Structural Analysis and Design of Residential Buildings Using Staad.Pro, Orion, and Manual Calculations

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-

May 13, 2022

This should be a long post, but I am going to try and keep it as brief as possible. This post is more like an excerpt from the publication ‘Structural Analysis and Design of Residential Buildings using Staad Pro V8i, CSC Orion, and Manual calculations’. Here, we are going to briefly present some practical analysis and design of some reinforced concrete elements using Staad Pro software, Orion software, and manual calculations. Ultimately, we are going to make some comparisons of the results obtained based on the different methods adopted in the analysis and design. To learn how to model, design, and detail buildings from the scratch using Staad Pro, Orion, and manual methods, see the link at the end of this post.

To show how this is done, simplified architectural floor plans, elevations, and sections, for a residential two-storey building have been shown below for the purpose of structural analysis and design (see the pictures below).

The first step in the design of buildings is the preparation of the ‘general arrangement’, popularly called the G.A. The G.A. is a drawing that shows the disposition of the structural elements such as the slabs and their types, the floor beams, the columns, and their interaction at the floor level under consideration. For the architectural drawings above, the adopted G.A. is shown in Figure 7. below. There are no spelt out rules about how to prepare G.A. from architectural drawings, but there are basic guidelines that can guide someone on how to prepare a buildable and structurally efficient G.A.

New%2BPicture%2B%25284%2529 — Fig.7: General Arrangement

Design data:
F_ck = 25 N/mm², F_yk = 460 N/mm², C_nom (slabs) = 25mm, C_nom (beams and columns) = 35mm, C_nom (foundations) = 50mm
Thickness of slab = 150mm; Dimension of floor beams = 450mm x 230mm; Dimension of columns = (230 x 230mm)

DESIGN OF THE FLOOR SLABS
PANEL 1: MANUAL ANALYSIS

The floor slab (PANEL 1) is spanning in two directions since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is less than 2.
Hence, k = Ly/Lx = 3.825/3.625 = 1.055 (say 1.1)

Moment coefficients (α) for two adjacent edges discontinuous (pick from table);
Short Span
Mid-span = 0.042
Continuous edge = 0.056
Long Span
Mid-span = 0.034
Continuous edge = 0.045

Design of short span
Mid span
M = αnLx² = 0.042 × 10.9575 × 3.6252 = 6.0475 KN.m
M_Ed = 6.0475 KNm
Effective Depth (d) = h – Cc – ϕ/2
Assuming ϕ12mm bars will be employed for the construction
d = 150 – 25 – 6 = 119mm; b = 1000mm (designing per unit width)

k = M_Ed/(f_ckbd² )
= (6.0475 × 10⁶)/(25 × 1000 × 119² ) = 0.0171
Since k < 0.167 No compression reinforcement required
z = d[0.5+ √(0.25 – 0.882k)] = z = d[0.5+ √(0.25 – (0.882 × 0.0273)] = 0.95d
A_s1 = M_Ed/(0.87f_yk z)
A_s1= (6.0475 × 10⁶)/(0.87 × 460 × 0.95 × 119) = 133.668 mm²/m
Provide Y12mm @ 250mm c/c BOT (A_Sprov = 452 mm²/m)

A little consideration will show that this provided area of steel will satisfy serviceability limit state requirements. To see how to carry out deflections and crack control verifications, see the the link at the bottom of this post.

Result from Orion showing the Short Span (mid span) design moments (Wood and Armer effects inclusive) (PANEL 1)

Result from Staad showing the Short Span (mid span) design moments (Wood and Armer effects inclusive) (PANEL 1)

A little observation will show that the design moment values from the different methods are very similar. The full detailing of the floor slabs is as shown below.

Figure 9: Bottom Reinforcement Detailing

Figure 10: Top Reinforcement Detailing

Figure 11: Section of the floor slab

DESIGN OF THE BEAMS
Let us take Beam No 1 from our GA as a design case study:
The loading of the beam has been carried out as shown below. The beam is primarily subjected to load from the slab, the weight of wall, and its own self-weight. To see how to manually calculate the loading on beams, follow the link at the end of the post.

The internal forces from the loading is as shown below;

The internal forces from Orion software for Beam No 1 is as shown below. Load decomposition using finite element analysis was used for the load transfer.

The internal forces from Staad software for Beam No 1 is as shown below.

A_s1 = M_Ed/(0.87f_ykz)
= (36.66 × 10⁶)/(0.87 × 460 × 0.95 × 399) = 241.667 mm²
Provide 2Y16 mm BOT (A_Sprov = 402 mm²)
The detailing of Beam No 1 is as shown below;

DESIGN OF THE COLUMNS
Loads from slabs and beams are transferred to the foundations through the columns. In typical cases, columns are usually rectangular or circular in shape. Normally, they are usually classified as short or slender depending on their slenderness ratio, and this, in turn, influences their mode of failure. Columns are either subjected to axial, uniaxial, or biaxial loads depending on the location and/or loading condition. Eurocode 2 demands that we include the effects of imperfections in the structural design of columns. The design of columns is covered in section 5.8 of EC2.

The column axial loads have been obtained by summing up the reactions from all the beams supported by the columns, including the self weight of the column.

Let us use column A1 as example.

At the roof level, the column is supporting beam No 2 (Support Reaction V₁ = 13.27 KN) and Beam No 3 (Support Reaction V_A = 12.99 KN). At the first floor level (see Analysis and Design of Beam No 1 and 2), the column is supporting Beam No 1 (Support Reaction V₁ = 41.38 KN), and Beam No 2 (Support Reaction V_A = 42.49 KN). Therefore the summation of all these loads gives the axial load transferred from the beams. For intermediate supports, note that the summation of the shear forces at the support gives the total support reaction (neglect the signs and use absolute value. Another method of calculating Column Axial Load is by Tributary Area Method. This method has not been adopted in this work.

COLUMN A1
Total Columns Self weight = 12.14 KN
Load from roof beams = 13.27 + 12.99 = 26.26 KN
Load from floor beams = 46.21 + 42.49 = 88.70 KN
Total = 127.13 KN

Axial Load from Orion (A1) = 126.6 KN
Axial Load from Staad (A1) = 130.684 KN

COLUMN A3
Total Columns Self weight = 12.14 KN
Load from roof beams = 35.41 + 11.46 = 46.87 KN
Load from floor beams = 105.33 + 60.85 = 166.18 KN
Total = 225.19 KN

Axial Load from Orion (A3) = 202.3 KN
Axial Load from Staad (A3) = 201.632 KN

COLUMN A5
Total Columns Self weight = 12.14 KN
Load from roof beams = 17.19 + 5.70 = 22.89 KN
Load from floor beams = 83.64 + 37.91 = 121.55 KN
Total = 156.58 KN

Axial Load from Orion (A5) = 155.9 KN
Axial Load from Staad (A5) = 163.207 KN

COLUMN A7
Total Columns Self weight = 12.14 KN
Load from roof beams = 43.15 + 9.48 = 52.63 KN
Load from floor beams = 38.26 + 62.45 = 100.71 KN
Total = 165.48 KN

Axial Load from Orion (A7) = 133.9 KN
Axial Load from Staad (A7) = 140.392 KN

As you can see, for design purposes, the axial loads from the three methods are very comparable. To see how to obtain the column design moments from the use of sub-frames, follow the link at the end of the post.

Design of Column E5
Reading from chart; d₂/h = 0.2;
M_Ed/(f_ck bh² )
= (10.002 × 10⁶)/(25 × 230 × 230² ) = 0.03288
N_Ed/(f_ckbh)
= (399.88 × 10³)/(25 ×230 × 230) = 0.302
From the chart:
(A_sF_yk)/(bhf_ck ) = 0.05
Area of longitudinal steel required (As) = (0.05 × 25 × 230 × 230)/460 = 143.75 mm²
A_s,min = 0.10 NEd/fyd
= (0.1 × 399.887)/400 = 0.099 mm² < 0.002 × 230 × 230 = 105.8 mm²
Provide 4Y16mm (Asprov = 804 mm²)

Links
Minimum size = 0.25ϕ = 0.25 × 16 = 4mm < 6mm
We are adopting Y8mm as links
Spacing adopted = 200mm less than min{b, h, 20ϕ, 400mm}

Result from Orion for column E5

Result from Staad for column E5

Staad Provided Y8@225mm links
The column detailing is as shown below;

DESIGN OF FOUNDATIONS
All loads from the superstructure of a building are transferred to the ground. If the foundation of a building is poorly designed, then all the efforts input in designing the superstructure is in vain. It is therefore imperative that adequate care is taken in the design of foundations. Foundation design starts with a detailed field and soil investigation. It is very important to know the index and geotechnical properties of the soil, including the soil chemistry, so that the performance of the foundation can be guaranteed.

Analysis and Design of footing E8

Bearing Capacity of the foundation = 150 KN/m²;

Effective depth
Concrete cover = 50mm
AssumingY12mm bars,
d = 400 – 50 – 6 = 344mm
The ultimate limit state design moment can be obtained by considering the figure below;

k = M_Ed/(f_ck bd²)
= (37.518 × 10⁶)/(25 × 1000 × 344² ) = 0.01268 (designing per metre strip)
Since k < 0.167 No compression reinforcement required
z = d[0.5+ √(0.25 – 0.882k)] = z = d[0.5+ √(0.25 – (0.882 × 0.0273)] = 0.95d
A_s1 = M_Ed/(0.87f_ykz)
= (37.518 × 10⁶)/(0.87 × 460 × 0.95 × 344) = 286.869 mm²/m

To calculate the minimum area of steel required;
f_ctm = 0.3 × (f_ck)^(2⁄3) = 0.3 × 25 ^(2⁄3) = 2.5649 N/mm² (Table 3.1 EC2)
A_Smin = 0.26 × f_ctm/F_yk × b × d = 0.26 × 2.5649/460 ×1000 × 344 = 498.7 mm²
Check if A_Smin < 0.0013 × b × d (447.2 mm²)
Since, A_Smin = 498.7 mm², the provided reinforcement is adequate.
Provide Y12 @ 200mm c/c (A_Sprov = 565 mm²/m) each way

Shear at the column face
Ultimate Load on footing from column = 399.887 kN
Design shear stress at the column perimeter v_Ed = βV_Ed/(u₀d)
β is the eccentricity factor (see section 6.4.3 of EC2)
β = 1+ 1.8√[(16.48/230)²+(8.99/230)²] = 1.146
Where u_o is the column perimeter and d is the effective depth
v_Ed = βV_Ed/(u₀d)
= (1.15 × 399.887 × 10³)/(4(230) × 344) = 1.452N/mm²
V_Rd,max = 0.5vf_cd
v = 0.6[1 – (f_ck/250) ] = 0.6[1 – (25/250) ] = 0.54 N/mm²
f_cd = (α_cc f_ck)/γ_c = (0.85 × 25)/1.5 = 14.167 N/mm²
V_Rd,max = 0.5 × 0.54 × 14.167 = 3.825 N/mm²v_Ed < V_Rd,max. This is very ok

