Thursday, 25 May 2017

SLOPE AND DEFLECTION OF BEAMS


Now we will start here, in this post, another important topic i.e. Deflection of beams and its various terms. If a beam will be loaded with point load or uniformly distributed load, beam will be bent or deflected from its initial position.

We will first discuss here the terminologies and various terms used in deflection of beam with the help of this post and after that we will see the amount by which beam will be deflected from its initial state and also slope of the deflected beam under the action of load in our next post.

 
So let us come to the main topic without wasting the time and let us see here various terms used in the concept of deflection of beam. Because without understanding the terms used in the deflection of beams, we will not be able to secure the amount by which beam will be deflected from its initial state and also slope of the deflected beam under the action of load.

Let us consider a beam AB of length L is simply supported at A and B as displayed in following figure. Let us think that one load P is acting at the midpoint of the beam. 
As beam is loaded with load P at the center of the beam and therefore beam will be subjected with bending moment and hence beam will be bent or deflected from its initial position under this load P.

Deflection

As we have seen above that beam is loaded with a load P at the midpoint of the beam and hence beam will be subjected with a bending moment and therefore beam will be bent in the form of a curvature or in to a circular arc.

We have shown the initial position of the beam i.e. before loading as ACB and deflected position of the beam i.e. after loading as ADB.

Now the vertical distance CD will be termed as the deflection of the beam under the action of the load P.

 

Let us define the deflection of the beam

Deflection of beam is basically defined as the vertical distance of beam measured before and after loading. Deflection of beam will be denoted by “y” and its unit of measurement will be mm.

Elastic curve

As we have seen above that beam is loaded with a load P at the midpoint of the beam and hence beam will be subjected with a bending moment and therefore beam will be bent in the form of a curvature or in to a circular arc.

Curvature or circular arc of the beam, formed under the action of load, will be termed as elastic curve.

Therefore, curvature ADB of the beam will be termed here as elastic curve.

Slope

Slope in deflection of beam will be defined as the angle turned by the beam under the action of load and hence under the action of bending moment.

 
Slope in deflection of beam will be basically defined as the angle made between the tangent drawn at the elastic curve and original axis of the beam.

Slope will be measured in radian and will be indicated by dy/dx or θ.

We will see the amount by which beam will be deflected from its initial state and also slope of the deflected beam under the action of load 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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Saturday, 20 May 2017

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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Friday, 19 May 2017

MIDDLE THIRD RULE FOR RECTANGULAR SECTION


Now we will be concentrated here a very important topic i.e. Middle third rule for rectangular section with the help of this post.

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 stress developed 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 that is Middle third rule for rectangular section

Let us consider a rectangular section of area A and of width b and Depth d as displayed in following figure. Let us consider that an eccentric load P is acting over the rectangular section with eccentricity e with respect to axis YY.
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 along the width of the column.
Let us analyze the above equation and we will conclude that in order to not develop any tensile stress at any point in the section along the width of the column, eccentricity of the load must be less than or equal to (b/6) with respect to axis YY.

Therefore we can say that if load will be applied with an eccentricity equal to or less than b/6 from the axis YY and on any side of the axis YY then there will not be any tensile stress developed in the column.

Hence range within which load could be applied without developing any tensile stress at any point of the section along the width of the column will be b/3 or middle third of the base.

Similarly in order to not develop any tensile stress at any point in the section along the depth of the column, eccentricity of the load must be less than or equal to (d/6) with respect to axis XX.

Therefore we can say that if load will be applied with an eccentricity equal to or less than d/6 from the axis XX and on any side of the axis XX then there will not be any tensile stress developed in the column.

 
Hence range within which load could be applied without developing any tensile stress at any point of the section along the depth of the column will be d/3 or middle third of the depth.

Now let us consider that load is eccentric with respect to axis XX and axis YY both, in this situation load must be applied anywhere within the rhombus ABCD whose diagonals AC = b/3 and BD= d/3 in order to not develop any tensile stress at any point in the column.

Area ABCD 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 discuss Middle quarter rule for circular section 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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Wednesday, 10 May 2017

DIFFERENCE BETWEEN DIRECT STRESS AND BENDING STRESS

In our previous topics, we were discussing the basic concept of direct stresses and bending stresses, where we have discussed that if a body is subjected to axial tensile or compressive loads and also to bending moments, in that situation there will be produced direct stress and bending stress in the body.

We have determined there the resultant stress when a column will be subjected with an eccentric load and load was eccentric with respect to single axis i.e. YY axis.

Today we will continue here with one very important topic in strength of material i.e. difference between direct stress and bending stress and also we will see here resultant stress when a column will be subjected with an eccentric load and load will be eccentric with respect to both axes i.e. XX and YY axis.

Let us go ahead step by step for easy understanding, however if there is any issue we can discuss it in comment box which is provided below this post.

 
Before going ahead, let us first brief here the basics of direct stresses and bending stresses for easy understanding and after that we will see here the determination of resultant stress when a column will be subjected with an eccentric load and load will be eccentric with respect to both axes i.e. XX and YY axis.

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.

