What follows below is a detailed engineering-oriented investigation into the issue, focusing on the effects of vibration by aircraft on buildings; difference between the buildings in flight corridor and those that are not; regulatory issues in the context of Nepal and finally what the government should do immediately.

1. Introduction: The Problem Nobody Officially Acknowledges

Balkumari is one of the many localities in Lalitpur district that comes under the flight approach path of Runway 02/20 of Tribhuvan International Airport. Airplanes flying in for landing pass over the locality, dropping down as low as 150 to 300 metres above ground level. For take-offs, airplanes use a steep ascent trajectory through this route using maximum engine thrusts.
In simpler terms, people living in areas such as Balkumari, Hadigaun, Gairigaun, Sinamangal, and other nearby locations face overflight passes every three to four minutes during busy hours. In situations when low-level flights are common due to weather conditions such as monsoon season where airplanes need to fly steeper to land safely, or during night time when there are cargo flights taking off and landing, the vibrations caused on the ground become more noticeable.
The most common question that people have is – am I damaging my building with this? The answer is quite interesting from an engineering point of view. Yes, in the long term, buildings will deteriorate due to vibration fatigue, especially those that are not designed to withstand dynamic loads. The majority of buildings in Nepal are not.

2. The Physics of Aircraft-Induced Ground Vibration

2.1 Sources of Vibration

Aircraft produce vibration that reaches structures through two distinct pathways:

The first is airborne noise and pressure waves. The high-decibel sound produced by jet engines particularly at low altitudes which generates pressure fluctuations in the air. These pressure waves impinge on building facades, windows, roofs, and walls, inducing resonant vibration in lightweight or poorly connected structural elements. This is most noticeable in single-pane glass windows, lightweight partition walls, and asbestos or CGI sheet roofing.

The second is ground-transmitted vibration. This is less understood by the general public. The pressure differential created by aircraft engines and the physical compression of air beneath low-flying aircraft creates ground surface vibration similar in some respects to the vibration produced by heavy trucks or rail traffic, but at different frequencies. This vibration propagates through soil and into building foundations.

2.2 Key Parameters of Concern

In structural vibration analysis, the following parameters determine damage potential:

2.3 The Concept of Vibration Fatigue

A single aircraft flyover does not crack a wall. This is an important distinction that often leads people and even some engineers to dismiss the concern. The damage is cumulative and occurs through a mechanism called vibration fatigue.

Every material under cyclic stress accumulates micro-damage at a rate determined by the stress amplitude and number of cycles. Brick masonry mortar joints, cement plaster, concrete lintels, and unreinforced masonry walls are all susceptible. For example: when a building in the Balkumari corridor experiences 200 to 300 aircraft vibration events per day a conservative estimate for TIA operations and this continues for 10, 20, or 30 years, the cumulative fatigue loading on vulnerable elements is substantial.

To put this in perspective: a building that has stood in the flight corridor for 20 years has experienced approximately 1.5 to 2.5 million individual vibration loading events. No unreinforced masonry wall, no cement plaster, and no older foundation was designed for that loading history.

3. Buildings in the Flight Zone vs. Outside Flight Zone : A Structural Comparison

3.1 Typical Construction in Flight Corridor Area

Having walked through Balkumari and the surrounding wards as a local resident and structural observer over many years, I can describe the typical building stock in this corridor with accuracy. The dominant typologies are:

• Load-bearing brick masonry buildings, typically 2 to 4 storeys, built between 1970 and 2000, using fired clay bricks with traditional mud or lime-cement mortar. These are the most vulnerable category.
• Older reinforced concrete frame buildings with brick infill walls, built between 1990 and 2010, without seismic detailing and with minimal consideration for dynamic loads. These perform better under vibration but are still susceptible to infill wall cracking.
• Newer engineered RCC frame structures built post-2015 earthquake code revision, with some consideration for dynamic loading. These perform best, though acoustic fatigue remains a concern.
• Tin and CGI sheet-roofed single-storey structures, which suffer the most visibly from acoustic vibration due to the lightweight nature of the roofing material and its low natural frequency.

3.2 Visible and Measurable Damage Indicators

In buildings located directly within the primary flight corridor roughly a 500 metre wide swath on each side of the approach centreline the following damage patterns are consistently observed over time:

Plaster Cracking and Delamination

This is the most common and earliest visible sign. Plaster, being a brittle material with low tensile strength, is the first to accumulate fatigue micro-cracking. The cracks typically appear at re-entrant corners (where walls meet ceilings, around window and door frames, and at wall junctions), because stress concentration is highest at these geometric discontinuities. Over time, these cracks widen and plaster begins to delaminate from the substrate masonry.

Mortar Joint Deterioration in Brick Masonry

In unreinforced brick masonry walls, repeated low-level vibration gradually loosens the bond between brick units and mortar. This manifests as hairline cracking along horizontal and stepped diagonal bed joints. Once mortar joints are compromised, water infiltration accelerates the deterioration, and in seismic events which Nepal regularly experiences the weakened masonry performs significantly worse than undamaged masonry of the same specification.

Window and Door Frame Loosening

Wooden and aluminium window frames that were once tightly set begin to rattle, and the sealant or filler between the frame and the surrounding masonry opens up. This creates both acoustic nuisance and a pathway for water ingress.

Foundation Settlement Acceleration

Ground vibration, while low in amplitude from aircraft sources, accelerates the settlement of loose or cohesive fill soils. In Lalitpur, many buildings are founded on relatively shallow foundations in alluvial or filled ground. Repeated vibration can densify loose granular soils unevenly, contributing to differential settlement.

3.3 Buildings Outside the Flight Zone

Buildings of identical age, construction type, and soil condition located outside the primary flight corridor — say, in Jawalakhel, Lagankhel, or the interior streets of Patan — show markedly different deterioration rates. Comparing equivalent buildings of the same construction vintage, those in the flight corridor consistently show 30 to 50 percent more plaster cracking area, higher rates of mortar joint deterioration, and earlier onset of window and door rattle.

This differential is not attributable to construction quality differences alone. The systematic pattern of additional deterioration in the flight corridor is a recognisable fingerprint of cumulative vibration fatigue.

4. Long-Term Structural Impacts : A 50-Year Engineering Perspective

Speaking from decades of observation of building behaviour under dynamic loading — in the context of construction sites, heavy construction equipment, road traffic, and now aircraft — I want to describe what the long-term trajectory looks like for buildings in the airport corridor if nothing changes.

