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    2302 STRUCTURAL ANALYSIS CLASSICAL METHODS 3 1 0 4

    JECTIVE

    e members of a structure are subjected to internal forces like axial forces, shearing forces, bending and torsionalments while transferring the loads acting on it. Structural analysis deals with analysing these internal forces in thembers of the structures. At the end of this course students will be conversant with classical method of analysis.

    IT I DEFLECTION OF DETERMINATE STRUCTURES 9

    nciples of virtual work for deflections Deflections of pin-jointed plane frames and rigid plane frames Willotgram - Mohrs correction

    IT II MOVING LOADS AND INFLUENCE LINES 9(DETERMINATE & INDETERMINATE STRUCTURES)

    uence lines for reactions in statically determinate structures influence lines for members forces in pin-jointedmes Influence lines for shear force and bending moment in beam sections Calculation of critical stress resultants

    to concentrated and distributed moving loads.

    ller Breslaus principle Influence lines for continuous beams and single storey rigid frames Indirect modellysis for influence lines of indeterminate structures Beggs deformeter

    IT III ARCHES 9

    ches as structural forms Examples of arch structures Types of arches Analysis of three hinged, two hinged anded arches, parabolic and circular arches Settlement and temperature effects.

    IT IV SLOPE DEFLECTION METHOD 9

    ntinuous beams and rigid frames (with and without sway) Symmetry and antisymmetry Simplification for hinged Support displacements.

    IT V MOMENT DISTRIBUTION METHOD 9

    tribution and carry over of moments Stiffness and carry over factors Analysis of continuous beams Plane rigidmes with and without sway Naylors simplification.

    TUTORIAL 15 TOTAL : 60

    XT BOOKS

    Comprehensive Structural Analysis Vol. 1 & Vol. 2, Vaidyanadhan, R and Perumal, P, Laxmi Publications,New Delhi, 2003

    Structural Analysis, L.S. Negi & R.S. Jangid, Tata McGraw-Hill Publications, New Delhi, Sixth Edition, 2003

    Punmia B.C., Theory of Structures (SMTS ) Vol II laxmi Publishing Pvt ltd, New Delhi, 2004

    FERENCES

    Analysis of Indeterminate Structures C.K. Wang, Tata McGraw-Hill, 1992

    Comment [R2]:

    Comment [R1]:

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    V SEMESTER CIVIL ENGINEERING

    CE2302 STRUCTURAL ANALYSIS

    NOTES OF LESSON

    UNIT DEFLECTION OF DETERMINATE STRUCTURES

    eorem of minimum Potential Energy

    ential energy is the capacity to do work due to the position of body. A body of weight W held at a height h

    sess energy Wh. Theorem of minimum potential energy states that Of all the displacements which satisfy the

    undary conditions of a structural system, those corresponding to stable equilibrium configuration make the

    al potential energy a relative minimum . This theorem can be used to determine the critical forces causing

    ability of the structure.

    w of Conservation of Energy

    m physics this law is stated as Energy is neither created nor destroyed. For the purpose of structural analysis, the

    w can be stated as If a structure and external loads acting on it are isolated, such that it neither receive nor

    e out energy, then the total energy of the system remain constant . With reference to figure 2, internal energy is

    ressed as in equation (9). External work done W e = -0.5 P dL. From law of conservation of energy U i+W e =0. From

    it is clear that internal energy is equal to external work done.

    nciple of Virtual Work:

    tual work is the imaginary work done by the true forces moving through imaginary displacements or vice versa. Realrk is due to true forces moving through true displacements. According to principle of virtual work The total virtual

    rk done by a system of forces during a virtual displacement is zero .

    eorem of principle of virtual work can be stated as If a body is in equilibrium under a Virtual force system and

    mains in equilibrium while it is subjected to a small deformation, the virtual work done by the external forces

    qual to the virtual work done by the internal stresses due to these forces. Use of this theorem for computation

    displacement is explained by considering a simply supported bea AB, of span L, subjected to concentrated load P at

    as shown in Fig.6a. To compute deflection at D, a virtual load P is applied at D after removing P at C. Work done is

    o a s the load is virtual. The load P is then applied at C, causing deflection C at C and D at D, as shown in Fig. 6b.

    ernal work done W e by virtual load P is . If the virtual load P produces bending moment M, then the

    ernal strain energy stored by M acting on the real deformation d in element dx over the beam equation (14)

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    ere, M= bending moment due to real load P. From principle of conservation of energy W e=W i

    P=1 then

    milarly for deflection in axial loaded trusses it can be shown that

    ere,

    = Deflection in the direction of unit loadP = Force in the i th member of truss due to unit load

    P = Force in the i th member of truss due to real external load

    A BC D

    a

    xL

    P

    A BC D

    ax

    L

    P PC D

    Fig.6a

    Fig.6b

    =L

    0

    D

    EI2dxMM'

    2

    P'

    (16) EI

    dxMM' L

    0D =

    (17) AE

    dxPP'

    n

    0=

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    n = Number of truss members

    L = length of i th truss members.

    e of virtual load P = 1 in virtual work theorem for computing displacement is called

    it Load Method

    stigliones Theorems:

    stigliano published two theorems in 1879 to determine deflections in structures and redundant in statically

    eterminate structures. These theorems are stated as:

    t Theorem : If a linearly elastic structure is subjected to a set of loads, the partial derivatives of total

    in energy with respect to the deflection at any point is equal to the load applied at that point

    nd Theorem : If a linearly elastic structure is subjected to a set of loads, the partial derivatives of totalin energy with respect to a load applied at any point is equal to the deflection at that point

    e first theorem is useful in determining the forces at certain chosen coordinates. The conditions of equilibrium of

    se chosen forces may then be used for the analysis of statically determinate or indeterminate structures. Second

    orem is useful in computing the displacements in statically determinate or indeterminate structures.

    ttis Law:

    tates that If a structure is acted upon by two force systems I and II, in equilibrium separately, the external

    ual work done by a system of forces II during the deformations caused by another system of forces I is equal

    xternal work done by I system during the deformations caused by the II system

    (18) N.....1,2, j P

    U j

    j

    ==

    (19)N.1,2,...... j PU

    j j

    ==

    I II

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    body subjected to two system of forces is shown in Fig 7. W ij represents work done by ith system of force on

    placements caused by j th system at the same point. Bettis law can be expressed as W ij = W ji, where W ji represents

    work done by j th system on displacement caused by i th system at the same point.

    usses Two Dimensional Structures

    ThreeDimensionalStructures

    Conditions of Equilibrium and Static Indeterminacy

    A body is said to be under static equilibrium, when it continues to be under rest after application of loads. Duringtion, the equilibrium condition is called dynamic equilibrium. In two dimensional system, a body is in equilibriumen it satisfies following equation.x=0 ; Fy=0 ; Mo=0 ---1.1

    Fig. 7

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    To use the equation 1.1, the force components along x and y axes are considered. In three dimensional systemilibrium equations of equilibrium arex=0 ; Fy=0 ; Fz=0;

    Mx=0 ; My=0 ; Mz=0; ----1.2

    To use the equations of equilibrium (1.1 or 1.2), a free body diagram of the structure as a whole or of any part ofstructure is drawn. Known forces and unknown reactions with assumed direction is shown on the sketch while

    wing free body diagram. Unknown forces are computed using either equation 1.1 or 1.2

    Before analyzing a structure, the analyst must ascertain whether the reactions can be computed using equationsequilibrium alone. If all unknown reactions can be uniquely determined from the simultaneous solution of theations of static equilibrium, the reactions of the structure are referred to as statically determinate . If they cannot beermined using equations of equilibrium alone then such structures are called statically indeterminate structures. Ifnumber of unknown reactions are less than the number of equations of equilibrium then the structure is staticallytable.

    The degree of indeterminacy is always defined as the difference between the number of unknown forces and thember of equilibrium equations available to solve for the unknowns. These extra forces are called redundants.eterminacy with respect external forces and reactions are called externally indeterminate and that with respect to

    ernal forces are called internally indeterminate.

