Page 1 of 105 TVET CERFICATE V in CARPENTRY TIMBER STRUCTURAL DESIGN CAPTD501. Perform timber structural design Credits: 10 Learning hours: 100 Sector: Construction Sub-sector: Carpentry Module Note Issue date: November, 2020 Purpose statement This module describes knowledge and skills required to design timber structures. It describes the skills, knowledge and attitudes required for the trainee to perform structure planning, analysis, design of structural elements and present drawing details on structures designed.
Transcript
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TVET CERFICATE V in CARPENTRY
TIMBER STRUCTURAL DESIGN
CAPTD501. Perform timber structural design
Credits: 10 Learning hours: 100
Sector: Construction
Sub-sector: Carpentry
Module Note Issue date: November, 2020
Purpose statement
This module describes knowledge and skills required to design timber structures. It describes the skills, knowledge and attitudes required for the trainee to perform structure planning, analysis, design of structural elements and present drawing details on structures designed.
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Table of Contents
Elements of competence and performance criteria Page No.
Learning Unit Performance Criteria 1.Perform structure planning
1.1.Proper determination position and
orientation of the structure elements
3
1.2 Accurate identification of span of the slab
1.3. Proper selection of footing type according to the loads against the structure.
2. Perform analysis
2.1. Methodical determination of force type,
force action and computation of loads for each
structure element.
20
2.2. Proper determination of bending moment position, deflection position, and shear force zone using appropriate analysis method
2.3. Appropriate determination of influence lines for bending moment and shear for sections.
3.Design structure elements
3.1. Proper determination of young modulus for the recommended materials
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3.2Adequate calculation of section according to the shape of structure element.
3.3. Appropriate presentation of the design in
terms of details, drawings and schedule
preparation
LU & PC is linked in LO inside the content
Total Number of Pages: 106
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Introduction
Structural design is the methodical investigation of the stability, strength and rigidity of structures. The basic objective in structural analysis and design is to produce a structure capable of resisting all applied Loads without failure during its intended life. The primary purpose of a structure is to transmit or Support loads. If the structure is improperly designed or fabricated, or if the actual applied loads exceed the design specifications, the device will probably fail to perform its intended function, with possible serious consequences. A well-engineered structure greatly minimizes the possibility of costly Failures.
Structural design process A structural design project may be divided into three phases, i.e. planning, design and construction. Planning: This phase involves consideration of the various requirements and factors affecting the General layout and dimensions of the structure and results in the choice of one or perhaps several Alternative types of structure, which offer the best general solution. The primary consideration is the function of the structure. Secondary considerations such as aesthetics, sociology, law, economics and the environment may also Be taken into account. In addition there are structural and constructional requirements and limitations, which may affect the type of structure to be designed.
Design This phase involves a detailed consideration of the alternative solutions defined in the planning Phase and results in the determination of the most suitable proportions, dimensions and details of the Structural elements and connections for constructing each alternative structural arrangement being Considered.
Construction: This phase involves mobilization of personnel; procurement of materials and Equipment, including their transportation to the site, and actual on-site erection. During this phase, Some redesign may be required if unforeseen difficulties occur, such as unavailability of specified Materials or foundation problems.
Philosophy of designing The structural design of any structure first involves establishing the loading and other design Conditions, which must be supported by the structure and therefore must be considered in its Design. This is followed by the analysis and computation of internal gross forces, (i.e. thrust, and shear, Bending moments and twisting moments), as well as stress intensities, strain, deflection and Reactions produced by loads, changes in temperature, shrinkage, creep and other design conditions.
Finally comes the proportioning and selection of materials for the members and connections to Respond adequately to the effects produced by the design conditions. The criteria used to judge whether particular proportions will result in the desired behavior reflect Accumulated knowledge based on field and model tests, and practical experience. Intuition and Judgment are also important to this process.
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The traditional basis of design called elastic design is based on allowable stress intensities which are chosen in accordance with the concept that stress or strain corresponds to the yield point of the material and should not be exceeded at the most highly stressed points of the structure, the selection Of failure due to fatigue, buckling or brittle fracture or by consideration of the permissible deflection of the structure.
Design codes Many countries have their own structural design codes, codes of practice or technical documents Which perform a similar function. It is necessary for a designer to become familiar with local Requirements or recommendations in regard to correct practice.
Learning Unit 1 – Perform structure planning
LO 1.1 – Determine structural elements position and orientation
● Content/Topic 1: Structural elements
Slab: a large, thick, flat piece of stone or concrete, typically square or rectangular in shape.
Columns: A column or pillar in architecture and structural engineering is a structural element That transmits, through compression, the weight of the structure above to other structural elements below.
Columns are frequently used to support beams or arches on which the upper parts of walls or Ceilings rest.
Beams: a long, sturdy piece of squared timber or metal used to support the roof or floor of a Building.
Footings: a stable position or placing of the feet
A footing is a foundation unit constructed in brick work, masonry or concrete under the base of a wall or a column for the purpose of distributing the load over a large area. A footing or a shallow foundation is placed immediately below the lowest part of the Superstructure supported by it.
Trusses: A timber roof truss is a structural framework of timbers designed to bridge the space
Above a room and to provide support for a roof. Trusses usually occur at regular intervals, linked by Longitudinal timbers such as purlins. The space between each truss is known as a bay.
Stairs: a set of steps leading from one floor of a building to another, typically inside the building.
A lintel: is a beam placed across the openings like doors, windows etc. in buildings to support the
load from the structure above. The width of lintel beam is equal to the width of wall, and the ends of it are built into the wall. Horizontal lintels are easy to construct as compared to arches
Arches: An arch is a vertical curved structure that spans an elevated space and may or may not
Support the weight above it ● Content/Topic 2 : Elements to consider for structural positioning
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Loads: acting on structures for buildings and other structures can be broadly classified as vertical loads, horizontal loads and longitudinal loads.
Forces: is any interaction that, when unopposed, will change the motion of an object.
Shape: is the form of an object
End of LO1.1
LO 1.2 – Identify span of the slab
● Content/Topic 1: Types of slabs in construction
Slabs: are constructed to provide flat surfaces, usually horizontal, in building floors, roofs, Bridges, and other types of structures. The slab may be supported by walls, by reinforced Concrete beams, by structural steel beams, by columns, or by the ground There are 16 different types of Slabs in Construction but let us look on some of them. They are Some which are outdated and some of them are frequently used everywhere. In this article I will Give a detailed explanation about an each slab where to use particular slab.
1. FLAT SLAB:
Flat slab is a reinforced concrete slab supported directly by concrete columns or caps. Flat Slab don’t have beams. They are supported on columns itself. Loads are directly transferred to columns. In this type of Construction a plain ceiling is obtained thus giving attractive appearance from architectural point Of view. The plain ceiling diffuses the light better and is considered less vulnerable in the case of fire Than the usual beam slab construction. The flat slab is easier to construct and requires less Formwork. The thickness of Flat slab is minimum 8″ or 0.2m. THERE ARE FOUR DIFFERENT TYPES OF FLAT SLABS: 1. Slab without drop and column without column head (capital). 2. Slab with drop and column without column head. 3. Slab without drop and column with column head. 4. Slab with drop and column with column head.
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2. The solid slab or conventional slab is a customized, loosely reinforced, full concrete slab Which is used in residential and industrial construction. ... The solid slab has a smooth formwork Underside with a high surface quality and can therefore be wallpapered or painted The main difference between flat slab & conventional slab-beam system is that the one Is directly supported on the column while another system has a beam for support. The load Is transferred directly from slab to column in the flat slab. 3.Rubbed slab: are made up of wide band beams running between columns with narrow ribs spanning the orthogonal direction. Normally the ribs and the beams are the same depth. 4. A waffle slab: foundation, also called a ribbed slab foundation, is an above-ground type of Foundation used to provide load-bearing capacity in expansive, rocky or hydro collapsible soils. 5. Pre-stressed and post-stressed Slabs
Pre-stressed: is a form of structure used in construction. It is substantially "pre-stressed"
(compressed) during its fabrication, in a manner that strengthens it against tensile forces, which will
exist when in service. In order to improve the performance of the concrete in service
Post-stressed Slabs is a combination of conventional slab reinforcement and additional
protruding high-strength steel tendons
Lintels:
Lintels are provided inside building above the doors and windows to re direct the top Load. There are two types of lintels. Pre cast Lintels: Lintels, which are manufactured in factories, is called Pre cast Lintels. Cast in situ: Lintels are casted at the site it is called Cast in situ lintels. The length of the lintel is more than door length and has a width of the wall, thickness Of lintel is 0.1m
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Accurate identification of span of the slab Span of both either slab or beam are classified into 2 types such as: clear and effective spans. ● Content/Topic 2: Spanning slabs
‘Clear Span’ is an architectural term to describe the distance between the two inside surfaces Of the span supports; the distance that is unsupported, if you will Effective span of beams and slabs The span of a slab is the clear length of the slab between supporting columns
End of LO1.2
LO 1.3 –Identify types of footing/foundation
● Content/Topic 1: Types of Foundation
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Shallow foundation (3m depth) Individual Footing or Isolated Footing. Combined Footing. Strip Footing. Mat Footing or Raft Foundation.