Transverse shear at ‘d’ from the face of column

Width of shaded area = a – d = 0.635 – 0.344 = 0.291m
Area of shaded area = (1.5m × 0.291m) = 0.4365 m²
Therefore, ΔV_Ed = (189.386 + 175.939)/2 × 0.4365 m² = 79.077 KN
v_Ed = V_Ed/bd = (79.077 × 10³)/(1500 × 344) = 0.15325 N/mm²
V_Rd,c = [C_Rd,c k (100ρ₁ f_ck )^(1/3) + k¹.σ^cp] × (2d/a) ≥ (V_min + k₁.σ,sub>cp) bw.d
C_Rd,c = 0.18/γ_c = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1+√(200/344) = 1.7624 > 2.0, therefore, k = 1.7624
V_min = 0.035k^(3/2) f_ck^(1/2)
V_min = 0.035 × (1.7624)^(3/2) × (25)^(1/2) = 0.4094 N/mm²
ρ₁ = As/bd = 565/(1000 × 344) = 0.001642 < 0.02;
V_Rd,c = [0.12 × 1.7624 (100 × 0.001642 ×25 )^(1/3) ] = 0.3386 N/mm² × (2d/a) < V_min
But in this case, d = a
Hence, V_Rd,c = 2 × 0.3386 = 0.6772 N/mm²
Since V_Rd,c (0.6772) > V_Ed (0.1503 KN), No shear reinforcement is required.

Punching Shear at 2d from the face of column
Punching shear lies outside the footing dimensions. No further check required.

Design Result from Orion

The detailing of the footing is as shown below;

The structural analysis and design of all members have been fully done, including a step by step tutorial on how to model and design on Orion and Staad Pro, and how to manually design. See the completed models below;

New%2BPicture%2B%252836%2529 — Fully completed model on Orion

New%2BPicture%2B%252837%2529 — Fully completed model on Staad Pro

To download the simplified e-book where all the members have been designed and completely detailed, including bar bending schedule and quantification of materials, click HERE.

Thank you, and God bless you.

ABOUT THE BOOK (Structural Analysis and Design of Residential Buildings)

By

Ubani Obinna Uzodimma

-

February 13, 2020

This book is primarily intended for practising civil engineers and civil engineering students in universities and polytechnics. The main aim of this publication is for it to serve as a simplified practical guide when carrying out design using Staad.Pro or Orion software; and also as a first hand exposure to the manual design of reinforced concrete residential buildings. On the other hand, for engineers who are already familiar with BS 8110-1:1997, but intend to switch or have a good idea of the Eurocodes, this book can serve as an initial exposure, before considering more technical textbooks or reference manuals. I strongly believe that with this simple publication by your side, you can carry out full analysis and design of reinforced concrete structures manually, or by using computer programs like Orion and Staad.

Chapter 1 of this publication talks about the problems and challenges of housing in Africa, the influx of many design software and programs in the market, the position of engineers, and the implications of such programs. Furthermore, the basis of structural design based on Eurocode 0 (EN 1990) was introduced; including the concept of limit state design. Finally, the building case study we intend to analyse and design was briefly introduced.

Chapter 2 talks about the materials that are employed in reinforced concrete design – steel and concrete. The major ingredients of concrete and their properties were briefly discussed for the purpose of adequately specifying concrete. The behaviour of fresh concrete and the mechanical properties of hardened concrete were also reviewed. For reinforcement, the mechanical properties, types, and available sizes were reviewed. Loading of buildings to Eurocode 1 (EN 1991) was reviewed, and the self weights of very common construction materials were also presented.

Chapter 3 talks about how to prepare the general arrangement (GA) of buildings from architectural drawings. Here also, our case study building was fully presented; including the floor plans, elevations, and GA. Furthermore, a step by step guide to modelling, analysis, and design on Orion and Staad Pro were also presented in a well detailed manner. The results of the analyses and design from both programs were also briefly presented.

Chapter 4 presented the reinforced concrete section analysis and design formulae in Eurocode 2 for flexural design, shear design, check of deflection and crack control.

In Chapter 5, the full manual analysis of the floor slabs were presented, including design of the staircase. Some references to the results from the computer programs were also highlighted.

The Chapter 6 of this book extensively handled the loading, analyses, and design of floor beams. As a matter of fact, all the floor beams in the building plan were manually analysed and designed. Also, the internal stresses obtained from computer aided design were compared with the results from manual analysis for the entire thirteen beams.

In Chapter 7, the full loading and analyses of the columns were carried out manually starting from the roof to the ground floor. Also, the processes of obtaining column design moments at ultimate limit state using sub-frames were presented. The column axial loads obtained from manual analysis were subsequently compared with the results from Staad Pro and Orion. At the end, the designs of the columns based on the recommendations of Eurocode 2 were carried out.

Chapter 8 talks about the design of foundations. The design processes recommended by Eurocode 7 (EN 1997) were presented, including all the checks necessary for adequate structural performance of the foundation. The result obtained from the manual analysis was compared with result from Orion Software.

In Chapter 9, all the detailing guides according to Eurocode 2 were presented. Subsequently, the practical detailing of the floor slabs, beams, columns, and foundations of the case study building were presented in full.

In Chapter 10, the bar bending schedule and quantification of all the detailed reinforcements were presented.

Chapter 11 talked about how to quantify materials and cost a building, using our case study building as an example. This detailed chapter talked about how to estimate the quantity of concrete needed to complete a certain phase of construction like foundations, slabs etc. Also, presented is how to obtain the quantity of cement required (in 50 kg bags), the quantity of sand and granite required (in metre cube, tonnes, and number of tipper trips). The chapter also talks about how to estimate the quantity of excavation and fill, and how to estimate excavation and filling cost. Furthermore, the process of calculating the number of blocks required for completing a project was presented. Also included in this section is how to estimate the quantity of cement and sand needed to mould blocks (based on the number of blocks to be moulded from one bag of cement). Also, guidance is given on how to estimate the quantity of formwork needed.

In Chapter 12, I discussed the discrepancies I discovered in manual analysis and computer aided analysis. This was followed by conclusion based on the findings from the work.

As you can see, this is a handbook that you can keep by your side that will enable you to carry out full design and presentation using any method of your choice. I called it a notebook because there are some details that were omitted that can only be found in more standard and specific textbooks. You can see that this book is not specifically devoted to any special topic, but to almost everything about a residential building. In the appendices, I presented how we can write MATLAB programs for flexural design, check of deflection, shear design, calculation of section properties, and analysis of sub-frames for analysis of column moments.

To purchase this book, contact;
E-mail: rankiesubani@gmail.com
WhatsApp: +2347053638996

Yours,
Ubani Obinna Uzodimma

Practical Analysis and Design of Steel Roof Trusses

By

Ubani Obinna Uzodimma

-

December 4, 2020

The most widespread alternative for roof construction in Nigeria is the use of trusses, of which timber and steel are the primary choices of materials. An advantage of using trusses for roofs is that ducts and pipes that are required for the operation of the building’s services can be installed through the truss web. However, careful attention must be paid to the design of the truss members and their connection (which may be welded or bolted) since their failure can be catastrophic both in terms of loss of life and economy.

The roof of a church building in Uyo, Akwa Ibom State, Nigeria collapsed on the 10th of December 2016 and left more than 60 worshippers dead, and many injured. This is to show how important, and why engineers must pay careful attention to such design situations. The aim of this article is to show in the clearest manner, how steel truss members can be designed according to EN 1993-1-1-:2005 (Eurocode 3) design code.

A truss is basically a system of triangulated members that are connected and designed to carry load. A truss is inherently stable in shape and resists load by developing primarily axial forces which might be tensile or compressive in nature. The connections in trusses are always assumed to be nominally pinned. When the connection of trusses is stiff, secondary effects such as bending moment and shear force are induced in the truss members.

For a good structural performance of roof trusses, the ratio of span to truss depth should be in the range of 10 to 15. However, it should be noted that the architectural design of the building determines its external geometry and governs the slope given to the top chord of the truss.

For the design of a compression member in a roof truss, several buckling modes need to be considered. In most truss members, only flexural buckling of the compressed members in the plane of the truss structure and out of the plane of the truss structure need to be evaluated. The flexural buckling in Eurocode 3 is achieved by applying a reduction factor to the compression resistance.

Example on the Design of Steel Roof Trusses

To illustrate this, a simple design example has been presented. The skeletal structure of a roof system (18.0m long and 7.2m wide) is as shown in the Figure below. The truss is made up of Howe Truss configuration spaced at 3m intervals. It is desired to specify the appropriate angle sections that will safely carry the anticipated loading using Eurocode design code (Specified steel grade S 275).

The idealised 2D model of the roof truss typical loading configuration is as shown below;

Load Analysis
Span of roof truss = 7.2m
Spacing of the truss = 3.0m
Nodal spacing of the trusses = 1.2m

Permanent (dead) Loads
Self-weight of long span aluminium roofing sheet (0.55mm gauge thickness) = 0.019 kN/m²
Weight of ceiling (adopt 10mm insulation fibre board) = 0.077 kN/m²
Weight of services = 0.1 kN/m²
Weight of purlin (assume CH 150 x 75 x 18 kg/m) = (18 x 3m)/(1.2 x 3) = 15 kg/m² = 0.147 kN/m²
Self weight of trusses (assume) = 0.2 kN/m²
Total deal load (g_k) = 0.536 kN/m²
Therefore the nodal permanent load (g_k) = 0.536 kN/m² × 1.2m × 3m = 1.9296 kN

Variable (Imposed) Load
Category of roof = Category H – Roof not accessible except for normal maintenance and repairs (Table 6.9 EN 1991-1-1:2001)
Imposed load on roof (q_k) = 0.75 kN/m²
Therefore the nodal variable load (Q_K) = 0.75 kN/m² × 1.2m × 3m = 2.7 kN

Wind Load
Wind velocity pressure (dynamic) is assumed as = q_p(z) = 1.5 kN/m²
When the wind is blowing from right to left, the resultant pressure coefficient on the windward and leeward slopes with positive internal pressure (c_pe) is taken as −0.9
Therefore the external wind pressure normal to the roof is;
q_e = q_pc_pe = −1.5 × 0.9 = 1.35 kN/m²
Vertical component p_ev = q_e cos θ = 1.35 × cos 36.869 = 1.08 kN/m² acting upwards ↑
Therefore the nodal wind load (W_k) = 1.08 kN/m² × 1.2m × 3m = 3.888 kN
To see how wind load is analysed using Eurocode, click HERE

