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*

## Comments

## 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.