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class 13 notes - " PROBLEM 4.14 For the given...

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Unformatted text preview: " PROBLEM 4.14 For the given loading of the beam AB, determine the range of values of the mass of the crate for which the system will be in equilibrium, knowing that the maximum allowable value of the reactions at each support is 2.5 kN and that the reaction at E must be directed downward. 1,2 RV ' I15 [iN D _ _ C ‘1 _ . . . ' 75‘? V'” ' , F12 111‘108 leB “1472.2 m SOLUTION Free-Body Diagram: Note that W = mg is the weight of the crate in the flee-body diagram, and that 0 S Ey S 2.5 kN i. 2F; = O: Ax = O +‘) MA = 0: —(1.2 m)(1.2 kN) — (2.0 m)(1.6 kN) — (3.8 m)Ey + (6 m)W = 0 or 6W = 4.64 kN + 3.8Ey (1) +12F=0: Ay—1.2kN—1.6kN—Ey+W=O or Ay =2.8kN+Ey—W (2) Considering the smallest possible value of E y: For Ey = O, W = Wm = 0.77333 kN . From (2) the corresponding value of Ay is: Ay = 2.02667 kN S 2.5 kN, which satisfies the constraint on Ay. For the largest allowable value of E y: Ey =2.5kN, W=Wm =2.3567kN PROBLEM 4.14 CONTINUED From (2) the corresoonding value of Ay is: Ay = 2.9433 kN 2 2.5 kN which violates the constraint on Ay. Thus (A y)m = 2.5 kN. Solving (1) and (2) for Wwith (Ay) = 2.5 kN, HEX W = Wmax = 1.59091 kN Therefore: 773.33 N S W S 1590.91 N, or 773.33 N S m(9.81 m/sz) S 1590.91 N, and 78.8 kg s m s 162.2 kg 4 4.60 0E7E/ZM/Nz: All/£72m (a) Begum-r I: Cowsmflme‘r), (5) 25m mm fizz; DFmM/N/W'fi, (c) /F pOSS/EZE, F/No Rfdcr/o/‘U. P: Ava/b, B 131;— l C 2 3 A P P P 2R 2R \‘\5\ 6 7 8 \ IF I L L I? I, 77%;? NM-CWfl/zm NM‘MILLEL Raver/mg (a) 80.001971 CON/REV; CMWm/W @ (A) EEflcr/a/viz 0575mm» 7; (c) fool: @210»? M/H/V 77}//va=p —‘*' : A / 5' D : I / “"' fly fly [OD/b fl 4202 54 53, g 47$ 2. Foo/2 Cawueuwr Benn 770M; (Ty/2w“ fl) (a)82m71/MP&2Pa2 awn/24m; C (6) la»: 71mg: //YD£~’7P/ZM/~#7Zv‘ 31—,- Z. (c) N0 EQUIL/eielum (FMA 9T0) as Woo/b _.__—._____..___r__.____ 3. 7740 IZEKCT/a/yj (a) 32mm- Pfizer/m GOA/Sfflfll/Y7 (é) @flcnw: 067R/vfl1v/m? 3 may, 9 (c) EGau/L/ez/uln MAI/ymllvk’p (5) Pine 770m”: fiifé‘FM/AW/‘E 3,. (c) flaw/LIKED!” Mfl/NTA/l/aw {3:50 In; §;33;3u,bxg,9)‘ g:gg,)/b—r _ — mama—1mm; NA‘V—‘Pflrmr 255707 (A) gem/<5 7: WPZéW coms‘r/zA/n7 (b) EEficWaM‘.’ /ND£7£/2M/N4-7£ (c) fat/1415210»? mfl/flfinwzp (SN/C'-O) 55: sic/bf 5y TEA! (5. Fou/z Mw~mvcurzm=~z ISM-Pm»);ch Qaficmwf (03 BEN/<57; cola/>427: WM”; (5) Beyer/mg: M/DéffflM/h'fl T; _, loo/b (Kc) EQUIL/EFZ/UM Max/ym/m—‘p a): 54m ~5 5,545,7/54— (575 +53 = Mao/M) 7. 77/25; AM—(MCWMZ MIN-Mala @75wa (A) ZE/K/(EW cow/2572; dag-real” (b) /2£fl677065‘.' DEW/~47; 6, wow 9 (c) EQU/uge/Um W/flrym’p 8 = 9 =30 a; 1‘ 8, 77/259 Carycuelff/vf Efflt7/LW5 [meow/r A) (6)3;QflC/4E72 wraps/2 Cows-7224m7 A 1" :C» (5) 2575776W5! W057£2 Mum/7 [9 -3 [DO/b (c) No [av/ugmw» (27494- b) ...
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class 13 notes - &amp;quot; PROBLEM 4.14 For the given...

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