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Exam 3 practice

Exam 3 practice - SOME EXAM 3 PRACTICE PROBLEMS A Mont i9...

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Unformatted text preview: SOME EXAM 3 PRACTICE PROBLEMS A ,. Mont i9- ce 24so snmcs smpv suooiisnous roe eiouoav fig EXAM? a Go back over all class notes and reworl< //7 C/ass Prob/ems: Eémcdxres/ Centre/31’s? “mat i; a Review all homeworks (throudh fig? and C'oiuizzes. Be able to work all problems ”on uour own” H COEEECTLY l l a Heed all su estions made in class lectures ll 9 BE ABLE l0. * Eecodnize Zel’orce (2’?) and Multi'Force (MP) members * Draw proper l’fiv's 010 truss parts and Frame/I machine parts 5* show proper joint WU? ' 3* show proper section Ffip's * show proper Ml‘ member F59? with internal forces atjoints in equilibrium - * Write out proper equilibrium equations that match the F139 s 3*" 5olve For internal. forces in a truss usinoi bow method otjoinrts and method of sections * Eecodnize bu observation what members are zero’torce members 3* 5olve For required internal Forces in frame/ machine members usind the most efficient pathwau _ 3““ Locate the centroid/ center of dravitu usind the ”shale inteoiration techniqueH and the ”composite area/ volume method” 2* Calculate surFace area and volume of solids usinoi Pappus’Quldinus Theorem ‘3“ Chande Distributed Loads (UL) to Concentrated Loads (Ct) l EV“ Solve dam problems iii-3:10 lnllLL L‘E’f \[019 WM li‘l CLAS§ CE 2450: STATICS EXAM 3 TOPICS Chapter 6: Structural Analysis 1) Trusses: A) Characteristics: a) all external loads are at joints b) all members are straight, 2~Force (2F) members c) all joints are treated as “pin” connections (no moment) d) all forces in members are axial, i.e. either tension or compression B) Analysis Methods: a) method of joints for SIMPLE TRUSS ONLY : # members 2 # joints-3 (use particle statics—concurrent force system at every joint—therefore, 2 force equilibrium eqns./ joint) i) FBD of each joint, with unknown forces shown as arrows in tension ii) must start at joint with only 2 unknowns b) method of sections (use rigid body statics—nonconcurrent force system—w 2 ) Frames: therefore, 3 equileqns. for each complete section) i) cut through 3 members only to separate the truss into 2 complete parts (a left and right section); ii) one of the cut members must be the member in question; iii) choose the section with least number of applied and reaction forces (if possible, choose a section that doesn’t have support reactions, in order to eliminate having to find reactions first, but this is not always possible); iv) FBD of chosen section with internal force in cut members shown as arrows in tension and all applied forces drawn in their respective places; v) use one equileqn. to find the one unknown force in recluirecl member; vi) if more than one unknown is required in the section, use eqns. independent of each other to avoid compounding errors if they occur. A) Characteristics: _ a) external forces can be anywhere on the members b) members can be bent 0) members can be “continuous” and therefore can be Z—Fome or Multi-Force (MP) (1) “rigid” frame usually has 3 reactions at supports 6) “non—rigi ” frame usually has 4 reactions at supports B) Analysis Method: a) rigid frame: i) can solve for all reactions first (if necessary); ii) take members apart; check for statical determinacy (#eqns=#unks); iii) solve for internal forces by using FBD of MP members and 3equil.eqns./MF member; iv) number of internal forces at joints depends on the types of members connected there: MF to NIP or M? to 2F v) applied forces at joints must be placed on MF member only; if joint is MP to MP, then the applied load may be placed on either one of the MF members, but NOT both. b) nonrigid frame: i) solve for as many reactions as can be found with whole frame together; ii) FBD of MP members and 3 equileqns/MF member; check for statical determinacy; iii) solve for required forces. 