Question

A M14 x 2 hexagonal head bolt is used to clamp together two 15 mm steel plates. Young's modulus of the bolt and the plates can be taken as 207 GPa. (i)-Determine a suitable length for the bolt. (ii)-Determine the bolt stiffness (iii)-Determine the stiffness of the members

Answer 1

(i) 50 mm

(ii) 874.62 m N/m

(iii) 3116.45 m N/m

Step-by-step explanation:

Given data

hexagonal head bolt = M14 x 2

steel plate = 15 mm

Young's modulus = 207 Gpa

Solution

1st part

dia of bolt (D) = 14 mm and height (H) = 12.8 mm from table

we know grip length = thickness of this plate i.e 30 mm , i.e (15mm+15mm)

so hexagonal bolt length = grip length + H

hexagonal bolt length = 30 mm + 12.8 mm = 42.80 mm i.e = 45 mm (round off)

2nd part

bolt stiffness =
`(Ad*At*young modulus)/(Ad*Lt*At*Ld)`

here Lt is length of thread = 2d +6mm

d is 14 so length of thread = 2*14 +6 = 34 mm

At from table 8-L i.e. = 115 mm2

Ad area of without thread part =
`\pi /4`×
`d^(2)`

Ad=
`\pi /4`×
`14^(2)` = 153.94 mm2

Ld is length of bolt without thread = length of bolt - Lt = 45 -34 = 11 mm

last Lt thread part length = length of bolt - Ld = 30-11 = 19 mm

put all these value in bolt stiffness i.e.

bolt stiffness =
`(153.94*115*207)/(153.94*19+115*11)` = 874.62

3rd part

stiffness of member =
`(0.5774 \pi  Ed)/(2 ln (5(0.5774L +0.5d)/(0.5774L +2.5d)))`

here l is 30 and d is 14

so

stiffness of member =
`(0.5774 \pi  (207) 14)/(2 ln (5(0.5774(30) +0.5(14))/(0.5774(30) +2.5(14))))`

stiffness of member = 3116.45 m N/m