How to Load Columns in RC Building Design

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June 25, 2022

When carrying out the manual design of reinforced concrete structures, column loads are usually assessed by considering the support reactions from beams they are supporting, or by the tributary area method. The latter is more popular due to its simplicity and speed, but usually fails to capture all the loads that are imposed on the columns (such as wall loads), while the former is more complex and time-consuming, it usually very representative of all the possible loads that are imposed on the column.

Steps on how to load a column from beam support reaction

The following steps can be adopted when using beam support reactions to obtain the axial load on columns.

(1) Load the floor slab adequately, and factor the loads at ultimate limit state, using the appropriate load combination.

(2) Load all the beams that are connected to the column.

(3) Transfer the loads from slab to the beam using the appropriate relationship. Based on yield lines, the loads are usually triangular or trapezoidal, but this is cumbersome to analyse manually. An equivalent UDL (reasonably accurate) can be used to transfer slab loads to beams by employing the formulas below.

One way slabs
Long span q = nl_x/2
Short slab q = nl_x/5

Two way slabs
Long span q = nl_x/2(1 – 1/3k²)
Short span q = nl_x/3

Where;
q = Load transferred from slab to the beam
n = load at ultimate limit state
1.4g_k + 1.6q_k(BS 8110)
1.35g_k + 1.5q_k (Eurocode 2)
k = L_y/L_x
L_y= Length of long span of slab
L_x = Length of short span of the slab

(4) Analyse the floor beams completely using any suitable method of your choice, while also considering any additional load that may be on the beam such as wall load and finishes.

(5) Obtain the support reactions of the beam, which represents the load that is transferred from the floor to the column.

Design Example

In this post, the floor plan shown below is for a shopping complex, and it desired to obtain the column axial loads at ultimate limit state.

Design Data
Size of all columns = 230 x 230mm
Size of all beams = 450 x 230mm
Thickness of slab = 200mm
Unit weight of concrete = 25 kN/m³
Unit weight of sandcrete block = 3.47 kN/m²
f_ck= 25 N/mm²
f_yk = 500 N/mm²

Load combination = 1.35g_k+ 1.5q_k
Design variable load (q_k) = 4 kN/m²

k = L_y/L_x = 6/5 = 1.2 (in all cases)
Only external beams are carrying block work load.

Load Analysis

Roof beam
Permanent load on roof beams g_k = 6 kN/m
Variable load on roof beam q_k = 1.5 kN/m
(These values are assumed)
At ultimate limit state, n = 1.35(6) + 1.5(1.5) = 10.35 kN/m

Slab Load Analysis
Concrete own weight = 25 kN/m³× 0.2m = 5.0 kN/m²
Screeding and Finishes (say) = 1.35 kN/m²
Partition allowance = 1.5 kN/m²
Total (g_k) = 7.85 kN/m²
Variable action (q_k) = 4 kN/m²

n = 1.35gk + 1.5qk = 1.35(7.85) + 1.5(4) = 16.6 KN/m²

Wall load on beam
Unit weight of sandcrete block = 3.47 kN/m²
Height of wall = 3.5m
Wall load on beam = 3.47 kN/m²× 3.5m = 12.145 kN/m

Load on Beams
External Longitudinal beams (Axis 1:A-D and Axis 3:A-D)
Self weight of beam (factored) = 1.35 × 0.23m × 0.25m × 25 = 1.94 kN/m
Load from slab = nl_x/2(1 – 1/3k²) = [(16.6 × 5)/2] × (1 – 1/(3 ×1.2²)) = 31.893 kN/m
Load from block work = 12.145 kN/m
Total load = 45.978 kN/m

External Transverse beams (Axis A:1-3 and Axis D:1-3)
Self weight of beam (factored) = 1.35 × 0.23m × 0.25m × 25 = 1.94 kN/m
Load from slab = nl_x/3 = (16.6 × 5)/3 = 27.667 kN/m
Load from block work = 12.145 kN/m
Total load = 41.752 kN/m

Internal Longitudinal Beam (Axis 2:A-D)
Self weight of beam (factored) = 1.35 × 0.23m × 0.25m × 25 = 1.94 kN/m
Load from slab = nl_x/2(1 – 1/3k²) = [(16.6 × 5)/2] × (1 – 1/(3 ×1.2²)) = 2 × 31.893 kN/m = 63.786 kN/m (we multiplied by two because this beam is receiving slab load from two directions.)
Load from block work = 0 (there are no block works inside the building)
Total load = 65.726 kN/m

Internal Transverse beams (Axis B:1-3 and Axis C:1-3)
Self weight of beam (factored) = 1.35 × 0.23m × 0.25m × 25 = 1.94 kN/m
Load from slab = nlx/3 = (16.6 × 5)/3 = 2 × 27.667 kN/m = 55.334 kN/m (slab load is coming from two directions)
Load from block work = 0
Total load = 57.274 kN/m

Self weight of Columns
Self weight of column (factored) = 1.35 × 0.23m × 0.23m × 3.75m × 25 kN/m³ = 6.695 kN

Structural Analysis
A little consideration will show that the support reactions for the floor beams (equal spans) can be obtained by considering the shear force coefficients given below;

**Fig 3:** Shear force coefficient (2 – span beam)

Fig 4: Shear force coefficient (3 span beam)

Load Analysis of Column A1
1st Floor
Load from longitudinal roof beam (1:A-D) = 0.4 × 10.35 kN/m × 6m = 24.84 kN
Load from transverse roof beam (A:1-3) = 0.375 × 10.35 kN/m × 5m = 19.41 kN
Self weight of column = 6.695 kN
Total = 50.945 kN

Ground Floor
Load from above = 50.945 KN
Load from longitudinal floor beam (1:A-D) = 0.4 × 45.978 kN/m × 6m = 110.347 kN
Load from transverse floor beam (A:1-3) = 0.375 × 41.752 kN/m × 5m = 78.285 kN
Self weight of column = 6.695 kN
Total = 246.272 kN

Load Analysis of Column A2
1st Floor
Load from longitudinal roof beam (2:A-D) = 0.4 × 10.35 kN/m × 6m = 24.84 kN
Load from transverse roof beam (A:1-3) = 2(0.625 × 10.35 kN/m × 5m) = 64.687 kN
Self weight of column = 6.695 kN
Total = 96.222 kN

Ground Floor
Load from above = 96.222 KN
Load from longitudinal floor beam (2:A-D) = 0.4 × 65.726 kN/m × 6m = 157.742 kN
Load from transverse floor beam (A:1-3) = 2(0.375 × 41.752 kN/m × 5m) = 156.57 kN
Self weight of column = 6.695 kN
Total = 417.229 kN

Load Analysis of Column B1
1st Floor
Load from longitudinal roof beam (1:A-D) = (0.5 × 10.35 kN/m × 6m) + (0.6 × 10.35 kN/m × 6m) = 68.31 kN
Load from transverse roof beam (B:1-3) = 0.375 × 10.35 kN/m × 5m = 19.41 kN
Self weight of column = 6.695 kN
Total = 94.415 kN

Ground Floor
Load from above = 94.415 KN
Load from longitudinal floor beam (1:A-D) = (0.5 × 45.978 kN/m × 6m) + (0.6 × 45.978 kN/m × 6m) = 303.455 kN
Load from transverse floor beam (B:1-3) = 0.375 × 57.274 kN/m × 5m = 107.388 kN
Self weight of column = 6.695 kN
Total = 511.953 kN

Load Analysis of Column B2
1st Floor
Load from longitudinal roof beam (2:A-D) = (0.5 × 10.35 kN/m × 6m) + (0.6 × 10.35 kN/m × 6m)= 68.31 kN
Load from transverse roof beam (A:1-3) = 2(0.625 × 10.35 kN/m × 5m) = 64.687 kN
Self weight of column = 6.695 kN
Total = 139.692 kN

Ground Floor
Load from above = 139.692 kN
Load from longitudinal floor beam (2:A-D) = (0.5 × 65.726 kN/m × 6m) + (0.6 × 65.726 kN/m × 6m) = 433.7916 kN
Load from transverse floor beam (A:1-3) = 2(0.625 × 57.274 kN/m × 5m) = 357.9625 kN
Self weight of column = 6.695 kN
Total = 938.141 kN

As you can see, that was very straightforward with the use of shear force coefficients, given the fact that the beams were of equal span. When the beams are not equal, you have to carry out the analysis, and transfer the shear forces accurately. Now, let us compare load analysis of column B2, by the tributary area method.

