Bridge Design Processes

Concrete Rate Analysis Software By CivStruct Dynamics, Port Harcourt, Nigeria

CivStruct Dynamics, Port Harcourt, Nigeria, is introducing Concrete Rate Analysis Software for the rapid cost breakdown analysis of concrete.

This software, the first of its kind will assist Quantity Surveyors, Civil Engineers, Building and Civil Engineering contractors in carrying out their work.

Concrete Rate Analysis By CivStruct Dynamics

If you are interested in the use of the software you can reach us through our contact below:

Email: monomaklass@gmail.com
Phone: +2348172318965



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

Loading

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; g_k         = 5.1 kN/m²

Imposed load

Imposed load                          = 1.5 kN/m²

Characteristic imposed load; q_k   = 1.5 kN/m²

Design load, n = 1.4 g_k + 1.6 q_k = 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.

β_y = (24 + 2N_d + 1.5 N_d²)/1000……………………………………..equation 16 

where N_d is the number of discontinous edges

Here the number of discontinouos edges N_d is equal to two

Therefore β_y = (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(l_x/l_y) {Ö(β_y+ β₁) + Ö(β_y + β₂)}]…………………… equation 17 

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

we find that β₁ = 0 and β₂ = 4β_y/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 β_x, the value of the short span bending moment coefficient,

 we make use of equation 18

Ög = Ö(β_x+ β₃) + Ö(β_x + β₄)………………………………………….. equation 18 

from inspection of the support conditions of the slab

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

substituting for these values

Ö0.27171 = Ö(β_x+ 0) + Ö(β_x + 4β_x/3)

0.52125 = Ö β_x (Ö1+Ö(1+(4/3))

0.52125 = Ö (β_x) (2.527525)

 Ö β_x = (0.52125/2.527525)

Squaring both sides

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

Span moments

Short span = β_xnl_x² = 0.042 × 9.54 × 4.5² = 8.11 kNm

Long span = β_ynl_x² = 0.034 × 9.54 × 4.5² = 6.57 kNm

Support edge moments

Short edge = β₂nl_x² = 0.045 × 9.54 × 4.5² = 8.69 kNm

Long span = β₄nl_x² = 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 l_y/l_{x .}
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.

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

M_A’ (0) + 2M_A (0+l) + M_B (l) =wl³/4

2M_A (l) + M_B (l)   = wl³/4

2M_A + M_B    = (wl²)/4 -------------------------------------------(1)

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

M_A + 2M_B    = (wl²)/4 -------------------------------------------(2)

Equating (1) and (2)

2M_A + M_B   = M_A + 2M_B    

M_A   = M_B    

Substituting for M_A in (1)

3M_B = (wl²)/4

M_B = M_A = (wl²)/12

This is the fixed end moment for M_A and M_B

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

M_A’ (0) + 2M_A (0+l) + M_B (l) =wl³/4

2M_A (l) = wl³/4

M_A = (wl²)/4

This is the fixed end moment M_A

Structural Design of a Reinforced Concrete Balcony Slab to BS 8110

Design of Reinforced Concrete Balcony with Dwarf  Wall









Plan of Balcony Slab





Section through Balcony Slab

Moment and Shear

Slab Geometry

Span of slab = 1200 + (225/2)  = 1312.5 mm = 1.3125 m

Design width                            = 1000 mm = 1 m

Slab Details

Thickness of slab = 150 mm

Characteristic strength of concrete; f_cu = 20 N/mm²

Characteristic strength of reinforcement; f_y = 460 N/mm²

Material safety factor; γ_m = 1.05

Cover to bottom reinforcement; c = 20 mm

Cover to top reinforcement; c’ = 20 mm

Loading details

Slab loading

Dead load

Self weight of slab = 0.15 × 24                       = 3.6 kN/m²

Finishes @ 0.6 kN/m² =                       = 0.6 kN/m²

Characteristic dead load; g_k                  = 4.2 kN/m²

Dwarf wall loading

Wall is 150 mm hollow block wall (BS 648:1964, Schedule of weights of building materials)

Load per m run (Point load) = 1.52 × 1            = 1.52 kN/m

Characteristic dead load; g_k                  = 1.52 kN/m

Imposed load     (Office general use = 2.5 kN/m² ,BS 6399-1:1996, Table 1)

