Answer:
To determine the maximum deflection in region AB of the overhang beam, we can use the following equation:
δmax = (5wL^4)/(384EI)
where δmax is the maximum deflection, w is the distributed load, L is the length of the beam, E is the modulus of elasticity, I is the moment of inertia, and I is the distance from the fixed end to the point of maximum deflection.
In this case, the beam is fixed at one end and has an overhang at the other end, so we need to consider two separate regions: the region from the fixed end to the end of the overhang (region AB), and the region from the end of the overhang to the free end (region BC).
For region AB, the length of the beam is L = 10 ft, the modulus of elasticity is E = 29(10^3) ksi, the moment of inertia is I = 204 in^4, and the distance from the fixed end to the end of the overhang is x = 6 ft. We also need to find the distributed load w that produces the maximum deflection.
To find w, we can use the fact that the total load on the beam is equal to the weight of the beam itself plus any additional loads. Let's assume that the beam has a weight of 100 lb/ft and that there is a concentrated load of 500 lb at the end of the overhang. Then the total load on the beam is:
wtotal = wbeam + wload = (100 lb/ft)(10 ft) + 500 lb = 1500 lb
The distributed load is then:
w = wtotal/L = 1500 lb/10 ft = 150 lb/ft
Now we can plug in the values and solve for δmax:
δmax = (5wL^4)/(384EI) = (5)(150 lb/ft)(10 ft)^4 / (384)(204 in^4)(29(10^3) ksi)
δmax = 0.256 in
Therefore, the maximum deflection in region AB of the overhang beam is 0.256 inches.