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Design of Pile Foundation Using Pile Load Test (Eurocode 7)

By

Ubani Obinna Uzodimma

-

December 13, 2020

Pile load test is the most reliable method of estimating the load carrying capacity of a pile, but it is rather expensive. Load tests are performed on-site on test piles to verify the design capacity of the piles. A pile load test consists of applying increments of static load to a test pile and measuring the settlement of the pile. The load is usually jacked onto the pile using either large dead weight or a beam connected to uplifts anchor piles to supply reaction for the jack. Load tests could be by constant rate of penetration test (CRP) or maintained load Test (ML).

In arriving at the ultimate load of a pile from pile load test, several definitions of failure have been given some of which are as follows;

(1) Terzaghi (1942) says that a failure load is that which causes a settlement equal to 10% of the pile diameter.
(2) The CP 2004 defines failure load as that load at which the rate of settlement continues undiminished without further increase increment of load unless this rate is so slow as to indicate that this settlement may be due to consolidation of the soil. This definition is a bit difficult to apply in practice especially in cohesive soils.
(3) German DIN 4026 defines failure load as that load which causes irreversible settlement at the pile head equal to 2.5% of the pile diameter.

Design Example
It is desired to estimate the number of piles in a group of piles to carry the following loads;

Permanent load Gk = 3600 KN
Variable Load Qk = 1740 KN

Pile load test has been carried out on 3 piles on the site and the result is as shown below. The piles are 750mm diameter and 15m long. Assume settlement of 10% of the pile diameter as failure criterion.

Solution

(1)The ultimate resistance of the pile is the load at settlement of 10% of the pile diameter.

Settlement = 750 × (10/100) = 75mm

(2) From the load settlement graph for each pile;
Pile 1 R_m = 4156.25 KN
Pile 2 Rm= 4318.325 KN
Pile 3 R_m = 4887.8 KN

(3) The mean and minimum measured pile resistances are;
R_m,mean = 4454.125 KN
R_m,min = 4156.25 KN

(4) The characteristic pile resistance is obtained by dividing the mean and minimum measured pile resistances by the correlation factors ξ₁and ξ₂and choosing the minimum value. The equation below is given by equation 7.2 of Eurocode 7.

For 3 number of test piles (Table A9, EC7);
ξ₁ = 1.20
ξ₁ = 1.05

R_c,k = min{4454.125/1.2 , 4156.25/1.05} = 3711.77 KN

Design Approach 1

Combinations of sets of partial factors
DA1.C1 —– A1 + M1 + R1
DA1.C2 ——- A2 + M1 or M2 + R4

Partial factors for actions;
A1 γ_G = 1.35 γ_Q = 1.5
A2 γ_G = 1.0 γ_Q = 1.30

Partial factors for materials
M1 and M2 not relevant (γϕ’ = 1.0, not used)

Partial Resistance factors
R1 γ_t = 1.15 (Total/combined compression)
R4 γ_t = 1.5

DA1.C1 F_c,d = 1.35Gk + 1.5Qk = (1.35 × 3600) + (1.5 × 1740) = 7470 KN
DA1.C2 F_c,d = 1.0Gk + 1.3Qk = (1.0 × 3600) + (1.3 × 1740) = 5862 KN

For a single pile;
DA1.C1 R_c,d = R_c,k/γ_t = 3711.77/1.15 = 3227.62 KN
DA1.C2 R_c,d = R_c,k/γ_t = 3711.77/1.5 = 2474.513 KN

Assuming no pile group effect, for n piles,
Resistance = n × R_c,d
Hence,
DA1.C1 n ≥ F_c,d/R_c,d= 7470/3227.62 = 2.31
DA1.C2 n ≥ F_c,d/R_c,d= 5862/2474.513 = 2.36

Therefore, DA1.C2 controls, and 3 number of piles will be required.

Therefore
R_c,d= 3 × 2474.513 = 7423.539
F_c,d = 5862 KN

F_c,d/R_c,d= 5862/7423.539 = 0.789 < 1.0 Ok

Design Approach 2

Combinations of sets of partial factors
DA2 —– A1 + M1 + R2

Partial factors for actions;
A1 γ_G = 1.35 γ_Q = 1.5

Partial factors for materials
M1 not relevant (γϕ’ = 1.0, not used)

Partial Resistance factors
R2 γ_t = 1.1 (Total/combined compression)

F_c,d = 1.35Gk + 1.5Qk = (1.35 × 3600) + (1.5 × 1740) = 7470 KN

For a single pile;
R_c,d = R_c,k/γ_t = 3711.77/1.1 = 3374.336 KN

Assuming no pile group effect, for n piles,
Resistance = n × R_c,d
Hence,
n ≥ F_c,d/R_c,d= 7470/3374.336 = 2.21

Therefore number of piles required = 3 piles

Design Approach 3

Combinations of sets of partial factors
DA3 —– A1 + M1 + R3

Partial Resistance factors
R3 γ_t = 1.0 (Total/combined compression)

Since the R3 recommended partial resistance factor is 1.0, there is no margin for safety on the resistance provided. Therefore this cannot be used for the design.

Summary

3 Number of 750mm diameter piles is suitable for the load at ultimate limit state.

Thank you for visiting Structville today, and we will look at other aspects of pile design in our subsequent posts.

Meanwhile, have you read…
Solved example on elastic settlement of shallow foundations

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Analysis of Continuous Beams with Partially Distributed Load Using Force Method and Clapeyron’s Theorem

By

Ubani Obinna Uzodimma

-

February 14, 2020

For the continuous beam loaded as shown above, it is desired to find the bending moments at the critical points using force method (method of consistent deformations) and Clapeyron’s theorem (3 Moment Equation). We should however note that both methods are force methods (flexibility method) since we generally solve for unknown forces.

Solution

(1) By force method
Degree of static indeterminacy neglecting horizontal forces and reactions.

D = 2m + r – 2n
D = 2(3) + 4 – 2(4) = 2
Therefore the structure is indeterminate to the 2nd order.

Basic System
A basic system is a system that is statically determinate and stable. This obtained by removing the redundant supports at A and B, and replacing them with unit loads. See the figure below.

Case 1

Bending moment on the basic system due to vertical unit virtual load at support A

Case 2
Bending moment on the basic system due to unit virtual load at support B

Case 3
Bending moment due to externally applied load on the basic system;

Influence Co-efficients

δ₁₁

δ₁₁= [1/3 × 2L × 2L × 2L] + [1/3 × 2L × 2L × L] = 4L³

δ₂₂

δ₁₁= 2[1/3 × L × L × L] = 2L³/3

δ₂₁= δ₁₂

δ₁₁= [1/6 × L × L(4L + L)] + [1/3 × L × L × L] = 3L³/2

δ_1P

δ_1P= [1/6 × qL²/16 × (2L + 2L) × L/2]] + [15/12 × qL²/16 × L × L/2] = 13qL⁴/384

δ_2P

δ_2P= [1/6 × qL²/16 × (L + L) × L/2] + [15/12 × qL²/16 × L/2× L/2] = 13qL⁴/768

The appropriate cannonical equation is given by;

δ₁₁X₁ + δ₁₂X₂ + δ_1P = 0
δ₂₁X₁ + δ₂₂X₂ + δ_2P = 0

Therefore;

(4L³)X₁ + (3L³/2)X₂ = -13qL⁴/384
(3L³/2)X₁ + (2L³/3)X₂ = -13qL⁴/768

On sloving the above equations simultaneously;

