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

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December 21, 2016

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

-

November 1, 2016

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

-

September 28, 2016

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

-

September 24, 2016

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

-

August 22, 2016

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

-

May 12, 2016

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

-

April 28, 2016

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

Modal Analysis of Rigid Multi-Storey Frames

By

Ubani Obinna Uzodimma

-

April 24, 2016

A structure undergoes free vibration when it is brought out of static equilibrium and can then oscillate without any external dynamic excitation. Free vibration of structures occurs with some frequencies which depend only on the parameters of the structures such as the boundary conditions, distribution of masses, stiffnesses within the members etc, and not on the reason for the vibration.

At each natural frequency of free vibration, the structure vibrates in simple harmonic motion where the displaced shape (mode shape) of the structure is constant but the amplitude of the displacement is varying in a sinusoidal manner with time. The number of natural frequencies in a structure coincides with the number of degrees of freedom in the structure. These frequencies are inherent to the given structure and are often referred to as eigenfrequencies. Each mode shape of vibration shows the form of an elastic curve which corresponds to a specific frequency.

A method of obtaining the natural frequencies and mode shapes of vibration is modal analysis. This is a technique by which the equations of motion, which are originally expressed in physical coordinates, are transformed to modal coordinates using the eigenvalues and eigenvectors gotten by solving the undamped frequency eigenproblem. The transformed equations are called modal equations.

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

Mü + ku = 0 ———-(1)

From Figure 1.1, the following equations can be developed;
M₁Ü₁ + K₁U₁ + K₂ (U₁ − U₂) = 0
M₂Ü₂ + K₂(U₂ − U₁) = 0

This can be expanded to give;
M₁Ü₁ + U₁(K₁+ K₂) + U₂(−K₂) = 0
M₂Ü₂ + U₁(-K₂) + U₂(K₂) = 0

Arranging it in matrix form we obtain;

The characteristic polynomial equation is thus;

The solution of the equation is;
U₁ = A₁sin(𝜔𝑡+ 𝜑), U₂ = A₂sin(𝜔𝑡+ 𝜑)

Where A_iis the amplitude of the displacement of mass M_i, and 𝜑 is the initial phase of vibration.
[(K₁ + K₂) − M₁ω₂]A₁ − K₂A₂=0
−K₂A₁ + (K₂−M₂ω₂) A₂ = 0

To obtain the frequency equation, a non-trivial solution exists (non-zero amplitudes A_i), if the determinant of the coefficients to the amplitude is zero. This is also called the characteristic polynomial equation and is thus;

The solution of equation (3) presents the eigenfrequencies of the system. The system does not allow us to determine amplitudes directly, but we can find the ratio between these amplitudes. Hence for equation 3 above;

If we assume A₁ = 1.0, then entries [1 𝐴₂]^𝑇 defines for each eigenfrequency, the column 𝜑 of the modal matrix Φ. The shape of each mode of free vibration is unique but the amplitude of the mode shape is undefined. The mode shapes are usually normalised such that the largest term in the vector is 1.0. The mode shapes have the important property of being orthogonal with respect to the mass and stiffness matrix of the structure.

Worked Example

A frame with an infinitely rigid floor is supported by 300 x 300 mm columns. If it is loaded as shown below, carry out the full modal dynamic analysis of the structure. (Take EI = 2.1 × 10⁷ kN/m²)

In the paper downloadable in this post, modal dynamic analysis was carried out on the three-storey frame shown above. The results obtained regarding the three assumed degrees of freedom are shown in the pictures below;

(a) Mode 1 vibration parameters and displaced shape

(b) Mode 2 vibration parameters and displaced shape

(c) Mode 3 vibration parameters and displaced shape

The displacement time history response is therefore;

u₁(t) = 0.00128448 sin(5.0261t) + 0.00044346 sin(13.7654t) + 0.0002408175 sin(19.0459t)

u₂(t) = 0.00107836 sin(5.0261t) – 0.00009 sin(13.7654t) – 0.000313845 sin(19.0459t)

u₃(t) = 0.00061299 sin(5.0261t) – 0.00046104 sin(13.7654t) + 0.0002157225 sin(19.0459t)