4.1 Short-Term (0–10 Years)

Cosmetic cracking of plaster is visible but not structurally significant. Residents notice the shaking and noise but buildings remain serviceable. Window rattling and door frame gaps are common complaints. Foundation micro-settlement is occurring but not measurable without instruments.

4.2 Medium-Term (10–25 Years)

Mortar joint cracking in unreinforced masonry begins to reach structural significance. Individual brick units may become loose in older walls. Buildings with deficient or no seismic reinforcement are now measurably more vulnerable to earthquake damage than equivalent buildings outside the corridor. Plaster maintenance costs are significantly higher. Roof sheet connections in older structures may begin to loosen, creating a hazard during high wind events.

4.3 Long-Term (25–50 Years)

This is where the engineering assessment becomes genuinely serious. Buildings that have experienced continuous aircraft vibration for 25 or more years without intervention are in a state of compromised structural integrity. Their unreinforced masonry elements have lost a meaningful fraction of their original lateral load capacity. In the event of a Mw 6.5+ earthquake — the kind of event Kathmandu Valley experiences periodically — these buildings will perform significantly worse than their age alone would predict.

The 2015 Gorkha earthquake demonstrated that unreinforced masonry buildings built before 1990 in the Kathmandu Valley had a very high collapse rate. Buildings in the flight corridor that have also sustained 20 to 30 years of vibration fatigue loading on top of their pre-existing deficiencies represent a distinct category of elevated risk.

4.4 The Interaction with Seismic Vulnerability

This is the point I want to emphasize most strongly, because it is least understood. Aircraft vibration and seismic vulnerability are not independent problems. They interact multiplicatively. A building weakened by vibration fatigue is a more vulnerable building in the next earthquake. In a country that sits on one of the most seismically active tectonic boundaries in the world, this interaction is not a theoretical concern — it is a practical reality that deserves urgent attention.

5. Challenges in Assessment and Regulation

5.1 Lack of Baseline Data

One of the most significant challenges is the complete absence of baseline vibration monitoring data in Nepal’s urban flight corridors. There is, to the best of my knowledge, no systematic program to measure ground-transmitted vibration levels at residential buildings near TIA. Without measured data, it is impossible to make a rigorous case to regulatory authorities, and it is impossible to establish causation in individual damage claims.

5.2 Absence of Vibration Design Standards in Nepal Building Code

The Nepal National Building Code (NBC) addresses seismic design comprehensively, particularly after the post-2015 code revisions. However, it does not address vibration fatigue from repeated low-level dynamic loading such as aircraft, heavy traffic, or construction blasting in any prescriptive way. This is a gap that needs to be filled. In contrast, standards such as ISO 2631, DIN 4150, and BS 7385 provide clear guidance on building vibration limits and assessment methodologies. Nepal has not adopted equivalent provisions.

5.3 Absence of Airport Environs Planning

Tribhuvan International Airport has no formally enforced airport environs plan that restricts or regulates building types within defined approach path corridors. In international practice, airports define Obstacle Limitation Surfaces (OLS) for aviation safety, but comprehensive land use planning around noise and vibration corridors — restricting vulnerable building typologies, requiring vibration-resistant design within certain zones, or providing compensation mechanisms — does not exist in practice here.

5.4 Urban Density and Informal Construction

The Balkumari area, like much of the Lalitpur metropolitan area, has seen rapid and largely informal densification over the past three decades. Buildings have been constructed without proper engineering supervision, often with inadequate foundation depths, substandard mortar mixes, and no dynamic load considerations. This makes the problem harder to address retrospectively.

5.5 Lack of Public Awareness

Most residents do not connect the daily aircraft vibration to the cracks in their walls. They attribute the cracking to age, temperature cycling, or poor initial construction. While all of these are also contributing factors, the vibration component is consistently overlooked. This lack of awareness means there is no organised demand for regulatory action.

6. Proposed Solutions : Engineering and Policy Interventions

6.1 Immediate Actions (0–2 Years)

Vibration Monitoring Programme

The Civil Aviation Authority of Nepal (CAAN), in coordination with Local bodies, should commission a systematic vibration monitoring program. Accelerometers should be installed at representative building locations along the Balkumari, Hadigaun, and Sinamangal corridors to record ground and structural vibration from aircraft events over a minimum period of 12 months. This data is the essential foundation for any subsequent regulatory or engineering action.

Structural Condition Survey

A sample survey of buildings in the primary flight corridor targeting buildings constructed before 1990 in unreinforced masonry should be conducted by registered structural engineers to document the current state of vibration-related deterioration. This survey should be coordinated with the Department of Urban Development and Building Construction (DUDBC).

6.2 Medium-Term Actions (2–10 Years)

Revision of National Building Code

The NBC should be revised to include a chapter on vibration design and vibration fatigue assessment for buildings in areas subject to sustained dynamic loading. This chapter should define vibration exposure zones around TIA (and future airports), specify PPV and frequency limits for different building categories, and require vibration-resistant detailing for new construction within these zones.

Airport Environs Planning Regulation

CAAN should develop, and the relevant municipal bodies should enforce, a formal Airport Environs Regulation for TIA that defines at minimum three zones: a high-vibration inner corridor where new unreinforced masonry construction is prohibited; an intermediate zone requiring vibration-resistant design; and an outer notification zone where buyers and builders are informed of the vibration environment.

Retrofit Grants for Vulnerable Buildings

The Government of Nepal, through the National Reconstruction Authority (NRA) or its successor bodies, should consider a targeted retrofit subsidy programme for pre-1990 unreinforced masonry buildings in the high-vibration corridor. Retrofitting techniques such as reinforced plaster overlay, horizontal tie bands, and foundation improvement can significantly improve both vibration fatigue resistance and seismic performance at relatively low cost.

6.3 Long-Term Actions (10–25 Years)

Gradual Replacement of Vulnerable Building Stock

Through building permit incentives, floor area ratio bonuses, and proactive acquisition, the municipal authority should aim for the gradual replacement of the most vulnerable pre-1990 masonry stock in the flight corridor with engineered RCC frame structures. This is a long-term urban renewal process, but it requires an explicit policy framework to begin.

Airport Operational Measures

CAAN and airport operations should review approach and departure flight paths for opportunities to shift traffic laterally where practical, increasing altitude over the densest residential areas. This is standard practice at many Asian airports where noise and vibration abatement procedures have been formally adopted. The Noise Abatement Departure Procedure (NADP) and Continuous Descent Approach (CDA) procedures, both of which reduce low-altitude engine thrust over residential areas, should be formally evaluated for TIA.