Deep Foundation (20 – 65m depth) Pile Foundation. End Bearing Pile. Friction Pile Making minor mistakes causes the whole structure collapse. But the reality is very simple if you read this guide. Obviously, this is not a one stop shop to gain all the knowledge about designing foundation and implementation. But this comprehensive guide will give you basic knowledge how the design of Foundation works and how it supports the whole structure. Before beginning types of foundation. We have a deep sense of responsibility for explaining you the basic mechanism of foundation design.
What is a Foundation in Building Construction? The substructure (beneath the ground soil) that transfers the load from the column to the soil is Known as Foundation. The main function of building elements such as columns, beams is transferring the load (stress) from Slab to beam, beam to the column and then the column to the foundation of the earth (soil). First, they inspect the soil where the buildings is going to be built. (Soil Investigation) Second, they calculate the live and dead load that the foundation needs to transfer.
Third, they will design the structural member’s specifications such as size, reinforcement And mix ratio to support the load for a longer period. Safe Bearing Capacity of Soil SBC (Safe Bearing Capacity) is defined as the capacity of earth soil that supports the load Applied to the ground without any shear failure or settlement. It is denoted in T/m2 or KN/m2. It means safe bearing capacity is the maximum average pressure between the foundation and Earth soil which should not produce shear failure or settlement. In Simple, It is the maximum capability of soil to support without any failure or settlement. It’s theoretical calculation also known as ultimate bearing capacity.
Types of Foundation Generally, there are two broad categories of types of foundation
Individual Footing or Isolated Footing
Combined Footing
Strip Footing
Mat Footing or Raft Foundation
Deep Foundation (20 – 65m depth)
Pile Foundation
End Bearing Pile
Friction Pile
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Types of Shallow Foundation
A shallow foundation is typically 3 metre in depth. This types of Foundation mostly built for lightweight buildings where the condition of the soil is Extremely good with uniform layer and low water table.
Open Foundation or Spread Footings Shallow foundations also referred as spread footings or open foundation since the earth soil Will be completely removed to construct the footing and later filled by backfilling.
Open Foundation or Spread Footings Shallow foundations also referred as spread footings or open foundation since the earth soil Will be completely removed to construct the footing and later filled by backfilling.
It is referred as Spread footing because the footing base width is wider than the typical Load bearing wall.
Individual Footings or Isolated Footings (Square, Rectangular, Circular, Continuous Spread, Combined & Ring Spread)
It i s the most common type of foundation widely used for buildings.
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In this foundation, each column has its own footing.
These foundations connected with plinth beam just below the ground level. It comes in Various shapes such as rectangular, circular and continuous according to the load distribution. Note – You can get a rough idea of the foundation shape using SBC of soil. If the soil bearing Capacity is 10 T/m2 and the column has a vertical load of 10 T/m2 then the size of the footing will be 1 m2 . Footing Area = Column Load / Safe Bearing Capacity You can also design different geometrical shapes of foundation such as circular, square or rectangular. There are no limits. Also, You can find the Safe Bearing Capacity of soil in this Method if you know the column load.
Combined Footing
Are constructed for two or more columns when they are close to each other and their foundations overlap
Sometimes two or more footing may be combined because the adjacent footing might be
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Closer where shuttering is not possible at all for the given procedure like below one.
Strip Footings Strip foundations are continuous footing which will be constructed at the load bearing masonry Construction where walls will be acted as load bearing structure.
The footing will be constructed continually under the walls to support the load developed on the walls. These types of footings constructed in old masonry construction still these are in Existence in local construction.
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Spread footing is a type of the shallow foundation. It is defined as the structural members
used to support the column and walls as well as transmit and distribute the load coming on
the structure to the soil beneath it.
Strap footing: is a component of a building's foundation. It is a type of combined footing,
consisting of two or more column footings connected by a concrete beam. This type of
beam is called a strap beam
Mat or Raft Foundation You mostly see these type of foundation for High Residential Building construction where it Has a basement.
Are a large concrete slab which can support a number of columns and walls. The slab is spread out under
the entire building
This type of foundation will be constructed under two situations, Inconsistent Earth Soil Layer – The condition of the soil extremely differs in various places
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Under the building layout where designer can’t able to calculate the load and design the foundation as we done in spread footings. So they come up with Mat or Raft Foundation where the total load of the building will be spread throughout the foundation Cost Effective – Once the designer designs the foundation they analysis the cost of the foundation construction. If the foundation has a number of columns very adjacent to others, it will likely to increase the stuttering cost and concrete cost, Comparing these conditions the designer will design mat or Raft foundation where the construction will be easy and cost effective.
Types of Deep Foundation Deep foundation means as the name suggests the foundation will be penetrated through the weak soil layer at a great depth and supported on top of strong soil layer or on the rock. The foundation shall be greater than the shallow foundation (>3m). Normally all deep foundations are referred us pile foundation. But there are other types, normally all deep foundations are referred us pile foundation. But there are some other types, Caisson or Pier Foundation Caisson or pier foundations are often used in large construction under water bodies such as river, sea.
Drilled Shaft Foundation These type of foundations are used in bridges and large structures (metro train projects) where a long cylindrical casing will be inserted and then poured with concrete along with reinforcement. Once the concrete hardens the shape will be removed from the hole.
Step 1 – Driving the Casing
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Step 2 – Drilling the shaft Step 3 – Placing the cage Step 4 – Pouring the concrete
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Pile Foundations
Pile means the vertical member which will be penetrated through great depth. As the function of the pile is to carry the load to the deep there are different types according to construction methods and materials used.
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● Content/Topic 2: Application area
Stair cases: Sequence of steps which provides a means of ascending and descending
between two different level of floors or landing
There are three main stair types:
Curved stair
Straight stair
Turning stair:
They include quarter-turn stairs, Half-turn stairs, Three-quarter turn stairs, Newel stairs and Geometrical
stairs
Warehouse: is a commercial building generally used for storage of goods and warehousing
is the process of proper storage and handling of goods
Different Types of Warehouses:
Distribution Center. Distribution warehouses are used to store and sell large
quantities of goods.
Climate-controlled Warehouse. Climate-controlled warehouses are most often used
to store and ship perishable goods.
Private Warehouse.
Public Warehouse.
Automated Warehouse.
Fulfillment Center
Buildings is a shelter to enable dwelling of place. So, an essential purpose of a building (for
the human being in this case)
Types of buildings:
Residential Buildings: means any public building which is used for sleeping or lodging purposes and
includes any apartment house, rooming house, hotel, children's home, community−based residential
facility or dormitory but does not include a hospital or nursing home.
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Educational Buildings: a building designed for various activities in a primary, secondary, or higher
educational system and often including living areas for students, such as dormitories. The
development of education in medieval Europe and the establishment of colleges gave rise to
specialized buildings
Institutional Buildings: simply refers to any structure that fulfills a role related to healthcare,
education, recreation, or public works. Construction services teams that specialize in this type of
work build everything from hospitals and elementary schools to athletic facilities and university
building
Assembly Buildings: is a new way of real estate and hospitality firm designed especially for office buildings.
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Business Buildings: a building or a part of it is primarily used for keeping records of business
transactions, maintaining accounts, bookkeeping purposes or managing other types of records then
it can be classified as a business building
Industrial Buildings: means a building or part thereof wherein products or, material is fabricated,
assembled or processed, such as assembly plants, laboratories, power plants, refineries, gas plants,
mills dairies and factories.
