Analysis of Three-Hinged Arch Structures

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December 3, 2020

Arches are important structural elements in engineering that provide economical solutions in buildings and bridges. Three-hinged arch structures are pinned at the supports (springings) and somewhere along the barrel, which is usually at the crown. The structural analysis and design of three-hinged arches involve the determination of the internal stresses (bending moment, shear force, axial force, and torsion in the structure due to externally applied load, and providing adequate sections to resist the applied load.

In horizontal beams supporting uniformly distributed load, the bending moment increases with the square of the span and hence they become uneconomical for long-span structures. In such situations, arches could be advantageously employed, since they would develop horizontal reactions, which in turn reduce the bending moment.

A three-hinged arch, which is usually made from steel or timber, is statically determinate. Unlike statically indeterminate arches, they are not affected by differential settlement or temperature changes. Three-hinged arch structures have three natural hinges as the name implies. The two supports are hinged, and another internal hinge is usually located at the crown.

A three-hinged arch has four unknown reactions, i.e., two vertical reactions and two horizontal reactions at the supports. For their determination, three equilibrium equations can be formulated considering the whole of the structure.

Since it is well known that the bending moment at any internal hinge is zero, the internal hinge in the barrel of the arch provides an additional equation for the equilibrium of any part of the system. This means that the sum of the moments of all external forces, which are located on the right (or on the left) part of the structure with respect to the internal hinge is zero. Therefore, a three-hinged arch is a geometrically unchangeable and statically determinate structure. The figure below shows a three-hinged Bolt-laminated Ekki Timber bridge at Finowfurt, Germany.

The peculiar feature of arched structures is that horizontal reactions are induced even when the structure is subjected to vertical load only. These horizontal reactions under vertical loading A_x = B_x = H are called the thrust of the structure. At any cross-section of the arch, bending moments, shear, and axial forces are developed. However, the bending moments and shear forces are smaller than corresponding internal forces in a simply supported beam covering the same span and subjected to the same load.

This fundamental property of the arch is thanks to the thrust developed. Thrusts in both supports are pointed towards each other and consequently reduce the bending moments that would arise in beams of the same span and load configuration. The two parts of an arch may be connected by a tie. In this case in order for the structure to remain statically determinate, one of the supports of the arch should be supported on a roller.

Solved Example on Analysis of a Three-hinged Arch

For the parabolic arch that is loaded as shown below, compute the support reactions and plot the internal stresses diagram for the identified sections. The arch is hinged at points A, B, and C.

SOLUTION
Geometrical properties of the arch
The ordinate (y) at any point along a parabolic arch is given by;

y = [4y_c (Lx – x²)] / L²
Where;
y_c = Height of the crown of the arch from the base
L = Length of arch
x = Horizontal ordinate of interest
Hence, y = [4 × 10 (45x – x²)] / 45²

The general equation of the arch now becomes;
y = (8/9)x – (8/405)x² —————– (1)

Differentiating equation (1) with respect to x
dy/dx = y’ = (8/9) – (16/405)x —————— (2)

From trigonometric relations, we can verify that;
Sin θ = y’/[1 + (y’)²]^0.5 —————- (3)
Cos θ = 1/[1+ (y’)²]^0.5 —————- (4)

From the above relations, we can carry out the calculations for obtaining the geometrical properties of the arch structure.

Let us consider point A (support A of the structure);

We can verify that at point A, x = 0, and y = 0;
From equation (2) above, y’ = 8/9;
Thus,
Sin θ = (8/9)/[1 + (8/9)²]^0.5 = 0.664
Cos θ = 1/[1 + (8/9)²]^0.5 = 0.747

Similarly, let us consider point 3 of the structure;

At point 3, x = 27.5m
From equation (1), we can obtain the value of y as; y = [(8/9) × 27.5] – [(8/405) × 27.5²] = 9.5061m;
The tangent at that point can be obtained from from equation (2); dy/dx = y’ = (8/9) – [(16/405) × 27.5] = – 0.1975
Thus,
Sin θ = (-0.1975)/[1 + (-0.1975)²^0.5 = -0.1937
Cos θ = 1/[1 + (-0.1975)²]^0.5 = 0.9813

For the entire section, it is more convenient to set out the geometrical properties in a tabular form. See the picture below;

Support Reactions

Let ∑M_B = 0; anticlockwise negative
(Ay × 45) – (12 × 22.5 × 33.75) – (25 × 10) – (15 × 6.913) = 0
Therefore, Ay = 210.36 kN

Let ∑M_A = 0; clockwise negative
(By × 45) – (25 × 35) + (15 × 6.913) – (12 × 22.5 × 11.25) = 0
Therefore, By = 84.64 kN

Let ∑M_C^L = 0; anticlockwise negative
(Ay × 22.5) – (Ax × 10) – (12 × 22.5 × 11.25) = 0
22.5Ay – 10Ax = 3037.5 ——— (a)

Substituting, Ay = 210.36 kN into equation (a)
Hence, Ax = 169.56 kN

Let ∑M_C^R = 0; clockwise negative
(By × 22.5) – (Bx × 10) – (25 × 12.5) – (15 × 3.087) = 0
22.5By – 10Bx = 358.8 ——— (b)