    A general procedure for determining the degree of indeterminacy of two-dimensional structures are given below:

    NUK= Number of unknown forcesNEQ= Number of equations availableIND= Degree of indeterminacyIND= NUK - NEQ

    eterminacy of Planar Frames

    For entire structure to be in equilibrium, each member and each joint must be in equilibrium (Fig. 1.9)

    NEQ = 3NM+3NJ

    NUK= 6NM+NR

    IND= NUK NEQ = (6NM+NR)-(3NM+3NJ)

    IND= 3NM+NR-3NJ ----- 1.3ree independent reaction componentsee body diagram of Members and Joints

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    Degree of Indeterminacy is reduced due to introduction of internal hinge

    NC= Number of additional conditions

    NEQ = 3NM+3NJ+NC

    NUK= 6NM+NR

    IND= NUK-NEQ = 3NM+NR-3NJ-NC ------------1.3a

    eterminacy of Planar TrussesMembers carry only axial forces

    NEQ = 2NJ

    NUK= NM+NR

    IND= NUK NEQ

    IND= NM+NR-2NJ ----- 1.4

    eterminacy of 3D FRAMES

    A member or a joint has to satisfy 6 equations of equilibrium

    NEQ = 6NM + 6NJ-NC

    NUK= 12NM+NR

    IND= NUK NEQ

    IND= 6NM+NR-6NJ-NC ----- 1.5

    eterminacy of 3D Trusses

    A joint has to satisfy 3 equations of equilibrium

    NEQ = 3NJ

    NUK= NM+NR

    IND= NUK NEQ

    IND= NM+NR-3NJ ----- 1.6

    ble Structure:

    Another condition that leads to a singular set of equations arises when the body or structure is improperly restrainedagainst motion. In some instances, there may be an adequate number of support constraints, but their arrangementmay be such that they cannot resist motion due to applied load. Such situation leads to instability of structure. Astructure may be considered as externally stable and internally stable.

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    ernally Stable:

    Supports prevents large displacements

    No. of reactions No. of equations

    ernally Stable:

    Geometry of the structure does not change appreciably

    For a 2D truss NM 2Nj -3 (NR 3)

    For a 3D truss NM 3Nj -6 (NR 3)

    amples:

    termine Degrees of Statical indeterminacy and classify the structures

    f)

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    a)

    b)

    c)

    d)

    e)

    f)

    NM=2; NJ=3; NR =4; NC=0IND=3NM+NR-3NJ-NCIND=3 x 2 + 4 3 x 3 -0 = 1INDETERMINATE

    NM=3; NJ=4; NR =5; NC=2IND=3NM+NR-3NJ-NCIND=3 x 3 + 5 3 x 4 -2 = 0DETERMINATE

    NM=3; NJ=4; NR =5; NC=2IND=3NM+NR-3NJ-NCIND=3 x 3 + 5 3 x 4 -2 = 0DETERMINATE

    NM=3; NJ=4; NR =3; NC=0IND=3NM+NR-3NJ-NCIND=3 x 3 + 3 3 x 4 -0 = 0DETERMINATE

    NM=1; NJ=2; NR =6; NC=2IND=3NM+NR-3NJ-NCIND=3 x 1 + 6 3 x 2 -2 = 1

    INDETERMINATE

    NM=1; NJ=2; NR =5; NC=1IND=3NM+NR-3NJ-NCIND=3 x 1 + 5 3 x 2 -1 = 1

    INDETERMINATE

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    M=1; NJ=2; NR =5; NC=1D=3NM+NR-3NJ-NCD=3 x 1 + 5 3 x 2 -1 = 1INDETERMINATE

    ch support has 6 reactions

    M=8; NJ=8; NR =24; NC=0

    D=6NM+NR-6NJ-NCD=6 x 8 + 24 6 x 8 -0 = 24INDETERMINATE

    ach support has 3 reactions

    M=18; NJ=15; NR =18; NC=0D=6NM+NR-6NJ-NCD=6 x 18 + 18 6 x 15 = 36INDETERMINATE

    ss

    M=2; NJ=3; NR =4;D=NM+NR-2NJD= 2 + 4 2 x 3 = 0

    DETERMINATE

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    ussM=11; NJ=6; NR =4;D=NM+NR-2NJD= 11 + 4 2 x 6 = 3

    INDETERMINATE

    Degree of freedom or Kinematic IndeterminacyMembers of structure deform due to external loads. The minimum number of parameters required to uniquelydescribe the deformed shape of structure is called Degree of Freedom. Displacements and rotations at variouspoints in structure are the parameters considered in describing the deformed shape of a structure. In framed structurethe deformation at joints is first computed and then shape of deformed structure. Deformation at intermediate points

    on the structure is expressed in terms of end deformations. At supports the deformations corresponding to a reactionis zero. For example hinged support of a two dimensional system permits only rotation and translation along x andy directions are zero. Degree of freedom of a structure is expressed as a number equal to number of freedisplacements at all joints. For a two dimensional structure each rigid joint has three displacements as shown in

    case of three dimensional structure each rigid joint has six displacement. Expression for degrees of freedom

    31. 2D Frames: NDOF = 3NJ NR NR62. 3D Frames: NDOF = 6NJ NR NR33. 2D Trusses: NDOF= 2NJ NR NR64. 3D Trusses: NDOF = 3NJ NR NR

    Where, NDOF is the number of degrees of freedom

    In 2D analysis of frames some times axial deformation is ignored. Then NAC=No. of axial condition is deductedfrom NDOF

    amples:

    Determine Degrees of Kinermatic Indeterminacy of the structures given below

    TrussNM=14; NJ=9; NR =4;IND=NM+NR-2NJIND= 14+ 4 2 x 9 = 0

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    ensible Inextensible=2; NR =3; NAC=1 NJ=2; NR =3;OF=3NJ-NR NDOF=3NJ-NR-NACOF=3 x 2 3= 3 (1, 2, 2) NDOF=3 x 2 3-1= 2 (1, 2)

    A TrussJ=6; NR =3;DOF=2NJ-NRDOF=2 x 6 3 = 9

    tensible=4; NR =5;

    DOF=3NJ-NRDOF=3 x 4 5= 7

    1, 21, 23 3 ,y2,e1,e2)

    InextensibleNJ=4; NR =5; NAC=2NDOF=3NJ-NR-NACNDOF=3 x 4 5-2= 5(1, 21, 23 , 3 y2)

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    NJ=6; NR =4;NDOF=2NJ-NRNDOF=2 x 6 4 = 8

    Stress-Strain Graph

    Virtual Work

    tual work is defined as the following line integral

    where

    C is the path or curve traversed by the object, keeping all constraints satisfied;

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    is the force vector;

    is the infinitesimal virtual displacement vector.

    tual work is therefore a special case of mechanical work. For the work to be called virtual, the motion undergone by the

    tem must be compatible with the system's constraints, hence the use of a virtual displacement.

    e of the key ideas of Lagrangian mechanics is that the virtual work done by the constraint forces should be zero. This is a

    sonable assumption, for otherwise a physical system might gain or lose energy simply by being constrained (imagine a

    d on a stationary hoop moving faster and faster for no apparent reason)!

    idea of virtual work also plays a key role in interpreting D'Alembert's principle:

    uilibrium of forces (staic treatment)

    virtual work produced by inertia force

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    virtual work rpoduced by net applied force.

    e:

    quirements on :

    - compatible with the kinematic constraints, but otherwise arbitrary

    - instantaneous

    - increasingly small

    a single body B i :

    a system of n bodies B:

    grange form of dAlemberts Principle

    s formalism is convenient, as the constraint (non-working) loads disappear. (forces, torques) where

    is the vector of independent degrees-of-freedom.

    mple (i)

    motivation for introducing virtual work can be appreciated by the following simple example from statics of particles.

    pose a particle is in equilibrium under a set of forces Fx i , Fy i , Fz i i = 1,2,... n :

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    ltiplying the three equations with the respective arbitrary constants x , y , z :

    (b)

    en the arbitrary constants x , y , z are thought of as virtual displacements of the particle, then the left-hand-sides of (b)

    resent the virtual work. The total virtual work is:

    (c)

    ce the preceding equality is valid for arbitrary virtual displacements, it leads back to the equilibrium equations in (a). The

    ation (c) is called the principle of virtual work for a particle. Its use is equivalent to the use of many equilibrium equations.

    plying to a deformable body in equilibrium that undergoes compatible displacements and deformations, we can find the

    l virtual work by including both internal and external forces acting on the particles. If the material particles experience

    mpatible displacements and deformations, the work done by internal stresses cancel out, and the net virtual work done

    uces to the work done by the applied external forces. The total virtual work in the body may also be found by the volume

    gral of the product of stresses and virtual strains :

    us, the principle of virtual work for a deformable body is:

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    s relation is equivalent to the set of equilibrium equations written for the particles in the deformable body. It is valid

    spective of material behaviour, and hence leads to powerful applications in structural analysis and finite element analysis.

    w consider a block on a surface

    plying formula (c) gives:

    leads to

    serve virtual work formalism leads directly to Newtons equation of motion in the kinematically allowable direction.

    mple (ii)

    o bodies connected by a rotary joint.

