to be different in machine parts than frames Why n 25m AD is a two force member 2 MB0 FAD sin 70 5 606 250 7 0 ABC FAD 44908 lb 2 ME0 44908 2 cos 20 55 M 0 ED M 314 bft Problem 6100 Pin connection at B Z MB0 N 3 50 25 0 N 417 lb Which Force actually stops the Bike Applied Force vs Friction Force Problem 6101 Engr210 7 Spring 2005 Lesson 15 Rigid Body Equilibrium 3 Instructor Ahmed Abdel Rahim Page 1 of 2 Today s Obiectives 1 Identify support reactions in 3D and draw a free body diagram and 2 Aapply the equations of equilibrium in 3d Type of supports Ball and socket joints Journal bearings mechanical systems Smooth surface support Single hinge Single bearing Cables Reactions Remember our Rule Reactions develop where motion are prevente Allow motion NO REACTION Type of Reactions 2 smooth surface Support iF ball and socket Type of Reactions single hinge Mfg1 Single oumal bearing Refer to the table in the Text book to determine the type of support and support reactions Eguilibrium in 3D when a body is in equilibrium the net force and the net moment equal zero ie 2F 0 and 2M0 0 These two vector equations can be written as six scalar equations 2 Fx ZFY ZFZ0 MX 2 MY 2 M2 The moment equations can be determined about any point Usually choosing the point where the maximum number of unknown forces are present simplifies the solution Important note In some cases there may be as many unknown reactions as there are equations of equilibrium However if the supports are not properly constrained the body may become unstable Engr210 7 Spring 2005 Lesson 15 Rigid Body Equilibrium 3 Instructor Ahmed Ab delRahim e 2 of 2 Example 564 Example 565 Determine Xy and 2 components of Reactions where the wing is xe Determine The tention in each of the three cables Determine The greatest vertical force F that can be applied if e ca Ies can sustain a max force of 800 lb Aso Find the Support eaction Engr210 7 Spring 2005 Instructor Ahmed AbdelRabim Lesson 27 Moment Shear and Normal Force diagrams Page 1 of 1 Today s Obiective 1 Develop shear and moment equations for beams Draw Shear Normal and Moment Equations for beams N Things to consider Normal forces in Beams Define x in the shear and moment equations No change in loading a Supports b start and end of distributed laods c Concentrated loads Ign conveiv VT 1 Vl39l39lv Fosilive shear M M PosiLive moment M M Beam Sign conveniion Depends on which side you are doing the analysis Procedure for anal is a FBD for the entire structure b Support Reactions c Set x y axis left side of the beam d De ne x1 x2 x3 e Find V and M at sextion 1 2 and 3 f Draw the shear and bending moment diagrams Example 748 3m Example 749I 752 amp 754 50 lbl39l 200 lb Ii 21 kN 40 kNm Engr210 7 Spring 2005 Lesson 3Force vectors in 3D Instructor Ahmed AbdelRahim Page 1 of 2 Today s Obiectives 1 Represent a 3D Force vectors 2 Determine the magnitude components and coordinate angles of a 3D vector 3 Add Force vectors in 3D Review Discussion Examples of reallife 3D Forces Forces in cables that support power towers U n it Vectors For a vector A with a magnitude of A an unit vector is de ned as UA A A Characteristics of a unit vector Its magnitude is 1 It is dimensionless It points in the same direction as the original vector A The unit vectors in the Cartesian axis system are iLandk along the positive Mandi axes respectively Position Vector of a Point For Point A with coordinates AX Ay and A2 The position vectors is AAx1Ay1Azr The magnitude of the vector A A AX2 Ay2 A2212 The magnitude of the projection ofvector A in the X y plane Axy sz Ay212 The magnitude of the vector A can also be represented by A Axy2 A2212 Similarly A AX22 Ay212 A Ayz2 AX212 Unit vector The direction or orientation ofvector A is de ned by the angles or 3 and y The magnitude of the angles can be determined as follows Ax A 1 cosoz A cosB A cosy A The anagles should satisfy the relationship coszoc c0s2 c0szy1 proove Engr210 7 Spring 2005 Lesson 3Force vectors in 3D Instructor Ahmed AbdelRahim Page 2 of 2 Representing unit vector in terms of the angles m gagA AAtiAyiAzks Thus A A A y A 39 w Zk quotA A AHA J A or gAcosui cosBjcosyk Adding 3D vectors AijAvjAz and g ijBv1sz then AxBxi AvBvL AZBZK AxBxi AV39BVj AzBZK Defining a force vector Vector is defined by its magnitude and two angles 1 Using the relationship cos 2 a cos 2 i cos 2 y 1 find the third angle 2 Find Ax Ay and A2 using the relationship Ax A cos on Ay A cos 391 A2 A cos y 3 Write the force vector A Axi ij Azk Vector is defined by its magnitude and two components 4 Using the relationship A Ax2 Ay2 A22 2 find the third component 5 Find the three angles using the cusa A cos 3 Av cos A s e e 7 7 relationship A A A 6 Write the force vector A Axi ij Azk Vector is defined by three components 1 Using the relationship A Ax2 Ay1 A22 2 find the vector magnitude N v Find the three angles using the cos a i A cos B A39 cos 1 e z y relationship A A A 3 Write the force vector A Axi ij Azk Engr210 7 Spring 2005 Instructor Ahmed AbdelRahim Lesson 23 Friction Dry Friction Page 1 of 2 Today s Obiective Analysis of A Wedge Determine the forces on a wedge 9 r h P gti 39 F fr Impending Impending niche N 1050 FBDs of the wedge De nitions the friction forces WHEN force p to lift a large always in the direction opposite to the motion weight W or impending motion of the wedge applied along the contacting surfaces Other uses a adjust the elevation the