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

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January 3, 2018

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

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December 29, 2017

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

-

November 30, 2017

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

-

November 17, 2017

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

-

November 13, 2017

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

-

November 11, 2017

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

-

November 1, 2017

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

How to Succeed as a Civil Engineering Student

By

Ubani Obinna Uzodimma

-

October 24, 2017

The role of a civil engineer in the advancement of civilisation and the cause of mankind cannot be overemphasized. If for any reason you find yourself studying civil engineering in an institution of higher learning, you should feel lucky provided you are convinced that you are where you are supposed to be.

Civil engineers work very closely with people, and are involved in the development and maintenance of public infrastructures and utilities. There are varieties of clients like private individuals, government, corporate bodies, organisations, etc. for civil engineering projects. Therefore, the sole responsibility of a civil engineer is to conceptualise, design, and build infrastructures and utilities that will meet and satisfy the intended use of the structure in an economical manner.

This activity must have safety as top priority, and must offer minimal disruption to public/individual daily activities. That is why sometimes civil engineers will have to schedule their work at night or work during the weekends in order to keep other people happy and comfortable. So as you can see, the career path you are about take has ‘people’ as top priority, and you should be proud to be engaged in such a noble profession.

Civil engineering is a very broad field of knowledge. That is the main reason why many important aspects of the course cannot be covered at undergraduate level. If all relevant topics were to be fully treated, civil engineering students should be spending a minimum of seven years at undergraduate level.

There are specialties in the field of civil engineering, and students normally choose an area of specialisation during their Masters Degree program. The prominent specialties are;

Structural Engineering
Geotechnical engineering
Water Resources Engineering
Highway and Transportation Engineering
Environmental/Sanitary Engineering
Construction Management

However, whenever you graduate from the university as a civil engineer, there is a minimum level of knowledge expected of you in all these fields of knowledge, and whenever you fail to meet it, it is very disappointing. For example, basic training in the university should give you the capacity to understand, explain, and recommend structural concepts, carry out simple-medium complicity designs, carry out simple field measurements, produce and interpret working drawings, have knowledge of civil engineering materials, and be able to make technical decisions. A lot of knowledge is also gained from the field. As a matter of fact, practice and experience makes a complete civil engineer; and that is why professional license can only be issued after some years of practise.

To succeed as civil engineering student, you must keep the following in mind;

(1) You must be passionate
To excel in this field, you must love it. It is only passion that will make the entire hard work associated with studying civil engineering seem effortless. If you are interested in just obtaining the degree and moving over to the banking industry, it will appear as so much work to you. But if you are determined to practise civil engineering after graduation, then the natural willingness to study and acquire knowledge will make the efforts seem so simple.

(2) Do not be far away from like minded people
To succeed as a civil engineering student, you have to make friends with those in the same department with you, and if possible live close by. There will be numerous assignments, term papers, laboratory reports etc. and sometimes you might need help. It is good these days that social media has bridged a lot of communication gap. So you must belong to the ‘WhatsApp’ group of your class, and make sure you always have internet data in order to be updated with the latest information. If you are close with like minded people, there will always arise discussions that will make you stay on your toes and work hard.

(3) Understand from first principles and follow up
Most civil engineering courses usually build up as you progress. At every stage, you should try and understand what is being taught. For instance, you must understand statics (rigid body mechanics) before you start looking at strength of materials (deformable body mechanics). If you do not understand how to calculate support reactions when you should, then it means you basically cannot progress in structural analysis. In my 5th year, we were permitted to plot bending moment diagrams without showing calculations or steps; and we still get our full marks because at that level, we were dealing with advanced topics. But when we were in the 2nd year, the whole emphasis was on how to carry out such calculations for plotting moment diagram. So never miss any step.

(4) Read widely
Do not rely on classroom notes and exercises only. They can help you pass your exams, but beyond the classroom walls, you will definitely see deficiencies. To be technically excellent, study variety of textbooks with different backgrounds (American, British, Indian, Scandinavian etc). This will give you confidence, power, and speed especially in the exams and after your graduation. It is important to be familiar with more than one method of solving a problem.

