How to Carry Out Concrete Mix Design

By

-

August 10, 2017

Concrete mix design is the process of specifying and proportioning the quantity of concrete ingredients (cement, sand, gravel, and water) required to produce concrete with a specified fresh and hardened properties. Nowadays, engineers should have a deeper understanding of concrete mixes, than just specifying say 1:2:4 concrete mix ratio for their construction works.

Under normal and well-controlled conditions, 1:2:4 mix ratio should yield concrete with 28 days compressive strength greater than 20 N/mm². However, there are variations in results when batching by weight and when batching by volume. This situation demands that to produce concrete of a specified compressive strength, you have to accurately carry out concrete mix design, which incorporates the tendency of failure during tests.

A little consideration from concrete mix design results will show that the mix ratio for batching by weight, and for batching by volume are not basically the same for obtaining a specified grade of concrete.

Ingredients for concrete mix design — Figure 1: Basic ingredients for concrete production.

The compressive strength of concrete is a very variable quantity. Therefore, when carrying out a concrete mix design, it is important to target a higher average strength so that every part of the structure will meet the specified strength. Statistical standard deviation which is a measure of scatter or dispersion of strength about the mean is normally used to take care of this.

The target strength of concrete (f_m) during mix design is given by;

f_m = f_min + KSD ————– (1)

Where;

f_m is the mean compressive strength
f_min is the minimum compressive strength of the concrete. In the Eurocodes, this is called the characteristic strength of the concrete (f_ck), while in the US, it is called the design strength (f_c‘).
K is the probability factor which is usually taken as 1.64 or 2.33 to express the probability of 1 in 20 and 1 in 100 respectively, for the strength to fall below the minimum strength.
SD is the standard deviation which is best obtained by considering the previous test results obtained using the same materials, the same procedure, and under the same supervision.

The term KSD is normally referred to as the margin.

The probability of strength values in the range f_ck ± KSD and below f_ck – KSD (risk) for normal distribution is shown in Table 1.0 below;

Table 1: The probability of strength values in the range f_ck ± KSD and below f_ck – KSD (risk) for normal distribution (Neville and Brooks, 2010)

Probability factor (k)	Probability of strength values in the range f_ck ± KSD (%)	Probability of strength values below f_ck – KSD (%)
1.00	68.2	15.9 (1 in 6)
1.64	90.0	5.0 (1 in 20)
1.96	95.0	2.5 (1 in 40)
2.33	98.0	1.0 (1 in 100)
3.00	99.7	0.15 (1 in 700)

In the Eurocodes and in many other codes, the range of the risk of 1 in 20 is recommended for concrete tests. This means that in 20 concrete cubes, there is a probability of only one cube not meeting the required strength. When statistical data is not available for obtaining the standard deviation, the values in table 2 according to ACI code could be used.

Table 2: Required average compressive strength when data is not available to establish a standard deviation

Specified Compressive strength f’_c(MPa)	Required Average Compressive Strength f’_cr (MPa)
Less than 21	f’_c+ 7.0
21 to 35	f’_c+ 8.5
Over 35	1.10f’_c + 5

The process of carrying out concrete mix design are as follows;

Test the materials to be used for the concrete mix design. For the aggregates, it is important to obtain the oven-dry relative density, fineness modulus, absorption, and bulk density.
Establish the target strength of the concrete at 28 days by considering the characteristic strength and the margin which is based on the standard deviation.
Establish the required water-cement cement ratio for the target strength using relevant tables
Calculate the water requirement for the desired slump using relevant tables
Use the calculated water content to calculate the cement content using the water-cement ratio relationship
Calculate the mass of coarse aggregates using the bulk volume of coarse aggregate per unit volume of concrete
Calculate the mass of fine aggregates
Summarise your results

Worked Example

Let us consider the trial mix design for a concrete of minimum specified strength of 25 N/mm², to be employed in the construction of the floor beams and slab of a building.

Materials Analysis
Coarse aggregate: Crushed granite of nominal maximum size of 20mm
Oven dry relative density = 2.68
Fineness modulus = 2.60
Absorption = 0.4% (saturated surface dry)
Bulk density = 1650 kg/m³

Specific gravity of cement = 3.15

Fine aggregate: Sharp sand from river
Relative density = 2.64
Absorption = 0.60 %
Bulk density = 1600 kg/m³

The target strength can be obtained from the relation below;
f_m = f_ck + KSD
f_m = 25 + 8.5 = 33.5 N/mm²

Water-Cement Ratio

The relationship between the water to cement ratio for non-air-entrained concrete (normal concrete) and the 28 days compressive strength is given in Table 3.

Table 3: Relationship between compressive strength and water to cement ratio

Compressive strength at 28 days (MPa)	Water to cement ratio by mass
45	0.38
40	0.42
35	0.47
30	0.54
25	0.61
20	0.69
15	0.79

Therefore, Water – Cement ratio for non-air entrained 33.4 N/mm² concrete = 0.491 (interpolating from Table 3)

Cement and Water Content
Water content is normally estimated from workability requirements, which is guided by the slump. The range of slump required for different types of construction is given in Table 4;

Table 4: Different slumps for different types of construction works

Concrete Construction	Maximum slump (mm)	Minimum slump (mm)
Reinforced foundation walls and footings	75	25
Plain footings, caissons, and substructure walls	75	25
Beams and reinforced walls	100	25
Columns in buildings	100	25
Pavements and slabs	75	25
Mass cocrete	75	25

For maximum size of aggregate of 19mm, and a slump of 75mm, the Table 5 gives a water demand of 205 kg/m³

Table 5: Approximate mixing water and target air content requirements for different slumps and nominal maximum sizes of aggregate

Slump	Water content (kg/m³) for 9.5 mm aggregate size	Water content (kg/m³) for 12.5 mm aggregate size	Water content (kg/m³) for 19 mm aggregate size	Water content (kg/m³) for 25 mm aggregate size	Water content (kg/m³) for 37.5 mm aggregate size	Water content (kg/m³) for 50 mm aggregate size	Water content (kg/m³) for 75 mm aggregate size	Water content (kg/m³) for 150 mm aggregate size
25 – 50	207	199	190	179	166	154	130	113
75 – 100	228	216	205	193	181	169	145	124
150 -175	243	228	216	202	190	178	160	–
Approximate amount of entrapped air (%)	3	2.5	2	1.5	1	0.5	0.3	0.2

Therefore cement content;
205/C= 0.491; Therefore the cement content (C) = 205 / 0.491 = 417.515 kg/m³

Mass of Coarse Aggregates

For 19mm aggregate with a fineness modulus of 2.60, the bulk volume of dry rodded coarse aggregate per m³ of concrete is 0.64 (see Table 6).

Table 6: Bulk volume of coarse aggregate per unit volume of concrete

Nominal maximum size of aggregate (mm)	Bulk volume (2.40 fineness modulus)	Bulk volume (2.60 fineness modulus)	Bulk volume (2.80 fineness modulus)	Bulk volume (3.0 fineness modulus)
9.5	0.50	0.48	0.46	0.44
12.5	0.59	0.57	0.55	0.53
19	0.66	0.64	0.62	0.60
25	0.71	0.69	0.67	0.65
37.5	0.75	0.73	0.71	0.69
50	0.78	0.76	0.74	0.72
75	0.82	0.80	0.78	0.76
150	0.87	0.85	0.83	0.81

Therefore, mass of coarse aggregate (M_c) per m³ of concrete = 0.64 × 1650 = 1056 kg/m³
Approximate air content = 2%

Mass of fine aggregates
The mass of coarse aggregate can be estimated using the relationship below;

Mass of fine aggregate M_f = γ_f [1000 – (W – C/γ + Mc/γ_c + 10A)] ————- (2)

Where;
γ_f = Specific gravity of fine aggregate (saturated surface dry)
W = Mixing water requirement
C = Cement Content
γ = Specific gravity of cement (take value as 3.15 unless otherwise specified)
Mc = Coarse Aggregate content
γ_c = Specific gravity of coarse aggregate (saturated surface dry)
A = Air content (%)

Therefore:
M_f = 2.64 [1000 – (205 + (417.515/3.15) + (1056/2.68) + (10 × 2))] = 655.843 kg/m³

Final Volume computations
Water = 205 / (1 × 1000) = 0.205 m³
Cement = 417.515 / (3.15 × 1000) = 0.1325 m³
Air = 2/100 = 0.02
Coarse aggregate = 1056 / (2.68 × 1000) = 0.394 m³
Fine aggregate = 655.843 / (2.64 × 1000) = 0.248 m³

SUMMARY OF TRIAL MIX DESIGN
By weight
Water = 205 kg/m³
Cement = 417.515 kg/m³
Coarse Aggregate = 1056 kg/m³
Fine Aggregate = 655.843 kg/m³
Yield of concrete = 2334.36 kg/m³
Mix ratio by weight (Cement : Fine Aggregate : Coarse Aggregate) = (1:1.57:2.53)

By volume
Water = 0.205 m³
Cement = 0.1325 m³
Coarse Aggregate = 0.394 m³
Fine Aggregate = 0.248 m³

Mix ratio by volume (Cement : Fine Aggregate : Coarse Aggregate) = (1:1.87:2.97)

MIX DESIGN WITHOUT CONSIDERING TEST MARGIN
However, carrying out mix design for 25 N/mm² grade of concrete without considering the margin, we can obtain the following result using the steps described above;

Water-cement ratio
Water – Cement ratio for non-air entrained 25 N/mm² concrete = 0.61 (see Table 3 above)

Water and Cement Demand
For maximum size of aggregate of 19mm, and a slump of 75mm, gives a water demand of 205 kg/m³.
Therefore the cement content C = 205/0.61= 336.06 kg/m³

Mass of Coarse Aggregate
For 19mm aggregate with fineness modulus of 2.60, the bulk volume of dry rodded coarse aggregate per m³ of concrete is 0.64. Therefore;
Weight per m³ = 0.64 × 1650 = 1056 kg/m³

Approximate air content = 2%

Mass of Fine Aggregate
Mass of fine aggregate M_f = γ_f [1000 – (W – C/γ + Mc/γ_c + 10A)]
M_f = 2.64 [1000- (205 + (336.06/3.15) + (1056/2.68) + 10(2))] = 724.11 kg/m³

Volume computations
Water = 205 / (1 × 1000) = 0.205 m³
Cement = 336.06 / (3.15 × 1000) = 0.1066 m³
Air = 2/100 = 0.02%
Coarse aggregate = 1056 / (2.68 × 1000) = 0.394 m³
Fine aggregate = 724.11 / (2.64 × 1000) = 0.274 m³

Summary of trial mix design without considering the margin
By weight
Water = 205 kg/m³
Cement = 336.06 kg/m³
Coarse Aggregate = 1056 kg/m³
Fine Aggregate = 724.11 kg/m³
Yield of concrete = 2321.17 kg/m³
Mix ratio by weight (Cement:Fine Aggregate:Coarse Aggregate) = (1: 2.15: 3.142)

By volume
Water = 0.205 m³
Cement = 0.1066 m³
Coarse Aggregate = 0.394 m³
Fine Aggregate = 0.274 m³
Mix ratio by volume (Cement:Fine Aggregate:Coarse Aggregate) = (1:2.57:3.696)

Thank you for visiting Structville…….
We love you, and you can like our facebook fan page on;
www.facebook.com/structville

Design of Biaxial Reinforced Concrete Columns

By

Ubani Obinna Uzodimma

-

August 9, 2017

A biaxial column is a column that is subjected to compressive axial force and bending moment in the two planes. They are usually found at the corners of a building or at locations where the beam spans and/or loading are not equal. In the design of biaxial reinforced concrete columns, a non-linear analysis method is required, which should take into account second-order, imperfection, and biaxial bending effects.