X₁ = 13qL/1920 KN

X₂ = -13qL/320 KN

The final moment values can now be computed;

M_f = M₁X₁ + M₂X₂ + M_P

M_B = (13qL/1920 × L) + 0 + 0 = 13qL²/1920
M_C = (13qL/1920 × 2L) – (13qL/230 × L) + 0 = -13qL²/480
M_D = (13qL/1920 × L) – (13qL/230 × L/2) + qL²/16 = 47qL²/960

(2) By Clapeyron’s Theorem;

First of all, we draw the free bending moment diagram

By Clapeyron’s three moment equation (EI = constant, no sinking of support);

M_AL₁ + 2M_B + M_CL₂ + 6A₁X₁ + 6A₂X₂= 0

Geometrical Properties of the free moment diagram (centroid)

SPAN A – C
M_A = 0
2M_B(L + L) + M_CL = 0
4M_B L + M_CL = 0 ——————– (1)

SPAN B-D
M_D = 0
M_BL + 2M_C(L + L) = [-6 × 7qL³/192 × 13L/128]/L
M_B L + 4M_CL = -13qL³/128 ——————– (2)

Solving (1) and (2) simultaneously;
M_B = 13qL²/1920
M_C = -13qL²/480

Therefore, the bending moment due to externally applied load is given below;

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9 Guidelines for Quality Tiling | Principles of Floor and Wall Tiling

By

Ubani Obinna Uzodimma

-

January 5, 2021

Tiles are hard-wearing thin plates that are used for covering walls, floors, roofs, countertops, etc in a building. They can be manufactured from a wide variety of materials such as ceramics, wood, natural stones, baked clays, glass, etc. The process of arranging and laying tiles on a surface is known as tiling. The aim of this article to provide guidelines that must be followed in order to achieve good and quality tiling in a building. This applies to both walls and floors.

Tiles usually provide final finishes to a surface such that the surface no longer requires any additional screeding, plastering, or painting. Therefore, tiles as a final surface finish should be elegant, neatly arranged, beautiful, and durable. As a matter of a fact, the type and quality of tiles installed in a building play a huge role in the final aesthetics of a building.

Read Also …
Cost of tiling a 3 bedroom flat in Nigeria

Tile is a good choice of finish for floors, walls, countertops, etc for a lot of reasons. They come in a variety of colours, shapes, sizes, and quality. This gives architects and interior designers enough flexibility to obtain a balanced and pleasing appearance.

neat tiling work — **Fig 1:** Neat tiling work around a floor drain

Let us look at the most common types of tiles available;

Ceramic tiles
Ceramic tiles are produced by taking mixtures of clay, earthen elements, powders, and water and pressing them into the desired shape at a high temperature. There are two popular types of ceramic tiles, and they are porcelain and vitrified tiles.

Porcelain tiles
Porcelain tiles are hard, dense, ceramic tiles with low water absorption. The most popular type is the glazed porcelain tiles, even though it may be unglazed. Porcelain tiles can be polished and made to shine without glazing due to their hardness and density. Furthermore, due to their higher density when compared with other tiles, they are usually heavier and more difficult to cut and drill through. Porcelain tiles can be used in dry or wet areas of a building.

polished porcelain tile — **Fig 2**: Polished porcelain tile

Vitrified tiles
Vitrified tiles are less dense ceramic tiles with higher water absorption. They are produced by the hydraulic pressing of a mixture of clay, quartz, feldspar, and silica, which make the vitreous surface. It however very important that the type of vitrified tile be clearly specified.

Full-body vitrified tiles have the same pigment throughout the entire thickness. Therefore, when the tiles are chipped or begin to wear out, the colour remains the same. Therefore, full-body vitrified tiles are suitable for high traffic areas. Vitrified tiles can also be glazed, with a lot of designs and surface texture printed on them.

Granite tiles
Granite tiles are produced from the natural stone ‘granite’ and are very durable and aesthetically-pleasing tiles. Due to the fact that they are produced using natural materials, they have natural flecks and minor differences in shades and pattern. The structure of granite is crystalline, and it contains quartz, mica, and feldspar minerals. This particular mix of minerals creates the unique colors, textures, and pattern movement found in all the different varieties of granite. Granite tiles are very dense and resistant to wear and unnecessary damages.

Marble tiles
Marble is a natural stone that is made by limestone’s metamorphic crystallization that results in the conversion of calcium carbonate into calcite crystals. Their natural composition makes them not to appear uniform but rather have ‘veins’ that are running in irregular patterns across the surface of the tiles. Marbles have high porosity but can be polished to shine, or the surface treated with a sealant. They are luxurious tiles and are actually more expensive. They require more maintenance and cleaning than other types of tiles.

marble tiles floor — **Fig 5**: Marble floor tiles

Guidelines for quality tiling

A poorly done tiling work is an economical disaster. Therefore it is very necessary that adequate planning be put in place before any tiling work commences.

poor tiling work — **Fig 6**: Terrible tiling work

Have you ever gone into a building and instead of your mind being at rest, you will be wondering why the tiler did not follow a definite/regular pattern, or why the grout lines are misaligned? You could spend your whole time there trying to use your mind to fix the tiles so that they will be properly aligned. But that will not be possible. To fix the tiles properly, the existing one will have to be broken and a new one reinstalled at an extra cost.

poor tiling — **Fig 7**: Ugly tiling work

This is not a good thing because your mind should be relaxed – life has enough puzzles already! On the other hand, once you look at a tiling work and discover a regular pattern/balance, your mind stops evaluating the work in a distressing manner.

Here are the guidelines to follow in order to do good tiling work;

(1) Right quality of materials
This cannot be overemphasized. You should select and insist on the best materials. For instance, the tiles to be used should not be chipped at the edges or cracked prior to installation. Depending on the usage of the building and the anticipated human traffic, the appropriate tile thickness and type of tile should be specified.

Full-body vitrified tiles or porcelain tiles will be fantastic for private sitting rooms, while granite tiles could be better for the entrance porch and other areas exposed to the weather or high traffic. Do not use tiles that would wear out easily or get discoloured when subjected to heavy traffic.

granite tiles entrance to a building — **Fig 8**: Granite entrance step to a building

Also, it should be ensured that there are no factory defects on the tiles. It is possible to have unequal tiles (with about 1 – 2 mm difference in dimensions) in the same pack. Such tiles are usually problematic to tilers. Also, some tiles are curved in plan or discoloured when supplied. All defective tiles should be rejected. As a matter of fact, you should insist on rectified tiles. Tiles are also classified in grades, usually from grade 1 to grade 5. The proper grade of tile should be specified.

Furthermore, the tile gum (adhesive), fasteners, and cement used for laying the tiles should be of superior quality. The mixing of the cement/adhesive should be of adequate strength, the floor screeding should be standard, and fasteners (when employed) should be well installed.

(2) Workmanship
Tilers are in grades. Some tilers are more skilled than others in terms of carefulness, use of tools/equipment, experience, and commitment to quality work. Some tilers are so careless when cutting tiles that they cause misalignment of tiles. It takes a lot of patience and commitment to throw down a straight line of tiles along a corridor without any misalignment or uneven grouting space. It is possible to have a straight space of 3mm groove (grouting space) for a length of 50m in the hands of a good tiler.

The tiler should also ensure that the tiles are very level, and not kicking at the joints. Where are openings or recesses for utilities, the tiler should ensure that he cuts neatly. For instance, the use of diamond tile hole cutters can be used for making holes neatly on tiles. The tiler should be patient on the job.