Substituting, By = 84.64 kN into equation (b)
Hence, Bx = 154.56 kN

Internal Stresses

Bending Moment
M_A = 0 (hinged support)
M₁ = (210.36 × 7.5) – (169.56 × 5.555) – (12 × 7.5 × 3.75) = 298.294 kN.m
M₂ = (210.36 × 15) – (169.56 × 8.889) – (12 × 15 × 7.5) = 298.181 kN.m
M_C = (210.36 × 22.5) – (169.56 × 10.000) – (12 × 22.5 × 11.25) = 0

Coming from the right hand side;

M_C = (84.64 × 22.5) – (154.56 × 10) – (25 × 12.5) – (15 × 3.087) = 0
M₃ = (84.64 × 17.5) – (154.56 × 9.506) – (25 × 7.5) – (15 × 2.593) = -214.442 kN.m
M₄ = (84.64 × 10) – (154.56 × 6.913) = -222.073 kN.m
M_B = 0 (hinged support)

Shear
Q = ∑V cosθ – ∑H sinθ

Q_A = (210.36 × 0.747) – (169.56 × 0.664) = 44.551 kN
Q₁ = [210.36 – (12 × 7.5)] 0.860 – (169.56 × 0.509) = 17.203 kN
Q₂ = [210.36 – (12 × 15)] 0.959 – (169.56 × 0.284) = -19.040 kN
Q_C^R = [210.36 – (12 × 22.5)] 1.000 – (169.56 × 0.000) = – 59.64 kN
Q₃ = [210.36 – (12 × 22.5)] 0.981 – (169.56 × – 0.193) = – 25.782 kN
Q₄^L = [210.36 – (12 × 22.5)] 0.897 – (169.56 × – 0.443) = – 21.618 kN
Q₄^R = {[210.36 – (12 × 22.5) – 25] × 0.897} – [(169.56 – 15) × (- 0.443)] = 7.452 kN
Q_B = (- 84.64 × 0.747) – [154.56 × (- 0.664)] = 39.402 kN

Axial
N = -∑V sinθ – ∑H cosθ

N_A = (-210.36 × 0.664) – (169.56 × 0.747) = – 266.340 kN
N₁ = – [210.36 – (12 × 7.5)] 0.509 – (169.56 × 0.860) = – 207.085 kN
N₂ = – [210.36 – (12 × 15)] 0.284 – (169.56 × 0.959) = -171.230 kN
N_C = – [210.36 – (12 × 22.5)] 0.000 – (169.56 × 1.000) = -169.560 kN
N₃ = -{[210.36 – (12 × 22.5)] × – 0.193} – (169.56 × 0.981) = -177.849 kN
N₄^L = -{[210.36 – (12 × 22.5)] × – 0.443} – (169.56 × 0.897) = – 178.516 kN
N₄^R = -{[210.36 – (12 × 22.5) – 25] × – 0.443} – [(169.56 – 15) × (0.897)] = – 176.136 kN
N_A = – (-84.64 × – 0.664) – (154.56 × 0.747) = – 171.657 kN

Internal Stresses Diagram (Not to scale)

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Deflection of Statically Indeterminate Frames

By

Ubani Obinna Uzodimma

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May 23, 2022

The deflection of statically indeterminate frames can occur under the action of externally applied loads, temperature difference, or differential settlement of support. These deformations are often expressed in the form of deflections (translations) and rotations (slope). The knowledge about the deformation of structures is very important in that we normally verify the serviceability limit state of structures through them. By being able to limit the deflections in a building, we can make the appearance pleasant, and protect the supported partitions and finishes from cracking.

The definition of slope and deflection is represented in the picture below, showing a simply supported beam deflecting under the action of a concentrated load P.

One of the methods of computing the deflection of statically indeterminate frames is the force method (virtual work method). In this approach, the frame is analysed using the force method, and the bending moment diagram due to the externally applied load obtained. Furthermore, a unit virtual load is placed on the basic system of the structure at the point where the deflection is sought, and analysed to also obtain the bending moment diagram. Using Verecschagin’s rule, it is possible to combine the bending moment diagram from the externally applied load with the moment diagram due to the virtual load to obtain the deflection at that point.

To learn how to apply force method to the analysis of indeterminate frames, click HERE
To learn about Vereshchagin’s rule, click HERE

In this post, we are going to calculate the vertical deflection at point B of the frame loaded as shown below. A little consideration will show that the frame is indeterminate to the 3rd order. See previous posts on how to verify the determinacy of structures. Take EI = Constant

(1) Step 1
Reduce the structure to a basic system. A basic system is a system that is statically determinate and stable. The adopted basic system is as shown below;

(2) Step 2
Replace the support reactions with unit loads and analyse each load case independently. Furthermore, plot the bending moment due to the individual virtual loads. These diagrams can be plotted without finding support reactions. The different load cases and their corresponding bending moment diagrams are shown below;

CASE 1

CASE 2

CASE 3

(3) Step 3

Place the external load on the basic system, and plot the bending moment diagram.

(4) Step 4
Combine the various bending moment diagrams using the Verecschagin’s rule, and obtain the influence coefficients.

