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MIDDLE QUARTER RULE FOR CIRCULAR SECTION


Now we will be concentrated here another very important topic i.e. Middle quarter rule for circular sections with the help of this post.

Cast iron and Cement concrete columns are weak under tensile load and therefore we must be sure that there should not be any tensile load anywhere in the section and hence load must be applied in such a way that there will be no tensile load in the section of cement concrete columns.

As we have discussed that when a body will be subjected with an axial tensile or axial compressive load, there will be produced only direct stress in the body. Similarly, when a body will be subjected to a bending moment there will be produced only bending stress in the body.

Now let us think that a body is subjected to axial tensile or compressive loads and also to bending moments, in this situation there will be produced direct stress and bending stress in the body.

If a column will be subjected with an eccentric load then there will be developed direct stresses and bending stresses too in the column and we will determine the resultant stress developed at any point in the column by adding direct and bending stresses algebraically.

Principle used

We will consider here compressive stress as positive and tensile stress as negative and we will have the value of resultant stress at any point in the column section. There will be maximum stress and minimum stress in the section of column as mentioned here.

σMax = Direct stress + Bending stress
σMax = σd + σb

σMin = Direct stress - Bending stress
σMin =  σd - σb

If minimum stress σMin = 0, it indicates that there will be no stress at the respective point in the section

If minimum stress σMin = Negative, it indicates that there will be tensile stress at the respective point in the section

If minimum stress σMin = Positive, it indicates that there will be compressive stress at the respective point in the section

Let us come to the main subject i.e. Middle quarter rule for circular sections

Let us consider a circular section of area A and of diameter d as displayed in following figure. Let us consider that an eccentric load P is acting over the circular section with eccentricity e with respect to axis YY.
Let us find here first direct stress and it could be written as displayed here in following figure.
Now we will determine here the bending stress and we can easily determine bending stress by considering the following steps as displayed here.
Minimum stress at any point in the section will be given by following formula as mentioned here.
As we have seen above the various conditions of minimum stress values and their importance and therefore we can easily say that minimum stress (σMin) must be greater or equal to zero for no tensile stress at any point and on any side of the centre of the circle.
Let us analyze the above equation and we will conclude that in order to not develop any tensile stress at any point and on any side of the centre of the circle, eccentricity of the load must be less than or equal to d/8.

Therefore we can say that if load will be applied with an eccentricity equal to or less than d/8 from the axis YY and on any side of the axis YY then there will not be any tensile stress developed in the circular section.

Similarly, we can also say that if load will be applied with an eccentricity equal to or less than d/8 from the axis XX and on any side of the axis XX then there will not be any tensile stress developed in the circular section.

Hence range within which load could be applied without developing any tensile stress at any point of the section will be d/4 or middle quarter of the main circular section.

Area of the circle of diameter d/4 within which load could be applied without developing any tensile stress at any point of the section will be termed as Kernel of the section.

We will start new topic in the category of strength of material in our next post.

Please comment your feedback and suggestions in comment box provided at the end of this post.

Reference:

Strength of material, By R. K. Bansal
Image Courtesy: Google

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