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Two steel plugs fit freely into the ends of steel tubular distance piece 400 mm long and are drawn together by a steel bolt (500 mm long) and nut, the nut being tight fit in the beginning. The nut is further tightened by 1/4 turn to draw the pieces together, the pitch of the bolt thread being 2 mm. The pieces A and B are then subjected to forces of 50 kN tending to pull them apart. Calculate the stresses in the bolt and the tube. The area of cross section of the bolt is 700 sq. mm and that of the tube is 500 sq. mm. E for steel = 200 GN/m2 .

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Let p1 (tensile) and p2 (compressive) be the final stresses in the bolt and tube respectively. 

For equilibrium,

(tensile force in 1) - (compressive force in 2) = external tensile force

\(\therefore\) p1A1 -  p2A2 = 50000

700p1 -  500p2 =  50000

p1 - 0.714p2 = 71.43  (1)

From compatibility,

(total extension of bolt) + (total compression of tube) = movement of nut

\(\Delta\)1 + \(\Delta\)2 = movement of nut

Solving (1) and (2)

p1 = 138.62 N/mm2 (tensile) and p2 = 94.1 N/mm2 (compressive)

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