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, cmin, plus an allowance in design for deviation, Δcdev;
cnom = cmin + Δcdev ———- (1)
Minimum concrete cover, cmin, 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 cmin satisfying the above mentioned requirements shall be used. This is given by;
where;
cmin,b is the minimum cover due to bond requirement
cmin,dur is the minimum cover due to environmental conditions
Δcdur,γ is for additive safety element
Δcdur,st is the reduction of minimum cover for use of stainless steel
Δcdur,add is 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 cmin,b given in Table 4.2 of EN 1992-1-1:2004 and given in Table 2 below;
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.
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),
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
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….
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 IC = (0.23 × 0.233)/12 = 2.332 × 10-4 m4
EIC = (2.332 × 10-4 m4) × (2.1 × 107 KN/m2) = 4897.2 KN.m2
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.
We will now have to evaluate the basic system for different cases of a unit rotation Zi = 1.0, applied at the fixed nodes. Analysis of Case 1 Z1 = 1.0; Z2 = Z3 = Z4 = 0
This article is aimed at providing the procedure for the design of column base plates subjected to axial and shear forces according to Eurocode 3. Find a design example below.
Problem Statement
It is required to specify the appropriate thickness of a base plate to support a UC 203 x 203 x 60 subjected to the following loads. The connection is assumed to be pinned with four bolts outside the column profile. The supporting concrete is to have a grade fck = 25 N/mm2
(Image Copyright belongs to Albion Sections Limited, UK)
Roof purlins are members used to directly support roof sheeting materials, and could be made of timber or steel. In timber construction, purlins are nailed to the rafter or supporting trusses, while in steel roof construction, they are welded or bolted to the rafters or trusses by the means of cleats. As structural members, they resist loads, and provide lateral restraints for truss members, therefore it is important to design them properly against forces such as bending, shear, torsion, buckling etc.
In their design life, purlins are subjected to dead load (e.g self weight of sheeting materials and accessories), live load (e.g. during maintenance services and repairs), and environmental loads (e.g. wind and snow load). Therefore, a purlin should be adequately strong to withstand the loads it will encounter during its design life, and should not sag in an obvious manner thereby giving the roof sheeting an undulating and/or unpleasant appearance. This article will be focusing on design of steel purlin using cold formed sections.
Arrangement of Roof Purlins
By default, purlin sections assume the slope of the roof they are supporting. The spacing of roof purlins on rafters usually calls for careful arrangement, in the sense that it should follow the nodal pattern of the supporting trusses. By implication, purlins should ideally be placed at the nodes of trusses and not on the members themselves so as not to induce secondary bending and shear forces in the members of the truss. Furthermore, if manual analysis is employed to analyse a truss loaded in such a manner, such secondary stresses may not be captured especially if nominally pinned connections are assumed.
Cold formed Z (Zed) and C (channel) sections are normally specified for roof purlins in steel structures (see their form in image below).
When compared with thicker hot-rolled sections, cold-formed sections normally offer the advantages of lightness, high strength and stiffness, easy fabrication and installations, easy packaging and transportation etc. The connection of purlins can be sleeved or butted depending on the construction method adopted.
In terms of arrangement of roof purlins, we can have single spans with staggered sleeved/butt arrangement, single/double span with staggered sleeve arrangement, double span butt joint system, and single span butt joint system. The choice of the arrangement to be adopted can depend on the supply length of the sections as readily available in the market, the need to avoid wasteful offcuts, the loading and span of the roof, the arrangement of the rafters etc.
Therefore the roof designer must plan from start to finish. However, single and double span butt joint system are the most popular in Nigeria, due to their simplicity, and the culture of adopting shorter roof spans in the country. However, they are less structurally efficient than sleeved connections.
Design Exampleof Roof Purlins
We are to provide a suitable cold formed channel section for the purlin of the roof arrangement shown below.
Partial Factor for loads (BS EN 1990 NA 2.2.3.2 Table NA.A1.2(B)) Permanent action γG = 1.35 (unfavourable) Variable action γG = 1.5 Combination factor for Roofs, ψ0 = 0.7 Wind loads, ψ0 = 0.5 Reduction factor ξ = 0.925
Initial Sizing of Sections The maximum spacing of the trusses is at 3000 mm c/c Limiting the deflection to L/200; For continuous purlins, minimum depth of section (preliminary guide) = L/45 3000/45 = 66.667 Minimum width = 0.5L/60 = (3000/2)/60 = 25 Try C120-15 section (C – purlin) (Section picked from Albion Technical Manual, 2010).
Section Properties In the design of purlins using EN 1993-1-3:2006, we normally utilise the effective section properties. This calculation of effective section properties can be very tedious and prone to error, hence it is very advisable to obtain information from the manufacturer’s details or you can write a program using Microsoft Excel or MATLAB for such calculations. However, we are going to make a sample calculation for the section that we are considering.
