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How to Succeed as a Civil Engineering Student

By
Ubani Obinna Uzodimma
-

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.

civil engineering student

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;

IMG 20171006 122919


(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….

Screenshot 2016 06 08 21 48 34
Screenshot 2016 06 08 21 48 14

Be inspired for excellence today!!!

Thank you for visiting Structville.

Free Vibration of Water Tank Stand When Filled With Water

By
Ubani Obinna Uzodimma
-

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.

drt555

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

The mass matrix of the structure is given below;

 
Determination of the stiffness matrix
unit%2Bdispalcement

(12EIC)/L3 = (12 × 4897.2)/33 = 2176.533 KN/m
K11 = (12EIC)/h3 + (12EIC)/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;

equation%2Bfor%2Bmotion
The characteristic polynomial equation is given by;
characteristic%2Bpolynomial%2Bequation

On solving;

DEEEEEE

((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

DERRRR

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

MODE%2BI%2BVIBRATION
MODE%2B2%2BVIBRATION
 
The generalised mass and stiffness of the structure (modal mass and stiffness);
MODAL%2BMASS%2BAND%2BSTIFFNESS

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;

ϕ1T Mϕ1 = 1.0

The conversion factor;

bghh

Therefore, C1 = 1/√(9379.8) = 0.0103

Hence;

phi%2B1

Similarly;

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

Hence;

phi2

We can therefore obtain;

bhyy

Let χi(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;

bfrr

Substituting ü and u into equation 1 and multiplying by ΦT , we obtain;

ΦT MΦẍ + ΦT KΦx = 0
Which finally leads to;

ẍ + Kgx = 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)

Missing%2Bout

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;

Displacement%2Btime%2Bhistory%2Bgraph
Thank you for visiting Structville. God bless.

Analysis of Sub-Frames Using the Stiffness Method

By
Ubani Obinna Uzodimma
-

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 Zi = 1.0, applied at the fixed nodes.

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

CASE%2B1

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

CASE%2B2

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

CASE%2B3

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

CASE%2B4

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;

K1P = -(q1L12/12) = -(26.148 × 3.8252) / 12 = -31.880 kNm
K2P = (q1L12/12) – (q2L22/12) = [(26.148 × 3.8252) / 12] – [(25.437 × 2.802) / 12] = 15.261 kNm
K3P = (q2L22/12) – (q3L32/12) = [(25.437 × 2.82) / 12] – [(27.345 × 3.3252) / 12] = -8.5741 kNm

K4P = (q3L32/12) = (27.345 × 3.3252) / 12 = 25.193 kNm

The appropriate cannonical equation;

K11Z1 + K12Z2 + K13Z3 + K14Z4 + K1P = 0
K21Z1 + K22Z2 + K23Z3 + K24Z4 + K2P = 0
K31Z1 + K32Z2 + K33Z3 + K34Z4 + K3P = 0
K41Z1 + K42Z2 + K43Z3 + K44Z4 + K4P = 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;

Mi = M1Z1 + M2Z2 + M3Z3 + M4Z4 + P

Bottom column support moments
MA = (6.3102/EI) × (2EI/4) = 3.1551 kNm
MB = (-2.9701/EI) × (2EI/4) = -1.485 kNm
MC = (2.2384/EI) × (2EI/4) = 1.1192 kNm
MD = (-4.6346/EI) × (2EI/4) = -2.3173 kNm

Beam Support Moments
M1R = [(6.3102/EI) × (13.6EI/3.825)] – [(2.9701/EI) × (6.8EI/3.825)] – 31.880 = -14.7239 kNm

M2L = [(6.3102/EI) × (6.8EI/3.825)] – [(2.9701/EI) × (13.6EI/3.825)] + 31.880 = 32.537 kNm

M2R = [-(2.9701/EI) × (13.6EI/2.8)] + [(2.23841/EI) × (6.8EI/2.8)] – 16.619 = -25.6090 kNm

M3L = [-(2.9701/EI) × (6.8EI/2.8)] + [(2.2384/EI) × (13.6EI/2.8)] + 16.619 = 20.278 kNm

M4R = [(2.2384/EI) × (13.6EI/3.325)] – [(4.6346/EI) × (6.8EI/3.325)] – 25.193 = -25.5157 kNm

M3L = [(2.2384/EI) × (6.8EI/3.325)] – [(4.6346/EI) × (13.6EI/3.325)] + 25.193 = 10.814 kNm

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Design of Column Base Plate Connections (EC3)