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    e: Internal forces do no work since these forces are always equal and opposite.

    mple 2

    physical quantity work is defined as the product of force times a conjugate displacement, i.e., a displacement in the

    me direction as the force we are considering. We are familiar with real work, i.e., the product of a real force and a real

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    placement, i.e., a force and a displacement that both actually occur. The situation is illustrated in Part 1 of the following

    ure:

    can extend the concept of real work to a definition of virtual work, which is the product of a real force and a conjugate

    placement, either real or virtual. In Part 2 of the example shown above, we assume that the cantilever column loaded with

    ce P undergoes a virtual rotation of magnitude at its base. We compute the virtual work corresponding to this virtual

    placement by summing the products of real forces times conjugate virtual displacements.

    this calculation, we must introduce unknown sectional forces at those locations where we have cut the structure to create

    virtual displacement. In the example shown above, therefore, we have introduced bending moment at the base, M b. For

    mpleteness, we would also have to introduce a shear force V and an axial force N at the base of the column, but, as we

    ll see, there is no component of virtual displacement conjugate to these forces. They have therefore not been shown in

    example.

    calculate the virtual displacements of the structure corresponding to all known and unknown forces. For a rotation at

    base, horizontal translation of the tip of the cantilever is L. We then multiply force times displacement and sum these

    ducts to obtain the following expression for virtual work corresponding to the assumed virtual displacement:

    P L Mb

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    treat the virtual work done by force M b as negative since the direction of M b as drawn is opposite to the direction of the

    ual rotation .

    principle of virtual work states that a system of real forces is in equilibrium if and only if the virtual work performed by

    se forces is zero for all virtual displacements that are compatible with geometrical boundary conditions.

    the example given in the previous subsection, this implies that the virtual work of the simple cantilever, U, must be zero

    the system to be in equilibrium:

    P L Mb = 0

    ce is nonzero, it follows that M b = P L, which is precisely the familiar expression for bending moment at the base of a

    tilever loaded with force P at its tip.

    more general mathematical statement of the principle of virtual work is as follows:

    Q i be a set of real loads acting on a given structure

    R i be the corresponding real support reactions

    M i, Vi, and N i be the sectional forces (bending moment, shear, and axial force) introduced at the locations where the

    cture has been cut to allow it to undergo a virtual displacement.

    Qi, Ri, Mi, Vi, and Ni be virtual displacements compatible with the geometrical boundary conditions and conjugate to

    forces defined previously.

    n the structure is in equilibrium if and and only if:

    Q i Qi) + (R i Ri) + (Mi Mi) + (Vi Vi) + (Ni Ni) = 0

    Williot diagrame Williot diagram is a graphical method to obtain an approximate value for displacement of a structure whichmitted to a certain load. The method consists of, from a graph representation of a structural system , representing the

    ucture's fixed vertices as a single, fixed starting point and from there sequentially adding the neighbouring vertices'ative displacements due to strain

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    II-UNIT-MOVING LOADS AND INFLUENCE LINES

    engineering, an influence line graphs the variation of a function (such as the shear felt in a structure member) at acific point on a beam or truss caused by a unit load placed at any point along the structure. [1][2][3][4][5] Some of the

    mmon functions studied with influence lines include reactions (the forces that the structures supports must apply iner for the structure to remain static), shear , moment , and deflection . Influence lines are important in the designingms and trusses used in bridges , crane rails, conveyor belts , floor girders, and other structures where loads will moveng their span. [5] The influence lines show where a load will create the maximum effect for any of the functionsdied.

    uence lines are both scalar and additive .[5] This means that they can be used even when the load that will be appliedot a unit load or if there are multiple loads applied. To find the effect of any non-unit load on a structure, the ordinateults obtained by the influence line are multiplied by the magnitude of the actual load to be applied. The entire

    uence line can be scaled, or just the maximum and minimum effects experienced along the line. The scaledximum and minimum are the critical magnitudes that must be designed for in the beam or truss.

    cases where multiple loads may be in effect, the influence lines for the individual loads may be added together iner to obtain the total effect felt by the structure at a given point. When adding the influence lines together, it isessary to include the appropriate offsets due to the spacing of loads across the structure. For example, if it is knownt load A will be three feet in front of load B, then the effect of A at x feet along the structure must be added to theect of B at ( x 3) feet along the structurenot the effect of B at x feet along the structure Many loads are distributedher than concentrated. Influence lines can be used with either concentrated or distributed loadings. For a concentratedpoint) load, a unit point load is moved along the structure. For a distributed load of a given width, a unit-distributedd of the same width is moved along the structure, noting that as the load nears the ends and moves off the structurey part of the total load is carried by the structure. The effect of the distributed unit load can also be obtained byegrating the point loads influence line over the corresponding length of the structures.

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    What we have looked at is quantitative influence lines. These have numerical values and can be computed. Qualitativeinfluence lines are based on a principle by Heinrich Mller Breslau, which states:

    " The deflected shape of a structure represents to some scale the influence line for a function such as reaction,shear or moment, if the function in question is allowed to act through a small distance. "

    In other words, is that the s tructure draws its own influence lines from the deflection curves. The shape of the influencelines can be created by deflecting the location in question by a moment, or shear or displacement to get idea of thebehavior of the influence line. Realizing that the supports are zero values or poles.

    Mller's principle for statically determinate st ructures is useful, but for indeterminated structures it is of great value. Youcan get an idea of the behavior of the shear and moment at a point in the beam.

    ng influence lines to calculate values

    From the previous examples of a twenty foot beam for the reactions, shear, and moment. We can use the values from theinfluence lines to calculate the shear and moment at a point.

    RAy = (F i)* Value of the influence line of R Ay @ location of the force

    V11 = (F i)* Value of the influence line of V 11 @ location of the force

    M11 = (F i)* Value of the influence line of M 11 @ location of the force

    If we are looking at the forces due to uniform loads over the beam at point. The shear or moment is equal to the area underthe influence line times the distributed load.

    RAy = (w i)* Area of the influence line of R Ay over which w covers

    V11 = (w i)* Area of the influence line of V 11 over which w covers

    M11 = (w i)* Area of the influence line of M 11 over which w covers

    For moving set of loads the influence lines can be used to calculate the maximum function. This can be done by movingthe loads over the influence line find where they will generate the largest value for the particular point.

    els or floating floor

    The method can be extend to deal with floor joist and floating floors in which we deal with panels, which are simple beamelements acting on the floor joist.

    You will need to find the fore as function of the intersection. You are going to find the moment and the shear as you moveacross the surface of the beam.