Biaxial Reinforced Concrete Columns — The columns circled in red are biaxially loaded

EN 1992-1-:2004 (Eurocode 2) did not give an express method of designing biaxial columns other than working from the first principles. This may not be very easy to achieve without the use of charts or computer programs. To develop such a program, one can divide the compression zone into strips that are parallel with the neutral axis of the section, and calculate the stress in each strip using the parabolic-rectangular diagram. The force and moment at each strip in the x and y-axis can be summed up in the ultimate limit state to find the moment and axial force developed by the concrete in compression.

There are however simplified methods of dealing with biaxial bending in reinforced concrete structures. An example is the approach given in clause 5.8.9(4) of EN 1992-1-1 for the design of biaxially bent sections in slender columns. This is based on the observation that the form of the M_x – M_y interaction diagram can conveniently be represented by a super-ellipse. A super-ellipse has an equation of the form;

x^a + y^a = k —— (1)

If a = 2, this equation becomes a circle, while if a = 1 it describes a straight line. At loads approaching the squash load, the M_x – M_y interaction diagram approaches a circle, while in the region of the balance point it is close to a straight line. Clause 5.8.9(4) adopts the equation below as a means of describing the complete interaction surface;

(M_E,dz/M_R,dz )^a + (M_Ed,y/M_Rd,y )^a ≤ 1.0 —— (2)

The proximity to the squash load can be defined using the N/N_uz ratio parameter, and EN 1992-1-1 assumes the relationship between this parameter and the exponent ‘a‘ given in the Table below. Intermediate values may be interpolated.

N_Ed/N_Rd	0.1	0.7	1.0
a	1.0	1.5	2.0

In the Table above, N_Rd is the squash load of the column, and may be calculated from

N_Rd = A_cf_cd + A_sf_yd —— (3)

The difficulty with the approach from the practical point of view is that it cannot be used as a direct design method since N_Rd can only be established once the reinforcement area has been found. It, therefore, has to be used iteratively. An initial estimate is made of N/N_uz, the section is designed, a corrected value of N/N_uz can then be estimated, and the process repeated until a correct solution is obtained.

A much simpler, but considerably more approximate method to the design of biaxial columns has been adopted in BS 8110. The design is carried out for an increased uniaxial moment, which takes account of the biaxial effects. The required uniaxial moment is obtained from whichever is appropriate of the two relationships set out below:

if M_x/h’ > M_y/b’ then M’_x = M_x + βh’M_y/b’ —— (4)
if M_x/h’ < M_y/b’ then M’_y = M_y + βb’M_x /h’ —— (5)

In the above relationships, M_x and M_y are the design moments about the x and y axes, respectively, while M’_x and M’_y are the effective uniaxial moments for which the section is actually designed. b’ and h’ are the effective depths of the column section (see image below). The factor β is defined in BS 8110 as a function of N/bhf_cu. In terms of f_ck, it can be obtained from the relationship;

β = 1 – N/bhf_ck (0.3 < β < 1.0) —— (6)

BIAXIALLY%2BBENT%2BCOLUMN%2B%2528BS%2B8110%2529

This approach has the great advantage of being very simple. It is, however, an approximate approach.

Example on the Design of Biaxial Reinforced Concrete Columns to BS 8110 and Eurocode 2

A reinforced concrete column fixed at both ends is subjected to the loading condition shown below. We are required to obtain the appropriate longitudinal reinforcement for the column using BS 8110-1:1997 and Eurocode 2.

The column is carrying longitudinal and transverse beams of depth 600mm and width 300mm. It is also supported by beams of the same dimension. The centre to centre height of the column is 3500mm. The plan view of the arrangement of the beams and column is as shown below.

DESIGN ACCORDING TO BS 8110-1:1997
N = 716.88 kN;
TOP: M_x-x = 175.87 kNm; M_y-y = 35.52 kNm
BOTTOM: M_x-x= -85.832 kNm; M_y-y = -25.269 kNm

Concrete grade f_cu = 30 N/mm²
Yield Strength of reinforcement f_y = 460 N/mm²
Concrete cover = 40mm
L_o = 3.5m, Effective length L_e = 0.75 × (3500 – 600) = 2175 mm;
Size of column = 400 x 300mm;
Slenderness = 2175/300 = 7.25 < 15. Thus column is short.

Effective depth about x-x axis
h’ = (400 – 40 – 12.5 – 10) = 337.5 mm

Effective depth about y-y axis
b’ = (300 – 40 – 12.5 – 10) = 237.5 mm

M_x/h’ = 521.109 kN; M_y/ b’ = 149.558 kN

Axial-load ratio
N_ratio = (N × 1000) / (F_cu × b × h) = (716.88 × 1000) / (30 × 400 × 300) = 0.1991

From Table 3.22 of BS 8110-1:1997, for N_ratio = 0.1991
Therefore, β = 1 – 1.1644(0.1991) = 0.7681

As M_x/h’ exceeds M_y/b’
M_x‘ = M_x + βh’/b’M_y
M_x‘ = 175.87 + [0.7681 × (0.3375)/(0.2375 ) × 35.52] = 214.64 kNm

Minimum Eccentricity in Columns
According to clause 3.8.2.4 of BS 8110-1:1997, at no section in a column should the design moment be taken as less than that produced by considering the design ultimate axial load as acting at a minimum eccentricity, e_min equal to 0.05 times the overall dimension of the column in the plane of bending considered but not more than 20 mm.

Where biaxial bending is considered, it is only necessary to ensure that the eccentricity exceeds the minimum about one axis at a time.

In the x-x direction = e_min = 0.05 × 400 = 20mm, therefore adopt 20mm, M_x = 716.88 × 0.02 = 14.3376 kNm < 214.64 kNm

In the y-y direction = e_min = 0.05 × 300 = 15mm, therefore adopt 15mm, M_y = 716.88 × 0.015 = 10.7532 kNm < 214.64 kNm

Section design ratios for chart entry
Axial load ratio N_ratio = (N × 1000)/(F_cu × b × h) = (716.88 × 1000)/(30 × 400 × 300) = 0.1991

M_ratio = M/(F_cu × b × h²) = (214.64 × 10⁶) / (30 × 300 × 400²) = 0.149

With d’/h = 337.5/400 = 0.84375

From chart, (ρ × F_y)/(F_cu) = 0.286 Therefore, ρ = (0.286 × 30) / 460 = 0.01865
A_SCreq = 0.01865 × 400 × 300 = 2238.26 mm²

Provide 6Y25mm (A_sprov = 2946 mm²)
Maximum area of reinforcement = 0.06bh = 0.06 × 400 × 300 = 7200 mm²

DESIGN ACCORDING TO EUROCODE 2
Clause 5.8.9(2) of EN 1992-1-1:2004 permits us to perform separate designs in each principal direction, disregarding biaxial bending as a first step. Imperfections need to be taken into account only in the direction where they will have the most unfavourable effect. However, in this example, we have carried out imperfection analysis in both directions.

N_Ed = 716.88 KN

Elastic Moments
Y – direction: M₀₁ = 175.87 kNm; M₀₂ = -85.832 kNm
Z – direction: M₀₁ = 35.52 kNm; M₀₂ = -25.269 kNm

Clear column height (L) = 3500 – 600 = 2900 mm

Calculation of the effective height of the column (L_o)
Let us first of all calculate the relative stiffnesses of the members in the planes of bending.

In the y-direction;
Second moment of area of beam 1 (I₁) = bh³/12 = 0.3 × 0.6³/12 = 0.0054 m⁴
Stiffness of beam 1 (since E is constant) = 4I₁/L = (4 × 0.0054) / 6 = 0.0036

Second moment of area of beam (I₂) = bh³/12 = 0.3 × 0.6³/12 = 0.0054 m⁴
Stiffness of beam 2 (since E is constant) = 4I₂/L = (4 × 0.0054) / 3.5 = 0.00617

Second moment of area of column (I_c) = bh³/12 = 0.0016 m⁴
Stiffness of column = 4I_c/L = (4 × 0.0016) / 3.5 = 0.001828

For compression members in regular braced frames, the slenderness criterion should be checked with an effective length l₀ determined in the following way:

L_o = 0.5L √[(1 + k₁)/(0.45 + k₁)) × (1 + k₂)/(0.45 + k₂))]

Where;
k₁, k₂ are the relative flexibilities of rotational restraints at ends 1 and 2 respectively
L is the clear height of the column between the end restraints

k = 0 is the theoretical limit for rigid rotational restraint, and k = ∞ represents the limit for no restraint at all. Since fully rigid restraint is rare in practise, a minimum value of 0.1 is recommended for k₁ and k₂.

In the above equations, k₁ and k₂ are the relative flexibilities of rotational restraint at nodes 1 and 2 respectively. If the stiffness of adjacent columns does not vary significantly (say, the difference not exceeding 15% of the higher value), the relative flexibility may be taken as the stiffness of the column under consideration divided by the sum of the stiffness of the beams (or, for an end column, the stiffness of the beam) attached to the column in the appropriate plane of bending.

