Q#16 (Past Exam Paper – June 2015 Paper 12 Q23)
The diagram represents a steel tube with wall thickness w which is small in comparison with the diameter of the tube.
The tube is under tension, caused by a force T, parallel to the axis of the tube. To reduce the stress in the material of the tube, it is proposed to thicken the wall.
The tube diameter and the tension being constant, which wall thickness gives half the stress?
A w / 2 B √2 w C 2w D 4w
Solution:
Answer: C.
Stress = Force / Area
The steel tube is hollow in the centre with its wall having thickness w. When we calculate stress, we need to consider the area of the steel ‘material’, not the hollow space.
Suppose the length of the tube is L and the diameter is d. Imagine that we have cut the steel tube (along the length) so that it is now in the form of a cuboid (a rectangular sheet with thickness w).
Area of annulus = (circumference of tube) × w
Area of annulus = 2π(d/2) × w = πdw
Area = πdw
Since π and d are constant, the area of the annulus is proportion to w.
Because the wall is thin in comparison with the diameter of the tube, the area of the annulus (representing the wall thickness) is proportional to w.
Stress = Force / Area
The tension (force) is kept constant.
So, the stress is inversely proportional to the area.
To obtain half the stress, the area should be doubled. But we have seen above that the area is directly proportional to w.
Thus, if the area is doubled, w also needs to be twice greater.
Post a Comment for " The diagram represents a steel tube with wall thickness w which is small in comparison with the diameter of the tube."