Direct stress

Let us consider one column, as displayed in following figure, which is fixed at one end and let us apply one load P axially to the other end of the column. In simple, we can say that column will be subjected to compressive load and as load is applied axially, there will be developed direct compressive stress only in the body and intensity of this direct compressive stress will be uniform across the cross section of the column.
We have following information from above figure
P = Axial applied compressive load which is acting on the column through its axis
A= Area of cross section of the given column
Area of cross section of the given column = b x d, (For rectangular cross section)
σd = Direct compressive stress developed in the column due to axial applied compressive load
b= Width of the cross section of the given column
d= height or depth of the cross section of the given column

 
Direct compressive stress developed in the column = Axial applied compressive load/ Area of cross section of the given column

σd = P/ A
σd = P/ (b x d)
Unit of Direct compressive stress = N/mm2

Bending stress

Now let us consider one column, as displayed in following figure, which is fixed at one end and let us apply one load P to the other end of the column at a distance e from the axis of the column. In simple, we can say that column will be subjected to an eccentric load and line of action of this load will be at a distance e from the axis of the column. 
Distance between the axis of the column and line of action of load i.e. e will be termed as eccentricity of the load and such load will be termed as eccentric load. There will be produced direct stress and bending stress in the column due to this eccentric load.

Unit of bending stress = N/mm2

Recall the concept of bending stress and we will write here the expression for the bending stress developed in the body. 
Where,
I is the area moment of inertia  of the column rectangular section across the axis YY
I = db3/12
M = Moment formed by the load P
M= P x e 
P = Load applied with an eccentricity e
y = Distance of the point from neutral axis where bending stress is to be determined

Let us come to the main topic i.e. resultant stress when a column will be subjected with an eccentric load and load will be eccentric with respect to both axes i.e. XX and YY axis.

Let us see the following figure, where one column is subjected with a load which is eccentric with respect to XX axis and YY axis. We have following information from this figure as mentioned here.
P = Eccentric load applied on column
b = Width of the column
d = depth of the column

ex = Eccentricity of load with respect to XX axis
eY= Eccentricity of load with respect to YY axis 

σd = Direct stress = P/ (b x d)
σbX = Bending stress due to eccentricity of load with respect to XX axis
σbY = Bending stress due to eccentricity of load with respect to YY axis

MX, Moment of eccentric load about XX axis = P x ex
MY, Moment of eccentric load about YY axis = P x eY

IX, Moment of inertia about XX axis = bd3/12
IY, Moment of inertia about YY axis = db3/12

 
We will calculate here the direct stress, bending stress due to eccentricity of load with respect to XX axis and bending stress due to eccentricity of load with respect to YY axis and finally we will add these stresses algebraically to secure the resultant stress.

Direct stress

Direct stress is already calculated above and it is as mentioned here
σd = P/ (b x d)

Bending stress due to eccentricity of load with respect to XX axis

σbX = (MX . y )/IX
σbX = (P . ex . y )/IX
Value of IX is already determined above, bd3/12
Value of y will be in the range of (- d/2) to (+ d/2)

Bending stress due to eccentricity of load with respect to YY axis

σbY = (MX . x )/IY
σbY = (P . eY . x )/IY
Value of IY is already determined above, db3/12
Value of y will be in the range of (- b/2) to (+ b/2)

Resultant stress (σR) will be calculated as mentioned here

Let us determine the resultant stress at each point

Point A, resultant stress

As we can observe here that value of co-ordinates X and Y will be negative here and therefore resultant stress will be minimum at this point

Point B, resultant stress

As we can observe here that value of co-ordinate X will be positive and value of co-ordinate Y will be negative here and therefore resultant stress will be written as mentioned here
R)B = σd + σbX - σbY

Point C, resultant stress

As we can observe here that value of co-ordinates X and Y will be positive here and therefore resultant stress will be maximum at this point

Point D, resultant stress

As we can observe here that value of co-ordinate X will be negative and value of co-ordinate Y will be positive here and therefore resultant stress will be written as mentioned here
R)B = σd - σbX + σbY

Please comment your feedback and suggestions in comment box provided at the end of this post. We will discuss another topic in our next post.

Reference:

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

Also read

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Tuesday, 9 May 2017

CONCEPT OF DIRECT AND BENDING STRESSES

In our previous topics, we have seen some important concepts such as bending stress in beams, basic concept of shear force and bending moment, strain energy stored in body, beam bending equation, bending stress of composite beam, shear stress distribution diagram for various sections etc.

Today we will see here one very important topic in strength of material i.e. direct stresses and bending stresses with the help of this post. Let us go ahead step by step for easy understanding, however if there is any issue we can discuss it in comment box which is provided below this post.

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.

Here we will see this situation where a body will be subjected to axial loads and bending moments and we will analyze here the stresses developed in the body i.e. bending stresses and direct stresses.