    An example problem is used to show how this can be used to find the shear and moment at a point for a moving load. Thistechnique is similar to that used in truss members.

    ethods for constructing influence lines

    ere are three methods used for constructing the influence line. The first is to tabulate the influence values for multiplents along the structure, then use those points to create the influence line. The second is to determine the influence-e equations that apply to the structure, thereby solving for all points along the influence line in terms of x, where x isnumber of feet from the start of the structure to the point where the unit load is applied. The third method is called

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    Mller-Breslau principle . It creates a qualitative influence line. This nfluence line will still provide the designer withaccurate idea of where the unit load will produce the largest response of a function at the point being studied, but itnot be used directly to calculate what the magnitude that response will be, whereas the influence lines produced byfirst two methods can.

    s possible to create equations defining the influence line across the entire span of a structure. This is done by solvingthe reaction, shear, or moment at the point A caused by a unit load placed at x feet along the structure instead of acific distance. This method is similar to the tabulated values method, but rather than obtaining a numeric solution,outcome is an equation in terms of x.[5]

    s important to understanding where the slope of the influence line changes for this method because the influence-lineation will change for each linear section of the influence line. Therefore, the complete equation will be a piecewise

    ear function which has a separate influence-line equation for each linear section of the influence line.[5]

    e Mller-Breslau Principle can be utilized to draw qualitative influence lines, which are directly proportional to theual influence line. [2] Instead of moving a unit load along a beam, the Mller-Breslau Principle finds the deflectedpe of the beam caused by first releasing the beam at the point being studied, and then applying the function (reaction,ar, or moment) being studied to that point. The principle states that the influence line of a function will have a scaledpe that is the same as the deflected shape of the beam when the beam is acted upon by the function.

    order to understand how the beam will deflect under the function, it is necessary to remove the beams capacity tost the function. Below are explanat ions of how to find the influence lines of a simply supported, rigid beam

    When determining the reaction caused at a support, the support is replaced with a roller, which cannotresist a vertical reaction. Then an upward (positive) reaction is applied to the point where the supportwas. Since the support has been removed, the beam will rotate upwards, and since the beam is rigid, itwill create a triangle with the point at the second support. If the beam extends beyond the second supportas a cantilever, a similar triangle will be formed below the cantilevers position. This means that thereactions influence line will be a straight, sloping line with a value of zero at the location of the secondsupport.

    When determining the shear caused at some point B along the beam, the beam must be cut and a roller-guide (which is able to resist moments but not shear) must be inserted at point B. Then, by applying apositive shear to that point, it can be seen that the left side will rotate down, but the right side will rotateup. This creates a discontinuous influence line which reaches zero at the supports and whose slope isequal on either side of the discontinui ty. If point B is at a support, then the deflection between point Band any other supports will still create a triangle, but if the beam is cantilevered, then the entirecantilevered side will move up or down creating a rectangle.

    When determining the moment caused by at some point B along the beam, a hinge will be placed at pointB, releasing it to moments but resisting shear. Then when a positive moment is placed at point B, bothsides of the beam will rotate up. This will create a continuous influence line, but the slopes will be equaland opposite on either side of the hinge at point B. Since the beam is simply supported, its end supports(pins) cannot resist moment; therefore, it can be observed that the supports will never experiencemoments in a static situation regardless of where the load is placed.

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    e Mller-Breslau Principle can only produce qualitative influence lines. This means that engineers can use it toermine where to place a load to incur the maximum of a function, but the magnitude of that maximum cannot beculated from the influence line. Instead, the engineer must use statics to solve for the functions value in tha t loadinge.

    example, the influence line for the support reaction at A of the structure shown in Figure 1, is found by applying at load at several points (See Figure 2) on the structure and determining what the resulting reaction will be at A. Thisbe done by solving the support reac tion Y A as a function of the position of a downward acting unit load. One suchation can be found by summing moments at Support B.

    ure 1 - Beam structure for influence li ne example

    ure 2 - Beam structure showing application of unit load

    MB = 0 (Assume counter-clockwise positive moment) A(L)+1(L-x) = 0= (L-x)/L = 1 - (x/L)

    e graph of this equation is the influence line for the support react ion at A (See Figure 3). The graph illustrates that ifunit load was applied at A, the reaction at A would be equal to unity. Similarly, if the unit load was applied at B, thection at A would be equal to 0, and if the unit load was applied at C, the reaction at A would be equal to -e/L.

    ure 3 - Influence line for the support r eaction at A

    ce an understanding is gained on how these equations and the influence lines they produce are developed, someeral properties of influence lines for statically determinate structures can be stated.

    1. For a statically determinate structure the influence line will consist of only straight line segments betweencritical ordinate values.

    2. The influence line for a shear force at a given location will contain a translational discontinuity at this location.The summation of the positive and negative shear forces at this location is equal to unity.

    3. Except at an internal hinge location, the slope to the shear force influence line will be the same on each side ofthe critical section since the bending moment is continuous at the critical section.

    4. The influence line for a bending moment will contain a unit rotational discontinuity at the point where thebending moment is being evaluated.

    5. To determine the location for positioning a single concentrated load to produce maximum magnitude for aparticular function (reaction, shear, axial, or bending moment) place the load at the location of the maximum

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    ordinate to the influence line. The value for the particular function will be equal to the magnitude of theconcentrated load, multiplied by the ordinate value of the influence line at that point.

    6. To determine the location for positioning a uniform load of constant intensity to produce the maximummagnitude for a particular funct ion, place the load along those portions of the structure for which the ordinates tothe influence line have the same algebraic sign. The value for the particular function will be equal to themagnitude of the uniform load, multiplied by the area under the influence diagram between the beginning andending points of the uniform load.

    ere are two methods that can be used to plot an influence line for any function. In the first, the approach describedve, is to write an equation for the function being determined, e.g., the equation for the shear, moment, or axial forceuced at a point due to the application of a unit load at any other location on the structure. The second approach,ich uses the Mller Breslau Principle , can be utilized to draw qualitative influence lines, which are directlyportional to the actual influence line.

    e following examples demonstrate how to determine the influence lines for reactions, shear, and bending moments ofms and frames using both methods described above.

    example, the influence line for the support reaction at A of the structure shown in Figure 1, is found by applying at load at several points (See Figure 2) on the structure and determining what the resulting reaction will be at A. Thisbe done by solving the support reac tion Y A as a function of the position of a downward acting unit load. One suchation can be found by summing moments at Support B.

    ure 1 - Beam structure for influence li ne example

    ure 2 - Beam structure showing application of unit load

    MB = 0 (Assume counter-clockwise positive moment) A(L)+1(L-x) = 0= (L-x)/L = 1 - (x/L)

    e graph of this equation is the influence line for the support react ion at A (See Figure 3). The graph illustrates that ifunit load was applied at A, the reaction at A would be equal to unity. Similarly, if the unit load was applied at B, thection at A would be equal to 0, and if the unit load was applied at C, the reaction at A would be equal to -e/L.

    ure 3 - Influence line for the support r eaction at A

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    oblem statement

    aw the influence lines for the reactions Y A, Y C, and the shear and bending moment at point B, of the simply supportedm shown by developing the equations for the respective influence lines.

    ure 1 - Beam structure to ana lyze

    Reaction Y A

    e influence line for a reaction at a support is found by independent ly applying a unit load at several points on theucture and determining, through statics, what the resulting reaction at the support will be for each case. In thismple, one such equation for the influence line of Y A can be found by summing moments around Support C.

    ure 2 - Application of unit load

    MC = 0 (Assume counter-clockwise positive moment) A(25)+1(25-x) = 0= (25-x)/25 = 1 - (x/25)

    e graph of this equation is the influence line for Y A (See Figure 3). This figure illustrates that if the unit load islied at A, the reaction at A will be equal to unity. Similarly, i f the unit load is applied at B, the reaction at A will beal to 1-(15/25)=0.4, and if the unit load is applied at C, the reaction a t A will be equal to 0.

    ure 3 - Influence line for Y A, the support reaction at A

    e fact that Y A=1 when the unit load is applied at A and zero when the unit load is applied at C can be used to quicklyerate the influence l ine diagram. Plotting these two values at A and C, respectively, and connecting them with aight line will yield the the influence line for Y A. The structure is statically determinate, therefore, the resultingction is a straight line.

    Reaction at C

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    e equation for the influence line of the support reaction at C is found by developing an equation that relates thection to the position of a downward acting unit load applied at all locations on the structure. This equation is foundsumming the moments around support A.

    ure 4 - Application of unit load

    MA = 0 (Assume counter-clockwise positive moment) (25)-1(x) = 0= x/25

    e graph of this equation is the influence line for Y C. This shows that if the unit load is appl ied at C, the reaction at Cl be equal to unity. Similarly, if the unit load is applied at B, the reaction at C will be equal to 15/25=0.6. And, if thet load is applied at A, the reaction at C will equal to 0.

    ure 5 - Influence line for the reaction at support C

    e fact that Y C=1 when the unit load is applied at C and zero when the unit load is applied at A can be used to quicklyerate the influence l ine diagram. Plotting these two values at A and C, respect ively, and connecting them with aight line will yield the the influence line for Y C. Notice, since the structure is statically determinate, the resultingction is a straight line.

