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ws 22 - MEEN 221 Summer 2009 Worksheet 1 Dr.A Palagolo(1...

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Unformatted text preview: MEEN 221 Summer 2009 - Worksheet 1] Dr.A. Palagolo (1) HWSII Ch 8 (6, 15, 36, 40, 55) (2) Attendance Mandatory for 10:00 am — 12:30pm, unless excused by instructor. Multiple Quizzes may be given anytime during this period. (3) Today’s Material: Internal Forces , and Shear and Bending Moment Diagrams, 8.1 — 8.5 Sections EXAM II WILL BE HELD ON View row a a a 43 PM INTERNAL ACTIONS IN STRUCTURAL MEMBERS Machines and structures often fail because the stress levels at certain locations exceed allowable values for the material of the parts. Stress is the force per area that acts locally within any machine or structural member . For example SAE 1040 may yield with a stress as low as 40,000 psi while SAE 4340 may not yield with a stress as high as 230,000 psi. Guess which one costs more ??? This discussion emphasizes the need for an engineer to be able to predict the stress level within any structural member of a machine or structure . A necessary step in this process is determining the forces/and or moments at all locations where stress is to be calculated. The stresses are quite often determined directly from the internal actions (forcesI moments and torques) , which are appropriately called “ stress resultants “. For example axial , bending , shear and torsional stresses in a beam or rod may be calculated from : 0 axial stress = axial force / area 0 bending stress = bending moment * distance from the neutral axis / area moment of inertia o shear stress = shear force / effective shear area 0 torsional stress = torque / effective torsional area The procedure for determining the internal actions is summarized by the following steps from Riley section 8.1 . These steps first apply the Static Equilibrium Conditions to determine “ external “ forces ,moments and torques on the part of the machine or structure (MOS) where internal actions are desired. These “externa ” terms may actually include reactions between the part and ground or between the part and other parts of the MOS . The second step applies the Static Equilibrium Conditions to the part treating the internal actions that are present on a “cut “ through the part as unknowns . The internal actions are thus determined and the stresses are then solved for ( the stress step is not a part of this course but will be a critical step in your MOS design courses) (1) Decide on the path of least work (work smarter not harder ! ) to obtain the internal actions of interest . (2) Identify and determine the “ground “ reactions required to execute your strategy in (1) . (3) Identify and determine the reactions between parts in the system required to execute your strategy in (1) . (4) Cut the part with a plane at the location where internal actions are desired , show the actions on this plane as unknowns , and use Static Equilibrium Conditions to solve for them . ENGINEERING MECHANICS - STATICS, 2nd. Ed. ' W. F. RILEYA ND L. D. STURGES 7'“ ‘8-5* For the steel shaft shown in Fig. P8—5, ‘ ~ (a! Determine the toques transmitted by transverSe eross'sections at points 'TA§ B, and C of the shaft. '(b) Draw a torque diagram for the shaft.' _uomm a ,2 ORSon SIGN 9 Pos\T\\r€ de cvuvegnrr105J ffi3<MJvr'TH~vmi5 @0 EfiGINEERING MECHANICS —'.STATICS‘, 2nd. Ed. w. F. RILEY AND L. 1).. STURGES _ 8‘17 'Determine the internal resisting I forges and moment transmitted by section aa in the curved bar shown in Fig. P8-17. P: Axid Load. Shawn PoSl-R-Ive Ow‘i'wwok q) Tq,‘ 31°“ (+3 ) CoMpv‘ esfianC") ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. E. RILEY AND L. D. STURGES 8-27* A'threerbar frame is loaded and supported as shown in Fig. P8-27. Determine the internal