Remember that we will have to reduce the stiffness of the beams by half to account for cracking;

k₁ = k₂ = 0.001828 / (0.0018 + 0.003085) = 0.3743

L_o = 0.5 × 2900√[((1 + 0.3743)/(0.45 + 0.3743)) × (1 + 0.3743)/(0.45 + 0.3743)] = 2647.77 mm

Compare with BS 8110’s 0.75L = 0.75 × 2900 = 2175 mm

In the z-direction;
Second moment of area of beam 3 (I₃) = bh³/12 = 0.3 × 0.6³/12 = 0.0054 m⁴
Stiffness of beam 3 (since E is constant) = 4I₁/L = (4 × 0.0054) / 3.5 = 0.00617

Second moment of area of column (I_c) = bh³/12 = 0.0009 m⁴
Stiffness of column = 4I_c/L = (4 × 0.0009) / 3.5 = 0.00102857

k₁ = k₂ = 0.00102857/(0.003085) = 0.3334

l_o = 0.5 × 2900 √[(1 + 0.3334) / (0.45 + 0.3334) × (1 + 0.3334) / (0.45 + 0.3334)] = 2675.293 mm

Radius of gyration

i_x = h/√12 = 400/√12 = 115.47
i_z = b/√12 = 300/√12 = 86.602

Slenderness in the x-direction (λ_x) = 2647.77/115.47 = 22.930
Slenderness in the z-direction (λz) = 2675.293/86.602 = 30.892

Critical Slenderness for the y-direction
λ_lim = (20.A.B.C)/√n
A = 0.7
B = 1.1
C = 1.7 – M₀₁/M₀₂ = 1.7 – [(-85.832)/175.87] = 2.188
n = N_Ed / (A_c f_cd)
N_Ed = 716.88 × 10³ N
A_c= 400 × 300 = 120000 mm²
f_cd = (α_cc f_ck)/1.5 = (0.85 × 30)/1.5 = 17 N/mm²
n = (716.88 × 10³) / (120000 × 17) = 0.3514
λ_lim = (20 × 0.7 × 1.1 × 2.188 )/√0.3514 = 56.842

22.930 < 56.842, second order effects need not to be considered in the y-direction

Critical Slenderness for the z-direction
A = 0.7
B = 1.1
C = 1.7 – M₀₁/M₀₂ = 1.7 – [(-25.269)/35.52] = 2.411
n = N_Ed / (A_c f_cd)
N_Ed = 716.88 × 10³ N
Ac = 400 × 300 = 120000 mm²
fcd = (α_cc f_ck)/1.5 = (0.85 × 30)/1.5 = 17 N/mm²
n = (716.88 × 10³) / (120000 × 17) = 0.3514

λ_lim = (20 × 0.7 × 1.1 × 2.411 )/√0.3514 = 62.634

30.892 < 62.634, second-order effects need not be considered in the z-direction

Design Moments (y-direction)

Y – direction: M₀₁ = 175.87 kNm; M₀₂ = -85.832 kNm

e₁ is the geometric imperfection = (θ_i l₀/2) = (1/200) × (2647.77/2) = 6.619 mm

Minimum eccentricity e₀ = h/30 = 400/30 = 13.333 mm. Since this is less than 20mm, take minimum eccentricity = 20mm (clause 6.1(4) EC2).

Minimum design moment = e₀N_Ed = 20 × 10^-3 × 716.88 = 14.3376 kNm

First order end moment M₀₂ = M_Top + e_iN_Ed
e_iN_Ed = 6.619 × 10^-3 × 716.88 = 4.745 KNm
M₀₂ = M_Top + e_iN_Ed = 175.87 + 4.754 = 180.624 KNm

Longitudinal Steel Area
d₂ = C_nom + ϕ/2 + ϕ_links = 40 + 12.5 + 10 = 62.5 mm

d₂/h = 62.5/400 = 0.156
Let us read from chart; d₂/h = 0.15;

M_Ed/(f_ck bh²) = (180.624 × 10⁶)/(30 × 300 × 400² ) = 0.125
N_Ed/(f_ck bh) = (716.88 × 10³) / (30 × 300 × 400) = 0.199

From the chart, (A_s F_yk)/(bhf_ck) = 0.23

Area of longitudinal steel required A_s,req = (0.23 × 30 × 400 × 300)/460 = 1800 mm²

A_s,min = 0.10 N_Ed/f_yd = (0.1 × 716.88)/400 = 0.179 mm² < 0.002 × 400 × 300 = 240 mm²
A_s,max = 0.04bh = 4800 mm²
Provide 4Y25mm (Asprov = 1964 mm²)

Design Moments (z-direction)z – direction: M₀₁ = 35.52 kNm; M₀₂ = -25.269 kNm

e₁ is the geometric imperfection = (θ_i l₀/2) = (1/200) × (2675.293/2) = 6.688 mm

Minimum eccentricity e₀ = h/30 = 400/30 = 13.333 mm. Since this is less than 20mm, take minimum eccentricity = 20mm (clause 6.1(4) EC2)

Minimum design moment = e₀N_Ed = 20 × 10^-3 × 716.88 = 14.3376 kNm

First order end moment M₀₂ = M_Top + e_iN_Ed
e_iN_Ed = 6.688 × 10^-3 × 716.88 = 4.794 KNm
M₀₂ = M_Top + e_iN_Ed = 35.52 + 4.794 = 40.314 KNm

Longitudinal Steel Area
d₂ = C_nom + ϕ/2 + ϕ_links = 40 + 12.5 + 10 = 62.5 mm
d₂/h = 62.5/400 = 0.156

Reading from chart No 1; d₂/h = 0.156;
M_Ed/(f_ck bh²) = (40.314 × 10⁶)/(30 × 300 × 400²) = 0.0279
N_Ed/(f_ck bh) = (716.88 × 10³) / (30 × 300 × 400) = 0.199

From the chart, (A_s F_yk)/(bhf_ck) = 0.00 (Nominal reinforcement required)

Biaxial Effects

Check if λ_y / λ_z ≤ 2.0 and λ_z/λ_y ≤ 2.0

17.203/18.159 = 0.9473 < 2.0, and 18.159/17.203 = 1.0556 < 2.0

Furthermore, let us also check;
(e_y/h_eq) / (e_z/b_eq) ≤ 0.2 or (e_z / b_eq) / (e_y / h_eq) ≤ 0.2

The definition of eccentricity is given in Figure 5.8 of EC2

e_y = M_Ed,y/N_Ed = (180.624 × 10⁶) / (716.88 × 10³) = 251.958 mm
e_z = M_Ed,z/N_Ed = (40.314 × 10⁶) / (716.88 × 10³) = 56.235 mm

h_eq = i_z.√12 = 300 mm
b_eq = i_y.√12 = 400 mm

(e_y/h_eq) ÷ (e_z/b_eq) = 251.958/300 ÷ 56.235/400 = 5.9739 > 0.2

Therefore we have to check for biaxial bending interaction;

(M_E,dz/M_R,dz )^a + (M_Ed,y/M_Rd,y )^a ≤ 1.0

(A_s F_yk)/(bhf_ck) = (1964 × 460) / (30 × 300 × 400) = 0.251

Therefore from the chart; M_Rd/(f_ckbh²) = 0.13

M_Rd = (0.13 × 30 × 300 × 400²) × 10^-6 = 187.2 KNm

N_Rd = A_cf_cd + A_sf_yd
N_Rd = [(300 × 400 × 17) + (1964 × 400)] × 10^-3 = 2825.6 kN
N_Ed/N_Rd = 716.88 / 2825.6 = 0.2537

To evaluate the value of a, let us look at the table below as given in Clause 5.8.9(4) of EC2

By linear interpolation, a = 1.0 + [(0.2537 – 0.1 )/(0.7 – 0.1)] × (1.5 – 1.0) = 1.128

(M_E,dz/M_R,dz )^a + (M_Ed,y/M_Rd,y )^a ≤ 1.0

(40.314/187.2)^1.128 + (180.624/187.2)^1.128 = 0.1769 + 0.9606 = 1.1375 > 1.0.
This is not ok, and this shows that 4Y25 is inadequate for the biaxial action on the column.

Let us increase the area of steel to 6Y25mm (A_sprov = 2946 mm²)

Let us check again;
(A_s F_yk)/(bhf_ck) = (2964 × 460) / (30 × 300 × 400) = 0.3787
Therefore from the chart; M_Rd/(f_ckbh²) = 0.175
M_Rd = (0.175 × 30 × 300 × 400²) × 10^-6 = 252 kNm

N_Rd = A_cf_cd + A_sf_yd
N_Rd = [(300 × 400 × 17) + (2964 × 400)] × 10^-3 = 3225.6 kN

N_Ed/N_Rd = 716.88 / 3225.6 = 0.222

By linear interpolation, a = 1.0 + [(0.222 – 0.1 )/(0.7 – 0.1)] × (1.5 – 1.0) = 1.102

(M_E,dz/M_R,dz )^a + (M_Ed,y/M_Rd,y )^a ≤ 1.0

(40.314/252)^1.102 + (180.624/252)^1.102 = 0.132 + 0.6928 = 0.825 > 1.0.

This shows that 6Y25mm is adequate for the column, and shows some agreement with result from BS 8110-1:1997.

Thank you for visiting Structville

Visit our facebook fan page;
www.facebook.com/structville

How to Estimate the Cost of Tiling a 3 Bedrooms Flat

By

Ubani Obinna Uzodimma

-

August 7, 2017

Using tiles to cover the surfaces of floors, walls, countertops, etc can be a very attractive option for finishes in a building. Tiles can provide hard, durable, and aesthetically pleasing surfaces in a building provided that the tiling job is properly done. There are some basic principles that must be followed in order to achieve quality, beautiful and durable tiling.

For a homeowner and/or contractor, cost is a major factor that influences the decision and quality of tiling to be done. The basic factor that can influence the final cost of tiling a house are;

Cost of the tiles
Cost of screeding
Cost of adhesives
Cost of grouting cement
Cost of tiling accessories such as edge trimmers
Cost of labour
Size of area to be tiled

In this article, we are going to show how to estimate the cost of tiling a house in Nigeria using a three bedrooms apartment as an example. The floor plan of the three bedrooms apartment is shown in Figure 1. It is desirous to tile all the floors of the building according to the following specifications;

(1) All bedrooms, kitchen, and rear balcony/sit-out: 40cm × 40cm made in Nigeria ceramic tiles

(2) The sitting room, dining, and front balcony/sit-out: (30cm × 60cm) imported Virony glazed tiles;

(3) All the toilets and bathrooms: (30cm × 30cm) made in Nigeria ceramic tiles.

We are required to estimate the total cost of tiling the entire floor of the building, considering materials and labour.

FLOOR%2BPLAN — **Figure 1:** Floor plan of a three bedrooms apartment

Material Prices (informative)

It is very important for you to make an adequate market survey, in order to know the exact prices of materials in your area. This is the only chance you have of making a reasonably accurate estimate. The prices stated below are a bit representative, but merely informative. Note that tiles with designs may be more expensive than plain tiles or tiles with a single colour.

