Civil Engineering Reference
In-Depth Information
F a ¼ 256 24 : 53
ð
Þ¼ 6279 : 68N
Hence:
N net = F c -6279.68
f = 0.3 9 (F c -6279.68)
So,
0.3 9 (F c -6279.68) [ 11847.68
Leading to: F c [ 45771.95 N
This value gives the minimum total longitudinal compressive force to be
generated by the eight studs. Therefore, applying a preload tension force in each
stud equal to 6,000 N can be considered a suitable amount to prevent lateral
sliding in all cases. Fastening studs are made of titanium alloy VT-16, which is the
suitable material to manufacture a precise long rod with a relatively small diameter
and minimum creep requirements. The manufacturing process applies several
constraints on stud fabrication, and the selection of stud diameter is affected by
these restrictions. The eight studs are identically 6 mm diameter with external
thread M6 at both ends with 20 mm depth. The design stress (r) is calculated using
the formula:
r ¼ F
A
where A is the stud cross-section area. This gives:
r ¼ 212 : 2 MPa
The yield margin of safety (MS y ) is calculated from the formula:
MS y ¼ Allowable yield stress ð r Y Þ
Design yield stress ð r Þ
1
For
titanium
alloy,
VT-16
the
allowable
yield
stress
(r y )
is
903 MPa.
Therefore,
MSy ¼ 3 : 25
The stud elongation (d) due to preload (F) can be computed using the formula:
d ¼ Fl
AE
where: (l) is the length of stud, and (E) is the modulus of elasticity of titanium
alloy VT-16. The length of the active part of the stud is equal to 1,030 mm from
the preliminary configuration of Small Sat. Therefore,
d ¼ 2mm
The torque (T) applied to each stud to achieve the required pretension force is
calculated using the formula:
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