Final Exam - ESM 2104 Final Exam Study Guide Professor...

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Unformatted text preview: ESM 2104 Final Exam Study Guide Professor Hendricks Prepared by Andrew Roberson ESM 2104 Final Exam Study Guide Force -Law of Sines: bsin0=AsinB=Csina -Law of Cosines 9 b2=A2+C2-2ACcosG A C a must be between A and at B C in -Force is a vector: Made up of a magnitude and direction -Use the parallelogram method to find the components of a vector Rsin0=Bsino<=AsinB x" f R! {513 Exam| x the magnitude of the resultant is 2000 lb. 2/21 Findn’ma 5011114 11- inldeofresul 2d>:3c,o-'2.9 (1): lac—e Lami at.th HO” Zm‘: 3002+ lfiwl—Z-m'lHOOCcSJJ "\e 4-106=Q‘i'iOStlfié'lOG‘ZZ‘f'lOCcas'Ii _ l . Cos¢=(é‘1+|€6"¥)lo%,2~11m" Cl): 128.682" l ESQSZ”: [80- 9 J, — .7 4'8” -Finding x-direction and y-direction components of a force vector -Right triangle method X=Lcose Y=Lsin6 Example: The unstretched length of the spring of modulus k=1.42 kN/m is 100 mm. When pin P is in the position 0 = 30°, determine the (a)x- and (b)y-components of the force which the spring exerts on the pin. X=Bosm3zf =- “i0 l#80 Got: 30": (31.282 5 SL2 @Ow‘ft (60%? $1=LD°i18f+Zoz ‘31: am: 54.23 8= Moon F: Hts—106): LL’J’ZB'tDCS) F: [HZULLOQ'J'rLOaJ E: F5056 F= mflnmfi 1;: 5.7.58 c.03701é 20! 1215.758» :F t ’7 Q37 E: Fame 13:51:13 sm7fiiézfi F_i1-: Vector Projection Projection of vector A onto B P=ABcosGB i3 Moments -The moment around a point is made up of a Force and Radius Arm -Radius arm must be perpendicular to Force Mc=FxR R Example: A force of 150 N is applied to the end of the wrench to tighten a flange bolt which holds the wheel to the axle. Determine the moment M produced by this force about the center 0 of the wheel for the position of the wrench shown. Consider counterclockwise to be positive and clockwise to be negative. HON 651] mm 5; \U E -’ F50 c9570” T” E5 51.303“ 4-13 :13 iii: 1303mm” “ x F; H0991" X: “450 @326 (j: 0150 srnZ’O” Xrgzz'géa El : [53. cEO‘DI Zflor‘Fg-g F F.1 (x-QZS) E M;"5!,3oa «53-01001 _ [‘10‘3’15‘1-(H223él- ($2.5) 2M0: *79‘15Fi‘73 4 507q4-71L1é5 EMU: 58661:).L158N: F552 r0: bu -Couples replace two or more equal forces with one Equivalent Force and a Moment All three pictures are equal -The sum of forces in the direction is equal. -the moment around the top of the beam is equal 1- 1m Remember to draw a 2"“| picture with the equivalent forces, and moments Example: For the thrust configuration shown, determine the equivalent force—couple system at point 0. Then replace this force—couple system by a single force and specify the point on the x- axis through which the line of action of this resultant passes. US I IS"F WE..- EM; -ZT + ZTwsi’O“—~ lSTleD" 2' ME azgasr 2nd [35M mtfi=o ZMH=E*X+E{'D= .i‘M-Tx —2.i;357‘: .i7H7-x XS'Zés’S/iw—Iriin/S Example: Determine the height h above the base B at which the resultant of the three forces acts. 