The shelf AB can be considered as a simply supported beam. This beam can be simplified by splitting it into two sub-beams, 1) a simply supported beam with uniform distributed load and 2) a simply supported beam with a point load acting at the center. The point load is actually the unknown force in the rod.
Since the rod will also extend due to the load F, the total deflection at point B will be a combination of the two simplified beams and the rod. This can be summarized as
δs = δ1 + δ2
The deflections are shown in the diagram on the left.
The moment of inertia for the wood beam is
Iw = 11.25(3.5)3/12 in4
= 40.20 in4
Remember, the actual size of a 4x12 wood member is 11.25 in by 3.5 in.
The deflection equation of the beam and the hanger can be found in the
Beams Equation appendix.
δ1 = 5woLw4/(384EwIw)
= 5 (400/12) [10(12)]4/ [(384) (40.20) (1.5×106)]
= 1.493 in
δ2 = -FLw3/(48EwIw)
= -F [10(12)]3/ [(48) (40.20) (1.5×106)]
= -5.970×10-4 F
δs = FLs/AsEs
= F (3) (12)/ [30×106 As]
= 1.2×10-6 F / As
These deflections can now be substituted in the deflection compatibility equation to give
1.2×10-6 F/ As = 1.493 - 5.970×10-4 F
F / As (1.2×10-6 + 5.970×10-4 As) = 1.493
F/ As = 1.493 / (1.2×10-6 + 5.970×10-4 As)
Using factor of safety of 3, the allowable stress is reduced as
Allowable stress = Strength / Factor of safety
= 36 / 3 ksi
= 12 ksi
= 12×103 psi
Equating the allowable stress with developed stress, F/As, gives
12×103 = 1.493 / (1.2×10-6 + 5.970×10-4 As)
5.970×10-4 As = 1.244×10-4 - 1.2×10-6
As = 0.2064 in2
The cross-sectional area of steel hanger can be calculated as
As = πD2 / 4
0.2064 = πD2 / 4
D2 = 0.2628
D = 0.5126 in |