Solution to Problem 596 | Spacing of Rivets or Bolts in Built-Up Beams | Strength of Materials Review at MATHalino

Problem 596
Three planks 4 in by 6 in., arranged as shown in Fig. P-596 and secured by bolts spaced 1 ft apart, are used to support a concentrated load P at the center of a simply supported span 12 ft long. If P causes a maximum flexural stress of 1200 psi, determine the bolt diameters, assuming that the shear between the planks is transmitted by friction only. The bolts are tightened to a tension of 20 ksi and the coefficient of friction between the planks is 0.40.

Solution 596

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$M_{max} = \frac{1}{4}PL = \frac{1}{4}P(12) = 3P \, \text{ lb}\cdot\text{ft}$

$V_{max} = \frac{1}{2}P$

$I = \frac{1}{12} (6)(12^3) = 864 \, \text{ in}^4$

From allowable flexural stress
$f_b = \dfrac{Mc}{I}$

$1200 = \dfrac{(3P \times 12)(6)}{864}$

$P = 4800 \, \text{ lb}$

Strength of bolt
$R = \dfrac{VQ_\text{1st plank}}{I}s$

$R = \dfrac{\frac{1}{2}(4800) \, [ \, 4(6)(4) \, ]}{864}(12)$

$R = 3200 \, \text{ lb}$

Normal force
$R = \mu N$

$3200 = 0.40N$

$N = 8000 \, \text{ lb}$

From tensile stress of bolt:
$\sigma = \dfrac{\text{Force}}{\text{Area}}$

$20\,000 = \dfrac{8000}{\frac{1}{4}\pi d^2}$

$d = 0.7136 \, \text{ in}$ answer

Solution to Problem 596 | Spacing of Rivets or Bolts in Built-Up Beams

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