Analysis of the Truss for Internal Forces
N/B: Please note that the internal forces in the members are denoted by F_i-j which is also equal to F_j-i e.g. F_2-3 = F_3-2 ; so kindly distinguish this from other numeric elements

Dead Load

JOINT 1

θ = tan^-1⁡(2.7/3.6) = 36.869
Let ∑F_y = 0
5.79 – 0.956 + F_1-2 (sin θ) = 0
F_1-2 = (-4.834)/sin⁡36.869 = -8.0568 kN (COMPRESSION)

Let ∑F_x = 0
F_1-2 (cos θ) + F_1-3 = 0
F_1-3 = -(-8.0568 ×(cos⁡36.869)) = 6.445 kN (TENSION)

JOINT 3

Let ∑F_y = 0
F_{3 – 2} = 0 (NO FORCE)

Let ∑F_x = 0
F_3–5 – F_3–1 = 0
F_3–1 = F_3–5 = 6.445 kN (TENSION)

JOINT 2

ϕ = tan^-1⁡(0.9/1.2) = 36.869 = θ
Let ∑F_y = 0
-1.93 + F_2-4(sin θ) – F_2–3 – F_2-5(sin θ) – F_2-1(sin θ) = 0
-1.93 + F_2-4(sin 36.869) – 0 – F_2-5(sin 36.869) – [-8.0568(sin 36.869)] = 0
0.6 F_2–4 – 0.6F_2–5 = -2.904 ———- (1)

Let ∑F_x = 0
F_2-4(cos θ) + F_2-5(cos θ) – F_2-1(cos θ) = 0
F_2-4(cos 36.869) + F_2-5(cos 36.869) – [-8.0568(cos36.869)] = 0
0.8 F_2–4 + 0.8 F_2–5 = -6.4455———- (2)

Solving equations (1) and (2) simultaneously;
F_2-4 = – 6.448 kN (COMPRESSION)
F_2-5 = -1.608 kN (COMPRESSION)

JOINT 5

Let ∑F_y = 0
F_5-2(sin ϕ) + F_5–4 = 0
-1.608 (sin 36.869) + F_5–4 = 0
F_5–4 = 0.9646 kN (TENSION)

Let ∑F_x = 0
–F_5–3 – F_5–2 (cos ϕ) + F_5–7 = 0
-6.445– [–1.608 (cos 36.869)] + F_5–7 = 0
F_5–7 = 5.1586 kN (TENSION)

JOINT 4

α = tan^-1(1.8/1.2) = 56.309°
Let ∑F_y = 0
-1.93 – F_4–2 (sin θ) – F_4–5 – F_4–7 (sin α) + F_4–6(sin θ) = 0
-1.93 – [-6.448(sin 36.869)] – 0.9646 – F_4-7 (sin 56.309) + F_4–6 (sin 36.869) = 0
-0.832 F_4–7 + 0.6 F_4–6 = -0.9742 ——– (3)

Let ∑F_x = 0
F_4–7 (cos α) + F_4-6(cos θ) – F_4-2(cos θ) = 0
F_4-7(cos 56.309) + F_4-6(cos 36.869) – (-6.448(cos36.869)) = 0
0.5547 F_4–7 + 0.8 F_4–6 = – 5.1584 ——– (4)

Solving equations (3) and (4) simultaneously;
F_4-7 = – 2.319 kN (COMPRESSION)
F_4-6 = -4.8398 kN (COMPRESSION)

JOINT 6

Let ∑F_x = 0
– F_4-6(cos θ) + F_6-8 (cos θ) = 0
-(-4.8398 cos 36.869) + F_6-8 (cos 36.869) = 0
F_6-8 = (-3.87184)/cos ⁡36.869 = – 4.8398 kN (COMPRESSION)

Let ∑F_y = 0
-1.93 – F_6–4 (sin θ) – F_6–7 – F_6–8 (sin θ) = 0
-1.93 – [-4.8398(sin 36.869)] – F_6–7 – [–4.8398(sin 36.869)] = 0
F_6–7 = 3.8777 kN (TENSION)

SUMMARY OF RESULTS FOR DEAD LOAD (G_K)

BOTTOM CHORD
F_1-3 = 6.445 kN (T)
F_3-5 = 6.445 kN (T)
F_5-7 = 5.158 kN (T)

TOP CHORD
F_1-2 = -8.0568 kN (C)
F_2-4 = -6.448 kN (C)
F_4-6 = -4.839 kN (C)

VERTICALS
F_2-3 = 0 (NO FORCE)
F_4-5 = 0.9646 kN (T)
F_6-7 = 3.877 kN (T)

DIAGONALS
F_2-5 = -1.608 kN (C)
F_4-7 = – 2.319 kN (C)

Similarly, the summary of analysis results for imposed load (Q_k) is given below;

BOTTOM CHORD
F_{1 – 3} = 8.992 kN (T)
F_{3 – 5} = 8.992 kN (T)
F_{5 – 7} = 7.198 kN (T)

TOP CHORD
F_{1 – 2} = -11.241 kN (C)
F_{2 – 4} = – 8.998 kN (C)
F_{4 – 6} = -6.748 kN (C)

VERTICALS
F_{2 – 3} = 0 (NO FORCE)
F_{4 – 5} = 1.346 kN (T)
F_{6 – 7} = 5.391 kN (T)

DIAGONALS
F_{2 – 5} = -2.242 kN (C)
F_{4 – 7} = – 3.238 kN (T)

Summary of Results for Wind Load (W_k)

BOTTOM CHORD
F_{1 – 3} = -12.948 kN (C)
F_{3 – 5} = -12.948 kN (C)
F_{5 – 7} = -10.365 kN (C)

TOP CHORD
F_{1 – 2} = 16.187 kN (T)
F_{2 – 4} = 12.957 kN (T)
F_{4 – 6} = 9.717 kN (T)

VERTICALS
F_{2 – 3} = 0 (NO FORCE)
F_{4 – 5} = – 1.938 kN (C)
F_{6 – 7} = -7.763 kN (C)

DIAGONALS
F_{2 – 5} = 3.228 kN (T)
F_{4 – 7} = 4.662 kN (T)

In all cases, (T) – Tensile force; (C) – Compressive force

Structural Design of Roof Trusses to Eurocode 3

All structural steel employed in the design has the following properties;
f_y (Yield strength) = 275 N/mm²
f_u (ultimate tensile strength = 430 N/mm²)

Design of the bottom chord (considering maximum effects)

LOAD CASE 1: DEAD LOAD + IMPOSED LOAD only
Fu = γ_GjGk + γ_QkQk

Ultimate design force (N_Ed) = 1.35G_k + 1.5Q_k
N_Ed = 1.35(6.445) + 1.5(8.992) = 22.189 kN (TENSILE)

LOAD CASE 2: DEAD LOAD + WIND LOAD acting simultaneously

Partial factor for permanent actions (DK) = γGj = 1.0 (favourable)
Partial factor for leading variable actions (W_k) = γW_k = 1.5

Therefore ultimate design force in the member = Fu = γ_GjGk + γ_WkWk = G_k + 1.5W_k.
N_Ed = 1.0(6.445) – 1.5(12.894) = -12.896 kN (COMPRESSIVE)

Therefore, all bottom chord members should be able to resist an axial tensile load of 22.189 kN and a possible reversal of stresses with a compressive load of 12.896 kN

Length of longest bottom chord member = 1200mm

Consider EQUAL ANGLES UA 50 X 50 X 6
Gross Area = 5.69 cm²
Radius of gyration (axis y-y) r_i = 1.5 cm
Considering one M12 bolt (14mm diameter allowance) – Equivalent tension area = 3.72 cm²
Equivalent tension area for welded connection = 4.88cm²

N_t,Rd is the lesser of;
(A_net × Fy)/γ_M0 and (0.9A_net × fu)/γ_M2
f_u = 430 N/mm²; f_y = 275 N/mm²
N_t,Rd = (3.72 × 10² × 275)/1.0 × 10^-3) = 102.3 kN
Also check; (0.9 × 3.72 × 10² × 430)/1.25 × 10^-3 = 115.17 kN

Therefore;

N_Sd/N_t,Rd = 22.189/102.3 = 0.216 < 1.0 (Section is ok for tension resistance)

Compression and buckling resistance
Thickness of section t = 6 mm. Since t < 16mm, Design yield strength f_y = 275 N/mm² (Table 3.1 EC3)

Section classification
ε = √(235/f_y) = √(235/275) = 0.9244
h/t = 50/6 = 8.33.
Referring to Table 5.2 (sheet 3) of Eurocode 3, Part 1-1, for class 3 classification,
h/t ≤ 15ε and (h + b)/2t ≤ 11.5ε. In our case,
5ε = 15 × 0.92 = 13.8 > h/t (8.3) OK
(h + b)/2t = 8.33 < 10.8 (11.5 × 0.92) OK

Thus, the section satisfies both of the conditions.

Resistance of the member to uniform compression
N_C,Rd = (A × Fy)/γ_M0 = (5.69 × 10² × 275)/1.0 = 156475 N = 156.475 kN
N_Ed/N_C,Rd = 12.896/156.475 = 0.0824 < 1 Therefore section is ok for uniform compression.

Buckling resistance of the member
Since the member is pinned at both ends, critical buckling length is the same for all axis; L_cr = 1200mm
Slenderness ratio λ = L_cr/(r_i × λ₁)
λ₁= 93.9ε = 93.9 × 0.9244 = 86.801
In the planar axis (z-z and y-y)
λ = 1200/(15 × 86.801) = 0.9216

Buckling curve b is appropriate for all angle sections according to Table 6.2 of Eurocode 3
α = 0.34 for buckling curve b
Φ = 0.5 [1 + α(λ – 0.2) + λ²]
Φ = 0.5 [1 + 0.34(0.9216 – 0.2)+ 0.9216²] = 1.0473

X = 1/[Φ + √(Φ² – λ²)]
X = 1/[1.0473 + √(1.047² – 0.9216²)] = 0.6473 < 1

Therefore N_b,Rd = (X × A × f_y)/γ_m1 = (0.6473 × 5.69 × 10²× 275)/1.0 = 101286.2675 N = 101.286 kN
N_Ed/N_b,Rd = 12.869/101.286 = 0.127 < 1 Therefore the section is ok for buckling

Therefore, UA 50 x 50 x 6 is ok to resist all axial loads on the bottom chord of the truss.
Following the method shown above in section 4.0, other members of the truss can be efficiently designed.