normal forces b provide Stability for heavy objects the normal forces are perpendicular to the such as this large steel vessel contaCt39ng surfaces Analysis of A Wedge Analysis of A Wedge FBDs of the object on top of the Wedge More than One Wedge 7 v F Start by FBD forthe Wedge with the 3 Applied Force N1 7 i To determine the unknowns Stir we must apply Equilibrium Equations f o the wedge and the object N2 at the contacting surfaces between the wedge and the object the forces are equal in magnitude and opposite in direction to those on the wedge friction forces between the object and other surfaces always in the direction opposite to the motion or impending motion of the object which one should we start analyzing rst Example Engr210 7 Spring 2005 Instructor Ahmed AbdelRahim Lesson 23 Friction Dry Friction Page 2 of 2 Example N 100 load weighs 100 lb us between surfaces AC and BD is 03 Smooth rollers are placed between wedges A and B rollers and the wedges have negligible weights FBD for Wedge B Applying Equilibrium Equations for Wedge B Why Wedge B rst ZFX N2sin10 N30 ZFY N200510 100 03 N3 0 Solving the above two equations N2 1072Ib and N3 186Ib At wedge A we get 2 FY N1 1072 cos 10 0 N1 1056 lb ZFX P 1072in10 03 N1 O P 503Ib What if there were no rollers between A and B Engr210 7 Sp ng 2005 Lesson 4Position and Force VectorsDot Product Instructor Ahmed AbdelRahim Page 1 of 2 Today s Ob39ectives 1 Represent a position vector in M 00 5 Cartesian coordinate form Represent a force vector directed along a line Determine an angle between two vectors and Determine the projection of a vector along a specified line Definition A position vector is defined as a fixed vector that locates a point in space relative to another point Position Vector relative to point 0 A position vector is defined as a fixed vector that locates a point in space relative to another point Position Vector Directed from point A to is xB XA1YB YA1zB 2Am Note that B is the ending point and A is the starting point So ALWAYS subtract the tail coordinates from the tip coordinates Force Vector Directed Along A Line Plan 1 Find the unit vector in the line s direction Find the position vector AB r Bquot E F7 Find the magnitude of the position vector AB rAB Determine the unit vector HAS 5quot B I rAB Multiply the unit vector by the AAxiAy1Ak 2 magnitude of the force F F gAB Example Plan 1 Find the forces vectors along CA Given Two forces are acting on a pipe Find as shown in the figure The magnitude and the coordinate direction angles of the resultant force and CB 2 Add the two forces to get the resultant force FR Determine the magnitude and the coordinate angles of FR 0quot Question How to determine the force vectors Step 1 Coordinates of Points C 0 0 4 A 3 sin 40 3 cos 400 B 4 7 0 Step 2 position vectors 3 sin 40 3 cos 401 4 IcA 1 93 1 231 4 L 5 gt I Lea 41 714 Engr210 7 Spring 2005 Lesson 4Position and Force Vectors Dot Product Instructor Ahmed AbdelRahim Page 2 of 2 Step 3 magnitude of position vectors rCA 49312 232 4 21 2 rCA CB 412 72 4 211 2 CB 9 Step 4 unit vectors cA 1931 231 4 ns Hes 41 7i 4L9 Step 5 force vectors ECA 100 ECA EcA 3857 1 45961 80 5 lb ECB 81 quotGB ECB 361 631 36qub Step 6 Resultant B EcB ECA 3 FDA FEB 257 i 1704j 116 k lb Now we are ready to find the magnitude of the resultant R25712 17o42 116 21 2 R11727 The final step is to find the coordinate angles using the relationships A A A cosazXt cos j cosyiz or cos1quoti l3953971173 91 3 3 cos11704I1173 984 39y cos1116l1173 172 The DOT product of two vectors 39239 The dot product of vectors A and B is defined as AB A B cos 0 39239 Angle 0 is the smallest angle between the two vectors and is always in a range of 0 to 180 0 00 The result of the dot product is a scalar a positive or negative number 0 00 The units of the dot product will be the product of the units of the A and B vectors The DOT product of two vectors Examples ij 0 i i 1 A 39B Ax1 Ayl A2 39 Bxi BM 32 K Ax 13x AyBy AZBZ Using the Dot Product To Determine The Angle Between Two Vectors 1 Finding the dot product A B AxBx AyBy Asz 2 Finding the magnitudes A amp B of the vectors A amp B and 3 Using the definition of dot product and solving for 9 ie 4 e cos1 A BA 3 Where 0 5 0 5 180 Engr210 7 Spring 2005 Lesson 4Position and Force Vectors Dot Product Instructor Ahmed AbdelRahim Page 3 of 2 Determining the Proiection Of A Vector A cn 9 II You can determine the components of a vector parallel and perpendicular to a line using the dot product Steps Find the unit vector Uaa along line aa39 Find the mprojection of A along line aa39 by A AU Axe AyUy AzUz the projection can be written as a vector A by using the unit vector Uaa and the magnitude found in step 2 A All Uaa EnnglO 7 Spring 2005 Lesson 6 3D Equilibrium Instructor Ahmed AbdelRahim Today s Ob39ectives 1 Apply equations of equilibrium to solve a 3D problem 2 Review HW format and guidelines For a particle in equilibrium in the 3D 2Fx EFy j EFZ 5 0 A vector equation Or written in a scalar form 2Fx0EFy0 and 2Fz0 These are three scalar equations of equilibrium They can be used to solve for up to three unknowns Examgle 346 Force in AB 700 N Tension vs comp Find Force in AC and AD and F Example 347 Find Stretch in each ofthe springs L2 m and K300 Nm Engr210 7 Spring 2005 Lesson 18 Truss 7 Introduction and Joint Method Instructor Ahmed AbdelRahim Page 1 of 2 Today s Objectives 1 Define a Truss 2 3 4 List all