# Solution to Problem 128 Bearing Stress

**Problem 128**

A W18 × 86 beam is riveted to a W24 × 117 girder by a connection similar to that in Fig. 1-13. The diameter of the rivets is 7/8 in., and the angles are each L4 × 3-1/2 × 3/8 in.. For each rivet, assume that the allowable stresses are τ = 15 ksi and σ_{b} = 32 ksi. Find the allowable load on the connection.

## Summary of the problem

**Given:**

Shape of beam = W18 × 86

Shape of girder = W24 × 117

Shape of angles = L4 × 3-1/2 × 3/8

Diameter of rivets = 7/8 inch

Allowable shear stress = 15 ksi

Allowable bearing stress = 32 ksi

**Required:**

Allowable load on the connection

**Solution 128**

## Click here to expand or collapse this section

Designation |
Web thickness |

W18 × 86 | 0.480 in |

W24 × 117 | 0.550 in |

**Shearing strength of rivets:**

There are 8 single-shear rivets in the girder and 4 double-shear (equivalent to 8 single-shear) in the beam, thus, the shear strength of rivets in girder and beam are equal.

$V = \tau A = 15 [ \, \frac{1}{4} \pi (\frac{7}{8})^2 (8) \, ]$

$V = 72.16 \, \text{ kips}$

**Bearing strength on the girder:**

The thickness of girder W24 × 117 is 0.550 inch while that of the angle clip L4 × 3-1/2 × 3/8 is 3/8 or 0.375 inch, thus, the critical in bearing is the clip.

$P = \sigma_b \, A_b = 32 [ \, \frac{7}{8}(0.375)(8) \, ]$

$P = 84 \, \text{ kips}$

**Bearing strength on the beam:**

The thickness of beam W18 × 86 is 0.480 inch while that of the clip angle is 2 × 0.375 = 0.75 inch (clip angles are on both sides of the beam), thus, the critical in bearing is the beam.

$P = \sigma_b \, A_b = 32 [ \, \frac{7}{8}(0.480)(4) \, ]$

$P = 53.76 \, \text{ kips}$

The allowable load on the connection is

$P = 53.76 \, \text{kips}$ *answer*

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