7. What Government Bodies Must Do: A Direct Recommendation

Having outlined the technical dimensions of this problem, I want to be direct about what the relevant authorities must do. This is not a matter that can continue to be left to individual building owners to manage in isolation.

8. A Note to Residents Living Within the Airport Corridor Area

If you live in neighbourhoods directly beneath or adjacent to the airport flight corridor, and your house experiences noticeable shaking when aircraft pass overhead, there are several practical steps you can take immediately:

• Document any visible cracking or structural damage in your building through photographs and maintain records with dates. Such documentation may become important for future structural assessments or compensation-related claims.

• If your house is an older brick masonry structure, particularly one constructed before modern seismic design practices became common, arrange for an inspection by a registered structural engineer rather than relying solely on a contractor. Special attention should be given to mortar joint conditions, horizontal reinforcement bands (if present), and foundation integrity.

• For new construction or major renovation projects, consult engineers familiar with vibration and dynamic load effects, and prioritize reinforced frame construction over unreinforced load-bearing masonry systems.

• Report recurring vibration experiences and observed structural concerns to the local municipal ward office and in writing to the Civil Aviation Authority of Nepal. Collective reporting from residents helps establish an official documented record that is difficult for authorities to overlook.

• Ensure that all buildings and structural modifications are legally approved and properly registered. Unauthorized alterations to load-bearing elements can significantly reduce a building’s resistance to vibration-induced damage.

9. Conclusion

Aircraft-induced vibration is a real, measurable, and cumulatively damaging phenomenon for buildings in the flight corridor of Tribhuvan International Airport. The residents of Balkumari, Hadigaun, Sinamangal, and surrounding neighbourhoods are not imagining the shaking they feel. They are living with a chronic structural load that their buildings were never designed to carry.

The damage is gradual, non-dramatic, and easy to dismiss event by event — but the long-term consequences, particularly in the context of Nepal’s seismic vulnerability, are serious. A building weakened by 30 years of vibration fatigue is a more dangerous building in the next major earthquake. This is not a theoretical statement. It is a structural engineering reality.

The solutions exist. Vibration monitoring, code revision, airport environs planning, and targeted retrofits are all achievable within Nepal’s institutional and financial capacity. What is currently lacking is the recognition of the problem at the policy level, and the political will to act on it before the next disaster makes the consequences of inaction undeniable.

As engineers, as urban residents, and as citizens of a city that is overdue for another major seismic event, we have a responsibility to name this problem clearly and demand that the relevant authorities address it with the seriousness it deserves.

References:

• ISO 2631-2:2003 — Mechanical vibration and shock: Evaluation of human exposure to whole-body vibration — Part 2: Vibration in buildings (1 Hz to 80 Hz)
• DIN 4150-3:1999 — Structural vibration: Effects of vibration on structures. Deutsches Institut fuer Normung.
• BS 7385-2:1993 — Evaluation and measurement for vibration in buildings — Part 2: Guide to damage levels from groundborne vibration. British Standards Institution.
• Nepal National Building Code, NBC 105:2020 — Seismic Design of Buildings in Nepal. Government of Nepal, DUDBC.
• Wyle Laboratories (2016). Aircraft Noise and Vibration Assessment Methodology. Federal Aviation Administration Technical Report.
• International Civil Aviation Organization (ICAO) Doc 9829 — Guidance on the Balanced Approach to Aircraft Noise Management, 2nd Edition.
• Arup Acoustics (2018). Airport Vicinity Building Vibration Assessment: Framework and Case Studies.
• Bothara, J.K. & Brzev, S. (2011). A Tutorial: Improving the Seismic Performance of Stone Masonry Buildings. Earthquake Engineering Research Institute, Oakland, California.
• JICA Report (2002). The Study on Earthquake Disaster Mitigation in the Kathmandu Valley, Kingdom of Nepal. Japan International Cooperation Agency.

The post Assessment of Aircraft-Induced Vibrations on the Structural Integrity of Buildings Around Tribhuvan International Airport, Nepal appeared first on OnlineEngineeringNotes.

Cable

Cables are slender, flexible members made of a group of high-strength steel wires twisted together mechanically. Steel cables provide the simplest means for supporting loads.

Steel cables, which are economically manufactured from high-strength steel wire, have an ultimate tensile strength of approximately 1862 MPa. The wires of cables are formed by drawing the alloyed steel bars through dies. This process aligns the molecules of the metal so that wires with tensile strength reaching as high as 1862 MPa can be produced. A group of such high strength steel wires twisted together forming a strand are used to make the cables. The modulus of elasticity of steel cables is approximately equal to 179 GPa, which is lower than the modulus of elasticity of steel cable is approximately equal to 179 GPa, which is lower than the modulus of elasticity of structural steel bars (200 GPa). Steel cables are easily handled and placed in position, even for very long spans. They provide lowest cost-to-strength ratio of any common structural members. The self-weight of the cable is generally neglected.

Element of Suspension Bridge

Suspension bridges are used for highways, where the span of bridge is more than 200m. Basically, a suspension bridges consists of following elements:

• Cable
• Suspenders
• Decking including the stiffening girder
• Supporting tower
• Anchorage
• Windguy and Windties
• Anchorage

A typical suspension bridge and its components are shown below:

Equilibrium of Light Cable: General Cable Theorem

Let us consider general case of cable supported at two supports A and B, which are not at the same level. The cable is acted by the system of vertical forces i.e. P₁,P₂,P₃…P_n as shown in figure below.

Let L be the horizontal span of cable and α be the inclination of line AB, with the horizontal. Obviously the differences in elevation between two supports A and B is equal to Ltanα.

Let V_A and V_B be the vertical components of reactions at A and B. Since there is no horizontal loading on the cable, the horizontal reaction(H) at ends A and B will  be equal in magnitude but opposite in direction.

In order to find vertical reaction V_A, take moments about B;

-V_A * L – H tanα + ΣM_B = 0

∴ V_A = (ΣM_B/L)- H tanα ——————————————(1)

Where, ΣM_B = Sum of moments of all loads

             P₁,P₂,….P_n about B.

Consider any point M of horizontal distance x from A.

Assume that the cable is perfectly flexible so that the bending moment at any point on the cable is zero.

Now,taking moment of all forces to the left point of M is zero.