By
Ubani Obinna Uzodimma
-

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

base%2Bplate%2Bprofile
Continue

Design of Roof Purlins

By
Ubani Obinna Uzodimma
-
(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).

roof purlin sections

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.

PURLIN%2BCONNECTION

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 Example of Roof Purlins

We are to provide a suitable cold formed channel section for the purlin of the roof arrangement shown below.

PURLINS

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.

C SECTION

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

Checking of geometrical proportions
b/t ≤ 60     b1/t = 48.5/1.45 = 33.448 < 60 – OK
c/t ≤ 50     c/t = 15/1.45 = 10.344 < 50 – OK
h/t ≤ 500   c/t = 120/1.45 = 82.758 < 500 – OK
0.2 ≤ c/b ≤ 0.6       c/b1 = 15/48.5 = 0.309 – OK

The effect of rounding corners due to root radius has been neglected in this design.

Gross section properties
Abr = t(2cp + bp1 + bp2 + hp) = 1.45[(2 × 14.25) + 48.5 + 48.5 + 118.5)] = 353.8 mm2

Position of the neutral axis with respect to the flange in compression;

ZB1

Zb1 = 1.45[14.25(118.5 – 7.125) + (48.5 × 118.5) + 118.52/2 + 14.252/2] / 353.8 = 59.253mm

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

dttyy

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;

Buckling%2Bfactor

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;

kk

Where K is the spring stiffness per unit length;

k

b1 is the distance of the web to the centre of the effective area of the stiffener in compression flange (upper flange);

b1

kf = 0 for bending about the y-y axis

Therefore;

kaa

= 175882.211 × 0.00000491305 = 0.8641 N/mm

Is
Isa

So the elastic critical buckling stress for the edge stiffener is;

σcr,s = (2√(0.8641 × 210000 × 1007.801))/(54.629 ) = 495.901 N/mm2

The relative slenderness factor for the edge stiffener;

sf

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.

Therefore;
χd = 0.862
be2 = 23.4255 mm
ceff = 14.25mm

hc
hca
ghyy

The buckling factor kσ = 7.81 – 6.29ψ + 9.78ψ2
kσ = 7.81 – 6.29(-0.9346) + 9.78(-0.9346)2 = 22.231

vfgg

Therefore, take ρ as 1.0 since 1.137 > 1.0.

Therefore, the effective width of the zone in compression of the web is;
heff = ρhc = 1.0 × 61.252 = 61.252mm

Near the flange in compression;
he1 = 0.4heff = 0.4(61.252) = 24.5mm
he2 = 0.6heff = 0.6(61.252) = 36.7512mm

The effective width of the web is;
Near the flange in compression;
h1 = he2 = 24.5mm

Near the flange in tension;
h2 = hp – (hc – he2) = 118.5 – (61.252 – 36.7512) = 93.992mm

Effective section properties;

Aeff

Aeff = 1.45 × [14.25 + 48.5 + 24.5 + 93.992 + 23.4255 + (23.4255 + 14.25) 0.862]
Aeff = 343.858 mm2

Position of the neutral axis with regard to the flange in compression;

zc

zc = 1.45[(14.25 × 111.375) + (48.5 × 118.5) + (93.992 × 71.504) + 300.125 + 87.5199] / 343.858 = 60.903 mm

Position of the neutral axis with regard to the flange in tension;
Zt = hp – Zc

Second moment of area:

Ieff%252Cy
ieffanse


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%2Bmodel%2B1

Static Model 2

static%2Bmodel%2B2

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

INTERNAL%2B1
SAVEE

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

hgty

Model 2

ert

 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

hyuuu

Model 2

cfrtt

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]

From our calculations;
Weff,y = min(Weff,y,c , Weff,y,t)  = 12767 mm3

Therefore;
MC,Rd = Weff,y.Fy/γm0 = [(12767 mm3) × (350) × 10-6] / 1.0 = 4.46845 KNm

Check of span and single support;
MEd/MC,Rd = 2.31/4.46845 = 0.5169 < 1.0 OK!