If velocity (v) = du/dt; The velocity time history response is therefore;
The velocity time history response is obtained by direct differentiation of the displacement time history equations;

v₁(t)= 0.0064559 cos(5.0261t) + 0.0061044 cos(13.7654t) + 0.0045865cos(19.0459t)

v₂(t)= 0.0054199 cos(5.0261t) – 0.0012388 cos(13.7654t) – 0.00597746 cos(19.0459t)

v₃(t)= 0.0030809 cos(5.0261t) – 0.0063464 cos(13.7654t) + 0.0041086 cos(19.0459t)

The acceleration (a) time history response is obtained by direct differentiation of the velocity time history equations;

a₁(t) = -0.032448 sin(5.0261t) – 0.0840295 sin(13.7654t) – 0.087354 sin(19.0459t)

a₂(t) = -0.02724 sin(5.0261t) + 0.0170525 sin(13.7654t) + 0.113846 sin(19.0459t)

a₃(t) = -0.0154849 sin(5.0261t) + 0.0873607 sin(13.7654t) – 0.0782519 sin(19.0459t)

To download the full calculation sheet, click HERE.

Analysis of Internal Stresses in Frames Due to Settlement of Support | Sinking of Supports

By

Ubani Obinna Uzodimma

-

April 18, 2016

Internal stresses are induced in statically indeterminate frames when one support settles relatively to another. This is one of the major detrimental effects of differential settlement in civil engineering structures, and it is capable of inducing cracks and structural failures in buildings. Therefore, wherever it matters or where it is anticipated, the effects of the differential settlement must be taken into account during the design of civil engineering structures.

In this article, an indeterminate frame whose support has been subjected to an indirect action of differential settlement has been presented and fully solved. As stated above, if any statically indeterminate structure experiences differential settlement of supports, internal stresses will be developed in the members of the structure.

Analysis of such structures may be effectively performed by the force (flexibility) or stiffness (displacement) method in canonical form. In the paper downloadable in this article, the frame loaded as shown below is subjected to a differential settlement of 25mm at fixed support A. The internal stresses induced were calculated using the force method, stiffness method, and slope deflection method.

Canonical Equations

Consider a structure that is unable to be solved solely based on equilibrium equations (statically indeterminate). This structure has additional constraints (redundant constraints) that number “n.” Additionally, certain supports of this structure can undergo linear or angular movements (displacements) denoted by “d_i.”

The following set of equations are known as the Canonical equations and will be used to analyze this structure:

δ₁₁X₁ + δ₁₂X₂ + … + δ_1nX_n + δ₁∆ = 0
δ₂₁X₁ + δ₂₂X₂ + … + δ_1nX_n + δ₂∆ = 0
. . . . . . . . . . . . . . . . . . . .
δ₃₁X₁ + δ₃₂X₂ + … + δ_3nX_n + δ₃∆ = 0

where free terms δ_k∆ (k = 1, 2,…, n) represent displacements of the primary system in the direction of primary unknowns X_k due to settlements of the supports. For the calculation of these terms, we need to use the theorem of reciprocal unit displacements and reactions (Rayleigh’s second theorem). This is given by;

δ_i∆ = -EI_C (∆.R)

Where:
EI_C = Flexural rigidity of the column
∆ = Settlement at the point under consideration
R = Support reaction at the point under consideration

Steps in the Analysis of Structures Due to Sinking of Supports

This section outlines the step-by-step procedure for analyzing redundant structures experiencing support settlements:

1. Kinematics and Redundancy Determination:

Carry out a kinematic analysis of the structure.
Identify the degree of redundancy (number of redundant constraints, n).
Select a primary system based on the force method.
Formulate the canonical equations using the chosen primary unknowns (X_i).

2. Unit Load Analysis:

Construct unit bending moment diagrams for each redundant member.
Calculate the unit displacements (δ_i) corresponding to each redundant constraint (d_i).