3) Machines: A) Characteristics: a) all are “nonrigid” b) made up of MF and 2F members 0) applied forces are considered the “input” force, reactions forces are considered the “output” force; B) Analysis method: a) take members apart; b) FBI) of MP members, 3 equil.eqns.MF member similar to frame analysis method; C) can use the object the machineftool is acting on to simplify the analysis of forces on the machine parts. Chapter 5: Centroids and Uses 1) Locating Centroids: A) Single integration technique: 9.) lines: is i) define element properties: dL, 76:1 9; ii) Given Fflx), L 2 I dL 2 I x] (1- (3;); dx 1 5:?) d 9‘ 9 ’ a! - =ljelV(1—gx’)1dx%wa ‘ [yflLl (gndx h 3&2“ (9)019 ‘—“ .. L wearer—re) fag? 9) “a Yr ‘0) areas: M i) define element strip (vert. or horiz. rectangular strip) L ii) define element properties: dA, Xe} , Ye: iii) use eqns: A = I dA “ : Enid A 37 “7 12min A B) Composite LinelAreafVolume Method: i) equations: LZZLE A3213: V=2Vi Kazan tins: i=__"aa _ L A V wants Fill-LA: Vegan L A v 2:221Vi V ii) break up the composite into recognizable parts, keeping the parts oriented to the original reference axes of the whole; iii) find the individual centroids of each part in relation to the reference axes of the Whole, as well as the individual L’s, A's or V’s; iv) plug into eqns. and solve. 2. Uses of Centroids: A) Surface areas and volumes of revolution by Pappus—Guldinus Theorems: i) S.A. = 2 63 L (composite method) units2 Where 9 = revolutions in radians a = distance centroid of line is to generating axis L 2 length of generating line ii) V = 2 ed A (composite method) units 3 where A = generating area B) Distributed loads (D.L) to concentrated loads (C.L): i) magnitude of C.L. —~> “area” of D.L. shape ii) location of CL. -—> centroid of D.L. shape C) Hydrostatics forces on straight and curved surfaces: i) estimated D.L. (straight surfaces) -—> w = byw h where b :- “breadth” of submerged body (dimension not seen in 2D drawing) 7“. = 62.4 pcf (lb/ft 3) or 9810 Win 3 h = depth below the free surfaCe; ii) D.L. shapes (straight surfaces) —+ usually triangular, if object n free surface, trapezoidal, if object begins below free surface; iii) for curved surface, treat water as a “block” (area of shape) with W == Vyw , and located at the centroid of that shape; include as D.L. water acting on the side of the “block” (then change it to CL); iv) types of problems: resultant problems or equil. problems CE2450-3 STATICS QUIZ 4 NAM: 1.. (10 pts) Describe each member as either 'rnulti-force (MF) or two-force (Z-F): 7:- J AE BK DI FH 2. {19 pts) List the 5 zere ~force members in the following truss: STATE 1F P 0*? é: 4001b 4001b 4001b Extra 4 points! I ! E Of the 2 truss analysis methods taught in class, name the method that uses the principle of: a) rigid body statics b) particle statics CE 2450—2 TRUSS QUIZ (15 minutes, max) NAME: ”1". . 'VJ re" 3 5 ii i i p-IH i i . g ! i . {‘5 ii {3 Iii {i {-1 {i ii; For the truss shown, a) determine the primary (P) and secondary (S) zero—force members if there are any; P: S' — ‘ .—J b) the length of member DG. DG = it 0) Draw the proper section FBD to find the force in member DE. FBD : (1) Write the one equation needed to find this one internal force (in member DE). EQUIL EQN: e) Knowing the reaction at support L is Ly = +0.7 51;, describe (show pertinent FBDs) how you would obtain the force in member GI using method of joints and only ; equilibrium eguation calculations (write what they are). Ifyou do not believe this is possible= explain why. CE 2450—3 TRUSS QUIZ (15 minutes, max) NM: For the truss shown, a) determine the primaiy (P) and secondary (S) zero-force members if there are any; P: , S' ‘ ____._._......,__? b) the length of member BH. BR 3 m (3) Draw the proper section FBD to find the force in member AF. (Consider: do you need to solve for reactions?) FBD: d) Write the one equation needed to find this one internal force (in member AF). EQUIL EQN: e) Describe the process you wouid use to obtain the force in member BH using method of joints and only 1 eguilibrium equation calculation. Show joint FBDs and what the one mm is. Ifyou do not believe this is possible, explain why. CE 2450 STATICS PRACTICE STRATEGIES FOR ANALYSIS: TRUSSES In the following truss, describe your step-by—step strategy for determining the internal forces and types that act in all the membexs. Is this a, simpie truss? Why or why not? 5.06 ft 4.14 ft CE 2459 STATICS PRACTICE STRATEGIES FOR ANALYSIS: FRAMES In the foliowing flame, describe your step—by-step strategy for determining the reactions at the supports. 13 this a rigid flame? Why or why not? 4.5 in. 3 in. CE 2450 STATICS PRACTICE STRATEGIES FOR ANALYSIS: MACIfiNES In the following DRUM LIFTER, describe your step-by—step strategy for determining the forces acting at F and H on member DFH; The weight ofthe drum and its contents is llO-Ib. CE 2450 STATICS PRACTICE STRATEGIES FOR ANALYSIS: TRUSSES What are the angles in the cross members of this truss? Do you have enough information to determine them? -~<———*———— 8 panels @ 7 Ft 2 56 f mm”, 5% A CE 2658 STATICS Determine the axial forces in members CF and DF using method of sections onlv. State ifT GFC- Were” can" we ;;«t:z‘~* "enemy” r.v_..