lOMoARcPSD|15618290 Lecture notes, lectures 1-4 Structural Analysis I (University of Sheffield) Studocu is not sponsored or endorsed by any college or university Downloaded by Govindah Oke (govindahoke2@gmail.com) lOMoARcPSD|15618290 • CV2102: Structures 1 Scope of Structures 1 Topics Section I: Basic structural analysis Section I (Week 1 to Week 7) Types of structures and loadings Basic concepts and analysis techniques Analysis of Trusses Analysis of beams and frames Section II: Deflection of structures Deflection of beams Integration method Moment area method Deflection of structures using energy methods Principle of virtual work Lecture 1 1 CV2102: Structures 1 - Lecture 1 2 CV2102: Structures 1 Why we study structural analysis? Structural Engineering is about the conception, design and construction of the structural systems that are needed in support of human activities. Civil engineering projects include bridges, building, dams, storage facilities etc. involves structural engineering. Lecture 1 Answer: To prevent this to happen! 3 Downloaded by Govindah Oke (govindahoke2@gmail.com) Lecture 1 4 1 lOMoARcPSD|15618290 CV2102: Structures 1 Compassvale School, Singapore •Structural analysis is rarely an end in itself. Rather, it is a tool used by structural designers to assist in the creation of design concepts and to demonstrate that these concepts satisfy project requirements. Lecture 1 CV2102: Structures 1 5 CV2102: Structures 1 Downloaded by Govindah Oke (govindahoke2@gmail.com) 2 lOMoARcPSD|15618290 CV2102: Structures 1 CV2102: Structures 1 Lecture 1 CV2102: Structures 1 10 CV2102: Structures 1 Introduction Structural forms and classifications What is a structure? • Structural forms and classifications • Structures refer to a system of connected parts used to support a load (e.g. self-weight, occupants/furniture for building, traffics for bridges) – Structural members and their characteristics – Different forms of structures and their properties • Factors to consider – Safety (most important!) – Serviceability (e.g. extensive deflection or vibration is not allowed) – Esthetics – Economic and environmental constraints (cost to build and use, energy consumption etc.) • Loads – Different loadings type and their characteristics • Structural analysis and design – An introduction to the basic concepts Lecture 1 11 Downloaded by Govindah Oke (govindahoke2@gmail.com) Lecture 1 12 3 lOMoARcPSD|15618290 CV2102: Structures 1 Structural members and their characteristics Tie Rod and Struts • Tie rods normally refer to structural members subjected to tensile force only • Normally tie rods are rather slender • Struts are members that could subject to both tensile and compressive forces • They are commonly used in trusses and bridges • For the ease of design and construction, structures are often composing of a number of structural elements or members • In this course, the following commonly encountered member types shall be discussed: – Tie rods – Struts – Beams – Columns More on tie rod in Wikipedia: http://en.wikipedia.org/wiki/Tie_rod More on tie strut in Wikipedia: http://en.wikipedia.org/wiki/Strut Lecture 1 CV2102: Structures 1 •Compression Structures Strut 13 Lecture 1 14 CV2102: Structures 1 •Force transfer mechanism in an arch •Stable 3-hinge arch continues to support the load Downloaded by Govindah Oke (govindahoke2@gmail.com) 4 lOMoARcPSD|15618290 CV2102: Structures 1 CV2102: Structures 1 CV2102: Structures 1 CV2102: Structures 1 •Tension Structures Beams Columns • Beams are usually straight horizontal members used primarily to carry vertical loads • They are classified to how they are supported • Design to resist bending moment and shear forces • In steel beam, wide flange cross section are commonly used • Columns usually straight vertical members to resist vertical compression • Tubes and wide flange cross section are often used • A column is called a beam-column if it is subjected to both vertical