(3) Use the right tools
From experience, a lot of poor tiling jobs result from not having the right tools. For instance, there are some recesses that cannot be achieved properly using a manual tile cutter, but electric cutter. The tiler should ensure he has the right tools such as plumbs, spirit levels, laser levels, blue lines, spacers, cutting machines, diamond cutting discs/blades, diamond drill bits, tile trowels, rubber grout float, etc.

manual tile cutter — **Fig 10**: Manual tile cutter

(4) Quality of supervision
No matter how good a tiler is, his work still needs to be supervised and adequately monitored. Tilers may be forced to do ‘management work’ such as going ahead to install a poorly cut tile due to fatigue or need for speed. The supervisor must watch out for situations like this, and ensure that the tiler sticks to the highest quality.

(5) Tiling Pattern/Layout Drawing
There are different tiling patterns as may be recommended by the architect. The common types of tiling pattern are;

Herringbone pattern
Brickbond (draft) pattern
Block/stacked pattern
Diagonal pattern
Modular pattern, etc

**Fig 11:** Typical diagonal tiling layout of a kitchen

There should be a drawing that captures the tiling pattern, relative to the shape of the floor or wall being tiled. This gives the tiler and the contractors a good idea of what to do. The drawing should show the size and orientation of the tiles.

typical tiling pattern drawings — **Fig 12**: Typical stacked and herringbone tiling layout of a kitchen

herringbone tiling of a toilet floor — **Fig 13**: Herringbone tiling of a toilet floor

(6) Check for straightness of wall before commencing tiling work
This is particularly important because tiling exposes defective works such as lack of straightness of walls. Also, when the edges of the walls are not square, the tiling may not come out well. After checking for the straightness of the wall, the tiler and the supervisor should determine the margin of error, and adopt a corrective approach.

wall straightness — **Fig 14**: Check for straightness of tiling work

If the margin is small, it can be corrected by varying the thickness of the adhesive. But if it is too large, additional plastering works or demolition can be done to obtain the desired straightness. It is not good to try to manage the situation when the error is too large, because the finished tiling work will be adversely affected.

(7) Avoid small offcuts at the edges
It is usually very unpleasant to have offcuts less than 100 mm at the edge of walls or floors. In the first place, small offcuts are usually problematic and may lead to excess wastage. On the other hand, they are not pleasing to the eyes.

To achieve this, the tiler has some calculations to do, which usually involves finding the centrelines of the floor/wall to be tiled. Two things are involved, it is either the centreline of the tile plate coincides with the centreline of the floor (Figure 15b), or the groove (spacer/joint) between the adjacent tiles coincides with the centreline of the floor/wall (Figure 15a). Anyone that gives the bigger tile size at the edge should be adopted.

tiling arrangement — **Fig 15***: Effects of small off-cuts at the edges*

As you can see from Figure 15, option A gave an off-cut of 100 mm at the edges, while option B gave an offcut of 400 mm at the edges. Option B obviously looks better than option A. Tiny off-cuts at the edges do not usually look beautiful. Another advantage of this is that you will have equal offcuts at the two ends. This gives the tiling work a balanced look.

(8) Grouting
Grouting is almost as important as the tiling work itself. The grouting material should be non-expansive, and the grouter must be careful and skilled. For wall tiles, it is usually more beautiful when the grouting material does not fill up the groove completely. When the edge of the tiles can be seen, it is actually more attractive. Peradventure you desire to fill the groove completely, it must be very neat. The neatness of grouting is very important if the tiling must be beautiful.

Furthermore, the colour of the grouting cement must be carefully selected.

grouting of floor tiles — **Fig 16***: Grouting of floor tiles*

When tiles are to be closely joined (especially in granite or marble tiles), it is very important that the tiles close/join tightly. A little quantity of grouting cement can be used to cover other small gaps that might exist.

(9) Mitring and use edge trims
At exterior edges of elements like columns and walls, it always a good practice to mitre the tiles meeting at the edges so that the joint will close properly. The thickness of the tiles determine whether they can be mitred or not. When the tiles will not be mitred, suitable edge trims should be used. In areas where the edge of tiles must be exposed, it is better to expose the factory edge than the cut edge.

Edge trims can be made of aluminium, stainless steel, or plastics.

tile edge trim — **Fig 17**: Installation of edge trim

tile trim — **Fig 18***:Neatly installed edge trim in a convenience room*

These guidelines when followed usually results in a good tiling work. Feel free to add yours at the comment section.

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Influence Lines for Beams With Overhangs

By

Ubani Obinna Uzodimma

-

December 20, 2022

Influence lines are plots that are used to obtain the static effects of moving loads on structures such as bridges and cranes. In practice, they show the response of a structure as a unit load moves across it.

The unit load effects are later related to the actual load on the structure during the design. The action effects that can be evaluated at any section of a structure using influence lines are support reactions, internal stresses (bending, shear, axial), and deflections.

If the influence line for a structure is adequately plotted with respect to a particular section, you can obtain the desired effect of the load on any part of the structure once the moving load is at that section. This article will be dedicated to simple and compound statically determinate beams with overhangs.

Influence Lines for Statically Determinate Beams

Let us consider the beam loaded as shown below. It is desirous to obtain the influence line for the support reactions, and for the internal stresses with respect to section 1-1.

In all cases, we will be taking P as unity (i.e 1.0)

(1) Influence line for support reactions

Support A
Support reaction at point A (F_A) = (L – x)/L
At x = -L₁;
F_A = (L + L₁)/L

At x = 0;
F_A = 1.0

At x = L + L₂;
F_A = (L – L – L₂)/L = – L₂/L

Support B
Support reaction at point B (F_B) = x/L
At x = -L₁;
F_B = -L₁/L

At x = 0;
F_B = 0

At x = L;
F_B = 1.0

At x = L + L₂;
F_B = (L + L₂)/L

(2) Influence line for bending moment with respect to section 1-1

(0 ≤ x ≤ a)
M_1-1 = F_A.a – P(a – x)
M_1-1= [P(L – x).a]/L – P(a – x)
But taking P = 1.0;
= [(L – x).a]/L – (a – x)

At x = -L₁;
M_1-1 = [(L + L₁).a]/L – (a + L₁) = [L₁(a – L)/L] = -L₁.b/L

At x = 0;
M_1-1 = [(L – 0).a]/L – (a – 0) = [L₁(a – L)/L] = 0

At x = a;
M_1-1 = [(L – a).a]/L – (a – a) = [L₁(a – L)/L] = a.b/L

(a ≤ x ≤ L)
M_1-1 = [P(L – x)a]/L

At x = a;
M_1-1 = [(L – a)a]/L = a.b/L

At x = L;
M_1-1 = [(L – L)a]/L = 0

At x = L + L₂;
M_1-1 = [(L – L – L₂)a]/L = – L₂a/L

(2) Influence line for shear with respect to section 1-1

(0 ≤ x ≤ a)
Q_1-1 = P(L – x)/L – P = – F_B = –x/L

At x = -L₁;
Q_1-1 = L₁/L

At x = 0;
Q_1-1 = 0

At x = a;
Q_1-1 = -a/L

(a ≤ x ≤ L)
Q_1-1 = -(P.x)/L + P = (L – x)/L

At x = a;
Q_1-1 = b/L

At x = L;
Q_1-1 = 0

At x = L + L₂;
Q_1-1 = [(L – L – L₂)]/L = -L₂/L

Influence Line for Compound Beams

In the influence line analysis of compound beams, it is always advantageous to decompose the beams into simple beams at the location of the internal hinges. A little consideration of the analysis below will show that it largely built off from the analysis of the simple versions as was done above.

The simple logic is to pick the section under consideration and plot the influence line disregarding other adjacent beams. After drawing the diagram accurately, you can now use your ruler and draw your straight line towards the next support or internal hinge. All other values can be determined from the similar triangles rule. Note that whenever your influence line gets a zero value at an internal hinge, the influence line terminates.

The figure below shows the influence line for internal stresses at section 1-1 when the load travels through the compound beam.

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Design of Reinforced Concrete Slanted / Raker Beams

By

Ubani Obinna Uzodimma

-

February 14, 2020

Inclined beams (often called raker beams) are often found in structures like pedestrian bridges, ramps, staircases, stadiums, etc. Due to their geometry, these beams are often subjected to bending moment, shear force, and axial force. For indeterminate raker beams, it is not uncommon to see the axial forces varying from tension to compression.