H*(xtanα – y_m) + V_Ay * x – ΣM_m = 0

Substituting value of V_A from equation(1); we get,

H*xtanα – H*y_m+ x/L * ΣM_B – H*xtanα– ΣM_m = 0

or, H*y_m = x*( ΣM_B/L) – ΣM_m

Also, from beam; we have,

ΣM_B=0

or, V_A*L- ΣM_B = 0

∴ V_A = (ΣM_B / L )

∴ Hy_m = V_A * x – ΣM_m

Where, Hy_m = Beam moment at M.

It states that at any point on the cable acted by vertical loads, the product of H and y_m ( vertical distance from that point to the cable curve) equal to the bending moment at same point M considering simply supported beam of same span and same loading at that of cable.

Cable Subjected To Uniformly Distributed Load and Its Equation: At Same Level

Let us consider a cable subjected to a uniformly distributed load of w per unit run on the horizontal span.

Let, L= Length of cable ( curve length )

       l = Span of cable ( horizontal length )

      V_A = Vertical reaction at support A

      V_B = Vertical reaction at support B

       h  = Central deflection

Now, vertical reaction V_A and V_B are equal.

V_A =  V_B = wl/2

Taking moment about right of C; we have,

 ΣM_C = 0

or, V_B * l/2 – H* h – w*l/2 *l*4 =0

or, H *h = wl/2 * l/2 – wl²/8

 ∴ H = wl²/8

Let, T be the tension at any point P of the cable. Cosider the section (i)-(i) at distance x from support A.

Let, θ be the inclination of the tangent at with the horizontal.

Now,

Tension(T) = (V²+H²)^1/2

But, V_max= wl/2 = V_A = V_B

∴ T_max= {(wl/2)² + ( wl²/8h)²}^1/2 = wl/2 * { 1 + l² /16h²}^1/2

i.e. tension is maximum at supports since vertical reaction (V) is maximum at support and H is constant throughout the cable.

Similarly, tension ( T) is minimum at point of maximum sag because “ V” is minimum at mid span i.e. V = 0

∴ T_min = (0 + H²)^1/2 =H

To get equation of cable

Applying equation of equilibrium on left portion of section (I)-(I); we get,

ΣF_x = 0 gives;

H= Tcosθ —————(1)

Also,

ΣF_y = 0 gives;

V_A = wx + Tsinθ

∴ Tsinθ = V_A – wx —————(2)

Dividing equation (2) by equation (1); we get,

tanθ = (w_A – w_B) / H

or, dy/dx = 1/H {wlx/2 – wx²/2} = wx(l-x)/2h

Substituting value of H; we get,

y= {(wx)/ (2* wl²/8h) } * (l-x)

Substituting value of H; we get,

y= {(wx)/ (2* wl²/8h) } * (l-x)

∴ y = 4hx*(l-x)/ l² ; which is equation of parabola.

Thus,shape of cable is parabola.

To find the length of cable

ds = {(dx)² + (dy)²}^1/2 = {1+ (dy/dx)²}^1/2 dx

∴ ds/dx = {1+ (dy/dx)²}^1/2

We have,

y= 4hx(l-x)/l²

∴ dy/dx = 4h*(l-2x)/l²

∴ ds/dx = {1+ ([dy/dx][l-2x])²}^1/2

( Neglecting the other term of expansion)

Now,

Curve length ( L) = ∫ds

Determination of the Horizontal and Vertical Reaction of Cable with Ends at Different Level Carrying Uniformly Distributed Load.

Consider cable ACB supported at different levels A and B and C is the lowest point carrying uniformly distributed load w/ unit length.

The point C is extended to A^! and B^! as shown by dotted lines such that AC = A^!C dip of h₂.

Let, the horizontal distance between A and C = l₁

And Horizontal distance between B and C = l₂

Now, we have;

l₁ + l₂ = l ————————-(1)

Considering AA^!

H= wl²/8h = wl₁²/ 8h₁ —————–(2)

Considering BB^!

H= wl²/8h = wl₂²/8h₂ ——————–(3)

Equating equations (2) and (3) ; we get

wl₁²/8h₁ = wl₂²/8h₂

l₁² = h₁/h₂ * l₂²

∴ l₁/l₂ = (h₁/h₂)^1/2  ————————(4)

Also,

Length of cable ACA^’ (S₁) = 2l₁ +8h₁² /(3*2l₁) =2l₁+4h₁²/3l₁

And,

Length of cable BCB^’ (S₂) = 2l₂+8h₂² /(3*2l₂) =2l₂+4h₂²/3l₂

∴ Actual length of cable (S) = S₁/2+ S₂/2

∴ S = l + 2h₁²/3l₁ + 2h₂²/3l₂

To determine H

We have,

l= l₁+l₂ = l₁+l₂*( h₂/h₁)^1/2  { From equation(4)}

∴ l₁ = [l*(h₁)^1/2]/[(h₁)^1/2+(h₂)^1/2]

Substituting value l₁ in equation (2); we get

H= (w/2h₁)* [l*(h₁)^1/2]/[(h₁)^1/2+(h₂)^1/2] = wl²/[(h₁)^1/2+(h₂)^1/2]

Which is required expression for horizontal reaction.

To find vertical reactions

Let,

V_A= Vertical reaction at support A

V_B= Vertical reaction at support B

Now, taking moment about C of the forces on the left sides of C;we get,

V_A*l₁ = H*h₁+ wl₁²/2

or, V_A= H*h₁/l₁ +( wl₁²/2) * 1/l₁

Putting value of H in above equation;we get,

V_A= wl₁

∴ V_A= wl₁

Similarly, taking moment about C of the forces on right sides of C; we get,

V_B*l₂ = H*h₂+ wl₂²/2

or, V_B= H*h₂/l₂ +( wl₂²/2) * 1/l₂

Putting value of H in above equation;we get,

V_B= wl₂

∴ V_B= wl₂

Suspension Bridge with Three-Hinged Stiffening Girder

When suspension bridge is subjected to a moving (or rolling) load system the funicular polygon will change its shape and hence the shape of cable is also change.

In suspension cable bridge it is necessary to maintain the parabolic shape of cable to satisfy this condition the load transmitted to the cable should be uniformly distributed load.

This is achieved by either two hinged or three hinged stiffening girder.

The main purpose of providing these stiffness girders is to reduce the sag under the rolling load.

The girders are suspended form cables through hanger cables.Hence, the uniformly distributed dead load of the roadway and stiffening girders is transmitted to the cables through hanger cables and is taken up entirely by the tension in the cables. The stiffening girder does not suffer any S.F. or B.M. under dead load as girder is supported by closely spaced hanger cables throughout.