resisting forces and moment transmitted by - (a) Section aa in bar BEF. (b) Section bb_in bar ABCD. 7T" H ' '1: I '52:: 33.1 ‘5 u. v H SHEAR AND BENDING MOMENT DIA GRAMS FOR BEAMS As shown in section 8.5 of Riley , static equilibrium of the differential length beam shown below requires that dV/dx = w(x) which implies V(x) = [w(xyzx (I7 and dM/dx=V(x) which implies M(x)=jV(x)dx C2} where w(x) 2 load per unit length applied to the beam at position x M (x) _=bending moment existing on the beam cross section at position x V(x)= shear force existing on the beam cross section at position x w dx . V+dV L— dx ——>l Figure 8-16 Free-body diagram for a differential length of beam supporting a distributed load. The above relationships provide a means to plot the shear and bending moment along the entire length of the beam in contrast to the Worksheet 11!" “approach that only yielded M ,V and P at specific locations (cuts) . 8-42* A beam is loaded and supported as shown in Fig. P8—42. Using the coordinate axes shown, . . _ 1' write equations tor 25 RN the shear V and bending moment M for a section of the beam in the interval 3 m < x < 8 m. WC>0= cXUW‘xkd'eoL or Qanaean‘OQ‘Ed i, flood» Pew we: Slang-h § “[60: W) ‘m-‘cerwol d‘k 2.3% Fe pg? i SVOO: '\t~+erv\o.°. bendxg momma!- X (“a—ML: gzavmaxcv ma=ml+£aw>dx I ' (10.52. R w w W W: C°M3rawk> x. V} X“ X x\ xz X V , v V , V“ /4 "‘ x V/ X\ Xa x‘ x2 x m m7/Mz m Ma " /% X M‘ [II x\ ><a X. xaL NO‘VE‘VKBA' in EGG-er Cafe 1 Or &: Vac—W 4- Osreex umciew WCX) CMV‘VQ Mas; .‘_ are“ “AAEV‘ QAY‘VQ SHEAR AND BENDING MOMENT DIAGRAMS - SIGN CONVENTIONS FOR THE SHEAR AND BENDING MOMENT DIAGRAMS FOLLOW RILEY’S FIG. 814- . \ MIN/on?) V? Mth 441—8 ht,» OJI‘h" ‘- fonee ir 'ap ‘011 44%" Acme 4‘: CW 0'} oflaam A 1—49“ o f, _ M been a I A. W l ,. ' I I a Figure 8-14 Sign conventions for shear and moment. 0 THE MAXIMUM OR MINIMUM MOMENT OCCURS FROM CALCULUS WHERE dM/dx=0 THEREFORE M IS A MAXIIVIUM OR MINE/[UM AT THE LOCATION WHERE V = 0 . o A SMART LOCATION TO LOCATE A SPLICE ON A BEAM IS WHERE M = 0 , SINCE BENDING STRESS IS PROPORTIONAL TO .‘ THE BENDING MOMENT . 0 THE TABLE ON THE NEXT PAGE PROVIDES A REFERENCE FOR OBTAINING THE CHANGES (A ) IN THE BENDING MOMENT OR SHEAR FORCE AS THE INTEGRALS OF THE SHEAR OR w(x) , RESPECTIVELY . “ ORDER“ IN THIS TABLE REFERS TO THE ORDER OF THE QUANTITY h(x) OBTAINED FROM INTEGRATION . CASE “0 “ IS NOT SHOWN BUT CORRESPONDS TO g(x) = 0' AND THEREFORE A = 0 . . E ear of centre*zo\ orce or own v, Din om .v v Wm“, H $:’I<—I———»IW Farx—J: I"(,E:AV:'I _ = m [ +\F- :0 V Au- M‘F; (3‘: AL—T' I V‘- A+ F : flECTA/IGLE: rgin—44’ fi’eh' : VCxJ flew In: MCX) X, xz ' I . ‘3 Dr. A. Pa/aaao/o 6A3 8-53 Draw complete shear '* . “mm , and moment diagrams A zsoon-lb_ . for the beam shown a . ..f’: ,.¢.»:..-,,a; I. I — I. >' I . 1n F1g. P8 53._ V a? . 3n 1 3“ Fig. P8-53 SOLUTION Fr0m_alfree-body diagram for the complete beam: + T 2F = V7 - 600 = 0 y 0 V0 = 600 lb = 600 lb T -+ C 2M0 = — Mo + 2500 - 600(6) =‘o M0 = -1100 ft-lb = 1100 ft-lb C . Load; shear, and moment diagrams for the beam are_shown belOwE 8-58 (Draw complete shear and moment diagrams {of the beam shown in Fig. P8458; SOLUTION Erom a free—body diagram for the complete beam: + c EMA = -4(10)(2) - 16(4) + 3(7) — 19(9) B = 45 kN + c 2MB = rA(7) + 4(10)(5) + 16(3) — 19(2) A = 30 kN *Efififififififigfifl x 1 Fig. P8-58 ‘JGKN )qu u C: II 0 Load, shear, and moment diagrams for the beam are shown below: 3-63 Draw complete shear ‘ Q4), meo ’ £64Cf?orw9_ ., 59 2% : -.20Qo,+ 3009c) g- 4? (Jaw-4:] — 662ch + zoos + 9 8 vac/oases $5: 3/95“) A!“ "'1‘ ZF= —/o<>M-,4 —4€“°‘) '30“ 7 ' +fl,/oeo . so and moment diagrams, for the beam shown LUSVfiQS' 95 '37 Ith‘t‘er/goflmb" g): ‘75. $ "E9751 4*.875: 3.50 a? wo+e 1/:0 cué' .ro 114,; a. “MAMA at a? a? ...
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