Characteristic imposed load; q_k                        = 2.5 kN/m²

Design loading factors

Dead load factor; γ_G = 1.4

Imposed load factor; γ_Q = 1.6

Moment redistribution ratio; βb = 1.0

Design loads

Dead loads

Slab

Slab load     = 1.4 × 4.2× 1                   = 5.88 kN/m

Dwarf wall load  = 1.4 × 1.52              = 2.13 kN

Imposed loads

Slab

Slab load     = 1.6 × 2.5× 1                   = 4.00 kN/m

Moment and Shear

Service Moment

M_udl   = (0.5 × 4.2 × 1.3125²) + (0.5 × 2.5 × 1.3125²)  = 5.77kNm

M_point  = 1.52 × 1.3125                                                 = 2.00 kNm

Design Moment

M_udl   = (0.5 × 5.88 × 1.3125²) + (0.5 × 4.00 × 1.3125²)          = 8.51kNm

M_point  = 2.13 × 1.3125                                                 = 2.80 kNm

Design Shear force

V   = (5.88 × 1.3125) + (4.00 × 1.3125) + 2.13           = 15.10 kN

Slab Design (Per metre run of balcony)

Using 12 mm main bars and 10 mm distribution bars

Effective depth of reinforcement; d = 150 – 20 + (12/2)  =  124 mm

Support moment; m’ = 8.51 + 2.80 = 11.31 kNm/m

Design reinforcement

(3.4.4.4)

Lever arm; K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4)² = 0.176

K = m / (d² × f_cu) = 0.036778 < 0.176

Compression reinforcement is not required

(3.4.4.4)

z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d = 117.8 mm

Area of reinforcement designed; As_reqd = m / (z × f_y / γ_m) = 219.70 mm²/m

Minimum area of reinforcement required; As_min = 0.0013 × h = 195 mm²/m

Area of reinforcement required; As_req = max(As_reqd, As_min) = 219.70 mm²/m

Provide 12 mm dia bars @ 200 mm centres

Area of main reinforcement provided; As_prov = 566 mm²/m

Provide 10 mm dia bars @ 200 mm centres

Area of distribution reinforcement provided; As_prov = 393 mm²/m

Shear Check

Maximum allowable shear stress; v_max = min(0.8 × √(f_cu), 5) = 3.58 N/mm²

shear stress; v = V / (b × d) = 15.10 ×10³ / (1000 × 124)  = 0.122 N/mm²

Shear capacity of Slab; v_c = (min(fcu,40)/25)1/3×0.79×min(100×A_sprov/(b×d),3)1/3×max(400/d,1)1/4/1.25

Shear capacity of Slab; v_c = 0.756 N/mm² > 0.122 N/mm²

Shear Capacity Okay

Check deflection

Basic span/d  ratio = 7

K_udl = 0.25

K_point  = 0.33

Adjusted basic ratio = Basic ratio× (M_udl + M_point × K_udl/ K_point)/( M_udl + M_point )

Adjusted basic ratio = 7× (5.77 + 2.00 × 0.25/ 0.33)/( 5.77 + 2.00 )  =  6.56

Design service stress; f_s = 2 × f_y × As_req / (3 × As_prov × β_b) = 119.04 N/mm²

Modification factor; k₁ = min(0.55+(477N/mm²-f_s)/(120×(0.9N/mm²+(m/d²))),2) = min (2.38, 2.00)

Modification factor; k₁ =  2.00

Allowable span to depth ratio; = 6.56 × k1 = 13.12

Actual span to depth ratio; L / d = 1312.5/124 = 10.58

L/d ratio Okay

Check Cracking

Clear spacing of bars = 200 – 12 = 188 mm

3d = 3 × 124 = 372 mm

47000/f_s = 47000/119.04 = 394 mm

Bar spacing Okay

Anchorage length

Ultimate anchorage bond stress, f_bu = β√f_cu = 0.5√20

f_bu = β√_fcu = 2.24 N/mm²

L = (0.95fy) × f/ (4 f_bu)f = (0.95×460×12)/ (4× 2.24) = 585.26 mm

L = 0.58526 m

However, ISTRUCTE detailers’ manual recommends

L = (1.5 × span) + 0.1125 = (1.5 × 1.3125) + 0.1125

L = 2.08125 m

Also L = 0.3 × span preceding cantilever span

In this case we take 4.5 m

L = 0.3 × 4.5 = 1.35 m

Therefore we adopt L = 2.08125m

i.e. 2.1 m from the centerline of the support

Bending Moments and Shears: Equal Span Coefficients

For two span Beam

Using the three moment equation

M_AL + 2M_B(L+L) + M_CL = (wL³/4) + (wL³/4)

Since M_A = M_B = 0

4M_BL = wL³/2

M_B = wL²/8 = 0.125wL²………………………………….......…..1

Reaction R_AB

Note: S. S. R. means simple suport reaction
          C. C. means continuity correction

S.S.R =wl/2

C.C = (|M_A| - |M_B|)/L = (-M_B)/L = wl/8                  

 Reaction R_AB = ∑R = S.S.R + C.C.