Thickness (tnom) = 1.5 mm Depth = 120mm; Flange width = 50 mm; Lips/Edge Fold = 15 mm; Steel core thickness (t) = 1.5 – 0.05 = 1.45 mm (Note that EN 1993-1-3:2006 recommends a thickness of 0.04 mm for zinc coated sections, but we are using 0.05 mm here) Unit weight = 2.8 kg/m2; Web height hp = h – tnom = 120 – 1.5 = 118.5 mm Width of flange in tension = Width of flange in compression bp1 = bp2 = b – tnom = 50 – 1.5 = 48.5 mm Width of edge fold cp = C – tnom/2 = 15 – 1.5/2 = 14.25 mm
Effective section properties of the flange and lip in compression (clause 3.7.2) Effective width of the compressed flange; The stress ratio ψ = 1.0 (uniform compression) kσ = 4 for internal compression element (clause 3.7.2 Table 3.5)
ε = √(235/fyb) = √(235/350) = 0.819
The effective width is; beff = ρbb1 = 0.966 × 48.5 = 46.851 mm be1 = be2 = 0.5beff = 0.5 × 46.851 = 23.4255 mm
Effective Width of the edge fold (lip) Clause 3.7.3.2.2 Equation 3.47 The buckling factor is;
cp ⁄ bp1 = 14.25/48.5 = 0.2938 < 0.35 So kσ = 0.5
Therefore since 1.149 < 1.0, take reduction factor as 1.0
The effective width is therefore ceff = ρcp = 1.0 × 14.25 = 14.25mm
The effective area of the edge stiffener; As = t(be2 + ceff) = 1.45 (23.4255 + 14.25) = 54.629 mm2
We now have to use the initial effective cross-section of the stiffener to determine the reduction factor, allowing for the effects of the continuous spring restraint. The elastic critical buckling stress for the edge stiffener is;
Where K is the spring stiffness per unit length;
b1 is the distance of the web to the centre of the effective area of the stiffener in compression flange (upper flange);
kf = 0 for bending about the y-y axis
Therefore;
= 175882.211 × 0.00000491305 = 0.8641 N/mm
So the elastic critical buckling stress for the edge stiffener is;
The relative slenderness factor for the edge stiffener;
In our own case; χd = 1.47 – 0.723(0.840)= 0.862
As the reduction factor for buckling is less than 1.0, we can optionally iterate to refine the value of the reduction factor for buckling of the stiffeners according to clause 5.5.3.2(3). But we are not iterating in this post.
Position of the neutral axis with regard to the flange in tension; Zt = hp – Zc
Second moment of area:
Effective section modulus; With regard to the flange in compression; Weff,y,c = Ieff,y/zc = (777557.517) / 60.903 = 12767 mm3
With regard to the flange in tension; Weff,y,t = Ieff,y/zt = (777557.517) / 57.697 = 13476 mm3
LOAD ANALYSIS Permanent loads Employing long span aluminium roofing sheet (gauge thickness = 0.55mm) Load due to sheeting = 0.019 kN/m2 Other permanent accessories and fittings = 0.15 kN/m2 Total = 0.169 kN/m2
At a spacing of 1.2m, = 0.169 kN/m2 × 1.2m = 0.2028 kN/m Self weight of purlin = 2.8 kg/m = 0.0275 kN/m
Total Gk = 0.2028 KN/m + 0.0275 KN/m = 0.230 kN/m
Live load For a roof with 20° slope and no access except for normal repairs and maintenance, let us adopt a live load of 0.75 kN/m2 At a spacing of 1.2m, Qk = 0.75 kN/m2 × 1.2m = 0.9 kN/m
Wind Load Taking a dynamic wind pressure of 1.5 kN/m2 When the wind is blowing from right to left, the resultant pressure coefficient on a windward slope with positive internal pressure is; cpe = −0.90 upwards
Therefore the external wind pressure normal to the roof is; pe = qpcpe = 1.5 × − 0.90 = −1.35 kN/m2 The vertical component of the wind pressure is; pev = pecosθ = −1.35 × cos 20° = −1.268 kN/m2 acting upwards. At a spacing of 1.2m; Wk = −1.268 kN/m2 × 1.2m = 1.522 kN/m
STATIC SYSTEMS We are adopting two possible systems that will offer us continuous and single span systems. The 6m span is based on supply length.
Static Model 1
Static Model 2
Load Case 1 When Dead load and live load are acting alone; q = 1.35Gk + 1.5Qk = 1.35(0.230) + 1.5(0.9) = 1.6605 kN/m
Model 1
Load Case 2 When Dead load, live load and wind load are acting together; q = 1.35Gk + 1.5Qk + 0.9Wk Where the live load is the leading variable action q = 1.35(0.230) + 1.5(0.9) – 0.9 (1.522) = 0.2907 kN/m
Model 1
Model 2
Load Case 3 When Dead load and wind load are acting alone; q = 1.0Gk – 1.5Wk Where dead load is favourable q = 1.0(0.230) – 1.5(1.522) = -2.053 kN/m
Model 1
Model 2
Maximum span design moment MEd = 2.31 kNm Maximum shear force VEd = 3.85 kN
Verification of Bending Design moment resistance MC,Rd = Weff,y.Fy / γm0 [Clause 6.1.4.1(1) of EN 1993-1-3:2006]
Check for shear (using maximum shear force); VEd / Vb,Rd = (3.85)/(28.867) = 0.1333 < 1.0 Shear is ok
Deflection Check Maximum deflection under SLS (1.0gk + 1.0qk) = 0.07 mm Limiting deflection = L/200 = 3000/200 = 15 mm Since 0.07 < 15mm, deflection is Ok!
