The diagram presents a truss structure with members connected at joints labeled A, B, C, and D. The truss is subject to external forces, including a 12 kN force acting to the left at joint D and a 9 kN force acting to the right at joint C.Key dimensions and elements of the truss are as follows:- The horizontal distance between points A and B is 1.5 m.- The vertical height between points A and D, and points B and C, is 2 m.- All members of the truss have a square cross-section with side dimensions of 100 mm × 100 mm.- The modulus of elasticity (E) is given as 200 GPa.You are tasked with determining:(a) The force in member AC.(b) The forces in all members of the truss.(c) The horizontal displacement at point C.This analysis involves using the force method, a structural analysis technique to calculate internal forces and displacements.

A truss with hinged support at A and roller support at B is given. Horizontal 9 kN and 12 kN loads are applied at C and D respectively. E = 200 GPa Cross-sectional area, A = 104 mm2 To determine (a)The force in member AC (b)Forces in all members (c)Horizontal displacement at point CDegree of static indeterminacy, Ds DS= m +r-2j=6+3-2×4=1 Thus, the truss is indeterminate of degree 1. Assume force in member AC as redundant Using Castigliano's theorem At joint C FCA=RFCD = 9 -Rcos a=9-0.6RFCB=-Rsin a=-0.8R At joint D FDB=-FDC-12cos a =-9+0.6R-120.6 =-35+RFDA=-FDBsin a = 28-0.8R At joint B FBA=-FDBcos a=21-0.6R Member Length AE Fi ∂Fi∂R Fi∂Fi∂RLi CA 2.5 2000000 R 1 2.5R CD 1.5 2000000 9-0.6R -0.6 -8.1+0.54R CB 2 2000000 -0.8R -0.8 1.28R DB 2.5 2000000 -35+R 1 -87.5+2.5R DA 2 2000000 28-0.8R -0.8 -44.8+1.28R BA 1.5 2000000 21-0.6R -0.6 -18.9+0.54R ∆CA=∑Fi∂Fi∂RLiAE0=8.64R -159.3R =18.44 kN Thus, the force in member AC is 18.44 kNThe member forces due to applied external loads are Member Fi∂Fi∂RLi For R = 18.44 kN Member forces CA 2.5R 46.1 CD -8.1+0.54R 1.8576 CB 1.28R 23.603 DB -87.5+2.5R -41.4 DA -44.8+1.28R -21.20 BA -18.9+0.54R -8.94At joint C FCA=RFCD = 1-Rcos a=1-0.6RFCB=-Rsin a=-0.8R At joint D FDB=-FDC-12cos a =-1+0.6R0.6 =-1.67+RFDA=-FDBsin a = 1.33-0.8R At joint B FBA=-FDBcos a=1-0.6R Member Length Fi ∂Fi∂R Fi∂Fi∂RLi For R = 1.153 kN Member forces CA 2.5 R 1 2.5R 2.88 CD 1.5 1-0.6R -0.6 -0.9+0.54R -0.277 CB 2 -0.8R -0.8 1.28R 1.476 DB 2.5 -1.67+R 1 -4.175+2.5R -1.293 DA 2 1.33-0.8R -0.8 -2.128+1.28R -0.6523 BA 1.5 1-0.6R -0.6 -0.9+0.54R -0.277 ∆CA=∑ Fi∂Fi∂RLi = -5.603+4.86RR = 1.153 kNMember Length AE Member forces due to external loading, Fi Member forces due to unit load, fi FifiLiAE CA 2.5 2000000 46.1 2.88 0.00016596 CD 1.5 2000000 1.8576 -0.277 -0.00000038591 CB 2 2000000 23.603 1.476 0.000034838028 DB 2.5 2000000 -41.4 -1.293 0.00006691275 DA 2 2000000 -21.20 -0.6523 0.00001382876 BA 1.5 2000000 -8.94 -0.277 0.000001857285 ∆C = ∑FifiLiAE=4.7168×10-5 m=0.0472 mm Thus, the horizontal deflection at C is 0.0472 mm