___.u.n_;'_:"r- 2mg first; get em 13% it y...“ n ,1 i .» i —r ‘ Liz-re if i ‘ r 1 Dita/“t 'Tfii‘we’i/ii i Mum “we“, _ 600 mm m» w mum. “We. g 300mm [300 mm 300 mm‘ Why didinearly- 60% ofstatics studentsget—tkiswmng'on an exam? *Got stuck finding angles Lost a lot of time and therefore didn’t have anything written down to get points. M“Assumed all the cross-bracing were at the same angle. ** *Couldn’t find the lengths ofthe verticai members. *****Chose moment points that obliterated the very unknowns required to find. *Test strategy: It‘you find yourself stuck on some intermediate ‘hnknown”, substitute a variable in its place and finish writing out the equiiibrium equations. When you have time, continue trying to obtain the necessary info to solve the equations. Even if you run out of time, you wilt still get some partial credit, rather than a zero! CE 2450 STATICS QUIZ (15 min.) STRATEGIES FOR ANALYSIS: TRUSSES A roof truss is loaded as shown. Determine the force in members 1K, GI, and FE using method of sections only. Can you use one section for ail of them? What alternate strategy can you suggest? . e i 12ft 12ft 12ft 12ft 12ft 12ft flyélésu WW3 i"- * ‘3' “7" 2" (-a‘ '\ 6%}: 9‘) in Cat) 1%: W: {if 353'»; it’s-<- (if) CE 2450-1 STATICS QUIZ 3 7/12/04 NM: 20 pts. 15 min Determine than centroid of the shaded area with respect to the given 3 and y axes. ' (12.4734é2fl Determine the forces on member ABC of the frame. Determine the reactions on member AB at A and B. CE 2458 STATICS EXABVI 3 FALL 2006 N0 QUESTIONS WILL BE ALLOWED once the exam begins. Ifyou need to make an assumption, write it down clearly in the blue book. Show all work in Blue Book for full credit. Show proper Free Body Diagrams (FBD’s) "and necessary equations for full credit. 1. (25 points) a) Determine by observation W3: A) {WWW- 2 j the zero force members in the truss shown. 1” fl «.7 _, State whether they are either primary (P), b) it}? a: 4"; FEW (T) or secondary (S)- , . (9 Bi): 42% 6:) Given: Gy = SkN b) B}: method of sections only, determine the force in member EF. State whether tension (T) or compression(C). Fg- ' ' er / . V“ 1"" . L126 [11 1.26 m—v1+-1.26 m—J c) By method of ioints only, determine the force in member Bl). State whether T or C. Z. (25 points) Arm BCD is connected by pins to ' crank AB at B and to a collar at C. a) Draw FBI) of each member. State what type of member (multi—force (MF) or twoeforce (2~F)). e5 :3; we: EDD we flirt: b) Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when 3 m 0. - ‘7 ‘ Mr ‘flifldt‘wfl 3. (25 points) a) By direct integration only 9 Without-use of the integral functions on yourkealcnlator, prove that the y- eentroidal coordinate of the area . underneath the parabola, which has an area A of 40 —in2, is 3?, = 3.6 in . b) For the entire composite shaded area shownin the figure, use the composite , method to determine the y—centroida! coordinate i7. ‘37:; j, 5%{39’9 STATiCS GRADING RUBRlC -_ __—___ _ all forceslmoments with arrowheads - all forceslmoments with proper labels for unknowns _ all known magnitudes all climensionsl’anglesislopeslcoordinates reference axes (3D especially, 213 if not conventional) Recognizing Pertinent examples: Conce nts “at” , shrew Pulley _J 4 1/ --:r .6” M ~r~ fit! 2-F members at a support _ “x _m g ' . U . .- _ ‘ . couple shown as 2 parallei equal and opposite forces ‘ Hireotion and shape of Water forces as distrilgued-loaaf "" Ww—“fivw ”W" ‘7 ‘ 'l‘u“ immr esrandomagnitude ofrnormal'foroes identify type of problem: resultant or equilibrium, 2D or 3D ——-- _ coteoorree toroo system: corrcrrrerrt or rrorr-errt ——-- ecrorrorrre number one tree otororreetc oootrorroo _--- ne ermine OW many un “OWNS ere are an- OW many are required to be found proper equations written out proper vector notations used when not scalar orooororooeoocetorroterrowo —--- labels used in equations consistent with labels in FBD _ Arrowctte) _——-_; t moorrtoeeo. errertrorreeoe oorto —--- must use another valid method or independent equation(s) to prov- initial answer correct Legibility _—-—— rooroor rrorror presentation _--- root.oroocooeor,rrooreooroorrrooooo ---- clear legible writingliettering —-_- ...
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