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Storage Buildings: means a building or part thereof used primarily for storage or shelter of goods,
merchandise
Learning Unit 2 – Perform analysis LO 2.1 – Determine force type, action and compute load on structure
The objective of a structural engineer is to design a structure that will be able to withstand all the loads to
which it is subjected while serving its intended purpose throughout its intended life span. In designing a
structure, an engineer must, therefore, consider all the loads that can realistically be expected to act on
the structure during its planned life span. The loads that act on common civil engineering structures can be
grouped according to their nature and source into three classes or categories
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● Content/Topic 1: load categories
1.Dead load : The dead load includes loads that are relatively constant over time, including the weight of
the structure itself, and immovable fixtures such as walls, plasterboard or carpet. The roof is also a dead
load.
Due to the weight of the structural system itself and any other material permanently attached to it; Dead
loads are gravity loads of constant magnitudes and fixed positions that act permanently on the structure.
Such loads consist of the weights of the structural system itself and of all other material and equipment
permanently attached to the system For example, the dead loads for a building structure include the
weights of frames, framing and bracing systems, floors, roofs, ceilings, walls, stairways, heating and air
conditioning systems, plumbing, electrical systems, and so forth .ead loads are also known as permanent
or static loads
2.Live load: Refers to loads that do, or can, change over time, such as people walking around a building
(occupancy) or movable objects such as furniture. People repairs the roof structure,
Live load which are movable or moving loads due to the use of the structure; Live loads are loads of varying
magnitudes and/or positions caused by the use of the structure. Sometimes, the term live loads is used to
refer to all loads on the structure that are not dead loads, including environmental loads, such as snow
loads or wind loads.
Live loads, or imposed loads, are temporary, of short duration, or a moving load. These dynamic
loads.
3. Wind load: is the “load” placed by the wind speed and its air density onto a building. With high velocity
winds, low pressure areas are created on the building which creates suction pressure
Wind loads are produced by the flow of wind around the structure. The magnitudes of wind loads that may
act on a structure depend on the geographical location of the structure, obstructions in its surrounding
terrain, such as nearby buildings, and the geometry and the vibrational characteristics of the structure
itself.
4. Seismic load (Earthquakes) :is one of the basic concepts of earthquake engineering which means
application of an earthquake-generated agitation to a building structure or its model. It happens at contact
surfaces of a structure either with the ground, or with adjacent structures, or with gravity waves from
tsunami.
In addition to estimating the magnitudes of the design loads, an engineer must also consider the
possibility that some of these loads might act simultaneously on the structure.
The structure is finally designed so that it will be able to withstand the most unfavorable combination of
loads that is likely to occur in its lifetime. The minimum design loads and the load combinations for which
the structures must be designed are usually specified in building codes. Building codes vary from country to
country and also, owing to geographical variations, from region to region within a country.
● Content/Topic 2: Types of load
Usual types of loadings on the beams are discussed here.
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1.Concentrated It is a kind of load which is considered to act at a point on structureIf a load is acting on a
beam over a very small length, it is approximated as acting at the midpoint of that length and is
represented by an arrow as shown in Figure bellow.
2. Distributed load: A uniformly distributed load (UDL) is a load that is distributed or spread across the
whole region of an element such as a beam or slab.
Over considerably long distance such load has got uniform intensity. It is represented as shown in Fig. 2.3 (a) or as in (b). For finding reaction, this load may be assumed as total load acting at the Centre of gravity of the loading (middle of the loaded length). For example, in the beam shown in Fig. 2.3, the given load may be replaced by a 20 × 4 = 80 kN concentrated load acting at a distance 2 m from the left support.
fig. 2.3
Uniformly Varying Load: The load shown in Fig. 9.14 varies uniformly from C to D. Its Intensity is zero at C and is 20 kN/m at D. In the load diagram, the ordinate represents the load intensity and the abscissa represents the position of load on the beam.
Hence the area of the triangle represents the total load and the centroid of the triangle Represents the Centre of gravity of the load. Thus, total load in this case is 1/2× 3 × 20 = 30 kN And the Centre of gravity of this loading is at 1/3 × 3 = 1 m from D, i.e., 1 + 3 – 1 = 3 m from A. For finding the reactions, we can assume that the given load is equivalent to 30 kN acting at 3 m from A.
3. Couple load: consists of two parallel forces that are equal in magnitude, opposite in sense and do not
share a line of action
Content/Topic 3: Resultant of forces and their Applications:
1. Shear force: Shear forces are unaligned forces pushing one part of a body in one specific direction, and
another part of the body in the opposite direction
2. Compressive force is the application of power, pressure, or exertion against an object that causes it to
become squeezed, squashed, or compacted. Objects routinely subjected to compression forces include
columns, gaskets, disc brakes, and the components of fuel cells.
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3. Resultant force: is the single force and associated torque obtained by combining a system of forces and
torques acting on a rigid body.
The defining feature of a resultant force, or resultant force-torque, is that it has the same effect on the
rigid body as the original system of forces. The point of application of the resultant force determines its
associated torque.
LO 2.2 – Determine bending moment, deflection and shear force zone
● Content/Topic 1 : Types of structure supports
1. Fixed end: is a beam with one end fixedly tied to the main structure. The main structure is capable
to resist
2. Roller supports beam: are free to rotate and translate along the surface upon which the ... A roller
support cannot provide resistance to a lateral forces
3. Hinged support: can resist both vertical and horizontal forces but not a moment
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Type of supports
1. simple, 2. roller-type support 3. pinned support 4. fixed support
● Content/Topic 2:Analysis method for:
1. Components of forces
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Components of a Force. Forces acting at some angle from the coordinate axes can be resolved into
mutually perpendicular forces called components. The component of a force parallel to the x-axis is called
the x-component, parallel to y-axis the y-component, and so on.
How to decompose a force into x and y components
It is often useful to decompose a force into x and y components.
Exampleꓽ Find two forces such that one is in the x direction, the other is in the y direction, and the vector
sum of the two forces is equal to the original force.
Let's see how we can do this. Suppose we have a force F that makes an angle of 30° with the positive x axis,
as shown below:
And we want to decompose F into x and y components.
The first thing we need to do is to represent the two components on the xy-plane. We do this by dropping
two perpendiculars from the head of F: one to the x axis, the other to the y axis.
And we join the origin of the xy-plane with the x-intercept to represent the x component of F:
F
x
30 °
y
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And again, we join the origin with the y-intercept to represent the y component of F:
Fx and Fy are two vectors, i.e. they both have a magnitude and a direction. However, since Fx and Fy are in
the directions of the x and y axes, they are commonly expressed by the magnitude alone, preceded by a
positive or negative sign: positive when they point in the positive directions, and negative when they point
in the negative directions of the x and y axes.
In our example Fx and Fy are positive because both point in the positive directions of the x and y axes.
The positive values of Fx and Fy can be found using trigonometry:
Fx = F cos 30°
Fy = F sin 30°
To keep it simple, just remember that if a component is adjacent to the angle, then it is cos, otherwise it is
sin.
Often Fx will be the component adjacent to the angle, so it will be cos, and Fy will be sin.
The resultant force is in the same direction as the two forces, and has the magnitude equal to the sum of the two magnitudes:
Two forces acting in opposite directions Let's consider the case in which an object is subject to two forces that act in opposite directions. If the two forces are equal in magnitude
The resultant force will be zero because two opposite forces cancel each other out. On the other hand, if the two forces are not equal in magnitude:
The resultant force will be in the same direction as the force with the larger Magnitude (the 5 N force in the example), and have the magnitude equal to the Difference between the magnitudes of the two forces (in the example that would be 2 N):
More than 2 forces parallel to one another Let's now consider the case in which an object is subject to more than two parallel forces:
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To find the resultant force in this case, we first sum all the forces that go in one direction, and then all the
forces that go in the other direction: At this point, we have two forces that are in opposite directions, which is a case That we already know how to solve: the resultant force has the same direction as The force with the larger magnitude (the 11 N force), and its magnitude is equal To the difference between the two magnitudes (4 N):
Two forces that are not parallel In the previous cases, we have forces that are all parallel to one another. It's time to Consider the case in which an object is subject to two forces that are not parallel. For example, let's assume that we have a block subject to two forces, F1 and F2. F1 has magnitude 50 N and is applied at a 45° angle, whereas F2 has magnitude 60 N and is applied horizontally, as shown in the free-body diagram below:
The first step is to draw coordinate axes on our free-body diagram. Since one of the two forces is horizontal, for convenience, we choose the x-axis Horizontal, and the y-axis vertical, and we place the origin at the center of our block.