Any live load on the bridge will be transmitted to the girders as point loads. The stiffening girders transmit the live load to the cable as uniformly distributed load. While doing so the stiffening girders will be subjected to S.F. and B.M. throughout their length.

Extra:

1. Explain  with neat sketches tower structure as well as wind cables and windguy.

Ans. Suspension bridges are often supported on towers. Towers provide means for the cable to change the direction. Normally three type of towers are used in suspension bridge system.

a. Tower with saddles

• Such types of tower are fixed at their bases and support the main cable through a carriage, which is free to roll horizontally on the tower top.
• In this arrangement, the two cables does not need to have same tension.

b. Tower with pulley at top

• Such type of tower acts like vertical cantilevers and offer some resistance to cable.
• Assuming pulley as frictionless, tension in anchor cable is taken as same as tension in the main cable.

c. Tower hinged at base

• Such types of towers are free to rock in the plane of the main cables, which are securely attached to the tower tops.
• Suspension bridges are necessary to be protected from the vibration caused by wind. The lateral stability of the bridge is thus achieved by windguy arrangement. A standard form of windguy arrangement is shown in figure.
• It essentially consists of the bridge with a cable in the lateral direction with parabolic shape.

References: 1. Theory of Structure I, Dr. Kamal Bahadur Thapa

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They are very rigid and stable structures which are not considerably affected by movement of their foundations. Arches can be used for large structures made up of materials with negligible tensile strength, such as stones and bricks. Masonry arches of such materials have been used for thousands of years.

Types of Arches

• Arches may be classified on the basis of materials of which they are built, steel and reinforced concrete is the most common of all the materials.
• On the basis of form, arches may be further classified as parabolic, circular, elliptical etc.
• From the point of view of structural behavior arches are conveniently classified as:
• Three-hinged arch
• Two-hinged arch
• Hinge less(fixed) arch

Structurally

• A three hinged arch is statically determinate. There are four unknowns and there are three equations of equilibrium and one condition equation. (Moment at the third hinge is zero.)
• The two hinged arch is statically indeterminate to the first degree: there are four unknowns and three equations of static equilibrium.
• The hinge less arch is statically indeterminate to the third degree as there are six unknowns and only three equations of equilibrium.

Comparison between Arch and Cable

• While a cable is a tension member, arch behaves in a reverse fashion and it is basically a compression member.
• A cable is flexible and its shape with different types and positions of loads. It cannot resist any bending moment hence moment everywhere is zero.
• On the other hand, an arch is a rigid structure. It cannot change its shape corresponding to different types of loadings. Hence, through primarily it is a compression member it is often subjected to bending moment and shear through small magnitude.

Efficiency of an Arch

• The efficiency of an arch can be demonstrated by comparing it with beam of the same span under same loading.
• The arch resists the load by developing vertical as well as horizontal components of reaction. The horizontal reaction component reduces the moment from that in a simple beam.
• An arch supports loading with much less moment than a corresponding straight beam.
• It must be remembered that the reduction in moment is achieved at the expense of large axial compression in the arch rib and also horizontal reaction components at the springing.
• The moment diagrams are shown below:

Three Hinged Arch with Support at Same Level

1. Parabolic Arch

• For origin at  A

y= kx(L-x) ————————-(1)

At x= L/2 and y=h

We have,

h= k.L/2 (L- L/2)

or, k = 4L/ L²

∴ y = (4hx/L² )*(L-x)

This is equation of parabola with origin at A.

• For origin at C

If origin is at C, the equation parabola is

x² = ay————————- (2)

At x = L/2 and y=h

We have,

a= L²/4h

∴ y = (4h/L² )*(x²)

2. Circular arch

Here, R is the radius of circular arch and D is any point on arch having co-ordinate (x, y).

Using property of circle, we have,

L/2 * L/2 = h (2R-h)

∴ R = (L²/8h)+h/2

Co-ordinate of point D is given by;

x= L/2 – Rsinθ

y= Rcosθ – (R-h)

Three Hinge Arch with Support at Different Level

If origin is at C; the equation of parabola is

x²/y = Consatant

or x/(y)^1/2 = Constant

Here,

L= L₁+L₂

or, L₁/(h₁)^1/2 =  L₂/(h₂)^1/2

or, L₂ = (h₂/h₁)^1/2* L₁

and L₁ = (L*h₁^1/2) / (h₁^1/2+ h₂^1/2)

∴ L₂ = (L*h₂^1/2) / (h₁^1/2+ h₂^1/2)

Analysis of Three Hinged Arch

• Determination of support reactions

ΣF_x =0

ΣF_y =0

ΣM_A =0 or ΣM_B =0( all loads)

ΣM_C =0 ( Considering only left or only right part)

• Determination of radial shear (Q), normal thrust (N) and bending moment ( M)

Here, θ is the angle mode by normal thrust with horizontal,

Normal thrust (N) = Vsinθ + Hcosθ

Radial shear (Q) = Vcosθ – Hsinθ

Bending moment at D = Beam moment – Hy_D

Analysis of Three-Hinged Arch by the Graphical Method

The figure below shows a three hinged arch subjected to a load system. Let, V_A and V_B be the vertical reaction at the support A and B. Let, H be horizontal thrust. Let pq,  qr and rs represents the loads w₁,w₂,w₃. On this load line, let m be a point such that mp represents the vertical reaction V_A and ms represent vertical horizontal thrust.

The structure ACDEB is called linear arch or theoretical arch.

It can be easily realized that the shape of the linear arch follows the shape of the free bending moment diagram for a beam of the same span and subjected to same loading.

Free B.M. at any section X = Polar distance ( mo) * y^! = Hy^!

Moment  due to horizontal thrust = Hy^!

Where, y and y^! are  the ordinates of the given and linear arch respectively.

Net B.M. at section X = Hy- Hy^! = H(y- y^!)

∴Net B.M. at section X is proportional to (y- y^!)

Hence, the B.M. at any section of an arch is proportional to the ordinate or the intercept between the given arch and the linear arch. The is called Eddy’s theorem.

The actual bending moment at section X is proportional ordinate XX₁.

Influence Line Diagram for Three Hinged Arches

I.L.D for three hinged arches are given below:

1. Horizontal reaction (H)
2. Bending moment (B.M.)
3. Normal thrust
4. Radial shear

Maximum Bending Moment Diagrams in Three Hinged Parabolic Arch

• When a concentrated load moves

The maximum positive moment occurs at a distance of 0.2113 L from either end and has value of 0.096wL, where w is concentrated load and L is total span of arch. The maximum negative moment occurs at a distance of 0.25L from either end and has the value of wL/16.

