Answer:
Normal stress = 66/62.84 = 1.05kips/in²
shearing stress = T/2 = 0.952/2 = 0.476 kips/in²
Explanation:
A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in
4 -in.-thick
angle of 25°
Axial force P = 66 kip axial force
determine the normal and shearing stresses
Normal stress б = force/area = P/A
= 66/ (П* (d₂²-d₁²)/4
=66/ (3.142* (12²-8²)/4
= 66/62.84 = 1.05kips/in²
Tangential stress T = force* cos ∅/area = P/A
= 66* cos 25/ (П* (d₂²-d₁²)/4
=59.82/ (3.142* (12²-8²)/4
= 59.82/62.84 = 0.952kips/in²
shearing stress = tangential stress /2
= T/2 = 0.952/2 = 0.476 kips/in²
Answer:
i dont understand, can you put it in word form please.
Explanation:
1.a source of ignition
2.fuel
3.oxygen
Explanation:
In first case, the forces on LHS and on RHS is the same i.e. 3 N. The force acting on the car is balanced force. As a result, the car will not move at all.
In second case,
Force on RHS = 2000 N
Force on LHS = -6000 N
Net force acting on it is given by :
F = 2000+(-6000)
= -4000 N
Hence, this is the required solution.
