In the previous section, we calculated the effective spans. Now we will calculate the loads coming on the beam:
We have seen the difference between one-way slabs and two-way slabs. It was discussed in the presentation given at the beginning of chapter 5. Based on that we calculate ly/lx for each of the slab panels of our present case. It can be seen that all the four panels have the same ly and lx. = 4078 / 3586 =1.137
This is less than 2, and so each panel is a two-way slab. Thus some loads are transferred to the supporting beams, and the rest are transferred to the walls.
The fig. 10.17 given below helps us to visualize the calculation of loads coming on to the middle beam CF.
Fig.10.17
Transfer of loads
The slab panels are separated into triangular areas and trapezoidal areas by drawing lines at 45o from the corners.
From the fig. we can see that the load in the green colored triangular portions will be transferred to the walls and the load from the yellow colored trapezoidal portions will be transferred to the beams. The beam CF will carry two trapezoidal areas, one on each side. One of them is shown in the fig. 10.18 below:
Fig.10.18
Load on beam CF from one side
This trapezoidal portion is divided into two triangles and a rectangle. Base of the triangles in the above fig. is same as half of the base of the green triangle in the previous fig.10.17. And this will also be equal to the height of the triangle (∵ lines are drawn at 45o from the corners of slab panels). Thus we get (for the triangles in fig.10.18), height = base = 3586/2 =1793 mm. From the above discussion, we get the following results:
• Area of one triangle = 1/2 x1793 x1793 = 1607424.5 mm2
• Horizontal distance of the c.g of this triangle from the center of the beam = 492/2 + (1/3) x1793 = 843.67 mm
• Total area of the trapezoid = [(492 +4078)/2] x1793 = 4097005 mm2
So reactions at the supports are:
RC =RF = 4097005 / 2 = 2048502.5
Taking moments about the center of the beam:
[2048502.5 x (4078 / 2)] -[1607424.5 x 843.67] -[1793 x(492/2) x(492/4)] = 4176896597.5 -1356130469.83 -54252594 =2766513533.67
Equating this to the central moment of a simply supported beam, acted upon by a udl, we get:
wl2/8 = 2766513533.67
From this we get w = 1330.84
This is the udl that produces the same effect as produced by the above trapezoidal loading on the beam. In other words, we can remove the trapezoid, and place a rectangle with ht. 1330.84 mm (1.33 m) on the beam. The following fig.10.19 shows this derived udl.
Fig.10.19
UDL on beam CF
So now we know the load coming on the beam from the slab: Each one meter length of the beam CF will carry the load coming from [(1.33 + 1.33) x 1] m2 area of the slab. This is shown in fig.10.20 below.
Fig.10.20
Load on 1m length of the beam
Based on this, we can now calculate all the loads coming on one meter length of the beam CF.
• Self wt. of slab coming on the beam =1 x2.66 x0.12 x25 =7.98 kN/m
• Wt. of finishes of the slab = 1 x2.66 x1.3 =3.458 kN/m
• LL from the slab = 1 x2.66 x4.0 =10.64 kN/m
• Self wt. of beam = 1 x 0.25 x 0.23 x 25 =1.44 kN/m
So total load per meter length = 7.98 +3.458 +10.64 +1.44 =23.518 kN/m
Applying the load factor of 1.5, we get the factored load =wu = 1.5 x 24.05 =35.277 kN/m. So Mu = wul2/8 = 35.277 x 4.232/8 =78.9 kNm
Now we must calculate the 'most probable' d:
Eq.10.5
z is the lever arm which is taken as the larger of the following:
10.6
♦ 0.9d
♦ d - Df/2
For an initial value of d, we can deduct 50 mm from D. So we get d = 320 mm. So z is the larger of:
♦ 0.9d = 0.9 x 320 = 288 mm
♦ d - Df/2 = 320 - 120/2 =260 mm
Thus z = 288 mm. Substituting the values in 10.5 we get Ast,reqd = 758.78 mm2. Assuming 3 - 18#, we get an area of 763.41 mm2. These bars can be arranged in a single layer as shown in the fig.10.21 below:
Fig.10.21
Bars in single layer
The exposure condition is given as 'moderate'. So we can give a concrete cover of Cc = 30 mm. Thus d = 370 -30 -8 -9 =323 mm. This is the 'most probable' d.
Earlier, we assumed d = 370 -50 =320. Even though the d has changed, the effective span of the beam will not change because the lesser value is still the c/c distance between supports.
Check for minimum horizontal distance between bars (details here):
The the total width of the beam at the bottom most layer of bars can be written as:
230 = 2 x Cc + 2 x Φl + 3 x Φ + 2 x Sh.
⇒ 230 = 2 x 30 + 2 x 8 + 3 x 18 + 2 x Sh.
From this we get Sh = 50 mm
• Bar diameter = 18 mm
• Max. size of aggregate + 5 = 20 + 5 = 25 mm
Larger of the above = 25 mm < 50 mm. Hence OK
Next we have to calculate the effective flange width bf.
Eq.9.5
l0 =4230 mm; bw =230 mm; Df =120 mm; s1 = s2 =3500 +230 =3730 mm. Substituting these values we get: (a) 1655 mm & (b) 3730 mm. bf = smaller of the two = 1655 mm.