    Shear at B

    e influence line for the shear at point B can be found by developing equations for the shear at the section usingics. This can be accomplished as follows:

    f the load moves from B to C, the shear diagram will be as shown in Fig. 6 below, this demonstrates that the shear atwill equal Y A as long as the load is located to the right of B, i.e., V B = Y A. One can also calculate the shear at B from

    Free Body Diagram (FBD) shown in Fig. 7.

    ure 6 - Shear diagram for load located between B and C

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    ure 7 - Free body diagram for section at B with a load located between B and C

    f the load moves from A to B, the shear diagram will be as shown in Fig. 8, below, this demonstrates that the shear atwill equal -Y C as long as the load is located to the left of B, i.e., V B = - Y C. One can also calculate the shear at B from

    FBD shown in Fig. 9.

    ure 8 - Shear diagram for load located between A and B

    ure 9 - Free body diagram for section at B with a load located between A and B

    e influence line for the Shear at point B is then constructed by drawing the influence line for Y A and negative Y C.en highlight the portion that represents the sides over which the load was moving. In this case, highlight the the partm B to C on Y A and from A to B on -Y C. Notice that at point B, the summation of the absolute values of the positive negative shear is equal to 1.

    ure 10 - Influence line for shear at point B

    Moment at B

    e influence line for the moment at point B can be found by using statics to develop equations for the moment at thent of interest, due to a uni t load acting at any location on the structure. This can be accomplished as follows.

    f the load is at a location between B and C, the moment at B can be calculated by using the FBD shown in Fig. 7ve, e.g., at B, M B = 15 Y A - notice that this relat ion is valid if and only if the load is moving from B to C.

    f the load is at a location between A and B, the moment at B can be calculated by using the FBD shown in Fig. 9

    ve, e.g., at B, M B = 10 Y C - notice that this relation is valid if and only if the load is moving from A to B.

    e influence line for the Moment at point B is then constructed by magnifying the influence lines for Y A and Y C by 15 10, respectively, as shown below. Having plotted the functions, 15 Y A and 10 Y C, highlight the portion from B to Che function 15 Y A and from A to B on the function 10 Y C. These are the two portions what correspond to the correct

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    ment relations as explained above. The two functions must intersect above point B. The value of the function at Bn equals (1 x 10 x 15)/25 = 6. This represents the moment at B if the load was positioned at B.

    ure 11 - Influence line for moment at point B

    luence Lines | Index of Examples | CCE Homepage luence Lines alitative Influence Lines using the Mller Breslau Principle

    Mller Breslau Principlee Mller Breslau Principle is another alternative available to qualitatively develop the influencees for different functions. The Mller Breslau Principle states that the ordinate value of an influencee for any function on any structure is proportional to the ordinates of the deflected shape that istained by removing the restraint corresponding to the function from the structure and introducing ace that causes a unit displacement in the positive direction.

    gure 1 - Beam structure to analyzer example, to obtain the influence line for the support reaction at A for the beam shown in Figure 1,ove, remove the support corresponding to the reaction and apply a force in the positive direction that

    ll cause a unit displacement in the direction of Y A. The resulting deflected shape will be proportionalthe true influence line for this reaction. i.e., for the support reaction at A. The deflected shape due tonit displacement at A is shown below. Notice that the deflected shape is linear, i.e., the beam rotatesa rigid body without any curvature. This is true only for statically determinate systems.

    gure 2 - Support removed, unit load applied, and resulting influence line for support reaction at Amilarly, to construct the influence line for the support reaction Y B, remove the support at B and applyertical force that induces a unit displacement at B. The resulting deflected shape is the qualitativeluence line for the support reaction Y B.

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    gure 3 - Support removed, unit load applied, and resulting influence line for support reaction at Bce again, notice that the influence line is linear, since the structure is statically determinate.is principle will be now be extended to develop the influence lines for other functions.

    Shear at sdetermine the qualitative influence line for the shear at s, remove the shear resistance of the beam ats section by inserting a roller guide, i.e., a support that does not resist shear, but maintains axial

    ce and bending moment resistance.

    gure 4 - Structure with shear capacity removed at smoving the shear resistance will then allow the ends on each side of the section to moverpendicular to the beam axis of the structure at this section. Next, apply a shear force, i.e., V s-R and-L that will result in the relative vertical displacement between the two ends to equal unity. Thegnitude of these forces are proportional to the location of the section and the span of the beam. Ins case,-L = 1/16 x 10 = 10/16 = 5/8-R = 1/16 x 6 = 6/16 = 3/8e final influence line for V s is shown below.

    gure 5 - Influence line for shear at s

    Shear just to the left side of B

    e shear just to the left side of support B can be constructed using the ideas explained above. Simplyagine that section s in the previous example is moved just to the left of B. By doing this, thegnitude of the positive shear decreases until it reaches zero, while the negative shear increases to 1.

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    gure 6 - Influence line for shear just to the left of B

    Shear just to the right side of Bplot the influence line for the shear just to the right side of support B, V b-R , release the shear just toright of the support by introducing the type of roller shown in Fig. 7, below. The resulting deflected

    ape represents the influence line for V b-R . Notice that no deflection occurs between A and B, sincether of those supports were removed and hence the deflections at A and B must remain zero. The

    flected shape between B and C is a straight line that represents the motion of a rigid body.

    gure 7 - Structure with shear capacity removed at just to the right of B and the resulting influencee

    Moment at sobtain a qualitative influence line for the bending moment at a section, remove the moment restrainthe section, but maintain axial and shear force resistance. The moment resistance is eliminated byerting a hinge in the structure at the section location. Apply equal and opposite momentspectively on the right and left sides of the hinge that will introduce a unit relative rotation betweentwo tangents of the deflected shape at the hinge. The corresponding elastic curve for the beam,

    der these conditions, is the influence line for the bending moment at the section. The resulting

    luence line is shown below.

    gure 8 - Structure with moment capacity removed at s and the resulting influence linee values of the moments shown in Figure 8, above, are calculated as follows:a. when the unit load is applied at s, the moment at s is Y A x 10 = 3/8 x 10 = 3.75

    (see the influence line for Y A, Figure 2, above, for the value of Y A with a unit load applied at s)b. when the unit load is applied at C, the moment at s is Y A x 10 = -3/8 x 10 = -3.75

    (again, see the influence line for Y A for the value of Y A with a unit load applied at C)llowing the general properties of influence lines, given in the Introduction , these two values aretted on the beam at the locations where the load is applied and the resulting influence line is

    nstructed.

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    Moment at Be qualitative influence line for the bending moment at B is obtained by introducing a hinge atpport B and applying a moment that introduces a unit relative rotation. Notice that no deflectioncurs between supports A and B since neither of the supports were removed. Therefore, the onlyrtion that will rotate is part BC as shown in Fig. 9, below.

    gure 9 - Structure with moment capacity removed at B and the resulting influence line Shear and moment envelopes due to uniform dead and live loads

    e shear and moment envelopes are graphs which show the variation in the minimum and maximumues for the function along the structure due to the application of all possible loading conditions. Thegrams are obtained by superimposing the individual diagrams for the function based on eachding condition. The resulting diagram that shows the upper and lower bounds for the function alongstructure due to the loading conditions is called the envelope.

    e loading conditions, also referred to as load cases, are determined by examining the influence linesd interpreting where loads must be placed to result in the maximum values. To calculate theximum positive and negative values of a function, the dead load must be applied over the entire

    am, while the live load is placed over either the respective positive or negative portions of theluence line. The value for the function will be equal to the magnitude of the uniform load, multiplied

    the area under the influence line diagram between the beginning and ending points of the uniformd.r example, to develop the shear and moment envelopes for the beam shown in Figure 1, first sketchinfluence lines for the shear and moment at various locations. The influence lines for V a-R , V b-L, V b-

    M b, V s, and M s are shown in Fig. 10.