compression and bending More about beam in Wikipedia: http://en.wikipedia.org/wiki/Beam_% 28structure%29 Lecture 1 More about column in Wikipedia: http://en.wikipedia.org/wiki/Column 19 Downloaded by Govindah Oke (govindahoke2@gmail.com) Lecture 1 20 5 lOMoARcPSD|15618290 CV2102: Structures 1 CV2102: Structures 1 Structural forms and classifications Trusses • Used when long span is needed with not much depth constraints • Consist of slender struts/ties arranged in triangular or other fashions • Planer trusses (all members lie in same plane) used in bridges and roof • Space truss (members lie in different planes) used in derricks and towers • The combination of structural elements (rods, strut, beam, column) and materials (concrete, steel, timber) generate different structural forms or structural systems • In this course, focus shall be given to the following two most popular types of skeleton structural systems – Trusses – Frames (including simple beams) Lecture 1 CV2102: Structures 1 21 Lecture 1 22 CV2102: Structures 1 Structural Analysis and Design Selection of structural form (trusses, frames) and material (structural steel, concrete) Find out the loads on structures Rechecking Main study objective of this course Structural analysis (calculate the forces in the structure) Rechecking Design (to provide strong enough members to resist the forces Build (to construct the structures) Lecture 1 Downloaded by Govindah Oke (govindahoke2@gmail.com) 24 6 lOMoARcPSD|15618290 CV2102: Structures 1 Next Lecture • Some essential tools for structural analysis – Principle of superposition – Equilibrium – Internal forces – Free body diagrams Lecture 1 25 Downloaded by Govindah Oke (govindahoke2@gmail.com) 7 lOMoARcPSD|15618290 CV2102: Structures 1 CV2102: Structures 1 Idealization structures • In real sense, an exact analysis is almost impossible due to uncertainty in members dimensions and loading location (especially before the structure was built!) • Hence, in practice, in order to carry out force analysis, structural engineers must know how to idealize a structures to a simple form/geometry for analysis • Structural prismatic elements like rod/beam/column members are normally idealized as “line elements” with appropriate support conditions • At the same time an appropriate idealized loading and supports and connections should also be applied to the structure Lecture 2 •Idealization structures Geometry Loading Supports and connections •Principle of superposition •Equilibrium •Internal forces •Free body diagrams Lecture 2 CV2102: Structures 1for •Simplification 1 the purpose of analysis Lecture 2 CV2102: Structures 1 •All structures have to be simplified in a convenient form to carry out the structural analysis; actual structures often have complicated details •Experience and good judgement of structural behaviour are required in carrying out such simplification Support connections • Real and actual supports/connections could come with many different details • In structural analysis, they are idealized into a few supports and connection types Pin connection Lecture 2 2 3 Downloaded by Govindah Oke (govindahoke2@gmail.com) Fixed connection Lecture 2 4 1 lOMoARcPSD|15618290 CV2102: Structures 1 CV2102: Structures 1 Idealized connection types Idealized structures Roller support: no relative vertical movement, allows some freedoms for slight rotation Pin connection: no relative horizontal and vertical movement, allows some freedoms for slight rotation Fixed joint: no relative horizontal, vertical movement and rotation • When selecting the model for each support, the engineer must be aware how the assumptions will affect the actual performance • The assumptions must be reasonable for the structural design • The simplified loadings should give results that closely approximate the actual loadings • Common types of connections on coplanar structures are given in Table 2.1 of Textbook (Page 37) 90 Roller support Pin support