'F 2901b ‘9‘ ‘1 J3 53- 25011: F50 E: 630 ~2‘10-2 to = 80 ZN; Ego-aswzo-za-Z +2010 ~18-E Z ME=7280-352 80+ Z‘fo—BO ZMgr'ZI—Iflo 2m! PQJ'N‘E/ Free-Body Diagram Problems -In order for an object to be in static equilibrium, the sums of all forces and moments will be zero. Fx=0 Fy=0 Mc=0 -lnteractions with surfaces 430.19, support There will be a force (n) on any object resting on a surface that is perpendicular to the surface Q % There is both a vertical and horizontal force at any pin connection 5. ley sliding guide At built in connections there is a moment, vertical and horizontal tiff Ea Example: A woodcutter wishes to cause the tree trunk to fall uphill, even though the trunk is leaning downhill. With the aid of the winch W, what tension T in the cable will be required? The 1000-lb trunk has a center of gravity at G. The felling notch at O is sufficiently large so that the resisting moment there is negligible. Horizontal Z M: O= wig gem ea—Tfl- laws—TX- lam-.9“ O: lOCD- 83% 6° +11% '8? - [Ssoné LTmflBO- l8cosé° T(.581- 0,02534' loan l3, gm (5":(3 l6.‘f"f‘/T: USE—83:7 T1|358.‘87/lla.‘1‘1‘-1 : 82 :33? Example: To test the deflection of the uniform 200-lb beam the 128-Ib boy exerts a pull of 30 lb on the rope rigged as shown. Compute the force supported by the pin at the hinge O. Tm P :N’fifioflfiT-E +w~(é+2.8~tr1)+3w.+ 6+T+T+LE+233 O = 2.8P+l7.8("1-8)+ 200(3) + 3D (63 + 30(83) r2513: GHH + QOO+18© +2 .94 ‘P= HEB-Li /-2,g= 6612.286 3D Forces FZ=Fsin<p FY= FcoscpsinB FX=Fcos<pcosG -3D forces can also be represented as a F=Xi+Yj+Zk Example: The cable BC carries a tension of 820 N. Write this tension as a force T acting on point B in terms of the unit vectors i, j, and k. The elbow at A forms a right angle. 3‘8 U 8 n .34; i: If}. ~.8Los"58°=.éé0tc. X=O—i7=-L7T Q? .7+33m383: IRIS T“ T/J flan? < fhfihi kt; LT”: 820/2.i81<‘ |-7E.+ I.l0|1gj+-.£a?l0 ‘ -5”. ._ HUN”. LEVER? .r -Vector Projections: A force can be projected onto a line; the projection of F onto AB is the dot product between the force vector and the unit vector of A to B FAB=FoAB 3D Moments -The moment around a point can also be represented as a vector. Each component is the moment around each axis. -If a force is parallel to an axis, no moment due to the force around the axis exists -Always split the Force into X-Y—Z components, and multiply by the proper moment arm - Don't Forget to Find Direction of Moment, Clockwise or Counterclockwise, the direction is found by looking down towards the origin of the axis (from positive to negative) Imagine you are placing your eyeball at the letter x, y, or Z and looking down the axis Example: The 5-lb force is applied at pointA of the crank assembly. Determine the moment of this force about point 0. Mdfli‘fl- 2 m3 isms 5°52}? $3.220; Mowrtfl+ 2 M0,, 5555/1 33b'.7752.0017 3D Couples and Wrenches BOON -Find the sum of the forces in the X, Y, 2 directions, to find equivalent forces Rx, Ry, R2 -Find the moments around the X, Y, Z axis to find equivalent moments Mx, My, M2 Remember to draw a 2"“ picture with the equivalent forces, and moments - Rx, RY, Rz Can be written as a vector that creates a single equivalent force Wrenches -Wrenches are an equivalent force with a moment around the force -To find the equivalent wrench (M) Find the R unit vector M=m*R . uo Sum the moments in the first picture and compare to the second picture ~ , . 