                        = wL/2-wL/8 =3wL/8

            R_AB =3wL/8 = 0.375wL…………………………………2

Reaction R_BA

S.S.R, = wL/2

C.C = (|M_B|-|M_A|)/L = (M_B)/L  

Reaction R_AB = ∑R = S.S.R + C.C. = wL/2 + wL/8 = 5wL/8

R_BA =5wL/8 = 0.625wL................................................................3

Due to symmetry, R_AB =R_CB = 3wL/8

R_BA = R_BC =5wL/8

Maximum Span Moment on Span A-B

M_x = R_AB - wx²/2...........................................................................4

At maximum moment

dM_x/d_x = R_AB - wx = 0

x = R_AB/w = 3L/8 = 0.375L

Substituting for x = 3L/8 in equ (4)

M_max = (3wL/8).(3L/8) - (w/2). (3L/8)²

              = 9wL²/64( 1-1/2) = 9wL²/128

 M_max = 9wL²/128 = 0.0703wL²...................................................5

Alternatively,

M_x = R_BA - wx²/2 - M_B

At maximum moment

dM_x/d_x = R_BA - wx = 0

x = R_BA/w = 5L/8

Substituting for x = 5L/8 in equ (4)

M_max = (5wL/8).(5L/8) - (w/2). (5L/8)² -wL²/8

= 9wL²/128

M_max = 9wL²/128 = 0.0703wL².................................................6

For three span Beam

Using the three moment equation

SPANS A-C

M_AL + 2M_B(L+L) + M_CL = wL³/4 + wL³/4

M_A = 0

4M_BL + M_CL = wL³/2………………………………………………7

SPANS B-D

M_BL + 2M_C(L+L) + M_DL = wL³/4 + wL³/4

M_D = 0

M_BL + 4M_CL = wL³/2………………………………………………8
Equating 7 and 8

4M_BL + M_CL = M_BL + 4M_CL

4M_B + M_C  = M_B + 4M_C

3M_B = 3M_C

M_B = M_C

Substituting for M_B = M_C in equation 7
4M_CL + M_CL = wL³/2

5M_CL = wL³/2

M_C = wL²/10 = 0.1wL²....……………………………………………9

Since M_B = M_C

M_B = M_C = wL²/10 = 0.1wL²

S.S.R = wL/2

C.C = (|M_A|-|M_B|)/L = (- M_B)/L = wL/10
R_AB = ∑R = S.S.R + C.C.  = wL²/2-wL/10

R_AB = 4wL/10 = 0.4wL......................................................................10
Similarly

R_BA = wL/2 + wL/10 = 6wL/10

R_BA = 6wL/10 = 0.6wL……………………………………………..11
S.S.R. = wL/2
C.C = (|M_B|-|M_C|)/L = 0

R_BC =R_CB = wL/2 = 0.5wL.................................................................12

Maximum Span Moment on Span A-B

M_x = R_AB - wx²/2...........................................................................13

At maximum moment

dM_x/d_x = R_AB - wx = 0

x = R_AB/w = 4L/10 = 0.4L

Substituting for x = 4L/10 in equ (13)

M_max = (4wL/10).(4L/10) - (w/2). (4L/10)²

              = 16wL²/100( 1-1/2) = 16wL²/200

 M_max = 16wL²/200 = 0.08wL²......................................................14

Maximum Span Moment on Span B-C

M_x = R_BC - wx²/2 - |M_B| ...............................................................15

At maximum moment

dM_x/d_x = R_BC - wx = 0

x = R_BC/w = 5L/10

Substituting for x = 5L/10 in equ (15)

M_max = (5wL/10).(5L/10) - (w/2). (5L/10)² -wL²/10

= 25wL²/200 - wL²/10

M_max = wL²/40 = 0.025wL²............................................................16