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Correct Problem 6.102 The concentrated force P = 2.6 kN is applied at the free end of the overhanging beam. 150 mm 200 mm 25m
(Omax ) =
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P=2.6kN .m imec 10. E Fx = 0 , { fy=0 * RB-RA-2.6=0 » RB = RA+2.6 Ē MA=0 =D (Rox 2) - (2.6 kw XL+ 2)) = 0 2 RB = 2.643 - RB =- MA=0 -2.6 kp MB-MA = area of S. fuD blw B and A MB-o = (-1.3 kN x 2 m) = -2.6 kW MB=-2.6 kN | Hence maximum bending momentcentroid of cross-section, - ū = ý = AYI + A2Y2+A343 + Az + A3 = (3750 X 212.5 ) + ( sovoxlvo) + sooo x lvo) 3750 + 5000 + 50similarly 1 = 13 12 = (A2 d 2² ) + ( bd ² ) = sooo x (30.68)2 + (251(20013 = I=12= 21372978,66 mmt total moment of inertia ab= 3.60 MPa maximum compressive stress (oc) max = (^^) max (92) max I ore = (2.68100 nmm) x 130.68 mm 6.8039558 107 mmt o = 4.

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