Mechanics of Solids 232 Assignment 2012 °Show workings °Credit will be awarded for neatness Question 'I 60° 5 kN i 150 mm 200 mm P kN 10 mm 6 mm B Hinge support elevation 8 mm diameter pin Side view...

TTs260712_SSunS13470_2
Question 1:

Solution:
Take moment of all forces about B.
∑











Support reactions at B: VB and HB
∑
∑
√






Bearing stress is defined as the stress that results from the contact of two members. The equation
to determine average bearing stress would be b = F/A where F is the force applied to the
member, and A is the contact area of the object resisting the force.
For 6 mm thick
acket support, projected contact area becomes product of dia of pin and
thickness i.e. 8*6 = 48 mm
2
. And force taken by each support plate is RB/2 i.e. 4669.05 N.
Hence, bearing stress in a support plate = 4669.05/48 = 97.272 MPa. Ans.(iv)
For 10 mm thick crank, projected contact area becomes product of dia of pin and thickness i.e.
8*10 = 80 mm
2
. And force taken by each support plate is RB i.e. 9338.1 N.
Hence, bearing stress in crank = 9338.1 /80 = 116.726 MPa. Ans.(v)
Solution:
P = 1800 kPa = 1.8 MPa D = 1.2 m = 1200 mm t = 20 mm
In...