The beam AB is 5.0 inches wide and 10.0 inches high. Under a concentrated load [tex] F_1 [/tex] of 20 kips at mid-span, the beam has a maximum deflection ([tex] \Delta_1 [/tex]) at mid-span of 3.0 inches.

Beam CD, which is 40 inches wide and 8.0 inches high, is subjected to a concentrated load [tex] F_2 [/tex] of 20 kips (equal to [tex] F_1 [/tex]).

Determine the maximum deflection ([tex] \Delta_2 [/tex]) of beam CD at mid-span.

The maximum deflection of beam CD under a centrally located load of 20k is 0.59 inches. This is calculated using the principles of beam deflection in materials science.

Explanation:

This question relates to the deflection of beams under loading. It encompasses topics found in materials science and engineering and can be solved using principles from the area of elasticity. The deflection (delta) of a beam subjected to a centrally located concentrated load (P) can be calculated using the equation delta = PL^3/48EI, where L is the length of the beam, E is the Young's modulus, and I is the moment of inertia of the beam's cross-section.

In this case, the beam geometry has altered but the load and length remain the same. However the moment of inertia (I), which is dependent on the beam's geometry, has changed. For a rectangle, I is calculated using the formula I = bh^3/12, where b is the width of the rectangle and h is the height. According to this, the moment of inertia for beam AB is 5(10^3)/12 = 416.67 in^4. For beam CD, it is 40(8^3)/12 = 2133.33 in^4.

Deflection is proportional to the inverse of the moment of inertia. Therefore, it can be calculated that the deflection (A2) of beam CD will be A1 * (I_AB/I_CD) = 3(416.67/2133.33) = 0.59 inches.

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