Check of shear resistance at ULS
 The design shear resistance is given by;

vbrd


Vb,Rd = (hw/sinφ.t.fbv) / γM0 (Clause 6.1.5)

The shear buckling strength (fbv) which is based on the relative web slenderness can be obtained from the table below (Table 6.1 of EN 1993-1-3).

Table

Where λw ̅    is the relative slenderness for webs without longitudinal stiffeners.

bnhy

Since 1.154  < 0.83 but less than 1.40;

fbv = 0.48fyb = 0.48 × 350 = 168 N/mm2

Vb,Rd = [(118.5 / sin90°) × 1.45 × 168] / 1.0 = 28866.6 N = 28.867 KN

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.

Analysis of Trusses Using Direct Stiffness Method: A Solved Example

By
Ubani Obinna Uzodimma
-

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.

Global%2BCoordinates

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;

Member%2BStiffness

Where;
l = cos θ;
m = sin θ;
L = Length of member

The global stiffness of each member is given below;

Member%2B1
member%2B2
Member%2B3
Member%2B4
Member%2B5
Member%2B6

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).

GENERAL%2BSTIFFNESS%2BMATRIX




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;

Internal%2BForces
This is then given below;
Solving%2Bfor%2Bdisplacement

On solving;

Displacements

In the same vein, this same relationship can help us compute the support reactions;

Support%2BReactions
REACTIONS

On solving;

Values%2Bof%2Bsupport%2BReactions

You can also realise that the obtained support reactions satisfy equilibrium requirements in the structure.

∑Fy = 0 ( i.e 12 + 13 = 25 KN)
∑Fx = 0 ( i.e 7.5662 + 2.4338 = 10 KN)

Internal Axial Forces
For computing the internal forces in the members, the following relationship is used;

[N] = [T][K’][u]

Which when expanded yields;

INTERNAL%2BSTRESSES

The axial forces in the members are now as given below;

N1
N2
N3
N4
N5
N6
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Stress Increment in Soil under a Pad Foundation

By
Ubani Obinna Uzodimma
-

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.

Boussinesq%2Bproblem%2Bquestion

The increment in the stress at any point below a pad foundation can be obtained by following the relationship below.

Boussinesq%2BEquation

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.

Division%2Bof%2Bpad%2Bfooting%2Binto%2Brectangles

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;

I4

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);

Table%2Bof%2BI4

Therefore, for the example under consideration;
The service pressure (q) = P/A = (1350 kN) / (1.5m × 1.5m) = 600 kN/m2

b = B/2 = 1.5/2 = 0.75m
m1 = L/B = 1.5m/1.5m = 1.0
n1 = z/b = 3m/0.75m = 4.0

Reading from the table above;
I4 = 0.108

∆σz = q I4
∆σz = 600 × 0.108 = 64.8 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.

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99.99% of people will be clueless about this puzzle….

By
Ubani Obinna Uzodimma
-

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….

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Design of Cantilever Retaining Wall Supporting Lateritic Earth Fill

By
Ubani Obinna Uzodimma
-

Cohesive soil profiles due to their nature are partially self supporting up to an extent called the critical depth (the critical depth is given by 2Zo, kindly see figure above). As a result, such soils would not exert much pressure on a retaining wall as a granular material would (especially considering active earth pressure). Tropical laterites (especially those found in Nigeria) usually possess angle of internal friction and some cohesion. Their usage for construction purposes is also widespread.

The walls retaining such soils are subjected to active and passive pressure. In this example, we are going to consider active pressure only (all passive pressure neglected), but note that passive pressure could be more critical and hence govern the design (I recommend you read standard geotechnical engineering textbooks for more knowledge on this subject).

