Chapter 3 (cont..2) Compressive force contributed by parabolic portion

In the previous section we derived the complete details about the rectangular portion. In this section we will see the parabolic portion. The parabolic portion has an area of 0.447f_ck (8/21)x_u (see derivation here). This area multiplied by the width 'b' of the section will give the volume of the parabolic stress block. And this volume is the compressive force C₂ contributed by the parabolic portion. So we get

Eq.3.4

C₂ = 0.447f_ck b (8/21)x_u

The distance of the centroid of this block (the point of application of C₂) from the top most compression fibre = (9/14)x_u

Taking moments of the forces C₁ and C₂ about the top most compression fibre, we get:

Eq.3.5

Where

Eq.3.6

C_u = C₁ + C₂ , the resultant compressive force, and, 

x₃ is the point of application of C_u from the top most compression fibre.

Substituting 3.3 and 3.4 in 3.6 we get

C_u = 0.447f_ck b (3/7)x_u + 0.447f_ck b (8/21)x_u

⇒ C_u  = 0.447f_ck b (17/21)x_u

⇒ Eq.3.7

C_u  = 0.362 f_ck b x_u

It may be noted that the above eq.3.7 is given in another form in the assumption c of cl.38.1 co the code. There, the area of the stress block is given as 0.362 f_ck x_u . Multiplying this area by b, we get the volume, which is same as eq.3.7

Now we will derive an expression for x₃:

Substituting 3.3, 3.4 and 3.7 in 3.5 we get:

0.447f_ckb(3/7)x_u(3/14)x_u+ 0.447f_ckb(8/21)x_u(9/14)x_u  = 0.447f_ck b (8/21)x_u x₃

⇒ 0.447f_ck b (99/21x14)x²_u  = 0.447f_ck b (17/21)x_u x₃

⇒ (99/14)x_u  = 17 x₃

⇒ Eq.3.8

x₃  = (99/238)x_u  = 0.416x_u

This is the same expression for 'depth of center of compressive force from the extreme fibre in compression' in assumption c of the code.

So we will get the lever arm z as:

z = d - 0.416x_u 

Thus we derived all the expressions required for the calculations of the quantities above the NA. Fig.3.12 below shows these details:

Fig.3.12
Complete details of stress block above NA

Stress-Strain curve for steel

Note that the quantity x_u (the depth of NA) is still an unknown. We have not derived an expression for it yet. We will now discuss about the steel. We want to know how the steel will be behaving at the point of impending failure. While we discuss about it, we will get the expression for x_u also.

We want the stress in steel. We must calculate it using the graph in fig.23 of the code. So let us first discuss about fig.23. Fig.23A gives the stress strain curve for cold worked bars like Fe415 and Fe500. And Fig.23B gives the stress strain curve for mild steel like Fe250.

These graphs are plotted by conducting tension tests on steel. Strains are plotted along the X-axis, and stresses are plotted along the Y-axis. For mild steel, there is a definite yield point. That is., there is a particular stress after which the steel will begin to yield. So after this point the steel will continue to elongate at constant stress. Thus the graph after this point will be horizontal. The stress at the yield point is taken as the characteristic strength f_y of steel. We have to apply the partial safety factor to this characteristic value to get the 'design value'. So we divide f_y by 1.15 giving f_y/1.15 = 0.87f_y. The lower curve is obtained by dividing all the values by 1.15, and so, it is the design curve. If we know the strain in steel, we can calculate the design value of stress in that steel by using the design curve.

As we did in the case of concrete, we can now note down the effect of reducing the stress in steel from f_y to 0.87f_y: In an analysis problem, the steel is considered to have yielded if the stress in it is 0.87f_y. If it was f_y instead of 0.87f_y, we can expect the steel to take more load before it yields.  

In a design problem, we can apply only that load which cause a stress of 0.87f_y. If it was f_y instead of 0.87f_y, we can apply a greater load. But that is not the case. We must consider only 0.87f_y.  

For cold worked bars, there is no definite yield point. That is., there is no particular stress after which the steel will elongate at constant stress. So the change from the inclined graph to the horizontal graph do not take place at a particular point. Instead, the change takes place gradually through some distance. Because of this, it is not easy to point out a particular value as the characteristic strength. So we follow the procedure given by the code to obtain the characteristic strength from the graph.

