The bolts that attach a bracket to an industrial machine must each carry a static tensile load of 4 kN.
(a) Using a safety factor of 5 (based on proof strength), what size of class 5.8 coarse‐thread metric bolts is required?
(b) If the nuts are made of a steel with two‐thirds the yield strength and proof strength of the bolt steel, what is the least number of threads that must be engaged for the thread shear strength to be equal to the bolt tensile strength?

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

answer:

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