Thank you for visiting
TO DOWNLOAD THE FULL DESIGN PAPER IN PRINTABLE PDF FORMAT, CLICK HERE

Analysis and Design of a Network of Interacting Primary and Secondary Beams

By

Ubani Obinna Uzodimma

-

July 29, 2022

Table of Contents

In some construction cases, it is desirable to have large uninterrupted floor areas, and in such cases, the presence of internal columns has to be minimal. This can be achieved to some extent by using a network of interacting reinforced primary and secondary beams.

This feature is desirable in buildings like conference halls, auditoriums, stadiums, churches, dance halls, and all buildings where there is a need for a performing stage, and spectators. The structural implication of such features is usually the presence of long spans elements, and a lot of solutions already exist for such constructions.

It is already a well-known fact that the use of conventional reinforced concrete beams becomes more uneconomical as the construction span increases. This is primarily because the predominant in beams are bending moment and shearing forces, which are all functions of the length of the beam.

To accommodate these internal forces during design usually calls for an increase in the size of the member sections to satisfy ultimate and serviceability limit state requirements. This further adds to the total dead load of the structure, which is conventionally undesirable for economical reasons. A better solution for handling the problem of long-span construction is the use of structural forms like trusses and arches.

Trusses are arrangements of straight members connected at their ends. The members are arranged to form a triangulated system to make them geometrically unchangeable, and hence will not form a mechanism. They resist loads by developing primarily axial forces in their members especially if the ends of the members are pinned together. In typical trusses, loads are applied only at the joints.

Trusses provide practical and economical solutions to engineering problems, and can efficiently span greater lengths than beams due to the development of predominantly axial forces in the members. Trusses can be found on the roof of buildings, bridges etc. The picture above shows a model of a truss bridge.

Arches are also widely used in modern engineering due to their ability to cover large spans and their attractiveness from an aesthetic point of view. The greater the span, the more an arch becomes more economical than a truss. Materials of the modern arches are concrete, steel, and timber. Arches are mainly classified as three-hinged, two-hinged, and arch with fixed supports.

Arches carry most of their loads by developing compressive stresses within the arch itself and therefore in the past were frequently constructed using materials of high compressive strength and low tensile strength such as stones and masonry. They may be constructed in a variety of geometries; semi-circular, parabolic or even linear where the members comprising the arch are straight. The picture below shows the New River Gorge Arch steel bridge near Fayetteville, West Virginia USA.

Network of Primary and Secondary Beams

However, when it is desirable to have a relatively large hall devoid of internal columns, a network of interacting reinforced primary and secondary beams can be employed as alternatives to other solutions. A little consideration will show that in such cases, it is possible that the cost of adopting such a method of construction can be cheaper than that of trusses or arches.

No serious expertise is required on the part of the contractor when constructing a system of primary and secondary beams since very familiar construction processes are adopted. The logistics and expertise associated with assembling trusses and arches can offset the cost of larger concrete sections and reinforcing bars. Therefore, the knowledge of how this analysis and design can be carried out is necessary for the engineering community.

In the simplest term, a network of primary and secondary beams involves supporting a beam (called a secondary beam) on another beam (called a primary beam) instead of a column or wall. The position of support can be at the ends or at any intermediate location of the secondary beam. By implication, it is very usual for the secondary beam to be shallower than the primary beam (but sometimes this may not the case).

The choice of selecting the axis of the primary beam usually depends on the length. The beam with the shorter span is preferably the primary beam, so as to easily control strength and deflection. The analysis often involves loading and analysing the secondary beams first, and then the support reactions are transferred to the primary beam as concentrated loads. A primary beam is often supported on a column or wall.

In the analysis of a secondary beam, it should be borne in mind that the supports of the beam are not entirely rigid, since the primary beams supporting the secondary beams will undergo some deflection. This can be taken into account by representing the supports of secondary beams with springs, whose stiffness is equal to the stiffness of the primary beams.

Design Example

In the example below, a hall of 12m x 20m with the general arrangement shown below has been analysed and designed. In the example below, we are going to analyse only the internal beams, and we are going to consider all the spans to be fully loaded at ultimate limit state.

From the figure below, the secondary beams are shown in red (dotted lines) while the primary beams are shown in green. Every beam under consideration is supported by columns at the first and last supports. No internal column exists anywhere in the hall, and it is also assumed that the internal beams are not carrying block work loads, but light moveable partitions (drywalls).

With the general arrangement shown above, let us attempt to design some of the secondary and primary beams.

Design Data
Thickness of slab = 150 mm
All secondary beams = 450mm x 230mm
All primary beams = 900 mm x 400mm
Density of concrete = 25 kN/m³
Design compressive strength of concrete = 35 N/mm²
Yield strength of all reinforcements = 460 N/mm²
Concrete cover to slab = 25mm
Concrete cover to beams = 30mm
ULS Combination = 1.35g_k + 1.5q_k
SLS Combination = 1.0g_k + 1.0q_k
Imposed load = 5 kN/m² (NA to BS EN 1991-1-1:2002)
Building Category = Category C4

Structural Design of the secondary beams

Flexural design of span 1-2 and span 5-6
M_Ed = 53.95 kNm
Effective depth (d) = h – C_nom – ϕ/2 – ϕ_links
Assuming ϕ16 mm bars will be employed for the main bars, and ϕ8mm bars for the stirrups (links)
d = 450 – 30 – 8 – 8 = 404 mm

Effective flange width of the beams (T-beams)
b_eff = b_w + b’
Where b’ = 0.2(a_w + l_o) ≤ 0.4l_o ≤ 1.0a_w (for T-beams)
b_w = web width;
a_w = Clear distance between the webs of adjacent beams = 3000 – 230 = 2770 mm
l_o = The distance between points of zero moment on the beam = 0.85L = 0.85 × 4000 = 3400mm
Therefore in this case, b’ = 0.2(2770 + 3400) = 1234 mm ≤ 0.4l_o
Therefore b_eff = 230 + 1234 = 1464mm

k = M_Ed/(f_ckb_effd²) = (53.95 × 10⁶)/(35 × 1464 × 404²) = 0.00645
Since k < 0.167, no compression reinforcement required
z = d[0.5 + √(0.25 – 0.882K)] = z = d[0.5+ √(0.25 – 0.882(0.00645))] = 0.95d
A_s1 = M_Ed/(0.87f_yk z) = (53.95 × 10⁶)/(0.87 × 460 × 0.95 × 404) = 351.244 mm²
Provide 2X16 mm BOT (AS_prov = 402 mm²)

To calculate the minimum area of steel required;
f_ctm = 0.3 × f_ck^(2⁄3) = 0.3 × 35^(2⁄3) = 3.20996 N/mm² (Table 3.1 EC2)
A_Smin = 0.26 × f_ctm/f_yk × b × d = 0.26 × (3.20996/460) × 230 × 404 = 168.587 mm²
Check if AS_min < 0.0013 × b × d (120.796 mm²)
Since, AS_min = 168.587 mm², the provided reinforcement is adequate.
Deflection checks were found to be satisfactory.

Flexural Design of Support 2
M_Ed = 72.89 kNm
k = M_Ed/(f_ckb_wd²) = (72.89 × 10⁶)/(35 × 230 × 404²) = 0.055
Since k < 0.167 No compression reinforcement required
z = d[0.5+ √(0.25 – 0.882K) ] = z = d[0.5+ √(0.25 – 0.882(0.055)) ] = 0.949d

A_s1 = M_Ed/(0.87f_yk z) = (72.89 × 10⁶)/(0.87 × 460 × 0.949 × 404) = 475 mm²
Provide 3X16mm TOP (A_Sprov = 603 mm²)

Shear Design Support 1
Ultimate shear force V_Ed = 68.33 kN
V_Rd,c = [C_Rd,c.k.(100ρ₁ f_ck)^(1/3) + k₁.σ_cp]b_w.d ≥ (V_min + k₁.σ_cp) b_w.d

Where;
C_Rd,c = 0.18/γ_c = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/404) = 1.704 > 2.0, therefore, k = 1.702
V_min = 0.035k^(3/2) f_ck^(1/2) = V_min = 0.035 × 1.702^(3/2) × 35^(1/2) = 0.4598 N/mm²
ρ₁ = As/bd = 402/(230 × 404) = 0.004326 < 0.02; K₁ = 0.15

σ_cp = N_Ed/Ac < 0.2f_cd (Where N_Ed is the axial force at the section, Ac = cross sectional area of the concrete), f_cd = design compressive strength of the concrete.)
Take N_Ed = 0

V_Rd,c = [0.12 × 1.703(100 × 0.004326 × 35 )^(1/3)] 230 × 404 = 46977.505 N = 46.977505 KN
Since V_Rd,c (46.977505 KN) < V_Ed (68.33 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 = [(230 × 0.9 × 404 × 0.516 × 23.333)/(2.5 + 0.4)] × 10^-3 = 347.195 kN
Since V_Rd,c < V_Ed < V_Rd,max
Hence, A_sw/S = V_Ed/(0.87F_ykzcot θ) = 68330/(0.87 × 460 × 0.9 × 404 × 2.5) = 0.1878

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)/460 = 0.0010289
A_sw/S (min) = 0.0010289 × 230 × 1 = 0.2366
Since 0.2366 > 0.16809, adopt 0.2366
Maximum spacing of shear links = 0.75d = 0.75 × 404 = 303mm
Provide X8mm @ 275mm c/c (A_sw/S = 0.36556) Ok
The detailing sketches of the sections designed are shown below;

Loading, Analysis, and Design of Primary Beams

We can observe from the general arrangement of the structure that the primary beams are parallel to the short span direction of the slab. Therefore, the equivalent load that is transferred from the slab to the beam can be represented by;

p = nL_x/3 = (16.3425 × 3)/3 = 16.3425 kN/m

Since the beams are receiving loads from both sides, we can multiply by two (to account for the slab loads at the adjacent sides of the beam);

Hence p = 16.3425 × 2 = 32.685 kN/m

Self weight of the beam (ULS) = 1.35 × 0.9m × 0.4m × 25 kN/m³ = 12.15 kN/m
Therefore total uniformly distributed load on the primary beams = 32.685 + 12.15 = 44.835 kN/m

For beams on grid lines 2 and 5, the total load transferred from the secondary beams are the summation of the shear forces on supports 2 and 5 of the secondary beam. This is given by;

P = 104.78 kN + 91.11 kN = 195.89 kN

Structural Design
M_Ed = 1982.37 kNm
Effective depth (d) = h – C_nom– ϕ/2 – ϕ_links
Assuming ϕ32 mm bars will be employed for the main bars, and ϕ10mm bars for the stirrups (links)

d = 900 – 30 – 16 – 10 = 844 mm;

Effective flange width of the beam (T-beams)
b_eff = b_w + b’
Where b’ = 0.2(a_w + l_o) ≤ 0.4l_o ≤ 1.0a_w (for T-beams)
b_w = web width;
a_w = Clear distance between the webs of adjacent beams = 4000 – 315 = 3685 mm
l_o = The distance between points of zero moment on the beam (simply supported beam) = L = 12000 mm
Therefore in this case, b’ = 0.2(3685 + 12000) = 3137 mm ≤ 0.4l_o
Therefore b_eff = 400 + 3137 = 3537 mm
k = M_Ed/(f_ckb_wd²) = (1982.37 × 10⁶)/(35 × 3537 × 844²) = 0.0224
Since k < 0.167, no compression reinforcement required
z = d[0.5+ √(0.25 – 0.882K)] = z = d[0.5+ √(0.25 – 0.882(0.0224) ] = 0.95d
A_s1 = M_Ed/(0.87f_yk z) = (1982.37 × 10⁶)/(0.87 × 460 × 0.95 × 844) = 6177.91 mm²
Provide 8X32mm BOT (A_Sprov = 6432 mm²)

The minimum area of steel required;
f_ctm = 0.3 × f_ck^(2⁄3) = 0.3 × 35^(2⁄3) = 3.20996 N/mm² (Table 3.1 EC2)
AS_min = 0.26 × f_ctm/F_yk × b × d = 0.26 × 3.20996/460 × 400 × 844 = 612.458 mm²
Check if A_Smin < 0.0013 × b × d (438.88 mm₂)
Since, A_Smin= 612.458 mm², the provided reinforcement is adequate.