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The next step is to determine the x and y components of all the forces that act on the block:
If we sum all the x components, we will get the x component of the Resultant force:
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Similarly, if we sum all the y components, we will get the y component of the resultant force:
At this point, we know the x and y components of R, which we can use to find the Magnitude and direction of R:
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2. Determination of moment of forces and reaction to supports
Moments of forces and supports reaction The effect of a force on a rigid body depends on its point of application, as well as its magnitude And direction. It is common knowledge that a small force can have a large turning effect or leverage. In mechanics, the term ‘moment’ is used instead of ‘turning effect’. The moment of force with a magnitude (F) about a turning point (O) is defined as: M = F × d, where d is the perpendicular distance from O to the line of action of force F. The distance d is often called lever arm. A moment has dimensions of force time’s length (Nm). The direction of a moment about a Point or axis is defined by the direction of the rotation that the force tends to give to the body. Loading systems before any of the various load effects (tension, compression, bending, etc.) can be Considered, the applied loads must be rationalized into a number of ordered systems. Irregular loading is difficult to deal with exactly, but even the most irregular loads may be reduced and approximated to a number of regular systems. These can then be dealt with in mathematical terms using the principle of Superposition to estimate the overall combined effect. Concentrated loads are those that can be assumed to act at a single point, e.g. a weight hanging from a ceiling, or a person pushing against a box. Concentrated loads are represented by a single arrow drawn in the direction, and through the point of Action, of the force. The magnitude of the force is always indicated. Uniformly distributed loads, written as UDL, are those that can be assumed to act uniformly over an area or along the length of a structural Member, e.g. roof loads, wind loads, or the effect of the weight of water on a horizontal surface. For the Purpose of calculation, a UDL is normally considered in a plane. In calculating reactions, uniformly Distributed loads can in most, but not all, cases be represented by a concentrated load equal to the total distributed load passing through the centre of gravity of the distributed load.
a) According to the manner in which are supported
If a beam is fixed at one end and is free at the other end, it is called cantilever beam.
When both end of a beam are simply supported it is called simply supported beam. Such a beam can support load in the direction normal to its axis.
If a beam is projecting beyond the support, it is called an over-hanging beam. The overhang may be only on one side or may be on both sides.
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A beam is said to be continuous, if it is supported at more than two points.
A beam is said to be fixed ended, if it is fixed at both ends.
A beam is said to be cantilever simply supported if a beam is fixed at one end and simply Supported at another end.
REACTIONS FROM SUPPORTS OF BEAMS A beam is in equilibrium under the action of the loads and the reactions. Hence the equilibrium May be written for the system of forces consisting of reactions and the loads. Solutions of these Equations give the unknown reactions.
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Example1. Determine the reactions at supports A and B of the overhanging beam shown in
figure bellow.
Solution: As supports A and B are simple supports and loading is only in vertical direction, the
Reactions RA and RB are in vertical directions only.
2. The distribution of loads in a simply supported beam is as given in the diagram below. Determine the reactions at the supports a clockwise moment is usually considered as having a positive sign and an anticlockwise moment a negative sign. Beam reactions
The magnitude of the reactions may be found by the application of the third condition for equilibrium, i.e. the algebraic sum of the moments of the forces about any point must be zero. Take the moments around point A, then: (80 × 2) + (70 × 4) + (100 × 7) + (30 × 10) - (RB× 12) = 0; Giving RB = 120 kN. RA is now easily found with the application of the second condition for equilibrium. RA - 80 - 70 - 100 - 30 + RB=0; with RB = 120 kN gives:
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RA=160 kN.
2. Find the supports reaction for the situation bellow:
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Example 4.
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Example 5
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Moment diagrams by the method of sections.
We also consider qualitative deflected shapes of beams and the relationships between loads,
shears, and bending moments.
Internal forces were defined .as the forces and couples exerted on a portion of the structure by the rest of
the structure. Consider, for example, the simply supported beam shown in Fig. (a). The free body diagram
of the entire beam is depicted in Fig. (b), which shows the external loads, as well as the reactions Ax and
Ay, and By at supports A and B, respectively. , the support reactions can be computed by applying the
equations of equilibrium to the free body of the entire beam. In order to determine the internal forces
acting on the cross section of the beam at a point C, we pass an imaginary section cc through C, thereby
cutting the beam into two parts, AC and CB, as shown in Figs. (c) and (d). The free-body diagram of the
portion AC (Fig(c)) shows, in addition to the external loads and support reactions acting on the portion AC,
the internal forces, Q;S, and M exerted upon portion AC at C by the removed portion of the structure.
Note that without these internal forces, portion AC is not in equilibrium. Also, under a general coplanar
system of external loads and reactions, three internal forces (two perpendicular force components and a
couple) are necessary at a section to maintain a portion of the beam in equilibrium. The two internal force
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components are usually oriented in the direction of, and perpendicular to, the centroidal axis of the beam
at the section under consideration, as shown in Fig. (c). The internal force Q in the direction of the
centroidal axis of the beam is called the axial force, and the internal force S in the direction perpendicular
to the centroidal axis is referred to as the shear force (or, simply, shear).
The moment M of the internal couple is termed the bending moment. Recall from mechanics of materials
that these internal forces, Q; S, and M, represent the resultants of the stress distribution acting on the
cross section of the beam. The free-body diagram of the portion CB of the beam is shown in Fig. (d).
Note that this diagram shows the same internal forces, Q; S, and M, but in opposite directions, being
exerted upon portion CB at C by the removed portion AC in accordance with Newton’s third law. The
magnitudes and the correct senses of the internal forces can be determined by simply applying the three
equations of equilibrium, ∑Fx = 0, ∑ Fy =0, and∑ M = 0, to one of the two portions (AC or CB) of the beam.
It can be seen from Figs. (c) and (d), that in order for the equilibrium equation ∑Fx =0 to be satisfied for a
portion of the beam, the internal axial force Q must be equal in magnitude (but opposite in direction) to
the algebraic sum (resultant) of the components in the direction parallel to the axis of the beam of all the
external forces acting on that portion.
Since the entire beam is in equilibrium—that is, ∑ Fx = 0 for the entire beam—the application of ∑ Fx =0
individually to its two portions will yield the same magnitude of the axial force Q. Thus, we can state the
following:
The internal axial force Q at any section of a beam is equal in magnitude but opposite in direction
to the algebraic sum (resultant) of the components in the direction parallel to the axis of the beam
of all the external loads and support reactions acting on either side of the section under
consideration.
Using similar reasoning, we can define the shear and bending moment as follows
The shear S at any section of a beam is equal in magnitude but opposite in direction to the algebraic
sum (resultant) of the components in the direction perpendicular to the axis of the beam of all the
external loads and support reactions acting on either side of the section under consideration.
The bending moment M at any section of a beam is equal in magnitude but opposite in direction to
the algebraic sum of the moments about (the centroid of the cross section of the beam at) the
section under consideration of all the external loads and support reactions acting on either side of
the section.
Beam convection: axial force, shear force and bending moment
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Fig 2
Procedures of analysis
The procedure for determining internal forces at a specified location on a beam can be summarized
as follows:
I. Compute the support reactions by applying the equations of equilibrium and condition (if
any) to the free body of the entire beam. (In cantilever beams, this step can be avoided by
selecting the free, or externally unsupported, portion of the beam for analysis;
II. Pass a section perpendicular to the centroidal axis of the beam at the point where the
internal forces are desired, thereby cutting the beam into two portions.
III. Although either of the two portions of the beam can be used for computing internal forces,
we should select the portion that will require the least amount of computational effort,
such as the portion that does not have any reactions acting on it or that has the least
number of external loads and reactions applied to it.
IV. Determine the axial force at the section by algebraically summing the components in the
direction parallel to the axis of the beam of all the external loads and support reactions
acting on the selected portion. According to the sign convention adopted in the preceding
paragraphs, if the portion of the beam to the left of the section is being used for computing
the axial force, then the external forces acting to the left are considered positive, whereas
the external forces acting to the right are considered to be negative (see Fig. .2(b)). If the
right portion is being used for analysis, then the external forces acting to the right are
considered to be positive and vice versa.