Worked Example

The cantilever retaining wall shown below is backfilled with tropical lateritic earthfill, having a unit weight, ρ, of 18 kN/m3, a cohesion C of 8 kN/m3 and an internal angle of friction, φ, of 26°. The allowable bearing pressure of the soil is 150 kNm/2, the coefficient of friction is 0.5, and the unit weight of reinforced concrete is 24 kN/m3. Water level is at a great depth and adequate drainage is provided at the back of the wall.

Retaining%2Bwall%2Bproblem%2Bquestion

We are to;
1. Determine the factors of safety against sliding and overturning for the active pressure.
2. Calculate ground bearing pressures.
3. Design the wall and base reinforcement assuming fcu = 30 N/mm2, fy = 460 N/mm2 and the cover to reinforcement in the wall and base are, respectively, 40 mm and 50 mm.

Geotechnical Design

Wall Pressure Calculations

Retaining wall with lateritic earth fill

Coefficient of active pressure KA
Using Rankine’s theory
KA = (1 – sinφ) / (1 + sinφ);
KA = (1 – sin 26°) / (1 + sin 26°) = 0.39046;

Pressure at the top of wall = 2C√KA = 2 × 8 × √0.39046 = 9.998 kN/m2
Z0 = 2 C/γ√KA = (2 × 8)/(18 × √0.39046) = 1.423m

Pressure at the bottom of retaining wall = H.γ.KA – 2C√KA = (5.5 × 18 × 0.39046) – (2 × 8 × √0.39046) = 32.408 kN/m2

Earth%2BPressure%2BDiagram

Resistance to Sliding

Consider the forces acting on a 1 m length of wall. Horizontal force on wall due to backfill, PA, is;
PA = 0.5 × 32.408 × 4.077 = 66.063 kN/m

Weight of wall (Ww) = 0.35 × 5.1 × 24 = 42.84 kN/m
Weight of base (Wb) = 0.4 × 4 × 24 = 38.4 kN/m
Weight of soil (Ws) = 2.85 × 5.1 × 18 = 261.63 kN/m
Total vertical force (Wt) = 324.87 kN/m

Friction force, FF, is
FF = μWt = 0.5 × 324.87 = 171.435 kN/m

Neglecting all passive pressure force (FP) = 0.
Hence factor of safety against sliding is;
F.O.S = FF/PA = 171.435/66.063 = 2.595
The factor of safety 2.595 > 1.5. Therefore the wall is very safe from sliding.

Resistance to Overturning

Taking moments about the toe, sum of overturning moments (MO) is;
MO = (66.063 kN/m) × (4.077/3) = 89.779 kN.m

Sum of Restoring Moment MR;
MR = (Ww × 0.975m) + (Wb × 2m) + (Ws × 2.575m)
MR = (42.84 × 0.975m) + (38.4 × 2m) + (261.63 × 2.575m) = 792.266 kNm

F.O.S = MR / MO = 792.266/89.779 = 8.824
The factor of safety 8.824 > 2.0. Therefore the wall is very safe from overturning.

Bearing Capacity Check

Bending moment about the centre line of the base;
M = [66.064 × (4.077/3)] + (Ww × 1.025m) – (Ws × 1m)
M = [66.064 × (4.077/3)] + (42.84 × 1.025m) – (261.63 × 0.575m) = -16.745 kNm

Total vertical load = 324.87 kN

The eccentricity (e) = M/N = (16.745/324.87) = 0.0515m

Let us check D/6 = 4/6 = 0.666
Since e < D/6, there is no tension in the base

The maximum pressure in the base;
qmax = P/B (1 + 6e/B) = 324.87/4 [1 + (6 × 0.0515)/(4)] = 87.491 KN/m2
87.491 KN/m2 < 175 KN/m2

Therefore, the bearing capacity check is satisfied.

