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*

Bakit hinde ni minus ang area sa may BE katulad nang ginawa sa BC na 864-19(6)?