Therefore, the channel Z120-15 section is adequate for the applied load.
An indeterminate truss is supported and loaded as shown above, using the direct stiffness method, obtain the displacements, support reactions, and internal forces that are induced in the members due to the externally applied loads, (EA = Constant, dimensions in mm).
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Summary of Procedure
(1) Establish the x and y global coordinate system. The origin is usually located at the joint for which the coordinates for all the other joints are positive.
(2) Identify each joint and member numerically, and arbitrarily specify the near and far ends of each member symbolically by directing an arrow along the member with the head directed towards the far end.
(3) Specify the two code numbers at each joint, using the lowest numbers to identify unconstrained degrees of freedom, followed by the highest numbers to identify the constrained degrees of freedom.
For the truss that is loaded as shown above, there are four unconstrained degrees of freedom which occur at nodes ② and ③. The displacements are the horizontal and vertical translations which have been labelled from 1 – 4 and represented by the arrow heads at those nodes. The members have also been labelled accordingly with their numbers enclosed in a box (see figure above). Also notice that the unconstrained degrees of freedom have labelled first. The aim of this is to simplify the arrangement of the structure’s stiffness matrix.
Computation of member global stiffness matrix
Without much attention to the derivation, the stiffness matrix is given by;
[k] =[TT][k’][T] ————— (1)
Where;
K’ = member stiffness matrix which is of the same form as each member of the truss. It is made of the member stiffness influence coefficients, k’ij
T = Displacement transformation matrix
TT = This transforms local forces acting at the ends into global force components and it is referred to as force transformation matrix which is the transform of the displacement transformation matrix [T].
When the equation is solved, the global member stiffness matrix is obtained which is given by;
Where; l = cos θ; m = sin θ;
L = Length of member
The global stiffness of each member is given below;
The general stiffness matrix of the structure [KT] is given by;
[KT]= [K1] + [K2] + [K3] + [K4] + [K5] + [K6]
This now yields an 8 x 8 matrix which represents all the degrees of freedom in the truss both unconstrained (1-4) and constrained (5-8).
Since the unconstrained degrees of freedom are at points 1-4, we can therefore compute the deformation at such nodes using the relation below;
[P] = [K][u]
Where [P] is the vector of joint loads acting on the truss, [u] is the vector of joint displacement and [k] is the global stiffness matrix. On partitioning the above stiffness matrix, the relationship for this problem is as given below;
This is then given below;
On solving;
In the same vein, this same relationship can help us compute the support reactions;
On solving;
You can also realise that the obtained support reactions satisfy equilibrium requirements in the structure.
When foundations are constructed, they increase the net pressure in the soil. The knowledge of this stress increment is very important for estimating the settlement of foundations, and for the evaluation of lateral pressure on adjacent structures.
Boussinesq in 1883 solved the problem of stresses produced at any point in a homogeneous, elastic, and isotropic medium as the result of a point load applied on the surface of an infinitely large half-space. This has been extended into solving most of the problems of stresses that foundations impose on soils and for different types of loading, despite the fact that soils are practically neither homogeneous nor elastic.
In this example, it is required to obtain the stress that a soil mass experiences at a depth of 3m under the pad foundation shown below.
The increment in the stress at any point below a pad foundation can be obtained by following the relationship below.
For example, let us consider the plan of a pad footing shown below. The loaded area can be divided into four rectangles A, B, C, and D, as shown. It is desired to determine the stress at a point below point 1 at depth z. Note that this point is very common to all four rectangles. The increment in stress at a depth z below point 1 due to each rectangular area can be calculated using the equation above.
The total stress increase caused by the entire loaded area is given by equation (2) below;
∆σz = q [I3(A) + I3(B) + I3(C) + I3(D)] ———- (2)
At the centre of a pad footing, this is given by;
∆σz = q I4 ————– (3)
Where;
For the increment in stress at the centre of a rectangular footing, we can obtain the value of I4 from the Table below (derived from the relationship above);
Therefore, for the example under consideration; The service pressure (q) = P/A = (1350 kN) / (1.5m × 1.5m) = 600 kN/m2
Therefore, apart from the overburden pressure (geostatic stress), the stress at a depth of 3m under the pad foundation increased by 64.8 kN/m2 due to the foundation load.
The values given for the following triangles were obtained using a consistent relationship. I really do not expect a lot of people to be able figure it out….
