Solution to Problem 597 | Spacing of Rivets or Bolts in Built-Up Beams | Strength of Materials Review at MATHalino

Problem 597
A plate and angle girder similar to that shown in Fig. 5-32 is fabricated by riveting the short legs of four 125 × 75 × 13 mm angles to a web plate 1000 mm by 10 mm to form a section 1020 mm deep. Cover plates, each 300 mm × 10 mm, are then riveted to the flange angles making the overall height 1040 mm. The moment of inertia of the entire section about the NA is I = 4770 × 10⁶ mm⁴. Using the allowable stresses specified in Illustrative Problem 591, determine the rivet pitch for 22-mm rivets, attaching the angles to the web plate at a section where V = 450 kN.

From Illustrative Problem 591

τ = 100 MPa shear stress
σ_b = 220 MPa bearing stress for single shear rivet
σ_b = 280 MPa bearing stress for double shear rivet

Solution 597

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Rivet capacity in terms of shear (double shear)
$R_s = 2(A_s \, \tau) = 2 \, [ \, \frac{1}{4}\pi (22^2)(100) \, ]$

$R_s = 24\,200\pi \, \text{ N } = 24.2\pi \, \text{ kN } = 76.03 \text{ kN}$

Rivet capacity in terms of bearing (use σ_b = 280 MPa)
$R_b = A_b \, \sigma_b = [ \, 22(10) \, ] (280)$

$R_b = 61\,600 \, \text{ N } = 61.6 \, \text{ kN}$

Use R = 61.6 kN for safe value of R

Moment of area
$Q_A = 2(2430)(491.1) + 300(10)(515)$

$Q_A = 3,931,746 ~ \text{mm}^3$

Spacing of rivets
$s = \dfrac{RI}{VQ_A}$

$s = \dfrac{61.6(4770 \times 10^6)}{450(3,931,746)}$

$s = 166 ~ \text{mm}$ answer

Solution to Problem 597 | Spacing of Rivets or Bolts in Built-Up Beams

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