Pin connection =90 Fixed connection Lecture 2 5 CV2102: Structures 1 Lecture 2 6 CV2102: Structures 1 •Reactions •Different types of idealised support conditions for planar structures: Pin support with Fx and Fy Tie support with Fn Generally known as roller support with only one reaction force Fy Partially fixed or fixed support with Fx , Fy and M Downloaded by Govindah Oke (govindahoke2@gmail.com) 2 lOMoARcPSD|15618290 CV2102: Structures 1 or a combination of bodies is to be analysed and it is Quite often, a body isolated as a single body from the surrounding bodies. Hence, a free body diagram is a diagrammatic representation of the isolated body or combination of bodies treated as a single body, showing ALL forces applied to it and by mechanical contact with other bodies that are imagined to be removed. CV2102: Structures 1 •Example: Draw the free body diagram for the beam. Neglect the weight of the beam. Idealized structures: An example Lecture 2 CV2102: Structures 1 •Example: A cylinder is supported on a smooth inclined surface by a two-bar frame . Draw the free body diagram for the cylinder, the two bar frame and the pin at C. 9 CV2102: Structures 1 •Example: Draw the free body diagram for the pulley, the post and the beam CD. Downloaded by Govindah Oke (govindahoke2@gmail.com) 3 lOMoARcPSD|15618290 CV2102: Structures 1 CV2102: Structures 1 Equations of Equilibrium • For the general 3D cases, six equations: Fx 0 M x 0 Fy 0 M y 0 Fz 0 M z 0 • For planar structures (and problems appear in this course), reduces to 3 equations: Fx 0 , F y 0 , M o 0 Lecture 2 CV2102: Structures 1 14 CV2102: Structures 1 Principle of Superposition Principle of Superposition • Forms the basis for much theory of structural analysis in this course • The total displacements or internal loadings P (stress) at a point in a structure subjected to several external loadings can be determined by adding together the displacements or internal loadings (stress) caused by each of the external loads acting separately • Linear relationship exist among loads, stresses and displacements Lecture 2 2P 2 • 2 requirements for the principle to apply: – Material must behave in a linear-elastic manner, Hooke’s Law is valid – The geometry of the structure must not undergo significant change when the loads are applied, small displacement theory 15 Downloaded by Govindah Oke (govindahoke2@gmail.com) Lecture 2 16 4 lOMoARcPSD|15618290 CV2102: Structures 1 CV2102: Structures 1 More examples on FBD D More examples on FBD P P 1 C F 2 y x 1 FxB FxA B A FyA 2 P FyB FBD for the whole structure FxA Fxc FyA Fxc M Fyc Fyc FxA + 0 = 0 FyA – F = 0 Moment above A: M + F3 = 0 FBD for ACD FxA Lecture 2 FyB 17 CV2102: Structures 1 •12kN •18kN.m •K •1m •1m Lecture 2 •18kN.m •8kN/m •10kN •22kN.m •18kN.m •12kN •8kN/m •10kN •22kN.m •K •4m •1m•1m •2m •29kN •8kN/m •22kN.m •10kN •25kN •17 • • •x=17/8 •K •29kN 18 CV2102: Structures 1 •29 •12kN FxB FBD for CB FyA •15 •18 •25kN •11 •28 •46.0625 •32 •17 •10 • •Q(kN) •20 •M(kN·m) 19 / 47 Downloaded by Govindah Oke (govindahoke2@gmail.com) 5 lOMoARcPSD|15618290 CV2102: Structures 1 •80kN.m •160kN •1m •1m •40kN/m •2m •80kN.m •40kN •130 •80kN.m •160kN •40kN/m •310kN •120 • •30 •130 •130kN •310kN • •40kN • •210 •40kN •40kN/m •130kN •2m •4m •160kN •340 • •190 •160 •Q(kN) •M(kN·m) •280 22 / 47 哈工大 土木工程学院 CV2102: Structures 1 CV2102: Structures 1 Some hints on drawing FBD and equilibrium equations (EE) Internal forces • FBD should never contain any support • When removing the supports to the structures, replaced them by the corresponding reactions • The number of reactions should always equal to the numbers of constrains removed. • Away fix the global x and y axes before writing down the EE, this shall fix the positive directions of forces • In general, one could always assume that all reactions are acting in the positive direction • When writing down the EE, always use the condition of “Sum-To-Zero”. That is, instead of writing: FxB = -Pcos(45) one shall write: • Internal forces are forces developed inside a member under the action of external loadings and reactions from the supports • In structural analysis, to find out the internal forces of members are very important as such member forces information are critical to the design of the structures so that engineers could design a member section size strong enough to resist the internal forces • Hence, it is very important that in structural analysis all the member forces are calculated correctly. Otherwise, it could lead to error in design and eventually failure of members and structures FxB + Pcos(45) = 0 Lecture 2 23 Downloaded by Govindah Oke (govindahoke2@gmail.com) Lecture 2 24 6 lOMoARcPSD|15618290 CV2102: Structures 1 CV2102: Structures 1 Next Lecture • To obtain member force, one needs to make a “cut” to the member under concern and expose the corresponding internal forces P • Determinacy and stability • Simple applications of equilibrium equations P B A A C A simple beam ACB N N M M V B V C A cut before C exposes the internal forces N, V and M P FxA A N N M M V FyA B V C FxB FyB FBD for RHS FBD for LHS Lecture 2 25 Downloaded by Govindah Oke (govindahoke2@gmail.com) Lecture 2 26 7 lOMoARcPSD|15618290 Determinacy • Equilibrium equations (EE) provide sufficient conditions for equilibrium • When all forces (internal and reactions) can be determined strictly from the EE, the structure is referred as statically determine • If the structure has more unknown forces than the numbers of EE , the structure is a statically indeterminate structure • If a structure is statically indeterminate, addition equations, known as compatibility equations are needed to solve for the forces • In this course, we shall only work on statically determine planar (2D) structures Lecture 3 • Determinacy • Stability • Simple application of equilibrium equations •For stability, a minimum of 3 reaction components is needed: 1 Lecture 3 2 3 Lecture 3 4 Stability • To ensure equilibrium of a structure or its members: – Must satisfy EEs – Members must be properly “held” or constrained by their supports • If the structure is not properly supported, instability may be resulted and lead to collapse of the (partially or whole) structure • Instability could be caused by – Partial constraints (i.e. no enough numbers of constraints / support reactions) – Improper Constraints (i.e. enough numbers of constraints but put in wrong positions/ directions) Lecture 3 Downloaded by Govindah Oke (govindahoke2@gmail.com) 1 lOMoARcPSD|15618290 Lecture 3 5 Lecture 3 6 Lecture 3 7 Lecture 3 8 Downloaded by Govindah Oke (govindahoke2@gmail.com) 2 lOMoARcPSD|15618290 •Examples of external statical classification: •Computation of reactions using Equation of Equilibrium •Example: Determine the reactions for the structure. Lecture 3 Downloaded by Govindah Oke (govindahoke2@gmail.com) 12 3 lOMoARcPSD|15618290 •Example: A pin-connected two-bar frame is supported and loaded as shown below. Determine the reactions at supports A and B. •And they must be properly arranged: In some cases, the number of reaction forces may equal EEs. However, instability or movement of structure could still occur if support reactions are • Concurrent (i.e. pass through) at a point • All parallel and acting along one direction only Simple application of EEs Simple application of EEs Procedure of analysis using EEs Some useful hints (1) Identify all the supports and hinge connections (if any) (2) Remove the supports and re-apply the reactions (3) Draw FBD for the whole structures or FBDs for separate parts of the structure. Remember that 3 EEs could by applied to each separated FBD (4) Apply EEs in turns (to different separate parts if needed) to solve for the unknown reaction and forces (5) Always try to solve one unknown at a time (1) There is no need to know in advance the directions of the reaction before the analysis. One could always assuming the reactions