Shear Design
Support A
Ultimate shear force V_Ed = 562.84 kN

V_Rd,c = [C_Rd,c.k.(100ρ₁ f_ck)^(1/3) + k₁.σ_cp]b_w.d ≥ (V_min + k₁.σ_cp) b_w.d
Where;
C_Rd,c = 0.18/γ_c = 0.18/1.5 = 0.12
k = 1+√(200/d) = 1+√(200/844) = 1.486 > 2.0, therefore, k = 1.486 V_min = 0.035k^(3/2) f_ck^(1/2) = V_min = 0.035 × 1.486^(3/2) × 35^(1/2) = 0.375 N/mm₂
ρ₁ = As/bd = 3216/(400 × 844) = 0.009526 < 0.02; (Assuming only 4X32mm bars will be fully anchored at the supports. This is on the safer side for shear design)
σ_cp = 0

V_Rd,c = [0.12 × 1.486 (100 × 0.009526 × 35 )^(1/3)] 400 × 844 = 193759.0667 N = 193.759 kN
Since V_Rd,c (193.759 kN) < V_Ed (562.84 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 = [(400 × 0.9 × 844 × 0.516 × 19.8333)/(2.5 + 0.4)] × 10^-3 = 1072.239 KN
Since V_Rd,c < V_Ed < V_Rd,max

Hence, A_sw/S = V_Ed/(0.87F_ykzcot θ) = = 562840/(0.87 × 460 × 0.9 × 844 × 2.5 ) = 0.7405

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)/460 = 0.0010289

A_sw/S (min) = 0.0010289 × 400 × 1 = 0.41156
Since 0.7405 > 0.41156, adopt 0.7405
Maximum spacing of shear links = 0.75d = 0.75 × 844 = 633mm
Provide 2 Legs of X10mm @ 200mm c/c (A_sw/S = 0.785) Ok

Reinforcement Details of Primary Beams

According to clause 9.2.5 of EC2, where a beam is supported by a beam instead of a wall or column (primary and secondary beam interaction), reinforcement should be provided and designed to resist the mutual reaction. This reinforcement is in addition to that required for other reasons. The supporting reinforcement between two beams should consist of links surrounding the principal reinforcement of the supporting member. Some of these links may be distributed outside the volume of the concrete, which is common to the two beams.

In this case, additional links of X10@100mm c/c have been distributed at a length of 900mm on the primary beam, within the interaction zone (see detailing sketches below.

To download the full design paper in printable PDF format, click HERE.

Analysis of Statically Indeterminate Frames using Force and Displacement Methods

By

Ubani Obinna Uzodimma

-

February 17, 2021

Structural analysis is an important aspect of any structural design. It is simply the process of determining the effect of direct and indirect actions on structures. The effects of actions usually obtained from structural analysis are the displacements, bending moment, shear forces, axial forces, torsion, etc. Linear first-order analysis of statically indeterminate frames can be carried out using the force method, displacement methods, or other approximate solutions.

It is a well-known fact that there are two basic methods of analysing statically indeterminate structures which are;

Flexibility methods (also known as force methods, compatibility methods, or the method of consistent deformations), and
Displacement methods (also known as stiffness or equilibrium method)

Each method involves the combination of a particular solution which is obtained by making the structure statically determinate and a complementary solution in which the effect of each individual modification is assessed. In the force methods, the behaviour of the structure is considered in terms of unknown forces, while in the stiffness method, the behaviour of the structure is considered in terms of unknown displacements.

By implication, both analysis methods always involve reducing the structure to a basic system (a determinate system). In the force method, the basic system involves the removal of redundant forces, while the stiffness method involves restraining the joints of the structure against displacement.

analysis of statically indeterminate frames

In the solved example downloadable from this post, the frame shown above has been analysed for bending moment due to the externally applied load using force method and displacement method. In the force method, the graphical method was employed (Vereschagin’s rule of diagram multiplication).

To get yourself familiar with Vereschagin’s rule, click HERE
To see an elaborate example of its application, click HERE;

Highlights of the solved problem
NOTE: Diagrams are not to scale. Read assigned values

Analysis of Frames using the Force Method

Step 1: Calculate the degree of static indeterminacy of the structure
D = (3M + R) – 3N
Where; M = Number of members; R = Number of support reactions; N = Number of nodes
In this case; D = (3 x 3 + 5) – (3 x 4) = 2
Therefore the structure is statically indeterminate to the 2nd order

Step 2: Reduce the structure to a basic/primary system by removing redundant supports and replace them with unit loads (X₁ and X₂)

Step 3: Analyse the primary system for the various unit loads, treating each as a separate load case.

Case 1

Case 2

Step 4: Analyse the primary system when subjected to the external load, and plot the internal forces diagram.

Step 5: Compute the influence coefficients by diagram multiplication
A few examples of how the bending moment diagrams from above are combined are shown in the figures below.

δ₁₁ – Deformation at point 1 due to unit load at point 1. In this case, the bending moment diagram from case 1 is combining with itself;

δ₁₂ = δ₂₁ – Deformation at point 1 due to unit load at point 2 (this is equal to the deformation at point 2 due to unit load at point 1). In this case, the bending moment from case 1 is combining with the diagram from case 2.

δ₂₂ – Deformation at point 2 due to unit load at point 2. Here, case 2 bending moment diagram is combining with itself;

Δ_1P – Deformation at point 1 due to externally applied load. Here, the bending moment from case 1 is combined with the bending moment from the externally applied load.

Δ_2P – Deformation at point 2 due to externally applied load. Here, the bending moment from case 2 is combining with the bending moment from the externally applied load.

Step 6: Formulate the appropriate cannonical equation, and solve for the reactive forces X₁ and X₂

δ₁₁X₁ + δ₁₂X₂ + Δ_1P = 0
δ₂₁X₁ + δ₂₂X₂ + Δ_2P = 0

Step 7: Compute the final values of the internal forces
M_def = M₀ + M₁X₁ + M₂X₂

Step 8: Plot the internal stresses diagram

Analysis of Frames using the Displacement Method

analysis of statically indeterminate frames 1

Step 1: Calculate the degree of kinematic indeterminacy. This is the number of unknown nodal displacements that the structure can undergo. They are generally nodal rotations and linear translations. In this example, the structure has unknown displacements at joint C (unknown rotation), joint E (unknown rotation), joint F (unknown rotation), and a lateral displacement (side sway) that is represented at joint F. Therefore it is kinematically indeterminate to the 4th degree.

Step 2: Reduce the structure to the primary system by restraining all unknown displacements. Apply unit displacements at the joints so restrained (Z₁, Z₂, Z₃ and Z₄)

Step 3: Analyse the structure for the various unit displacements, treating each as a separate load case.

Case 1; Z₁ = 1.0

Case 2; Z₂ = 1.0

Case 3; Z₃ = 1.0

Case 4; Z₄ = 1.0

Step 4: Compute the stiffness coefficients for each load case by considering reactions due to unit displacement at the fixed ends. In all cases, K_ij stands for the force at point i due to unit displacement at point j.

Step 5: Compute the free terms of the canonical equation for each node from the fixed end moments and reactive forces due to the externally applied load.

Step 6: Formulate the appropriate canonical equation, and solve for the real nodal displacements
Solve for Z₁, Z₂, Z₃, and Z₄

Step 7: Substitute back and obtain the final values of the bending moment
M_def = M_O + M₁Z₁ + M₂Z₂ + M₃Z₃ + M₄Z₄

Step 8: Plot the internal stresses diagram

To download the full solved paper in printable PDF format, click HERE

Lateral Deflection of Multi-storey Rigid Frame Buildings under the Action of Wind Load

By

Ubani Obinna Uzodimma

-

October 21, 2020

Rigid frames, rigid planar shear walls, coupled shear walls, and cores are usually employed for the bracing system of a multi-storey building. These different units contribute to the overall resistance of the system, but their contributions can be very different both in weight and in nature, so it is essential for the designer to know their behaviour in order to achieve an optimum bracing system (Zalka, 2013).

The effect of wind becomes very influential as a building gets taller. Under the action of wind, vertical structures are subjected to load regimes which are often modelled as lateral (horizontal) loads, and as a result, lateral deflections are induced in the building (sway). It is very important to consider this behaviour in a tall building from both statics and dynamics perspective, in order to guarantee the performance of the structure while in use.

Rigid frames are very significant in the structural behaviour of buildings. They possess all the three basic stiffness characteristics, i.e., they have local bending stiffness, global bending stiffness and shear stiffness. Under lateral loads, the behaviour of frames can be complex, because they undergo both bending and shear deformations. Hence, the behaviour of frames in resisting lateral loads may be characterised by three types of stiffnesses and corresponding deflection types which are;

(a) Shear deformation
(b) Global bending
(c) Local bending

In a paper downloadable from this post, a twenty storey multi-storey rigid frame was subjected to a uniformly distributed wind load of 5.05 kN/m calculated by using the Eurocodes. The deflection behaviour of the rigid frame was investigated using a manual method proposed by Zalka (2013), and computer-based method (finite element analysis).

