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1067_Mechanics Homework Mechanics of Materials Solution

# 1067_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: PROBLEM 9_57 9.57 For the beam shown and knowing that P = 40 kN, determine (a) the reaction at E, (b) the deﬂection at C. Use E = 200 GPa. SOLUTION I Uant Forces in {LN} jQW3+LS in WI. W200 X 46.1 +13?! :0 gruo-QOJO-r Pa =' O 0.5 m 0.5 m 0.5 m 0.5 m [yuggo] . Lx=2_, 3:03 RN: 120- 25 14“ {WOJAWTI +DM,=0 —M,- 20*”0'60 +295 = O M 4" 4° ”° MA: 2:25- :zo m...» 3 Read“ was cute. 5+ «Hod/£91 i v: a) e+e~r M- node. . (W 'P P D I? 'l’i R; 93 gr:- = V = P..- “°<‘<"°-5>°‘ 4”<>4"‘>u'* q°<*“'5>° E133 = M” MI“ +1?” ' ”(3'05)" 400‘"). " 40¢“ L53. EI '9 = rm 4 a‘i?.x'-Zo<x-°-5>‘- 20(x- I>‘ —2c><><-I.S>z +0. EIJ 9 EliMAXI 4- tng‘ " %9<X—05>3- é2<x‘ I); _ '.%D<x~ LS): + c')‘ + c" O+o+O+O+O+Q=O o+0+O+o+0+O+CL=O I» M=°1 '3'M,(21‘+-;P.(233— 35(._5)3..%9(m - 5530.5? + o + o I! 0 'g'g'C293-l20)(2f+ ﬂazo—Rng)’ = so 2.5ccc7 P; = so + 2'40- 190 = Ho 1?; -= 41.25 WT -- M, = (2)6«140-120 = -37.5 mm R; = 12.0- qms = 72.15 In: Dmlat Er 200::qu Pa. , 1' .- ”5.5xlo‘ M” = qs.sx)o“‘m* £1 = (200¥[0°)(455310") '3 ‘7JOHD‘ Noam." : cN00 kM-ML (JD) Deuediam ail” C (X w+ X9 I M) Ely; = -}(‘-37.s}(\\‘+ \$633530? - 3§(o.s)3_o _ 0 +0 +0 3-" '6.‘1583 ku.m3 h _ c.4512. y“ ’ ‘uoo = —. 0.710 No" m y; 2 0.710 m. L --I [' ___ ‘ V T'" | ...
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