Structural Design

Design of the Wall

Conservatively taking the moment at the base of the wall due to the active force;
M = [66.064 × (4.077/3)] = 89.781 kN.m
At ultimate limit state M = 1.4 × 89.781 = 125.693 kN.m

Thickness of wall = 350mm

Effective depth = 350 – 40 – (8) = 302mm (assuming Y16mm bars will be employed for the construction)

k = M/(Fcubd2) = (125.693 × 106) / (30 × 1000 × 3022) = 0.0459
k < 0.156, no compression steel needed
la = 0.5 + (0.25 – k/0.9)0.5 = 0.5 + (0.25 – 0.0459/0.9)0.5 = 0.946
ASreq = M/(0.95Fy.la.d ) = (125.693 × 106) / (0.95 × 460 × 0.946 × 302) = 1006 mm2/m
ASmin = (0.13bh)/100 = (0.13 × 1000 × 350) / 100 = 455 mm2
Provide Y16 @ 150mm c/c (Asprov = 1340 mm2/m)

Check for shear
Design shear force on wall at ultimate limit state V = (1.4 × 66.064) = 92.4896 kN/m
Shear stress v = V/bd = (92.4896 × 1000) / (1000 × 302) = 0.3062 N/mm2

Vc = 0.632 (100As/bd)1/3 (400/d)1/4
Vc = 0.632 × [(100 × 1340)/(1000 × 302)]1/3 × (400/302)1/4
Vc = 0.632 × 0.7627 × 1.0727 = 0.517 N/mm2
For Fcu = 30N/mm2, Vc = 0.517 × (30/25)1/3 = 0.5494 N/mm2
Since v < Vc, no shear reinforcement required.

Design of the Base

The pressure distribution diagram on the base at serviceability limit state is as shown below;
qmax = P/B (1 + 6e/B) = 324.87/4 [1 + (6 × 0.0515)/(4)] = 87.491 kN/m2
qmin = P/B (1 – 6e/B) = 324.87/4 [1 – (6 × 0.0515)/(4)] = 74.927 kN/m2

Base%2BEarth%2BPressure

At ultimate limit state;
qmax = 1.4 × 87.491 = 122.487 KN/m2
qmin = 1.4 × 74.927 = 104.898 KN/m2

qC = 104.898 + [2.85 (122.487 – 104.898)]/4 = 117.430 kN/m2
qB = 104.898 + [3.2 (122.487 – 104.898)]/4 = 118.968 kN/m2

On investigating the maximum design moment about point C;

Back fill = 1.4 (18 × 5.1 × 2.85 × 2.85/2) = -521.952 kNm
Heel Slab = 1.4 (24 × 0.4 × 2.85 × 2.85/2) = -54.583 kNm
Earth Pressure = (104.898 × 2.85 × 2.85/2) + [(117.430 – 104.898) × 0.5 × 2.85 × 2.85/3] = 426.017 + 16.965 = 442.982 KNm
Net moment = -521.952 – 54.583 + 442.982 = -133.553 kNm

On investigating the maximum design moment about point B;

Back fill = (conservatively neglected)
Heel Slab = 1.4 (24 × 0.4 × 0.8 × 0.8/2) = – 4.3008 kNm
Earth Pressure = (118.968 × 0.8 × 0.8/2) + [(122.487 – 118.968) × 0.5 × 0.8 × (2 × 0.8/3)] = 37.9776 + 0.751 = 38.727 kNm
Net moment = -4.3008 + 38.727 = 34.426 kNm

Design of the toe
MB = 133.553 kN.m

Thickness of base = 400 mm

Effective depth = 400 – 50 – (8) = 342 mm (assuming Y16mm bars will be employed for the construction)

k = M/(Fcubd2) = (133.553 × 106) / (30 × 1000 × 3422) = 0.0385
k < 0.156, no compression steel needed
la = 0.5 + (0.25 – k/0.9)0.5 = 0.5 + (0.25 – 0.0385/0.9)0.5 = 0.95
ASreq = M/(0.95Fy.la.d ) = (133.553 × 106) / (0.95 × 460 × 0.95 × 342) = 941 mm2/m
ASmin = (0.13bh)/100 = (0.13 × 1000 × 400) / 100 = 520 mm2
Provide Y16 @ 175 mm c/c Top (Asprov = 1148  mm2/m)