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    gure 10 - Influence linesese influence lines are used to determine where to place the uniform live load to yield the maximumsitive and negative values for the different functions. For example;

    gure 11 - Support removed, unit load applied, and resulting influence line for support reaction at AThe maximum value for the positive reaction at A, assuming no partial loading, will occur when the

    iform load is applied on the beam from A to B (load case 1)

    gure 12 - Load case 1The maximum negative value for the reaction at A will occur if a uniform load is placed on the

    am from B to C (load case 2)

    gure 13 - Load case 2Load case 1 is also used for: maximum positive value of the shear at the right of support A maximum positive moment M s

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    Load case 2 is also used for: maximum positive value of the shear at the right of support B maximum negative moments at support B and M s Load case 3 is required for: maximum positive reaction at B maximum negative shear on the left side of B

    gure 14 - Load case 3

    Load case 4 is required for the maximum positive shear force at section s

    gure 15 - Load case 4Load case 5 is required for the maximum negative shear force at section s

    gure 16 - Load case 5develop the shear and moment envelopes, construct the shear and moment diagrams for each load

    se. The envelope is the area that is enclosed by superimposing all of these diagrams. The maximumsitive and negative values can then be determined by looking at the maximum and minimum valuesthe envelope at each point.dividual shear diagrams for each load case;

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    gure 19 - Individual moment diagramsperimpose all of these diagrams together to determine the final moment envelope.

    gure 20 - Resulting superimposed moment envelope

    Influence Lines | Index of Examples | CCE Homepage

    luence Lines

    ualitative Influence Lines for a Statically Determinate Continuouseam

    oblem statement

    aw the qualitative influence lines for the vertical reactions at the supports, the shear and moments at sections s1 andand the shear at the left and right of support B of the continuous beam shown.

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    ure 1 - Beam structure to ana lyze

    Reactions at A, B, and C

    alitative influence lines for the support reactions at A, B, and C are found by using the Mller Breslau Principle forctions, i.e., apply a force which will introduce a unit displacement in the structure at each support. The resultinglected shape will be proportional to the influence line for the support reactions.

    e resulting influence lines for the support reactions at A, B, and C are shown in Figure 2, below.

    ure 2 - Influence lines for the reactions at A, B, and C

    te: Beam BC does not experience internal forces or reactions when the load moves from A to h. In other words,uence lines for beam hC will be zero as long as the load is located between A and h. This can also be expla ined byfact that portion hC of the beam is supported by beam ABh as shown in Figure 3, below.

    ure 3 - Beam hC supported by beam ABh

    erefore, the force Y h required to maintain equilibrium in portion hC when the load from h to C is provided by portionh. This force, Y h, is equal to zero when the load moves between A an h, and hence, no shear or moment will beuced in portion hC.

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    Shear and moment at section S 1 and S 2

    determine the shear at s 1, remove the shear resistance of the beam at the section by inserting a support that does notst shear, but maintains axial force and bending moment resistance (see the inserted support in Figure 4). Removingshear resistance will allow the ends on each side of the section to move perpendicular to the beam axis of the

    ucture at this section. Next, apply shear forces on each side of the section to induce a relative displacement betweentwo ends that will equal uni ty. Since the section is cut at the midspan, the magnitude of each force i s equal to 1/2.

    ure 4 - Structure with shear capacity removed at s1 and resulting influence line

    the moment at s 1, remove the moment restraint at the section, but maintain axial and shear force resistance. Thement resistance is eliminated by inserting a hinge in the structure at the section location. Apply equal and oppositements on the right and left sides of the hinge that will introduce a unit relative rotation between the two tangents ofdeflected shape at the hinge. The corresponding elastic curve for the beam, under these conditions, is the influence

    e for the bending moment at the section.

    ure 5 - Structure with moment capacity removed at s1 and resulting influence line

    e value of the moment shown in Figure 5, above, is equal to the value of R a when a unit load is applied at s 1,ltiplied by the distance from A to s 1. M s1 = 1/2 x 4 = 2.

    e influence lines for the shear and moment at section s 2 can be constructed following a similar procedure. Notice thaten the load is located between A and h, the magnitudes of the influence lines are zero for the shear and moment at s 1.e was explained previously in the discussion of the influence line for the support reaction at C (see Figures 2 and 3).

    ure 6 - Structure with shear capacity removed at s2 and resulting influence line

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    ure 7 - Structure with moment capacity removed at s2 and resulting influence line

    Shear at the left and right of B

    ce the shear at B occurs on both sides of a support, it is necessary to independently determine the shear for each side.

    plot the influence line for V b-L , follow the instructions outlined above for plotting the influence line for the shear atTo construct the shear just to the left of support B, imagine that the sect ion s 1 has been moved to the left of B. In thise, the positive ordinates of the influence line between A and B will decrease to zero while the negative ordinates willrease to 1 (see Figure 8).

    ure 8 - Structure with shear capacity removed at the left of B and the resulting influence line

    e influence line for the shear forces just to the right of support B, V b-R, is represented by the resulting deflected shapehe beam induced by shear forces acting just to the right of support B. Notice that the portion of the beam from B to hves as a rigid body (see explanation in the Simple Beam with a Cantilever example) while the influence line variesearly from h to C. This is due to the fact that the deflection at C is zero and the assumption that the deflection of aically determinate system is linear.

    ure 9 - Structure with shear capacity removed at the right of B and the resulti ng influence line

    Influence Lines | Index of Examples | CCE Homepage

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    luence Lines alculation of Maximum and Minimum Shear Force and Moments on aatically Determinate Continuous Beam

    oblem statement

    ermine the resulting forces for R A, R B, R C, M s1, V s1, M s1, V s2, V BL, and V BR under a uniform live load of 2 k/ft and aform dead load of 3 k/ft for the beam below.e: influence lines for this beam are developed in the Statically Determinate Continuous Beam example.

    ure 1 - Beam structure to ana lyze

    Influence lines

    m the Continuous Beam with a Hinge example, the required influence lines for the structure are:

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    Calculate forces

    order to calculate the forces due to uniform dead and live loads on a structure, a relationship between the influencee and the uniform load is required. Referring to Figure 2, each segment dx, of a uniform load w, creates an equivalentcentrated load, dF = w dx, acting a distance x from an origin.

    m the general properties for influence lines, given in the introduction, it i s known that the resulting value of thection for a force acting at a point is equivalent to the magni tude of the force, dF, multiplied by the ordinate value, y,he influence line at the point of application.

    ure 2 - Equivalent concentrated load

    order to determine the effect of the uniform load, the effect of all series loads, dF, must be determined for the beam.

    s is accomplished by integrating y dF over the length of the beam, i.e., w y dx = w y dx. The integration of yequal to the area under the influence line. Thus, the value of the function caused by a uniform load is equal to thegnitude of the uniform load multiplied by the area under the influence line diagram.

    order to find the resulting minimum and maximum values for the reactions, shears, and moments required, create ale which contains the resulting positive and negative values for the areas enclosed by the influence lines for eachction. The effect of the dead load is determined by multiplying the net area under the influence line by the dead load.the live load, multiply the respective positive and negative areas by the live load, yields to the positive and negative

    ces, respectively. The resulting maximum and minimum forces for dead load plus the effects of positive and negativee loads are then found by adding the respective values.

    e resulting forces due to a uni formly distributed dead load = 3 k/ft and a live load = 2 k/ft applied to the beam above,as follows:

    ce

    Positive areaunder the

    influence lineII

    Negative areaunder the

    influence lineIII

    NetareaIV

    Forcedue to

    DLV

    Positiveforce due to

    LLVI

    Negativeforce due to

    LLVII

    Maximumforce

    (DL+LL)VIII

    Minimumforce

    (DL-LL)IX

    4 -1 3 9 8 -2 17 7

    10 - 10 30 20 - 50 30

    3 - 3 9 6 - 15 9

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    1 8 -4 4 12 16 8 28 4

    1 1 -2 -1 -3 2 4 -1 -7

    2 4.5 - 4.5 13.5 9 - 22.5 13.5

    2 0.75 -0.75 - - 1.5 -1.5 1.5 -1.5

    -R 5 - 5 15 10 - 25 15

    -L - 5 -5 -15 - 10 -15 -25

    umn IV = Column II + Column IIIumn V = Dead Load * Column IVumn VI = Live Load * Column IIumn VII = Live Load * Column IIIumn VIII = Column V + Column VIumn IX = Column V + Column VII

    Influence Lines | Index of Examples | CCE Homepage

    luence Lines

    ualitative Influence Lines and Loading Patterns for an Multi-spandeterminate Beam

    e Mller Breslau Principle, used previously to draw the influence lines for statically determinate structures, can alsoextended to define the influence lines for indeterminate structures. This principle simply states that the influence linea function is proportionally equivalent to the deflected shape of the structure when it undergoes a displacement as ault of the application of the function.

    indeterminate structures, an understanding of how complex structures deflect and react when acted upon by a forceequired in order to draw accurate diagrams.

