Please don't provide handwritten solution ....  For the loading scenario in Figure , determine the maximum normal stress. Show the steps in this determination clearly using FBDs and equations

Q6/ A material has a true stress -strain curve given by o=Kɛ", derive and calculate the true and engineering ultimate tensile strength of this material. (use : k=689.47kpa , n=0.5)

1. A part made of Aluminum 6061-T6 has a yield strength  = 400 MPa. For each stress state below, draw all 3 Mohr's circles, find the principal stresses, and calculate the safety factor against yield using both the distortion-energy (von Mises) and maximum shear stress (Tresca) criterions. (If relevant) A clearly labeled diagram (or diagrams) clearly pertaining to your analysis with a coordinate system and relevant labels. Final answer with appropriate units and significant figures. You can use the fprintf() command in MATLAB to format numerical results A 2-3 sentence reflection on your answer. Does it make sense? Why or why not? What are some implications?

PROBLEM SOLVING: A rigid bar is supported by three non-rigid steel rods, as shown. Rods AB and FG are 0.8 m long with a cross-sectional diameter of 30 mm. Rod CDE is made up of equally-long 0.40 – m CD and DE whose cross-sectional diameters are 60 mm and 30 mm, respectively. A 45 kN downward load is applied to the rigid bar. There is no initial strain before the application of the load. Let Esteel = 200 GPa Determine the internal forces on rods AB, CDE, and FG Determine the displacement of the rigid bar. If the allowable displacement of the horizontal rigid bar is 0.071 mm downwards, what maximum allowable vertical load can be applied for a factor of safety of 3? А F 0.40 m 60mm¢ D 0.8 m 0.8 m 30mm¢ 30mmф 0.40 m 30mm E G 45 kN