The code specifies that the 'stress corresponding to 0.002 strain offset' should be taken as the characteristic strength f_y. This can be explained as follows: Consider a tensile test being conducted on a steel specimen. When the applied stress increases, the strain also increases. We plot the strain along the X axis and the stress along the Y axis. Initially this graph is a straight line. The slope of this line = Stress/Strain = E_s, the modulus of elasticity of steel. But as the stress increases, it will begin to take a curved shape. With further increase of stress, the graph will become horizontal. After becoming horizontal, it will continue to elongate at constant stress. So when we reach the horizontal portion, we can be sure that the yielding have begun. But, because of the curved region between the inclined and horizontal graphs, we cannot find a particular value of stress at which the yielding began.

To solve this, we make use of a special property of steel: The property of 'giving a residual strain after yielding'. That is., while doing the tension test, if we release the load while in the initial inclined region, the steel will contract back to it's original length. This is because, the initial inclined region is the 'elastic' region. But after yielding, if the load is released, the steel will not attain it's original length. There will be a residual strain. So it follows that, if we release the load at a particular value of stress, and if on release, a residual strain is obtained, we can be sure that the steel has yielded. But still, we do not know the particular value of stress upon the release of which, the steel will begin to give residual strains. Also, the residual strain can be any value. It can be 0.001, 0.002, 0.0025, or any similar value. Whatever be the value, if there is a residual strain, we can be sure that yielding has begun. In order that all of us use a same standard value, the code specifies the required value of residual strain as 0.002. So we want the 'value of the stress', upon the release of which, the steel will give a residual strain of 0.002.

This can be easily determined from the stress strain curve of steel. We continue the test as usual. There is no need for unloading at any point. After plotting the graph, we draw a line parallel to the initial inclined line. And this parallel line should pass through the 0.002 point on the X-axis. The point of intersection of this line with the orginal graph gives the 'characteristic yield strength' f_y of steel. As usual, we divide all the stress values by the partial safety factor 1.15, to get the design curve. The point of intersection of the parallel line with the design curve will give the 'design yield strength' of steel.

For analysis and design purposes, we can use the graph in fig.23A of the code. We will try to plot a design curve on our own. While trying to plot it, we will understand it's salient features. Let us plot the design curve for Fe415 steel.

We know that the initial inclined portion and the final horizontal portions are straight lines. So they can be easily plotted. For the inclined line, any two points on it is sufficient, and for the horizontal portion, any one point on it is sufficient. The curved portion is the difficult part. It requires more points. The fig.23A gives us these points. We can see a dashed line parallel to the initial inclined line, starting from 0.002 on the X axis. It meets the characteristic curve at a certain point. A horizontal dashed line is drawn from this meeting point towards the left up to the Y axis. At the point of intersection of the horizontal dashed line and the Y axis is marked: 'f_y'. This means that the 'y coordinate of the point of intersection' of the inclined dashed line with the characteristic curve is f_y. (It will indeed be f_y, the characteristic yield strength, because 0.002 is the 'proof strain' as discussed earlier). So the y coordinate of the corresponding point on the design curve will be equal to f_y/1.15 = 415/1.15 = 360.9 N/mm². Now we want the x coordinate. For this we will use the fig.3.13 given below:

Fig.3.13
Last point of the curved portion

In the fig., the initial inclined line is drawn in red colour, the final horizontal line is drawn in magenta colour, and the intermdiate curve is drawn in blue colour. A triangle PQR is drawn. PR corresponds to the inclined dashed line in the code, drawn from the strain value of 0.002. So we can see that the x coordinate will be equal to 0.002 + PQ. Thus our next step is to calculate PQ.

In the ΔPQR, tan∠RPQ = ^RQ⁄_PQ  

⇒ PQ = ^RQ⁄_tan∠RPQ    

But tan∠RPQ is the slope of PR. So it is also the slope of the initial inclined line of the design curve. 

But the slope of this initial inclined line = 

^Stress⁄_Strain   = Elastic modulus of steel = E_s = 2 x 10⁵ N/mm².

We know that RQ = the y coordinate of the point R, which we already calculated as 360.9 

Thus we get PQ = ^360.9⁄₂₀₀₀₀₀ = 0.0018045

So the x coordinate of point R = 0.002 + 0.0018045 = 0.0038045 = 0.00380

Thus the coordinates of the last point of the curved portion are (0.00380,360.9)

This is for Fe415 steel. For the general case, for the last point of the curved portion, we can write the following:

Eq.3.9

x coordinate = 0.002 + [f_y/ 1.15 x 200000] = 0.002 + f_y/230000

y coordinate = f_y/ 1.15

The above discussion was based on fig.3.13 above. It is a complete fig. with all the required points of the design curve of Fe415 steel. Such a complete fig. is used for discussion purpose only. Based on that discussion, we have derived the general form in Eq.3.9. So, on a new empty graph paper, with only the axes drawn to proper scale, we can directly mark the last point of the curved portion using Eq.3.9. This is shown in the animation below:

This is also the first point of the final horizontal portion. As the portion is horizontal, it requires only one point to plot. So with this just one point, the calculations related to the final horizontal line is complete.

