3 3.For the statically indeterminate cantilever beam (EI is constant) shown in Figure 3 and using givencoordinates x, and x2 as shown:a) draw the appropriate segment FBDs and determine the bending moment equations, M₁(xi) and M₂(x2), ofsegments AC and BC, respectively,b) applying Methods of Integration, determine the equations of the slopes (-dv₁(x₁) dv₂(x₂)dx2dx₁(v₁ (x₁), V₂ (x₂)) of the two segments of the beam in terms of P, L, Cy, El and the integration constants,c) determine:i. all integration constants (possibly in terms of the redundant support force Cy, applied load P, and length L),ii. redundant support force Cy in terms of P,iii. fixed support forces, Ax and Ay (in terms of P) and moment, M₁ (in terms of P and I)Ad)using L= 3 m, P = 25 kN and EI= 1 x 107 N.m², determine the maximum deflection, Vmax andthe maximum slope, Omax of the beamEl is constantL)Figure 3L) and elastic curvesB

Given Propped cantilever beam with point load acting on it. Length=3mLoad,P=25 kNEI=10*106 N-m2 =104 kN-m2 It is asked determine the following, 1. Bending moment equations for span AC and BC 2. Using method of Integration, Determine the slope and deflection equation 3. Support reactions and moment at the fixed support 4. Maximum slope and deflection of the beam Let the support reactions at C be RC and at A be RA. Using deflection equations, Deflection in downward=deflection in upwardPL33EI=RCl33EI25*333EI=RC*233EIRC=84.375 kN 3. Support reactions and moment at the fixed support: By using equations of equilibrium, ∑Fy=0⇒RA+RC=25RA=25-84.375 =-59.375 kNMoment at fixed support:MA=25*3-(84.375*2) =-93.75 kN-m1. Bending moment equations for span AC and BC: BM equation for span AC:MAC=RA*x (0≤x≤2)⇒From leftBM equation for span BC:MBC=25*x (0≤x≤1)⇒From right 2. The slope and deflection equation Using method of Integration: EId2ydx2=Mx=-RAx+RC*(x-2)EId2ydx2=-59.375*x+84.375*(x-2)On integrating both sides,EIdydx=θx=-59.375*x22+84.375*(x-2)22+C1Again integrating on both sides,EI(y)=-59.375*x32*3+84.375*(x-2)32*3+C1x+C2 Applying boundary conditions: At x=0, y=0⇒C2=0At x=2, y=0⇒0=-59.375*232*3+C1*2+0C1=39.5834. Maximum slope and deflection of the beam: The slope equation,dydx=-59.375*x22+84.375*(x-2)22+39.583The deflection equation,y(x)=-59.375*x32*3+84.375*(x-2)32*3+39.583x Maximum deflection occurs in beam where slope will be zero. EIdydx=-59.375*x22+84.375*(x-2)22+39.583=0By solving,x=1.175 mBy substituting the value of x in delfection, EI(y1.175m)=-59.375*1.17532*3+39.583*1.175 = 30.457104=3.045*10-3 m =3.045 mmMaximum slope will be at the support where x=0,EIdydx=-59.375*022+84.375*(x-2)22+39.583 EIdydx=39.583dydx=θmax=39.583EI=39.583104θmax=3.958*10-3 radians Therefore, Maximum slope at the Fixed support, θmax=3.958*10-3 radians Maximum deflection at x=1.175m is y=3.045 mm