δ₁₁ = Deflection at point 1 due to unit load at point 1

δ₁₁ = 1/3(3 × 3 × 5) + (3 × 3 × 6) + 1/3(5 × 5 × 5) + 1/3(3 × 3 × 3) = 119.667/EI

δ₁₂ = δ₂₁ Deflection at point 1 due to unit load at point 2

δ₁₂ = 1/3(3 × 4 × 5) + 1/2 [4 × (4 +10) × 6] – 1/2(5 × 5 × 10) – 1/2(3 × 3 × 10) = 66/EI

δ₁₃ = δ₃₁ Deflection at point 1 due to unit load at point 3

δ₁₃ = -(3 × 3 × 6) + 1/6 [5 × (22 + 6) × 5] – 1/6 [3 × (6 + 6) × 3] = 44.667/EI

δ₂₂ = Deflection at point 2 due to unit load at point 2

δ₂₂ = 1/3(4 × 4× 5) + 1/6 [4(8 + 10) + 10(20 + 4) ] × 6 – (8 × 8× 10) = 1138.667/EI

δ₂₃ = δ₃₂ Deflection at point 2 due to unit load at point 3

δ₂₃ = – 1/2 [3 × (4 +10) × 6] – 1/2 [10 × (11 + 3) × 8] = -686/EI

δ₃₃ = Deflection at point 3 due to unit load at point 3

δ₃₃ = 1/3(3 × 3× 3) + (3 × 3× 6) + 1/6 [11(22 + 3) + 3(6 + 11) ] × 8 = -497.667/EI

δ₁₀ = Deflection at point 1 due to externally applied load

δ₁₀ = 1/3(160 × 8) × 3 = -1280/EI

δ₂₀ = Deflection at point 2 due to externally applied load

δ₂₀ = 1/3(160 × 8 × 10 ) = -128/EI

δ₃₀ = Deflection at point 3 due to externally applied load

δ₃₀ = 1/12 [160 × (33 + 3) × 8] = – 384/EI

(5) Step 5
Insert the influence coefficients into the appropriate canonical equation and solve for X₁, X₂, and X₃ which are the support reactions at the points where they are required.

119.667X₁ + 66X₂ + 44.667X₃ = 1280
66X₁ + 1138.667X₂ – 686X₃ = – 4266.667
44.667X₁ – 686X₂ + 497.667X₃ = 3840

On solving;
X₁ = 8.196 kN;
X₂ = -0.09887 kN;
X₃ = 6.844 kN

(6) Step 6
Obtain the final bending moment values and plot the diagram

Final Moments

M_i = M₀ + M₁X₁ + M₂X₂ + M₃X₃

M_A = -160 + (8.196 × 5) – (-0.09887 × 10) + (6.844 × 11) = -42.7473 kNm
M_B = (8.196 ×- 3) – (-0.09887 × 10) + (6.844 × 3) = -3.0673 kNm
M_C^B = (8.196 ×- 3) – (-0.09887 × 4) + (6.844 × 3) = -3.660 kNm
M_C^L = (8.196 × 3) + (-0.09887 × 4) =24.19252 kNm
M_C^R = (6.844 × 3) = 20.532 kNm

The final bending moment diagram is shown below.

(7) Step 7
Place a unit vertical load at point B of the basic system and plot the bending moment diagram

(8) Step 8
Combine the resulting bending moment diagram with the final moment diagram form the externally applied load to obtain the deflection at point B. However, you should know that the complex bending moment on our final bending moment diagram can actually be split as shown below;

The subsequent combination to obtain the vertical deflection at point B is as shown below;

Deflection of statically indeterminate frames

Therefore, the vertical deflection at point B;
δ_B = 1/3(42.747 × 8 × 8) – 1/3(40 × 8 × 8) + 1/6(3.067 × 8 × 8) = 91.318/EI metres

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Blast Resistant Structures: 21 Things You Should Know

By

Ubani Obinna Uzodimma

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February 13, 2020

Explosives continue to be the terrorist’s preferred weapon, since they are destructive, relatively easy to obtain or fabricate, and still comparatively easy to move surreptitiously on the ground and by sea. Terrorists are also well aware that explosives produce fear in the general population far beyond the geographical location of their intended target.

On the 26th of August, 2011, a car bomb attacked the UN Building in Abuja, the capital of Nigeria, after breaking through two security barriers. The attack claimed about 21 lives, and wounded about 60. A wing of the building collapsed, and the ground floor was severely damaged.

There seems to be a sense of anxiety about the vulnerability of our buildings, bridges, tunnels, and utilities in the midst of numerous recognized international social and political instabilities, and given the potential for domestic groups and individuals to seek influence and create disruption by resorting to violent means. As a result, consultants designing buildings may be expected to provide advice and sometimes specific enhancements in response to quantifiable threats, as well as perceived vulnerabilities.

This post explains 21 basic things you should know about the design of blast resistant structures.

(1) An explosion is a violent thermochemical event

It involves supersonic detonation of an explosive material, violently expanding hot gases, and radiation of a shock front that has peak pressures that their magnitude is higher than those that buildings normally experience under any other loadings.

(2) Building an effective security and risk assessment team

A comprehensive risk assessment for blast-protection design involves close collaboration among city planners, architects, engineers, blast consultant subject matter experts, and law enforcement security professionals. In this group effort, collaborating professionals assess and select the security measures needed to detect, deter, prevent, defeat, mitigate, or recover from terrorists’ bomb attacks.