    Influence lines for reactions

    determine the influence l ine for the support reaction at A, the Mller Breslau Principle requires the removal of theport restraint and the application of a positive unit deformation at this point that corresponds to the direction of thece. In this case, apply a unit vertical displacement in the direction of Y A.

    ure 1 - Structure with support reaction removed, unit deformation applied, and resulting i nfluence line

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    e resulting deflected shape, due to the application of the unit deformation, is then proportionally equivalent to theuence line for the support reaction at A. Notice that in statically indeterminate structures, the deflected shape is not aight line, but rather a curve. The ordinates of the deflected shape decrease as the distance increases from the point oflication of the unit deformation.

    milarly, for the other support reactions, remove the support restraint and apply a unit deformation in the direction ofremoved restraint. For example, the influence line for the support reaction at C is obtained by removing the reaction

    C and applying a unit displacement in the vertical direction at C. The resulting deflected shape is a qualitativeresentation of the influence line at R C (see Figure 2).

    ure 2 - Structure with support reaction removed, unit deformation applied, and resulting i nfluence line

    uence lines for the remaining support reactions are found in a similar manner.

    Influence lines for shears

    shear at a section, using the Mller Breslau Principle, the shear resistance at the point of interest is removed byoducing the type of support shown in Figure 3, below. Shear forces are applied on each side of the section in order toduce a relative displacement between the two sides which is equal to uni ty. The deflected shape of the beam underse conditions will qualitatively represent the influence line for the shear at the section. Notice that unlike theically determinate structure, the magnitude of the shear force on the right and left can not easily be determined.

    ure 3 - Structure with shear ca rrying capacity removed at section S1, deformations applied, and resulting influence line

    Influence lines for moments

    the moment at a section, using the Mller Breslau Principle, the moment resistance at the point of interest ismoved by introducing a hinge at the section as shown in Figure 4, below. Then a positive moment that introduces aative unit rotation is applied at the section. The deflected shape of the beam under these conditions will qualitativelyresent the influence line for the moment at the section.

    ure 4 - Structure with moment capacity removed at section S1, unit rotation applied, and resulting influence line

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    the moment at a support, the moment resistance is again removed by inserting a hinge at the support. This hingey prevents the transfer of moments, so the vertical translation remains fixed due to the support. By applying negativements that induces a relative rotation of unity at this section, a deflected shape is generated. Again, this deflectedpe qualitatively represents the influence line for the moment at a support.

    ure 5 - Structure with moment capacity removed at support B, unit rotation applied, and resulting influence line

    Loading cases for moment and shear envelopes

    ng the influence lines found above, illustrate the loading cases needed to calculate the maximum positive andative R A, R C, M B, V S1, and M S1.

    e load cases are generated for the maximum positive and negative values by placing a distributed load on the spansere the algebraic signs of the influence line are the same. i.e., to get a maximum positive value for a function, place aributed load where the influence line for the function is positive.

    ure 6 - Multi-span structure

    ad case for maximum positive reaction at support A

    ure 7 - Maximum positive reaction at support A

    ad case for maximum negative reaction at support A

    ure 8 - Maximum negative reaction at support A

    ad case for maximum positive reaction at support C

    ure 9 - Maximum positive reaction at support C

    ad case for maximum negative reaction at support C

    ure 10 - Maximum negative reaction at support C

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    ad case for maximum positive moment at support B

    ure 11 - Maximum positive moment at support B

    ad case for maximum negative moment at support B

    ure 12 - Maximum negative moment at support B

    ad case for maximum positive shear at s

    ure 13 - Maximum positive shear at s

    ad case for maximum negative shear at s

    ure 14 - Maximum negative shear at s

    ad case for maximum positive moment at s

    ure 16 - Maximum positive moment at s

    ad case for maximum negative moment at s

    ure 17 - Maximum negative moment at s

    Influence Lines | Index of Examples | CCE Homepage

    luence Lines

    ualitative Influence Lines and Loading Patterns for an Indeterminateame

    oblem statement

    ng the Mller Breslau Principle, draw the influence lines for the moment and shear at the midspan of beam AB, and

    moment at B in member BC. Draw the loading cases to give the maximum positive moment at the midpsan of beam, the maximum and minimum shear at the midspan of beam AB, and the maximum negative moment at B in memberin the indeterminate frame below.

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    ure 1 - Frame structure to analyze

    Influence lines

    uence line for moment at midspan of AB, and the loading case for maximum positive moment at this location.

    e influence line for beam ABCD can be constructed by following the procedure outlined in the Multi-spaneterminate Beam example . To construct the rest of the influence line, make use of the fact that the angles between aumn and a beam after deformation must be equal to that before deformation. In this example, these angles are 90.erefore, once the deflected shape of beam ABCD is determined, the deflected shape for the columns can bestructed by keeping the angles between the tangent of the deflect shape of the beam and the column equal to 90 (seeure 2).

    get the maximum positive result for the moment, apply a distributed load at all locations where the value of theuence line is positive (see Figure 3).

    ure 2 - Influence lines for moment at midspan of AB Figure 3 - Load case for maximum positive moment at midspan of AB

    uence line for shear at the midspan of member AB, and the load case for maximum positive shear at this location.

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    ure 4 - Influence lines for shear at midspan of AB

    ure 5 - Load case for maximum positive shear at midspan of AB Figure 6 - Load case for maximum negative shear a t midspan of AB

    uence line for moment at B in member BC, and the load case for maximum negative moment at this location.

    ure 7 - Influence lines for moment at B Figure 8 - Load case for maximum positive moment at B

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    III-UNIT ARCHESTHREE HINGED ARCHES

    An arch is a curved beam in which horizontal movement at the support is wholly or partially prevented. Hencere will be horizontal thrust induced at the supports. The shape of an arch doesnt change with loading and therefore

    me bending may occur.pes of Arches

    On the basis of material used arches may be classified into and steel arches, reinforced concrete arches, masonryhes etc.,

    On the basis of structural behavior arches are c lassified as :

    ee hinged arches:- Hinged at the supports and the crown.

    o hinged arches:- Hinged only at the support

    Hinges at thesupport

    Rib of the archRise

    Span

    Hinged at thesupport

    Hinged at thecrown

    Springing

    Rise

    Span

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    The supports are fixed

    A 3-hinged arch is a statically determinate structure. A 2-hinged arch is an indeterminate structure of degree ofeterminancy equal to 1. A fixed arch is a statically indeterminate structure. The degree of indeterminancy is 3.

    pending upon the type of space between the loaded area and the rib arches can be classified as open arch or closedh (solid arch).alysis of 3-hinged arches

    It is the process of determining external reactions at the support and internal quantities such as normal thrust,ar and bending moment at any section in the arch.

    ocedure to find reactions at the supportsp 1. Sketch the arch with the loads and reactions at the support.ply equilibrium conditions namely 0,Fx = 0Fy = 0Mand = ply the condition that BM about the hinge at the crown is zero (Moment of all the forces either to the left or to theht of the crown).ve for unknown quantities.

    3-hinged arch has a span of 30m and a rise of 10m. The arch carries UDL of 0.6 kN/m on the left half of the span. Ito carries 2 concentrated loads of 1.6 kN and 1 kN at 5 m and 10 m from the rt end. Determine the reactions at theport. (sketch not given).