assumptions required to Analyze a truss Analyze a truss using the joint method Identify zeroforce members Trusses Structure elements commonly used in covering long spans and railroad bridges Load carrying capacity will depend on the type of material used Steel aluminum or wood Trusses general assumptions a Twoforce members b Members connected at joints through frictionless pins c Load are applied at joints ONLY d Weights of the members are neglected e With these two assumptions the members act as twoforce members They are loaded in either tension or compression Analyzing Trusses using the method of Joints a Forces applied at each joints are concurrent forces in the direction of members and intersects at the joint b Each Joint is under equilibrium thus two equations can be applied Sum Fx0 and Sum Fy0 Solution Plan 1 lfthe support reactions are not given draw a FBD ofthe entire truss and determine all the support reactions using the 3 equations of equilibrium Draw the freebody diagram ofa joint with one or two unknowns Assume that all unknown member forces act in tension ullin the m unless you can determine by inspection that the forces are compression loads 3 Apply the scalar equations of equilibrium 2 FX O and Z Fy 0 to determine the unknowns If the answer is positive then the assumed direction tension is correct otherwise it is in the opposite direction compression 4 Repeat steps 2 and 3 at each joint in succession until all the required forces are determined Engr210 7 Spring 2005 Lesson 18 Truss 7 Introduction and Joint Method Instructor Ahmed AbdelRahim Page 2 of 2 ZeroForce Members D If a joint has only two noncollinear members and there is no external load or support reaction at that joint then those two members F E 9 K are zeroforce members In this example 0 members DE CD AF and AB are zero force members You can easily prove these results j by applying the equations of equilibrium to A C joints D and A Zeroforce members can be removed as 3 shown in the gure when analyzing the truss P r E a m r 2 Truss equilibrium equation DC B I P Zrero Force Members Example lfthree members form a truss joint for which two of the members are collinear F F and there is no external load or reaction at that joint then the third noncollinear gt F member is a zero force member Wh Do we use zero force members Stability especially for truss members under compression Different Loading conditions 57s Determine the number onero Force members TwoGroup Activities Engr210 7 Spring 2005 Lesson 12 Distributed Loads Instructor Ahmed AbdelRahim Page 1 of 2 Today s Obiectives 1 Determine an equivalent force for a distributed load Concem Two types of Loads Concentrated Point Force Distributed Loads Examples Book Shelves Winds load Fluids pressure Weight of items on the body s surface How can we determine a single equivalent resultantforce and its location Magnitude line of Action and point of application Equivalent means Have the same effect General Case w in wu 39 m J I dF wx dx The net force on the beam is given by t FR ILdF ILwxdx A A is the area under the loading curve wx The force dF will produce a moment of xdF about point 0 The total moment about point 0 is given as MRO ILxdF Iwaxdx Assuming that FR acts at it will produce the moment about point 0 as MRO x FR x IL wx dx Eguivalent systems Engr210 7 Spring 2005 Instructor Ahmed AbdelRahim Lesson 12 Distributed Loads Page 2 of 2 Example Example Engr210 7 Spring 2005 Lesson 25 Friction Belt Friction Instructor Ahmed AbdelRahim Page 1 of 1 Today s Obiective Determine the tension forces in a belt Definitions Belts transmit power from one shaft to another Machines Brakes etc Design Issues a Slipping friction b breaking tension two Types flat belts and Analysis of A belt Contact Angle i in radians How to change from degrees to radians the belt slips or is just about to slip Motion Impending then T2 gt T1 T2T1e p T2gtT1 Example 885 50 lb g s mom aw 39 T1 50 and T2 150 i 11337 rad 113372 u 18 turns use n 2 turns Example 891 SumMa0 F475P T2T1equotl T2gtT1 F 1953P Sum Mb O P423 N Engr2 10 7 Spring 2005 Lesson 1 1 CouplesForce Couple Systems Instructor Ahmed Abdel Rabin Page 1 of Todax s Obiectives 1 De ne a couple and determine the moment of a couple 2 Determine the effect of moving a force 3 Find an equivalent forcecouple system for a system of forces and couples Con ce at F Cl 1 3 quot 5 F X A couple is two parallel forces with the same magnitude but opposite in direction separated by a perpendicular distance d M0 M0 F d using a scalar analysis r x F using a vector analysis Here r is any position vector from the line of action of F to the line of action of The net external effect of a couple is that the net force eguals zero and the magnitude of the net moment equas F d Since the moment of a couple depends only on the distance between the forces the moment of a couple is a free vector It can be moved anywhere on the body and have the same external effect on the body Moments due to couples can be added using the same rules as adding any vectors AN EgzUIVALENT SYSTEM Two force and couple systems are called eguivalent systems if they have the same external effect on the body MOVING A FORCE ON ITS LINE OF ACTION Moving a force from A to 0 when both points are on the vectors line of action does not change the external effect Hence a force vector is called a sliding vector MOVING A FORCE OFF ITS LINE OF ACTION Moving a force from