(3) Nature of blast impact

When the shock front radiating from a detonation strikes a building component, it is instantaneously reflected. This impact with a structure imparts momentum to exterior components of the building. The associated kinetic energy of the moving components must be absorbed or dissipated in order for them to survive. Generally, this is achieved by converting the kinetic energy of the moving facade component to strain energy in resisting elements.

(4) Mmmm… Not the whole structure may collapse

Structures as a whole generally are not pushed over by a common explosion. The overall mass of a structure is usually large enough to keep the kinetic energy imparted to the structure as a whole small enough that it can be absorbed by the multiple elements that would need to fail before the building topples.

(5) Stand-off distance (assumed location of explosion) is very important during assessment

Distance is the single most important asset to a structural engineer designing for blast resistance. The farther the explosion is from the structure, the lower are the effects that the structure must resist. Further, there often is merit to the construction of blast walls or line-of-sight barriers to add protection to a facility.

(6) Keep it confidential

When blast-resistant designs are for the security and safety of a facility/building in response to a threat of a malevolent attack, information about the assumed size and location of an explosion should be kept confidential. This information could be useful to an aggressor because it can reveal a strategy to overwhelm the designed defenses.

(7) Facades and fenestration can turn to missiles

The catastrophic fracturing of building materials such as facades, windows, glazing, and leakage of pressure in response to a blast often can be destructive to the interiors of buildings, even when the facades of those buildings are designed to resist the effects of an explosion. These can turn to missiles and occupant injuries are very possible should the design base explosion occur. Except when the most restrictive approaches to blast-resistant design are employed (e.g., with elastic response, so a building can remain functional). Airborne glass fragmentation continues to be identified as one of the most lethal aspects of building component responses to explosive events.

(8) Mass of a building and occupancy load can be beneficial

Blast effects usually are impulsive, meaning that they impart velocity to objects through development of momentum. With momentum being proportional to the product of mass and velocity, and kinetic energy being proportional to the product of mass and velocity squared, the larger the mass, the smaller the velocity and, hence, the smaller the energy that must be dissipated through strain. Therefore, additional mass to a building can be beneficial during an explosion.

(9) Provision for column failure and redistribution of stresses

Consultants designing for blast often provide for the possibility that a column will be severely damaged by an explosion, in spite of our best efforts at prevention. When consultants assume that a column has lost its strength, they must develop alternative load paths to prevent a collapse from progressing from the initially damaged column through the structure. One form of alternative resistance involves making beams strong and ductile enough to span over the area of damage, thereby redistributing the load on the damaged column to adjacent columns.

(10) Emergency response considerations

Designers working to enhance blast resistance must consider occupant exit and the needs of emergency responders. Blast resistance invariably includes windows with blast-resistant glass. By definition such glass is difficult to break. Firefighters will need to use special tools and engage unusual tactics to fight a fire in a building that is difficult to enter and vent, and that has features that inhibit extraction of trapped occupants. Designers might need to compensate for blast-resistance features, or enhance fire resistance.

(11) Magnitude of design pressure

Designers need to understand that the magnitudes of the pressures that an explosion imparts to a structure are highly dependent on the nature of the explosive material, the shape and casing of the device, the size and range of the explosion, the angle of incidence between the advancing shock front and the impacted surface, the presence of nearby surfaces that restrict the expansion of hot gases or that reflect pressure fronts, and the robustness of the impacted surface itself.

(12) Extensive damage of structures during a blast

In many explosions that cause extensive destruction, the damage develops in two phases:
(i) the energy released by the explosion degrades or destroys important structural elements, and
(ii) the damaged structure is unable to resist gravity and collapses beyond the area of initial damage.

(13) Individual elements are often treated as nonlinear SDOF elements

It is common in blast-resistant design to treat individual elements as single degree-of-freedom nonlinear systems. Performance is judged by comparison of response to limiting ductility factors (i.e., the ratio of peak displacement to displacement at yield) or support rotations, with the response calculated as though the structural element were subjected to a pressure function while isolated from other structural influences.

(14) Column integrity is the key

Typically, columns are the component whose integrity is key to sustaining the capability of a building’s structural system to survive a blast load. As such, attention to ensuring their survival should have the highest priority when the blast resistance of a building is to be upgraded.

(15) Beware of fragmentation

Injury and structural damage during an explosion may result not only from the direct pressure and impulse of an air blast, but also from the impact of flying objects called debris and fragmentation. Military weapons are typically explosive charges with some type of metal casing. Upon detonation, this case is ruptured and expelled as fragmentation at high velocity. Similarly, terrorist devices may be embedded with objects such as ball bearings or nails that will be ejected at high velocity. Other fragmentation may occur as objects interact with the blast wave created from a detonation and become airborne.

(16) Blast design is different from seismic design

The conventional approach to blast design is similar to that for seismic design, in two important ways:

(i) both loadings clearly are dynamic and, hence, solutions are energy-based, and

(ii) the way we detail structural elements determines the effective loads for which structures must be designed (meaning, we limit the strength we need to supply by allowing post-elastic behavior to dissipate energy).