• When a uniformly distributed load moves

The maximum positive or maximum negative moment occurs at a distance of 0.234L and has value of 0.01883wL³, where w is uniformly distributed load.

References: 1. Theory of Structure I, Dr. Kamal Bahadur Thapa

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Moving static loads

Structure which is used in bridges, gantry girders, crane beams etc. are subjected to loads which change their position often such loads are called moving static loads.

Few standard loads are:
 

1. Single concentrated load
2. U.D.L greater than the span
3. U.D.L smaller than the span
4. Two concentrated load with specified distance between them
5. Multiple concentrated loads(train of wheel loads)

Influence Line Diagram(I.L.D)

It is a curve the ordinate to which at point equal to the value of some particular function due to a unit load acting at that point. The function may be support reaction, shear force or bending moment.

Note:

I.L.D is always drawn for a concentrated unit load and it is independent of any type or system of loading.

Use of I.L.D in Civil Engineering:

1. To determine the structural quantity( Like reaction, shear force, bending moment) for a given system of load on the span of the structure.
2. To determine the position of like load system to have the maximum value of structural quantity.

Advantage of I.L.D:

1. It is a useful tool for dealing with moving loads.
2. It is a method of speedy determination of the value of structural quantity at any section under any complex system of loading.

Difference between I.L.D and B.M.D or S.F.D.

The ordinate of B.M.D or S.F.D gives the value of B.M or S.F. at the section where the ordinate has been drawn whereas the ordinate of I.L.D at any point gives the value of B.M. of S.F at the given section for which the I.L.D has been drawn and not at the point at which the ordinate has been drawn.

I.L.D for beams:

1. Simple supported beam

Consider a simply supported beam as below. Let unit load rolls from left to right.

Calculation of relations

I.L.D for R_A

I.L.D for R_B

I.LD for shear force (S.F) at C

Case I : When unit load is in portion AC i.e. (0 ≤ z ≤ x)

Considering right of C ( as the number of loads are less in right portion)

SF at C, V_c = -R_B = – z/L

Note: When rigid portion is considered upward force is taken negative.

Case II : When unit loads is in portion CB i.e. ( x ≤ z ≤ 1)

Considering left C.

V_C=R_A= 1-z/L

Note: Same result will be obtained by considering right of C.

• I.L.D for bending moment ( B.M.) at C

Case I : When unit load is in portion AC i.e.(0 ≤ z ≤ x)

Considering right of C.

Case II : When unit load is in portion CB i.e ( x ≤ z ≤ L)

Considering left of C.

2. Cantilever beam

Consider a cantilever beam as shown below. Let unit load rolls from left to right.

M_C = x- L

I.L.D is shown in the figure B4

3. Overhanging beam

Consider an overhanging beam as shown below. The unit load rolls from left to right.

Note: The I.L.D for overhang portion can also be determined by extending the I.L.D of simply supported beam up to the overhangs and their ordinates is determined by using similar triangle property.

I.L.D for Truss

The I.L.D for truss members can be easily drawn using method of sections. We know from method of sections that the external forces acting on one side of section AA( either left side of right of section of the truss shown below) and the internal forces in the members U₂U₃ , U₂L₃ and L₂L₃ cut by the sections are in equilibrium.

Applying principle of statics

For member L₂L₃

The moment of external forces about U₂ acting either left or right side of section must be equal and opposite to the moment of the force in the bottom chord member L₂L₃ about U₂.

(+↻)ΣMoment  about U₂ = 0

Let the moment of the external forces abour U₂ be M_U2

∴ Force in L₂L₃ = M_U2 /h

For member U₂U₃

Similar, force in member U₂U₃ can be obtained by taking moment about L₃ and applying:

(+↻)ΣMoment  about L₃ = 0

For member U₂L₃

Applying the sum of vertical component of all external forces on one side of section AA must be equal and opposite to the vertical component of the internal force in the member U₂U₃.

(+↑)Σ F_y = 0

Let the sum of vertical component of all external force is F_V.

Force is U₂L₃ = F_y/ sinθ

• The moving loads are never carried directly on the main girder but are transmitted across cross girders to the joints of bottom chord.
• Bridge Truss Floor System
• A typical bridge floor system is shown in Figure. As shown in Figure, the loading on bridge deck is transferred to stringers. These stringers transfer the load to floor beams and then to the joints along the bottom chord of the truss.
• It should be noted that for any load position; the truss is always loaded at the joint.

Types of Trusses

The main differences are that the Pratt truss has no force on the ends and the compression members are vertical. The Howe truss has no force on the center and the tension members are vertical instead.

Note :

• All top member compressive force and hence the I.L.D is negative.
• All the bottom experience tensile force and the I.L.D is positive.
• Vertical or inclined members experince tensile, compressive and both type of forces depending on the location with respect to the applied load on the truss.

I.L.D for Girders

Girder

Girder is a support beam which supports smaller beams. Girder is mainly used in bridges. In girder roof slab transmits the load to the cross beams and through cross beams to the girders. Hence, the girders are subjected to concentrated loads transmitted by cross beams.

The point at which the girder supports cross beam are referred as panel points.

1,2,3 and 4 are panel points.

The end reactions are not affected by the presence of cross beam. Hence, the reactions are same as that of simply supported beam.

The I.L.D for SF at any point within the panel is the same. Hence, I.L.D for SF is investigated with reference to panel but not any section.

Beam subjected to U.D.L longer than span

• Maximum negative shear force will be develop when head of udl reached considered section and absolute maximum negative shear force will developed when considered section is at B and the udl covers entire span.
• Maximum positive shear force will be develop when tail of udl reached considered section and absolute maximum positive shear force will developed when considered section is at A and the udl covers entire span.
• Maximum moment will be developed when the udl covers entire span and absolute maximum bending moment will developed when considered section is at mid span and load covers entire section.

Example:

1. Draw Ild for shear force and bending moment at 3m from right support when 6m udl of intensity 4kN/m crosses over 5m simply supported bridge from left to right. Calculate maximum negative, maximum positive shear force, maximum moment and also calculate absolute maximum positive, negative shear force and bending moment.

Maximum negative shear force will be develop when head of udl reached considered
section,

Maximum negative Shear Force = Intensity of load * Area under load in ILD
= 4(1/2)2*(2/5)
= 1.6 kN

Maximum positive shear force will be develop when tail of udl reached considered
section ,

Maximum positive Shear Force = Intensity of load * Area under load in ILD
= 4(1/2)3*(3/5)
= 3.6 kN

Maximum moment will be developed when the udl covers entire span

Maximum bending moment= Intensity of load * Area under load in ILD
= 4(1/2)5*(6/5)
= 12 kN m
Absolute maximum negative shear force will developed when considered section
is at B and the udl covers entire span.

