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Solution' The given configuration is shown below & QP a La A E 2a ca С force & This shifted to point a can be system of ax explained below :-
vertical force at c=2p. Axial force at c= Moment of force at c = 2P xa - Px za 1 → CW CCW 은 o Hence equivalent eystem con be shown on below:- OP B qa ga A. ke с. RAH RAV RJ456 RB By vertical force balonce Rar + Ro Sin 45º = 2P A - By Horizontal force balance RAH = P + BaCOS 45° -1 Balancing moment about B -> RAV X (Qataa) = 2Px za => RAV — a By e equation P+ Rgsins = ap P. 52 By equation 6 RAH = P + 1.414 Pxt- 52 RB= 10414P 14 P : a P =
Finding Axial force in beam Between pointy as c Fac= RAH = 20 (tre for Right warde sove for left words) = 2 (03 +1) KN = 8 KN (conet) (ax P = student No.t1 KN) between C & B Fro = RAH -P = &p_p =P = 4 kN (Const) Findings shear forces in Beam" Between AIC SF |AC = RAV ( tue for upwarde in right side) P = 4 KN Cconstant) Between C & B SF leot RAN-&P = -p= - 4 kN (conston) Finding moment of Bending in beam :- BM: between AIC
B.m. I Ac = Ravox = P. x (x is from a fur any section blu A&c . for sagging & -ve for (tve hogging) B.mila (fc. at 1 =o) Pyo = 0 x = ga i e at С. B.m. la PxQ a = 4 x 2 ( 03+ 100) cm = 8 24 kncm Between C & Bi- c = at Bim! Rs. sin45° x x x from B for BC Рус port) X=0 i.e. cet B B.M.l=0 at xeda , Bimilc = 2 px= 8 24 kN CM.
Here beom the different diagroma of internal loading in a val a ga АА da qa B 1824 KN cm TB. m. 0 ga a a 4KN stb da SED -4kn & KN 4 kn Axial forke / Diagrom / / / 9 а da E (a Anxwery Reation x at A in Horizontal = QP= 8 KN vertical =p=4 KN Reaction 45° from vertical = J2 P = 5.656 KN 6 Diagrom are shown above 9 B