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The analysis and design of such elements have been presented here using Eurocode 2. For the inclined rectangular raker beam loaded as shown below, we are to fully analyse and design the structure at ultimate limit state.

Design Information
Dimensions = 600 × 300mm
Concrete cover = 40mm
Yield strength of reinforcement = 500 N/mm²
Grade of concrete = 35 N/mm²

Geomterical Properties of the structure

Length of AB = Length of BC = 7/cos 25 = 7.7236m

cos 25 = 0.9063
sin 25 = 0.4226

At ultimate limit state, load on beam
P_Ed = 1.35Gk + 1.5Qk = 1.35(25) + 1.5(5) = 41.25 kN/m

Analysis of the structure by slope deflection method

In all cases;
L = 7.0m
L’ = 7.7236m

Fixed End Moments
F_BA = qcosθL’²/8 = (41.25 × 0.9063 × 7.7236²) / 8 = 278.769 kNm
F_BC = -qcosθL’²/12 = (41.25 × 0.9063 × 7.7236²) / 12 = -185.846 kNm
F_BC = qcosθL’²/12 = (41.25 × 0.9063 × 7.7236²) / 12 = 185.846 kNm

K₁₁ = 3EI/L’ + 4EI/L’
K₁₁ = (3EI/7.7236) + (4EI/7.7236) = 0.9063EI

K_1P = F_BA + F_BC = 278.769 – 185.846 = 92.923 KNm

For equilibrium and compatibility;
K₁₁Z₁ + K_1P = 0
0.9063Z₁+ 92.923 = 0

On solving;
Z₁ = -102.53/EI (radians)

Therefore;
M_BA = 278.769 + (-102.53/EI × 3EI/7.7236) = 238.944 KNm
M_BC = -185.846 + (-102.53/EI × 4EI/7.7236) = -238.944 kNm
M_CB = 185.846 + (-102.53/EI × 2EI/7.7236) = 159.296 kNm

Shear Forces and Span Moments

Span A-B

Support Reactions
∑M_B = 0
7.7236R_AB – (41.25 × 0.9063 × 7.7236²)/2 = -238.944
R_AB = 113.436 KN

A little consideration will show that;
Ay = 131.360 KN
Ax = 13.029 KN

∑M_A = 0
7.7236R_BA – (41.25 × 0.9063 × 7.7236²)/2 – 238.944 = 0
R_BA = 175.309 KN

Maximum span moment;
Mz = R_A.z – (41.25 × 0.9063 × z²)/2 = 0
Mz = 113.436z – 18.692z²

∂Mz/∂z = Qz = 113.436 – 37.385z

The maximum moment occurs at the point of zero shear;
Therefore, let ∂Mz/∂z = Qz = 113.436 – 37.385z = 0
On solving; z = 3.034m

Mmax = 113.436(3.034) – 18.692(3.034)²= 172.102 KNm

Span B – C

∑M_C = 0
7.7236R_BC – (41.25 × 0.9063 × 7.7236²)/2 – 238.944 = -159.296

R_BC = 154.685 KN

A little consideration will show that;
Ay = 131.360 KN
Ax = 13.029 KN

∑M_B = 0
7.7236R_CB – (41.25 × 0.9063 × 7.7236²)/2 – 159.296 + 238.944 = 0
R_CB = 134.060 KN

Maximum span moment;
Mz = R_BC.z – (41.25 × 0.9063 × z²)/2 – 238.944 = 0
Mz = 154.685z – 18.692z²– 238.944 = 0

∂Mz/∂z = Qz = 154.685 – 37.385z

The maximum moment occurs at the point of zero shear;
Therefore, let ∂Mz/∂z = Qz = 154.685 – 37.385z = 0
On solving; z = 4.1376m

Mmax = 154.685(4.1376) – 18.692(4.1376)²– 238.944 = 81.078 KNm

Kindly verify that the axial force diagram for this structure is given by;

Structural Design
Span
M_Ed = 172.102 KN.m

Effective depth (d) = h – C_nom – ϕ/2 – ϕlinks
Assuming ϕ16 mm bars will be employed for the main bars, and ϕ10mm bars for the stirrups (links)
d = 600 – 40 – 8 – 10 = 542 mm

k = M_Ed/(f_ckbd²) = (172.102 × 10⁶)/(35 × 300 × 542²) = 0.0557

Since k < 0.167, no compression reinforcement required
z = d[0.5+ √(0.25 – 0.882K) ]
z = d[0.5+ √(0.25 – 0.882(0.0557))] = 0.948d

A_s1 = M_Ed/(0.87_yk z) = (172.102 × 10⁶)/(0.87 × 500 × 0.948× 542) = 770 mm²
Provide 4H16mm BOT (AS_prov = 804 mm²)

Check for deflection

ρ = A_s,prov /bd = 804 / (300 × 542) = 0.004944
ρ₀ = reference reinforcement ratio = 10^-3√(f_ck) = 10^-3√(35) = 0.005916
Since if ρ ≤ ρ₀;

L/d = K [11 + 1.5√(f_ck) ρ₀/ρ + 3.2√(f_ck) (ρ₀ / ρ – 1)^(3⁄2)
k = 1.3 (One end continuous)

L/d = 1.3 [11 + 1.5√(35) × (0.005916/0.004944) + 3.2√(35) × [(0.005916 / 0.004944) – 1]^(3⁄2)
L/d = 1.3[11 + 10.619 + 1.650] = 30.250

β_s = (500 As_prov)/(F_yk As_req) = (500 × 804) / (500 × 770) = 1.0441

Therefore limiting L/d = 1.0441 × (7/7.7236) × 30.250 = 28.625
Actual L/d = 7723.6/542 = 14.25

Since Actual L/d < Limiting L/d, deflection is satisfactory.

Support B
M_Ed = 238.944 KN.m

Effective depth (d) = h – C_nom – ϕ/2 – ϕlinks
Assuming ϕ16 mm bars will be employed for the main bars, and ϕ8mm bars for the stirrups (links)
d = 600 – 40 – 8 – 10 = 542 mm

k = M_Ed/(f_ckbd²) = (238.944 × 10⁶)/(35 × 300 × 542²) = 0.0775
Since k < 0.167, no compression reinforcement required
z = d[0.5+ √(0.25 – 0.882K) ]
z = d[0.5+ √(0.25 – 0.882(0.0775))] = 0.926d

A_s1 = M_Ed/(0.87f_yk z) = (238.944 × 10⁶)/(0.87 × 500 × 0.926 × 542) = 1094 mm²
Provide 6H16 mm TOP (AS_prov = 1206 mm²)

Shear Design (Support A)
V_Ed = 113.436 kN
N_Ed = 67.323 KN (compression)

We are going to anchor the 4No of H16mm reinforcement provided fully into the supports.

V_Rd,c = [C_Rd,c.k.(100ρ₁ f_ck)^(1/3) + k₁.σ_cp]b_w.d

Where;
C_Rd,c = 0.18/γ_c = 0.18/1.5 = 0.12
σ_cp = N_Ed / bd = (67.323 × 1000) / (542 × 300) = 0.414 N/mm²
k = 1 + √(200/d) = 1 + √(200/542) = 1.607 > 2.0, therefore, k = 1.702
ρ₁ = As/bd = 804/(300 × 542) = 0.004944 < 0.02; K₁ = 0.15

V_Rd,c = [0.12 × 1.607(100 × 0.004944 × 35 )^(1/3) + (0.15 × 0.414)] × 300 × 542 = 91199.759 N = 91.199 KN

Since V_Rd,c (91.999 KN) < V_Ed (113.436 KN), shear reinforcement is required.