However, blast loading, with its extremely fast rise time and usually short duration, is either dynamic or impulsive, depending on the nature of the explosive, its distance from the subject structure, and the level of confinement that the structure creates for the expanding hot gases.

(17) You may have to tolerate more damage

While we tolerate some damage in earthquakes, to dissipate energy, we usually allow more damage for blast events. We expect facades to sustain severe damage. In fact, blast-resistant design often tolerates breaching of the building enclosure (with attendant risk of fatalities) and even sometimes partial collapse of buildings.

(18) Reversal of stresses on floor slabs

Blast-related upward impulses on floor slabs can reverse force distributions in these structural elements. In systems that are not strong and ductile enough for these reversed forces, blast-induced deflection can fracture structural elements that are required to resist gravity loads. Hence, floor systems can fail after the direct effects of the blast pass and the slab falls back downward under the influence of gravity.

(19) Consider the effectiveness of post-event evacuation functions

While the immediate devastating effects of explosive events can be limited by protective design strategies, post-event evacuation functions and the subsequent activities of rescue and recovery are dependent upon the integrity of stairs, vertical transportation systems, power, lighting, voice communication, smoke management, and other systems. These systems can be placed out of service throughout the building or suffer disproportionate damage if they are not specifically designed in response to the project design basis explosive threats.

(20) Engage only licensed and experienced professionals

Blast resistance designers must be licensed design professionals who are knowledgeable in the principles of structural dynamics and experienced with their proper application in predicting the response of elements and systems to the types of loadings that result from an explosion, or they must work under the direct supervision of licensed professionals with appropriate training and experience.

(21) Threat Elimination remains the overall best option

Finally, elimination of the explosive event threat and the reduction of building materials and contents that have the opportunity to participate as fuel sources in the fire development triad of oxygen, fuel load, and an ignition source remains the most attractive, competent, and cost-effective process in developing designs for facilities on any site. However, based on the threat assessment, it may be unreasonable to assume that the facility will not be exposed to explosive event effects.

Disclaimer:
I do not claim ownership to the pictures used in this post. All pictures used here belong to their respective owners, of which it is a little bit impractical to recognise them all.

Information for this article is extracted from: Donald O. Dusenberry (2010): Handbook for Blast-Resistance Design of Buildings. John Wiley and Sons Inc.

Thank you for visiting, and may we remain determined in our struggle to make the world free from hatred, discrimination, violence, terrorism, poverty, and crime. God bless you as you discuss and share.

Your friend,
Ubani Obinna Ranks

Application of Finite Difference Method to the Elastic Analysis of Simply Supported Thin Plates

By

Ubani Obinna Uzodimma

-

December 5, 2020

A plate is a structure whose thickness is comparatively smaller compared to its lateral dimensions. A typical example of a plate is the floor slab commonly found in our buildings. Other examples that you can think of are sheets of steel, the panes of glass windows in your homes, or even sheets of plywood. A typical concrete plate is shown in the picture below. In this post, we are interested in the linear analysis of isotropic and homogeneous plates.

Plates are used as structural members to support and transmit loads and as such, they can be subjected to bending moment, shear force, axial force, and twisting moments. The load-carrying action of a plate is similar, to a certain extent, to that of beams or cables and therefore, plates can be approximated by a gridwork of an infinite number of beams, or by a network of an infinite number of cables, depending on the flexural rigidity of the structure (Ventsel and Krauthammer, 2001).

Plates generally behave like two-dimensional structures. By implication, the loads applied on a plate are carried in two directions, and given the significant twisting rigidity of plates, a plate is stiffer than a beam of comparable span and thickness (Ventsel and Krauthammer, 2001). This two-dimensional structural action of plates results in lighter structures, and therefore offers numerous economic advantages. The bending properties of a plate depend greatly on its thickness than on any of its dimensions (Timoshenko and Woinowsky-Krieger, 1987).

Plates deform when loaded. If the deflections of a plate are small compared to its thickness, then the ‘small deformation theory’ applies, which makes the following assumptions:

1.There are no stresses in the middle plane of the plate. Hence, there are no stresses in the neutral axis during bending.
2.The points that are lying perpendicularly to the middle plane before bending remain perpendicular to the middle plane after bending.
3.The normal stresses in the transverse direction can be neglected.

By following these assumptions, the stresses on the plates can be expressed as a function of the deflection of the plate w, which is a function of the two coordinates (x,y) of the plate (Timoshenko and Woinowsky-Krieger, 1987). The governing differential equation for the deflection of thin plate under pure bending is based on the biharmonic equation shown below;

Before now, the common methods of analysis of thin plates have been the tedious classical solution using either the trigonometric or double series solution (Aginam et al, 2012). These days, numerical methods like Finite Element Method, Boundary Element Method, Finite Strip Method, Gridwork Method, Finite Difference Method, etc have become very popular due to their very easy modelling and programming into computer software.

However, finite element analysis is more prominent. In this article, the application of the finite difference method to the pure bending analysis of a thin plate simply supported on all sides has been presented. The result obtained has been compared with the results from other types of solutions.

Finite Difference Method of Analysis of Thin Plates

The finite difference method is a numerical solution to partial differential equations. When we have a function of two real variables f(x,y) that satisfies a given differential equation, we can evaluate f(x,y) numerically by laying a rectangular grid over the domain, and evaluate f(x,y) at the nodes – the points where the lines of the x-axis and y-axis intersect (For basic introduction you can see, Advanced Engineering Mathematics, Stroud and Booth, 2003).

