A condenser accepts steam from the turbine in problem 2 at a pressure of 2.34 kPa. Saturated water at the same pressure leaves t
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Answer:
M10 × 1.5
least number of threads is 4.3
Explanation:
given data
static tensile load = 4 kN = 4000 N
safety factor = 5
coarse‐thread metric = 5.8
solution
we know that proof load for 5.8 class is Sp = 380 MPa
so area of cross section is express as
area = \frac{force \times FOS}{Sp} ......................1
area = \frac{4000 \times }{380}
area = 52.6 mm²
so by table at 58 mm² we can say we resist M10 × 1.5
and
bolt tensile strength is express as
bolt tensile strength = nut shear strength ......2
so here bolt tensile strength = At × Sy (bolt) ...............3
and nut shear strength = πd ( 0.75 t ) Sys
nut shear strength = π × 10 × ( 0.75 t ) × 0.58 × × Sy(bolt) ............4
so from equation 3 and 4 we get
t = 6.37 mm
and
for the pitch is = 1.5 mm
least number of threads is 4.3