In the previous section we saw that the spacing of stirrups should not be greater than 226.85 mm at any part of the beam that we are now designing. In this section, we will calculate the spacing for the region from A to E.
For this we use the basic Eq.13.58
Vus =Vu -τcbd
Vu is the shear force at the critical section =310.5 kN
τc = 0.786 N/mm2
So we get Vus = 137668 N
But from Eq.13.39
Putting Asv = 100.53 mm2, (∵ 2-legged stirrups of 8 mm dia.) we get Sv = 145.008 mm. - - - (3)
Let us check this by using table 62 of SP 16
• Vus/d = 250.305 N/mm = 2.503 kN/cm
2.593 → 14
2.503 → 14.5199
2.420 → 15. So we get Sv =14.5199 cm =145.199 mm.
Thus at the point of greatest shear, the spacing required is 145.008 mm
Upto this point we have the following two informations:
1. At no part of the beam, the spacing should exceed 226.854 mm
2. The spacing at critical section should not exceed 145.008 mm - - - (4)
• So spacing starts from 145.008 mm
• Increases to 226.854 mm
• And there after remains constant at 226.854 mm
We will find the point at which the constant spacing of 226.854 mm begins. For this, we have to work in a sort of 'reverse order'. That is., we usually calculate the spacing required for a given value of Vu that occurs at a particular section. But now we are having the 'spacing', and we are going to calculate the section at which this spacing occurs:
We will put Sv =226.854 mm in the equation for Vus:
From this we get Vus =87,999.155 N
But Vus =Vu -τcbd . So we get Vu =260,831.155 N =260.83 kN
Now we find the point at which this much Vu is being applied. For this, we put the above value for 'y' , and solve for 'x' in the equation (1)
y =405 -135x. We get x = 1.068 m. To the left of this point, spacing required will be less than 226.854 mm. To the right of the point, the spacing required will be greater than 226.854 mm. But a spacing greater than 226.854 mm cannot be allowed for our beam. So we can say that, towards the right of x= 1.068, the spacing remains constant at 226.854 mm. This is the third information that can be added to the above two in (4)
1. At no part of the beam, the spacing should exceed 226.854 mm
2. The spacing at critical section should not exceed 145.008 mm
3. The spacing remains constant at 226.854 mm after 1.068 m from the support - - - (5)
With the above three information, we can design a satisfactory distribution of spacing along the length of the beam. For this, first we calculate the spacing at regular intervals of 0.2 m along the length of the beam, up to the point of constant spacing. So we have to calculate Sv at 0.2, 0.4, 0.6, 0.8..... m from the left support.
• The critical section at 0.7 m is an important point. • 1.068 m is also an important point because, the Sv remains constant after this point. So we will add these two points also.
The tabulation is shown below:
We do not need the value of Sv beyond a distance of 1.2 m, as the constant Sv begins at 1.068 m.
All the values in the Sv column are obtained by the same method that we used for calculatiing the Sv at the critical section, as we did in (3)
Now we will make a plot with distance on the X axis and Sv on the Y axis.
This is the plot of the spacing required. From the graph, we can see that when the distance increases, the spacing also increases. This means that we can provide stirrups at 'greater distances apart' at the regions away from supports. The portion beyond 226.883 cannot be considered as part of this plot because spacing greater than it are not acceptable. So it is shown in dotted line.
We will plot another curve on the same graph which will give the spacing that we actually provide.
• Unlike the above plot, the new plot will have 'horizontal straight line' segments, which will indicate that, spacing provided is constant for various distances along the length of the beam.
• The new plot should not rise above the above plot at any point. This is because, if it does, it will mean that the spacing provided is more than the spacing required at that point.
• Even though the new plot will have straight line segments, it should follow the above curve as closely as possible.
• If it is a simple horizontal line, it would be far too below the above curve at distances away from the support. This would mean that we will be placing the stirrups at smaller spacing unnecessarily at the portions away from the support.
• So the new plot will have some 'steps', so that it will follow the above curve as closely as possible.
The arrangement should also have ease of placement at the site. A possible plot is shown below:
In the above graph, the plot of 'spacing provided' is given in red colour. The first stirrup is usually placed at a distance of 5 cm from the face of the support. So point 'L' will have an X coordinate of 0.20 m. This is shown in the fig. below:
• In the above graph, the x coordinate of 'L' is 0.20, and the y coordinate is 140. This means that the spacing is 140 mm at the point of the first stirrup.
• The graph is horizontal from L to M. This means that the spacing of 140 mm, continues upto point 'M'.
• So the stirrups have the same spacing of 140 mm from L to M.
• M has an x coordinate of 0.900. So LM = 0.900-0.200 = 0.700 m. 140 x 5 = 700. So there will be five spaces of 140 mm, giving 6 stirrups from L to M.
• At M, there is a step. The plot jumps to N
Note that some portion of LM is above the 'plot of required spacing'. This means that the spacing provided here is more than the required spacing. This will appear to be 'unsafe'. But we know that the critical section is taken at a distance of d from the face of the support, and the region between the critical section and the face of the support need to be designed for the shear at critical section only.
Now we look at the segment NO. 'N' and 'O' have y coordinates 160. So the spacing in the region NO is 160 mm. 'O' have x coordinate 1.220. The distance NO = 1.220 -0.900 =0.320 m. 160 x 2 = 320. So there will be two spaces of 160 mm, giving 3 stirrups from N to O. (But the first stirrup among the three is already placed at M, which we discussed in the case of LM )
We have reached 'O' and so, we have crossed the 'section beyond which spacing is constant'. From 'P' onwards, the spacing is given a constant value of 220 mm, upto the center of the beam. 'Q' has an x coordinate of 2.980. So PQ =2.980 -1.220 =1.760 m. 220 x 8 =1760. So there will be 8 spaces of 220 mm, giving 8 stirrups from P to Q.
Q' has an x coordinate of 2.980. So we have covered a distance of 2.980 m from the center of the left support. This point is just 20 mm (∵ 3.000 -2.980 =0.020) from the center of the beam. We have to provide a mirror image of this arrangement from the right side support also. It will cover a distance of 2.980m on the right side of the beam. So the distance between the middle two stirrups will be 20 x 2 =40 mm.
40 mm is a very small distance which may obstruct the placing of concrete. So when we give the mirror arrangement from the right side support towards the interior, the last stirrup can be avoided.
A table can be formed showing the spacing in each segment on the left side as follows:
A diagram showing the complete arrangement is given below:
In the next section we will see another solved example.
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