(40cm × 40cm) Ceramics Tiles = ₦3,360 per carton (12 pieces per carton – 1.92 square metre)

(30cm × 60cm) Porcelain Glazed Tiles = ₦5,760 per carton (8 pieces per carton – 1.44 square metre)

(30cm × 30cm) Ceramic Tiles = ₦2,700 per carton (17 pieces per carton – 1.53 square metre)

1 tonne of sand = ₦4,000

1 bag of cement = ₦4,000
5 kg of white cement = ₦2,500
Water seal = ₦700 per kg
Tile gum (20kg) = ₦1,450

BILL NO 1: TILES AND TILING ACCESSORIES
(a) Tiling of the bedrooms, kitchen, backyard sit-out, and store with (40cm × 40cm) Time Ceramics Tiles

We have to calculate the floor areas where we are going to apply the (40cm × 40cm) tiles;

Bedroom 1 = (3.6m × 3.45m) = 12.42 m²
Bedroom 2 = (3.6m × 3.6m) = 12.96 m²
Bedroom 3 = (3.15m × 3.17m) = 9.9855 m²
Kitchen = [(3.35m × 2.35m) + (1.37m × 1.075m)] = 9.34525 m²
Store = (2.025m × 1.775m) = 3.59 m²
Kitchen Sit-out = (1.37m × 2.075m) = 2.84275 m²
Total Area = 51.148 m²
Area of a piece of tile = (0.4m × 0.4m) = 0.16 m²

The number of pieces of tiles required to complete the floor areas = 51.148/0.16 = 319.675 pcs

Therefore, the number of cartons required = 319.675/12 = 26.639 cartons.

We will therefore provide 30 cartons of (40cm × 40cm) ceramic tiles. Extra tiles will account for damages, and for skirting of the foot of the walls.

Therefore the cost of the (40cm x 40cm) tiles = 30 × ₦3,360 = ₦100,800

(b) Tiling of the sitting room, dining, and entrance porch with (60cm x 30cm) tiles
The floor areas where we are going to apply the (60cm × 30cm) tiles;

Sitting room = (5.65m × 4.625m) = 26.13125 m²
Dining = (2.35m × 3.35m) = 7.8725 m²
Entrance Porch = (3.78m × 1.475m) = 5.5755 m²
Total Area = 39.579 m²
Area of a piece of tile = (0.6m × 0.3m) = 0.18 m²

The number of pieces of tiles required to complete the floor areas = 39.579/0.18 = 219.883 pcs;

Therefore, the number of cartons required = 219.883/8 = 27.485 cartons

We will therefore also provide 30 cartons of (60cm × 30cm) ceramic tiles.

Therefore the cost of the (60cm x 30cm) tiles = 30 × ₦5,760 = ₦172,800

(c) Tiling of the bathroom and toilet floors with (30cm x 30cm) tiles
On calculating the floor areas where we going to apply the (30cm × 30cm) tiles;

Bathroom and Toilet 1 and 2 = 2(2.0m × 1.1m) = 4.4 m²
Bathroom and toilet 3 = (2.0m × 0.985m) = 1.97 m²
Visitor’s toilet = (2.0m × 0.875m) = 1.75 m²
Total Area = 8.12 m²
Area of a piece of tile = (0.3m × 0.3m) = 0.09 m²

The number of pieces of tiles required to complete the floor areas = 8.12/0.09 = 90.222 pcs

Therefore, the number of cartons required = 90.222/17 = 5.307 cartons

We will therefore provide 7 cartons of (30cm × 30cm) ceramic tiles

Therefore the cost of the (30cm x 30cm) tiles = 7 × ₦2,700 = ₦18,900

(d) Loading, transportation, and offloading at the site (say) = ₦20,000

Therefore, sub-total 1 = ₦100,800 + ₦172,800 + ₦18,900 + ₦20,000 = ₦312,500

BILL NO 2: CEMENT AND SAND
We are going to make 25mm thick screeding to receive the tiles.

From above, you can verify that the total floor area of the building = 51.148 m²+ 39.579 m²+ 8.12 m²= 98.847 m²

Volume of mortar required = 98.847 × 0.025 = 2.471 m³
Using a mix ratio of 1:5

Cement required = 6 bags/m³
Sand required = 1800 kg/m³

Cement required = 14.826 bags (provide 15 bags)
Sand required = 4.447 tonnes (provide 5 tonnes)

Alternative calculation by site experience
From experience, 2 bags of cement is sufficient to complete the entire tiling process of (3.6m × 3.6m) room

Therefore;
2 bags of cement = 12.96 m²
x bags of cement = 98.847 m²

On solving;
x = 15.254 bags of cement; say 16 bags of cement

From experience, about 20 – 22 head pans of sharp sand is sufficient to complete the entire screeding process of (3.6m × 3.6m) room;

22 head pans = 22 × 0.0175 m³= 0.385 m³

Therefore;
0.385 m³ of sand = 12.96 m²
x m³ of sand = 98.847 m²

On solving;
x = 2.936 m³ sand = 4.844 tonnes of sand

Therefore, provide;

5 tonnes of sharp sand = ₦20,000
The cost of cement = 16 × ₦4,000 = ₦64,600
Water seal = 16 packs = 16 × ₦700 = ₦11,200
Tile gum = 17 bags × ₦1450 = ₦24,650
Other tiling accessories (allow) = ₦20,000
Transportation (say) = ₦5,000

Sub total 2 = ₦145,450

BILL NO 3: LABOUR
Cost of tiling 1m²of floor = ₦500

Therefore;
1 m² = ₦350
98.847 m² = x

On solving;
x = ₦49,423 (rounding up to the nearest 100)

Therefore, cost of labour = ₦50,000

GRAND TOTAL = ₦ 312,500 + ₦145,450 + ₦50,000 = ₦507,950

Therefore the cost of tiling the floor of the building = ₦507,950 + VAT + Engineer’s/Architect’s Profits

Thank you for reading

Like our facebook page;
www.facebook.com/structville

Free Vibration of Trusses: A solved Example

By

Ubani Obinna Uzodimma

-

August 5, 2017

Dynamics of structures is a special branch of structural analysis, which deals with the behaviour of structures subjected to dynamic loads (loads that vary with time). Such loads develop dynamic reactions, internal forces, and displacements in the structure. These values all change with time, and maximum values often exceed static load reaction values. Dynamic analysis of any structure often begins with free vibration analysis.

A structure undergoes free vibration when it is brought out of static equilibrium and can then oscillate without any external dynamic excitation. Free vibration of structures occurs with some frequencies which depend only on the parameters of the structures such as boundary conditions, distribution of masses, stiffnesses within the members etc, and not on the reason for the vibration.

At each natural frequency of free vibration, the structure vibrates in simple harmonic motion where the displaced shape (mode shape) of the structure is constant but the amplitude of the displacement varies in a sinusoidal manner with time. The number of natural frequencies in a structure coincides with the number of degrees of freedom in the structure. These frequencies are inherent to the given structure and are often referred to as eigenfrequencies. Each mode shape of vibration shows the form of an elastic curve which corresponds to a specific frequency.

Read also…
Deflection of trusses

Worked Example

A truss is loaded as shown below with a lumped mass of 5000 kg. We are to obtain the eigenfrequencies and mode shapes by assuming linear behaviour, neglecting damping, and assuming that the stiffness and inertial effects are time-independent.

E = 205 kN/mm²
Cross-sectional area = 0.00548 m² (UB 254 x 146 x 43)

A little consideration will show that the structure has two degrees of freedom which are given by the vertical and horizontal translation at node C. The lumped mass at the node will definitely participate in those movements. The displacements are given by Y₁ and Y₂ below.

Geometrical Properties
θ = tan^-1(3/4) = 36.869°
cos θ = 0.8
sin θ = 0.6
L_AC = √(3² + 4²) = 5m

γ = tan^-1(4/4) = 45°
cos γ = 0.7071
sin γ = 0.7071
L_BC = √(3² + 3²) = 4.243m

ANALYSIS OF STATE 1; Y₁ = 1.0

∑F_y = 0
-0.6F_AC – 0.7071F_CB = 1.0 ———– (1)

∑F_x = 0
-0.8F_AC + 0.7071F_CB = 0 ———– (2)

Solving (1) and (2) simultaneously
F_AC = -0.7143
F_CB = -0.8081

ANALYSIS OF STATE 2; Y₂ = 1.0

∑F_y = 0
-0.6F_AC – 0.7071F_CB = 0 ———– (3)

∑F_x = 0
-0.8F_AC + 0.7071F_CB = -1.0 ———– (4)

Solving (3) and (4) simultaneously
F_AC = 0.7143
F_CB = -0.6061

The frequency equation of a 2 degree of freedom system in terms of displacement is given by;

frequency%2Bequation%2Bin%2Bterms%2Bof%2Bdisplacement

Where;
M is the lumped mass
A is the amplitudes (Note that the system does not allow us to find the amplitudes but we can find the ratio of the amplitudes)
ω is the angular velocity
δ_ijis the axial displacement given by;

For the system above, we can easily compute this;

δ₁₁ = 1/EA[(-0.7143 × -0.7143 × 5) + (-0.8081 × -0.8081 × 4.243)] = 5.3219/EA
δ₂₂ = 1/EA[(0.7143 × 0.7143 × 5) + (-0.6061 × -0.6061 × 4.243)] = 4.1098/EA
δ₂₁ = δ₁₂ = 1/EA[(-0.7143 × 0.7143 × 5) + (-0.6011 × -0.8081 × 4.243)] = -0.4729/EA

To simplify our calculations, let δ₀ = 1/EA
Also let λ = 1/Mδ₀ω²
Therefore;

(5.3219 – λ)A₁ – 0.4729A₂ = 0
0.4729A₁ + (4.1098 – λ)A₂ = 0

On expanding;
λ² – 9.4317λ + 21.6484 = 0

On solving the quadratic equation;
λ₁ = 5.4845
λ₂ = 3.9472

You can verify that ω² = EA/λM

Therefore;
ω₁² = (0.1823 × 205 × 10⁶ × 0.00548)/5000 = 40.959
ω₂² = (0.2533 × 205 × 10⁶ × 0.00548)/5000 = 56.911

Hence;
ω₁= 6.399 rad/s
ω₂= 7.544 rad/s

To obtain the mode shapes;

For the first mode of vibration; Let A₁ = 1.0
(5.3219 – λ)A₁ – 0.4729A₂ = 0
(5.3219 – 5.4845) – 0.4729 A₂ = 0
Therefore, A₂ = -0.1626/0.4729 = -0.344
Φ₁ = [1.000 -0.344]^T

For the second mode of vibration; Let A₁ = 1.0
(5.3219 – λ)A₁ – 0.4729A₂ = 0
(5.3219 – 3.9472) – 0.4729 A₂ = 0
Therefore, A₂ = 1.3737/0.4729 = 2.9069
Φ₂ = [1.000 2.907]^T

The modal matrix is then defined as;

Free Vibration of Trusses: A solved Example 26

Remember Structville….
We have an interesting Facebook interactive page
Visit us at www.facebook.com/structville

Understanding Sign Conventions in Structural Analysis

By

Ubani Obinna Uzodimma

-

August 4, 2017

Table of Contents

Understanding sign conventions is one of the most important aspects of structural analysis. As a matter of fact, it is always a pausing point for all civil engineering students being introduced to structural analysis for the first time. If you do not understand sign conventions properly, it will be difficult to make significant progress in mastering structural analysis calculations.

