In the previous section, we completed the design of a T-beam section. Now we will see another design problem:
Solved example 10.2
A room has internal dimensions 4000 x 14690 mm. (Figs.10.15 & 10.16) The walls of the room are of brick masonry 230 mm thick. The slab of the room has a thickness of 120 mm, and it rests on these outer walls and three equally spaced beams along the 4000 mm direction. The width of the web of these beams is 230 mm. The slab carries a Live load of 4.0 kN/m2. Assume wt of finishes as 1.3 kN/m2 in addition to the self weights. Assuming Fe 415 steel, design an interior beam of the room. Assume moderate exposure conditions.
Fig.10.15
3D view (The slab is not shown for clarity)
Fig.10.16
Plan view
Solution:
The slab of this room is a continuous one, with four spans. We will find the effective span of the slab for each of these spans. The calculations are given below:
Calculation of effective spans:
For this problem, it is convenient to mention before hand that, each of the spans have their support widths less than it's ln/12 (∵ 3500/12 = 291.67). So while using the cl.22.2(b) of IS 456, we will not have to look to the portion below the magenta colored dashed line of the chart. (Fig.7a.4)
Total depth of the slab is 120mm. Assume dia. of main bars of slab = 8 mm and Cc = 30 mm. So effective depth d = 120 -30 -4 = 86 mm
First we will consider the end span. The calculations based on IS 456 is given below:
Clear span ln =3500 mm.
ln/12 = 3500/12 =291.67. So t1 < ln/12 & t2 < ln/12
As mentioned above, we only need the portion above the magenta colored dashed line for all spans of the beam. This is shown in the fig. below. This fig. is applicable to all the spans.
Fig.7a.4
Application of chart to end span
Now we calculate the following:
• c/c distance between the supports = 3500 +115 +115 =3730
• clear span + effective depth = 3500 +86 =3586
Effective span = leff = Lesser of the above = 3586 mm
Thus we calculated leff of the end span.
Now we will consider the second span. The calculations based on IS 456 is given below:
Clear span ln =3500 mm.
ln/12 = 3500/12 =291.67. So t1 < ln/12 & t2 < ln/12
Fig.7a.4 is applicable here also. Now we calculate the following:
• c/c distance between the supports = 3500 +115 +115 =3730
• clear span + effective depth = 3500 +86 =3586
Effective span = leff = Lesser of the above = 3586 mm
Thus we calculated leff of the second span. From the plan view, we can see that the continuous slab is symmetric. Thus all the spans have leff = 3586 mm.
The above calculated value is the effective spans in the long direction. We must calculate leff in the 4000 mm direction also:
The effective depth of the slab in this direction = 120 -30 -8 -4 =78 mm. An additional 8 mm is coming here because the layer of bars which take up the loads in the 4000 mm direction will be above the layer of bars which take up the load in the 3586 mm direction. So the effective depth will be less by an amount equal to one bar diameter. We will learn more details about this when we discuss Two-way slabs.
As the slab is simply supported in this direction, the effective span is the lesser of the following:
(a) clear span + effective depth = 4000 +78 =4078 mm
(b) c/c distance between the supports = 4000 +230 =4230 mm
So the effective span in this direction = 4078 mm
This completes all the calculations regarding 'effective spans' of the slab in our problem.
Next we have to calculate the effective span of the beam:
For this calculation, we will require the effective depth d. But in this problem, the depth of the beam is not given. So we will have to assume some preliminary dimensions for our beam.
Let us adopt l/D ratio of 13. Where l is the clear span. So D= l/13 =4000/13 =307.7 mm. So the depth of the beam below the slab = 307.7 -120 =187.7 mm. This can be rounded off to 200 mm. So D = 320 mm. Also we have bw =230 mm.
Now we have to do the checks that have to be done immediately after fixing the preliminary dimensions. These are given in 4.2, 4.3 and 4.4
4.2: D/b ratio = 320/230 = 1.39. The recommended range is 1.5 to 2. If it is to be 1.5, D should be 1.5 x 230 =345 mm. So depth below slab =345 -120 =225. For ease of putting up the form work, we will adopt 250 mm. Thus we get D = 250 +120 =370 mm. Now D/b =370/230 =1.61. Hence OK.
4.3: l/D = 4000/370 = 10.81. This is greater than 2. So it is not a Deep beam
4.4: • 60b = 60 x230 =13800.
• 250b2/d = (250 x2302/320) =41328.125 (Assuming d = 370 -50 =320 mm)
Lesser of the above = 13800. The clear distance of 4000 mm available to the beam is less than 13800. Hence OK.
So we have done the three checks, and the preliminary D can be taken as 370 mm. d can be taken as 370 -50 =320 mm. The beams are simply supported. So the effective span is the lesser of the following:
(a) clear span + effective depth =4000 +320 =4320 mm
(b) c/c distance between supports = 4000 +230 =4230 mm
So the effective span of the beam = 4230 mm. Thus we have completed all the effective length calculations. It may be noted that the above results will change if we change any of the dimensions of the beam later in this design process.