Consider the 2-D network shown below

Let the vertical spacing of the grid be h, and the horizontal spacing be k.
From our knowledge of numerical approximation of derivatives using Taylor series expansion;

∂w/∂x = (Δx.w)/2h = (w_b – w_a)/2h = 1/2h [w (x + h, y) – w(x – h, y)]
∂w/∂y = (Δy.w)/2k = (w_d – w_g)/2k = 1/2k [w(x, y + k) – w(x, y – k)]

Analogously, for higher-order differentials;
∂²/∂x² = ∂/∂x (∂w/∂x) = (Δx² w)/h² = (w_b – 2w ₀ – w_a)/h² = 1/h² [w(x+h, y) – 2w(x,y) + w(x-h, y)]
∂²/∂y² = ∂/∂y (∂w/∂y) = (Δy² w)/k² = (w_d – 2w₀ – w_g)/k²= 1/h² [w(x, y + k) – 2w(x, y) + w(x, y – k)]

For the pure bending biharmonic equation;
(∂⁴ w)/∂x⁴ + (2∂⁴ w)/(∂x² ∂y² ) + (∂⁴ w)/∂y⁴ = P(x,y)/D, we have;

(20w₀ ) – 8(w_a + w_b + w_d + w_g) + 2(w_c + w_e + w_f ) + w_h ) + 1.0(w_i +w_j + w_k + w_l )= (P₀ α⁴)/D

These expressions are often represented as patterns or computational molecules for the easy computation of the numerical solutions. The biharmonic equation pattern for computing the displacement at the interior nodes are as shown as below.

Support Conditions
For simply supported and fixed supports, we know that the deflection at those points is equal to zero. In order to formulate the boundary conditions, let us consider the interior point 0 in the figure below.

You should know that at that interior point, the biharmonic equation holds true. Let us assume that points a, c, f constitutes the boundary of the plate that is simply supported, and let node i represent a fictitious node created outside the plate boundary. A little consideration will show that for w_a to be zero, we have to eliminate the deflection w_i, obtained by the continuation of the network beyond the boundary of the plate. This is easily done by setting w₀ = -w_i for simply supported plates (See David Johnson, Advanced Structural Mechanics, 2000).

Solved Example
Let us show how the finite difference method can be applied in the analysis of thin plates subjected to uniform lateral pressure of 5 kN/m². The plate is square with dimensions of 6m x 6m and simply supported on all sides. The data of the plate is as shown below;

The plate has been represented with gridlines that have 6 divisions on all the planar dimensions. Therefore, h = k = 1.0. This shows that the gridline aspect ratio (α) = 1.0. The gridlines have been extended by fictitious nodes, in order to capture the boundary conditions. The nodes have been numbered, and all the points where the deflections are expected to be the same due to the symmetry of the plate and the loading condition have been given the same numbering.

By implication, we have seven distinguished nodes, hence, we are going to solve a 7 x 7 matrix, which can be readily solved using Microsoft Excel or MATLAB. The computational molecule (biharmonic operator/pattern) used is the one shown below. We place the computational molecule node by node and generate the appropriate equation for each node. After the equations are generated, they are arranged as appropriate and solved. The results obtained from this method have been compared with the results from the classical solution and finite element analysis.

Since h = k, the representative computational molecule is given below;

P(x,y)/D = (P₀ α⁴)/D
D = (Eh³) / [12(1- v²)] = (21.7 × 10⁶ × 0.12³) / [12(1 – 0.2²)] = 37497.6/11.52 = 3255 kN/m²
(P₀ h⁴)/D= (5 × 1.0⁴/3255= 0.001536098

Node 1
20w₁ – 8(4w₂) + 2(4w₄) + 1(4w₃) = 0.001536098

Node 2
20w₂ – 8(2w₄ + w₁ + w₃) + 2(2w₂ + 2w₅) + 1(2w₇ + w₂ + 0) = 0.001536098

Node 3
20w₃ – 8(2w₅ + w₂ + 0) + 2(2w₄ + 0) + 1(2w₆ + w₁ + w₃) = 0.001536098

Node 4
20w₄ – 8(2w₄ + w₅ + w₇) + 2(2w₃ + w₁ + w₆) + 1(2w₇ + 0 + 0) = 0.001536098

Node 5
20w₅ – 8(w₃ + w₄ + w₆) + 2(w₇ + w₂) + 1(w₅ + w₂ – w₅ + 0) = 0.001536098

Node 6
20w₆ – 8(w₅ + w₇) + 2(w₄) + 1(2w₃ – w₆ – w₆) = 0.001536098

Node 7
20w₇ – 8(w₃ + w₄ + w₆) + 2(w₂ + w₅) + 1(w₂ + w₇ – w₇ + 0) = 0.001536098

Arranging in matrix form;

On solving;
w₁ = 0.008059 m = 8.059mm (maximum deflection at the midspan of the plate)
w₂ = 0.007062 m
w₃ = 0.004202 m
w₄ = 0.006192 m
w₅ = 0.00369 m
w₆ = 0.002211 m
w₇ = 0.00369 m

On comparing with other methods;

Classical Solution from (Timoshenko and Woinowsky-Krieger, 1987, Table 8, Page 120)
Maximum deflection = 0.00406qa⁴/D = (0.00406 × 5 × 6⁴)/3255 = 0.00808258 m

A little consideration will show that there is a percentage difference of 0.2917% between the answer from the classical method and FDM.

