A ten foot pole is dropped into a milling saw and randomly cut into three shorter poles. What is the probability that these three pieces will form a triangle?

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