point A to O as shown above requires creating an additional couple moment Since this new couple moment is a free vector it can be applied at any point P on the body Engr210 7 Spring 2005 Lesson 11 CouplesForce Couple Systems Instructor Ahmed AbdelRahim Page 2 of 2 Finding the resultant of a force and couple sEtem FR MR EMU FRX FR 2F MR0 EMF 1quot 2M0 Reducing a forcemoment to a single force gus r f FR and MR0 are perpendicular to each other then the system can be further reduced to a single force FR by simply moving FR from O to P Example Find The equivalent resultant force and couple moment acting at A Solution Plan 1 Sum all the X and y components of the forces to find FRA 2 Find and sum all the moments resulting from moving each force to A and add them to the 500 lb ft free moment to find the resultant MRA gtZFX 45 150 lb 50 lb sin 30 145 lb l ZFy 35 150 lb 50 lb cos 30 1333 lb FRA 145 2 1333 212 197 lb and 6 tan1 1333145 426 MRA 451502 5o cos30 3 50 sin30 6 500 760 Ibft Engr210 7 Spring 2005 Lesson 24 Frictional forces on Screws Instructor Ahmed AbdelRahim Page 1 of 2 Today s Obiective Analyze forces on screws Screws Functions transmit power or motion Automabile Jack as fasterners how would we analyze them Differenet type a Squarethreaded b Vthreaded Analysis Approach an inclided plane warpped around a cylinder Wedge 1 L 7 lh Analysis of a Screw Design Paremeters r mean radius ofthe thread I rise with a single revolution 360 motion the Lead of the screw 9 Lead angle tan391I2nr WAxial laod applied MMoment needed to turn the screw S Horizontal force on the screw caused by the applied moment M Sr N Normal force FFriction Force Rresultrant of N and F I Friction angle tan391u Static and Kinetic I39 kl V Anal sis ofa screw U ward Motion F action downwards ZFXS Rsine 0 ZFYW Rcose 0 Solving S Wtan e Then M Sr Wr tan e Analysis of a screw Dowward Motion F action upwards Two Cases 9 gt 4 and e lt 4 egt MWrtane e lt 1 Verv rouch screws Moment should be applied notice the change in the direction of S M Wrtan cl e Engr210 7 Spring 2005 Instructor Ahmed AbdelRahim Lesson 24 Frictional forces on Screws Page 2 of 2 Analysis of a screw Slef Lockingl Example 8 74 Under Equilibrium No moment applied I 2 9 W Given r 14 mm 6 mm p 02 Static M 15 Nm Note Unitsll Find F e tan391l2m 7768 stan391p5 4 1131 Moving Upward M W r tan e WF F 620 N Notice 5 gt e Self Locking force will remain even ig the moment was removed Engr210 7 Spring 2005 Lesson 22 Friction Dry Friction Instructor Ahmed AbdelRahim Page 1 of 2 Today s Obiectives a Review the characteristics of dry friction b Draw a FBD including friction c Solve problems involving friction Definitions Friction is defined as a force of resistance acting on a body which prevents or retards slipping of the body relative to a second body Frictional forces act tangent lparallell to the contacting surface in a direction opposing the relative motion or tendency for motion Questions to think about Smooth surfaces vs rough surfaces Are rough surfaces the same How can we reduce friction force Weather effect Eguilibrium of a body under friction forces VV ficx ulluni Nnrmul and l nuionul Forces Apply the 3 Equilibrium equations P F NW and Ph Wx If we increase P friction force F increases up to a certain point motion F No motion Molian Eguilibrium of a body under friction forces The maximum friction force is attained just before the block begins to move impending motion The value of the force is found using Fs us N where us is called the coefficient of static friction The value of us depends on the materials in contact Once the block begins to move the frictional force typically drops and is given by Fk W N The value of uk coefficient of kinetic friction is less than us Questions to think about How to determine the coefficient of friction How about traffic accident investigation O ZFy N Wcoses ZFxHsN Wsines 0 us WsinesWcoses taneS Engr210 7 Spring 2005 Lesson 22 Friction Dry Friction Instructor Ahmed AbdelRahim Solving Friction Problems Draw the necessary free body diagrams Make sure that you show the friction force in the correct direction it always opposes the motion or impending motion Determine the number of unknowns not assume F us N unless the impending motion condition is given Apply the equations of equilibrium and appropriate frictional equations to solve for the unknowns Page 2 of 2 Impending Tipping vs Sliding W Impending l N 31 bZ 7 x TIPng r l N N X 3 quot F u N Four unkwons make assumptions and check your answer against it Assume Slipping occurs Known F p5 N Solve x P and N Check 0 S x S b2 Assume Tipping occurs Known x b2 Solve P N and F Check F s ue N Example Given us05a3ftandb4ft Find The smallest magnitude of P that will cause impending motion tipping or slipping of the drum P 5 3715 ft115 ft 4 l A FBD ofthe drum There are four unknowns P N F and x Assume the drum slips Then the friction equation is Fus N 05 N gtZFX 45P 05N O TZFYN 35P 100O These two equations give P 100 lb and N 160 lb ZMO 351OO 15 45 1004 160 x 0 Check x 144 g 15 so OK Drum slips as assumed at P 100 lb Engr210 7 Spring 2005 Instructor Ahmed AbdelRahim Lesson 20 Frames Page 1 of 2 Today s Obiectives Example 1 a Draw the free body diagram of a frame and its members 2000 N b Determine the forces acting at the joints and supports of a frame or machine Frames are commonly used to support various external loads Frames are structures that have