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1278_Mechanics Homework Mechanics of Materials Solution

# 1278_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: L3 7 Le N” _: 76.2! Cu. 125. G6 Runaway = T) A ﬁnk-AW: ) ‘f P ( (Maya-.33 17mm" \QOXIO" PROBLEM 10.102 2 L = (23(055‘; .- i. to um = 0.6065 211l— at‘r—vﬁé“ Y1 = 6‘” (ﬁnnf‘ﬁ‘e ) t F. 3. I20 MP4. Pe Ix SWCHMJMQ) Jul _3_c_ I, I3) 3 37.534 (0‘3 __ 1.75718»: to“ 75.556: e. a: 10.102 The compression member AB is made of a steel for which a, = 250 MPa and E = 200 GPa. It is [ice at its top A and ﬁxed at its base B. Using the interaction method with an allowable bending stress equal to 120 MPa and knowing that the eocentricities e,r and e, are equal, determine the largest allowable common value. SOLUTION S+ee2i f 1%(75H0‘3X5wt0‘3f - 781.2910” m“ ‘97“? : [H.HBQHD" m TIE (Soulo”)(752t0'53-' L‘ISJSHO“ m“ magi—2 = 21.65: info“ we 6, = 250 MPa. E: 200000 Ha: - ﬁuﬂg .. f ZTT'z (Rococo) _ 2. 3750 x lo—L M Le/rm = {Jo/mosque" = 7&2] < c:c F.S.= g + %(0.é065)ﬂé(0.4065’)3 .— Lgcéz 250 £3662 [I 41-00. goes?) = {07.32 mpg x . +—Ega——— = 1 ML e = e I; Jame-la) ‘ ‘7 -.. 1- i” Amen-doth} zsxio" ) e _ I _ Homo“ 78!. 25 >2 lo“ ’ (3751-: >10" 30051.32 3: [0‘ = l — 0.4mm 7.7:;ﬁlo-5m = 7.75 mm ...
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