A steel plate subjected to a force of 5 kN and fixed to a channel by means of three identical bolts is shown in Fig. The bolts are made of plain carbon steel 30C8 (Syt = 400 N/mm2) and the factor of safety is 3. Determine the diameter of the bolt.
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Answer:
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Given:- Force= 5kN
factor of safety = 3
To find:- size of bolt
Solution:-
Step 1:
Permissible shear stress is given by
τ=Ssy/fs
Substituting the given set of values into the above equation we get,
=0.5*Syt/fs
=0.5∗380/3
=63.33N/mm2
Step 2:
Primary and secondary shear forces: The centre of gravity of three bolts will be at the centre of bolt-2
P′1=P′2=P′3
=P/3
=50003
=1666.67N
P′′1=P′′3=Pe∗r1/r1^2+r3^2
=5000∗305∗75/75^2+75^2
=10166.67N
Step 3: Resultant shear force.
The resultant shear force on the bolt 3 is maximum.
P3=P′3+P′′3=1666.67+10166.67=11833.34N
Step 4: Size of bolts:
τ=P3/A or 63.33=11833.34/π4dc^2
Calculating the above equation we get the values.
dc=15.42mm
Also,
d=dc/0.8 = 15.42/0.8
=19.28 or 20mm
The diameter of the bolts is 15.42mm and the size is M20.