Question 6.18: A box beam is made by nailing together four boards in the co......
A box beam is made by nailing together four boards in the configuration shown in Fig. 1a. The beam supports a concentrated load of 1000 lb at its midspan, and it rests on simple supports (Fig. 1b). If each nail can withstand an allowable shear force of 150 lb, what is the maximum spacing that can be used?
Plan the Solution Since the shear force V is constant over each half of the beam, we can use a constant nail spacing Δs. The shear flow along each row of nails can be computed using the shaded area between A and B in Fig. 2. Equations 6.84 and 6.86 can be combined to compute the maximum spacing. The finish dimensions of the boards are given in Appendix D.8.
V_a ≥ q_rΔs (6.86)
q_r = \frac{VQ} {I} = \frac{VA^′\bar{y}^′}{I} (6.84)
Since there are two rows of nails transferring shear between the shaded area in Fig. 2 and the vertical boards, we need to modify Eq. 6.84 to read
2q_r = \frac{VQ} {I} = \frac{VA^′\bar{y}^′}{I} (1)
in order to calculate the shear flow along a single line of nails. The spacing along each row of nails is then given by Eq. 6.86.
Δs_{max} =\frac{V_a} {q_r} (2)
The moment of inertia can be computed by taking a 6.5 in. × 7.25 in. rectangle and subtracting a 3.5 in. × 4.25 in. rectangle.
I = \frac{6.5(7.25)^3} {12} – \frac{3.5(4.25)^3}{12} = 184.0 in^4The shear flow is given by Eq. (1),
q_r = \frac{VA^′\bar{y}^′}{2I} = \frac{(500 lb)[(3.5 in.)(1.5 in.)] (2.875 in.)} {2(184.0 in^4)}q_r = 20.5 lb/in.
Then, from Eq. (2),
Δs_{max} = \frac{V_a}{q_r} = \frac{150 lb}{20.5 lb/in.} = 7.32 in.
Use Δs = 7 in.
Review the Solution The answer seems reasonable, and the calculations are easily rechecked and found to be correct.
Of course, factors other than shear strength should be considered in the design of joints. For example, nails cannot be driven too close to the edge of a piece of timber, and nails cannot be spaced too close together, or the wood is likely to split.
