Chapter 3 (cont..3) Stress-Strain curve for steel

In the previous section we calculated the last point of the curved portion. In this section, we will calculate the second last point.

We can see a dashed line parallel to the initial inclined line, starting from 0.001 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: '0.975f_y'. This means that the 'y coordinate of the point of intersection' of the inclined dashed line with the characteristic curve is 0.975f_y. So the y coordinate of the corresponding point on the design curve will be equal to 0.975f_y/1.15 = (0.975 x 415)/1.15 = 351.8 N/mm². Now we want the x coordinate. For this we will use the fig.3.14 given below:

Fig.3.14
Second last point of the curved portion

A triangle P'Q'R' is drawn. P'R' corresponds to the inclined dashed line in the code, drawn from the strain value of 0.001. So we can see that the x coordinate will be equal to 0.001 + P'Q'. Thus our next step is to calculate P'Q'.

In the ΔP'Q'R', tan∠R'P'Q' = ^R'Q'⁄_P'Q'  

⇒ P'Q' = ^R'Q'⁄_tan∠R'P'Q'    

But tan∠R'P'Q' is the slope of P'R'. 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 R'Q' = the y coordinate of the point R', which we already calculated as 351.8

Thus we get P'Q' = ^351.8⁄₂₀₀₀₀₀ = 0.001759

So the x coordinate of point R = 0.001 + 0.001759 = 0.002759 = 0.00276

Thus the coordinates of the second last point of the curved portion are (0.00276, 351.8)

As before, for the general case, for the second last point of the curved portion, we can write the following:

Eq.3.10

x coordinate = 0.001 + [(0.975f_y)/ (1.15 x 200000)] = 0.001 + 0.975f_y/230000

y coordinate = 0.975f_y/ 1.15

So on a new empty graph paper, on which we marked the last point earlier, we can mark the second last point also directly using Eq.3.10. This is shown in the animation below:

In this way, we can calculate the coordinates of the remaining points on the curved portion. We will do one more calculation, but this time for the first point of the curved portion.

For this point, there is no dashed line. This is because, we are now looking at the first point of the curved portion., that is., the first point of transition, which will be on the initial inclined line. We can see that there are 5 inclined dashed lines and 6 horizontal dashed lines. So for the present case, we can use the initial inclined line itself. 

The bottom most horizontal dashed line has a marking of 0.8fy. So the y coordinate is 0.8f_y. So we get the y coordinate of the corresponding point on the design curve as 0.8f_y/1.15 = (0.8 x 415)/1.15 = 288.7 N/mm². Now we want the x coordinate. For this we will use the fig.3.15 given below:

Fig.3.15
First point of the curved portion

A triangle P''Q''R'' is drawn. P''R'' corresponds to the initial inclined portion. P'' coincides with the origin. We can see that the x coordinate will be equal to P''Q''. Thus our next step is to calculate P''Q''.

In the ΔP''Q''R'', tan∠R''P''Q'' = ^R''Q''⁄_P''Q''  

⇒ P''Q'' = ^R''Q''⁄_{tan∠R''P''Q''}    

But tan∠R''P''Q'' is the slope of P''R''. 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 R''Q'' = the y coordinate of the point R'', which we already calculated as 288.7

Thus we get P''Q'' = ^288.7⁄₂₀₀₀₀₀ = 0.0014435

So the x coordinate of point R'' = 0.0014435 = 0.00144

Thus the coordinates of the first point of the curved portion are (0.00144, 288.7)

As before, for the general case, for the first point of the curved portion, we can write the following:

Eq.3.11

x coordinate =  (0.8f_y)/ (1.15 x 200000)] = 0.8f_y/230000

y coordinate = 0.8f_y/ 1.15

So now we have the first point of the curved portion. This is also the last point of the initial straight portion. So we have two points on the initial portion: First is the origin (0,0), and the second is the first point of the curved portion, which we calculated just now. We can draw a line between these two points and thus obtain the initial portion.

Thus we have all the details required to make a complete design curve. As mentioned earlier, when we calculate the details for the intermediate curved portion, we will get the details for the initial and final portions automatically. We can make separate tables giving all the coordinates of the design curve for Fe415 and Fe500 steel.

Table for Fe415:

Strain	Stress (N/mm2)	Calculation of Strain	Calculation of Stress
0.000	0.00
0.00144	288.7	(415x0.80)/230000	(415 x 0.80)/1.15
0.00163	306.7	0.0001+[(415x0.85)/230000]	(415 x 0.85)/1.15
0.00192	324.8	0.0003+[(415x0.90)/230000]	(415 x 0.90)/1.15
0.00241	342.8	0.0007+[(415x0.95)/230000]	(415 x 0.95)/1.15
0.00276	351.8	0.001+[(415x0.975)/230000]	(415 x 0.975)/1.15
≥ 0.00380	360.9	0.002+[(415x1.0)/230000]	(415 x 1.00)/1.15

Table for Fe500:

Strain	Stress (N/mm2)	Calculation of Strain	Calculation of Stress
0.000	0.00
0.00174	347.8	(500x0.80)/230000	(500 x 0.80)/1.15
0.00195	369.6	0.0001+[(500x0.85)/230000]	(500 x 0.85)/1.15
0.00226	391.3	0.0003+[(500x0.90)/230000]	(500 x 0.90)/1.15
0.00277	413.0	0.0007+[(500x0.95)/230000]	(500 x 0.95)/1.15
0.00312	423.9	0.001+[(500x0.975)/230000]	(500 x 0.975)/1.15
≥ 0.00417	434.8	0.002+[(500x1.0)/230000]	(500 x 1.00)/1.15

In the above tables, 

• The first row is the origin point (0,0). 
• The second row is the last point of the initial straight portion, which is also the first point of the intermediate curved portion. 
• The last row is the last point of the intermediate curved portion, which is also the first point of the final horizontal portion. A '≥' is given to the strain value. In this, the '=' part indicates the exact point. The '>' part indicates the strain values to the right of the point. For all those points to the right, the stress remains constant, giving a horizontal line.

Now we will have a quick look at the application of these curves: The main application of the design curve is to find the stress corresponding to any strain value. That is., during an analysis process, we may find that the steel reinforcement is strained to a particular value, and we will want to find the stress in that steel. In such a situation, we make use of the design curve. If the strain value fall in the initial or final straight portions, the calculation of stress is easy. But if it falls in the curved portion, then we will have to use linear interpolation. These regions are shown in the fig.3.16 below:

Fig.3.16
Strains in the three regions

We have seen that the table of each steel will give complete details (including that of the straight portions) of the design curve. So we don't need to draw the design curve. If the table is available, stress corresponding to any strain value can be calculated.

In the next section, we will have a detailed discussion on the role played by the design curves in the analysis of a beam section.

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