at least one multiforce member Recall that trusses have nothing but twoforce members How is a frame different than a truss How can you determine the forces at thejoints and supports of a frame i lt7 2 m e if 2 m iv Draw FBD for the Frame a Identify any twoforce members b Forces on contacting surfaces usually between a in and a member are equal and opposite and 2 Develop a strategy to apply the equations of equilibrium to solve forthe unknowns Example 1 2000 N Example 2 4n Engr210 7 Spring 2005 Lesson 20 Frames Instructor Ahmed AbdelRahim Page 2 of 2 FBD for each Element gt F 5 501h T I A FBD LilIIUUGV E39 ZFY 2T 700 0T ZFX CX 350 0 OX 350bC 0 350 lb E ZFY CY 350 CY 350bC ZFX BX350 3508in30 0B ZFY BY 350 08 30 0 B BX 175 lb BY 303m ZFY ZFX 7001 l39i 39 39 I 39 A FBD OI member ABC TBD sin 45 43031 4 700 8 0 TBD 2409 lb AY 2409 sin 45 3031 700 0 AY 700 lb AX 240900845O 175 350 0 AX 1880 lb A FBD of member BD 2409 lb D 2409 lb At D the X and Y component are DX 2409 008 45 1700 lb DY 2409 sin45 1700 lb Engr210 7 Spring 2005 Lesson 17 Equilibrium of RB ampExam2 Review Instructor Ahmed AbdelRahim Page 1 of 1 Today s Ob39ectives T er of suggorts Ballandsocket joints Journal bearings mechanical systems Smooth surface support 1 Exam1 review 2 Aapply the equations of equilibrium in 3d Example 586 Single hinge Single bearing Cables Determine Force P for equilibrium of the crankshaft Exam Qle 521 Determine Forces FA Fc and FE W5 lb Engr210 7 Spring 2005 Lesson 29Centroids Instructor Ahmed AbdelRahim Page 1 of 2 Today s Obiective a Understand the concepts of center of gravity center of mass and centroid b Determine the location of these points for a system of particles or a body Why center of gravity is important a Resultant of Distributed forces b Point of application of the weight Design and stability problems Definitions and Facts The center of gravity G is a point which locates the resultant weight of a system of particles or body Definitions and Facts Cont Similarly the center of mass is a point which locates the resultant mass of a system of particles or body Generally its location is the same as that of G The centroid C is a point which defines the geometric center of an obiect The centroid coincides with the center of mass or the center of gravity only if the material of the body is homogenous density or specific weight is constant throughout the body If an object has an axis of symmetry then the centroid of object lies on that axis CG CM For A System Of Particles WR 2W Moments about the y axis we get x WR lel szz onn The x y z coordinates of the CG CM ziw 29w 22W 2W y2W zEW f CG CM For A System Of Particles Using Integration assuming that a rigid body is made up of an infinite number of particles then I dW J39 y 1W J E dW Vvii Z jaw JdW JdW Engr210 7 Spring 2005 Instructor Ahmed AbdelRahim Lesson 29 Centroids Page 2 of 2 Steps for Determining Area Centroid Example 1 Choose an appropriate differential element dA at a general point xy Hint Generally if y is easily expressed in terms ofx eg y x2 1 use a vertical rectangular element 2 Express dA in terms ofthe element dx or dy 3 Determine coordinates x y of the centroid of the rectangular element in terms of the general point xy 4 Express all the variables and integral limits in the formula using either x or y depending on whether the differential element is in terms of dx or dy respectively and integrate Example Engr210 7 Spring 2005 Lesson 36 Moment of Inertia about inclined axes Instructor Ahmed AbdelRahim Page 1 of 2 Today s Obiectives a Determine Moment of inertia of an area about inclined axes b Determine the orientation of the principal axes c Determine Principal Moments of Inertia lnclided Axes Moment of Inertia about u and v axes x Ix I I y y cos26 I s1n26 2 3 x If I y ycos26I s1n26 2 W IX I I ys1n26I cos26 2 3 To Apply these equations X gt y 9 measured from the X axis counterclockwise Principal Moment of Inertia We are typically interested in the orientation ofthe axes that give us Imax principal axes where ma and minquot To find the orientation of the principal I 60 or axes IxII 2 y s1n26 2Ixy cos260 K 2 I 2I tanZ P 2 xy Ix y The equation has two roots which are 90 apart defining the inclination of the principal axes Principal Moment of Inertia 2 Imaxmin Ix Iy i Ix Iyi 12 2 K 2 I xy Facts a The product ofinertia with respect to the principal axes luv0 b Any symmetrical axis represents a principal axis of inertia for the area Engr210 7 Spring 2005 Instructor Ahmed AbdelRahim Lesson 36 Moment of Inertia about inclined axes Page 2 of 2 Example 1079 k8ina r 8ina MT 2in2in e 45 Determine lu IV luv ImaXmin and 6p Engr210 7 Spring 2005 Lesson 8 Moment ofa Force Instructor Ahmed AbdelRahim Page 1 of 3 Today s Obiectives 1 Define moment of a force and 2 Determine moments of a force in 2D and 3D cases Applications Steering wheel What is the net effect of the two forces on the wheel Changing a Tire What is the effect of the force on the lug nut De nition The moment of a force about a point provides a measure of the tendency for rotation sometimes called a torque 2D Momment In the 2D case the magnitude of the moment is Mo F d d is the perpendicular distance from point 0 to the line of action of the force The direction of the Moment i is either clockwise negative or counterclockwise positive depending on the tendency for rotation For example M0 F d and the direction is counterclockwise Often it is