The compression capacity of the compression strut (V_Rd,max) assuming θ = 21.8° (cot θ = 2.5)

V_Rd,max = (b_w.z.v₁.f_cd)/(cot⁡θ + tanθ)
V₁ = 0.6(1 – f_ck/250) = 0.6(1 – 35/250) = 0.516
f_cd = (α_cc ) f_ck)/γ_c = (1 × 35)/1.5 = 23.33 N/mm²
Let z = 0.9d

V_Rd,max = [(300 × 0.9 × 542 × 0.516 × 23.333) / (2.5 + 0.4)] × 10^-3 = 607.554 KN

Since V_Rd,c < V_Ed < V_Rd,max
Hence, A_sw/S = V_Ed/(0.87F_ykzcot θ) = 113436/(0.87 × 500 × 0.9 × 542 × 2.5 ) = 0.21383

Minimum shear reinforcement;
A_sw/S = ρ_w,min × b_w × sinα (α = 90° for vertical links)
ρ_w,min = (0.08 × √(f_ck))/f_yk = (0.08 × √35)/500 = 0.0009465
A_sw/S (min) = 0.0009465 × 300 × 1 = 0.2839
Since 0.2839 > 0.21383, adopt 0.2839 (i.e the minimum reinforcement)

Maximum spacing of shear links = 0.75d = 0.75 × 542 = 406.5mm
Provide 2H8mm @ 300mm c/c (A_sw/S = 0.335) Ok

Note that this link must be properly closed with adequate anchorage length, because it will assist in resisting torsion.

Shear Design (Support B just to the left)
V_Ed = 175.309 KN
N_Ed = 67.323 KN (tension)

We are going to anchor the 6No of H16mm reinforcement provided fully into the supports.

V_Rd,c = [C_Rd,c.k.(100ρ₁ f_ck)^(1/3) + k₁.σ_cp]b_w.d

Where;
C_Rd,c = 0.18/γ_c = 0.18/1.5 = 0.12
σ_cp = N_Ed / bd = (67.323 × 1000) / (542 × 300) = – 0.414 N/mm²
k = 1 + √(200/d) = 1 + √(200/542) = 1.607 > 2.0, therefore, k = 1.702
ρ₁ = As/bd = 1206/(300 × 542) = 0.00741 < 0.02; K₁ = 0.15

V_Rd,c = [0.12 × 1.607(100 × 0.00741 × 35 )^(1/3) – (0.15 × 0.414)] × 300 × 542 = 82716.452 N = 82.716 KN

Since V_Rd,c (82.716 KN) < V_Ed (175.309 KN), shear reinforcement is required.

The compression capacity of the compression strut (V_Rd,max) assuming θ = 21.8° (cot θ = 2.5)

V_Rd,max = (b_w.z.v₁.f_cd)/(cot⁡θ + tanθ)
V₁ = 0.6(1 – f_ck/250) = 0.6(1 – 35/250) = 0.516
f_cd = (α_cc ) f_ck)/γ_c = (1 × 35)/1.5 = 23.33 N/mm²
Let z = 0.9d

V_Rd,max = [(300 × 0.9 × 542 × 0.516 × 23.333) / (2.5 + 0.4)] × 10^-3 = 607.554 KN

Since V_Rd,c < V_Ed < V_Rd,max
Hence, A_sw/S = V_Ed/(0.87F_ykzcot θ) = 175309/(0.87 × 500 × 0.9 × 542 × 2.5 ) = 0.33047

Provide 2H8mm @ 250mm c/c (A_sw/S = 0.0.402) Ok

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Analysis of Statically Determinate Compound Frames (Part 3): A solved Example

By

Ubani Obinna Uzodimma

-

February 14, 2020

Question
A frame is loaded as shown above. There are internal hinges at nodes B and 2. Plot the internal stresses diagram due to the externally applied load.

Solution
Check for static determinacy.
A quick but non-general way of verifying degree of static indeterminacy of simple frames is;

D = R – e – S

where;
D = Degree of static indeterminacy
R = Number of support reactions = 5
e = Number of static equilibrium equations = 3
S = Number of special condition = 2 ( internal hinges)

D = 5 – 3 – 2 = 0
Therefore, the structure is statically determinate and stable.

Read also in this blog….
Static Determinacy of Rigid Frames
Understanding Sign Conventions in Structural Analysis

Support Reactions
Let ∑F_X = 0
Ax + 5 = 0
Ax = -5 KN ←

Let ∑M₂^R = 0
2Dy – (10 × 3) = 0
Dy = 15 KN

Let ∑M_B^R = 0
6Dy + 3Cy – (10 × 7) – ((2 × 4²)/2) = 0
6(15) + 3Cy – 70 – 16 = 0
Cy = -1.333KN↓

Let ∑M_B^B = 0
Note that the due to the direction of the horizontal reaction A, it will exert a positive moment about column AB. Therefore;
4Ax – (5 × 2) – M_A = 0
4(5) – 10 = M_A
M_A = 10 KN.m

Let ∑M₂^L = 0
4Ay + 4Ax + Cy – M_A – ((2 × 4²)/2) – (5 × 2) = 0
4Ay + 4(5) – 1.333 – 10 – 16 – 10 = 0
Ay = 4.333 KN

Equilibrium check;
∑F_Y ↓ = (2 × 8) + 10 = 18KN
∑F_Y ↑ = 15 – 1.333 + 4.333 = 18KN OK

Fig 1: Support Reactions Due to Externally Applied Load

Analysis of Internal forces

Section A – 1 (0 ≤ y ≤ 2.0m)
(i) Bending Moment
My = 5y – 10 ———— (1)
At y = 0;
M_A = -10 kNm
At y = 2.0m;
M₁^B = 5(2) – 10 = 0

(ii) Shear Force
The section is subjected to constant shear force;
Qx = ∂M_y/∂y = 5 KN

(iii) Axial Force
Ny + 4.333 = 0
N_A-1 = -4.333 KN (compression)

Section 1 – B (2.0m ≤ y ≤ 4.0m)
(i) Bending Moment
M_y = 5y – 5(y – 2) – 10 ———— (2)
At y = 2m;
M₁^UP = 5(2) – 10 = 0
At y = 4.0m;
M_B^B = 5(4) – 5(2) – 10 = 0
This shows that there is no bending moment at this section.

(ii) Shear Force
Differentiating equation (1) above yields zero. Therefore, there is shear force at the section also.

(iii) Axial Force
Ny + 4.333 = 0
N_1-B = -4.333 KN (compression)

Section B – C (0 ≤ x ≤ 3.0m)
(i) Bending Moment
M_x = 4.333x – x² ———— (3)
At x = 0;
M_B^R = 0
At x = 3.0m;
M_C^L = 4.333(3) – (3)2 = 4 KNm

Maximum moment occurs at the point of zero shear. Therefore;
Qx = ∂M_x/∂x = 4.333 – 2x = 0
On solving;
x = 4.333/2 = 2.1665m
Mmax = 4.333(2.1665) – (2.1665)2 = 4.693 KNm

(ii) Shear Force
The section is subjected to constant shear force;
Qx = ∂M_x/∂x = 4.333 – 2x

At x = 0;
Q_B_R = 4.333 KN
At x = 3.0m;
M_C^L = 4.333 – 2(3) = -1.667 KN

(iii) Axial Force
Nx = 0
No axial force

Section C – 2 (3.0m ≤ x ≤ 4.0m)
(i) Bending Moment
Mx = 4.333x – 1.333(x – 3) – x² ———— (4)
At x = 3m;
M_C^R = 4.333(3) – 32 = 4 KNm
At x = 4.0m;
M₂ = 4.333(4) – 1.333(1) – (4)2 = 0

(ii) Shear force
Expanding equation (4) above;
M_x = –x² + 3x + 4

Qx = ∂Mx/∂x = -2x + 3

At x = 3m
Q_C^R = -2(3) + 3 = -3 KN
At x = 4m
Q₂^L = -2(4) + 3 = -5 KN

(iii) Axial Force
No axial force

Section D – E (0 ≤ x ≤ 1.0)
(i) Bending Moment
M_x = -10x ————— (5)

At x = 0;
M_E = 0
At x = 1.0m;
M_C^L = -10(1) = -10 KNm

(ii) Shear Force
Qx = ∂M_x/∂x = -10 KN

But since we are coming from the right to the left, downward force is positive.
Therefore the section subjected to a positive constant shear force of 10 KN

(iii) Axial force
There is no axial force at the section

Internal Stresses Diagram
We can now plot the internal stresses diagram from the values obtained above.