The disposition of the multi-storey frame is shown below;

The properties of the building are as follows;

Property	Value
Storey height (h)	3 m
Total height of building (H)	60m
Dimension of beams (d x b)	750 mm x 400 mm
Dimension of columns (d x b)	400 mm x 400 mm
Width of each bay (L)	6 m
Modulus of elasticity of concrete (E)	25 kN/mm²

Geometrical Properties of the frame
Second moment of area of beams I_B = (bd³)/12 = (0.4 × 0.75³)/12 = 0.0140625 m⁴
Second moment of area of columns I_C = (bd³)/12 = (0.4 × 0.4³)/12 = 0.0021333 m⁴
Flexural rigidity of beams EI_b = 25 × 10⁶ × 0.0140625 = 351562.5 kNm²
Flexural rigidity of columns EI_C = 25 × 10⁶ × 0.0021333 = 53333.333 kNm²
Shear stiffness of beams = K_B = ∑(12EI_b)/(L_ih) = 3 × [(12 × 351562.5 )/(6 × 3)] = 703125 kN

The part of the shear stiffness associated with the columns is;
K_C = ∑(12EI_C)/h² = 4 × [(12 × 53333.333 )/3²] = 284444.443 kN

From the above, the reduction factor r can be defined as;
r = (K_C )/(K_B + K_C) = (284444.443 )/(703125 + 284444.443) = 0.2880

The shear stiffness of the frame work can now be defined as;
K = KB × r = 703125 × 0.2880 = 202500 kN

For the local bending stiffness (EI = EI_C.r), the sum of the moment of areas of the columns should be produced and multiplied by the reduction factor r. As the bays of the frame are identical, the second moment of area of one column is simply multiplied by n and r.

I = r∑I_C = 4 ×0.2880 × 0.0021333 = 0.002457216 m⁴

The second global moment of area I_g is given by; I_g = ∑A_c,it_i
Where A_c,i is the cross-sectional area of the ith column, and t_i is the distance of the ith column from the centroid of the cross-sections.

I_g = ∑A_c,it_i = 0.4 × 0.4 × (92 +32 + 32 + 92) = 28.8 m⁴

The total second moment of area for the bending stiffness (I_f) is given by;
I_f = I_C.r + I_g = 28.8 + 0.002457216 = 28.8024 m⁴

The parameters S, ?, and ?H are also needed for the calculation of the maximum deflection;

S = 1 + (I_C.r)/I_g = 1 + (0.002457216)/28.8 = 1.00008532 ≈ 1.0
? = √bs = √(K/EI) = √[202500/(25 × 10⁶ × 0.002457216)] = 1.8156
Therefore ?H = 1.9485 × 60 = 108.936

With the above auxiliary quantities, the maximum total deflection of the frame work can now be calculated;
y_max = y(H) = (wH⁴/8EI_f) + (wH²/2KS²) – wEI/(K²S³) × [((1 + ?Hsinh?H)/cosh⁡?H) – 1]

y_b(H) = wH⁴/8EI_f = (5.05 × 60⁴)/(8 × 25 × 10⁶ × 28.8024) = 0.01136 m
y_s(H) = wH²/2KS² = (5.05 × 60²)/(2 × 202500 × 1²) = 0.044889 m
y_i(H) = wEI/(K²S³) × [((1 + ?Hsinh?H)/cosh⁡?H) – 1] = (5.05 × 25 × 10⁶ × 0.002457216) / (202500² × 1³) × [(1+ 108.936 sinh(108.936)) /cosh⁡(108.936) – 1] = 0.00082413 m

y_max = 0.01136 + 0.044889 – 0.00082413 = 0.055424 m = 55.424 mm

From the results of the two methods employed, it is observed that the value obtained from the manual method gave a lower value for the top deflection of the building. The maximum deflection from the finite element was 60.441mm, while the result from manual calculation was 55.524mm. This shows that the result from the finite element analysis is about 8.135% greater than that of the manual method. This offers great insight for the speedy check of computer-based processes and results.

An excerpt picture from finite element analysis is shown below;

To download the paper in PDF format, click HERE

Nigeria Road Traffic Signs and their Meanings | Information and Pictures

By

Ubani Obinna Uzodimma

-

December 22, 2020

Table of Contents

Road traffic signs are signs that are erected by the side or above the road to give information or instructions to road users. We are all road users, and a thorough understanding of these signs will make our usage and interaction with highway facilities safer and more convenient. On the contrary, lack of knowledge of traffic signs can mean incompetency and disobedience to traffic regulations. This may result in your vehicle being impounded, fines/penalties, and in the worst case, a highway crash.

In Nigeria, whenever Federal Road Safety Corps (FRSC) Marshals are on highway patrol, penalty tickets often issued that are related to lack of road signs knowledge, and their penalties are briefly highlighted below;

1. Construction Area Speed Limit Violation (CAV) – Fine = N3,000 – 3 Penalty points
2. Light/Sign Violation (LSV) – Fine = N2,000 – 2 penalty points
3. Road Marking Violation (RMV) – Fine = N5,000 – 5 penalty points
4. Route Violation (RTV) – Fine = N10,000 – 10 penalty points
5. Speed Limit Violation (SLV) – Fine = N5,000 – 5 penalty points
6. Unauthorized Tampering / Removal of Road Signs (UTS) – Fine = N5,000 – 5 penalty points
7. Wrongful Overtaking (WOV) – Fine = N3,000 – 3 penalty points
8. Other offenses / Violations (OFV) – Fine = N3,000 – 2 penalty points

Penalty points are points allotted to traffic offenses accumulated in a driver’s record. The implications of such penalty points are given below;

Warning (10 – 14 penalty points)
Suspension of license (15 – 20 penalty points)
Withdrawal of license (21 and above penalty points)

Nigeria road traffic signs are divided into the following categories;

A. Warning Signs – Warning signs are usually triangular in shape, with yellow or white background, black inscription, and red border. Another variety is the ‘Give Way’ or ‘Yield’ sign, which is an upside-down triangle. As the name implies, they are signs that warn you about what is ahead. They can also give instructional warning about a certain situation.

B. Regulatory Signs – Regulatory signs can be prohibitive, or mandatory.
(i) Regulatory Signs (Prohibitive): These signs are usually circular in shape with yellow or white background, black inscription, and red border. Another variety is the STOP sign, which is octagonal in shape. The STOP sign requires that you stop your vehicle completely whenever you see it. Generally, prohibitive signs give instructions that are prohibitive in nature.
(ii) Regulatory Signs (Mandatory): Regulatory signs are usually circular with a blue background, white inscription, and sometimes with a white border. They give positive instructions.

C. Informative Signs – These signs are usually rectangular in shape with a green background, and they provide guidance or information while on the highway.

These signs are shown in full below. All traffic signs are labelled at the bottom;

(a) Warning Signs

GIVE%2BWAY — **FIG A1**: YIELD EVERY RIGHT OF WAY. GIVE WAY TO TRAFFIC ON YOUR RIGHT OR TRAFFIC ON YOUR LEFT

CARRIAGEWAY%2BNARROWS — **FIG A4**: CARRIAGEWAY NARROWS

NO%2BUTURN44 — **FIG A6**: INTERSECTION WITH MINOR ROAD

And many more …

Additional road signs — **FIG A7***: OTHER WARNING SIGNS*

(b) Regulatory Signs (Prohibitory)

NO%2BENTRY — **FIG B1**: NO ENTRY FOR VEHICLES

SPEED%2BLIMIT%2BMAX — **FIG B3**: SPEED LIMIT (MAXIMUM)

POLICE%2BCHECK — **FIG B6**: POLICE CHECK POINT – STOP

**FIG B7**: NO ENTRY FOR VEHICLES EXCEEDING 3m HEIGHT

67800 — **FIG B9**: NO ENTRANCE FOR LORRIES

And many more …

(c) Regulatory Signs (Mandatory)

T67777 — **FIG C3**: FORWARD – ONE DIRECTION

BNGHH — **FIG C5**: 4-LANE 2-WAY DIVIDED HIGHWAY

FFFGG — **FIG C6**: GIVE PRIORITY TO APPROACHING VEHICLE

And many more ….

regulatory mandatory traffic signs — **FIG C8***: VAROIUS REGULATORY (MANGATORY) TRAFFIC SIGNS*

(d) Informative Signs

MINNA — **FIG D1**: ROUTE ASSURANCE SIGN (Minna 25 km)

OWORONSHOKI — **FIG D2**: PLACE NAME (Oworonshoki)

And many more…

(e) Traffic Lights/Signals

These are electronic light signaling devices used to control traffic flow at intersections to avoid conflict of vehicles. The basic signals are shown in the pictures below.

TO DOWNLOAD ALL NIGERIAN ROAD SIGNS (A FULL PAPER) IN A PRINTABLE FORMAT, CLICK HERE

This post is dedicated to members of NYSC Federal Road Safety Club (Batch ‘B’ 2015 and Batch ‘A’ 2016), Ikot Ekpene L.G.A., Akwa Ibom State, and the FRSC Marshals at RS 6.32 Ikot Ekpene Unit Command. God bless you all for your efforts in preaching the message of safety.

Design of Reinforced Concrete (R.C.) Staircase | Eurocode 2

By

Ubani Obinna Uzodimma

-

December 26, 2020

Table of Contents

In multi-storey buildings, ramps, elevators, escalators, and stairs are often used to facilitate vertical circulation. Circulation refers to the movement of people and goods between interior spaces in buildings, and to entrances and exits. Stairs are important building elements that are used to provide vertical circulation and access across different floor levels in a building. It is also recommended that when an access height exceeds 600mm, a staircase should be provided. A staircase can be made of reinforced concrete, steel, timber, and other composite construction materials.

In modern architecture, stairs are designed to be aesthetically pleasing, and they contribute immensely to the interior beauty of a building. There are different types of stairs with different configurations. For stairs in a building, the recommended slope for comfort is 27°, but for practical purposes, this can sometimes be extended to 35°.

Types of staircase

Generally, stairs are usually of the following types:

straight flight staircase
circular staircase
helical staircase
spiral staircase, or
a combination of the above-mentioned types.

Straight stairs are stairs along which there is no curvature or change in direction on any flight between two successive floors or levels. There are several possible arrangements of straight stairs. For example, they may be arranged in a straight run with a single flight between floors, or a series of flights without a change in direction.

Also, straight stairs may permit a change in direction at an immediate landing. When the stairs require a complete reversal of direction, they are called parallel stairs or half-landing stairs (turning through 180°). When successive flights are at an angle to each other, (usually 90°), they are called angle stairs or quarter-turn stairs. In addition, straight stairs may be classified as scissors stairs when they comprise a pair of straight runs in opposite directions and are placed on opposite sides of a wall.

RC STAIRCASE — Straight staircase with half landing

Circular stairs when viewed from above appear to follow a circle with a single centre of curvature and large radius.

Curved stairs when viewed from above appear to follow a curve with two or more centres of curvature, such as an ellipse.

The picture below shows a helical staircase.