are acting in the positive directions, if eventually the answer is negative, it simply implies that the reaction is acting in the opposite direction (2) When apply the EEs, very often the solutions will become very straightforward by take moment (i.e. using the M=0 condition) at appropriate point (3) Always write the EE in “sum-to-zero” form, this could reduce the chance of making errors Lecture 3 15 Downloaded by Govindah Oke (govindahoke2@gmail.com) Lecture 3 16 4 lOMoARcPSD|15618290 Numerical examples The 6kN/m udl is converted to 36kN at its centroid Note: the 6.8kNm moment at the free end should be included when taking moment at A Fx 0; Ax 270 cos 600 0 Ax 135kN Segment BC : With anti - clockwise in the direction, With anti - clockwise in the direction, 0 0 M A 0; 270 sin 60 (3) 270 cos 60 (0.3) B y (4.2) 6.8 0 M c 0; 8 B y (4.5) 0 B y 1.78kN B y 159kN Fy 0; 1.78 C y 0 C y 1.78kN Fy 0; 270 sin 600 159 Ay 0 Fx 0; Bx 0 Ay 74.8kN Lecture 3 17 Lecture 3 18 •Example : •Compute the reactions for the structure shown below. Segment AB : With anti - clockwise in the direction, M A 0; M A 36(3) (1.78)(6) 0 M A 97.3kN.m Fy 0; Ay 36 1.78 0 Ay 34.2kN Fx 0; Ax 0 Lecture 3 19 Downloaded by Govindah Oke (govindahoke2@gmail.com) 5 lOMoARcPSD|15618290 •40kN/m •A •120kN •B •8m •2m •D •K •C •3m •3m •120kN •40kN/m •145kN •40kN/m •A •C •K •A •(kN·m) • •180 •263 •60 • • •175 •20kN •D •120 •145 •3m •10kN • •235kN •Home Works •120kN •B •60kN •60kN •60 (kN) •3m •B •3m •10kN •E •2m •2m •5kN/m •2m •2m Downloaded by Govindah Oke (govindahoke2@gmail.com) •2kN/m •C •2m •20kN·m •2m •2m •F •2m •D •4m •10kN •2m •5m 6 lOMoARcPSD|15618290 •q •B •A •B •x •D •l 1 q( l x ) 2 •A •D •C 1 q( l x ) 2 8 •B q( l x ) 1 1 2 x qx 2 ql x 2 2 8 •C 1 q( l x ) •D •C •l - x •l 1 1 q( l x ) x qx 2 2 2 •C •l - x •l •A •D •l •q •A •x 2 •B x (3 2 2)l 0.172l •F •q P •A •D •a •B •a •a •C •E •a •FPa/2 •FP/4 •FQ •F •A •C •B l l •FP a •FPa/4 •M •a •FP •FP/2 • 0.5ql 2 • 0.125ql 2 • 0.5ql Downloaded by Govindah Oke (govindahoke2@gmail.com) •M(kN·m) 0.5ql • •FQ(kN) 7 lOMoARcPSD|15618290 Past Years Exam Past Years Exam Past Years Exam Past Years Exam Downloaded by Govindah Oke (govindahoke2@gmail.com) 8 lOMoARcPSD|15618290 Past Years Exam Some remarks • More examples could be found in the textbook • When udl is involved, one needs to convert it to the equivalent point load acts at the centroid • In all exercises, calculations are always very simple but one need to draw FBDs and apply EEs correctly to get the correct answers • More exercises are given in Tutorial sheet 1 • More practice is useful to allow one to solve the this type of question in a faster speed with less chances to make error Lecture 3 34 Next Lecture • Introduction to planar truss • Determinacy and stability of truss Lecture 3 35 Downloaded by Govindah Oke (govindahoke2@gmail.com) 9 lOMoARcPSD|15618290 CV2102: Structures 1 CV2102: Structures 1 Introduction to planar trusses • A truss is a structure composed of slender members joined together at their end points • The joint connections are usually formed by bolting or welding the ends of the members to a common plate called gusset plate • Planar trusses lie in a single plane and is often used to support roof or bridges Lecture 4 • • • • Introduction to planar truss Roof and bridge trusses Internal forces Determinacy and stability of truss Lecture 4 1 CV2102: Structures 1 Lecture 4 CV2102: Structures 1 Roof trusses 2 Different forms of roof trusses • They are often used as part of an industrial building frame • Roof load is transmitted to the truss at the joints by means of a series of purlins • To keep the frame rigid and thereby capable of resisting horizontal wind forces, knee braces are sometimes used at the supporting column Lecture 4 3 Downloaded by Govindah Oke (govindahoke2@gmail.com) Lecture 4 4 1 lOMoARcPSD|15618290 CV2102: Structures 1 CV2102: Structures 1 Different forms of bridge trusses Bridge trusses • The load on the deck is first