Chapter 12 (cont..4) - Comparison of sections

In the previous section, we completed the 1^st cycle in the analysis of the beam section. We obtained the x_u to be used in the next cycle as x_u =212.335 mm. Now we will continue the iteration process:

Cycle 2:
When the NA is in the position x_u =212.335, ε_sc (Using Eq12.1) =0.00271 The corresponding stress f_sc is obtained from from the same table for Fe 415 steel:
0.00241  → 342.800
0.00271  → 350.483
0.00276  → 351.800. So we get f_sc =350.483 N/mm².

Similarly, when the NA is in this position, ε_st (using Eq.3.15) =0.00303. The corresponding stress f_st is obtained from the same table for Fe 415 steel: 
0.00276  → 351.800
0.00303  → 354.140
0.00380 → 360.900. So we get f_st =354.140 N/mm². With these stress values, we can calculate the forces:

C_u =C_us +C_uc =(f_sc -0.447f_ck)A_sc + 0.362f_ckbx_u = (350.483 – 0.447 x 20) x 402.12 + 2172.00 x212.335  = 598533.964 N

T_u =f_stA_st =354.140 x1722.38 = 609963.349 N

These forces are not equal. So we don't have an equilibrium. What is the value of x_u if these were indeed the stresses? We can calculate this by writing the equality: C_u = T_u. Thus we write:

(350.483 – 0.447 x 20) x 402.12 + 2172.00 x_u = 609963.349
From this we get = x_u =217.598 mm. We will take the average of this x_u and the value of x_u used in this cycle. It is equal to 0.5 x(217.598 +212.335) =214.966 mm. This can be used as the trial value for the next cycle.

Cycle 3:
When the NA is in the position x_u =214.966, ε_sc (Using Eq12.1) =0.00272
. The corresponding stress f_sc is obtained from from the same table for Fe 415 steel:
0.00241  → 342.800
0.00272  → 350.732
0.00276  → 351.800. So we get f_sc =350.483 N/mm².

Similarly, when the NA is in this position, ε_st (using Eq.3.15) =0.00295. The corresponding stress f_st is obtained from the same table for Fe 415 steel: 
0.00276  → 351.800
0.00295  → 353.441
0.00380 → 360.900. So we get f_st =353.441 N/mm². With these stress values, we can calculate the forces:

C_u =C_us +C_uc =(f_sc -0.447f_ck)A_sc + 0.362f_ckbx_u = (350.732 – 0.447 x 20) x 402.12 + 2172.00 x214.966  = 604348.790 N

T_u =f_stA_st =353.441 x1722.38 = 608759.311 N

These forces are not equal. So we don't have an equilibrium. What is the value of x_u if these were indeed the stresses? We can calculate this by writing the equality: C_u = T_u. Thus we write:

(350.732 – 0.447 x 20) x 402.12 + 2172.00 x_u = 608759.311
From this we get = x_u =216.997 mm. We will take the average of this x_u and the value of x_u used in this cycle. It is equal to 0.5 x(216.997 +214.966) =215.982 mm. This can be used as the trial value for the next cycle. 

Cycle 4:
When the NA is in the position x_u =215.982, ε_sc (Using Eq12.1) =0.00272
. The corresponding stress f_sc is obtained from from the same table for Fe 415 steel:
0.00241  → 342.800
0.00272  → 350.827
0.00276  → 351.800. So we get f_sc =350.827 N/mm².

Similarly, when the NA is in this position, ε_st (using Eq.3.15) =0.00292. The corresponding stress f_st is obtained from the same table for Fe 415 steel: 
0.00276  → 351.800
0.00292  → 353.176
0.00380 → 360.900. So we get f_st =353.176 N/mm². With these stress values, we can calculate the forces:

C_u =C_us +C_uc =(f_sc -0.447f_ck)A_sc + 0.362f_ckbx_u = (350.827 – 0.447 x 20) x 402.12 + 2172.00 x215.982  = 606592.039 N

T_u =f_stA_st =353.441 x1722.38 = 608302.526 N

These forces are approximately equal. So we have an equilibrium. The final values can be taken as:
• f_sc =350.827 N/mm²  • f_st =353.441 N/mm² • x_u =215.982 mm

[It may be noted that fig.12.5 is drawn only for getting a good understanding about the extreme positions of the NA. It is not necessary to draw such a fig. in an analysis problem.]