Spiral stairs are similar to circular stairs except that the radius of curvature is small and the stairs may be supported by a column.

Terminologies in staircase design

parts of staircase 1 — Functional parts of a staircase

Flight: A series of steps extending from floor to floor, or from a floor to an intermediate landing or platform.

Guard: Protective vertical barrier along edges of stairways, balconies, and floor openings.

Landings (platforms): Used where turns are necessary or to break up long climbs. Landings should be level, as wide as the stairs, and at least 1000mm long in the direction of travel.

Step: Combination of a riser and the tread immediately above.

Rise: Distance from floor to floor.

Run: Total length of stairs in a horizontal plane, including landings.

Riser: Vertical face of a step. Its height is generally taken as the vertical distance between treads.

Tread: Horizontal face of a step. Its width is usually taken as the horizontal distance between risers.

Nosing: Projection of a tread beyond the riser below.

Soffit: Underside of a stair.

Railing: Framework or enclosure supporting a handrail and serving as a safety barrier.

Baluster: Vertical member supporting the handrail in a railing.

Balustrade: A railing composed of balusters capped by a handrail.

Handrail: Protective bar placed at a convenient distance above the stairs for a handhold.

Ideas on the selection of staircase dimensions

Headroom
Ample headroom should be provided not only to prevent tall people from injuring their heads but to give a feeling of spaciousness. A person of average height should be able to extend his hand forward and upward without touching the ceiling above the stairs. The minimum vertical distance from the nosing of a tread to overhead construction should preferably never be less than 2100 mm.

Stairway Width
The width of a stairway depends on its purpose and the number of persons to be accommodated in peak hours or emergencies. Also, there are building codes that regulate the geometric design of stairways. The following can be used as guidelines;

For residential flats between two to four storeys, use a minimum width of 900 mm, for flats more than 4 storeys, use a width of 1000 mm.
For public buildings of under 200 persons per floor, use a width of 1000 mm, for buildings between 200 – 400 persons per floor, use a width of 1500 mm. For over 400 persons, use a width between 1500 – 3000 mm. However, when the width of a stairway exceeds 1800 mm, it is necessary to divide it using handrails.

Step Sizes
Risers and treads generally are proportioned for comfort and to meet accessibility standards for the handicapped, although sometimes space considerations control or the desire to achieve a monumental effect, particularly for outside stairs of public buildings. Treads should be at least 250 mm, exclusive of nosing.

The most comfortable height of riser is 175 mm. Risers less than 100 mm and more than 200 mm high should not be used. The steeper the slope of the stairs, the greater the ratio of the riser to tread. In the design of stairs, account should be taken of the fact that there is always one less tread than riser per flight of stairs. No flight of stairs should contain less than three risers.

Structural design of a staircase

The theoretical procedures employed in the structural analysis of stairs is the concept of an idealised line structure and when detailing the reinforcement for the resulting stairs, additional bars should be included to limit the formation of cracks at the points of high-stress concentration that inevitably occur. Typical detailing of corners (edge between flight and landing) of a staircase is shown below;

staircase detailing 1 — Detailing of end support of staircase with landing

staircase joint detailing — Detailing of end support of staircase with landing

The ‘three-dimensional’ nature of the actual structure and the stiffening effect of the triangular tread areas, both of which are usually ignored when analysing the structure, will result in actual stress distributions that differ from those calculated, and this must be remembered when detailing (Reynolds et al, 2008). The typical nature of internal stresses induced in a simply supported straight flight stair and reinforcement pattern is as shown in the picture below.

staircase in bending — Bending behaviour of a single flight staircase

Simple straight flights of stairs can span either transversely (i.e. across the flight) or longitudinally (i.e.along the flight). When spanning transversely, supports must be provided on both sides of the flight by either walls or stringer beams. In this case, the waist or thinnest part of the stair construction need be no more than 60 mm thick say, the effective lever arm for resisting the bending moment being about half of the maximum thickness from the nose to the soffit, measured at right angles to the soffit. When the stair spans longitudinally, deflection considerations can determine the waist thickness.

In principle, the design requirements for beams and slabs apply also to staircases, but designers cannot be expected to determine the deflections likely to occur in the more complex stair types. BS 8110 deals only with simple types and allows a modified span/effective depth ratio to be used. The bending moments should be calculated from the ultimate load due to the total weight of the stairs and imposed load, measured on plan, combined with the horizontal span. Stresses produced by the longitudinal thrust are small and generally neglected in the design of simple systems.

Sample design of reinforced concrete staircase

A section of a staircase is shown above. The width of the staircase is 1160mm. We are expected to carry out a full structural analysis and design of the staircase according to EC2 using the following data; Density of concrete = 25 kN/m³; Compressive strength of concrete (f_ck) = 30 N/mm²; Yield strength of steel (f_yk) = 460 N/mm²; Concrete cover = 25mm; Imposed load on staircase (q_k) = 4 kN/m² (category C3).

The structural idealisation of the staircase is shown below;

Loading of the staircase

Thickness of waist and landing = 200 mm
Depth of riser = 150mm
Load actions on the stairs
Concrete self weight (waist area) = 0.2 × 25 = 5 kN/m² (normal to the inclination)
Stepped area = 0.5 × 0.15 × 25 = 1.875 kN/m² (global vertical direction)
Finishes (say) = 1.2 kN/m²

We intend to apply all gravity loads purely in the global y-direction, therefore we convert the load at the waist of the stair from local to global direction by considering the angle of inclination of the flight area to the horizontal;

? = tan⁻¹(1.2/1.75) = 34.438989°

Therefore the UDL from waist of the stair in the global direction is given by = (5 × cos 34.438989) = 4.124 kN/m²

Total dead load on flight area (g_k) = 4.124 + 1.875 + 1.2 = 7.199 kN/m
Variable load on staircase (q_k) = 4 kN/m²
The load on the flight area at ultimate limit state = 1.35g_k + 1.5q_k
E_d = 1.35(7.199) + 1.5(4) = 15.719 kN/m²

On the landing;
g_k = 5 + 1.2 = 6.2 kN/m²; q_k = 4 kN/m²

The load on the landing at ultimate limit state = 1.35g_k + 1.5q_k
n = 1.35(6.2) + 1.5(4) = 14.370 kN/m²

The loading of the structure for dead and live loads at ultimate limit state is shown below;

The ultimate bending bending moment diagram due to ultimate loads is shown below;

The ultimate shear force diagram is shown below;

Flexural design of the staircase span

A little consideration will show that it is best to use the design moment M_Ed= 41.119 kNm to design the entire stairs.

M_Ed = 41.119 kNm
d = h – Cc – ϕ/2
Assuming ϕ12mm bars will be used for the construction
d = 200 – 25 – 6 = 169mm; b = 1000mm (designing per unit width)

k = M_Ed/(f_ckbd²) = (41.119 × 10⁶)/(30 × 1000 × 169²) = 0.0479

Since k < 0.167 No compression reinforcement required
z = d[0.5+ √((0.25 – 0.882k)] = z = d[0.5+ √(0.25 – 0.882 × 0.0479)] = 0.95d

A_s1 = M_Ed/(0.87f_ykz) = (41.119 × 10⁶)/(0.87 × 460 × 0.95 × 169) = 639.96 mm²/m

Provide Y12mm @ 150mm c/c BOT (A_s,prov = 753 mm²/m)

To calculate the minimum area of steel required;
fctm = 0.3 × f_ck^(2⁄3) = 0.3 × 30^2⁄3 = 2.896 N/mm²(Table 3.1 EC2)

A_s,min = 0.26 × f_ctm/f_yk × b × d = 0.26 × 2.896/460 ×1000 ×169 = 276.631 mm²/m
Check if A_s,min < 0.0013 × b × d (219.7 mm²/m)
Since, A_s,min = 276.631 mm², the provided reinforcement is adequate.

Check for deflection

k = 1.0 for simply supported beams and slab
ρ = As/bd = 753/(1000 × 169) = 0.004455 < 10^-3√30

Since ρ < ρ₀
L/d = k [11 + 1.5√(f_ck) ρ₀/ρ + 3.2√(f_ck) (ρ₀/ρ – 1)^1.5]

L/d = 1.0 [11 + 1.5√30 × (0.005477/0.004455) + 3.2√30 (0.005477/0.004455 – 1)^1.5] = 1.0 (21.1 + 1.9258) = 23.0258

Modification factor β_s = 310/σ_s
σ_s = (310_fykA_s,req)/(500A_s,prov) = (310 × 460 × 639.96)/(500 × 753) = 242.358 N/mm²

β_s = 310/242.358 = 1.2789
Taking the distance between supports as the effective span, L = 4.35m
The allowable span/depth ratio = β_s × 23.0258 = 1.2789 × 23.0358 = 29.460
Actual deflection L/d = 4350/169 = 25.739
Since 25.739 < 29.460, deflection is ok.

Shear design

Ultimate shear force V_Ed = 35.358 kN
v = V_Ed/bd = (35.358 × 1000)/(1000 × 169) = 0.209 N/mm²
V_Rd,c = [C_Rd,ck(100ρ₁f_ck)^1/3] ≥ (_Vmin)

C_Rd,c = 0.18/γ_c= 0.18/1.5 = 0.12

k = 1 + √(200/d) = 1 + √(200/169) = 2.087 > 2.0, therefore, k = 2

V_min = 0.035k^1.5f_ck^0.5
V_min = 0.035 × 2^1.5 × 30^0.5 = 0.5422 N/mm²

ρ₁ = As/bd = 753/(1000 × 169) = 0.0044556 < 0.02

V_Rd,c = [0.12 × 2(100 × 0.0044556 × 30)^1/3] = 0.569 N.mm²
Since V_Rd,c > V_Ed , no shear reinforcement is required. Shear is ok

Staircase reinforcement details

To download the full paper in PDF format, click HERE

Analysis of Bridge Deck Girders under Load Model 1 (EN 1991-2) Using Staad Pro

By

Ubani Obinna Uzodimma

-

February 13, 2020

Bridges are structures designed to span over obstacles, and as such provide access routes for humans, animals, vehicles, fluid, and other objects as may be desired. EN 1990 (Basis of Structural Design) classified bridges under consequence class 3 (CC3) which implies that the consequences are high if failure should occur. As a result, the design of bridges is taken very seriously and all possible load regimes that the structure will be subjected to in its design life are taken into account.

Bridges are important for adequate development, safety, and efficiency of road transportation system in any country including Nigeria. Nigerian engineers have stepped up to solve the engineering problems facing them internally and independently.

Loads from vehicles are one of the the most prominent loads that bridges are subjected to in their design life. Eurocode 1 Part 2 (EN 1991-2) handles the actions of traffic on bridges.

