**Interpreting the problem** I need to parse the user’s request carefully since they are looking for a step-by-step solution along with an explanation and a final answer. The problem mentions an aluminum spacer being measured for dimensional change under a unidirectional stretch of 750 kN. However, I find it confusing since "unidirectional" implies tension in just one direction, but it references both length and thickness. Maybe they intended for two separate tests: one for tension along the length and another for thickness. I’ll need to clarify that!**Clarifying the dimensional change** The phrase "unidirectional stretch along its length and thickness direction" raises some questions for me. It seems to imply that the bar might be pulled along its length, requiring us to determine dimensional changes in all directions. However, "unidirectional" suggests it's pulled in just one direction. I wonder if the engineer intends for separate tests: one for length and another for thickness. I'll compute changes for both using Poisson's ratio, starting with the bar's dimensions: length of 1 m, width of 95 mm, and thickness of 180 mm, with a force of 750 kN and E of 760 kN/mm².**Calculating axial tension and strain** So, I'm figuring this out with the load applied along the bar's length, meaning there's axial tension. The cross-sectional area resisting that is calculated as width times thickness: 95 mm times 180 mm gives me 17,100 mm²—got that right! Now, I need to calculate axial stress (σ) using the formula σ = P/A, where P is 750 kN. Converting kN to kN/mm², I find it's about 0.04386 kN/mm², or 43.86 MPa, which is useful since E is given as 760 kN/mm². Next, I compute strain (ε) using ε = σ/E, resulting in a strain of roughly 5.77e-5. The elongation (ΔL) then becomes delta L = ε * L, which is 0.05772 mm. Lateral strains in width and thickness are determined through Poisson's ratio, giving me ε_lat ≈ -1.270e-5.**Calculating changes in dimensions** I’m computing based on the lateral strains. So multiplying the strain by Poisson's ratio (0.22) gives me approximately 1.27e-5. With this, I find the change in width (Δb) is about -0.001206 mm, and the change in thickness (Δt) is around -0.002286 mm. For lengthwise tension, I have length increasing by 0.0577 mm, width decreasing by 0.00121 mm, and thickness decreasing by 0.00229 mm. Now, considering the phrase "along its length and thickness direction," it seems they might want to explore a case where the bar is stretched through thickness direction, possibly in compression as a spacer between plates. This leads me to think that if load is applied along that axis, I need to use the length multiplied by width for the resisting cross-sectional area.**Analyzing tensile stress and strain** If I'm pulling along the thickness, then the cross-sectional area I should consider is length times width (1 m by 95 mm) for calculating stress. This gives me 95,000 mm², which is correct. Since the tensile load applies in the thickness direction, I need to compute the stress based on that area. But pulling 750 kN across the area suggests a stress of about 7.895 MPa, leading to a small strain of approximately 1.039e-5. This change in thickness results in about 0.00187 mm. I also compute lateral contractions producing changes in length and width. The problem specifies “unidirectional stretch along its length and thickness direction,” indicating they likely want calculations for both orientations. I’ll present both cases, as well as compute for aluminum, noting the elastic modulus given as 760 kN/mm² seems too high based on common values. I suspect it should really be 76 kN/mm², likely due to typesetting errors in the problem statement. I'll proceed with that assumption, but I feel we should also double-check it for accuracy, especially since it affects our strain results.**Computing dimensional changes under load** Let's outline the steps clearly. First, I need to establish the dimensions, load, Young's modulus (E), and Poisson's ratio (nu) in consistent units—mm, kN, kN/mm². In Case A, I'm calculating the load along the length, leading to an area of 17,100 mm². Doing the stress calculation gives me around 43.86 MPa. Then, I can derive axial strain and changes in length, width, and thickness using the formulas for strains. For Case B, with load applied along the thickness, the area is 95,000 mm², resulting in a much lower stress of 7.895 MPa. I'll calculate strains and changes there too. Importantly, if E is 760 kN/mm², I'll note that it seems too high for aluminum and should be closer to 76 kN/mm². It’s good to present results for both values to be thorough.**Finalizing calculations for axial stress** I need to ensure that the axial stress is below the yield strength for aluminum. At 44 MPa, it's well below the typical 250 MPa yield limit, which confirms it's safe. Now I should produce the final answer with a clear, step-by-step explanation without heavy formatting. First, I'll clarify my assumption about E. Then, I'll convert units, move on to Case 1 calculations, followed by Case 2 calculations, and finally, I'll present the numeric values, being extra careful about the sign for Poisson's ratio!