Now we will calculate the second last point of the curved portion. We will do this in the next section.

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Chapter 3 (cont..1) Compressive force contributed by rectangular portion

In the previous section we saw the shape of the concrete stress block. In this section, we will discuss the compressive force produced by this block. The fig.3.4 below shows the elevation view of the beam with the stress block attached to it. 

Fig.3.4
Elevation view showing the Stress block attached to the beam

At the first glance, we might get the impression that the stresses are plotted on the X-axis, and distances along the beam section are plotted on the Y-axis. But this is not true. The graph given by the code is a Stress-Strain curve. Not a Stress-Distance curve. So we need a method to represent strain in terms of the distance. This can be made more clear using the following fig.3.5:

Fig.3.5
Relation between strain and distance

In the above fig., we can see that the strain is plotted as equal to the vertical distance, even though they are two different quantities. This is made possible by selecting a suitable scale for plotting the strain along the vertical axis. For selecting the scale, we make use of the 'manner in which the strain variation occurs in a beam section'. So we will have a brief discussion about this variation:

A beam will consist of numerous planes which are perpendicular to the axis of the beam. We need to know about the behaviour of these planes When the bending of the beam occurs. The X-ray views in fig.3.6 below shows some of the planes inside the beam.

Fig.3.6
View of some planes inside the beam

When the beam bends, our first impression would be that, the planes will bend into curved shapes as shown in fig.3.6(b). If a plane bend into a curved shape, it can no longer be called a 'plane' because, such a curved surface will have 3 dimensions. But a real plane have only 2 dimensions: length and width. It has no thickness, height, or curvature. In our case, all the plane sections which are perpendicular to the axis of bending, remain plane even after bending. This is shown in fig.3.6(c). An elevation view of the beam after bending is shown in the fig.3.7 below:

Fig.3.7
Elevation view of the planes after bending

This property of the beam makes our calculations a lot more easier. It may be noted that this property does not change even at very high stresses, when the beam section is at the point of impending failure. The fig.3.8 below shows the end of a section coloured in red. AB is the original edge, and PQ is the new edge. PQ is a straight line because, AB will remain plane even after bending.

Fig.3.8
End portion of a segment

We can see that the change in length Δ l of any fibre is proportional to the distance of the fibre from the NA. If Δ l is proportional to the distance, the strain (strain = Δ l / l ) will also be proportional to the distance. This proportionality would not be true if PQ is a curve. Also it is a direct proportion. That is., when distance increases, strain also increases and vice versa. So we can represent it using the equation: Strain = A constant x distance. Thus while plotting the graph, we can select a scale for the Y-axis in such a way that the strain and distance will be the same. 

Thus we see that the stress block of concrete can be attached to the beam section as shown in the fig.3.4 above. In fact, we do not have to plot this stress block along the beam section. We only need to understand the 'variation of stress' (ie., the stress at various points) along the section. The above discussion was to make clear how the 'stress-strain curve' given in the code appear like a 'stress-distance' curve in the fig.3.4.

Also, from the above discussion, we see that strain is directly proportional to distance. So the strain diagram will be a straight line.  

We can now proceed to calculate the area enclosed by the graph. That is., the total area of blue portion and red portion in fig.3.4. The fig.3.2 and 3.3 that we saw in the previous section are for f_ck =20 N/mm². We will derive the area for the general purpose where f_ck can take any value.

We must find the areas of the rectangular portion and parabolic portion separately. We know the length of the rectangular portion. It is equal to 0.447f_ck . But we do not know the height. To find the height, we make use of the same 'variation of strains' in the beam that we discussed above.

We have seen that Strain = A constant x distance. So we can represent the variation of strain with distance by a straight line. Also, if we know the distance of any fibre from the NA, we can easily calculate the strain in that fibre. We will use this property in reverse to find the height of our rectangle. That is., we know the strain at the beginning of the rectangular portion. We want to find out the distance at which this strain occurs. For this, we use the fig.3.9 below:

Fig.3.9
Distance at which the strain of 0.002 occur

In the fig., PQ is a straight line and it represents the strain across the section. We have two triangles above the NA.
• Base of the larger triangle = 0.0035. 
• Base of the smaller triangle = 0.002. 
• Altitude of the larger triangle = x_u. 
• Altitude of the smaller triangle = x₂. 