The code specifies four load models for vertical loads on highway bridges. Among these four load models, Load Model 1 (LM1) is the one used to represent the effect of normal traffic on bridges and it is used for both local and global verifications. LM1 consists of tandem axle loads (TS) which are concentrated loads applied in the longitudinal direction of the bridge, and uniformly distributed loads (UDL’s) which are applied both transversely and longitudinally. The values of the loads depends on the notional lanes on which they are applied unlike BS 5400 which uses lane factors. (See more on the paper uploaded). It is however pertinent to point out that LM1 is an artificial model and does not represent the weight of real vehicles. The models are created and calibrated based on traffic data collected in France between 1980 and 1994, and the effects they induce on bridge decks are similar to that induced by real traffic. See the form of LM1 below.

In the paper downloadable in this post, a 15.0m single span bridge simply supported between two abutments (the first picture on this post) is subjected to LM1 and analysed as static load using Staad Pro software. The bridge has a deck that is 10.1m wide and a carriageway that is 7.2m wide. The reinforced concrete slab is supported by five precast beams (girders) and the aim of the analysis is to find out the effects of the traffic load on the girders. The load was arranged to induce the worst effect on the first and second girders. The finite element analysis results obtained were very comparable to the ones obtained using other rational and computational methods.

Relevant Pictures from Analysis

(a) Another 3D view of the bridge (Modelled using Autodesk Revit 2009)

(b) A 2D elevation view of the bridge

(c) Section through the bridge deck

(d) Division of the carriageway into notional lanes

(e) Sectional view of the notional lanes, remaining area, and the pedestrian areas

(f) Finite element modelling of the bridge deck on StaadPro V8i

(g) Loading of the bridge with LM1 (including crowd load)

(h) Full 3D Model of the bridge deck on Staad Pro

(i) Bending moment diagram of the girders due to LM1

(j) Shear force diagram of the girders due to LM1

To download the full loading, analysis, and results paper, click HERE.

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Analysis and Design of a Continuous R.C. Raker Beam for Stadium

By

Ubani Obinna Uzodimma

-

March 13, 2023

Table of Contents

The most common construction concept of sports stadiums today is a composite type where usually precast concrete terrace units (seating decks) span between inclined (raker) steel or reinforced concrete raker beams and rest on each other, thereby forming a grandstand.

The raker beams are usually formed in-situ with the columns of the structure, or sometimes may be preferably precast depending on site/construction constraints. This arrangement usually forms the skeletal frame of a stadium structure.

A typical section through the grandstand and the L-shaped seating unit are shown in Figures 1 and 2 respectively.

Stadium%2BSection — **Figure 1:** Section through the grandstand

This process was used in the construction of Cape Town Stadium South Africa, for the 2010 World cup (see Figure 3);

In this article, a raker beam isolated from a double-tiered reinforced concrete grandstand (see Figure 1) has been presented for the purpose of structural analysis and design. Each grandstand frame has precast L-shaped seating terrace units that span in between reinforced concrete raker beams inclined at angles between 20° – 22° with the horizontal. The raker beams are spaced at 7m centre to centre.

Crowd live loads and other loads are transferred from the seating units to the raker beams, which then transfers them to the columns and then to the foundations. Loads from the service and concourse areas are also transferred using the same load path.

The details of the structure and the loading on the first tier are shown below;

Loading on the precast seating units

Permanent Actions
Self-weight of the 7m precast seating deck (see Figure 2)
(G_K1) = (25 × 0.25 × 0.15 × 7) + (25 × 0.95 × 0.15 × 7) = 31.5 kN
Weight of finishes, rails, seats (G_K2) = (2 × 0.95 × 7) = 13.3 kN

Variable Actions
Imposed load for structural class C5 (q_K) = 5 kN/m²
Total imposed load for seating unit (Q_K) = (5 × 0.95 × 7) = 33.25 kN

Total action on L-shaped seating terrace unit at ultimate limit state by Eurocode 2
(P_Ed) = 1.35Σ(G_Ki) + 1.5Q_K = 1.35(44.8) + 1.5(33.25) = 110.355 kN

Loading on the raker beams

Height of beam = 1200mm
Width of beam = 400mm

Self weight of raker beam
Concrete own weight (waist area) = 1.2m × 0.4m × 25 kN/m³ = 12.00 kN/m (normal to the inclination i.e. in the local direction)
Height of riser in the raker beam = 0.4m;
Width of tread in the raker beam = 0.8m;
Angle of inclination (𝛼) = 20.556°
Stepped area (risers) = ½ × 0.4 × 25 = 5 kN/m (in the global direction)

For purely vertical load in the global y-direction, we convert the load from the waist of the beam by;
UDL from waist of the beam = (12.00 × cos 20.556°) = 11.236 kN/m
Total self weight (G_k) = 11.236 + 5 = 16.235 kN/m
Self weight of raker beam at ultimate limit state;
n = 1.35Σ(G_Ki) = 1.35 × 16.235 = 21.917 KN/m

Load from precast seating units
End shear from precast seating unit = P_Ed/2 = 110.355/2 = 55.1775 kN
Total number of the precast seating units on the beam = 24/0.8 = 30 units
For intermediate beam supporting seating units on both sides;
Total number of precast seating units = 2 × 30 = 60 units
Therefore, the total shear force transferred from the seating units to the raker beam = 55.1775 × 60 = 3310.65 kN
Equivalent uniformly distributed load in the global direction at ultimate limit state = 3310.65/24 = 137.94 kN/m
Total load on intermediate raker beams at ultimate limit state in the global direction = 137.94 + 21.917 = 159.857 kN/m

Structural Analysis

A full 3D Staad Pro Model of the grandstand is shown in Figure 5;

stadium model on Staad Pro — **Figure 5:** 3D Modelling of the grandstand on Staad Pro

The raker beam was analysed for the loading obtained as shown above. Due to its inclination, it was subjected to significant bending, axial, and shear forces. The internal stresses are shown below;

The summary of the internal stresses in the raker beams is shown in the Table below;

Structural Design

The structural design of the raker beam using EN 1992-1-1 has been carried out and all the parameters used in the, and steps followed are shown below in the subsequent sections.

Design compressive of concrete f_ck = 35 N/mm²
Yield strength of steel f_yk = 460 N/mm²
b_w= 400mm; h = 1200mm; C_c = 40mm

Flexural Design of span AB (M_ABspan)
M_Ed = 948.078 kNm
d = h – Cc – ϕ/2 – ϕ_link
d = 1200 – 40 – 16 – 10 = 1134 mm
k = 𝑀_𝐸𝑑/𝑓_𝑐𝑘 𝑏𝑑² = (948.078 × 10⁶)/35 × 400 × 1134² = 0.0527
Since k < 0.167 No compression reinforcement required
z = d[0.5+ √(0.25−0.882𝐾)] = z = d[0.5 + √(0.25 − 0.882(0.0527)] = 0.95d

𝐴_𝑠1 = 𝑀_𝐸𝑑/0.87𝑓_𝑦𝑘𝑧 = (948.078 × 10⁶)/0.87 × 460 × 0.95 × 1134 = 2199 mm²
Provide 5Y25mm BOT (ASprov = 2450 mm²)

To calculate the minimum area of steel required; (Table 3.1 EC2)
f_ctm=3.2099 N/mm²
A_s,min= 0.26 × f_ctm/f_yk × b_w × d = 0.26 × 3.2099460 ×400 × 1134 = 822.962 mm²
Check if ASmin < 0.0013 × b_w × d (589.68 mm²)
Therefore, A_s,min = 822.962 mm²

Check for Deflection
Since the span is greater than 7m, allowable span/depth ratio = 𝛽𝑠 × 31.842 × 7000𝐿 = 1.11 × 31.842 × (7000/12816) = 19.374
Actual deflection L/d = 12816/1134 = 11.301
Since 11.301 < 19.374, deflection is ok.

Flexural Design of support A (MA);
M_Ed = 1967.54 kNm
k = 0.1093; la = 0.8919; A_S1 = 4861 mm²; A_Smin= 822.9785 mm²
Provide 4Y32mm + 4Y25mm TOP (A_Sprov = 5180 mm²)

Flexural Design of support B (MB);
M_Ed = 2283.18 kNm
k = 0.1268; la = 0.8717; A_S1 = 5772 mm²; A_Smin = 822.9785 mm²
Provide 6Y32mm + 4Y20mm TOP (A_Sprov = 6080 mm²)

Flexural Design of span BC (MBCspan)
M_Ed= 1249.787 kNm
k = 0.0694; la = 0.9345; A_S1 = 2947mm²; ASmin = 822.9785 mm²
Provide 5Y25mm + 2Y20mm BOT (A_Sprov= 3083 mm²)

Shear Design

Let us consider support B
V_Ed = 983.88 kN; N_Ed = 339.376 kN (Tension)
Note that due to the tensile axial force in the section, the second term of V_Rdequation assumes a negative value.
𝜌₁ = 0.0134; 𝜎_𝑐𝑝 = − 0.7070 N/mm₂; v_min = 0.3504 N/mm²; V_Rd = 230.6532 kN
Since V_Rd,c < V_Ed, shear reinforcement is required

Assuming that the strut angle 𝜃 = 21.8° v₁ = 0.5160; f_cd = 19.8450 N/mm²; z = 0.9d = 1020.6 mm;
V_RD,max = 1440.64 kN
Since V_RD,max > V_Ed
𝐴_𝑠𝑤/S = 0.9635
Trying 3Y10mm @ 200mm c/c (235/200 = 1.175)
1.175 > 0.9153 Hence shear reinforcement is ok.

The design results show that the effect of axial force was not very significant in the quantity of shear reinforcement required. Asv/Sv ratio of 1.175mm (3Y10mm @ 200 c/c) was found to satisfy shear requirements at all sections. The greatest quantity of longitudinal reinforcement was provided at the intermediate support with a reinforcement ratio of 1.3404%. The provided reinforcement was found adequate to satisfy ultimate and serviceability requirements.

It is worthwhile to observe the variation in the nature of axial forces on the beams (i.e from compressive to tensile). The effect can be more significant when using BS 8110-1:1997 for design than Eurocode 2 (EN 1992-1-1).

You can view some sections of the reinforced concrete raker beam below;

To download the full calculation sheet (PDF), click HERE

Example on the Design of Steel Roof Trusses

Structural Design of Roof Trusses to Eurocode 3

Network of Primary and Secondary Beams

Design Example

Structural Design of the secondary beams

Loading, Analysis, and Design of Primary Beams

Reinforcement Details of Primary Beams

Analysis of Frames using the Force Method

Analysis of Frames using the Displacement Method

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Check for deflection

Shear design

Staircase reinforcement details

Loading on the precast seating units

Loading on the raker beams

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Structural Design

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