Question 12.S-P.7: A 12-ft-long overhanging timber beam AC with an 8-ft span AB...
A 12-ft-long overhanging timber beam AC with an 8-ft span AB is to be designed to support the distributed and concentrated loads shown. Knowing that timber of 4-in. nominal width (3.5-in. actual width) with a 1.75-ksi allowable stress is to be used, determine the minimum required depth h of the beam.
Reactions. Considering the entire beam as a free body, we write
+\curvearrowleft \sum{M_{A}} = 0: B(8 ft) – (3.2 kips) (4 ft) – (4.5 kips) (12 ft) = 0B = 8.35 kips B = 8.35 kips↑
\underrightarrow{+} \sum{F_{x}} = 0: A_{x} = 0+↑ \sum{F_{y}} = 0: A_{y} + 8.35 kips – 3.2 kips – 4.5 kips = 0
A_{y} = -0.65 kips A = 0.65 kips ↓
Shear Diagram. The shear just to the right of A is V_{A} = A_{y} = -0.65 kips. Since the change in shear between A and B is equal to minus the area under the load curve between these two points, we obtain V_{B} by writing
V_{B} – V_{A} = -(400 lb/ft) (8 ft) = -3200 lb = -3.20 kips
V_{B} = V_{A} – 3.20 kips = -0.65 kips – 3.20 kips =-3.85 kips.
The reaction at B produces a sudden increase of 8.35 kips in V, resulting in a value of the shear equal to 4.50 kips to the right of B. Since no load is applied between B and C, the shear remains constant between these two points.
Determination of |M|_{max}. We first observe that the bending moment is equal to zero at both ends of the beam: M_{A} = M_{C} = 0. Between A and B the bending moment decreases by an amount equal to the area under the shear curve, and between B and C it increases by a corresponding amount. Thus, the maximum absolute value of the bending moment is |M|_{max} = 18.00 kip \cdot ft.
Minimum Allowable Section Modulus. Substituting into Eq. (12.9) the given value of σ_{all} and the value of |M|_{max} that we have found, we write
S_{min} = \frac{|M|_{max}}{σ_{all}} (12.9)
S_{min} = \frac{|M|_{max}}{σ_{all}} = \frac{(18 kip \cdot ft) (12 in./ft)}{1.75 ksi}= 123.43 in^{3}
Minimum Required Depth of Beam. Recalling the formula developed in part 4 of the design procedure described in Sec. 12.4 and substituting the values of b and S_{min}, we have
\frac{1}{6} bh^{2} ≥ S_{min} \frac{1}{6}(3.5 in.)h^{2} ≥ 123.43 in^{3} h ≥ 14.546 in.
The minimum required depth of the beam is h = 14.55 in.