easier to determine M0 by using the components of F as shown Using this approach M0 FY a FX b Note the different signs on the terms The typical sign convention for a We can determine the direction of rotation by imagining the body pinned at 0 and deciding which way the body would rotate because of the force Engr210 7 Sp ng 2005 Lesson 8 Moment ofa Force Instructor Ahmed AbdelRahjm Page 2 of 3 MOMENT IN 3D MW mw 3D Moment vector cross product 3D Moment can be obtained using the vector cross product ris the position vectorfrom point 0 to any point on the line of action of F CAXB In general the cross product of two vectors A and B results in another vector C C A xB The magnitude and direction of the resulting vector can be written as CAx BABsin0Uc U0 is the unit vector perpendicular to both A and B vectors as shown or to the plane containing the A and B vectors The right hand rule For example ixj k A vector crossed into itself is zero eg ix i CROSS PRODUCT i k A X B A Ay A BX B y B Z i Furulumun A 1 ilAJL AVBV 3 git Fm lemL nlj 1 A7 11411 7 mm Furuldmem k l krAJJ 74m Engr210 7 Spring 2005 Lesson 13 Rigid Body Equilibrium 1 Instructor Ahmed AbdelRahim Page 1 of 2 Today s Obiectives 1 Identify support reactions 2 Draw a free diagram Particle vs Rigid Body Analysis 1 In contrast to the forces on a particle the forces on a rigidbody are not usually concurrent 2 This will cause rotation of the body due to the moments created by the forces Rigid Body vs Flexible body No deformation a Forces on a particle vs forces on a Rigid Body For a rigid body to be in equilibrium the net force as well as the net moment about any arbitrary point 0 must be equal to zero 2D analysis ZFX Oand ZFy 0 Particle 2 Equs ZFXOZFyO and ZMOO RB 3 3D analysis 2 FX 0 Z Fy O and 2 F 0 Particle 3 ZFX 02Fy Oand ZFZ O ZMXO ZMXOand ZMZORB6 Free Body Diagrams For RB Eguilibrium Outlines shape ALL forces 1External Applied Forces 2 Weights Spring Forces 3 Support reactions Label Loads and all dimensions Role of Supports and Their Reaction General Rule support prevents translation in a given direction Reaction at that direction support prevents rotation Moment Reaction Table 51 Lists all supports and their Possible Reaction Common Support Types Roller Support Prevents y translation Pin Hinged support Prevents X and y translation Fixed Support Prevents X and y translation and Rotation Cables Reaction force in the cable direction Stability of 2D and 3D Structures of Equations vs of Support Reactions of Eq of R Stable and Determinate We can solve in Engr210 of Eq gt of R UnStabIe of Eq lt of R Stable but Indeterminate Engr210 7 Spring 2005 Instructor Ahmed AbdelRahim Lesson 13 Rigid Body Equilibrium 1 Page 2 of 2 Example Example An operator applies 20 lb to the foot pedal A spring with k 20 lblin is stretched 15 in Find Reactions at A and B ifi w mm 7400 mm 52 N 30 N i lii lb 5 m 7i b Engr210 7 Spring 2005 Lesson 10 Moment about a line Couples Instructor Ahmed AbdelRahim Page 1 of 2 Today s Obiectives 1 Determine the moment of a force about an axis using scalar analysis and 2 Determine the moment of a force about an axis using vector analysis Concept Recall that the moment of a force about any point A is MA F dA where dA is the perpendicular or shortest distance from the point to the force s line of action This concept can be extended to find the moment of a force about an axis In the figure above the moment about the yaxis would be My 20 03 6 Nm the moment about the xaxis would be Mx 20 04 8 Nm However this calculation is not always trivial and vector analysis may be preferable Moment about an axis Step 1 Compute the moment of the Force F about any arbitrary point 0 that lies on the aa axis using the cross product Mo r x F Step 2 Find the component of Mo along the axis aa using the dot product Ma Ua 39 Mo Facts Moment of a force about point 0 is a vector Moment of a force about a line is scalar Moment about an axis Triple scalar product 7 7 u aquot u I u u Muuu39er rx ry 7 F F F ua represents the unit vector along the axis a a axis r is the position vector from any point on the a a axis to any point A on the line of action of the force and F is the force vector Engr210 7 Spring 2005 Lesson 10 Moment about a line Couples Instructor Ahmed AbdelRahim Page 2 of 2 Example Given A force of 80 lb acts along the edge DB Find The magnitude of the moment of this force about the axis AC Solution Plan M Ac uAc 39 rAB X FDB 1 quotAc rAc IrAc 3 FDB quot03 r03 r03 4 Complete the triple scalar product W rAB20jft rAc13i16jft rDB 5i 10j 15kft Unit and Force vectors uAc 13i 16jftl132162 2 ft 06306i 07761 FDB 80rDBI5210215212lb 2138i 4276 j 6414k lb MAc 06306 20 6414 0 07706 0 0 lbft 809 Ibft Engr210 e Sp ng 2005 Lesson 30 Centroids for Composite Sections Instructor Ahmed AbdelRahim Page 1 of 2 Today s Ob39ective a Determine the location of centroid for composite sections Com Qosite sections m Coonsite sections made up of connected simpler shaped parts or holes rectange triangle circle semicircle etc Knowing the location of the centroid C or center of gravity G of the simpler shaped parts we can easily determine the location of the C or G for the more complex composite body Usingthe equations X 2XiAil2Ai yZYiAi2Ai The location of the centroid could be determined Note you could replace A with W M or L Procedures for the Analysis Divide the body into parts that are known shapes Holes are considered as parts