 

Direct stress

Let us consider one column, as displayed in following figure, which is fixed at one end and let us apply one load P axially to the other end of the column. In simple, we can say that column will be subjected to compressive load and as load is applied axially, there will be developed direct compressive stress only in the body and intensity of this direct compressive stress will be uniform across the cross section of the column.
We have following information from above figure
P = Axial applied compressive load which is acting on the column through its axis
A= Area of cross section of the given column
Area of cross section of the given column = b x d, (For rectangular cross section)
σd = Direct compressive stress developed in the column due to axial applied compressive load
 b= Width of the cross section of the given column
d= height or depth of the cross section of the given column

Direct compressive stress developed in the column = Axial applied compressive load/ Area of cross section of the given column
σd = P/ A
σd = P/ (b x d)
Unit of Direct compressive stress = N/mm2

Bending stress

Now let us consider one column, as displayed in following figure, which is fixed at one end and let us apply one load P to the other end of the column at a distance e from the axis of the column. In simple, we can say that column will be subjected to an eccentric load and line of action of this load will be at a distance e from the axis of the column. 
Distance between the axis of the column and line of action of load i.e. e will be termed as eccentricity of the load and such load will be termed as eccentric load. There will be produced direct stress and bending stress in the column due to this eccentric load.
Unit of bending stress = N/mm2

Recall the concept of bending stress and we will write here the expression for the bending stress developed in the body. 
Where,
I is the area moment of inertia  of the column rectangular section across the axis YY

I = db3/12

M = Moment formed by the load P
M= P x e

P = Load applied with an eccentricity e

y = Distance of the point from neutral axis where bending stress is to be determined

 

Let us analyze the above concept and figure out here that there will be produced direct stress and bending stress in the column due to this eccentric load.

Let us draw again the column but with some variation. We have applied two equal and opposite axial load P as displayed in following figure. In simple, we have now three forces acting on the column. 
One load is eccentric load which is acting at a distance e from the axis of the column and two loads of same magnitude are acting along the axis of the column but in opposite direction i.e. one load is acting axially in downward direction and second load is also acting axially but in upward direction.

So, what we have done so far? 

We have applied two equal and opposite loads P axially on the column. As these loads are equal in magnitude and in opposite directions and hence they can cancel out each other and we have left with one eccentric load P which is acting with eccentricity e as displayed in above figure.

Let us see the third and fourth figure displayed above, what we are looking here?

We have shown there column with single load P which is acting axially downward in third figure and hence we can say that there will be developed direct compressive stress here due to load P. 

Similarly, we have also shown the rest two loads on column in fourth figure and as these force will form a moment or couple (P x e) and therefore there will be developed one bending stress in the column due to this moment.

Therefore 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 in the column by adding direct and bending stresses algebraically.

Resultant stress when a column will be subjected with an eccentric load

As we have assumed here that we have column with rectangular cross-section and therefore let us draw here the rectangular cross-section of the column. We will see here that on which faces of rectangular cross-section, total stress will be maximum and similarly on which faces of rectangular cross-section, total stress will be minimum. 
Simultaneously we will also secure the value of maximum stresses and minimum stresses developed in the column due to eccentric loading. We have seen above with formula for bending stress that bending stress σb will be dependent over the value of y. 

 
If point, where resultant stress is to be determined, is lying on the same side of axis YY as the load, we will consider the value of y as positive and bending stress will be of same type as direct stress.

If point, where resultant stress is to be determined, is lying on opposite side of axis YY as the load, we will consider the value of y as negative and bending stress will be of opposite type as direct stress.

Let us consider our case where load P is acting on column with eccentricity e with respect to YY axis as displayed above in figure. We will have four layers of the rectangular section here that is AB, BC, CD and DA. 

If we will analyze here the stresses at layers, we can easily write here that stress will be maximum along the layer BC and will be minimum along the layer AD.

We will consider here compressive stress as positive and tensile stress as negative and we will secure the value of maximum stress and minimum stress over the column rectangular section.

Maximum stress

As we have seen that maximum stress will be along the layer BC and we can write here the expression for maximum stress as mentioned here.

As point, where resultant stress is to be determined, is lying on the same side of axis YY as the load hence bending stress will be of same type as direct stress. Here in this case, direct stress is compressive in nature and therefore bending stress will also be compressive in nature.

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

Minimum stress

As we have seen that minimum stress will be along the layer AD and we can write here the expression for minimum stress as mentioned here.

As point, where resultant stress is to be determined, is lying on opposite side of axis YY as the load hence bending stress will be of opposite nature as direct stress. Here in this case, direct stress is compressive in nature and therefore bending stress will be tensile in nature.

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

We have already determined above the expression for direct stress (σd) and bending stress (σb) and hence by putting these values in above equation we will have the exact formula for maximum and minimum stresses.

Here we will have to analyse the above equation

If minimum stress σMin = 0, it indicates that there will be no stress along the layer AD
If minimum stress σMin = Negative, it indicates that there will be tensile stress along the layer AD
If minimum stress σMin = Positive, it indicates that there will be compressive stress along the layer AD

Please comment your feedback and suggestions in comment box provided at the end of this post. We will discuss another topic in our next post i.e. Resultant stress when a column will be subjected with an eccentric load and load will be eccentric with respect to both axes i.e. XX and YY axis.

Reference:

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

Also read

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