21m: —Ho (w) =-Ho(¥) ? 2 dIanaJ 440K EMT: 4-10 ‘8) = no“) :2: n Solve for m using algebra and use m m P m— L—?fi)m‘]lm)\.,£) - to find M ~ my, a m: o “zoolpr Kim M filo- $=Uiox *fi‘fi, M O: L70X+EZO’ZOM/Lfiblf7 O: ‘ZOx-Ll'OM/bfL-IJ ’ZOX - 8516M : Lioxi 3102‘4‘1’WVI x (*20410) = 310 4 Ml-w-ntsfis) )(t*5-33 - .DD7M O r 20 S .33— .Dowm) ~ “lam/W7 Ovloem- ,|‘~1“lM-_%OISM l, M t - 1 Dad £7 M ? I 01' '63 Example: The resultant of the two forces and m t [(73 .1133 91 y _ . _ couple may be represented by a wrench. Determine — / U1 7 < 2'01} 010“ ) the vector expression for the moment M of the wrench and find the coordinates of the point P in the x-z plane through which the resultant force of the wrench passes. R: < lBOL + Hwy; fi:I/J70,3<\|30L+HOD EMJO T'Rg'E+n/i- E3C%7O-3 MT M'm'f’omfilmv OI‘IJO%+.7¢3M fl: €$§§<II~GWID~D “02:7 63m .2: ,OD7M EM; “1% JED-.3: M “‘9'?er Rx? ll =_C‘jém+ 00754431: IS‘i'SmfiH m=7|06 M = <6_HZ‘16+"1.SGIJ) 2: _OO'7-7 {0,5,3 {DE-1'93 3D Equilibrium 1 _ -Sum of forces in X,Y,Z directions must :Mt‘ [1021:34an ‘. 33 #123"): be zero : 10-. _ . ~_— .. -Sum of moments around X, Y, Z axis “'0 X 1 38 ED ' 53 WZ' 6" must be zero XGG'fLS' “fa-fibfllo : '. D 'i -Draw Free Body Diagram, and list Unknowns -Use moment axis that go through forces to eliminate forces that are used in calculation (see example) Example: The industrial door is a uniform rectangular panel weighing 910 lb and rolls along the fixed rail D on its hanger-mounted wheels A and B. The door is maintained in a vertical plane by the floor-mounted guide roller C, which bears against the bottom edge. For the position shown compute the horizontal side thrust on each of the wheels A and B, which must be accounted for in the design of the brackets. Detail of door hanger Z Vigor-tiw+ Marx l32CE"!"ll0 0.527.519 EMAE=D= Cy(7“i)ll+iS-I113,4 5635-3422757]; BHFGBLH EMBJOECgfllfli-fliz — 1341A): fang-27.964140 Ashlin? Example: A rectangular sign over a store has a mass of 148 kg, with the center of mass in the center of the rectangle. The support against the wall at point C may be treated as a ball-and- socket joint. At corner D support is provided in the y-direction only. Calculate (a) the tensions T1 and (b)T2 in the supporting wires, (c) the total force supported at C, and (d) the lateral force R supported at D. i T Isl—z. R “Cy GD 1.; :0 ll I s l Wfi-‘bfit-Lsyszh) =1L ('Z.SL+I.SJ‘+3,2K) Z ‘9'. 31a JLJ' XML; or Law/a —TI {391 41125 -T.,.- I “in Tt-BWZ-Efléz 41.3sz = Haas-3% Wme -'3>1+2.5.3-“2 H.szq*%£%+%%fi=2%ue 2 .0: i2 T| +2334 T1: 2466-48 XML; :Z‘STZS _3HT':‘| 25' I'STV-lfszq: 3.“1' I'STflrifilo-f 8643:1041? T1: lion, TL=LZOIMZSI3Q= ZQILTfiZHZfi' [.20IT.r2‘-165.68 —— _ 5.52L1T,=2%5é8 7]: 9233a 2 Plait): K'I“Tzfl‘ "P-rilfl'l Riff "5/1.qu ‘Tz'|-%-alq 1? = ims- I’m: = “#612 E ETD: Cw-Tlerzx CK=TI . 