3. Free Term Calculations:

Determine the free terms within the canonical equations, considering the external loading and support settlements.

4. Solving for Primary Unknowns:

Solve the system of canonical equations with respect to the primary unknowns (X_i).

5. Internal Force Determination:

Construct the internal force diagrams (axial forces, shear forces, bending moments) for all members using the obtained primary unknowns (Xi) and the principle of superposition.

6. Support Reaction Verification:

Calculate the reactions at all supports using equilibrium equations and the internal forces.
Verify the calculated support reactions through equilibrium checks.

Worked Example

The portal frame shown below is fixed at column base A, and hinged at point C as shown below. The column of the frame has a square cross-section of 30cm x 30cm while the beam has a depth of 60cm and a width of 30cm. The foundation of support A settles vertically by 25mm. Draw the bending moment, shearing force, and axial force diagram due to the differential settlement action on the frame (E = 2.17 × 10⁷ kN/m²).

SOLVED%2B%2BQUESTION%2BON%2BSETTLEMENT%2BOF%2BFRAME

SOLUTION
Geometrical properties
Moment of inertia of column I_C = (bh³) / 12 = (0.3 × 0.3³)/12 = 6.75 × 10^-4 m⁴
Moment of inertia of beam I_B = (bh³) / 12 = (0.3 × 0.6³)/12 = 5.4 × 10^-3 m⁴
We desire to work in terms of I_C such that I_C/I_B = 0.125

Hence flexural rigidity of the column, EI_C = (2.17 × 10⁷) × (6.75 × 10^-4) = 14647.5 KN.m²

The deformation of a structure at a point due to support settlement is given by;

δ_i∆ = -EI_C (∆.R) ———— (1)
Where;
EI_C = Flexural rigidity of the column
∆ = Settlement at the point under consideration
R = Support reaction at the point under consideration

Basic system
The next step in the analysis is to reduce the structure to a basic system, which is a system that must be statically determinate and stable. The frame is statically indeterminate to the second order, which means that we are going to remove two redundant supports.

The degree of static indeterminacy (R_D) is given by;

R_D = (3m + r) – 3n – s ————————— (2)

So that R_D = (3 × 2) + 5 – (3 × 3) – 0 = 2

By choice, I am deciding to remove the two reactive forces at support C of the frame to obtain the basic system as shown in Figure 1.2 below;

Analysis of Case 1;
X1 = 1.0, X2 = 0
The simple static analysis of the structure with case 1 loading is shown below in terms of the internal stresses diagram. Realise that there is a unit value support reaction acting at support A (pointed downwards) and a bending moment of 5.0 units.

CASE%2B1%2BINTERNAL%2BSTRESSES%2BDIAGRAM

Analysis of Case 2;
X1 = 0, X2 = 1.0
The simple static analysis of the structure loaded with case 1 loading is shown below in terms of the internal stresses diagram. It will be so important to realise that there is no vertical support reaction at point A. A unit horizontal reaction is developed at the support to counter the applied unit load for case 2.

CASE%2B2%2BINTERNAL%2BSTRESSES%2BDIAGRAM

The appropriate canonical equation for this structure is therefore;

δ₁₁X₁ + δ₁₂X₂ + δ₁∆ = 0
δ₂₁X₁ + δ₂₂X₂ + δ₂∆ = 0

Computation of the influence coefficients

Influence coefficients are based on Mohr’s integral such that δ_i = 1/EI∫Mmds.

We can compute this by using the graphical method (making use of bending moment diagrams), we directly employ Vereschagin’s rule which simply states that when we are combining two diagrams of which one must be of a linear form (due to the unit load) and the other of any other form, the equivalent of Mohr’s integral is given by the area of the principal combiner (diagram of arbitrary shape) multiplied by the ordinate which its centroid makes with the linear diagram. The rule can also work vice versa. This process has been adopted in this work.