Now we can obtain M_uR 
M_uR = 0.362f_ckbx_u(d - 0.416x_u) + ( f_sc - 0.447f_ck) A_sc(d-d')
= 143618673.976 + 47853116.410 =191471790.385 Nmm  =191.5 kNm

Also x_u = 215.982 mm. x_u,max = 0.4791 x 396 =189.724 mm. So we get x_u > x_u,max. So this is an over reinforced section. 
Thus we complete the analysis process.

Comparison between 'Under reinforced' and 'Over reinforced' Doubly reinforced sections.

Let us compare the sections in the last two solved examples:
• Solved example 2:
Given b =300 mm, d =396 mm, d' =50 mm, A_st =1722.38 mm², A_sc =628.32 mm², f_y =415 N/mm², f_ck =20 N/mm², x_u =188.527 mm, x_u,max =189.72 mm, M_uR =203.53 kNm, x_u < x_u,max, Under reinforced.

• Solved example 3:
b =300 mm, d =396 mm, d' =48 mm, A_st =1722.38 mm², A_sc =402.12 mm², f_y =415 N/mm², f_ck =20 N/mm², x_u =215.982 mm, x_u,max =189.72 mm, M_uR =191.5 kNm. x_u > x_u,max, Over reinforced.

We can see that all the properties except A_sc are the same. The second section have a lesser A_sc than the first. The second section is Over reinforced, while the first section is under reinforced. Let us now discuss the reason for this.

M_ulim of both the sections will be the same. It is given by:

Thus M_ulim =130.66 kNm 

A_st,lim of both the sections will be the same. It is given by 

Thus A_st,lim = 1141.67 mm².
As A_st is same for both sections, ΔA_st will also be the same (Eq.11.8):
A_st = A_st,lim + ΔA_st
Thus we get ΔA_st = 1722.38 -1141.67 =580.71 mm².

If the section under consideration is an under reinforced one, then all the A_st would have yielded at the ultimate state. So the stress in it = 0.87f_y, and the force ΔT_u = 0.87f_yA_st =209666.068 N. The quantity of compression steel A_sc must be such that, the force developed in it should balance ΔT_u. 
So, next we find the stress f_sc in A_sc: For this we can use Table 11.1. 
d'/d = 48/396 =0.121.

0.100  → 342.90
0.121  → 347.87
0.150  → 342.40. So we get f_sc =347.87 N/mm².

So the force in A_sc =( f_sc - 0.447f_ck) A_sc = (347.87 -0.447 x20)A_sc

[It may be noted that Table.11.1 can be used to find f_sc only in the condition when tension steel has yielded. If this steel has not yielded at the ultimate state, we have to use strain compatibility to find f_sc]

This must be equal to the force in ΔT_u. So we can write 209666.068 = (347.87 -0.447 x20)A_sc. From this we get A_sc = 618.61 mm². This is the 'ideal' quantity of A_sc. If we provide this exact quantity, x_u will be equal to x_u,max. But as explained based on fig.11.4 , this exact quantity cannot be provided. The actual quantity that we give should be greater than 618.61 mm². When we give this higher quantity, there will be 'more material' towards the top side of the beam, and so the NA will move up wards. Thus x_u will decrease from x_u,max, and so, it will be an under reinforced section. If A_sc is less than 618.61, then the NA will move downwards, and it will be an over reinforced section.

In the two solved examples, 
• The first one have an A_sc =628.32 >618.61 mm². So it is an under reinforced section.
• The second one have an A_sc =402.12 <618.61 mm². So it is an over reinforced section.

This completes the comparison between the two sections. In the next chapter, we will discuss a new topic: 'Shear in beams'.

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Chapter 12 (cont..3) - Analysis of an over reinforced doubly reinforced beam section

In the previous section, we analysed a doubly reinforced beam section. Now we will  analyse a doubly reinforced beam section which is 'over reinforced'. We analyse the same section which we saw in the previous example. With a minor difference: The compression steel now consists of 2-#16 instead of 2-#20.

Solved example 12.3
Determine the Ultimate moment of resistance M_uR of the beam section shown below. Given b =300 mm, d =396 mm, d' =48 mm, A_st =1722.38 mm², A_sc =402.12 mm², f_y =415 N/mm², f_ck =20 N/mm² 

Solution:
So we are going to analyse the given beam section to find the stresses, strains, forces and depth of NA at the ultimate state. Our main aim is to determine the M_uR. We will see the various steps involved in the analysis in detail:

Initially, we assume that 'both the tension steel and the compression steel will yield at the ultimate state'. 