transmitted to the stringers floor beams joints of supporting side truss • The top and bottom cords of these side trusses are connected by top and bottom lateral bracings resisting lateral forces • Additional stability is provided by the portal and sway bracing • In the case of a long span truss, a roller is provided at one end for thermal expansion Lecture 4 5 CV2102: Structures 1 Lecture 4 6 CV2102: Structures 1 Internal force of truss members Assumptions of analysis and design P1 1. The gusset plates provide limited amount of moment resistance 2. The members are joined together by smooth pins 3. No moment at the two ends of members 4. All loadings are applied at the joints 5. All supports are also located at the joints 6. The above assumptions lead to the conclusion that only axial forces (tension or compression) are acting in the all members regarding of their sizes, locations and dimensions y x Set the x and y axes parallel and perpendicular to the orientation of the member respectively V1 Assume that the two forces at the two ends are P1, V1 and P2 and V2 respectively V2 P1 P2 y x V1=0 Using the conditions Fx=0 and Fy=0, the forces must be equal and opposite at the two ends Now taking moment at one end since M=0 LV1=0 where L0 is the length of the member. Hence, V1=0 and only the equal and opposite axial force exist V1 =0 P1 Lecture 4 Since both end of any truss members are pinned end, there is no moment at the ends Thus, all truss members are axial force members 7 Downloaded by Govindah Oke (govindahoke2@gmail.com) Lecture 4 8 2 lOMoARcPSD|15618290 CV2102: Structures 1 CV2102: Structures 1 Classification of coplanar trusses Compound trusses Simple Truss • Simplest framework of rigid (stable) with the basic “stable” triangle element (ADC) • Addition members are added in to connect with the basic triangle • Note that simple trusses may not consist entirely of triangles Formed by joining two or more simple trusses together Type 1, Figure (a) : connected by a common joint and bar Type 2 , Figure (b) : connected by three bars Type 3, Figure (c): substituted bars of a truss by simple trusses Lecture 4 9 CV2102: Structures 1 Lecture 4 10 CV2102: Structures 1 Determinacy of planar trusses Complex truss A complex truss is one that cannot be classified as being either simple or compound trusses Lecture 4 •Overall classification 11 Downloaded by Govindah Oke (govindahoke2@gmail.com) 3 lOMoARcPSD|15618290 CV2102: Structures 1 CV2102: Structures 1 •Condition equations CV2102: Structures 1 CV2102: Structures 1 Internal stability of trusses Internal stability of trusses • To determine the internal stability of a compound truss, it is necessary to identify the way in which the simple truss are connected together • A truss is internally unstable if some of it components forms a collapsible mechanism • The internal stability can be checked by careful inspection of the arrangement of its members • If it can be determined that each joint is held fixed so that it cannot move in a “rigid body” sense with respect to the other joints, then the truss will be stable • However, if a truss is constructed so that it does not hold its joints in a fixed position, it will be unstable or have a “critical form” • A simple truss will always be internally stable • The truss at the top is unstable as there is no restraint between joints C and F or B and E • The truss at the bottom unstable since the inner simple truss ABC is connected to DEF using 3 bars which are concurrent at point O. • An external load can be applied at A, B or C and causes the truss to rotate slightly Lecture 4 15 Downloaded by Govindah Oke (govindahoke2@gmail.com) Lecture 4 16 4 lOMoARcPSD|15618290 CV2102: Structures 1 CV2102: Structures 1 •Examples CV2102: Structures 1 •Examples CV2102: Structures 1 Next Lecture • Analysis of planar trusses I: method of joints • Zero force members Lecture 4 19 Downloaded by Govindah Oke (govindahoke2@gmail.com) Lecture 4 20 5