Beam%2BDeflection%2BEquation%2BUsing%2BVirtual%2BWork%2BMethod

Where;
EI = Flexural Rigidity of the section
M = Bending moment due to externally applied load
̅M = Bending moment due to unit load at the point where the deflection is sought

Example on Analysis of Statically Determinate Frames (Part 2)

δ₁₁ (Deformation at point 1 due to unit load at point 1)
This is obtained by the bending moment of case 1 combining with itself. This is shown below.

INFLUENCE%2BCOEFFICIENT%2BDIAGRAM%2B%25281%2529

EI_Cδ₁₁ = (5 × 5× 4) + (1/3 × 5 × 5 × 5 × 0.125) = 105.208

δ₂₁ = δ₁₂ (Deformation at point 2 due to unit load at point 1 which is equal to deformation at point 1 due to unit load at point 2 based on Maxwell-Betti’s law). This is obtained by the bending moment diagram of case 1 combined with the bending moment diagram of case 2. This is shown below.

EI_Cδ₂₁ = (1/2 × 5 × 4 × 4) = 40

δ₂₂ (Deformation at point 2 due to unit load at point 2)
This is obtained by the bending moment diagram of case 2 combining with itself. This is shown below.

EI_Cδ₁₁ = (1/3 × 4 × 4 × 4) = 21.333

Influence coefficients due to support settlement

Case 1
(Take a good look at the support reactions)
δ_i∆ = -EI_C(∆.R)
R = Support reaction at point A = -1.0
∆ = Support settlement in the direction of support reaction = -25 mm = -0.025m
Hence δ₁ ∆ = – 14647.5 × (-1 × -0.025) = -366.1875

Case 2
(Take a good look at the support reactions again)
Horizontal support reaction at point A = R = 1.0
∆ = Support settlement/movement in the direction of support reaction = 0
Hence δ₂∆ = 0

The appropriate canonical equation now becomes;

105.208X₁ + 40X₂ = 366.187
40X₁ + 21.333X₂ = 0

On solving the simultaneous equations;
X₁ = 12.123 kN
X₂ = -22.730 kN

The final value of the internal stresses is given by the equations below;

M_def = M₁X₁ + M₂X₂
Q_def = Q₁X₁ + Q₂X₂
N_def = N₁X₁ + N₂X₂
Final Bending Moments (M_def)
M_A = (12.1226 × 5) + (-22.73 × 4) = -30.307 kNm
M_B = (12.123 × 5) = -60.613 kNm
M_C = hinged support = 0

Final Shear Force (Q_def)
Q_A – Q_B^B = (-22.73 × -1) = 22.73 kN
Q_B^R – Q_C^L = (12.123 × -1) = -12.123 kN

Final Axial Force (N_def)
N_A – N_B^B = (12.123 × 1) = 12.123 kN
N_B^R – N_C^L = (-22.73 × -1) = 22.730 kN

To download the full calculation sheet and see how the force method is compared with stiffness methods, click HERE
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Static Determinacy of Rigid Frames

By

Ubani Obinna Uzodimma

-

April 16, 2016

1.1 Introduction
A structure is stable when it maintains balance in force and moment. As a result, we know that from statics, if a structure is to be at equilibrium;

∑Fy = 0; ∑Fx = 0; ∑Mi = 0; ——————- (1.1)
Where;
∑Fy = Summation of the vertical forces
∑Fx = Summation of the horizontal forces
∑Mi = Summation of the moment of force components acting in the x-y plane passing through point i.

When the number of constraints in a structure permits the use of equation of statics (equation 1.1) to analyse the structure, the structure is said to be statically determinate. Otherwise, it is statically indeterminate, and additional equation which is derived from load-deformation relationship is used for analysis. For the records, there are two well known approaches to the analysis of indeterminate structures and they are;

1. Flexibility Methods – When the structure is analysed with respect to unknown forces
2. Stiffness Methods – when the structure is analysed with respect to unknown displacements

A structure may be indeterminate due to redundant components of reaction and/or redundant members. Note that a redundant reaction or member is one which is not really necessary to satisfy the minimum requirements of stability and static equilibrium. However, redundancy is desirable in structures because they are cheaper alternatives to determinate structures. The degree-of-indeterminacy (referred to as RD in this post) is equal to the number of unknown member forces/external reactions which are in excess of the equations of equilibrium available to solve for them.

