Force Statistics Essay - 275 Words

Force Statistics (Essay Sample) Content: Distributed Forces (Statics)Studentà ¢Ã¢â€š ¬s NameInstitutionDistributed Forces (Statics)Question 5/129258127552959000From the point of support Ax = 0, and Ra=Rb=W/2 where Ra is the reaction at point A, Rb is the reaction force at B and W is the applied force at C (Potter and Nash, 2010).25908002349500042386252921000010668002921000010668002330460 A W C DRa L/2 L/2 Rb1066800413385Positive (+)0Positive (+)1009650859155 W/22676525213995Negative (-)00Negative (-) 0 Shear Force Diagram-W/2 1 106680018034000Bending Moment Diagram26574751908175Wl/4Wl/410668003175 00 2943225393700Wl/20Wl/21495425269875Wl/2Wl/2267652531749Question 5/133. A beam with a uniformly distributed load at point l from point A (Potter and Nash, 2010)1619250570865002762242580640276225570230286702557023042672005708642933700469902571752184400257175494665 P/l N/m P8667752476500027336754935220P(L+L)/4P(L+L)/47715254620895Wl2 /8Wl2 /843053002325370-P/200-P/229527501449070P/200P/21333501296670Wl/20Wl/2333375349694516192503496310002755903487420923924198247000333375152527029337001982470016662401525270 L L L 1 Shear Force Diagram 1 Bending Moment DiagramAt Maximum bending moment with uniformly distributed load is given by wL2 /4 and the other point it's given by P(L+L)/4.2867025352425016192502571750091440033337500Question 5/137. From the diagram, Force M1 is used to counter the forces and make it zero. With the above condition Bending moment at C = B =0. Bending m oment at A which is the maximum moment = 2 * 4 =8kN and moment at D = 2*(4-2) = 4kN. The maximum shear force is given by wL2 / 2= 2*42 /2= 16kN30194258255002571751708150 2kN/m276225246380002571751714500A D C B16954492489201000125248920304800248920 2M 2 M 4M M 1304800268605 4M30480048895Wl= 22*4=8kN00Wl= 22*4=8kN35242559690008kN 1 276225-38100Shear Force Diagram304800431800016kN9239251771650027622526543000 1 Bending Moment DiagramQuestion 5/138. Since end A ix fixed then Bending moment = 0 and at point C the force acting is given by 4* F= 0.6*4/2= 0.3 0r 300 N.27622425806402762255702304267200570864257175494665 600 N/m3219450247015-285751714500022479002000250022479003810000 A c D B-54292477470Fixed End00Fixed End2294890172720031337251536702755903487420 2M 2M27622418288022955251828800 4M 4M2828926160020600*(4-2)= 1200NM00600*(4-2)= 1200NM231457423622006858007620Wl= 300*4= 1200NM00Wl= 300*4= 1200NM27622523622000 1 32194504762400 Shear Force Diagram430530067310600*(4-...