with negative weight or size Choose an origin and x and y coordinate system origin at the left bottom corner of the shape Make a table to summarize the properties of different parts A Xi Yi A Xi A Yi A1 A2 A3 Sum EA ZAX ZAy Using values at the last row of the table find the x and y for the whole section Exam Qles d h m we 5 m a Engr210 7 Spring 2005 Instructor Ahmed AbdelRahim Lesson 30 Centroids for Composite Sections Page 2 o Example Engr2lO Spring 2005 Lesson 2Addition of a System of Coplanar Forces Instructor Ahmed AbdelRahim Page 1 of 2 Today s Obiectives Today s Objective 1 Resolve a 2D vector into components 2 Add 2D vectors using Cartesian vector notations Review A RAB B Parallelo gram Law A B RBA RAB B A Triangle method always tip to tail RESOLUTION OF A VECTOR RbUILLmI Extend parallel lines Ewan lhu Item of R Hummus In l39unn unmpumnl Resolution of a vector is breaking up a vector into components It is kind of like using the parallelogram law in reverse CARTESIAN VECTOR NOTATION I The directions are based on the X and y aXes We use the unit vectors i and j to designate the X and v aXes ADDITION OF SEVERAL VECTORS 1 Step 1 is to resolve each force into its components 2 Step 2 is to add all the X components together and add all the y components together These two totals become the resultant vector 3 Step 3 is to nd the magnitude and angle of the resultant vector F2 bl You can also represent a 2D vector with a magnitude and angle FRy FR 6 FVPF2 FRI R R1 Ry FR 1 y 9 tan FRX Definition Concurrent Forces are forces that are applied at the same point Rule 6 is measured for the x axis counterclockwise Engr210 7 Sp ng 2005 Lesson 2Addition of a System of Coplanar Forces Instructor Ahmed AbdelRahim Page 2 of 2 Examples Given Three concurrent forces acting on a bracket Find The magnitude and angle of the resultant force Solution Plan Resolve the forces in their xy components 2 Add the respective components to get the resultant vector 3 incl magnitude and angle from the resultant components Force Components F115sin40 i15cos40 jkN 9642i1149jkN F2 121326 i 51326j kN 24i10jkN F336cos30 i 36sin 30 jkN 3118i 18jkN Resultant Summing up all the iand jcomponents respectively we get FR9642 243118i1149 10 18j kN 1682 H 349j kN Resultant Magnitude and Direction FR 16822 34921I2 172 kN tan1349I1682117 Engr210 7 Spring 2005 Instructor Ahmed AbdelRahim Lesson 28 Moment Shear and Normal Force diagrams Page 1 of 1 Today s Obiective 1 Draw Shear Normal and Moment Equations for beams using sections method Reminder Si n convention Vlrllv Posilivc shear Poxilive moment it Depends on which side you are doing the analysis For Shear Find the shear on the following sections a Start and end of the beam b Before and after supports 0 Before and after concentrated loads d Start and end of distributed laods For Moment Find the shear on the following sections a Start and end of the beam moment always equals to zero unless there is an applied momemt b At supports 0 At loads d Start and end of distributed laods e Before and after applied moments f At point of zero shear maximum momemt Procedure for analysis a FBD for the entire structure b Support Reactions c Define sections for shear d Determine the magnitude of the seaher at each section e Determine location of zero shear f Define sections for Moment g Determine the magnitude of the moment at each section h Draw the shear and bending moment diagrams Example 748 15 kNlm l D If B 2 1n 3 In Example 749I 752 amp 754 50 mm 1 200 lbft H Cl Kill quot 7 l0 fig 20 kN 4lkNm 14 l 3 c L 39lsn kN m 8 m rlti 1 m 41 w a C A B A Engr210 7 Spring 2005 Lesson 33 Moment of Inertia Instructor Ahmed AbdelRahim Page 1 of 2 Today s Obiectives a Define the moments of inertia for an area b Determine the Moment of Inertia for an area by integration Moment of Inertia Concept Area is not the only property that affects the structural property of the section Area Normal ForceShear Force Moment of Inertia Bending Moment Moment of Inertia Definition For the differential area dA shown in the figure dlx y2dA dly x2dA and dJo erA where JOisthe polar moment of inertia about the pole O or z axis Moments of inertia The moments of inertia for the entire area are obtained by integration Ix IAyZdA Iy IA x2 dA Jo IA r2 dA IAx2y2dA l IV The Moment of Inertia is also referred to as the second moment of an area and has units of length to the fourth power m or in Radius ofG ration of an Area A yL k x X For a given area A and its Moment of Inertia lx imagine that the entire area is located at distance kx from the x axis Then I kxA or k I lx A This k is called the radius of gyration of the area about the x axis Similarly kY myA and k0 1JoA The radius of gyration has units of length and gives an indication of the spread of the area from the axes Engr210 7 Spring 200 5 Lesson 33 Moment of Inertia Instructor Ahmed AbdelRahim Page 2 of 2 Moment of Inertia by Integration Example 1 2 9quot P Choose the element dA The element parallel to the axis about which the Mol is to be determined usually results in an easier solution For example we typically choose a horizontal strip for determining IE and a vertical strip for determining I If y is easily expressed in terms of x eg y x 1 then choosing a vertical strip with a differential element dx wide may be advantageous Integrate to find the Moment of Inertia my y14x EnnglO 7 Spring 2005 Lesson 5 2D Equilibrium Instructor Ahmed