3.91A1_q0|_1 Cf 2013.53+ 293.S3=587.oc; 21EE:O:TZ.EETIf+ Czfl LU. ‘ C%=375.723~27é.167'MSDH Cf magnum C1=C§+CE C: #172. Trusses -Truss has to be made up entirely of two force members -Tension forces pull out away from the member (positive sign) -Compression forces push in towards the center of the member (negative sign) -Use sectioning method to find unknown forces (see Example) Cut through the members that you need to find the force of Use moments to eliminate extraneous forces (if a force goes through a point, the moment around that point due to the force is zero) Example: Determine the forces in members (a)CH and (b)CF. (Note: compression forces must be negative.) 21n—-‘<—2 111T?! 113 2111 56.421113!) 1 r T A" n D '2; E CF, 2. A B T E ' u ' t c a: 23 GE I T: g F 231;}; £17113 2 l’lgofCir‘x-l — era-1+2}: 23m 0: -%SC¥: *i/IgC-FE at ‘1'); L1 fl-L ‘— CI warrikrk] 559% n2 %1“- H _ a Griz” a a “’13 1 :04 (gay in Machines E Fig: D :- —ZE -2LF23-C'H L3 -Machines are made up of both truss 3. -' CH {355% 5—3 =— £3 members and non-truss members 13 -_-_ _. -Draw free body diagram of each part of the CHCDEH‘S machine (split the pieces at the joints) CH: H -All internal forces must be equal and opposite, if a force goes to left on one FBD, on the other FBD with the same point the force will go to the right (see example) Example: A lifting device for transporting 135-kg steel drums is shown. Calculate the magnitude of the force exerted on the drum at E and F. EB 1‘10 CM GB,GH@ “an ,2"? mo 3?le ZE=U=GfliJQBX 3‘” 6A=Gg 12m SEED: 13+ GBHJHSHE 136 -‘l.8‘-’ -Zc:1 A girl 3%.05 290 290 mm mm \ C A = News 1 mEXEfiCQ an _ __ c- Ego-Gigi; llltl‘i U, "— E3: ‘HngSSmSUl.OSH=*QGlH£ EE1 EMgo:Gill,-log—.lZEgEflrza—zs) “El—“430:3 3%.00'..r_uc_—,1ZEK+.D‘1 Efl : o leg=nlza Egduz7sz E¥E3+E§'E4573l%3 Centroids -The centroid is the center of the area of an object. X=X dAdA Centroid X=X dmdm Center of mass -Remember to use the right bounds, and proper use of dx and dy for the double integral Example: Determine the x-coordinate of the mass center of the tapered steel rod of length L where the diameter at the large end is 6.30 and the diameter at the small end is 1.30. Din. = 6.3D 3.lSfl=.é5‘D+-L‘k 2.50 = KL Jazz—f” '2. T’flx- «:5 +145 0'“- L J 1 a { g 1..- ‘XS y 'f-flzgfllx +3150 “Kl-l 612.512??? 0L3: : IE! —— 1 K: C __ __ __ _ L.- L.— \ - 7— L. “[035” L3: “3 dx ] H113 01+ may $2331;de a (— L?- _. L .2111: :42 L “353 m 15:25 wi ~>Z_ "'-' if t _ 1|i15L1+LDBZL1HEEZSE .LTZZEX +|.Q15x1+2 oggugl" f’TZLELl- LEZEL +2,DEF5L. L, ' L} U 24. "I 2.657 E/f—Mm Lrt.C"lH;l_ Example: Calculate the coordinates of the mass center of the metal die casting shown ((a)x- coordinate; (b)z—coordinate). Dilnensic-na In 111 illilDIE-‘T em Distributed Loads -Find the equivalent force by integrating W between the boundaries. -Use centroid equation to find where equivalent force is applied. Example: A cantilever beam supports the variable load shown. Calculate (a) the magnitude of the supporting force RA and (b) the magnitude of the moment MA at A. i 'J E-i' = it?” + ELI." .. I ‘5‘ I l 1 26Ib9'fl $0.5M: ’ : J‘ H d): a? 3 A FEW-3.5 +.Ol?..