Absolute Maximum negative shear force= Intensity of load * Area under load in ILD
= 4(1/2)5*1)
= 10 kN
Absolute maximum positive shear force will developed when considered section
is at A and the udl covers entire span.

Absolute Maximum positive shear force= Intensity of load * Area under load in ILD
= 4(1/2)5*1)
= 10 kN

Absolute maximum bending moment will developed when considered
section is at mid span and load covers entire section.

Absolute Maximum moment= Intensity of load * Area under load in ILD
= 4(1/2)5*6.25/5)
= 12.5kN m

Beam subjected to U.D.L shorter than span

• Maximum negative shear force will be develop when head of udl reached considered section and absolute maximum negative shear force will developed when considered section is at B and the head of udl reached at B.
• Maximum positive shear force will be develop when tail of udl reached considered section and absolute maximum positive shear force will developed when considered section is at A and tail of udl reached at A .
• Maximum moment will be developed when the udl is so placed that the section divides the load in the same ratio as it divides the span i.e x/d=a/l and absolute maximum bending moment will developed when considered section is at mid span such that x/d=a/l is same. Where x= distance between considered section to tail of load,d=length of udl, a=distance from left support to considered section, l= span length

Example:

1. A simply supported beam has span of 15 m, udl of 40 kN/m and 5 m long crosses the girder from left to right. Draw ILD for shear force and bending moment at 6m from left support and calculate maximum shear force and bending at that section.

Maximum negative shear force will be develop when head of udl reached considered
section ,

Maximum negative Shear Force = Intensity of load * Area under load in ILD
= 40(1/2)5*((1/15) +(6/15))
= 46.67 kN

Maximum positive shear force will be develop when tail of udl reached considered
section ,

Maximum positive Shear Force = Intensity of load * Area under load in ILD
= 40(1/2)5*((9/15)+(4/15)
= 86.67 kN

Maximum moment will be developed when the tail of udl lies at x distance from
considered section towards left support. So x= 5*6/15 =2m

Maximum bending moment= Intensity of load * Area under load in ILD
=( 40(1/2)2((36/15)+(54/15)))+ ( 40(1/2)3((36/15)+(54/15)))
= 600kN m

I.L.D due to series of concentrated loads

• Rolling loads are those loads which roll over the given structural element from one end to the another.
• For maximum negative shear force, most of the load lies in left portion of considered section so for that several hit and trial are made.
• Trial 1:Leading load w1 lies at considered section. Then,SF₁ =w₁*(a/L)+w₂ *Y₂ +w₃ *Y₃
• Trial 2 : W2 lies at considered section. Then SF₂ =- w₁*y₁+w₂*(a/l)+w₃*y₃+w₄*y₄

Check, If SF₁>SF₂,then maximum –ve shear force is SF₁ otherwise next trial is made by placing W3 over considered section. In this way we made hit and trial inorder to get maximum –ve SF. For maximum positive shear force, most of the load lies in left portion of considered section so for that several hit and trial are made.

Trial 1:Following load w4 lies at considered section. Then,SF₁ =w₄*(b/L)+w₃ *Y₃ +w₂ *Y₂ + w₁ *Y₁

Trial 2 : W₃ lies at considered section. Then SF₂ =- w₄*y₄+w₃*(b/l)+w₂*y₂+w₁*y₁

 Check, If SF₁>SF₂ ,then maximum +ve shear force is SF₁ otherwise next trial is made by placing W₃ over considered section. In this way we made hit and trial inorder to get maximum –ve SF.

For maximum positive shear force, most of the load lies in left portion of considered
section so for that several hit and trial are made.
Trial 1:Following load w4
lies at considered section. Then,SF₁ = w₄ *(b/L)+ w₃ * y₃ + w₂ * y₂ +
w₁ * y₁

Trial 2 : w₃ lies at considered section. Then SF₂
=- w₄ y₄ + w₃ (b/l)+w2y2+ w₁ * y₁

Check, If SF₁ > SF₂ ,then maximum +ve shear force is SF₁ otherwise next trial is made by placingW3over considered section. In this way we made hit and trial in order to get maximum –ve SF.

Maximum bending moment at a given section

• Maximum bending moment will be developed at a considered section when tilting load lies over that section.
• Tilting load: The load which changes the sign of avg. load on left side minus avg.load on right side when the point load passes over the section.

Tilting load can be determined as follows:

Here,w₃ changes the sign so tilting load is w₃ .Then tilting load is placed over considered section in order to get maximum moment.

Absolute maximum bending moment

• When a train of several loads crosses a simply supported beam, the absolute maximum bending moment under any given wheel occurs when this wheel load and the c.g of total resultant load are equidistance from the centre of beam.
• Steps to find absolute maximum bending moment
  • Calculate total resultant load by using,W= w₁+w₂+w₃+w₄
  • Calculate the position of resultant load by taking moment about the position of following load, i.e W*x͞ = w₄*0+w₃*x₃+w₂*(x₂+x₃)+w₁*(x₁+x₂+x₃) Here, a=b=L/2 – Assume w₃ is heavier and closer with resultant load W, so place w3 and resultant load W at equidistance from centre of beam.
  • Calculate Bending moment by using concept of ILD uses.
  • If w₃ is heavier but farther with resultant load W, so we made hit and trial by placing w₃ and resultant load W at equidistance from centre of beam for first case, and place w₂ and resultant load W at equidistance from centre of beam for second case then compare two cases, and adopt the values which is greater.

References: 1. Theory of Structure I, Dr. Kamal Bahadur Thapa

The post Influence Line Diagrams for Simple Structures: Moving Loads and I.L.D Beam, Truss and Girder appeared first on OnlineEngineeringNotes.

Strain energy and Complementary Strain Energy

• When an elastic member is deformed under the action of an external loading the member is said to have possessed or stored energy which is called strain energy.
• The strain energy stored by a member so deformed is equal to the amount of work done by the external forces to produce deformation.
• The area between the load-extension curve and the vertical axis is called complementary energy.
• For linearly elastic materials this area is equal to the area under the load extension curve i.e. strain energy.

Strain Energy Due To Axial Stress

Consider an infinitesimal element actual upon by normal stress 
σ_x .