Design of the heel
MC = 34.426 KNm

Thickness of base = 400 mm
Effective depth = 400 – 50 – (6) = 344 mm (assuming Y12mm bars will be employed for the construction)

k = M/(Fcubd2) = (34.426 × 106) / (30 × 1000 × 3442) = 0.00981
k < 0.156, no compression steel needed
la = 0.5 + (0.25 – k/0.9)0.5 = 0.5 + (0.25 – 0.00981/0.9)0.5 = 0.95
ASreq = M/(0.95Fy.la.d ) = (34.426 × 106) / (0.95 × 460 × 0.95 × 344) = 242 mm2/m
ASmin = (0.13bh)/100 = (0.13 × 1000 × 400) / 100 = 520 mm2
Provide Y12 @ 175mm c/c (Asprov = 646 mm2/m)

Check for shear at the base
The maximum shear force at ‘d’ from the face of the wall is investigated at the either side;

Maximum shear force at point B
Earth fill = 1.4 × 18 × 5.1 × 2.85 = -366.282 kN/m
Wall base = 1.4 × 24 × 0.4 × 2.85 = -38.304 kN/m
Base Earth Pressure = 0.5 (104.898 + 117.430) × 2.85 = 316.817 kN/m
Net shear force = -366.282 – 38.304 + 316.817 = -87.769 kN/m

A little investigation will show that this is the maximum shear force at any section on the base.

Taking the maximum shear stress at the base  v = (87.769 × 1000) / (1000 × 342) = 0.2566 N/mm2

Vc = 0.632 (100As/bd)1/3 (400/d)1/4
Vc = 0.632 × [(100 × 1148) / (1000 × 342)]1/3 × (400/342)1/4
Vc = 0.632 × 0.694 × 1.0399 = 0.456 N/mm2
For Fcu = 30N/mm2, Vc = 0.456 × (30/25)1/3 = 0.484 N/mm2
Since v < Vc, no shear reinforcement required.

DETAILING
I will need my readers to do the detailing sketches. I will accept both manual (hand) and CAD details. But your manual sketches must be very neat and well scaled. You should show the plans and at least one section. If your detailing is good enough, I will publish it on this post with your Facebook, LinkedIn, or G+ URL attached.

So, I will keep updating the post with as many good pictures as I receive. Good knowledge for all to flourish @ Structville

Send your sketches and URL profile to info@structville.com


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Analysis of Arch-Frame Structure with Tie Element: A solved Example

By
Ubani Obinna Uzodimma
-
An arch is supported on frame with different levels of supports as shown above. The frame is supported on a hinge at support A and a roller at support C. A tie beam is used to connect node B and support C. Plot the bending moment, shear force, and axial force diagram die to the externally applied load.

 
Solution

Support Reactions

∑MC = 0
30Ay – 10Ax  –  (10 × 302)/2 = 0
30Ay – 10Ax = 4500 ————- (1)

∑FX = 0
Ax + 25 = 0
Ax = -25 kN (pointing in the reverse of assumed direction)

Plugging the value of Ax into equation (1);
30Ay – 10(-25) = 4500
30Ay = 4250
Ay = 141.667 kN

∑MA = 0
30Cy  – (25 × 10) – (10 × 302)/2 = 0
30Cy = 4750
Cy = 158.333 kN

∑MDR = 0

15Cy –  (H × 5) – (10 × 152)/2 = 0
15(158.333) – 5H – (10 × 152)/2 = 0
– 5H + 1250 = 0
H = 250 kN

Arch Frame%2BSupport%2BReactions
Geometric Properties of the Arch Section

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

y = [4yc (Lx – x2)] / L2

Where yc is the height of the crown of the arch
y = [(4 × 5) × (30x – x2)]/302 = (2/3)x – (x2/45)

dy/dx = y’ = 2/3 – 2x/45

At x = 0; y = 0
y’ = 2/3 – 0/8 = 0.667
sin⁡θ = y’/√(1 + y’2) = 1/√(1 + 0.6672) = 0.5546
cos⁡θ = 1/√(1 + y’2) = 1/√(1 + 0.6672) = 0.8319

