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 Ab = 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 Ab = 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 mm2 |
| L75 × 75 × 13 |
1780 mm2 |
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.