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Answer :
Final answer:
The maximum deflection of the beam CD under a concentrated load of 20k at mid-span is approximately 1.54 inches, calculated by comparing the ratio of the second moments of area for the two beams.
Explanation:
This problem can be solved using the concept of deflection in beams. The maximum deflection of a beam under a concentrated load occurs at the load and is given by FL⁴/384EI, where F is the force, L is the length, E is the Young's modulus, and I is the second moment of area. Given that the load F2 and the material properties (since they come into play through Young's modulus) are identical across both beams, the ratio of the deflections can be simplified to inversely proportional to their second moments of area (since dimensions other than the beam’s width and height remain constant).
The second moment of area I for a rectangular beam is given by (width*height⁴) /12. Therefore, the ratio of beam AB to beam CD deflection is ((5*10⁴) / (4*8⁴)) = 1.95. By setting up a ratio, we know that 3in/A2 = 1.95 → A2 = 3/1.95 = 1.54 inches.
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