At x = 7.5m;
y = (2/3)x – (x2/45) = (0.667 × 7.5) – (7.52/45) = 3.75m
y’ = 2/3 – 2(7.5)/45 = 0.333
sin⁡θ = y’/√(1 + y’2) = 0.333/√(1 + 0.3332) = 0.43159
cos⁡θ = 1/√(1 + y’2) = 1/√(1 + 0.3332) = 0.9487

At x = 15 m;
y = (2/3)x – (x2/45) = (0.667 × 15) – (152/45) = 5.00m
y’ = 2/3 – 2(15)/45 = 0
sin⁡θ = y’/√(1 + y’2) = 0/√(1 + 02) = 0
cos⁡θ = 1/√(1 + y’2) = 1/√(1 + 02) = 1.0

At x = 22.5m;
y = (2/3)x – (x2/45) = (0.667 × 22.5) – (22.52/45) = 3.75m
y’ = 2/3 – 2(22.5)/45 = -0.333
sin⁡θ = y’/√(1 + y’2) = -0.333/√(1 + 0.3332) = -0.43159
cos⁡θ = 1/√(1 + y’2) = 1/√(1 + 0.3332) = 0.9487

At x = 30m; y = 0
y’ = 2/3 – 2(30)/45 = -0.667
sin⁡θ = y’/√(1 + y’2) = -0.667/√(1 + 0.6672) = -0.5546
cos⁡θ = 1/√(1 + y’2) = 1/√(1 + 0.6672) = 0.8319


Internal Stresses in the Arch Structure
Bending Moment
MA = 0
MBB = (25 × 10)  = 250 kN.m
MBUP = (25 × 10)  = 250 kN.m
M1 = (141.667 × 7.5) + (25 × 13.75) – (25 × 3.75) – (10 × 7.52)/2 – (250 × 3.75) = 93.7525 kN.m
MD = (141.667 × 15) + (25 × 15) – (25 × 5) – (10 × 152)/2 – (250 × 5) = 0

Coming from the right
M2 = (158.333 × 3.75) – (250 × 5) – (10 × 122)/2 = -31.252 kN.m
MC = 0



 Shear force

Section A – B (Column)
QA – Ax = 0
QA –  25 = 0
QA – QBB  = 25  kN

Qi = ∑V cos⁡θ – ∑H sin⁡θ (for arch section)

Note that the horizontal forces of 25 kN will eliminate each other;

QBUP =  (141.667 × 0.8319) – (250 × 0.5546) = -20.797 kN
Q1R = Q1L = [141.667  – (10 × 7.5)] × 0.9487  – (250 × 0.43159) = -44.651 kN
QD =  [141.667 – (10 × 15)] × 1.0 – (250 × 0) = -8.333 kN
Q2R = Q2L =  [141.667 – (10 × 22.5)] × 0.9487 – (250 × –0.43159) = 28.839 kN
QC = [141.667 – (10 × 30)] × 0.8319 – (250 × -0.5546) = 6.932 kN

Axial force
Member AB = 141.667 kN (compression)
Member BC = 250 kN (Tension)

For the arch section;
Ni = -∑V sin⁡θ – ∑H cos⁡θ
NBUP = -(141.667 × 0.5546) – (250 × 0.8319) = -286.543 KN (Compression)
N1L = -[141.667 – (10 × 7.5)] × 0.43159 – (250 × 0.9487) = -265.848 KN
ND = -[141.667 – (10 × 15)] × 0 – (250 × 1) = -250 KN (Compression)
N2L = N3R = -[141.667 – (10 × 22.5)] × -0.43159 – (250 × 0.9487) = -273.141 KN
NC = -[141.667 – (10 × 30)] × -0.5546 – (250 × 0.8319) = -295.786 KN

Internal Stresses Diagram

Bending%2BMoment%2BDiagram
shear%2Bforce%2Bdiagram
Axial%2BForce%2BDiagram%2BArch Frame

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