T01: Net Areas (with units)

from utils import show, figure

Example T01u: Net Areas of Plates with Staggered Holes

Determine the net areas to be used when finding the factored tension resistance of the following plate lap joint. The net areas may be different for the central (20 mm) and side (10 mm) plates, so determine both.

figure("lap-plates-1.svg")

Lap Plate Connection

Please note that this type of hole pattern rarely occurs in practice -- practical patterns are more regular and 'grid-like'. This example illustrates:

how failure paths depend on the direction of the load relative to the hole group.
the calculations necessary to determine a net cross-sectional area for each potential failure path.

The figure shows an irregular bolt pattern in a lap tension splice. To determine the net areas of the plates, we must examine every possible failure path that has the following attributes:

it separates each plate into 2 complete parts.
it is of minimum length for that path.
there are no bolts or holes completely on the loaded side of the path; all of the bolt bearing areas are on the side opposite the load.

We then determine the area from the path with minimum width.

In this example, we will assume M20 bolts in punched holes, and thus the hole allowance to be used is $d = 20~\mathrm{mm}+2~\mathrm{mm}+2~\mathrm{mm} = 24~\mathrm{mm}$.

Compute with units

import pint                   # setup to use the module for computing with units
ureg = pint.UnitRegistry()
mm = ureg['mm']
ureg.default_format = '~P'

# if you do not want to use units, simply uncomment the following line:
# mm = 1

The 10mm Plates (outer plates)

The shows the paths appropriate for investigating the strength of the outside (10mm) plates. For this case, the loaded side of the connection is toward the right side, and so there are no complete holes on that side of any path.

figure("paths-1i.svg")

Failure Paths for Net Area Calculations, Outside (10mm) plates

Define the Data:

d = (20 + 2 + 2) * mm       # hole allowance: bolt dia. + 2mm clearance + 2mm for punching
s1 = 50 * mm
s2 = 55 * mm
s3 = 50 * mm
g1,g2,g3,g4,g5 = (35,50,45,50,30) * mm
t1 = 10 * mm                # thickness of one outside plate
t2 = 20 * mm                # thickness of inside plate
wg = g1+g2+g3+g4+g5
%show wg                         # gross width of plate

wg = 210 mm

In the following, we will compute the net width, $w_{ne}$, for each failure path, then use the minimum so computed to determine the cross-sectional area.

Path 1-1:

For path 1-1, it is only necessary to deduct the allowance for one hole from the gross width. In general, if no path segments are inclined to the load:

$w_{ne} = w_g - \sum d$

wne_11 = wg - d
%show wne_11

wne_11 = 186 mm

Path 2-2:

For paths with segments inclined to the load, we subtract all hole allowances, $d$, then add the $s^2/4g$ correction term for each inclined segment:

$w_{ne} = w_g - \sum d + \sum{s^2\over 4g}$

wne_22 = wg - 2*d + s3**2/(4*g3)
%show wne_22

wne_22 = 175.9 mm

Path 3-3:

$w_{ne} = w - \sum d + \sum{s^2\over 4g}$

wne_33 = wg - 3*d + s3**2/(4*g3) + s2**2/(4*g2)
%show wne_33

wne_33 = 167 mm

Paths 1-4, 2-4 and 3-4:

Adding the fourth bolt to each of the above paths will reduce the net width by $24~\mathrm{mm}$ for the hole, then increase it by $(s2+s3)^2/(4 g4)$ for the inclined segment. Paths that include this hole will not govern if the term for the inclined segment is less than 24. However, given that these variable values might change, its probably safest to compute them all.

delta = (s2+s3)**2/(4*g4) - d    # the amount wn increases by including the hole on path 4
wne_14 = wne_11 + delta
wne_24 = wne_22 + delta
wne_34 = wne_33 + delta
%show delta, wne_14, wne_24, wne_34

delta  = 31.12 mm
wne_14 = 217.1 mm
wne_24 = 207   mm
wne_34 = 198.1 mm

Summary

The path with the smallest $w_n$ governs, so $A_n$ for the pair of 10 mm plates is:

wne = min(wne_11,wne_22,wne_33,wne_14,wne_24,wne_34)
%show wne

wne = 167 mm

An_10 = wne * t1*2    # because there are 2 thicknesses of 10 mm plate
%show An_10

An_10 = 3340 mm²

20mm Plate (Inner Plate)

The following figure shows the possible failure paths for calculating the strength of the 20mm plate. For this case, the loaded side is toward the left.

figure("paths-2i.svg")

Failure Paths for Net Area Calculations, Inside (20mm) plate

Path 1-1:

wne_11 = wg - d
%show wne_11

wne_11 = 186 mm

Path 2-2:

wne_22 = wg - 2*d + s1**2/(4*g2)
%show wne_22

wne_22 = 174.5 mm

Path 2-3:

wne_23 = wg - 3*d + s1**2/(4*g2) + s1**2/(4*(g3+g4))
%show wne_23

wne_23 = 157.1 mm

Path 1-3:

By inspection this should not govern, as path 1-1 is longer than 2-2, therefore 1-3 will be longer than 2-3.

wne_13 = wg - 2*d + s1**2/(4*(g3+g4))
%show wne_13

wne_13 = 168.6 mm

Summary

The shortest path (minimum $w_n$) governs, and so the net area, $A_n$, of the 20 mm plate is:

wne = min(wne_11,wne_22,wne_23,wne_13)
%show wne

wne = 157.1 mm

An_20 = wne * t2
%show An_20

An_20 = 3142 mm²

The net area fracture strength of the plates will therefore be governed by the net area of the inner 20mm plate.

%show An_10, An_20

An_10 = 3340 mm²
An_20 = 3142 mm²