The three most prominent internal forces in structural analysis calculations are the bending moment, shear force, and axial force. It is very common for people to define and state their sign convention before proceeding with any structural analysis problem. This is mainly due to variations in the selection of positive and negative coordinates.

For instance, most Indian and American textbooks will adopt the conventional cartesian coordinates system, and plot all positive bending moment upwards, and negative moments downwards, while most British and Scandinavian textbooks will plot positive moment downwards, and negative moments upwards.

However, it is important to note that what matters most in sign conventions is consistency. You must be consistent with whatever sign convention you choose to adopt, otherwise, errors will come in. If properly done, the values will remain the same, but the signs will differ.

Is there any standard approach?

I will prefer to answer yes because it is more appropriate to plot the bending moment diagram in the tension zone/fibre/face of a structural member. This is especially important in reinforced concrete structures where there is a need to provide tension reinforcement.

For instance, if you consider a continuous beam subjected to uniformly distributed load with the bending moment diagram plotted in the tension zone, you can easily point out and provide bottom reinforcements at the spans, and top reinforcements at the supports.

See an example below;

BENDING%2BMOMENT%2BPLOT — Consistency of bending moment diagram with the arrangement of reinforcements

From the figure above, you can see the consistency of the bending moment diagram with the typical arrangement of the reinforcement. This is because the bending moment diagrams were plotted in the tension zone, and of course, it means that the positive moment was plotted downwards. However, if the cartesian coordinates system was strictly followed, we would have been reversing the placement of the reinforcements with the bending moment diagram.

The same thing also applies to frames. You should plot your bending moment diagram in the tension zone of the members of the frame.

However, there are procedures that you must follow in your calculations in order to be consistent. By consistent, what I mean is that your positive bending moment will imply a sagging moment, while negative will mean a hogging moment. With this, you can plot your diagram directly as you have calculated, without having to interchange signs.

The summary of the procedures is given below.

Sign Convention for Beams

Bending moment

When you are coming from the left-hand side of the structure, all clockwise moments are positive and vice versa. When you are coming from the right-hand side of the structure, all anti-clockwise moments are positive and vice versa.

Shear force

When you are coming from the left, upward forces are positive, while downward forces are negative. When you are coming from the right, downward forces are positive, while upward forces are negative.

Axial force

Compressive axial forces are negative, while tensile axial forces are positive.

Understanding Sign Conventions in Structural Analysis 39

All you have to do is to look at the direction of the force. When coming from the left-hand side of beams, axial forces pointing towards the right means that the beam/section is in compression. When coming from the right, axial forces pointing towards the right means that the beam/section is in tension and vice versa

Sign Convention for Frames

For the sign of convention of frames, we are going to use the frame below as an example.

Bending moment

When coming from the left-hand side, all clockwise moments are positive, and anti-clockwise moments negative. All you have to do is to look at the point where you are taking your moment and observe the nature of rotation the force will produce. This is valid for both vertical and horizontal forces.

When coming from the right, all anti-clockwise moments are positive.

For frames, we plot positive moments inside, and negative moments outside the frame. See the example below.

Shear force

When coming from the left, upward vertical forces are positive and downward forces are negative. For horizontal forces on columns, forces pointing towards the right produce negative shear forces.

When coming from the right, downward vertical forces are positive, while upward vertical forces are negative. Horizontal forces on columns pointing towards the left will produce positive shear forces.

We plot positive shear forces outside the frame, and negative shear forces inside the frame.

Axial Force

When coming from the bottom of columns, upward vertical support reactions means that the column is in compression whether you are coming from the left or right.

When coming from the right-hand side of beams, forces pointing towards the right means that the beam is in compression and vice versa.

To me, you are free to select any location to plot your axial forces. Some people draw axial force diagrams to coincide with the centerline of the structure. In the classroom where I was trained, we draw negative axial forces outside the frame, and positive axial forces inside the frame. So it’s your choice to make.

Understanding Sign Conventions in Structural Analysis 43

I hope you find this article useful. If so, I will like you to share it.

Continue to visit Structville and tell your colleagues about us.

Our facebook page is at
www.facebook.com/structville

Structural Analysis of Determinate Arch-Frame Compound Structure

By

Ubani Obinna Uzodimma

-

August 2, 2017

Table of Contents

A parabolic arch structure is hinged on two interacting trusses as shown in the image above. The arch is also hinged at the crown (point F). We are expected to obtain the internal forces acting on the structure due to the externally applied uniform load on the arch, and the horizontal
concentrated load at point D. Since the arch is hinged on the truss, we can decompose the structure, and analyse the arch as a three-hinged arch, after which we transfer the support reactions from the arch to the truss. The detached arch structure is as shown below;

Support Reactions

∑M_E = 0
16Dy – (10 × 16²)/2 = 0
Therefore, Dy = 1280/16 = 80 KN
A little consideration from symmetry will also show that Ey = 80 KN

∑M_F^L = 0
8Dy – 4Dx – (10 × 8²)/2 = 0
8(80) – 4Dx – (10 × 8²)/2 = 0
– 4Dx + 320 = 0
Therefore, Dx = 320/4 = 80 KN = Ex

Geometric Properties of the Arch Section

The ordinate of the arch at any given horizontal length section is given by;

y = [4y_c (Lx – x²)] / L²
Where y_c is the height of the crown of the arch
y = [(4 × 4) × (16x – x²)]/16² = x – (x²/16)

dy/dx = y’ = 1 – x/8

At x = 0; y = 0
y’ = 1 – 0/8 = 1
sin⁡θ = y’/√(1 + y’²) = 1/√(1 + 1²) = 0.7071
cos⁡θ = 1/√(1 + y’²) = 1/√(1 + 1²) = 0.7071

At x = 4m;
y = x – x²/16 = 4 – (4²/16) = 3m
y’ = 1 – 4/8 = 0.5
sin⁡θ = y’/√(1 + y’²) = 0.5/√(1 + 0.5²) = 0.4472
cos⁡θ = 1/√(1 + y’²) = 1/√(1 + 0.5²) = 0.8944

At x = 8m;
y = x – x²/16 = 8 – (8²/16) = 4m
y’ = 1 – 8/8 = 0
sin⁡θ = y’/√(1 + y’²) = 0/√(1 + 0.5² ) = 0
cos⁡θ = 1/√(1 + y’²) = 1/√(1 + 0) = 1.0

At x = 12m;
y = x – x²/16 = 12 – (12²/16) = 3m
y’= 1 – x/8 = 1 – 12/8 = -0.5
sin⁡θ = y’/√(1 + y’²) = (-0.5)/√(1 + 0.5²) = -0.4472
cos⁡θ = 1/√(1 + y’²) = 1/√(1 + 0.5² ) = 0.8944

At x = 16m;
y = x – x²/16 = 16 – (16²/16) = 0
y’= 1 – 16/8 = 1 – 12/8 = -1.0
sin⁡θ = y’/√(1 + y’²) = (-1.0)/√(1 + 1² ) = -0.7071
cos⁡θ = 1/√(1 + y’²) = 1/√(1 + 1² ) = 0.7071

Internal Stresses in the Arch StructureBending Moment
M_D = 0
M₁ = (80 × 4) – (80 × 3) – (10 × 4²)/2 = 0
M_F = (80 × 8) – (80 × 4) – (10 × 8²)/2 = 0
M₂ = (80 × 12) – (80 × 3) – (10 × 12²)/2 = 0
M_E = 0

Shear force
Q_i = ∑V cos⁡θ – ∑H sin⁡θ
Q_D = (80 ×0.7071) – (80 × 0.7071) = 0 (No shear)
Q₁^L = Q₁^R [80 – (10 × 4)] × 0.8944 – (80 × 0.4472) = 0 (No shear)
Q_F = [80 – (10 × 8)] × 1.0 – (80 × 0) = 0 (No shear)
Q₂^L = Q₃^R = [80 – (10 × 12)] × 0.8944 – (80 × -0.4472) = 0 (No shear)
Q_E = [80 – (10 × 16)] × 0.7071 – (80 × -0.7071) = 0 (No shear)

Axial force
N_i = -∑V sin⁡θ – ∑H cos⁡θ
N_D = -(80 × 0.7071) – (80 × 0.7071) = -113.136 KN (Compression)
N₁^L = -[80 – (10 × 4)] × 0.4472 – (80 × 0.8944) = -89.44 KN
N_F = -[80 – (10 × 8)] × 0 – (80 × 1) = -80 KN (Compression)
N₂^L = N₃^R = -[80 – (10 × 12)] × -0.4472 – (80 × 0.8944) = -89.44 KN
N_E = -[80 – (10 × 16)] × -0.7071 – (80 × 0.7071) = -113.136 KN

We now transfer the reactive forces from the arch section to the the supporting trusses below;

θ = tan^-1⁡(6/8) = 36.869°
cos⁡θ = 0.8
sin⁡θ = 0.6

Analysis of Joint D

∑F_X = 0
-35 – 0.8F_AD + 0.8F_DB = 0
– 0.8F_AD + 0.8F_DB = 35 —————- (1)

∑F_Y = 0
-80 – 0.6F_AD – 0.6F_DB = 0
– 0.6F_AD – 0.6F_DB = 80 —————- (2)

Solving (1) and (2) simultaneously;
F_AD = -88.542 KN (Compression)
F_DB = -44.792 KN (Compression)

Analysis of joint E

∑F_X = 0
80 – 0.8F_BE + 0.8F_EC = 0
– 0.8F_BE + 0.8F_EC = 80 —————- (3)

∑F_Y = 0
-80 – 0.6F_BE – 0.6F_EC = 0
– 0.6F_BE – 0.6F_EC = 80 —————- (4)

Solving (3) and (4) simultaneously;
F_BE = -16.667 KN (Compression)
F_DB = -116.667 KN (Compression)

Support Reactions

We will obtain the support reactions by resolving the joints.