(5) Learn computer programs with your free time or during your Industrial Training period
Do not wait until you graduate before you start learning how to use design software and computers effectively. Start from MS office and learn how to do important things like creating tables on MS Word, plotting graphs on Excel, carrying out regression analysis and ANOVA etc. This is very important for your studies and for your career. From there, you can advance to structural analysis and design software, programming softwares like MATLAB, and CAD softwares because a good engineer must be able to produce technical drawings. In this current dispensation, these skills are a must for you to land a job.

(6) Avoid Wild Social Activities
You must be focused and disciplined. Don’t always party all night and come back tired and wasted. However on some weekends, you can hang out with friends, share some drinks, and return to your apartment in good condition. Channel your energy on positive things, and always keep your mind refreshed.

(7) Ask Questions
Always make out time to discuss and ask questions whenever you are confused. Note that whatever you have learnt sticks to your memory, and all you will have to do is to refresh your memory periodically. I remember the first time I was taught analysis of framed structures by a senior colleague. This was after spending hours trying to figure out how to transition from vertical to horizontal members. All he did was to offer an explanation that made sense and solved an example. From that point, other classroom works were just like revision.

(8) Work on your speed in exams
For you to always land your A’s in exam and get your full marks, you have to be fast in your exams. Most structural analysis problems can be lengthy, but lecturers will often limit degree of indeterminacy for hand problems to just 3. This is to avoid you not solving more than 3 x 3 matrix in exam. So you will agree with me that you will still have to be fast because exams are usually 3 hours maximum depending on the weight of the course. One way to achieve this is to practice all examples extensively and plan ahead of every exam.

(9) Understand the requirements of your lecturers
Every lecturer has his/her interest as far academic prowess is concerned (popularly called academic interest). Once you tickle them briefly in your answer sheet by intelligently referring to their work or interest, they will be impressed and see you as a ‘research minded and serious undergraduate student’. Trust me that a lot of scholars are always obsessed with their works and feel so happy when someone refers to it. So you have to sit down and work hard, nothing good comes easy. It will be good to study their question patterns, policies, and the values they uphold in regards to their course. Do not violate this.

(10) Look towards the future
Academic excellence is not ultimate success in life. Find time to balance academic work with worthwhile social activity and volunteering. You can participate in student politics if you wish, and be religiously active. Do not be socially awkward because your social skills and lessons from human interaction will help you more after you leave school. You will later realise that a particular social activity you engaged in will give you more pleasant memories than making an ‘A’ in a 4 units course.

(11) Make your apartment your comfort zone
It will be very stressful for you if you cannot comfortably study in your apartment without distractions. Make sure you have all resources handy. In a country like Nigeria where there is no guaranteed 24 hours power supply, do not rely on soft copy of textbooks or e-books. Make sure you have all your study materials handy in your apartment. Do not make your apartment a movie theatre for your colleagues/neighbours, or a play station game room for your friends. Keep it as private as possible, and possibly make it a place where people can come and add value.

So believe in yourself as you take this giant step towards a career in engineering. It is not difficult but it can be challenging. It is can be stressful, but it won’t break you. Remain determined and persistent till victory roars….

Be inspired for excellence today!!!

Thank you for visiting Structville.

Free Vibration of Water Tank Stand When Filled With Water

By

Ubani Obinna Uzodimma

-

October 20, 2017

It is desired to find the eigenfrequencies and eigenvectors (modal parameters) of concrete water tank support when filled with water. This is a case of undamped free vibration, treating the stored water as a lumped mass on the tank stand. The tank will be treated as a system with 2 degrees of freedom, considering lateral displacement only.