So we have
0.0035/0.002 = x_u/x₂. 
⇒ x₂ = (2/3.5) x_u. 
x₁ = x_u –  x₂ = x_u – (2/3.5) x_u 
⇒
Eq.3.2
x₁ = (3/7)x_u

Thus we get the height of the rectangular portion. We can add this information to the fig.3.9 and get the modified fig.3.10 as shown below:

Fig.3.10
Ht. of rectangular portion in relation to the strain diagram

Knowing the length and height, we can find the area of the rectangular portion, which is equal to 0.447f_ck(3/7)x_u. This area multiplied by the width 'b' of the section will give the volume of the rectangular stress block. This volume is the compressive force C₁ contributed by the rectangular portion. So we get

Eq.3.3

C₁ = 0.447f_ckb(3/7)x_u 

This force acts at the centroid of the rectangular block. The centroid will be at a distance of  x₁/2 from the top surface of the beam. 

Substituting for x₁ from 3.2, we can write:

The point of application of C₁ from the top most compression fibre = (3/14)x_u

So now we have all the details about the rectangular portion. These details are shown in the fig.3.11 below:

Fig.3.11
Forces and distances related to the rectangular portion

In the above figs.3.10 and 3.11, a quantity ε_st is shown in the strains diagram. This is the strain in the reinforcing steel. The yellow arrow represents the stress in the steel. We will learn about them when we take up the discussion on steel. In our present discussion, these two quantities do not come in any calculations.

In the next section we will discuss about the parabolic portion.

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Chapter 3 - Analysis of Beam sections by Limit State Method

In the previous section we saw the analysis of sections by the working stress method. There we saw that after the end of phase 2, the stresses are higher and 'non-linear'. We discussed it based on fig.2.13 Here we are going to discuss the analysis at ultimate state. That is., the state when the failure of a section is impending. To know the behaviour of the section at the point of impending failure, we must load the section up to that point, and then analyse the results. As the materials reach the state of impending failure, the stresses will be very high. So the stress distribution will be very much non-linear. In other words, the stress distribution graphs will be curves.

We require such a graph (graph which gives the actual non linear variation of stress with strain) for analysis and design purposes. The purpose of the graph is: To find the area enclosed by the graph and then multiply the area with the width of the section (to get the volume of the stress block), and this volume will give the magnitude of the force. To obtain the graph, compression tests are done on concrete specimens at the lab, and in these tests, the concrete is compressed to the ultimate state. Based on these test readings, we can plot the graph which gives the stress at various strains. Such a plot is given in fig.21 of the code. In this fig., the top most curve shows the results (f_ck) obtained from the tests conducted in the lab. But the strength of concrete in an actual structure may not be as good as that obtained in the lab. So only 67 percent of f_ck is taken. This is shown in the middle curve. Then we have to apply the partial safety factor for the material, which is concrete here. As seen before in chapter 1, it's value is 1.5. So we get 0.67f_ck/1.5 = 0.447f_ck. Each of the y coordinate in the topmost curve is multiplied by 0.447 to get the corresponding y coordinate of the bottom most curve.

The 'effect' of reducing the value of f_ck to 0.447f_ck can be explained as follows: When we analyse a beam section by the limit state method, we are analysing it at the ultimate state. That is., we are trying to find the magnitude of the forces in the section when it is at the state of impending failure. According to the code, that state of impending failure is reached, if the stress in concrete is 0.447f_ck. We cannot take f_ck instead of 0.447f_ck. So the stress that concrete can be subjected to, is reduced from f_ck to 0.447f_ck. In effect, the contribution that can be expected from concrete is reduced. 

The reduction from f_ck to 0.447f_ck has effect on design also. When we are designing a new section, we must expect the concrete to take a stress of only 0.447f_ck. 

The maximum stress value of 0.447f_ck corresponds to the maximum strain of 0.0035. For the fibres with lower strains, the stress will also be lower. The bottom most curve in fig.21 is the curve in which the factor 0.447 has been applied to all it's points. So from that curve, we get the required stress for the fibres with lower strains also. Thus, it is the bottom most curve that we must use in analysis and design.

