The picture below Consista brass ( EB = 100 Gla and ABand a Aluminum [ EA₁ = 73GPa | AA!each red exists a non- deformable rigid plate. The rods are stress freeat 25°C. If the coefficient of thermal expansion for staat is taken13.106 [m/m°c ]ofa steel (E₂ -200CHPA and As = 1500mm ²= 2000 mind)JATaluminiscent2= 1500 mm ² ) kod betweenat dsand temperature increased toupre increased to 350°C.AsteelFind reactionary forces at A_and_D_ Considering RA = Rai andRp = - Rp₁" ? Also find the deflection of plate B.1.5 ma18 = 19,10`6 [m/m="c] α = 24,10 €ALBBrassImtDAluminum1.5m[m/m="c}Fant Fall=$6

Given: The modulus of elasticity of steel Es=200 GPa. The area of steel bar is As=1500 mm2. The length of a steel bar is Ls=1.5 m. The modulus of elasticity of brass is EB=100 GPa. The area of brass bar is AB=2000 mm2. The length of brass bar is LB=1 m. The modulus of elasticity of aluminum bar is EAl=73 GPa. The area of the aluminum bar is AAl=1500 mm2. The length of an aluminum bar is LAl=1.5 m. The coefficient of thermal expansion for steel is αs=13×10-6 m/m°C. The coefficient of thermal expansion for brass is αB=19×10-6 m/m°C. The coefficient of thermal expansion for aluminum is αAl=24×10-6 m/m°C.Consider the free body diagram at joint B and assume that both the internal axial forces will be in tension. ΣFx=-F1+F2=0F1=F2 ORFs=FB Consider the free body diagram at joint C and assume that both the internal axial forces will be in tension. ΣFx=-F2+F3=0F2=F3 ORFB=FAl Hence, Fs=FB=FAl By using the relationship of geometry of the deformation, we have: δs+δB+δAl=0FsLsAsEs+αs∆T·Ls+FBLBABEB+αB∆T·LB+FAlLAlAAlEAl+αAl∆T·LAl=0............(1) ∴∆Ts=∆TB=∆TAl=∆TArrange the equation (1) in terms of FB. FBLsAsEs+LBABEB+LAlAAlEAl=∆T-αsLs-αBLB-αAlLAlFB=-∆TαsLs+αBLB+αAlLAlLsAsEs+LBABEB+LAlAAlEAl Substitute the known numerical values in the above equation. FB=-350-2513×10-6×1500+19×10-6×1000+24×10-6×150015001500×2×105+10002000×1×105+15001500×0.73×105FB=-3250.07452.37×10-5FB=-1.021624×106 N×10-3 kN1 NFB=-1021.62 kN Hence, Fs=FB=FAl=-1021.62 kN So, the reactionary forces at A and D will be, Reactions at A and D=1021.62 kNTo find the deflections at plate B, we have: δs=FsLsAsEs+αs∆T·Ls Substitute the known numerical values in the above equation. δs=-1021.62×103150015002×105+13×10-6350-25·1500δs=-5.108+6.3375δs=1.23 mm