Friday 24 October 2014

Design of a Residential Building Floor by Direct Calculations of Bending Moments Coefficients using Equations 14 to 18 of BS 8110-1:1997


The bending moments of the following floor slab are computed using direct calculations of bending moments coefficients using equations 14 to 18 of BS 8110-1:1997

It should be noted that the bending moments coefficients of equations 16 to 18 of BS 8110-1:1997 were developed using the modified yield line methods to enable rapid analysis and design of reinforced concrete slabs


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Using slab of thickness = 150mm

Deadload load

Self weight = 0.15 × 24                        = 3.6 kN/m2

Finishes & partitions =                         = 1.5 kN/m2

Characteristic dead load; gk         = 5.1 kN/m2

Imposed load

Imposed load                          = 1.5 kN/m2

Characteristic imposed load; qk   = 1.5 kN/m2

Design load, n = 1.4 gk + 1.6 qk = 1.4 × 5.1 + 1.6 × 1.5 = 9.54 kN/m2

 
Considering panel A, it can be seen that two of its edges are discontinuous while the other two edges are continuous.

βy = (24 + 2Nd + 1.5 Nd2)/1000……………………………………..equation 16 

where Nd is the number of discontinous edges

Here the number of discontinouos edges Nd is equal to two

Therefore βy = (24 + (2 × 2) + (1.5 ×22)/1000 = 0.034

Therefore the long span bending moment coefficient = 0.034

And the short edge bending moment coefficient β2 = (4/3) × 0.034 = 0.045

For the remaining bending moment coefficients use is to be made of equations 17 and 18

 g = (2/9) [3-Ö18(lx/ly) {Öy+ β1) + Öy + β2)}]…………………… equation 17 

From our previous calculation, βy = 0.034 and from inspection of the support conditions of the slab

we find that β1 = 0 and β2 = 4βy/3 , substituting for these values

g = (2/9) [3-Ö18(4.5/5) {Ö(0.034+ 0) + Ö(0.034 + 0.045)}]

g = 0.27171

In order to calculate the value of βx, the value of the short span bending moment coefficient,

 we make use of equation 18

Ög = Öx+ β3) + Öx + β4)………………………………………….. equation 18 

from inspection of the support conditions of the slab

we find that β3 = 0 and β4 = 4βx/3 , and we now know that g = 0.27171

substituting for these values

Ö0.27171 = Öx+ 0) + Öx + 4βx/3)

0.52125 = Ö βx (Ö1+Ö(1+(4/3))

0.52125 = Öx) (2.527525)

 Ö βx = (0.52125/2.527525)

Squaring both sides

 βx = 0.042 and the long edge moment coefficient = β4 = (4/3) × 0.042 = 0.056

Span moments

Short span = βxnlx2 = 0.042 × 9.54 × 4.52 = 8.11 kNm

Long span = βynlx2 = 0.034 × 9.54 × 4.52 = 6.57 kNm

Support edge moments

Short edge = β2nlx2 = 0.045 × 9.54 × 4.52 = 8.69 kNm

Long span = β4nlx2 = 0.056 × 9.54 × 4.52 = 10.81 kNm

The above calculations will seem to be tedious to the first time observer, but it is not so. These calculations are the bases of the values presented in table 3.14 of BS 8110-1:1997. For the university student and the practicing engineer, they are very valuable tools for carrying out day-to-day work involving the design of slabs.


In the design office, these calculations could easily be resolved each time with the use of spreadsheet software that are installed in almost every computer. This can be achieved with very minimal computer programming knowledge. Also, there will be no need for carrying out interpolation for intermediate values of aspect ratios ly/lx .
All values of slab bending coefficients are computed directly and used immediately. For those engaged in civil structural design offices, automation of the above process will make life easier.

In the coming posts, we will learn how to implement this very easily with a spreadsheet.





Monday 13 October 2014

A Very Easy Way to Derive Fixed End Moments For Beams Fixed at Both Ends


In order to proceed with this problem, a zero length spans are added to the left and right hand sides
Applying the three moment theorem

MA’ (0) + 2MA (0+l) + MB (l) =wl3/4

2MA (l) + MB (l)   = wl3/4

2MA + MB    = (wl2)/4 -------------------------------------------(1)

Similarly it can be shown that by adding another zero span to the right of B

MA + 2MB    = (wl2)/4 -------------------------------------------(2)

Equating (1) and (2)

2MA + MB   = MA + 2MB    

MA   = MB    

Substituting for MA in (1)

3MB = (wl2)/4

MB = MA = (wl2)/12

This is the fixed end moment for MA and MB

A Very Easy Way to Derive Fixed End Moments For Propped Cantilvers


In order to proceed with this problem, a zero length span is added to the left hand side


Applying the three moment theorem
MA’ (0) + 2MA (0+l) + MB (l) =wl3/4

2MA (l) = wl3/4
MA = (wl2)/4

This is the fixed end moment MA