Respuesta :
Answer:
(i) 50 mm
(ii) 874.62 m N/m
(iii) 3116.45 m N/m
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 = [tex]\frac{Ad*At*young modulus}{Ad*Lt*At*Ld}[/tex]
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 = [tex]\pi /4[/tex]×[tex]d^{2}[/tex]
Ad= [tex]\pi /4[/tex]×[tex]14^{2}[/tex] = 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 = [tex]\frac{153.94*115*207}{153.94*19+115*11}[/tex] = 874.62
3rd part
stiffness of member = [tex]\frac{0.5774 \pi Ed}{2 ln (5\frac{0.5774L +0.5d}{0.5774L +2.5d})}[/tex]
here l is 30 and d is 14
so
stiffness of member = [tex]\frac{0.5774 \pi (207) 14}{2 ln (5\frac{0.5774(30) +0.5(14)}{0.5774(30) +2.5(14)})}[/tex]
stiffness of member = 3116.45 m N/m