We will now check if this really happens at the ultimate state. For making this check, first we find x_u the depth of NA. We have assumed that the tension steel has yielded. So the stress in tension steel =0.87f_y. Thus the force in tension steel =T_u =0.87f_yA_st = 0.87 x 415 x 1722.38 =621865.299 N

We have assumed that the compression steel has also yielded. So the stress in compression steel =f_sc =0.87f_y. Thus the force in compression steel =C_us = (f_sc -0.447f_ck)A_sc = (0.87fy – 0.447 x 20) x 402.12 =141590.473 N

Force in concrete = C_uc = 0.362f_ckbx_u = 2172.00x_u

Equating the tensile and compressive forces we get: 141590.473 + 2172.00xu = 621865.299 N - - - (1) 
So we get x_u = 221.121 mm - - - (2)

We have learned the method for calculating (from the strain diagram) the strain ε_sc  at the level of A_sc. We use the two similar triangles above the NA. The equation is Eq.12.1

From this we get ε_sc =0.00274 - - - (3)

Similarly, we know how to calculate the strain ε_st from the strain diagram. The equation  is Eq.3.14 :

Rearranging this, we get:

Note that we are using the equation for ε_st,u because we have assumed that the tension steel has yielded. Using this equation we get ε_st = 0.00277.

Now we compare the values with the yield strain ε*_st. We know that, for Fe415 steel, ε*_st = 0.00380. Comparing the values, we find that ε_sc < ε*_st and ε_st < ε*_st. 

[There is another method to prove that ε_st is less than ε*_st: We know that, for Doubly reinforced sections also, if x_u ≤ x_u,max, the tension steel would have yielded. In our case x_u,max =0.4791 x 396 =189.724 mm. In the above calculations, we have used x_u =221.121 which is greater than x_u,max. So for this value of x_u, the steel would not yield, and so, ε_st will indeed be less than ε*_st.]

So, if at the ultimate state, the NA is at the position x_u = 221.12 mm, the tension steel and compression steel would never yield. This means that:
• The stress in A_st is not 0.87f_{y
• The stress in Asc is not 0.87fy}
So the equality in (1) is not valid. Thus the value of x_u = 221.121 mm is not correct.

As x_u > x_u,max, it is an over reinforced section. We have to use iteration method using strain compatibility. We want an improved value of x_u. We will try to obtain the new value, based on the value that we are having.

Fig.12.5
Positions of NA

In the fig above, 'Line p' shows the position of the NA when x_u = x_u,max =189.724 mm. The NA of our beam section will never occupy this position because, it is an 'Over reinforced section'. The other 'Line q' shows the position when the tension steel has yielded. This position will also never occur because an over reinforced section will fail (by compression of concrete in the extreme fibre) before the yielding of it's tension steel occurs. The actual NA lies some where between these two lines. So we can take the average of these two values as the initial 'trial value'. Thus x_u for the 1^st cycle = 0.5 x(189.724 +221.121) =205.422 mm.

Cycle 1:
When the NA is in the position x_u =205.422, ε_sc (Using Eq12.1) =0.00268. The corresponding stress f_sc is obtained from from the table for Fe 415 steel: 
0.00241  → 342.800
0.00268  → 349.799
0.00276  → 351.800. So we get f_sc =349.799 N/mm².

Similarly, when the NA is in this position, ε_st is obtained using Eq.3.15 : 

Rearranging this we get:

[The equation involving x_u,o and ε_st,o is used here because the section is over reinforced. But it may be noted that Eq.3.14 and 3.15 are the same with only the subscripts changed from 'u' to 'o'.]

So we will get ε_st =0.00325. The corresponding stress f_st is obtained from the same table for Fe 415 steel: 
0.00276  → 351.800
0.00325  → 356.062
0.00380 → 360.900. So we get f_st =356.062 N/mm². With these stress values, we can calculate the forces:

C_u =C_us +C_uc =(f_sc -0.447f_ck)A_sc + 0.362f_ckbx_u = (349.799 – 0.447 x 20) x 402.12 + 2172.00 x205.422  = 583243.352 N

T_u =f_stA_st =356.062 x1722.38 = 613273.929 N

These forces are not equal. So we don't have an equilibrium. What is the value of x_u if these were indeed the stresses? We can calculate this by writing the equality: C_u = T_u. Thus we write:

(349.799 – 0.447 x 20) x 402.12 + 2172.00 x_u = 613273.929.
From this we get = x_u =219.249 mm. We will take the average of this x_u and the value of x_u used in this cycle. It is equal to 0.5 x(205.422 +219.249) =212.335 mm. This can be used as the trial value for the next cycle.

So we have completed one cycle. We will see the  rest of the cycles in the next section.


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