Fig 2: Final Bending Moment Diagram

Fig 3: Final Shear Force Diagram

Fig 4: Final Axial Force Diagram

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Imperfection Analysis of Structures

By

Ubani Obinna Uzodimma

-

December 9, 2022

Table of Contents

Imperfection in structures can be described as deviations and inconsistencies in structures and structural members, which can affect the behaviour of the structure from the theoretically assumed perfect state. Imperfections in structures are often envisaged during analysis, and their effects are taken into consideration, especially in cases where they could be critical.

Types of Imperfections

Different types of imperfections are considered in structural analysis and they are;

(i) Geometrical imperfection
(ii) Material imperfection
(iii) Structural imperfection

Geometrical Imperfection

As the implies, these are imperfections due to variance in the dimensions of the members, or lack of straightness or verticality in a structural member.

For example, when a beam member that is supposed to be 400mm depth is finished as 395mm, then this is geometric imperfection. According to clause 5.2(1)P of Eurocode 2, deviation in members cross-section is normally taken into account in material partial factor of safety, and may not need to be analysed for again.

Also, when a 2-storey framed building that is supposed to be perfectly straight/vertical is leaning by say an angle of 3 degrees, this is also geometric imperfection. This form of geometric imperfection is taken into account when analysing sensitive structures.

Material imperfection

This type of imperfection always arises from deviation in the assumed material properties. For instance, if the design engineer specified a concrete grade of M30, and grade 28.5 was achieved on site, then this is material imperfection. Also, when the modulus of elasticity of steel specified was 200 kN/mm² and 195 kN/mm² was fabricated, this is also material imperfection. Another variance in material imperfection is residual stresses, which occurs in steel members during welding or fabrication.

Material imperfection is often taken care of by introducing material factor of safety in design.

Structural Imperfection

Structural imperfection often arises in structures due to issues like eccentricity in joints, variations in support/boundary conditions, variation in stiffness/rotation of joints.

Analysis of Imperfections in Structures

In frames, all types of imperfections are usually introduced as equivalent geometrical imperfections with increased value of amplitude e_0d, while in plated structures, geometric imperfections and residual stresses are introduced to derive buckling factors.

According to clause 5.3.1(1) of Eurocode 3, the following imperfections should be taken into account;

a) global imperfections for frames and bracing systems
b) local imperfections for individual members

Global imperfections for frames or bracing systems cover for lack of verticality for frames or straightness of structures restrained by bracings, while local imperfections cover for lack of straightness or flatness of a member and residual stresses of the member.

We should note that other imperfections are covered by the partial factors in limit state design.

Global Imperfection Analysis for frames

In framed structures, the assumed shape of global imperfection may be derived from the elastic buckling mode of the the structure in the plane considered. For frames sensitive buckling in sway mode (e.g. tall buildings), the effect of imperfection is allowed for by means of equivalent imperfection in the form of an initial sway imperfection.

The effects of initial sway imperfection may be replaced by systems of equivalent horizontal forces. These initial sway imperfections are applied in all relevant horizontal directions, but need only be considered in one direction at a time. For multi-storey buildings, equivalent forces are used and they should be applied at each floor and roof level.

The imperfections may be determined from;

ϕ = ϕ₀α_hα_m
where;
ϕ₀ is the basic value = 1/200

α_h is the reduction factor for height applicable to columns
α_h = 2/√h but 2/3 ≤ α_h ≤ 1.0

α_m is the reduction factor for number of columns in a row

where m is the number of columns in the row.

A schematic diagram showing the representation of sway imperfections with equivalent horizontal loads is shown below;

According to clause 5.3.2(4)B, building frames sway imperfections may be disregarded whenever H_Ed ≥ 0.15 V_Ed

Solved Example

Let us carry out the global sway imperfection of the frame loaded as shown below.

BEAMS – UB 305 x 102 x 28

Solution

∑H_Ed = 15 + 23 + 45 = 83 kN
0.15∑V_Ed = 0.15(25 + 65 + 25 + 70 + 126 + 70 + 145 + 209 + 145) = 132 kN

Since H_Ed < 0.15 V_Ed sway imperfection will need to be considered.

α_h = 2/√h = 2/√12 = 0.577 since it is < 2/3, take 2/3 (check specified limits)

m = 3 (3 columns in a row)
α_m= sqrt [0.5 × (1 + 1/3)] = 0.8165

Therefore;
ϕ = ϕ₀α_hα_m= (1/200) × (2/3) × (0.8165) = 0.00272

Therefore,
At the roof level, imp1 = ϕV₁= 0.00272 × (25 + 65 + 25) = 0.3128 kN
At the second floor, imp2 = ϕV₂= 0.00272 × (70 + 126 + 70) = 0.7235 kN
At the first floor, imp3 = ϕV₃= 0.00272 × (145 + 209 + 145) = 1.357 kN

These imperfections belong to each cross frame (bay). For the double bay frame in our example, the imperfection will need to be doubled;

Analysing the structure using StaadPro;

Analysis Results
The moment diagram below is for the external load only without the effect of imperfections.

The moment diagram below is for external load with the effect of imperfection inclusive.

Analysis of results
Maximum Moment on ground floor column without the effect of imperfection (column at RHS) = 99.979 kNm
Maximum Moment on ground floor column with the effect of imperfection (column at RHS) = 109.048 kNm

This is about an 8.316% difference in the bending moment value.

A little consideration will show that for axial loads also;
Maximum axial load without the effect of imperfection (middle column) = 398.394 kN
Maximum axial load with the effect of imperfection (middle column) = 398.383 kN

This shows that these imperfections have more effect on the bending moment of the structure than on the axial load.

For shear force on the columns, a difference of about 8.123% was observed.

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Durability of Structures and How to Calculate Concrete Cover (Eurocode 2)

By

Ubani Obinna Uzodimma

-

February 14, 2020

According to clause 4.1 of EN 1992-1-1:2004, a durable structure should meet the requirements of serviceability, strength and stability throughout its design working life, without significant loss of utility or excessive unforeseen maintenance.

Fig. 1: Typical Carbonation Problem Leading to Lack of Durability in a Reinforced Concrete Structure

The required protection of any structure is established by considering the intended use of the structure, the design working life, maintenance programme, significance of possible direct and indirect actions, environmental conditions etc. Environmental conditions that can be considered may include storage of aggressive chemicals, chlorides in the concrete, acid solutions, alkali-aggregate reactions etc which are generally categorised as chemical attacks. Issues like abrasion, water penetration, temperature change etc are categorised as physical attacks.

According to clause 4.3 of Eurocode 2, in order to achieve the required design working life of a structure, adequate measures shall be taken to protect each structural element against the relevant environmental actions. The requirements for durability shall be included when considering the following:

Structural conception,
Material selection,
Construction details,
Execution,
Quality Control,
Inspection,
Verifications,
Special measures (e.g. use of stainless steel, coatings, cathodic protection).

An important aspect of durability of reinforced concrete structure is the anticipated exposure of the element under consideration. Some exposure conditions are given in the table below (Table 4.1 of EN 1992-1-1:2004);

2.0 Concrete Cover
Concrete cover is the distance between the surface of the reinforcement closest to the nearest concrete surface (including links and stirrups and surface reinforcement where relevant) and the nearest concrete surface.

Concrete cover is defined as a minimum cover, c_min, plus an allowance in design for deviation, Δc_dev;

c_nom = c_min + Δc_dev———- (1)

Minimum concrete cover, c_min, shall be provided in order to ensure the safe transmission of bond forces, protection of the steel against corrosion (durability) and for an adequate fire resistance.