Support A

∑F_Y = 0
Ay + 0.6F_AD = 0
Ay – (0.6 × 88.542) = 0
Ay = 53.1252 KN

∑F_X = 0
Ax + 0.8F_AD = 0
Ax – (0.8 × 88.542) = 0
Ax = 70.8336 KN

Support B

∑F_X = 0
Bx – 0.8F_BD + 0.8F_BE = 0
Bx – (0.8 × -44.792) + (0.8 × -16.667) = 0
Bx = -22.5 KN

∑F_Y = 0
By + 0.6F_BD + 0.6F_BE = 0
By – (0.6 × 44.792) – (0.6 × 16.667) = 0
By = -36.8754 KN

Support C

∑F_Y = 0
Cy + 0.6F_EC = 0
Cy – (0.6 × 116.667) = 0
Cy = 70 KN

∑F_X = 0
-Cx – 0.8F_EC = 0
-Cx – (0.8 × -116.667) = 0
Cx = 93.333 KN

Equilibrium Verification

∑F_Y ↓ = (10 × 16) = 160 KN
∑F_Y ↑ = Ay + By + Cy = 53.1252 + 36.8754 + 70 = 160 KN

∑F_X → = 45 + 70.833 = 115.833 KN
∑F_X ← = 22.5 + 93.333 = 115.833 KN

Internal Stresses Diagram

As you can see, there are no bending moment and shear forces on the structure, therefore, the axial force diagram is as given below;

Thank you for reading. We love you and keep visiting us.

Connect on our Facebook page

www.facebook.com/structville

Design of Steel Beams According to BS 5950 – 1: 2000

By

Ubani Obinna Uzodimma

-

August 1, 2017

Universal beam sections are normally employed in buildings to carry floor and wall load. Loads on beams may include the load from slab, walls, building services, and their own self-weight. It is necessary for structural beams to satisfy ultimate and serviceability limit state requirements. This post gives a solved design example of a laterally restrained beam according to BS 5950.

The structural design of steel beams to BS 5950 involves following specific guidelines and principles outlined in the British Standard. BS 5950 is a widely used code of practice for the design of steel structures in the United Kingdom, but has been replaced by Eurocode 3 (EN 1993-1-1).

When designing steel beams, several factors are considered to ensure structural integrity and safety. These factors include determining the appropriate loadings, selecting the appropriate section shape and size, analyzing the beam’s resistance to bending, shear, and deflection, and ensuring proper connection details.

The code specifies various load combinations, such as dead loads, live loads, wind loads, and imposed loads, that need to be considered during the design process. In terms of section selection, the code provides tables and charts to determine the suitable steel section based on the applied loads and required span. These sections include universal beams (UB), universal columns (UC), and parallel flange channels (PFC), among others. The appropriate section is chosen based on its moment resistance, shear capacity, and deflection limits.

Design calculations involve checking the beam’s capacity to resist bending, shear, and deflection. These calculations consider the applied loads, section properties, and material properties of the steel. The code provides formulas and design charts to assess these aspects.

Steel Beam Design Example

A laterally restrained beam 9m long that is simply supported at both ends support a dead uniformly distributed load of 15 kN/m and an imposed load uniformly distributed load of 5 kN/m. It also carries a dead load of 20 kN at a distance of 2.5m from both ends. Provide a suitable UB to satisfy ultimate and serviceability limit state requirements (P_y = 275 N/mm²).

Design of Steel Beams According to BS 5950 - 1: 2000 69

Solution
At ultimate limit state;
Concentrated dead load = 1.4G_k = 1.4 × 20 = 28 kN
Uniformly distributed load = 1.4G_k + 1.6Q_k = 1.4(15) + 1.6(5) = 29 kN/m

Support Reactions
Let ∑M_B = 0; anticlockwise negative
(9 × Ay) – (29 × 9 × 4.5) – (28 × 6.5) – (28 × 2.5) = 0
Ay = 158.5 kN

Let ∑M_A = 0; clockwise negative
(9 × By) – (29 × 9 × 4.5) – (28 × 6.5) – (28 × 2.5) = 0
By = 158.5 kN

Internal Stresses
Moment
M_A = 0 (Hinged support)
M_C = M_D = (158.5 × 2.5) – (29 × 2.5 × 1.25) = 305.625 kNm
M_midspan = (158.5 × 4.5) – (29 × 4.5 × 2.25) – (28 ×2) = 363.625 kNm

Shear
Q_A = Ay = 158.5 KN
Q_C^L = 158.5 – (29 × 2.5) = 86 kN
Q_C^R = 158.5 – (29 × 2.5) – 28 = 58 kN
Q_D^L = 158.5 – (29 × 6.5) – 28 = -58 kN
Q_D^R = 158.5 – (29 × 6.5) – 28 – 28 = – 86 kN
Q_B = 158.5 – (29 × 9) – 28 – 28 = – 158.5 kN

Internal Forces Diagram

Bending%2BMoment%2BDiagram%2BSteel%2BBeam

Structural Design to BS 5950
P_y = 275 N/mm²

Initial selection of section
Moment Capacity of section M_c = P_yS ——- (1)

Where S is the plastic modulus of the section
Which implies that S = M_c/P_y = (363.625 × 10⁶)/275 = 1320963.636 mm³ = 1320.963 cm³

With this we can go to the steel sections table and select a section that has a plastic modulus that is slightly higher than 1320.963 cm³
Try section UB 457 × 191 × 67 (S = 1470 cm³)

Properties of the section;
I_xx = 29400 cm⁴
Z_xx = 1300 cm³
Mass per metre = 67.1 kg/m
D = 453.4mm
B = 189.9mm
t = 8.5mm
T = 12.7mm
r = 10.2mm
d = 407.6mm

Strength classification
Since T = 12.7mm < 16mm, P_y = 275 N/mm²
Hence ε = √(275/P_y) = √(275/275) = 1.0

Section classification
Flange
b/T = 7.48 < 9ε; Flange is plastic class 1
Web
d/t = 48 < 80ε; Web is also plastic class 1

Shear Capacity
As d/t = 48 < 70ε, shear buckling need not be considered (clause 4.4.4)

Shear Capacity P_v = 0.6P_yA_v —– (2)

P_v= 0.6P_ytD = 0.6 × 275 × 8.5 × 453.4 = 635893.5 N = 635.89 kN

But design shear force F_v = 158.5 kN
Since F_v(158.5) < P_v(635.89), section is ok for shear.

Now, 0.6P_v = 0.6 × 635.89 = 381.54 kN
Since F_v(158.5 kN) < 0.6P_v(381.54 kN), we have low shear load.

Moment Capacity
Design Moment = 363.625 kNm

Moment capacity of section UB 457 × 191 × 67 (S = 1470 cm³ = 1470 × 10³ mm³)

M_c = P_yS = 275 × 1470 × 10³ = 404.25 × 106 N.mm = 404.25 kNm
1.2P_yZ = 1.2 × 275 × 1300 × 10³ = 429 × 106 N.mm = 429.00 kNm

M_c (404.25 kNm) < 1.2P_yZ (429.00 kNm) Hence section is ok

Evaluating extra moment due to self-weight of the beam
Self-weight of the beam S_w = 67.1 kg/m = 0.658 kN/m (UDL on the beam)
Moment due to self weight (M_sw) = (ql²)/8 = (0.658 × 9²)/8 = 6.66 kNm

(363.625 + 6.66) < M_c (404.25 kNm) < 1.2P_yZ (429.00 kNm) Hence section is ok for moment resistance.

Deflection Check
We check deflection for the unfactored imposed load; E = 205 kN/mm² = 205 × 10⁶ KN/m²; I_xx = 29400 cm⁴ = 29400 × 10^-8 m⁴

The maximum deflection for this structure occurs at the midspan and it is given by;
δ = (5ql⁴)/384EI = (5 × 5 × 9⁴) / (384 × 205 × 10⁶ × 29400 × 10^-8) = 7.087 × 10^-3 m = 7.087 mm

Permissible deflection; L/360 = 9000/360 = 25mm
7.087mm < 25mm. Hence deflection is satisfactory.

Web bearing
According to Clause 4.5.2 of BS 5950-1:2000, the bearing resistance P_bw is given by:

P_bw = (b₁ + nk)tP_yw —– (3)

Where;
b₁ is the stiff bearing length
n = 5 (at the point of concentrated loads) except at the end of a member and n = 2 + 0.6b_e/k ≤ 5 at the end of the member
b_e is the distance to the end of the member from the end of the stiff bearing
k = (T + r) for rolled I- or H-sections
T is the thickness of the flange
t is the web thickness
P_yw is the design strength of the web

Web bearing at the supports
Let us assume that beam sits on 200 mm bearing, and b_e = 20mm

k = (T + r) = 12.7 + 10.2 = 22.9mm;
Hence n = 2 + 0.6(20/22.9) = 2.52mm < 5mm.
P_bw = (b₁ + nk)tP_yw
P_bw = [200 + 2.52(22.9)] × 8.5 × 275 = 602392.45 N = 602.392 kN
P_bw (602.392 kN) > F_v (158.5 kN) Hence it is ok

Contact stress at supports
P_cs = [b₁ × 2(r +T )]P_y = [200 × 2(22.9)] × 275 = 2519000N = 2519 kN
P_cs (2519 KN) > F_v (158.5 KN) Hence contact stress is ok

Web buckling

According to clause 4.5.3.1 of BS 5950, provided the distance α_e from the concentrated load or reaction to the nearer end of the member is at least 0.7d, and if the flange through which the load or reaction is applied is effectively restrained against both;

(a) rotation relative to the web
(b) lateral movement relative to the other flange

The buckling resistance of an unstiffened web is given by;

P_x = [25εt/√(b₁ + nk)d] P_bw —– (4)
When α_e < 0.7d, the buckling resistance of an unstiffened web is given by;

P_x = [(α_e + 0.7d)/1.4d] × [25εt/√(b₁ + nk)d)] P_bw ————- (5)

Therefore, α_e = 20mm + (200/2) = 120mm
0.7d = 0.7 × 407.6 = 285.32mm
α_e(120mm) < 0.7d(285.32mm). Hence equation (5) applies;

P_x = [(120 + 285.32)/(1.4 × 407.6)] × [(25 × 1 × 8.5 )/√((200 + 2.52 × 22.9) × 407.6)] × 602.392 = 280.538 kN

P_x(280.538 kN) > F_v(158.5 kN) Hence it is ok.