1.2 Determinacy of rigid frames
In rigid frames, the applied load system is transferred to the supports by inducing axial loads, shear forces and bending moments in the members. Since three components of reaction are required for static equilibrium the total number of unknowns is equal to;

U = (3 × m) + r ————— (1.2)

Since we have three equations of equilibrium, we have (3 × n) equations, hence;

RD = (3m + r) – 3n – s ————— (1.3)

Where;
m = number of members
r = Number of support reactions
n = Number of nodes
S = Number of special conditions (e.g. internal hinge)

Another equation that can be used for calculation of degree of indeterminacy in frames is;

RD = R – e – S ———— (1.4)

Where;
R = Number of support reactions
e = Number of equations of equilibrium (i.e 3)
S = Number of special conditions (e.g. internal hinge)
Whenever;
RD = 0 (structure is statically determinate and stable
RD < 0 (structure is unstable)
RD > 0 (structure is statically indeterminate)

1.3 Solved examples

In the frames shown below, classify the following frames as statically determinate or indeterminate. All internal hinges are denoted by G.

Solution

Using equation 1.3
RD = (3m + r) – 3n – s

(a) m = 6, r = 6, n = 7, s = 1
RD = [3(6) + 6] – 3(7) – 1 = 2, Hence the frame is indeterminate to the 2nd order

(b) m = 5, r = 5, n = 6, s = 2
RD = [3(5) + 5] – 3(6) – 2 = 0, Hence the frame is statically determinate

(c) m = 4, r = 4, n = 5, s = 1
RD = [3(4) + 5] – 3(6) – 1 = 0, Hence the frame is statically determinate

(d) m = 3, r = 5, n = 4, s = 0
RD = [3(5) + 5] – 3(4) – 0 = 2, Hence the frame is statically indeterminate to the 2nd order

(e) m = 4, r = 5, n = 5, s = 1
RD = [3(4) + 5] – 3(5) – 1 = 1, Hence the frame is statically indeterminate to the 1st order

(f) m = 3, r = 3, n = 4, s = 1
RD = [3(3) + 3] – 3(4) – 1 = -1, Hence the frame is unstable

Alternatively using equation 1.4;
RD = r – e – s

(a) r = 6, e = 3, s = 1
RD = 6 – 3 – 1 = 2, Hence, the frame is indeterminate to the 2nd order

(b) r = 5, e = 3, s = 2
RD = 5 – 3 – 2 = 0, Hence, the frame is statically determinate

(c) r = 4, e = 3, s = 1
RD = 4 – 3 – 1 = 0, Hence, the frame is statically determinate

(d) r = 5, e = 3, s = 0
RD = 5 – 3 – 0 = 0, Hence, the frame is statically indeterminate to the 2nd order

(e) r = 5, e = 3, s = 1
RD = 5 – 3 – 1 = 0, Hence, the frame is statically indeterminate to the 1st order

(f) r = 3, e = 3, s = 1
RD = 3 – 3 – 1 = -1, Hence, the frame is unstable

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

(a) Warning Signs

(b) Regulatory Signs (Prohibitory)

(c) Regulatory Signs (Mandatory)

(d) Informative Signs

(e) Traffic Lights/Signals

Types of staircase

Terminologies in staircase design

Ideas on the selection of staircase dimensions

Structural design of a staircase

Sample design of reinforced concrete staircase

Loading of the staircase

Flexural design of the staircase span

Check for deflection

Shear design

Staircase reinforcement details

Loading on the precast seating units

Loading on the raker beams

Structural Analysis

Structural Design

Worked Example

Canonical Equations

Steps in the Analysis of Structures Due to Sinking of Supports

Worked Example

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