V. Determine the shear at the section by algebraically summing the components in the
direction perpendicular to the axis of the beam of all the external loads and reactions acting
on the selected portion. If the left portion of the beam is being used for analysis, then the
external forces acting upward are considered positive, whereas the external forces acting
downward are considered to be negative (see Fig..2(c)). If the right portion has been
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selected for analysis, then the downward external forces are considered positive and vice
versa.
VI. Determine the bending moment at the section by algebraically summing the moments
about the section of all the external forces plus the moments of any external couples acting
on the selected portion. If the left portion is being used for analysis, then the clockwise
moments are considered to be positive, and the counterclockwise moments are considered
negative (see Fig.2(d)). If the right portion has been selected for analysis, then the
counterclockwise moments are considered positive and vice versa.
VII. To check the calculations, values of some or all of the internal forces may be computed by
using the portion of the beam not utilized in steps 4 through 6. If the analysis has been
performed correctly, then the results based on both left and right portions must be
identical.
3. Determination of deflection of members and Presentation of bending moment and shear force
diagrams
In this section we first learn the types of support structure and are shown in the table below
However the analysis and procedures for shear forces, axial forces and bending moment are seen in
the previous outcome in this one we are going to perform calculation
4.Presentation of bending moment and shear force diagrams
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Diagrams
• The plot of the bending moment curve will be a straight line over those segments of the beam
where the shearing force has a constant value.
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• The change in bending moment between two points on a beam is equal to the area under the
shearing force curve between the same two points.
Example2
EX
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SHEARING FORCES AND BENDING MOMENTS FOR DISTRIBUTED LOADS
Example 1
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Example 1’
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We refer to the figure for a set of four free –body diagrams of portion of the beam in length
increments of 2m. (Any increment could have been chosen, ex: 1m).
Free-body diagrams used to find shearing forces and bending moments:
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Note:
-Over the length of a beam carrying a uniformly distributed load, the shearing curve is a straight
line.
-the change in shearing force between any two points on a beam where a uniformly distributed lad
acts is equal to the resultant of the distributed load over that length.
-the change in bending moment between any two points on a beam is equal to the area under the
shearing force curve between those two points.
Example 2
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SHEARING FORCES AND BENDING MOMENTS FOR CANTILEVER BEAMS
Note: the place where the beam is supported, it is fixed and can therefore resist moments. Thus, at the
fixed end of the beam, the bending moment is not zero, as it was for simply supported beams. In fact, the
bending moment of at the fixed end of the beam is usually the maximum.
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determine the shear force and bending moment at point B
5. Computation method of normal force
We can blame the normal force for the pain we feel when running into solid objects. The normal force is
the force that surfaces exert to prevent solid objects from passing through each other.
However, when two surfaces are in contact (e.g. a box and a table) they exert a normal force on each
other, perpendicular to the contacting surfaces. This normal force will be as large as necessary to prevent
the surfaces from penetrating each other
How do you solve for normal force?
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There isn't really a formula made specifically for finding the normal force. To find the normal force we
typically use the fact that we know something about the acceleration perpendicular to the surfaces (since
we assume the surfaces can't pass through each other). Therefore, we almost always use Newton's second
law to solve for normal force by using this strategy.
1. Draw a force diagram showing all forces acting on the object in question.
2. Choose the direction for Newton's second law in the same direction as the normal force (i.e.
perpendicular to the contacting surfaces)
Essentially we are solving for normal force by assuming the normal force will be as large or small as it
needs to be to prevent the surfaces from penetrating each other.
Let's apply this strategy to the following simple example. Consider the simple case of a box of
mass mmm that's sitting on a table at rest, as seen below.
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6. Slenderness ratio of column:
the ratio of the length of a column and the least radius of gyration of its cross section. It is used
extensively for finding out the design load as well as in classifying various columns in
short/intermediate/long
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The code classifies columns first as:
1. Short columns when the ratios lex/h and ley/b are both less than 15 for braced columns and less than 10
for unbraced columns and
2. Slender columns when the ratios are larger than the values given above
Here: b: is width of the column cross-section, h: is the depth of the column cross-section, lex: is the
effective height in respect of the major axis and ley: is the effective height in respect of the minor axis.
A braced column of 300*350 cross-section resists at the ultimate limit state an axial load of 1700kN and
end moments of 70kNm and 10kNm causing double curvature about the minor axis x-x as shown in
figure8.76. The column’s effective heights are lex=6.75m and ley=8.0m and the characteristic material
strengths fcu=30N/mm2 and fy 460N/mm2
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End of LO2.2
LO 2.3 – Present influence lines for bending moment and shear
● Content/Topic 1: Method of presenting influence lines
Definitions of influence line
In the literature, researchers have defined influence line in many ways. Some of the definitions of
influence line are given below.
An influence line is a diagram whose ordinates, which are plotted as a
Function of distance along the span, give the value of an internal force, a
Reaction, or a displacement at a particular point in a structure as a unit load
Move across the structure.
An influence line is a curve the ordinate to which at any point equals the value
Of some particular function due to unit load acting at that point.
An influence line represents the variation of either the reaction, shear,
Moment, or deflection at a specific point in a member as a unit concentrated
Force moves over the member.
2.3 Construction of Influence Lines
In this section, we will discuss about the construction of influence lines. Using
Any one of the two approaches (Figure 3.1), one can construct the influence line
At a specific point P in a member for any parameter (Reaction, Shear or Moment). In the present
approaches it is assumed that the moving load is having dimensionless magnitude of unity.
Classification of the approaches for construction of influence lines is given in Figure 3.1.
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2.3.1 Tabulate Values
Apply a unit load at different locations along the member, say at x. And these
Locations, apply statics to compute the value of parameter (reaction, shear, or
Moment) at the specified point. The best way to use this approach is to prepare a
Table, listing unit load at x versus the corresponding value of the parameter
Calculated at the specific point (i.e. Reaction R, Shear V or moment M) and plot
the tabulated values so that influence line segments can be constructed.
2.3.2 Sign Conventions
Sign convention followed for shear and moment is given below.
2.3.3 Influence Line Equations
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Influence line can be constructed by deriving a general mathematical equation to
Compute parameters (e.g. reaction, shear or moment) at a specific point under the effect of moving
load at a variable position x.
The above discussed both approaches are demonstrated with the help of simple
Numerical examples in the following paragraphs.
3.4 Numerical Examples
Example 1:
Construct the influence line for the reaction at support B for the beam of span 10m. The beam
structure is shown in Figure 3.2.
Figure 3.2: The beam structure
Tabulate values:
As shown in the figure, a unit load is places at distance x from support A and the
Reaction value RB is calculated by taking moment with reference to support A. Let
Us say, if the load is placed at 2.5 m. from support A then the reaction RB can be
Calculated as follows (Figure 3.3).
Σ MA = 0: RB x 10 - 1 x 2.5 = 0 ⇒ RB = 0.25
Figure 3.3: The beam structure with unit load
Similarly, the load can be placed at 5.0, 7.5 and 10 m. away from support A and
Reaction RB can be computed and tabulated as given below.
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Graphical representation of influence line for RB is shown in Figure 3.4.
Figure 3.4: Influence line for reaction RB.
Influence Line Equation:
When the unit load is placed at any location between two supports from support
A at distance x then the equation for reaction RB can be written as
Σ MA = 0: RB x 10 – x = 0 ⇒ RB = x/10
The influence line using this equation is shown in Figure 3.4.
Example 2:
Construct the influence line for support reaction at B for the given beam as
Shown in Fig 3.5.
Figure 3.5: The overhang beam structure
Solution:
As explained earlier in example 1, here we will use tabulated values and
Influence line equation approach.
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Tabulate Values:
As shown in the figure, a unit load is places at distance x from support A and the
Reaction value RB is calculated by taking moment with reference to support A. Let us say, if the
load is placed at 2.5 m. from support A then the reaction RB can be
Calculated as follows.
Σ MA = 0: RB x 7.5 - 1 x 2.5 = 0 ⇒ RB = 0.33
Figure 3.6: The beam structure with unit load
Similarly one can place a unit load at distances 5.0 m and 7.5 m from support A
And compute reaction at B. When the load is placed at 10.0 m from support A,
Then reaction at B can be computed using following equation.
Σ MA = 0: RB x 7.5 - 1 x 10.0 = 0 ⇒ RB = 1.33
Similarly a unit load can be placed at 12.5 and the reaction at B can be
Computed. The values of reaction at B are tabulated as follows.
Graphical representation of influence line for RB is shown in Figure 3.7.
