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1.6-4 a

# 1.6-4 a - SEBTIDN 1.6 Shear Stress and Strain 33 Soiution...

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Unformatted text preview: SEBTIDN 1.6 Shear Stress and Strain . 33 Soiution 1.6-3 NUMERICAL DATA {h} BEARING STRESS {)N PIN FROh-t FLANGE PLATE 2 E60 kips clI1 =3 2 in. P LE 1.: 111. ll 1 111. 0—H z r (TM :1 20 k5] (__ p i' (a) SHEAR STRESS ON PIN BEARING STRESS ON PIN FROM GUSSE’E' PLATE P V 4 P 'r : —-——_-— 7 x m .3 m :5" . (“in - ”do rrbg = ~1— ahg m 26.7 Itsi <— ‘ 4 4 _ Elvis 7 = 12.73 kSi "-— Prublem 1.5-4 The inciinﬁd ladder AB supports a house painter (82 kg} at C and the sell" weight (q = 36 Nl'm) of the iaddél‘ itself. Each ladder rail (r,. = 4 mm} is supported by a Shoe (IA. 2 5 mm) which is attached to the ladder rail by a bolt oi’diamclcr ‘77 = 8 mm; (3) Find support reactions at A and B. (h) Find the resultant force iii the Shoe holt at A. (c) Find maximum average Shear (T) and hearing (07,) Stresses in the Shoe bolt atA. . ' - ﬂ 7Typical rung x? H = 7 Iii .i-if / 15? ft: . . ,‘Q E Laddor ﬁll] (13.2 4 mm) Shot: bolt at A / Ifrh?‘ .7 ii: 'Shoe bolt ((11,, = 8 mm) ' ' " /- : f4/cf/ ii ‘4/, Ladder shoe (:3. = 5 mm') ‘ it? I ____.. .‘L 4mm _ : ‘ 1 ” ‘ ‘b m b = 0.7 m Assume no slip at A Section at base Solution 1.6-4 NUMERICAL DATA (:1) SUPPORT REAC’I‘IUNS 1r = 4 mm ll, : 5 mm L z \/i a + m3 + H2 L = 7.433171 (1 = 8 mm P : 32 kc {9.81 111/32) . ” ‘2 M» r— ‘1 LAC = 5 353 m P = 804.42 N a + b \I _ ' m ‘- a 21.8111 b : 0.7 m H z 7 in q x 36L LCB "“ a + “gL Lcn "" 3-0h1m m LAC 'i' LCB 3'"- 7.433 111 ...
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