The greater value for c_min satisfying the above mentioned requirements shall be used. This is given by;

c_min = max {c_min,b; c_min,dur + Δc_dur,γ – Δc_dur,st – Δc_dur,add; 10 mm} ———- (2)

where;
c_min,bis the minimum cover due to bond requirement
c_min,duris the minimum cover due to environmental conditions
Δc_dur,γis for additive safety element
Δc_dur,st is the reduction of minimum cover for use of stainless steel
Δc_dur,addis the reduction of minimum cover for use of additional protection

In order to transmit bond forces safely and to ensure adequate compaction of the concrete,
the minimum cover should not be less than c_min,b given in Table 4.2 of EN 1992-1-1:2004 and given in Table 2 below;

Minimum%2Bcover%2Bfor%2Bbond%2Brequirement

The cover due to environmental conditions (durability) is given in the table below; (Table 4.4 EN 1992-1-1:2004).

3.0 Solved Example
Design the concrete cover of an external reinforced concrete beam.. The concrete in use has resistance class C30/37.
Bottom longitudinal bars are H16; the stirrups are H8

The max aggregate size is: dg = 20 mm (< 32 mm).

The design working life of the structure is 50 years.
Normal quality control is put in place.

Solution

From table 4.1 of EN 1992-1-1:2004, the appropriate exposure class is XC3 (assuming the beam is sheltered from rain).

The recommended structural class (design life of 50 years) is S4.

First, the concrete cover for the stirrups is calculated.

c_min,b = 8 mm

We obtain from table 4.4N – EC2:

c_min,dur = 25 mm

Moreover:

Δc_dur,γ = 0 ;

Δc_dur,st = 0 ;

Δc_dur,add = 0 .

From relation (2):

c_min = max {c_min,b; c_min,dur + Δc_dur,γ – Δc_dur,st – Δc_dur,add; 10 mm}

c_min = max (8; 25 + 0 – 0 – 0; 10 mm) = 25 mm

Moreover:

Δc_dev = 10 mm (recommended value)

We obtain from relation (1):

c_nom = c_min + Δc_dev = 25 + 10 = 35 mm .

Secondly, the concrete cover for the longitudinal main bars is calculated.

c_min,b = 16 mm

We obtain from table 4.4N – EC2:

c_min,dur = 25 mm

Moreover:

Δc_dur,γ = 0 ;

Δc_dur,st = 0 ;

Δc_dur,add = 0 .

From relation (2):

c_min = max {c_min,b; c_min,dur + Δc_dur,γ – Δc_dur,st – Δc_dur,add; 10 mm}

c_min = max (16; 25 + 0 – 0 – 0; 10 mm) = 25 mm

Δc_dev = 10 mm.

We obtain from relation (1):

c_nom = c_min + Δc_dev = 25 + 10 = 35 mm.

Since the cover for stirrups is more critical, the actual concrete cover for the main longitudinal bars is 35 + 8 = 43mm

NOTE:

According to Clause 4.4.1.2(9), where in-situ concrete is placed against other concrete elements (precast or in-situ) the minimum concrete cover of the reinforcement to the interface may be reduced to a value corresponding to the requirement for bond provided that:

– the strength class of concrete is at least C25/30,

– the exposure time of the concrete surface to an outdoor environment is short (< 28 days),

– the interface has been roughened.

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Hope for Fresh Graduate Engineers

By

Ubani Obinna Uzodimma

-

February 14, 2020

On the of 22nd of August 2017, I woke up by 5:00am with the inspiration that there is hope for everybody, irrespective of life situation and circumstances. I started writing this post on my way to work that morning, and I feel like sharing this on this special day to my colleagues in the engineering profession. Irrespective of what you believe in and what you are passing through right now, I want you to share in the same inspiration, and be convinced that there is hope in that career situation.

Some established professionals are already living their dream lives. Some are very hopeful that they are not so far away from their desired life, while some are just comfortable with their lives and their job – living it as it comes.

On the other hand, there are a good number of people living in despair due to the current state of their career. They wake up in the morning without really knowing what to do with their day. They have tried a lot of things: applying for jobs and not getting called for interview, attending interviews and immediately knowing that the opportunity has been screwed, being hopeful after an interview and getting nothing afterwards etc.

You may have taken the bold step into self employment, business, entrepreneurship etc but without any successful outcome. You may have also had doors banged on your face by relatives, big uncles, companies, and firms, and it appears to you like you are the most unfortunate and incompetent person on earth. Do not give up or cast blames yet, because there is hope!!

Prior to my graduation, a lot of people promised to help me get a nice job but till this date, none of such promise has been fulfilled. However, I harbour no bitterness and anger towards such people (in fact I am not even disappointed) because such promises gave me hope when I needed something to look forward to. You can verify you actually feel better when you have something you are looking forward to; it relieves despair.

There are millions of fresh engineering graduates all over the world each year, with fewer getting hired. This situation is actually worse in developing countries. I attended a job examination by Department of Petroleum Resources (DPR) Nigeria, who was trying to recruit for various positions in the month of July, 2017. Instead of trying to focus on the ‘competitiveness’ of the situation, I was filled with wonder when I saw the number of youths who turned out for the test despite the fact that I wrote the exam on the last day (the exam was held for one week across three different locations in the country). But then, I knew right within me that despite the very limited positions available, there is hope for all the ambitious youths who turned out for the test. A lot of them will succeed somewhere else that is not DPR.

A lot of people will talk about ’employability issues’ of fresh graduates, but the honest truth is that there is so much crowd in the labour market, which makes it appear like basic qualifications ‘do not really count’ anymore, especially for fresh graduates. When I was an undergraduate, I equipped myself with knowledge of many civil engineering softwares (StaadPro, Orion, Prokon etc) with a view of having an outstanding CV. But guess what, those skills are the most common thing in every civil engineering graduate’s CV today. It no longer counts in a special way!!

Sometime ago, I attended an interview and by chance, I met another candidate, who had a Master’s Degree from a high ranking foreign university, competing for the same position with me. Immediately, I was seriously thinking that he was the most likely candidate for the job. But on the other hand, I was unmoved because I knew even if the opportunity passes by, there is something more beautiful that lies ahead of me. I am so confident of being destined for greatness. So I answered the interview questions honestly, and was called back the next day.

Sometimes, I forget my class of degree because I am currently focusing on who I really am, and on how I am improving as the day goes by. I can remember that at the same interview, the interviewer told me that I only made the interview because of my class of degree and creativity (a proof that academic credentials count), and that he loved my CV, even though they were hiring experienced engineers.

But then, believe me that I am no longer focusing on my class of degree, or the name attached to any qualification. I am now focusing on myself – who I really am. Your life will turn out to be beautiful when you realise who you are, and this begins when your interest shifts from mediocrity to adding value.

All I want to say is this, there is hope, especially when you believe and focus on yourself and in your dreams. The competition out there can be overwhelming, and it is not your fault – the crowd is too much. Some people are just lucky because they come from very wealthy families or are highly connected, while some get rewarded for their hard work. Possibly, you could not afford to study or advance your degree in developed countries, or scholarship opportunities may have eluded you severally, do not worry – completely focus on who you really are. Start from thinking in a productive and creative way!

Have you felt that you did not work hard enough during your undergraduate days, in that you are not really proud of your credentials? Do not worry, look deep within you and realise that there is hope for big success, based on your real content, and not what some piece of paper or a HR manager’s feedback says about you.

Your success can only come from who you truly are, and that is your source of hope. The question is this, do you know who are? Are you confident in what you can do? Once these questions are answered correctly, sit up and follow that dream of yours without giving up when it gets tough.

My declaration often goes this way,

“I know who I am
Made in the image of God
Crafted for greatness
I rule and reign in the midst of adversity
I know I am not ordinary
And so I triumph”

Happy new month of November to you!!

Love from,
Ranks Ubani

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