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Figure 3.7: Influence for reaction RB.
Influence line Equation:
Applying the moment equation at A (Figure 37.6),
Σ MA = 0: RB x 7.5 - 1 x x = 0 ⇒ RB = x/7.5
The influence line using this equation is shown in Figure 3.7.
Example 3:
Construct the influence line for shearing point C of the beam (Figure 3.8)
Figure 3.8: Beam Structure
Solution:
Tabulated Values:
As discussed earlier, place a unit load at different location at distance x from
Support A and find the reactions at A and finally computer shear force taking
Section at C. The shear force at C should be carefully computed when unit load is
Placed before point C (Figure 3.9) and after point C (Figure 3.10). The
Resultant values of shear force at C are tabulated as follows.
Figure 3.9: The beam structure – a unit load before section
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Figure 3.10: The beam structure - a unit load before section
Graphical representation of influence line for Vc is shown in Figure 3.11.
Figure 3.11: Influence line for shear point C
Influence line equation:
In this case, we need to determine two equations as the unit load position before
point C (Figure 3.12) and after point C (Figure 37.13) will show different shear
force sign due to discontinuity. The equations are plotted in Figure 3.11.
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Figure 3.12: Free body diagram – a unit load before section
Figure 3.13: Free body diagram – a unit load after section
Influence Line for Moment:
Like shear force, we can also construct influence line for moment.
Example 4:
Construct the influence line for the moment at point C of the beam shown in
Figure 3.14
Figure 3.14: Beam structure
Solution:
Tabulated values:
Place a unit load at different location between two supports and find the support
Reactions. Once the support reactions are computed, take a section at C and
Compute the moment. For example, we place the unit load at x=2.5 m from
Support A (Figure 3.15), then the support reaction at A will be 0.833 and support
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Reaction B will be 0.167. Taking section at C and computation of moment at C can be given by
Σ Mc = 0: - Mc + RB x 7.5 - = 0 ⇒ - Mc + 0.167 x 7.5 - = 0 ⇒ Mc = 1.25
Figure 3.15: A unit load before section
Similarly, compute the moment Mc for difference unit load position in the span.
The values of Mc are tabulated as follows.
Graphical representation of influence line for Mc is shown in Figure 3.16.
Figure 3.16: Influence line for moment at section C
Influence Line Equations:
There will be two influence line equations for the section before point C and after
Point C.
When the unit load is placed before point C then the moment equation for given
Figure 3.17 can be given by
Σ Mc = 0: Mc + 1(7.5 –x) – (1-x/15) x7.5 = 0 ⇒ Mc = x/2, where 0 ≤ x ≤ 7.5
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Figure 3.17: Free body diagram - a unit load before section
When the unit load is placed after point C then the moment equation for given
Figure 3.18 can be given by
Σ Mc = 0: Mc – (1-x/15) x 7.5 = 0 ⇒ Mc = 7.5 - x/2, where 7.5 < x ≤ 15.0
Figure 3.18: Free body diagram - a unit load before section
The equations are plotted in Figure 3.16.
Example 5:
Construct the influence line for the moment at point C of the beam shown in
Figure 3.19.
Figure 3.19: Overhang beam structure
Solution:
Tabulated values:
Place a unit load at different location between two supports and find the support
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Reactions. Once the support reactions are computed, take a section at C and
Compute the moment. For example as shown in Figure 3.20, we place a unit
Load at 2.5 m from support A, then the support reaction at A will be 0.75 and
Support reaction B will be 0.25.
Figure 3.20: A unit load before section C
Taking section at C and computation of moment at C can be given by
Σ Mc = 0: - Mc + RB x 5.0 - = 0 ⇒ - Mc + 0.25 x 5.0 = 0 ⇒ Mc = 1.25
Similarly, compute the moment Mc for difference unit load position in the span.
The values of Mc are tabulated as follows.
Graphical representation of influence line for Mc is shown in Figure 3.21.
Figure 3.21: Influence line of moment at section C
Influence Line Equations:
There will be two influence line equations for the section before point C and after
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Point C.
When a unit load is placed before point C then the moment equation for given
Figure 3.22 can be given by
Σ Mc = 0: Mc + 1(5.0 –x) – (1-x/10) x5.0 = 0 ⇒ Mc = x/2, where 0 ≤ x ≤ 5.0
Figure 3.22: A unit load before section C
When a unit load is placed after point C then the moment equation for given
Figure 3.23 can be given by
Σ Mc = 0: Mc – (1-x/10) x 5.0 = 0 ⇒ Mc = 5 - x/2, where 5 < x ≤ 15
Figure 3.23: A unit load after section C
The equations are plotted in Figure 37.21.
3.5 Influence line for beam having point load and uniformly
Distributed load acting at the same time
Generally in beams/girders are main load carrying components in structural
Systems. Hence it is necessary to construct the influence line for the reaction,
Shear or moment at any specified point in beam to check for criticality. Let us
Assume that there are two kinds of load acting on the beam. They are
Concentrated load and uniformly distributed load (UDL).
3.5.1 Concentrated load
As shown in the Figure 37.24, let us say, point load P is moving on beam from A
To B. Looking at the position, we need to find out what will be the influence line
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For reaction B for this load. Hence, to generalize our approach, like earlier
Examples, let us assume that unit load is moving from A to B and influence line
For reaction A can be plotted as shown in Figure 37.25. Now we want to know, if
Load P is at the center of span then what will be the value of reaction A? From
Figure 37.24, we can find that for the load position of P, influence line of unit load
Gives value of 0.5. Hence, reaction A will be 0.5xP. Similarly, for various load
Positions and load value, reactions A can be computed.
Figure 3.24: Beam structure
Figure 3.25: Influence line for support reaction at A
3.5.2 Uniformly Distributed Load
Beam is loaded with uniformly distributed load (UDL) and our objective is to find
Influence line for reaction A so that we can generalize the approach. For UDL of
W on span, considering for segment of dx (Figure 37.26), the concentrated load
dP can be given by w.dx acting at x. Let us assume that beam’s influence line
Ordinate for some function (reaction, shear, moment) is y as shown in Figure
3.27. In that case, the value of function is given by (dP) (y) = (w.dx).y. For
Computation of the effect of all these concentrated loads, we have to integrate
Over the entire length of the beam. Hence, we can say that it will be ∫ w.y.dx = w
∫ y.dx. The term ∫ y.dx is equivalent to area under the influence line.
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Figure 3.26: Uniformly distributed load on beam
Figure 3.27: Segment of influence line diagram
For a given example of UDL on beam as shown in Figure 3.28, the influence
Line (Figure 3.29) for reaction A can be given by area covered by the influence
Line for unit load into UDL value. i.e. [0.5x (1) XL] w = 0.5 w.l.
Figure 3.28: UDL on simply supported beam
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Figure 3.29: Influence line for support reaction at A.
37.6 Numerical Example
Find the maximum positive live shear at point C when the beam (Figure 37.30) is
Loaded with a concentrated moving load of 10 kN and UDL of 5 kN/m.
Figure 3.30: Simply supported beam
Solution:
As discussed earlier for unit load moving on beam from A to B, the influence line
For the shear at C can be given by following Figure 3.31.
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Figure 3.31: Influence line for shear at section C.
Concentrated load: As shown in Figure 3.31, the maximum live shear force at C
Will be when the concentrated load 10 kN is located just before C or just after C.
Our aim is to find positive live shear and hence, we will put 10 kN just after C. In
That case,
VC = 0.5 x 10 = 5 kN.
UDL: As shown in Figure 37.31, the maximum positive live shear force at C will
Be when the UDL 5 kN/m is acting between x = 7.5 and x = 15.
VC = [0.5 x (15 –7.5) (0.5)] x 5 = 9.375
Total maximum Shear at C:
(VC) max = 5 + 9.375 = 14.375.
Finally the loading positions for maximum shear at C will be as shown in Figure
3.32. For this beam one can easily compute shear at C using statics.
Figure 3.32: Simply supported beam
4.1 Müller Breslau Principle for Qualitative Influence Lines
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In 1886, Heinrich Müller Breslau proposed a technique to draw influence lines
Quickly. The Müller Breslau Principle states that the ordinate value of an
Influence line for any function on any structure is proportional to the ordinates of
The deflected shape that is obtained by removing the restraint corresponding to
The function from the structure and introducing a force that causes a unit
Displacement in the positive direction.
