The assembly shown consists of two15-mm diameter brass rods (members ABand CD), a 20-mm diameter steel rod(member EF), and a rigid plate at G.Determine:a) the horizontal displacement of pt. Fb) the horizontal displacement of pt. G - 450 mm -P1: The assembly shown consists of two15-mm diameter brass rods (members ABand CD), a 20-mm diameter steel rod(member EF), and a rigid plate at G.Determine:a) the horizontal displacement of pt. Fb) the horizontal displacement of pt. G- 300 mmBE4PDGP = 5 kNЕь %3 95 GPaEs = 180 GPa

Given: Diameter of the brass rod is 15 mm. Diameter of the steel rod is 20 mm. Load P is 5 kN. Eb=95 GpaEs=180 Gpa Calculation: (b) Rods AB and CD are of the same material and same load is applied. Both the rods are parallel to each other. Hence, there horizontal displacement will be the same. The horizontal displacement at point G must be the same as horizontal displacement of rods AB or CD. The free body diagram of the AB and CD is shown below: As the loads and beam is symmetric, the 4P load must be divided equally in rod AB and CD as shown below: Net load on the rods AB or CD is, PAB=PCD=2P-PPAB=PCD=P Horizontal displacement at point G is calculated as follows: δG=PABLABAABEb=5×10000.3π40.015295×109=8.935×10-5 mor,δG=0.08935 mm(a) The free body diagram of the steel rod is shown below: Horizontal displacement of the rod EF is calculated as follows: δEF=4PLEFAEFEb=45×10000.45π40.0202180×109=1.59155×10-4 mor,δEF=0.159155 mm Displacement of the point F is calculated as follows: δF=δEF+δG=0.159155+0.08935δF=0.2485 mm Answers: (a) δF=0.2485 mm (b) δG=0.08935 mm