Now we have all the required details to begin the actual design process. We will see this in the next section.
PREVIOUS
We have seen the difference between one-way slabs and two-way slabs. It was discussed in the presentation given at the beginning of chapter 5. Based on that we calculate ly/lx for each of the slab panels of our present case. It can be seen that all the four panels have the same ly and lx. = 4078 / 3586 =1.137
This is less than 2, and so each panel is a two-way slab. Thus some loads are transferred to the supporting beams, and the rest are transferred to the walls.
The fig. 10.17 given below helps us to visualize the calculation of loads coming on to the middle beam CF.
Fig.10.17
Transfer of loads
 |
From the fig. we can see that the load in the green colored triangular portions will be transferred to the walls and the load from the yellow colored trapezoidal portions will be transferred to the beams. The beam CF will carry two trapezoidal areas, one on each side. One of them is shown in the fig. 10.18 below:
Fig.10.18
Load on beam CF from one side
This trapezoidal portion is divided into two triangles and a rectangle. Base of the triangles in the above fig. is same as half of the base of the green triangle in the previous fig.10.17. And this will also be equal to the height of the triangle (∵ lines are drawn at 45o from the corners of slab panels). Thus we get (for the triangles in fig.10.18), height = base = 3586/2 =1793 mm. From the above discussion, we get the following results:
• Area of one triangle = 1/2 x1793 x1793 = 1607424.5 mm2
• Horizontal distance of the c.g of this triangle from the center of the beam = 492/2 + (1/3) x1793 = 843.67 mm
• Total area of the trapezoid = [(492 +4078)/2] x1793 = 4097005 mm2
So reactions at the supports are:
RC =RF = 4097005 / 2 = 2048502.5
Taking moments about the center of the beam:
[2048502.5 x (4078 / 2)] -[1607424.5 x 843.67] -[1793 x(492/2) x(492/4)] = 4176896597.5 -1356130469.83 -54252594 =2766513533.67
Equating this to the central moment of a simply supported beam, acted upon by a udl, we get:
wl2/8 = 2766513533.67
From this we get w = 1330.84
This is the udl that produces the same effect as produced by the above trapezoidal loading on the beam. In other words, we can remove the trapezoid, and place a rectangle with ht. 1330.84 mm (1.33 m) on the beam. The following fig.10.19 shows this derived udl.
Fig.10.19
UDL on beam CF
So now we know the load coming on the beam from the slab: Each one meter length of the beam CF will carry the load coming from [(1.33 + 1.33) x 1] m2 area of the slab. This is shown in fig.10.20 below.
Fig.10.20
Load on 1m length of the beam
Based on this, we can now calculate all the loads coming on one meter length of the beam CF.
• Self wt. of slab coming on the beam =1 x2.66 x0.12 x25 =7.98 kN/m
• Wt. of finishes of the slab = 1 x2.66 x1.3 =3.458 kN/m
• LL from the slab = 1 x2.66 x4.0 =10.64 kN/m
• Self wt. of beam = 1 x 0.25 x 0.23 x 25 =1.44 kN/m
So total load per meter length = 7.98 +3.458 +10.64 +1.44 =23.518 kN/m
Applying the load factor of 1.5, we get the factored load =wu = 1.5 x 24.05 =35.277 kN/m. So Mu = wul2/8 = 35.277 x 4.232/8 =78.9 kNm
Now we must calculate the 'most probable' d:
Eq.10.5
z is the lever arm which is taken as the larger of the following:
10.6
♦ 0.9d
♦ d - Df/2
For an initial value of d, we can deduct 50 mm from D. So we get d = 320 mm. So z is the larger of:
♦ 0.9d = 0.9 x 320 = 288 mm
♦ d - Df/2 = 320 - 120/2 =260 mm
Thus z = 288 mm. Substituting the values in 10.5 we get Ast,reqd = 758.78 mm2. Assuming 3 - 18#, we get an area of 763.41 mm2. These bars can be arranged in a single layer as shown in the fig.10.21 below:
Fig.10.21
Bars in single layer
The exposure condition is given as 'moderate'. So we can give a concrete cover of Cc = 30 mm. Thus d = 370 -30 -8 -9 =323 mm. This is the 'most probable' d.
Earlier, we assumed d = 370 -50 =320. Even though the d has changed, the effective span of the beam will not change because the lesser value is still the c/c distance between supports.
Check for minimum horizontal distance between bars (details here):
The the total width of the beam at the bottom most layer of bars can be written as:
230 = 2 x Cc + 2 x Φl + 3 x Φ + 2 x Sh.
⇒ 230 = 2 x 30 + 2 x 8 + 3 x 18 + 2 x Sh.
From this we get Sh = 50 mm
• Bar diameter = 18 mm
• Max. size of aggregate + 5 = 20 + 5 = 25 mm
Larger of the above = 25 mm < 50 mm. Hence OK
Next we have to calculate the effective flange width bf.
Eq.9.5
l0 =4230 mm; bw =230 mm; Df =120 mm; s1 = s2 =3500 +230 =3730 mm. Substituting these values we get: (a) 1655 mm & (b) 3730 mm. bf = smaller of the two = 1655 mm.
Now we have all the required details to begin the actual design process. We will see this in the next section.
PREVIOUS
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