Let us say, our objective is to obtain the influence line for the support reaction at
A for the beam shown in Figure 4.1.
Figure 4.1: Simply supported beam
First of all remove the support corresponding to the reaction and apply a force
(Figure 4.2) in the positive direction that will cause a unit displacement in the
Direction of RA. The resulting deflected shape will be proportional to the true
Influence line (Figure 4.3) for the support reaction at A.
Figure 4.2: Deflected shape of beam
Figure 4.3: Influence line for support reaction A
The deflected shape due to a unit displacement at A is shown in Figure 4.2 and
Matches with the actual influence line shape as shown in Figure 4.3. Note that
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The deflected shape is linear, i.e., the beam rotates as a rigid body without any
Curvature. This is true only for statically determinate systems.
Similarly some other examples are given below.
Here we are interested to draw the qualitative influence line for shear at section
C of overhang beam as shown in Figure 4.4.
Figure 4.4: Overhang beam
As discussed earlier, introduce a roller at section C so that it gives freedom to the
Beam in vertical direction as shown in Figure 38.5.
Figure 4.5: Deflected shape of beam
Now apply a force in the positive direction that will cause a unit displacement in
The direction of VC. The resultant deflected shape is shown in Figure 4.5.
Again, note that the deflected shape is linear. Figure 4.6 shows the actual
Influence, which matches with the qualitative influence.
Figure 4.6: Influence line for shear at section C
In this second example, we are interested to draw a qualitative influence line for
Moment at C for the beam as shown in Figure 4.7.
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Figure 4.7: Beam structure
In this example, being our objective to construct influence line for moment, we
Will introduce hinge at C and that will only permit rotation at C. Now apply
Moment in the positive direction that will cause a unit rotation in the direction of
Mc. The deflected shape due to a unit rotation at C is shown in Figure 4.8 and
Matches with the actual shape of the influence line as shown in Figure 4.9.
Figure 4.8: Deflected shape of beam
Figure 4.9: Influence line for moment at section C
4.2 Content/Topic 2: Maximum shear in beam supporting UDLs
If UDL is rolling on the beam from one end to other end then there are two
Possibilities. Either Uniformly distributed load is longer than the span or uniformly
Distributed load is shorter than the span. Depending upon the length of the load
And span, the maximum shear in beam supporting UDL will change. Following
Section will discuss about these two cases. It should be noted that for maximum
Values of shear, maximum areas should be loaded.
4.2.1 UDL longer than the span
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Let us assume that the simply supported beam as shown in Figure 4.10 is
Loaded with UDL of w moving from left to right where the length of the load is
Longer than the span. The influence lines for reactions RA, RB and shear at
Section C located at x from support A will be as shown in Figure 4.11, 4.12 and
4.13 Respectively. UDL of intensity w per unit for the shear at supports A and B
Will be given by
Figure 4.10: Beam Structure
Figure 4.11: Influence line for support reaction at A
Figure 4.12: Influence line for support reaction at B
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Figure 4.13: Influence line for shear at section C
Suppose we are interested to know shear at given section at C. As shown in
Figure 38.13, maximum negative shear can be achieved when the head of the
Load is at the section C. And maximum positive shear can be obtained when the
Tail of the load is at the section C. As discussed earlier the shear force is
Computed by intensity of the load multiplied by the area of influence line diagram
Covered by load. Hence, maximum negative shear is given by
And maximum positive shear is given by
4.2.2 UDL shorter than the span
When the length of UDL is shorter than the span, then as discussed earlier,
Maximum negative shear can be achieved when the head of the load is at the
Section. And maximum positive shear can be obtained when the tail of the load is
At the section. As discussed earlier the shear force is computed by the load
Intensity multiplied by the area of influence line diagram covered by load. The
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Example is demonstrated in previous lesson.
4.3 Content/Topic 3: Maximum bending moment at sections in beams supporting UDLs.
Like the previous section discussion, the maximum moment at sections in beam
Supporting UDLs can either be due to UDL longer than the span or due to ULD
Shorter than the span. Following paragraph will explain about computation of
Moment in these two cases.
4.3.1 UDL longer than the span
Let us assume the UDL longer than the span is traveling from left end to right
Hand for the beam as shown in Figure 4.14. We are interested to know
Maximum moment at C located at x from the support A. As discussed earlier, the
Maximum bending moment is given by the load intensity multiplied by the area of
Influence line (Figure 4.15) covered. In the present case the load will cover the
Completed span and hence the moment at section C can be given by
Figure 4.14: Beam structure
Figure 4.15: Influence line for moment at section C
Suppose the section C is at mid span, then maximum moment is given by
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4.3.2 UDL shorter than the span
As shown in Figure 38.16, let us assume that the UDL length y is smaller than
The span of the beam AB. We are interested to find maximum bending moment at
Section C located at x from support A. Let say that the midpoint of UDL is
Located at D as shown in Figure 4.16 at distance of z from support A. Take
Moment with reference to A and it will be zero.
Figure 4.16: Beam loaded with UDL shorter in length than span
Hence, the reaction at B is given by
And moment at C will be
Substituting value of reaction B in above equation, we can obtain
To compute maximum value of moment at C, we need to differentiate above
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Given equation with reference to z and equal to zero.
Using geometric expression, we can state that
The expression states that for the UDL shorter than span, the load should be
Placed in a way so that the section divides it in the same proportion as it divides
The span. In that case, the moment calculated at the section will give maximum
Moment value.
End of LO2.3
Learning Unit 3 – Design structure elements
LO 3.1 – Determine young modulus for recommended materials
Content/Topic 1Test methods for Young's modulus
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dynamic is constant change or motion
static characterized by a fixed or stationary condition. showing little or no change
tensile: Tensile testing, also known as tension testing, is a fundamental materials science and
engineering test in which a sample is subjected to a controlled tension
torsion, the twisting or wrenching of a body by the exertion of forces tending to turn one end or part
about a longitudinal axis while the other is held fast or turned in the opposite direction
bending test flexure testing or transverse beam testing, measures the behavior of materials
subjected to simple beam loading.
wavepropagation is any of the ways in which waves travel. With respect to the direction of the
oscillation relative to the propagation direction, we can distinguish between longitudinal wave and
transverse waves. For electromagnetic waves, propagation may occur in a vacuum as well as in a
material medium What are the types of wave propagation?
Example of types of Wave Propagation
Ground Waves.
Ionospheric Layer
Sky Waves (Ionospheric Wave)
nanoindentation called instrumented indentation testing, is a variety of indentation hardness tests
applied to small volumes.
Stress and strain of materials when a body is subjected to a deforming force, a restoring force occurs
in the body which is equal in magnitude but opposite in direction to the applied force. This restoring
force per unit area is known as stress. while Strain is simply the measure of how much an object is
stretched or deformed,
● Content/Topic 2: Application of formula for young modulus of the materials
1. STRAIN
When a body is subjected to external force there is some changes of dimension of a body. The ratio
of dimension change of a body to the original dimension of the body is known as strain .strain is
dimensionless
It can either be tensile strain, volumetric strain, shear strain and compressive strain.
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If there is increase in length due to external forces. The ratio of increase of length to the original
length of the body is known as tensile strain
If there is the decrease in length to the original length of the body the ratio of decrease of the
length to the original length of the body is known compressive strain
The ratio of change of volume of a body to the original volume is known as volumetric strain
The strain produced by shear stress is known as shear strain
2. STRESS
The force of resistance per unit area by a body against the deformation is known as stress .the
external force acting on a body is called load or force .the load applied to a body while the stress is
induced in the material of a body. A loaded member remain in equilibrium when the resistance
offered by the member against the deformation and the applied load are equal
Mathematically stress is written as ϭ =𝑃
𝐴
Where P=External force or load
A=cross section area
Ϭ=stress
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The unit of stress depend on the unit of force and cross section area. In S.I Stress is expressed in
N/m2 or N/mm2
3. MODULUS OF ELASTICITY OR YOUNG MODULUS
The ratio of tensile stress or compressive stress to the corresponding strain is a constant. This ratio is
known as young’s modulus or modulus of elasticity and is denoted by E
E=𝑻𝒆𝒏𝒔𝒊𝒍𝒆 𝒔𝒕𝒓𝒆𝒔𝒔
𝒕𝒆𝒏𝒔𝒊𝒍𝒆 𝒔𝒕𝒓𝒂𝒊𝒏 or
𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒗𝒆 𝒔𝒕𝒆𝒔𝒔
𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒗𝒆 𝒔𝒕𝒓𝒂𝒊𝒏
E=𝝈
𝒆
End of LO3.1
LO 3.2 – Calculate section according to the shape of structure element
● Content/Topic 1: Classification of various structure forms
Compression structures:
a) Trusses
Trusses are composed of straight members connected at their ends by hinged connections to form a stable
configuration (In the Fig below). When the loads are applied to a truss only at the joints, its members
either elongate or shorten. Thus, the members of an ideal truss are always either in uniform tension or in
uniform compression. Real trusses are usually constructed by connecting members to gusset plates by
bolted or welded connections. Although the rigid joints thus formed cause some bending in the members
of a truss when it is loaded, in most cases such secondary bending stresses are small, and the assumption
of hinged joints yields satisfactory designs. Trusses, because of their light weight and high strength, are
among the most commonly used types of structures. Such structures are used in a variety of applications,
ranging from supporting roofs of buildings to serving as support structures in space stations
The frame structures dealt with in this chapter are pin-jointed, determinate and plane frameworks.