AbdelRahim Page 1 0 Today s Ob39ectives 1 Draw a free body diagram FBD 2 Apply equations of equilibrium to solve a 2D problem What is FBD for a particle It is a drawing that shows all external forces acting on the particle How To Draw a FBD 1 isolated the particle free from its surroundin s Show all the forces that act on the particle a Applied forces want to move the particle b Reactive forces resist the motion Identify each force and show all known magnitudes and directions N 2 Example FBD at pointA Eguations of 2D Eguilibrium Newton s First law Remains at rest or move with a constant speed For a particle in equilibrium F 0 or Ein 2ij 0 A vector equation Or written in a scalar form 2Fx0 and EFy0 These are two scalar equations of equilibrium They can be used to solve for up to m unknowns For The example gtZFxTBcos30 TD 0 T2Fy TB sin 30quot 2452 kquot 0 Solving the second equation gives TB 490 k From the first equation we get TD 425 kquot Engr210 7 Spring 2005 Lesson 5 2D Equilibrium Instructor Ahmed AbdelRahim Page 2 of 2 Some Special Applications Springs Forces Spring Force spring constant deformation or F k S Frictionless Pulley Cuth ix in Ecllxlnll With a frictionless pulley T1 T2 Example A car with width 4 ft is towed at A constant speed by the 600 lb force The robe angle 9 is 25 Plan Step 1 FBD Step 2 apply 2D equilibrium equations 600 lb FBD at point A 25 30 FAB FAC Applying the scalar 2 D equilibrium equations at A we get ZFX FAC cos 30 FAB cos 25 0 ZFy FAC sin 30 FAB sin 25 600 0 Solving the above equations we get FAB 634 lb FAC 6641b Engr210 7 Spring 2005 Lesson 30 Theorems of Pappus and Guldinus Instructor Ahmed AbdelRahim Pagel of 2 Today s Obiective a Apply Pappus and Guldinus theorem to determine areas and Volumes of revolution AreasNolume of Revolution Areas of Revolution Area generated by revolving a plan curve about a fixed axis in the plane of the curve Volume of Revolution Volume generated by revolving a plan area about a fixed axis in the plane of the curve A9L V9A A Area of revolution V Volume of revolution r perpendicular distance from the axis of revolution to the centroid of the rotating curve or area 6 angle of revolution measured in radians O s e 5 2n Example 990 0 mm i i T20 mm 40 mm H K 20 mm Secuon A41 Example 996 i gls ti1 4111 wrliiliii EnnglO r Spnng 2005 Instructor Ahmed Ab deLRahxm Lesson 3o Theorems ofPappus and Guldmus Page 2 of 2 Example lt73m4i3ma Engr210 7 Spring 2005 Lesson lIntroduction Resolution and addition of a force vector Instructor Ahmed AbdelRahim Page 1 of 2 Today s Obiectives 1 2 Identify class objectives Identify course requirements and expectations Review general principals and definitions a Newton s Laws of Motion b Units of Measurements Define force vectors Determine resolution and addition of a force vector Class Obiectives h Analyze and Develop freebody diagrams for any system of forces in two and three dimensions 0 Analyze the equilibrium of rigid bodies under any system of forces a Analyze trusses beams frames and machines N Apply friction forces and analyze their different applications Calculate the moments of inertia for areas and masses Class requirements and expectations 1 4 midterm exam and onefinal exam a Exam1 15 b Exam2 15 C Exam1 15 Drop Lowest d Exam1 15 Grade e Final 20 35 f HWlCLass15 g Project 5 2 WebCT online quizzes 3 Class participation General Principals Statics Dynamic and Mechanics of Materials Particles vs Rigid Body No Deformation Concentrated loads at a point Newton s laws First law Motion and Force speed vs acceleration Second law F ma Third law Mutual forces Action and Reaction are gual opposite and collinear Mass vs Weight units Engr210 7 Spring 2005 Instructor Ahmed AbdelRahim Lesson lIntroduction Resolution and addition of a force vector Page 2 of 2 Unit of Measurements Force Vectors 1 International system of Units SI Vector What s a vector 2 US Customary units Magnitude and director point of Length m ft application Time s s What s a scalar then Mass kg slug Magnitude ONLY mass volume Force N lb length etc Could be ve or ve Newton Second Law N kg ms2 Multiplication of a vector by a scalar lb slug fts2 Change the magnitude of the vector Vector addition resultant vector Pre x for SI units RAB Nano 10399 micro106 mili 10393 Collinear vectors R AB Scalar giga 10399 mega106 kilo 1o3 Force Vectors Examples Graphical representation of vector addition Parallelogram Law Vector Subtraction AB AB B 1B Resolution of a Vector Components Kwon line of actions Adding more than two vectors Parallelogram law Sine Law Asin a Bsin b C sin c Engr210 7 Spring 2005 Instructor Ahmed AbdelRahim Lesson 7 3D Equilibrium Page 1 of 1 Today s Obiectives Example 352 1 Review Exam 1 Format and guidlines 2 Apply equations of equilibrium to solve a 3D problem For a particle in equilibrium in the 3D 2 F 0 or EFx EFy j EFz 5 0 A vector equation Or written in a scalar form 2Fx02 Fy0 and 2Fz0 These are three scalar equations of equilibrium They can be used to solve for up to three unknowns Find Tension in AB AC and AD Example 359 Find Maximum weight of the chandelier if the wires can sustain a maximum tension of 120 lb

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "I signed up to be an Elite Notetaker with 2 of my sorority sisters this semester. We just posted our notes weekly and were each making over $600 per month. I LOVE StudySoup!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.