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Staggered Fasteners

In tension member connections with bolts, the net area is largest when fasteners are placed in a single line. However, space constraints often require multiple rows of bolts. To reduce the loss of cross-sectional area, bolts can be arranged in a staggered pattern, where holes do not line up directly. This arrangement ensures that any potential failure plane passes through fewer bolt holes, helping maintain strength and efficiency compared to non-staggered layouts.

Staggered Connection Figures

Concept

Cochrane's Formula for Staggered Connections with Uniform Thickness

Cochrane (1922) proposed that when deducting the area corresponding to a staggered hole, a reduced diameter d′ should be used, given by:

$$ d' = d_h - \frac{s^2}{4g} $$

The net width, bn, is then:

$$ b_n = b - \Sigma d' $$
$$ b_n = b - \Sigma \left( d_h - \frac{s^2}{4g} \right) $$

The net area, An, is:

$$ A_n = t(b_n) $$
$$ A_n = t \left( b - \Sigma d_h + \Sigma \frac{s^2}{4g} \right) \quad \text{for uniform thickness} $$

For any possible failure path, compute the net area and use the smallest value.

$$ A_n = A_g-\sum d_h t+\sum \frac{s^2}{4g}t $$ $$ A_e = U A_n $$

Where:
s = pitch (horizontal distance between adjacent bolts)
g = gauge (vertical distance between adjacent bolts)
dh = diameter of hole

Cochrane's Formula for Staggered Connections with Non-Uniform Thickness

The tension member below illustrates a connection in which holes are made on different elements with different thicknesses.

Concept The equation of the net area becomes:
$$ A_n = A_g - t_1(d_{h1}) - t_2 \left( d_{h2}-\frac{s_1^2}{4g_1} \right) - t_3 \left( d_{h3}-\frac{s_2^2}{4g_2} \right) $$
Concept Concept Concept Concept Concept Concept Concept
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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q48

PSAD - Steel Design / Staggered Bolted Connections / Engr. Janclyde Espinosa (Clidez)

A plate having a width of 400mm and thickness of 12mm is to be connected with another plate by 34-mm⌀ bolts as shown. Given the following:
s1=100mm
s2=60mm
s3=150mm
Fy=248MPa
Fu=400MPa
Bolt hole diameter = 36mm
Strength reduction factor:
For gross area yielding, ϕ=0.9
For net area rupture, ϕ=0.75

q48

Determine the value of b, in mm, so that the net width along bolts 1-2-3-4 is equal to the net width along bolts 1-2-4.

  1. 19.71
  2. 18.43
  3. 20.52
  4. 21.64

Determine the ultimate load P, in kN, based on the rupture of the net area.

  1. 1160
  2. 1260
  3. 1680
  4. 1547

Determine the ultimate load P, in kN, that can be safely applied to the plate.

  1. 1071.36
  2. 1190.4
  3. 1428.48
  4. 1285.63
For each staggered path, use $b_n=b-\sum d_h+\sum(s^2/4g)$. Set the net width for path 1-2-3-4 equal to that for path 1-2-4.
Part 1. Solving the net-width equality gives $b=\boxed{19.71\text{ mm}}$.
Part 2. The controlling net area gives $\phi P_n=0.75F_uA_n=\boxed{1160\text{ kN}}$ for rupture.
Part 3. Gross yielding gives $\phi P_n=0.9F_yA_g=0.9(248)(400)(12)/1000=\boxed{1071.36\text{ kN}}$. This is lower than the rupture strength and therefore governs.