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. • Area of web : A, GX92 = 552 mm• Maximum Shear Stress (Toman ):(where, (width )aeb- 6 mm)Tman =Iz x (width)wrbDraw the loading, shear, and moment diagram ofthe beam shown in the figure. Draw the diagram inproper 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 NImmanswer box below--(Answes)...Tmax 52.885 MPamm*There fore,1210178.719mm 9IzVmaxkNVmax5KNQmaxmm376800mm 3TmaxMPatmax52. 885MPa|So lution:* Deler mine vertical Reaction affired support:100 mm• Area of Flange : A, e 10o oY- 700 mm6. Sino. 5m0.5le-(1+,d-)Z Fy =0* RA + 1 -3 –6 =0RA +1-9 eo:RA =5 EN:-: 9, = 223 kNGKN= 46 mm• Lo tatfon af centrojd G. frum bottom of wtb.: 9, = 92 + = 96 mm.92 mm%3DSechon plara• Shear Forte diayram Calcalatin:Take sechon plane from tho fixedsupport & start cal culatiny shealforre in earh part of the beam.-> VAR = R, = 5 EN• Locatiun of centruid G of whole T-sertjun:CEN4kNGmm= (552 x 46 ) + (800 x 96)(552 + 800)RA = 5 kNLoading Piagram- Vnc = Ra - 3 = 5 -3 = 2 kN5 = 75.5858 mm- Vep = 4 kN• Location of rentroid 6, frum centroidl axis z :-: hi = ý - 9, = 75.5858 - 46 = 29. 58 58 mm20. 4142 mm3KN5 kNBending moment diagramCalculat.: h2Y2 -9 = 96 - 75.58 58%3D: O, 5 )8.- At free end, Mp =0 EN M• Moment af inertia (Tz) about centroial z-axis :GEN1KN(Ta, + A, h,) + (Ia, + A,-h,² )?) + (- Me = 4 kN x o.5 m = 2 EN mIz =3Shear Forte Diogram- Me = 4 x 1 )- 6x0.5 ) = 1 kN m2.+ 552 X 29. 5858100x8+ 800 x 20.4 142 )12- MA = (4x1.5 ) -(6x1) - (3x0.5)Me =2 kN-m6 - 6 - 1-5Iz =1210178 , 714 mm(Answer )--- ....Me =1 EN m. MA = -1.5 KN-M• from the Shear fore diagram,Ma ximum shear fore Vmax occursat fixed end of tha beamQmax :Area moment of web : Qweb = A,9, = 552 x 46Area Moment of flange : Rmox = Qelame= 25392 mm= A2 92 = SUo X96 = 76800 mm: Qmax.Vman = RA= 5 kN--- Answer)= Maximum of { Q webo QAlonge }QflangeMA = -1.5 kN M::VmexBending Moment DiagrimQmox = 76800 mm---(Answee )

The steps with explanation are given below - The reaction force is calculated by balancing the…

• Locution of centroid G, from bottom of web • Lotadion of entroid G, from botten of wtb. • Maximum Shear Stress (Tmax ): : Tman (where,(width )meb = 6 mm ) %3D Iz x (width)wb 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 answer box below = (5x 10 N) × (76200 mm³) (121078.714 mmt) x(6 mm ) Provide complete computation = 52. 88475 NImm ---(answer) Tman 52.885 MPa mm* There fore, Iz 1210178. 719 Vmax kN Vmax KN Qmax 76800 Tmax MPa tmax 52. 885 MPa | Se lutien * Delermine vertical Reaction af fired support : • Area of web : A- GX92 • Area of Flange: A, e 100 y- 700 m - S52 mm 100 mm 6. Sin 6. Sm 0.5 le Z Fy =0 (1+,4-) 3 kN GKN = 46 mm RA + 4 -3-6 0 RA +1-9 eo 92 mm : [RA = 5 kN] * 9, = 92 + = 3Cmm Sechon plara • Shear Forte diayram Calcalatin: Take sechon plane from the fixed support & start calculating sheat fore in each part of the beum. - VAB = RA = 5 EN • Locatiun of rentruid G of whole T-sertiun: CEN Gmm 4kN = (562 x 46 ) + (800 x 96) ( 552 + 800) RA = 5 kN Loading Diagram - Vec RA-3 - 5-3 = 2 kN J = 75.5858 mm • Location of rentroid Gy from centroidal anis z : : hi = - 9, = 75.585 - 46 = 29. 58 58 mm : hz = Y2 -9 = - Vcp = 4 KN 3kN • Bending moment diagram Caleulation : (U,5 ) SKN 96 - 75.58 58 e 20. 4142 mm %3D 2EN B. - At free end, Mp = 0 EN M • Moment af inertia (Tz ) about centroidal z-axis: GEN 1KN Iz = (Ta, + A, h,² ) + ( Ja, + Ay-h,² ) - Me = 4 kN xo.5 m = 2 EN m - Ma = (4 x1 )– €x0.5 ) = 1 KN m - MA = (4x1.5 ) -(6x1) – (3x0.5) 6 - 6 - 1.5 Shear Forte Diagróm 6x 92 12 + 552 X 29. SE582) + ( 100x + 800 x 20.41422) 12 Me =2 kN-m Iz= 1210178. 714 mm (Answer ) -- .. M=1 EN m . MA = -1.5 KN M • from the Shear forre diagraum, Ma ximum shear foxe Vman ocrurs at fixed end of tha beam Omax: Area moment of web : Pweb = A,9, = 552 x 46 = 25392 mm3 Area Moment of flange : Qmx = Relange - Qnex. = Maximum of { qweb, Qflage 3 = Qflange A2 42 = svo x 96 = 76800 mm MA = -1.5 kN m Vmax = RA Vmex = 5 kN]-- (Answer) Bending Moment Diagrum Oe = 76800 mm ---(A nswee )