Data and Load Analysis;
Density of concrete = 24 KN/m³
Thickness of slab = 150mm
Dimension of columns = 230 x 230mm
Supporting beams = 300 x 230mm
Weight of water = 10 KN/m³
Size of tank = 2000 litres
Modulus of elasticity of concrete = 21.7 × 10⁶ KN/m²

Load Analysis
Self weight of slab = 24 KN/m³ × 0.15m × 2.5m × 2.5m = 22.5 KN = 2293.75 kg
Weight of supporting beams on the four sides = 4 × 24 KN/m³ × 2.27m × 0.23m × 0.3m = 15.036 KN = 1532.77kg
Weight of water = 10 KN/m³ × 2m³ = 20 KN = 2038.73 kg
Weight of tank (assume) = 75kg

At the 1st level = 2293.75 + 1532.77 + 2038.73 + 75 = 5940.25 kg
At the 2nd level = 2293.75 + 1532.77 + 2(2038.73) + 2(75) = 8053.98 kg

We are assuming that the floor slab is very stiff compared to the columns. So the flexural rigidity is taken as infinity.

Geometrical properties of the sections
Columns = 23cm x 23cm = 0.23m x 0.23m
Moment of inertia of column I_C = (0.23 × 0.23³)/12 = 2.332 × 10^-4 m⁴
EI_C = (2.332 × 10^-4 m4) × (2.1 × 10⁷ KN/m²) = 4897.2 KN.m²

The mass matrix of the structure is given below;

Determination of the stiffness matrix

(12EI_C)/L³ = (12 × 4897.2)/3³ = 2176.533 KN/m
K₁₁ = (12EI_C)/h³ + (12EI_C)/h³
K₁₁ = 2176.533 + 2176.533 = 4353.067
K₂₁ = -(2176.533 + 2176.533) = -4353.067
K₂₂ = 2(2176.533) + 2(2176.533) = 8706.132

The stiffness matrix is therefore;

For an undamped free vibration, the equation of motion is;

Mü + ku = 0 —————– (1)

Arranging this in matrix form, we obtain;

The characteristic polynomial equation is given by;

On solving;

((4353.067 × 10³ – 5940.25ω²) × (8706.132 × 10³ – 8053.98ω²) – (4353.067 × 10³ × 4353.067 × 10³) = 0

On simplifying;

(4.783 × 10⁷)ω⁴ – (8.677 × 10¹⁰)ω² + (1.8949 × 10¹³) = 0

On solving;

ω₁² = 253.9146 rad²/sec²
ω₂² = 1559.9 rad²/sec²

ω₁ = 15.9347 rad/sec
ω₂ = 39.4951 rad/sec

The natural frequencies;
F₁ = ω₁/2π = 2.5361 Hz
F₂ = ω₂/2π = 6.2858 Hz

Determination of the mode shapes

For ω₁² = 253.9146 rad²/sec²
Considering the first row of the matrix;
(2.8448 × 10⁶)A₁ – (4353.067 × 10³)A₂ = 0

Setting A₁ = 1.0
A₂ = (2.8448 × 10⁶)/(4353.067 × 10³) = 0.6780
Therefore;ϕ₁ = [1, 0.6535]^T

For ω₂² = 1559.9 rad²/sec²
Considering the first row of the matrix;
(-4.9129 × 10⁶)A₁ – (4353.067 × 10³)A₂ = 0

Setting A₁ = 1.0
A₂ = (-4.9129 × 10⁶)/(4353.067 × 10³) = -1.128
Therefore;ϕ₂ = [1, -1.128]^T

The generalised mass and stiffness of the structure (modal mass and stiffness);

As a verification;
K₁/M₁ = 2381668.751 / 9379.8 = ω₁² = 253.9146 rad²/sec²
K₂/M₂ = 25251112,921 / 9379.8 = ω₂² = 1559.9 rad²/sec²

Eigenvector Mass Orthonormalization
The idea is to obtain;

ϕ₁^T Mϕ₁ = 1.0

The conversion factor;

Therefore, C₁ = 1/√(9379.8) = 0.0103

Hence;

Similarly;

Therefore, C₂ = 1/√(16188.005) = 0.00785

Hence;

We can therefore obtain;

Let χ_i(t) be the generalised coordinate which represent the amplitude of the orthonormalised mode shape.