Method of sections
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Consider the following framed structure. It is required to calculate the force in the member 1-2 of the
bottom boom.
Figure page 21 recto
Solution
1) Imagine the structure to be cut completely through, along the section , a section passing through the
members 3-4 and 3-2.
2) Consider the portion of truss to the right of line s-s to be removed. The portion to the left of line
s-s would then, of course, collapse, because three forces (forces in members 1-2, 3-2 and 3-4)
which were necessary to retain equilibrium had been removed.
3) If three forces O, Ë and þ equal respectively to force 1-2, forces 3-4 and force 3-2, are now
applied to the portion of frame concerned, as shown in the previous figure, then the portion of
frame will remain in equilibrium under the action of the 50kNreaction, the two 20 kN loads and the
forces O, Ë and þ.
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4) These three forces (O,Ë ,þ ) are as yet unknown in amount and direction , but the force in
member 1-2, force O is to be determined at this stage , and it will be seen that the other two
unknown forces (Ë and þ ) intersect at and pass through the point 3 so that they have no moment
about that point.
5) Taking moments about the point 3, the portion of frame to the left of s-s is in equilibrium under
the moments of the two applied loads, the moment of the reaction and the moment of the force.
The arrow must act as shown in the figure.
Ex: Roof truss:
Calculate the forces in the members marked 2-3; 1-6 and 2-6 of the frame shown in the figure
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Method of resolution of forces at joints
The forces in the individual members of loaded frames may also be determined by considering the various
forces acting at each node point.
Example:
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b. beams is a structural element that primarily resists loads applied laterally to the beam's axis. Its
mode of deflection is primarily by bending
c. plane truss, is defined as a two- dimensional framework of straight prismatic members connected at
their ends by frictionless hinged joints, and subjected to loads and reactions that act only at the joints
and lie in the plane of the structure
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d. space truss, are three-dimensional structures with longitudinal members connected at
their ends by hinges assumed to be frictionless
e. plane frame, are two-dimensional structures constructed with straight elements connected
together by rigid and/or hinged connections
j. space frame, is analyzed by assuming rigid joints that cause internal torsions and moments in the
members
k.ARCH
An arch is a curved structure, with a shape similar to that of an inverted cable, as shown in Fig below. Such
structures are frequently used to support bridges and long-span roofs. Arches develop mainly compressive
stresses when subjected to loads and are usually designed so that they will develop only compression
under a major design loading. However, because arches are rigid and cannot change their shapes as can
cables, other loading conditions usually produce secondary bending and shear stresses in these structures,
which, if significant, should be considered in their designs. Because compression structures are susceptible
to buckling or instability, the possibility of such a failure should be considered in their designs; if necessary,
adequate bracing must be provided to avoid such failures.
BENDING STRUCTURES
i. beams
Bending structures develop mainly bending stresses under the action of external loads. In some structures,
the shear stresses associated with the changes in bending moments may also be significant and should be
considered in their designs. Some of the most commonly used structures, such as beams, rigid frames,
slabs, and plates, can be classified as bending structures. A beam is a straight member that is loaded
perpendicular to its longitudinal axis (Fig. below). Recall from previous courses on statics and mechanics of
materials that the bending (normal) stress varies linearly over the depth of a beam from the maximum
compressive stress at the fiber farthest from the neutral axis on the concave side of the bent beam to the
maximum tensile stress at the outermost fiber on the convex side. For example, in the case of a horizontal
beam subjected to a vertically downward load, as shown in Fig.below, the bending stress varies from the
maximum compressive stress at the top edge to the maximum tensile stress at the bottom edge of the
beam.
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ii. FRAMES
Rigid frames are composed of straight members connected together either by rigid (moment-resisting)
connections or by hinged connections to form stable configurations. Unlike trusses, which are subjected
only to joint loads, the external loads on frames may be applied on the members as well as on the joints
(see Fig. Below). The members of a rigid frame are, in general, subjected to bending moment, shear, and
axial compression or tension under the action of external loads. However, the design of horizontal
members or beams of rectangular frames is often governed by bending and shear stresses only, since the
axial forces in such members are usually small. Frames, like trusses, are among the most commonly used
types of structures. Structural steel and reinforced concrete frames are commonly used in multistory
buildings , bridges, and industrial plants. Frames are also used as supporting structures in airplanes, ships,
aerospace vehicles, and other aerospace and mechanical applications. It may be of interest to note that
the generic term framed structure is frequently used to refer to any structure composed of straight
members, including a truss. In that context, this textbook is devoted primarily to the analysis of plane
framed structures
Fig shows frame
● Content/Topic 2 : Calculation of degree of static and kinematic indeterminacy of structure
In order to determine the degree of indeterminacy we first understand what structure in equilibrium
means. A structure is considered to be in equilibrium if, initially at rest, it remains at rest when subjected
to a system of forces and couples. If a structure is in equilibrium, then all its members and parts are also in
equilibrium.
In order for a structure to be in equilibrium, all the forces and couples (including support reactions) acting
on it must balance each other, and there must neither be a resultant force nor a resultant couple acting on
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the structure. Recall from statics that for a space (three-dimensional) structure subjected to three-
dimensional systems of forces and couples, the conditions of zero resultant force and zero resultant couple
can be expressed in a Cartesian (xyz) coordinate system as
∑Fx =0, ∑Fy = 0, ∑Fz = 0
∑Mx =0, ∑My = 0, ∑Mz =0
These six equations are called the equations of equilibrium of space structures and are the necessary and
sufficient conditions for equilibrium. The first three equations ensure that there is no resultant force acting
on the structure, and the last three equations express the fact that there is no resultant couple acting on
the structure. For a plane structure lying in the xy plane and subjected to a coplanar system of forces and
couples, the necessary and sufficient conditions for equilibrium can be expressed as
∑Fx =0, ∑Fy =0, ∑Mz = 0 these three equations are referred to as the equations of equilibrium of plane
structures. The first two of the three equilibrium equations express, respectively, that the algebraic sums
of the x components and y components of all the forces are zero, thereby indicating that the resultant.
Static Determinacy of Internally Stable Structures
An internally stable structure is considered to be statically determinate externally if all its support reactions
can be determined by solving the equations of equilibrium. The degree of indeterminacy is determined by
r-3, where r is number of support reaction
If a structure is supported by more than three reactions, then all the reactions cannot be determined from
the three equations of equilibrium. Such structures are termed statically indeterminate externally. The
reactions in excess of those necessary for equilibrium are called external redundant, and the number of
external redundant is referred to as the degree of external indeterminacy. Thus, if a structure has r
reactions r > 3Þ, then the degree of external indeterminacy can be written as ie= r-3. The conditions of
static instability, determinacy, and indeterminacy of plane internally stable structures can be summarized
as follows:
r < 3 the structure is statically unstable externally
r =3 the structure is statically determinate externally
r > 3 the structure is statically indeterminate externally
Where r =number of reactions. It should be realized that the first of three conditions stated is both
necessary and sufficient in the sense that if r < 3, the structure is definitely unstable. However, the
remaining two conditions, r =3 and r > 3, although necessary, are not sufficient for static determinacy and
indeterminacy, respectively. In other words, a structure may be supported by a suffcient number of
reactions ðrb3Þ but may still be unstable due to improper arrangement of supports. Such structures are