|x1 618' “T2795 {)3 + *2? .913“ :w 4- ‘2 _. w 0 k): 1 ‘ii 1r—:QOZS.L’I_Z lZéTfi—h - L I 31% I Shear SH ~ KEOIZIl F Forces :rflho:Kq-F w3705+~0l21x1 F‘ lZIq: F; (2034-2 6 a ?..3 _— F XII Kudx 0 quto: Mq“ Mgcomz (mad 3 7H2) MA: ‘8320101 -Shear forces are internal forces within a loaded structure such as a beam. -There will be a shear (V) and moment (M) in every beam that is cut w whim! When W = The equation for the distributed force .1. w wax l a 5;: V=-de M=de ‘7 Don't forget to solve for constants after N integrating with initial conditions, V(0)=R, M<°>=° V Ax V+AV Example: What are the values of the (a) shear and (b) moment at x = 6.1 m? (c) Determine the maximum bending moment Mmax. rad m N/m A . B L2m+lx—3m4>L2m 21nd 1500 N Z MEEOHH (241+Z+3)+ “:00 "2+ F~LUFZTLSB U: A’ 830(Y‘ZX QA: 3100+ 53'2'4‘710 V: Ignza'aww mm A: U: 22*930x ZZZ—£5“ F: 850'3‘2‘1'510 Mthdx 1‘9 lav“ M13537. 22 x — Luz-mac. if t 0 : A + U _ F {Ma-‘Orwmm ‘Z'l-S)‘C°"A Liana =6.| (Ks—7.2a5-‘frslé-Iffi, "j 22-2q90(2_fi) C!:"ll57,8-5 l877.ZZ-Z“?”IO-V‘O -* L u.-.78 MW: 35 31120-12 £2341 IS (“I 26232—157. as x, M=3337.22x_=115x1— I15785 0: 3537.2? 830x 830x=363122 I -W*ogr'r-9b th'ZdZ—bb’lw Friction -Friction is equal to the coefficient of friction (static or kinetic) multiplied by the normal force of the surface on the object F=u*n Example: The P: 110 -lb wheel rolls on its hub up the circular incline under the action of the p=26 -|b weight attached to a cord around the rim. (a) Determine the angle 6 at which the wheel comes to rest, assuming that friction is sufficient to prevent slippage. (b) What is the minimum coefficient of friction which will permit this position to be reached with no slipping? . . g “ - g. ER=OZILHOSMB awe {flame lsasmnmmss mam-33%;; @= 30-65” 2611: 2‘ 153:0: {3— HO maQ-chme ml 3W3 30.9511“? WP: ME“ Friction on Pulleys -As a rope passes around a cylinder there is friction that causes two different tensions in the rope. -The relationship between the larger T and T1 Tl'1=eb*u b = the angle in radians of contact between the rope and pulley Example: A counterclockwise moment M = 1600 Ib-in. is applied to the flywheel. If the coefficient of friction 11 between the band and the wheel is 0.23, compute the minimum force P necessary to prevent the wheel from rotating. T,T.feo@ 1‘19 0' if" mpgM1 arm TI: T, fi-zsrr i Ti<.jxlT, TflogT J EMM:M+CIT—01‘rl:o | LIEOOZTM-GI-ZBQ FWD; EMU: O: ~ST+I3T, aw ZGP= BLDG-1L1? LrS-mn P=Ha53 Moments of Inertia -Similar to centroid but deal with mass (area) as well |x=y2da |y=x2da -Po|ar moment of inertia is the sum of IX and |y Example: Determine the moments of inertia of the rectangular area about the x- and y-axes and find the polar moment of inertia about point 0. h I'll—>— Parallel Axis -When finding the moment of inertia around a point that is not the center of mass of an object (lp). the parallel axis formula is used |p=|c+Ad2 -The original moment of inertia (lc), Area of the object (A), and distance between the center of mass and new point (d) ...
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Final Exam - ESM 2104 Final Exam Study Guide Professor...

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