The displacement time history response is therefore; u(t) = Φχ(t)

But the modal equation of motion

Mü + ku = 0 —————– (1)

Realise that;

Substituting ü and u into equation 1 and multiplying by Φ^T , we obtain;

Φ^T MΦẍ + Φ^T KΦx = 0
Which finally leads to;

ẍ + K_gx = 0

This leads to the following differential equations;

ẍ₁(t) + 253.9146x₁(t) = 0 ——————– (a)
ẍ₂(t) + 1559.9x₂(t) = 0 ——————– (b)

Let the initial condition be a unit velocity;
x₁(0) = 0; ẋ₁(0) = 0

Solving the differential equation by Laplace Transform;

For mode 1;
ẍ₁(t) + 253.9146x₁(t) = 0 x₁(0) = 0; ẋ₁(0) = 1.0

(s²x ̅ – sx₀ – x₁) + 253.9146x ̅ = 0
s²x ̅ – 1 + 253.9146x ̅ = 0
x ̅ (s²+ 253.9146) = 1
x ̅ = 1 / (s²+ 253.9146)

On solving by Laplace Transform;

x₁ = (1/√253.9146) sin(√253.9146)
x₁ = 0.0627 sin(15.935t)

For mode 2;
ẍ₂(t) + 1559.96x₂(t) = 0 x₁(0) = 0; ẋ₁(0) = 1.0

(s²x ̅ – sx₀ – x₁) + 1559.9x ̅ = 0
s²x ̅ – 1 + 1559.9x ̅ = 0
x ̅ (s²+ 1559.9) = 1
x ̅ = 1 / (s²+ 1559.9)

On solving by Laplace Transform;

x₂ = (1/√1559.9) sin(√1559.9)
x₂ = 0.0253 sin(39.495t)

Knowing that
u(t) = Φχ(t)

The displacement time history is therefore;

u₁(t) = 0.00064581 sin(15.935t) + 0.0001986 sin(39.495t)
u₂(t) = 0.0004219 sin(15.935t) – 0.000224 sin(39.495t)

The displacement time history graph is shown below;

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Analysis of Sub-Frames Using the Stiffness Method

By

Ubani Obinna Uzodimma

-

October 11, 2017

Several codes of practice in the world allow us to idealise structures into 2-dimensional frames for simplified analysis. For sub-frames, it is obvious that the force method becomes less handy due to the high number of redundants, and the next best alternative is the displacement method (stiffness method), where we solve for the unknown displacements.

The displacement method, also known as the stiffness method, has become a cornerstone of structural analysis. It offers a powerful and versatile framework for solving equilibrium problems in various structures, from trusses and beams to complex frames and continua.

At its core, the stiffness method focuses on determining the unknown displacements of points within a structure under the action of applied loads. This is achieved by establishing a relationship between the displacements and the internal forces generated within the structure. This relationship is expressed through the stiffness matrix, which encodes the inherent rigidity of the structure and its resistance to deformations.

Solved Example Using the Stiffness Method

In this article, we have a typical example where a problem that would have generated a (21 x 21) matrix using the force method, has been solved using a (4 x 4) matrix by the stiffness (displacement) method. Another approach to the solution of this problem is the moment distribution method. But in this case, the stiffness method remains the fastest.

For the frame loaded as shown above, we have to start by drawing the kinematic basic system of the structure. This has been achieved by fixing the nodes 1, 2, 3 and 4 against rotation as shown below.

sub-frame analysis using the stiffness method

We will now have to evaluate the basic system for different cases of a unit rotation Z_i = 1.0, applied at the fixed nodes.

Analysis of Case 1
Z₁ = 1.0; Z₂ = Z₃= Z₄= 0

K₁₁ = (4EI/3) + (4EI/4) + (13.6EI/3.825) = 5.889EI
K₂₁ = (6.8EI/3.825) = 1.778EI
K₃₁ = 0
K₄₁ = 0

Analysis of Case 2
Z₂ = 1.0; Z₂ = Z₃= Z₄= 0

K₁₂ = (6.8EI/3.825) = 1.778EI
K₂₂ = (4EI/3) + (4EI/4) + (13.6EI/3.825) + (13.6EI/2.8) = 10.7460EI
K₃₂ = (6.8EI/2.8) = 2.4286EI
K₄₂ = 0

