Question 12.7: 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.
STRATEGY: Draw the bending-moment diagram to find the absolute maximum bending-moment. Then, using this bending-moment, you can determine the required section properties that satisfy the given allowable stress.
MODELING and ANALYSIS:
Reactions. Consider the entire beam to be a free body (Fig. 1).
+\circlearrowleft \sum{M_A=0:} \quad \quad B(8 \text{ ft})-(3.2 \text{ kips})(4 \text{ ft})-(4.5 \text{ kips})(12 \text{ ft})=0 \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad B=8.35 \text{ kips} \quad \quad \pmb{\text{B}}=8.35 \text{ kips}\uparrow \\ \underrightarrow{+}\sum{F_x}=0: \quad \quad \quad \quad \quad A_x=0 \\ +\uparrow \sum{F_y}=0: \ A_y+8.35 \text{ kips}-3.2 \text{ kips}-4.5 \text{ kips}=0 \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad A_y=-0.65 \text{ kips} \quad \quad \quad \pmb{\text{A}}=0.65 \text{ kips}\downarrowShear Diagram. The shear just to the right of A is V_A = A_y = -0.65 \text{ kips}. Since the change in shear between A and B is equal to minus the area under the load curve between these two points, V_B is obtained by
\begin{matrix} V_B-V_A&=&-(400 \text{ lb/ft})(8 \text{ ft})=-3200 \text{ ft}=-3.20 \text{ kips}\\ V_B&=&V_A-3.20 \text{ kips}=-0.65 \text{ kips}-3.20\text{ kips} =-3.85 \text{ kips}\end{matrix}
The reaction at B produces a sudden increase of 8.35 kips in V, resulting in a 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 \pmb{\text{|M|}}_{\text{max}}. 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|_{\text{max}}=18.00 \text{ kip.ft}.
Minimum Allowable Section Modulus. Substituting the values of \sigma_{\text{all}} \text{ and } |M|_{\text{max}} into Eq. (12.9) gives
S_{\text{min}}=\frac{|M|_{\text{max}}}{\sigma_{\text{all}}}=\frac{(18 \text{ kip.ft})(12 \text{ in./ft})}{1.75 \text{ ksi}}=123.43 \text{ in} ^3
Minimum Required Depth of Beam. Recalling the formula developed in step 4 of the design procedure and substituting the values of b and S_{\text{min}} , we have
\frac{1}{6}bh^2\geq S_{\text{min}} \quad \quad \frac{1}{6} (3.5 \text{ in.})h^2\geq 123.43 \text{ in}^3 \quad \quad h\geq 14.546 \text{ in.}
The minimum required depth of the beam is h = 14.55 in.
REFLECT and THINK: In practice, standard wood shapes are specified by nominal dimensions that are slightly larger than actual. In this case, specify a 4-in. × 16-in. member with the actual dimensions of 3.5 in. × 15.25 in.
