Question 16
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
Reference: Past Exam Paper – June 2015 Paper 12 Q23
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.
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