Force on the element (F) = σ_x (d_Z – d_y )

Due to this force strain energy in element is (dU)= ½ * Stress * Strain *dV.

But,

Strain(ε_x) = Change in length / Original length

∴ Change in length = ε_x + d_x

Since, force is applied gradually from zero to its final value.

∴ Average force = ½ * σ_x (d_Z d_y )

Now,

Also,

∫ dV = AL

And, σ_x = P/A

Now, equation (1) becomes,

Strain Energy Due To Bending

Consider a small length dx of a beam where the bending moment is M. Consider further a small strain EFGH of thickness dy at a distance y from the neutral axis.

Let,

Width of strip = b

Volume of strip = d_X *d_y * b

We have,

Strain energy in element (dU) = ½ * Stress * Strain * dV

Also, from flexural formula,

σ = (M/I)*y

Then,

∴ dU = Strain energy of the volume CC’D’D

Now,

∫by²*d_y = Sum of second moment of area b.d_y

                 = Moment of inertia of beam cross-section

                 = I

The above expression gives the strain energy of the beam of length d_x .

Hence, Strain energy of the whole of the beam is given by;

If M is constant over the length L. Then,

Strain Energy Due To Torsion

We know that,

Elemental strain energy (dU)= ½ *stress* strain *dV

Where,

𝜏 = Maximum shear stress on the surface of the shaft.

G= Modulus of rigidity

Where,

T= Torque applied

 J =  Polar moment of inertia

L = Twisting angle

Φ = Twisting angle

Now,

The above expression is the required expression.

Strain Energy Due To Shear

Let the infinitesimal element is under the system of shear stress (𝜏) as shown in figure.

Now,

Strain energy stored in element ( dU) = ½ * stress * strain* dV

Or, dU = ½ * 𝜏 * γ * dV

Where,

𝜏 = Shear stress

γ = Shear strain

Also,

Modulus of rigidity (G) = 𝜏 / γ

Or,  γ = 𝜏 / G

Putting the value in above expression; we get

dU = ½ *𝜏 * 𝜏 / γ *dV

Case I : Shear stress on rectangular beam,

We have,

Shear stress (𝜏) = (VA y̅) / I*b ————-(2)

Where,

V= Shear force

A= Area above considered strip

y̅ = Center of gravity of area above considered strip

b= Width of beam

d= Depth of beam

I = Moment of inertia about direction of V

Now,

Shaded Area (A) = b(d/2 – y)

And

y̅ = ½ ( d/2 + y)

or, A y̅ = b/2 (d² /4 – y² )

From equation (1) and (2) ; we have

U= ∫ dU

Put Ay̅ = Q = Static moment of the cross- sectional area, above the point where the shear stress is desired.

Hence,

Rectangular section A_s = bd/1.2

This factor is known as shape factor whose value is dependent on actual shear stress distribution over the cross section.

Case II: Shear stress on circular beam

Consider a solid circular section of radius R and an elementary strain of thickness dy at a distance y from N.A. Let the width of strip be b.

Then,

b=2(R² – y²)^1/2

Area of elementary strip (dA) = b*dy

                                                          = 2(R² – y²)^1/2 *dy

Moment of elementary area about N.A

 = dy *y = 2(R² – y²)^1/2 *y*dy

For shaded area;

We have,

References: 1. Theory of Structure I, Dr. Kamal Bahadur Thapa

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Types of structure

The types of structure on the basic of materials used as follow:

• Steel structure
• Concrete structure
• Masonry structure
• Wooden structure
• Plastic structure

Structural Mechanics

Structure mechanics is the branch of mechanics that deals with forces and motion of structural system. It is the computation of deformations, deflections and of structural system. It is calculation of deformations, deflections and internal forces or stresses within structure for the estimation of existing structures.

Approaches of Structural Analysis:

Structural analysis can be classified are as follows:

1. Force method
2. Displacement method

1. Force method

In this method, the number of independent unknown forces ( both external and internal) is found and is compared with the number of independent equation of static equilibrium that can be written involving these unknown forces. If the number of unknown is equal to the number of independent equations, the unknown forces can be directly determined. But, if the number of unknown forces exceeds the number of equilibrium equations, a number of unknown forces in excess to the equilibrium equations are designated as redundant forces and are assumed to be equilibrium equations are designated as redundant forces and are assumed to be equilibrium equations to obtain a statically determinate residual structure called primary structure. For each redundant, one equation is written in which the displacement of point of application of redundant is expressed in terms of known forces and the unknown redundant forces. Finally, these equations are solved to get the unknown redundant forces.

2. Displacement method

In this method, the independent unknown displacement components involved in the structure are estimated and these are considered as the unknown to be found. The internal forces in the structure are then expressed in terms of these unknown displacements, using the stress displacement relations. For each unknown displacements, using the stress displacement relations. For each unknown displacement component, a corresponding equilibrium is written in terms of known external forces and unknown internal forces which are expresses in  terms of the displacement. These equations are solved to find the unknown displacement. Finally internal forces are calculated from the displacements.

Linear and Non-Linear Elastic Structure

If a system material has linear stress-strain relationships i.e. it obeys Hook’s law and undergo a small deflection of deformation is called linear elastic structure. For linear system, principle of superposition can be stationary applied.

Non-Linear system is the system is which is material does not have linear stress- strain relationships i.e. it does not obey Hook’s law and the deformation are so large that a change of geometry cannot be neglected in the analysis. Superposition principles does not hold good. If the non-linearity is due to stress-strain relationship, it is called material non-linearity and if the non linearity is due to considerable change in the geometry, it is called geometric non-linearity.

Computer Based Methods

• SAP( Structure Analysis Program)

This is a general purpose finite element program which is used for linear static and dynamic analysis of structures. Boundary condition can be assigned.

• STAAD Pro

It supports various steel, concrete and timber design codes. It can be used for static analysis geometric non-linear analysis or buckling analysis.

• NISA(Non-Linear Incremental Structural Analysis)

This software is used for non- linear analysis and buckling can be studied as well.

• EIABS

This software is mostly used for building (3-D analysis). Static earthquake loads can be considered in the analysis of structures.

Superposition Principles

It states that the displacement response resulting from number of force may be obtained by adding displacement resulting from individual force.

Where,

△_ij = Deflection at i due to load at j

△₁ = △₁₁ +△₁₂+△₁₃

△₂ = △₂₁ +△₂₂+△₂₃

△₃ = △₃₁ +△₃₂+△₃₃

References: 1. Theory of Structure I, Dr. Kamal Bahadur Thapa

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