In the next section, we will calculate the loads coming on the beam.
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Solved example 10.2
A room has internal dimensions 4000 x 14690 mm. (Figs.10.15 & 10.16) The walls of the room are of brick masonry 230 mm thick. The slab of the room has a thickness of 120 mm, and it rests on these outer walls and three equally spaced beams along the 4000 mm direction. The width of the web of these beams is 230 mm. The slab carries a Live load of 4.0 kN/m2. Assume wt of finishes as 1.3 kN/m2 in addition to the self weights. Assuming Fe 415 steel, design an interior beam of the room. Assume moderate exposure conditions.
Fig.10.15
3D view (The slab is not shown for clarity)
Fig.10.16
Plan view
Solution:
The slab of this room is a continuous one, with four spans. We will find the effective span of the slab for each of these spans. The calculations are given below:
Calculation of effective spans:
For this problem, it is convenient to mention before hand that, each of the spans have their support widths less than it's ln/12 (∵ 3500/12 = 291.67). So while using the cl.22.2(b) of IS 456, we will not have to look to the portion below the magenta colored dashed line of the chart. (Fig.7a.4)
Total depth of the slab is 120mm. Assume dia. of main bars of slab = 8 mm and Cc = 30 mm. So effective depth d = 120 -30 -4 = 86 mm
First we will consider the end span. The calculations based on IS 456 is given below:
Clear span ln =3500 mm.
ln/12 = 3500/12 =291.67. So t1 < ln/12 & t2 < ln/12
As mentioned above, we only need the portion above the magenta colored dashed line for all spans of the beam. This is shown in the fig. below. This fig. is applicable to all the spans.
Fig.7a.4
Application of chart to end span
Now we calculate the following:
• c/c distance between the supports = 3500 +115 +115 =3730
• clear span + effective depth = 3500 +86 =3586
Effective span = leff = Lesser of the above = 3586 mm
Thus we calculated leff of the end span.
Now we will consider the second span. The calculations based on IS 456 is given below:
Clear span ln =3500 mm.
ln/12 = 3500/12 =291.67. So t1 < ln/12 & t2 < ln/12
Fig.7a.4 is applicable here also. Now we calculate the following:
• c/c distance between the supports = 3500 +115 +115 =3730
• clear span + effective depth = 3500 +86 =3586
Effective span = leff = Lesser of the above = 3586 mm
Thus we calculated leff of the second span. From the plan view, we can see that the continuous slab is symmetric. Thus all the spans have leff = 3586 mm.
The above calculated value is the effective spans in the long direction. We must calculate leff in the 4000 mm direction also:
The effective depth of the slab in this direction = 120 -30 -8 -4 =78 mm. An additional 8 mm is coming here because the layer of bars which take up the loads in the 4000 mm direction will be above the layer of bars which take up the load in the 3586 mm direction. So the effective depth will be less by an amount equal to one bar diameter. We will learn more details about this when we discuss Two-way slabs.
As the slab is simply supported in this direction, the effective span is the lesser of the following:
(a) clear span + effective depth = 4000 +78 =4078 mm
(b) c/c distance between the supports = 4000 +230 =4230 mm
So the effective span in this direction = 4078 mm
This completes all the calculations regarding 'effective spans' of the slab in our problem.
Next we have to calculate the effective span of the beam:
For this calculation, we will require the effective depth d. But in this problem, the depth of the beam is not given. So we will have to assume some preliminary dimensions for our beam.
Let us adopt l/D ratio of 13. Where l is the clear span. So D= l/13 =4000/13 =307.7 mm. So the depth of the beam below the slab = 307.7 -120 =187.7 mm. This can be rounded off to 200 mm. So D = 320 mm. Also we have bw =230 mm.
Now we have to do the checks that have to be done immediately after fixing the preliminary dimensions. These are given in 4.2, 4.3 and 4.4
4.2: D/b ratio = 320/230 = 1.39. The recommended range is 1.5 to 2. If it is to be 1.5, D should be 1.5 x 230 =345 mm. So depth below slab =345 -120 =225. For ease of putting up the form work, we will adopt 250 mm. Thus we get D = 250 +120 =370 mm. Now D/b =370/230 =1.61. Hence OK.
4.3: l/D = 4000/370 = 10.81. This is greater than 2. So it is not a Deep beam
4.4: • 60b = 60 x230 =13800.
• 250b2/d = (250 x2302/320) =41328.125 (Assuming d = 370 -50 =320 mm)
Lesser of the above = 13800. The clear distance of 4000 mm available to the beam is less than 13800. Hence OK.
So we have done the three checks, and the preliminary D can be taken as 370 mm. d can be taken as 370 -50 =320 mm. The beams are simply supported. So the effective span is the lesser of the following:
(a) clear span + effective depth =4000 +320 =4320 mm
(b) c/c distance between supports = 4000 +230 =4230 mm
So the effective span of the beam = 4230 mm. Thus we have completed all the effective length calculations. It may be noted that the above results will change if we change any of the dimensions of the beam later in this design process.
In the next section, we will calculate the loads coming on the beam.
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