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

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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 - " PROBLEM 4.14 For the given loading...

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