Thank you for reading, and feel free to share.

Like Our Facebook page
www.facebook.com/structville

How to Calculate the Lap Length of Reinforcements

By

Ubani Obinna Uzodimma

-

July 29, 2017

Reinforced concrete structures rely on the effective transfer of stress between steel reinforcement bars. Lap splices, where two reinforcement bars overlap within the concrete element, are a common method to achieve this continuity. Lapping of reinforcement is a frequent occurrence in construction due to supply length limitations, construction joints, stage construction, and ease of handling of rebars.

Therefore, determining the appropriate lap length is important for ensuring structural integrity and preventing premature failure. This article discusses how to calculate the lap length for reinforcements in concrete structures.

Factors Influencing Lap Length

Several factors influence the required lap length:

Reinforcement diameter (Φ): Larger diameter bars require longer lap lengths for effective stress transfer.
Concrete grade (f_ck): Higher concrete strength allows for shorter lap lengths due to improved bond characteristics.
Reinforcement yield strength (f_yk): Higher yield strength steel requires longer lap lengths to develop its full capacity.
Concrete cover: The thickness of the concrete cover influences the bond transfer and hence, the anchorage length.
Bond condition: The orientation of the reinforcement (straight or bent) and the concrete cover can lead to a good or poor bond condition which can influence the anchorage length of reinforcements.
Confinement conditions: The presence of stirrups or ties around the lap splice can significantly reduce the required lap length.
Type of Lap Splice: Standard lap splices require longer lengths compared to hooked or welded lap splices.

Methods of Calculating Lap Length

Different design codes and standards provide methods for calculating lap length. Here’s a breakdown of two common approaches:

1. Basic Lap Length Formula:
This simplified method often serves as a preliminary estimate:

Lap Length (L) = 50d

Where:
d = Reinforcement diameter (mm)

This formula assumes standard lap splices in normal concrete conditions.

2. Code-Based Calculations:
Detailed calculations are typically based on specific design codes like Eurocode 2 (EC2) or ACI 318. These codes provide equations and tables that consider various factors influencing lap length:

EC2 Method:
EC2 employs a formula that incorporates the bar diameter, concrete grade, steel yield strength, and a factor accounting for the percentage of bars lapped in one section.

ACI 318 Method:
ACI 318 provides tables with tabulated lap lengths based on bar diameter, concrete strength, and type of splice (standard, hooked, or welded).

Note:

Minimum Lap Length: Codes often specify minimum lap lengths regardless of calculations.
Development Length: Lap length should be greater than or equal to the development length of the bar, which is the minimum length required for the bar to develop its full yield strength in the surrounding concrete.
Longitudinal Spacing of Bars: The spacing between lapped bars within the overlap zone can affect the required lap length.
Lightweight Concrete: Adjustments are often necessary for lap lengths in lightweight concrete due to its lower bond strength.

Anchorage Length

Apart from the lapping of steel reinforcements, reinforcing bars should be well anchored so that the bond forces are safely transmitted to the concrete to avoid longitudinal cracking or spalling. The length of rebar required to achieve adequate transfer of forces is known as the anchorage length. Transverse reinforcement shall be provided if necessary.

Apart from straight bars, other shapes that are specified in the code are;

(a) standard bend,
(b) standard hook, and
(c) standard loop.

The detailing rules and the equivalent anchorage length for each of these standard shapes are defined in EN1992-1-1 Figure 8.1. Types of anchorage are shown in the figure below (Figure 8.1 EC2).

For bent bars, the basic tension anchorage length is measured along the centreline of the bar from the section in question to the end of the bar, where:

l_bd = α₁ α₂ α₃ α₄ α₅ l_b,req ≥ l_b,min ———— (1)

where;
l_b,min is the minimum anchorage length taken as follows:
In tension, the greatest of 0.3l_b,rqd or 10ϕ or 100mm
In compression, the greatest of 0.6l_b,rqd or 10ϕ or 100mm

l_b,rqd is the basic anchorage length given by;
l_b,rqd = (ϕ/4) σ_sd/f_bd ————– (2)

Where;
σ_sd = The design strength in the bar (take 0.87f_yk)
f_bd = The design ultimate bond stress (for ribbed bars = 2.25η₁ η₂ f_ctd)
f_ctd = Design concrete tensile strength
f_ctd = 0.21f_ck^(2/3) for f_ck ≤ 50 N/mm²
η₁ is a coefficient related to the quality of the bond condition and the position of the bar during concreting
η₁ = 1.0 when ‘good’ conditions are obtained and
η₁ = 0.7 for all other cases and for bars in structural elements built with slip-forms, unless it can be shown that ‘good’ bond conditions exist
η₂ is related to the bar diameter:
η₂ = 1.0 for φ ≤ 32 mm
η₂ = (132 – φ)/100 for φ > 32 mm

‘Good’ bond conditions are applicable when any of the following conditions are fulfilled:

(a) Vertical bars or almost vertical bars inclined at an angle 45° ≤ α ≤ 90° from the horizontal,
(b) bars that are located up to 250 mm from the bottom of the formwork for elements with height h ≤ 600 mm, or
(c) bars that are located at least 300 mm from the free surface during concreting for elements with height h > 600 mm.

‘Poor’ bond conditions are applicable for all other cases and also for bars in structural elements built with slip-forms, unless it can be shown that ‘good’ bond conditions exist. The different bond regions are shown in the figure above that is reproduced from EN1992-1-1 Figure 8.2.

Description%2Bof%2Bbond%2Bconditins%2BEC2

α₁ is for the effect of the form of the bars assuming adequate cover.
α₂ is for the effect of concrete minimum cover.

α₃ is for the effect of confinement by transverse reinforcement
α₄ is for the influence of one or more welded transverse bars ( φt > 0.6φ) along the design anchorage length l_bd
α₅ is for the effect of the pressure transverse to the plane of splitting along the design anchorage length.

The values of these coefficients can be adequately obtained by following the Table below;

l_0,min = max{0.3α₆ l_b,rqd; 15ϕ; 200}

α₆ = √(ρ₁/25) but between 1.0 and 1.5
where ρ₁ is the % of reinforcement lapped within 0.65l₀ from the centre of the lap

percentage%2Bof%2Blapped%2Bbars%2Beurocode%2B2

Values of α₁, α₂, α₃ and α₅ may be taken as for the calculation of anchorage length but for the calculation of α₃, ΣA_st,min should be taken as 1.0As(σ_sd/f_yd), with As = area of one lapped bar.

SOLVED EXAMPLE FOR ANCHORAGE LENGTH
Calculate the design tension anchorage length of T16mm bar (f_yk = 460 N/mm², concrete cover = 35 mm, Concrete cylinder strength f_ck = 25 N/mm²) for;

(a) When it is a straight bar
(b) When it is bent into any other shape
Assume good bond conditions

Solution
l_bd = α₁ α₂ α₃ α₄ α₅ l_b,req ≥ l_b,min
l_b,rqd = (ϕ/4) σ_sd/f_bd
f_bd =2.25η₁ η₂ f_ctd
η₁ = 1.0 ‘Good’ bond conditions
η₂ = 1.0 bar size ≤ 32

f_ctd = (α_ct f_{ctk 0.05})/γ_c ————— (3)

where;
f_{ctk 0.05} = characteristic tensile strength of concrete at 28 days = 1.8 N/mm² (Table 3.1 EC2)
γ_c = partial (safety) factor for concrete = 1.5
α_ct = coefficient taking account of long-term effects on the tensile strength, this is an NDP with a recommended value of 1.

f_ctd = (1.0 × 1.8)/1.5 = 1.2 N/mm²
f_bd = 2.25 × 1.0 × 1.0 × 1.2 = 2.7 N/mm²

l_b,rqd = (ϕ/4) σ_sd/f_bd
σ_sd = 0.87 × 460 = 400.2 N/mm²>br/> l_b,rqd = (ϕ × 400.2 )/(4 × 2.7) = 37.05ϕ

Therefore;

l_bd = α₁ α₂ α₃ α₄ α₅(37.05ϕ)

(a) For straight bar
α₁ = 1.0
α₂ = 1.0 – 0.15 (C_d – ϕ)/ ϕ
α₂ = 1.0 – 0.15 (35 – 16)/16 = 0.8218
α₃ = 1.0 conservative value with K = 0
α₄ = 1.0 N/A
α₅ = 1.0 conservative value

l_bd = 0.8218 × (37.05ϕ) = 30.4ϕ = 30.4 × 16 = 486.4 mm
Say 500 mm

(b) For other shape bar
α₁ = 1.0 bC_d; = 35 is ≤ 3ϕ = 3 × 16 = 48
α₂ = 1.0 – 0.15 (Cd – 3ϕ)/ ϕ ≤ 1.0
α₂ = 1.0 – 0.15 (35 – 48)/16 = 1.121 ≤ 1.0
α₃ = 1.0 conservative value with K = 0
α₄ = 1.0 N/A
α₅ = 1.0 conservative value
l_bd = 1.0 × (37.05ϕ) = 37.05ϕ = 37.05 × 16 = 592 mm
Say 600mm

Compression anchorage (α₁ = α₂ = α₃ = α₄ = α₅ = 1.0)
l_bd = 37.05ϕ

For poor bond conditions
Anchorage for ‘Poor’ bond conditions = ‘Good’/0.7

	Good bond conditions		Poor bond conditions
Concrete class	• Straight bars in tension / compression • Other-than-straight bars in compression	• Other-than-straight bars in tension and adequate cover c_d > 3Φ	• Straight bars in tension/compression • Other-than-straight bars in compression	• Other-than-straight bars in tension and adequate cover c_d > 3Φ
C16/20	55Φ	39Φ	78Φ	55Φ
C20/25	47Φ	33Φ	67Φ	47Φ
C25/30	41Φ	29Φ	58Φ	41Φ
C30/37	36Φ	26Φ	52Φ	36Φ
C35/45	33Φ	23Φ	47Φ	33Φ
C40/50	30Φ	21Φ	43Φ	30Φ

Example on the calculation of lap length of 4X16mm bars of a column in a multi-storey building
Since the bars are in compression,
α₁ α₂ α₃ α₅= 1.0
As calculated above, l_bd = 37.05ϕ

Let us say that over 50% of reinforcement is lapped within 0.65l₀ from the centre of the lap
Hence, we will take α₆= 1.5

Lap length therefore = 1.5 × 37.05ϕ = 55.57ϕ = 55.57 × 16 = 889.2mm
Say 900 mm

Thank you so much for visiting. You can like our facebook page on
www.facebook.com/structville

How to Calculate the Number of Blocks Required to Complete a 3 Bedroom Flat

By

Ubani Obinna Uzodimma

-

July 26, 2017

Table of Contents

Sandcrete blocks are very popular construction materials for the construction of walls in buildings. They can serve as load-bearing elements or as mere partition panels. There are different types and sizes of sandcrete blocks that are employed in building construction. The size of the block (dimensions) can be a factor in determining the number of blocks required to successfully complete a building.

