Provide an explanation (in words) of how the problem was solved, the step-by-step process that you think was used, the strategy, the formula used, etc.
Extracted text: • Area of web : A, GX92 = 552 mm • Maximum Shear Stress (Toman ): (where, (width )aeb - 6 mm) Tman = Iz x (width)wrb Draw the loading, shear, and moment diagram of the beam shown in the figure. Draw the diagram in proper position. details. Include signs in the magnitude for the (5x 10 N) × (76z00 mm³) (121017 8. 714 mmt ) x(6 mm ) Provide complete computation = 52. 88475 NImm answer box below --(Answes) ... Tmax 52.885 MPa mm* There fore, 1210178.719 mm 9 Iz Vmax kN Vmax 5 KN Qmax mm3 76800 mm 3 Tmax MPa tmax 52. 885 MPa | So lution: * Deler mine vertical Reaction af fired support: 100 mm • Area of Flange : A, e 10o oY- 700 mm 6. Sin o. 5m 0.5 le -(1+,d-) Z Fy =0 * RA + 1 -3 –6 =0 RA +1-9 eo :RA =5 EN :- : 9, = 22 3 kN GKN = 46 mm • Lo tatfon af centrojd G. frum bottom of wtb. : 9, = 92 + = 96 mm. 92 mm %3D Sechon plara • Shear Forte diayram Calcalatin: Take sechon plane from tho fixed support & start cal culatiny sheal forre in earh part of the beam. -> VAR = R, = 5 EN • Locatiun of centruid G of whole T-sertjun: CEN 4kN Gmm = (552 x 46 ) + (800 x 96) (552 + 800) RA = 5 kN Loading Piagram - Vnc = Ra - 3 = 5 -3 = 2 kN 5 = 75.5858 mm - Vep = 4 kN • Location of rentroid 6, frum centroidl axis z : -: hi = ý - 9, = 75.5858 - 46 = 29. 58 58 mm 20. 4142 mm 3KN 5 kN Bending moment diagram Calculat .: h2 Y2 -9 = 96 - 75.58 58 %3D : O, 5 ) 8. - At free end, Mp =0 EN M • Moment af inertia (Tz) about centroial z-axis : GEN 1KN (Ta, + A, h,) + (Ia, + A,-h,² ) ?) + ( - Me = 4 kN x o.5 m = 2 EN m Iz = 3 Shear Forte Diogram - Me = 4 x 1 )- 6x0.5 ) = 1 kN m 2. + 552 X 29. 5858 100x8 + 800 x 20.4 142 ) 12 - MA = (4x1.5 ) -(6x1) - (3x0.5) Me =2 kN-m 6 - 6 - 1-5 Iz = 1210178 , 714 mm (Answer ) --- .... Me =1 EN m . MA = -1.5 KN-M • from the Shear fore diagram, Ma ximum shear fore Vmax occurs at fixed end of tha beam Qmax : Area moment of web : Qweb = A,9, = 552 x 46 Area Moment of flange : Rmox = Qelame = 25392 mm = A2 92 = SUo X96 = 76800 mm : Qmax. Vman = RA = 5 kN--- Answer) = Maximum of { Q webo QAlonge } Qflange MA = -1.5 kN M :: Vmex Bending Moment Diagrim Qmox = 76800 mm ---(Answee )