Analysis of Case 3
Z₃ = 1.0; Z₁ = Z₂= Z₄= 0

K₁₃ = 0
K₂₃ = (4EI/3) + (4EI/4) + (13.6EI/3.825) + (13.6EI/2.8) = 10.7460EI= 2.4286EI
K₃₃ = (4EI/3) + (4EI/4) + (13.6EI/3.325) + (13.6EI/2.8) = 11.2807EI
K₄₃ = (6.8EI/3.325) = 2.0451EI

Analysis of Case 4
Z₄ = 1.0; Z₁ = Z₂= Z₃= 0

K₁₄ = 0
K₂₄ = 0
K₃₄ = (6.8EI/3.325) = 2.045EI
K₄₄ = (4EI/3) + (4EI/4) + (13.6EI/3.325) = 6.4236EI

Stiffness coefficient due to externally applied load;

K_1P= -(q₁L₁²/12) = -(26.148 × 3.825²) / 12 = -31.880 kNm
K_2P= (q₁L₁²/12) – (q₂L₂²/12) = [(26.148 × 3.825²) / 12] – [(25.437 × 2.80²) / 12] = 15.261 kNm
K_3P= (q₂L₂²/12) – (q₃L₃²/12) = [(25.437 × 2.8²) / 12] – [(27.345 × 3.325²) / 12] = -8.5741 kNm

K_4P= (q₃L₃²/12) = (27.345 × 3.325²) / 12 = 25.193 kNm

The appropriate cannonical equation;

K₁₁Z₁ + K₁₂Z₂ + K₁₃Z₃ + K₁₄Z₄ + K_1P = 0
K₂₁Z₁ + K₂₂Z₂ + K₂₃Z₃ + K₂₄Z₄ + K_2P = 0
K₃₁Z₁ + K₃₂Z₂ + K₃₃Z₃ + K₃₄Z₄ + K_3P = 0
K₄₁Z₁ + K₄₂Z₂ + K₄₃Z₃ + K₄₄Z₄ + K_4P = 0

On substituting;

5.889Z₁ + 1.778Z₂     +    0Z₃    + 0Z₄ = 31.880
1.778Z₁ + 10.746Z₂   + 2.4286Z₃ + 0Z₄ = -15.261
0Z₁        + 2.4286Z₂   + 11.287Z₃ + 2.045Z₄ = 8.5741
0Z₁    + 0Z₂            + 2.045Z₃ + 6.4236Z₄ = -25.193

On solving;
Z₁ = 6.3102/EI (radians)
Z₂ = -2.9701/EI (radians)
Z₃ = 2.2384/EI (radians)
Z₄ = -4.6346/EI (radians)

Now that we have obtained the rotations, we can now substitute and obtain the moments at the critical points;

M_i = M₁Z₁ + M₂Z₂ + M₃Z₃ + M₄Z₄ + P

Bottom column support moments
M_A = (6.3102/EI) × (2EI/4) = 3.1551 kNm
M_B = (-2.9701/EI) × (2EI/4) = -1.485 kNm
M_C = (2.2384/EI) × (2EI/4) = 1.1192 kNm
M_D = (-4.6346/EI) × (2EI/4) = -2.3173 kNm

Beam Support Moments
M₁^R = [(6.3102/EI) × (13.6EI/3.825)] – [(2.9701/EI) × (6.8EI/3.825)] – 31.880 = -14.7239 kNm

M₂^L = [(6.3102/EI) × (6.8EI/3.825)] – [(2.9701/EI) × (13.6EI/3.825)] + 31.880 = 32.537 kNm

M₂^R = [-(2.9701/EI) × (13.6EI/2.8)] + [(2.23841/EI) × (6.8EI/2.8)] – 16.619 = -25.6090 kNm

M₃^L = [-(2.9701/EI) × (6.8EI/2.8)] + [(2.2384/EI) × (13.6EI/2.8)] + 16.619 = 20.278 kNm

M₄^R = [(2.2384/EI) × (13.6EI/3.325)] – [(4.6346/EI) × (6.8EI/3.325)] – 25.193 = -25.5157 kNm

M₃^L = [(2.2384/EI) × (6.8EI/3.325)] – [(4.6346/EI) × (13.6EI/3.325)] + 25.193 = 10.814 kNm

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