FEA result from Staad Pro (Mesh Size Division = 11 x 11)

Maximum deflection is 0.008228 m
A little consideration will show that there is a percentage difference of 2.0539% between the answer from FEM and FDM.

Conclusion
We can see that the finite difference method can be a handy solution to the analysis of thin plates. The method can be easily programmed on computer software, and the method generally yields lower bound solution.

REFERENCES
(1) Aginam C.H., Chidolue C.A., Ezeagu C.A.(2012): Application OF Direct Variational Method in the Analysis of Isotropic Thin Rectangular Plates. ARPN Journal of Engineering and Applied Sciences VOL. 7, NO. 9, ISSN 1819-6608 pp 1128 – 1138
(2) David Johnson (2000): Advanced Structural Mechanics. Thomas Telford Publishing ISBN: 0 7277 2860 1
(3)Timoshenko S. and Woinowsky-Krieger S. (1987): Theory of plates and shells (2nd Edition). McGraw Hill Book Company
(4)Ventsel E., Krauthammer T. (2001): Thin Plates and Shells Theory, Analysis, and Applications. Marcel Dekker, Inc New York. ISBN: 0-8247-0575-0

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Example on the Analysis of Statically Determinate Frames (Part 2)

By

Ubani Obinna Uzodimma

-

February 13, 2020

If you missed the PART 1 of this post, click HERE

Problem 1
A statically determinate frame is loaded as shown above. There are internal hinges at B and D, while member EF is hinged at point F. Draw the bending moment, the shear force, and the axial force diagrams.

SOLUTION

Notations
M_i^R – Bending moment at point i, just to the right
M_i^L – Bending moment at point i, just to the left
Q_i^R – Shear force at point i, just to the right
Q_i^L – Shear force at point i, just to the left

(a) Support Reactions

Let ∑M_E = 0 (clockwise positive)
10Ay + M_A – (25 × 5) – [(10 × 3²) / 2) + [(10 × 3²) / 2) + (15 × 5) = 0
10Ay + M_A = 50 ————– (1)

Let ∑M_D^L = 0 (clockwise positive)
7Ay + M_A – 4Ax – (25 × 2) = 0
7Ay + M_A – 4Ax = 50 ————— (2)

Let ∑M_B^Below = 0 (clockwise positive)
3Ay + M_A – 4Ax = 0
3Ay + M_A– 4Ax = 0 ————— (3)

From equation (1);
M_A = 50 – 10Ay —————— (1a)

Substituting the value of M_A in equation (1a) into equations (2) and (3), we obtain respectively;
7Ay – 4Ax + 50 – 10Ay = 50
– 3Ay – 4Ax = 0 ————– (4)

3Ay + 50 – 10Ay – 4Ax = 0
– 7Ay – 4Ax = – 50 —————– (5)

Solving equations (4) and (5) simultaneously;
Ay = 12.5 KN
Ax = – 9.375 KN

Hence;
M_A= 50 – 10(12.5) = – 75 KNm

At this point, it is very possible for us to sum up vertical and horizontal forces in order to obtain the rest of the reactive forces, but let us still obtain them by taking moments (this is very useful since it can serve as a check for correctness of results).

Let ∑M_D^R = 0 (anticlockwise positive)
3Ey – 4Ex – (15 × 8) – (10 × 6 ×3) = 0
3Ey – 4Ex = 300 —————- (6)

Let ∑M_B^R = 0 (anticlockwise positive)
7Ey – 4Ex – (15 × 12) – (10 × 6 ×7) – (25 ×2) = 0
7Ey – 4Ex = 650 —————– (7)

Solving equations (6) and (7) simultaneously;
Ey = 87.5 KN
Ex = – 9.375 KN

Equilibrium check;
∑Fy↓ = 25 + (10 × 6) + 15 = 100 KN
∑Fy↑ = 12.5 + 87.5 = 100 KN

This shows that the requirement for static equilibrium is satisfied,. You can also verify for the horizontal reaction at the supports.

Before we go ahead and obtain the bending moment and shearing forces, let us obtain the axial forces in members DE and DF. Since they are pinned at both ends, they are primarily trusses and will not develop any bending moment or shear force.