In order to properly know the cost of constructing a building, it is very important to know the number of blocks that will be able to complete the building. The number of blocks required to construct a building is largely dependent on the type of block, size of the building, area of walls to be built, etc.

on going bungalow construction — **Figure 1:** On-going bungalow construction

In this article, we are going to present a very simple approach to determining the number of blocks required to complete a three-bedroom flat. The floor plan of a three-bedroom flat is shown below, which can comfortably sit on a plot of land of 450 m². The building consists of a sitting room, three bedrooms (all en-suite), a visitor’s toilet, a dining, a kitchen, a store, and sit-out areas. It is required to estimate the quantity of 6 inches blocks (without holes) required to complete the project.

The data from the building is as shown below;

DOORS
Door type 1 (D1) = 1 number (Dimension = 2100 × 1200)
Door type 2 (D2) = 6 numbers (Dimension = 2100 × 900)
Door type 3 (D3) = 4 numbers (Dimension = 2100 × 700)

WINDOWS
Window Type 1 (W1) = 12 numbers (Dimension = 1200 × 1200)
Window Type 2 (W2) = 4 numbers (Dimension = 600 × 600)

Storey height (height of the building from floor to ceiling) = 3000 mm

Foundation details of the bungalow are given below.

Foundation%2Bto%2BDPC — **Figure 3:** Foundation details of the building

A section through the wall of the building is given below;

Section%2BThrough%2Bwall — **Figure 4:** Section of the wall

We are going to carry out the calculation in three main phases of construction of bungalow which are;

Foundation to DPC
DPC to lintel level, and
Lintel level to overhead level.

Estimation of the number blocks

The simplest approach is to estimate the number of blocks required in a metre square of wall.

The planar dimensions of a standard block in Nigeria is (450mm × 225mm).

6%2Binches%2Bblock%2Bwithout%2Bholes — **Figure 5**: Typical dimensions of 6 inches blocks in Nigeria

The planar area of 1 block = (0.225 × 0.45) = 0.10125 m²
Therefore one metre square of a wall will contain;
1/0.10125 = 9.876

Therefore, one metre square of wall will contain approximately 10 blocks.

PHASE 1: FOUNDATION TO DPC

At this stage, the total length of the walls of the building and the height of the walls at the substructure level are considered. There are no openings, and the entire foundation excavation length is considered. For the building we are considering, the foundation layout is as shown below;

FOUNDATION%2BPLAN — **Figure 6:** Foundation layout of a three-bedroom flat

Using any method of your choice, you can calculate the total length of the walls at the foundation level.
This is approximately equal to 100256 mm = 100.256m from my calculations.

The height of the wall from foundation to DPC (ignoring thickness of mortar) as shown in Figure 3 = 900mm = 0.9m

Therefore, the total area of wall = 100.256m × 0.9m = 90.2304m²
If 10 blocks are required for 1m2 of a wall, therefore (10 × 90.2304) = 903 blocks is required for 90.2304m² of wall.

The number of blocks required to raise the building from foundation to DPC = 903 blocks (disregarding damages/waste).

PHASE 2: DPC TO LINTEL LEVEL

In this phase, we will consider the effects of openings in the buildings which includes the doors, windows, and other openings as specified in the drawing. These openings are seen in the verandah areas and some of the internal spaces in the building. The same process of calculating the total length of walls is also adopted, after which we minus the area of the openings.

This process is shown below;

Total Length of walls (excluding areas with no walls like verandahs) = 87420 mm = 87.42m

Height of wall = 2100 = 2.1m
Therefore area of wall = (87.42 × 2.1) = 183.582 m²
Total area of all doors from building data = 19.74 m²
Total area of all windows from building data = 18.72 m²
Therefore the net area of walls = 183.582 m² – 19.74 m² – 18.72 m² = 145.122 m²
Hence, the total number of blocks required to take the building from DPC to lintel level = 145.122 × 10 = 1451.22 = 1452 blocks (disregarding wastes)

PHASE 3: LINTEL LEVEL TO OVERHEAD LEVEL

Here, we are going to assume that the building will be chained at the lintel level (this means that the entire level of the building will be cast and reinforced with steel at the lintel level. This offers the advantage of helping the building behave as a unit, and helps in reducing cracking in the building).

Obviously, the length of the wall in this case will not consider openings, so we are going to make use of 100.256m as calculated from the foundation level.

Hence, the length of walls = 100.256m
Height of wall = 675mm = 0.675m

Total area of walls = (100.256 × 0.675) = 67.6728 m²
Hence, the total number of blocks required to take the building from the lintel level to the roof level = 67.6728 × 10 = 676.728 = 677 blocks.

Ultimately, the total number of blocks required to complete the construction of a three-bedrooms flat = 903 + 1452 + 677 = 3032 blocks (disregarding wastage).

BB96016D 458D 4C07 A5AB C2A7BBC128F0 — **Figure 8:** Bungalow with completed block work

To consider wastage, you can make a 20% increase, so that the number of blocks to be moulded or bought = 1.2 × 3032 = 3638 blocks. However, when you are moulding and can guarantee the quality of blocks to be moulded, 10% increase will be satisfactory for the job.

READ ALSO IN THIS BLOG;

How to estimate the quantity of materials required for moulding blocks.

Thank you for visiting us, and God bless.

Like our Facebook page on www.facebook.com/structville

How to Estimate the Quantity of Sand and Cement Required for Moulding Blocks

By

Ubani Obinna Uzodimma

-

July 26, 2017

In Nigeria, the 9 inches hollow block (450mm x 225mm x 225mm) is normally used for the construction of blockwork of buildings exceeding one storey. The length of the block is 450mm (18 inches), the height is 225mm (9 inches), and the width is 225mm (9 inches). Contractors and builders are often faced with the challenge of either buying blocks for construction or moulding it on their own. When the conditions are right, builders will prefer to mould their own blocks, especially for the purpose of controlling the quality of the blocks, and to reduce the cost of transportation.

Let us assume that you have decided to mould your own blocks, and you have been faced with the challenge of estimating the quantity of cement and sand to purchase that will satisfy the construction requirement. This article will give you a guide on how to estimate the quantity of sand and cement required for moulding blocks in Nigeria. I am going to work through the steps, so that you will be able to make calculations just in case you are using any other size of block.

Initial data
Number of blocks required = 3000 pieces
Recommended production = 1 bag of cement to produce 35 blocks

Step 1: Calculate the volume of the block

For the block size shown above;
Volume of block without holes = (0.45 × 0.225 × 0.225) = 0.02278125 m³
Volume of holes = 2(0.225 × 0.125 × 0.15) = 0.0084375 m³
Therefore volume of the block = 0.02278 – 0.0084375 = 0.0143 m³

Step 2: Calculate the volume of the 35 blocks
If the volume of 1 block is 0.0143, the volume of 35 blocks = (0.0143 × 35) = 0.502 m³

Step 3: Calculate the volume sand required for the 35 blocks
The volume of 1 bag of cement is 34.72 litres = 0.03472m³.
This is obtained by knowing that the mass of 1 bag of cement = 50kg, and the density = 1440 kg/m³

The volume of a standard builder’s wheelbarrow is 0.065 m³ (unheaped). We assume that approximately 2 bags of cement (4 head pans of cement) will fill one builder’s wheelbarrow. Now, we can estimate the number of wheelbarrow trips of sand that the moulder should provide in order to make 35 blocks from one bag of cement.

Total volume of 35 blocks required = 0.502 m³

Let the number of wheel barrow trips of sand be x

Hence, (volume of 1 bag of cement) + (Total volume of sand) = Volume of 35 blocks
Hence, 0.03472 + x(0.065) = 0.502 m³
On solving, x = 7.1889 unheaped wheelbarrow trips of sand
The volume of sand required to make 35 blocks = 7.1889 × 0.065 = 0.46728 m³

However, note that this leads to a mix ratio of about 1:14 which will not be very good when subjected to a compressive strength test. The ideal mix ratio for blocks for proper strength should be about 1:10 or 1 bag to 25 blocks.

Step 4: Calculate the total volume of materials required
We can therefore estimate the quantity of materials to be purchased;

If 1 bag of cement is needed for 35 blocks, therefore 86 bags of cement is needed to mould 3000 blocks (gotten by 3000/35)

If 0.46728 m³ of sand is required to make 35 blocks, therefore, 41 m³ of sand (about 67.60 tonnes assuming the density of dry sand = 1600 kg/m³) is needed to make 3000 blocks.

For a 5 tonne tipper of 3.8 m³ capacity, we have to order for 11 trips of sharp sand.

You have to be very sure of the size of the tipper that will be delivering the sand to you in order to make your estimates more reasonable. On the other hand, you can make the considerations for variations between the weight of materials when wet and when dry. However, this will never be critical for block moulding based on experience. But in concrete it can be very considerable.

Therefore summarily, we need 86 bags of cement and 68 tonnes of sand to mould about 3000 pieces of 9 inches blocks with holes (for 1 bag = 35 blocks).

Thank you for reading, and God bless.

Visit our Facebook fan page on
www.facebook.com/structville

Worked Example

Example on the Design of Biaxial Reinforced Concrete Columns to BS 8110 and Eurocode 2

Material Prices (informative)

Worked Example

Is there any standard approach?

Sign Convention for Beams

Bending moment

Shear force

Axial force

Sign Convention for Frames

Bending moment

Shear force

Axial Force

Support Reactions

Geometric Properties of the Arch Section

Analysis of Joint D

Analysis of joint E

Support A

Support B

Support C

Internal Stresses Diagram

Steel Beam Design Example

Factors Influencing Lap Length

Methods of Calculating Lap Length

Anchorage Length

Estimation of the number blocks

PHASE 1: FOUNDATION TO DPC

PHASE 2: DPC TO LINTEL LEVEL

PHASE 3: LINTEL LEVEL TO OVERHEAD LEVEL

EDITOR PICKS

POPULAR POSTS

POPULAR CATEGORY

ABOUT US

FOLLOW US