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763_Mechanics Homework Mechanics of Materials Solution

# 763_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: PROBLEM 6.C5 - CONTINUED Foeu: l- = 714312271) TL. : VG ﬁx 1; /1-Za'mce~r)g .— VQ/Zptgw CQ = Q 74:50:: (Trimmw); x 0mm»; 5F .é‘ﬁgnﬂ GEN-arm. (ﬁLcw-AWF MIME/V7 an 912%? For: me: #13907 dram/Iv Fame. Z = / TB 7” 1%35MMZ(A&»QE F2512. Ill/#025 EECWOW’ NFWEN772(M0MLMT) y/ixm CEWEQ 1.: A7 6::- MOME/V7 V PWééE/AM 007F07— / Prob. 6.65 Tii) Mi) Mi) Mi) mm mm mm mm 1 10.00 70.00 10.00 40.000 2 6.00 70.00 50.00 70.000 3 10.00 0.00 50.00 50.000 4 10.00 0.00 0.00 Moment of inertia: Ix = 3759956 mm‘4 Shear = 50000 N Junction Q Tau Tau Force in of segments Before After segment: mm“3 MPa MPa kN 1 and 2 12000.000 15.96 26.60 2482.37 2 and 3 33000.000 73.14 43.88 20988.54 3 and 4 45500.000 60.51 60.51 27372.75 Moment of shear forces about origin: M = 2436.386 N‘m + counterclockwise Distance from origin to shear center: e 48.728 mm CONTINUED ...
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