γ = tan^(-1)⁡(4/3) = 53.130°
cos γ = 0.6
sin γ = 0.8

∑Fy = 0
87.5 + F_EDsin⁡ γ + F_EFsin⁡γ = 0
0.8F_ED + 0.8F_EF = – 87.5 —————– (8)

∑Fx = 0
9.375 – F_EDcos ⁡γ + F_EFcosγ = 0
-0.6F_ED + 0.6F_EF = – 9.375 ——————- (9)

Solving equations (8) and (9) simultaneously;
F_ED = – 46.875 KN (compression)
F_EF = – 62.5 KN (compression)

These forces are now resolved into their vertical and horizontal components at the points they are attached to the beam as shown below;

(b) Internal Stresses

MEMBERAB (0 ≤ z ≤ 5)
Bending Moment
Mz = (Ay.cos⁡γ.z) – (Ax.sin⁡γ.z) – M_A
Mz = (12.5 × 0.6 × z)- (-9.375 × 0.8 × z) – 75 = 15z – 75

At z = 0;
M_A = -75 KNm

At z = 5m
M_B^Below = 15(5) – 75 = 0

Shear Force
Qz = (Ay.cos⁡γ) – (Ax.sin⁡γ)
Qz = (12.5 × 0.6) – (-9.375 × 0.8) = 15 KN (constant shear force all through the member)

Axial Force
Nz = – (Ay.sin⁡γ) – (Ax.cos⁡γ)
Qz = – (12.5 × 0.8) – (-9.375 × 0.6) = – 4.375 KN (constant axial force all through the member)

MEMBER BD (3 ≤ x ≤ 7)

Bending Moment
The general moment equation for the member is given by;
Mx = Ay.x + (Ax × 4) – M_A – 25(x – 5)
Mx = 12.5x + (9.375 × 4) – 75 – 25(x – 5)
Mx = 12.5x – 37.5 – 25(x – 5)

You can however notice that for section BC, the last term of the general equation will not be involved since we cannot have a negative distance on the beam. Therefore it is not very wise to expand the equation yet.

At x = 3m;
M_B^R = 12.5(3) – 37.5 = 0

At x = 5m;
M_C = 12.5(5) – 37.5 – 25(5 – 5) = 25 KNm

At x = 7m;
M_D = 12.5(7) – 37.5 – 25(7 – 5) = 0

Shear Force
Q_B^R – Q_C^L = 12.5 KN
You can verify that the valid equation for moment at the section (B^R – C^L) is given by;

Mx = 12.5x – 37.5

If we differentiate the equation for moment, we obtain the equation for shear;
(∂Mx)/∂x = Qx = 12.5 KN

You can also verify that the valid equation for moment at the section (C^R – D^L) is given by;
Mx = 12.5x – 37.5 – 25(x – 5) = -12.5x + 87.5
(∂Mx)/∂x = Qx = -12.5 KN
Hence, Q_C^R – Q_D^L = -12.5 KN

Axial Force
You can verify that for section B^R – D^L
N_B – Ax = 0
Hence N_B = 9.375 KN (tension)

MEMBER DF (7 ≤ x ≤ 13)

Bending Moment
Mx = Ay.x + (Ax × 4) – M_A – 25(x – 5) + 37.5(x – 7) – [10(x – 7)²) / 2
Mx = 12.5x + (9.375 × 4) – 75 – 25(x – 5) + 37.5(x – 7) – [10(x – 7)²) / 2
Mx = 12.5x – 37.5 – 25(x – 5) + 37.5(x – 7) – 5(x – 7)²

At x = 7m;
M_B^R = 12.5(7) – 37.5 – 25(7 – 5) = 0

At x = 10m (mid-span);
M_mid = 12.5(10) – 37.5 – 25(10 – 5) + 37.5(10 – 7) – 5(10 – 7)² = 30 KNm

At x = 13m;
M_F^L = 12.5(13) – 37.5 – 25(13 – 5) + 37.5(13 – 7) – 5(13 – 7)² = – 30 KNm

Shear Force
You can verify that the valid equation for moment at section D just to the right (D^R) is;
Mx = 12.5x – 37.5 – 25(x – 5) + 37.5(x – 7) = 25x – 175
(∂Mx)/∂x = Q_D^R = 25 KN

The valid equation for moment at section F just to the left (F^L) is;
Mx = 12.5x – 37.5 – 25(x – 5) + 37.5(x – 7) – 5(x – 7)²
Mx = 12.5x – 37.5 – 25x + 125 + 37.5x – 262.5 – 5(x² – 14x + 49)
Mx = -5x² + 95x – 420
(∂Mx)/∂x = Qx = -10x + 95

Maximum span moment
The maximum moment at the span can be obtained by considering the bending moment equation unique for the section.

Mx = -5x² + 95x – 420
(∂Mx)/∂x = Qx = -10x + 95

Since the maximum moment occurs at the point of zero shear, let us equate the expression for the shear force to zero, and then solve for x;

Such that x = 95/10 = 9.5m
M_max = -5(9.5)² + 95(9.5) – 420 = 31.25 KNm

At x = 13m;
Q_F^L = -10(13) + 95 = -35 KN

Axial Force
For section D^L – F^R
N – Ax – 28.125 = 0
Hence N = 9.375 + 28.125 = 37.5 KN (tension)

MEMNER FG (coming from the right for simplicity) (0 ≤ x ≤ 2m)
Bending Moment (clockwise negative)
Mx = -15x

At x = 0;
M_G = 0 KNm

At x = 2m
M_F^R = -15(2) = – 30 KNm

Shear Force
Q_G^L – Q_F^R = 15 KN (downward force is positive when coming from the right)

Axial Force
No axial force in the member

Final Internal Stresses Diagram

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

By

Ubani Obinna Uzodimma

-

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.

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;

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.

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 Statically Indeterminate Frames

Analysis of Frames using the Displacement Method

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

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