So, in the Limit state method, we are considering sections at their point of impending failure, and at this point, the stress distribution in concrete is given by the graph in fig.21 of the code. And for analysis and design purposes, we use the lower most graph in that fig.

Let us now examine the various features of this curve. We can see that the curve has two portions. A parabolic portion, and a straight line portion. The parabolic portion starts from the origin (0,0) where stress = 0, and so strain is also equal to 0. From there, the stress begins to increase with strain. This increase is parabolic. The increase of stress continues upto the point where strain reaches 0.002. The maximum value of stress at this point is 0.447f_ck. When the strain exceeds 0.002, the stress remains constant at 0.447f_ck. The strain may continue to increase, but the stress will remain constant. However, the strain cannot increase indefenitely. When it reaches 0.0035, the code assumes that the concrete has reached the point of impending failure. So after this point, the concrete will fail by crushing. In a beam section, the topmost concrete fibres will be the ones that reach this strain first. The lower fibres will be having lower strains. But as the topmost fibres have reached the ultimate state, the whole section should be considered to have reached the ultimate state. This is because, the section will serve it's purpose only if strains in all the fibres in it are below the specified limits.

In a beam section which is subjected to a sagging moment, the Neutral axis NA of the beam, corresponds to the origin point (0,0) in fig.21. This is because, at the NA, both stress and strains are equal to zero. If this sagging moment is of such a magnitude that, the section is at the ultimate state, then, we can say that the stress variation in the section is given by the curve in fig.21 of the code. So, from the NA, as we move up, the stress will vary parabolically upto the level where strain equals 0.002, and then it will remain constant at 0.447f_ck. At the top most level of the beam section, the strain is equal to 0.0035. 

So now we know that the bottom most curve is our required graph. We must plot it from the NA towards the upper part of the beam section. But we already know two important points on the graph. Origin point (0,0) at the NA, and the final point (0.447f_ck, 0.0035) at the top edge. So there is no need to plot it. We can take it and directly attach it to the beam section. We attach it in such a way that the origin (0,0) is at the NA, and the strain 0.0035 is at the top most level. Then we give the graph a thickness 'b' which is equal to the width of the beam, and thus we get the 'Stress block'. This is shown in the animation below:

So the final position of the concrete stress block in the beam will be as shown in the fig.3.1 below:

Fig.3.1
Concrete stress block in compression for a beam


 














In the above stress block, the only unknowns are the total depth of the stress block (depth of NA), and the depth of the rectangular portion. In our later discussions we will see the methods to calculate these depths. 

There is also another point that we must note: We have attached the curve along the depth of the beam. The depth is a 'distance'. So it appears as if we plot the curve with 'stress' along one axis and 'distance' along the other axis. But the curve given by the code is a stress-strain curve. Not a stress-distance curve. Later in our discussions, we will see how they are related.

The graph of the stress block has a definite equation. We have seen it in chapter 1, and is given below again:

Eq.3.1

Where f_c is the stress and ε is the strain

We can make a plot of the design curve on our own, for any particular value of f_ck using Eq.3.1. For this we use the following steps:

• Choose a value for the constant f_ck, say 20 N/mm² for M20 grade concrete.

• For plotting the parabolic portion, choose values less than 0.002 for the strain ε (in convenient steps of say 0.00025) and for each of these values calculate the stress f_c, using the first part of Eq.3.1. The table 3.1 below shows the values of stress f_c when strain ε is less than 0.002

Table 3.1:

ε	fc
0	0.000
0.00025	2.095
0.00050	3.911
0.00075	5.448
0.00100	6.705
0.00125	7.683
0.00150	8.381
0.00175	8.800

• For plotting the straight line portion, we don’t have to choose values for strain and calculate the corresponding stress. This is because it is an equation of a horizontal straight line between strain (x) values ε = 0.002 and 0.0035. The y values of all the points on that straight line will be equal to 0.447 x 20 =8.94 . So we can easily draw this second part.

• plot ε along the X axis and f_c along the Y axis

Fig.3.2 below shows the plot, where f_ck = 20 N/mm² . In the plot, the parabolic portion is given blue colour and the straight line portion is given a red colour.

Fig.3.2
Design curve for M20 grade concrete

The coordinate points in the above graph are the same which are given in table 3.1. The area enclosed by the graph can be shown by shading the area between it and the X-axis. This is shown in the fig.3.3 below:

Fig.3.3
Area enclosed by the graph

In the next section we will discuss about the calculation of compressive force at a beam section by using this stress block.



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