At Joint

*C*:

$\Sigma F_V = 0$

$BC = 96 \, \text{kN}$ (Tension)

Consider the section through member *BD*, *BE*, and *CE*:

$\Sigma M_A = 0$

$8(\frac{3}{5}BE) = 4(96)$

$BE = 80 \, \text{kN}$ (Compression)

For Member *BC*:

Based on shearing of rivets:

$BC = \tau A$ Where A = area of 1 rivet × number of rivets, n
$96\,000 = 70 [ \, \frac{1}{4} \pi (19^2) n \, ]$

$n = 4.8$ say 5 rivets

Based on bearing of member:

$BC = \sigma_b \, A_b$ Where *A*_{b} = rivet diameter × thickness of *BC* × *n* rivets

$96\,000 = 140 [ \, 19(6)n \, ]$

$n = 6.02$ say 7 rivets

Use 7 rivets for member *BC*. *answer*

For member *BE*:

Based on shearing of rivets:

$BE = \tau \, A$ Where *A* = area of 1 rivet × number of rivets, *n*
$80\,000 = 70 [ \, \frac{1}{4} \pi (19^2)n \, ]$

$n = 4.03$ say 5 rivets

Based on bearing of member:

$BE = \sigma_b \, A_b$ Where A_{b} = rivet diameter × thickness of BE × n rivets

$80\,000 = 140 [ \, 19(13)n \, ]$

$n = 2.3$ say 3 rivets

Use 5 rivets for member *BE*. *answer*

Relevant data from the table (Appendix B of textbook): *Properties of Equal Angle Sections: SI Units*

Designation |
Area |

L75 × 75 × 6 |
864 mm^{2} |

L75 × 75 × 13 |
1780 mm^{2} |

Tensile stress of member *BC* (L75 × 75 × 6):

$\sigma = \dfrac{P}{A} = \dfrac{96(1000)}{864 - 19(6)}$

$\sigma = 128 \, \text{Mpa}$ *answer*

Compressive stress of member *BE* (L75 × 75 × 13):

$\sigma = \dfrac{P}{A} = \dfrac{80(1000)}{1780}$

$\sigma = 44.94 \, \text{Mpa}$ *answer*

## Good day. I would like to ask

Good day. I would like to ask a clarification max tensile/compressive stress on the members.

Why did we use A=Area-dt for member BC, but for member BE, we just used the given area of 1780?

Maraming salamat po.

## The member BE is in

In reply to Good day. I would like to ask by Mang Johnny (not verified)

The member BE is in compression, thus the whole cross-section is effective. The load transfers through the main area + bearing area, as the fastener is pressed against the hole area. Imagine as if the fastener was "clenched" between two semicircles representing a hole halves. The compression load transfers through the fastener further.

The member BC is in tension. In static analysis this check is commonly referred as "Net Area Tension stress" and accounts for the holes area (and usually stress concentration, but it's omitted here), that needs to be excluded from a gross (total) area. In this scenario, the fastener is pressed against either side, but the opposite hole side extends away from it and deforms in oval shape (it is in tension, remember?). Thus this hole area doesn't contribute into the total strength (imagine if there were no fasteners at all, but only holes), so we exclude it. It acts as a perforated toilet paper :D