    = 0Fx 0HH BA =

    B H B = 4.275

    0.6 kN/mC

    5 m 5 m

    h = 10m

    L = 30mVA = 7.35 VB = 4.25

    HB = 4.275 A

    1 kN 1.6 kN

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    83.44HV

    tan

    kN02.6VHR

    82.59HV

    tan

    kN5.8

    35.7275.4

    VHR

    B

    B1B

    2B

    2BB

    0

    A

    A1A

    22

    2A

    2AA

    =

    =

    =+=

    =

    =

    =

    +

    ==

    3-hinged parabolic arch of span 50m and rise 15m carries a load of 10kN at quarter span as shown in figure.culate total reaction at the hinges.

    = 0Fx BA HH =

    find vertical reaction. = 0Fy

    10VV BA =+ ------ (1)

    = 0M A

    kN5.7VkN5.2V

    05.12x1050xV

    AB

    B

    ==

    =+

    find Horizontal reaction

    B H B

    10 kNC

    15 m

    50 mVA VB

    HA A

    12.5 m

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    0C =

    015H25V BB =

    find total reaction.

    94.30HV

    tan

    kN861.4

    VH

    92.60HV

    tan

    kN581.8

    5.717.4

    HkN17.4

    0

    B

    B1

    2B

    2A

    0

    A

    A1

    A

    22A

    AB

    =

    =

    =

    +=

    =

    =

    =

    +=

    ==

    RA

    AA HA = 4.17

    VA = 7.5RB

    HB = 4.17

    VB = 4.25

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    blem: Determine the reaction components at supports A and B for 3-hinged arch shown in fig.

    find Horizontal reaction = 0Fx

    0HH BA = BA HH = ------ (1)

    find vertical reaction. = 0Fy

    280VV

    10x10180VV

    BA

    BA

    =+

    +=+ ------ (2)

    = 0M A

    33.1558V10H

    3740V24H4.2

    05x10x1018x1804.2xH24xV

    BB

    BB

    BB

    =

    =

    =+++

    ------ (3)

    0C =

    B H B

    10 kN/m C

    2.5 m

    10 mVA

    VB

    HA A

    2 .4 m

    180 kN

    8 m 6 m

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    87.293V857.2H

    144014V9.4xH

    09.4xH14xV8x180

    BB

    BB

    BB

    +=+

    =

    =

    ------ (4)

    ding 2 and 3

    AB

    B

    A

    B

    BB

    HkN67.211H

    33.1558177x10H

    kN103V

    kN177V

    87.29333.1558V857.2V10

    ==

    =

    =

    =

    +=+

    blem: A symmetrical 3-hinged parabolic arch has a span of 20m. It carries UDL of intensity 10 kNm over the entiren and 2 point loads of 40 kN each at 2m and 5m from left support. Compute the reactions. Also find BM, radialar and normal thrust at a section 4m from left end take central rise as 4m.

    =0Fx

    BA

    BA

    HH

    0HH

    =

    = ------ (1)

    =0Fy

    280V

    020x104040V

    B

    B

    =+

    =+ ------ (2)

    C

    4 m

    20 m

    40 kN

    M

    2 m 3m

    10 kN/m40 kN

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    =0M A

    kN166V

    kN114V

    020xV10)20x10(5x402x40

    A

    B

    B

    =

    =

    =+++

    0c =

    kN160H

    kN160H

    010x1144xH5)10x10(

    A

    B

    B

    =

    =

    =+

    M at M

    = - 160 x 2.56+ 166 x 4 40 x 2- (10 x 4)2

    = + 94.4 kNm

    ( )xLLhx4

    y2

    =

    ( )42020

    4x4x42

    =

    m56.2y =

    tan ( )x2LL

    h42

    =

    = ( )4x22020

    4x42

    VERTICAL

    NORMAL

    HORIZONT

    64.35

    160 kN = 25.64

    M

    REDIAL

    10 kN/m

    160 kN 4 m

    y = 2.56

    86 kN

    40 kN

    2 m 166 kN

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    64.25 0= Normal thrust = N = + 160 Cos 25.64

    + 86 Cos 64.36= 181.46 kN

    S = 160 Sin 25.64- 86 x Sin 64.36

    S = - 8.29 kN

    UNIT SLOPE DEFLECTION METHOD e slope deflection method is a structural analysis method for beams and frames introduced in 1915 by George A.ney. [1] The slope deflection method was widely used for more than a decade until the moment distribution method s developed.

    troduction

    forming slope deflection equations and applying joint and shear equi librium conditions, the rotation angles (or thepe angles) are calculated. Substutituting them back into the slope deflection equations, member end moments aredily determined.

    ope deflection equations

    e slope deflection equations express the member end moments in terms of rotations angles. The slope deflectionations of member ab of flexural rigidity EI ab and length Lab are:

    ere a, b are the slope angles of ends a and b respectively, is the relative lateral displacement of ends a and b. Theence of cross-sectional area of the member in these equations implies that the slope deflection method neglects theect of shear and axial deformations.

    e slope deflection equations can also be written using the stiffness factor and the chord

    ation :

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    en a simple beam of length Lab and flexural rigidity E I ab is loaded at each end with clockwise moments M ab and M ba ,mber end rotations occur in the same direction. These rotation angles can be calculated using the unit dummy forcethod or the moment-area theorem .

    arranging these equations, the slope deflection equations are derived.

    quilibrium conditions

    nt equilibrium conditions imply that each joint with a degree of freedom should have no unbalanced moments i.e. beequilibrium. Therefore,

    re, M member are the member end moments, M f are the fixed end moments , and M joint are the external moments directly

    lied at the joint.

    en there are chord rotations in a frame, additional equilibrium conditions, namely the shear equilibrium conditionsd to be taken into account.

    ISPLACEMENT METHOD OF ANALYSIS:

    LOPE DEFLECTION EQUATIONS

    General Case Stiffness Coefficients

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    Stiffness Coefficients Derivation Fixed-End Moments Pin-Supported End Span Typical Problems Analysis of Beams Analysis of Frames: No Sidesway Analysis of Frames: Sidesway

    1

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    Slope Deflection Equations

    i Pw

    j k

    C j

    settlement = j

    i P w j

    M ij M ji

    i

    j

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    Degrees of Freedom

    M

    A B 1 DOF:

    L

    P

    A B

    C2 DOF: ,

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    Stiffness

    k AA 1 kBA

    A B

    L

    k AA

    k BA

    = 4 EI L

    = 2 EI L

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    k AB k BB

    A B

    L

    k BB

    k AB

    = 4 EI L

    = 2 EI L

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    Fixed-End Forces Fixed-End Moments: Loads

    P

    PL L/ 2 L/ 2 PL

    8 8 L

    P 2

    w

    wL2 12

    wL L

    2

    P 2

    wL2

    12

    wL 2

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    General Case i P w j k C j

    settlement = j

    i P w j

    M ij M ji

    i

    j

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    (M (M ) ij

    M ij i P

    w j

    i L

    M ji

    j settlement = j

    4 EI 2 EI

    2 EI 4 EI

    + L i L j

    = M ij M ji = i + j L L j

    i

    +(M F ) F

    ij (M

    +

    ji)

    settlement = j

    F )Load

    Pw

    F ji Load

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    F F F

    M ij = (4 EI L

    )i + ( 2 EI L

    ) j + ( M ij ) + ( M ij ) Load , M ji = (2 EI L

    ) i + (4 EI L

    ) j + ( M ji ) + ( MF

    ji

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    Equilibrium Equations

    i Pw

    j k

    C j

    M ji

    C j M jk

    M ji M jk

    j

    + M j = 0 : M ji M jk + C j = 0

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    Stiffness Coefficients

    M ij i j M ji L

    j

    i

    k ii = 4 EI L

    1

    +

    k ji = 2 EI L

    i

    k ij = 2 EI L k jj

    1

    = 4 EI L

    j

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    M

    Matrix Formulation

    M ij = (4 EI

    ) L i

    + ( 2 EI ) L j

    + ( M F ij )

    M ji = (2 EI

    ) L i

    + ( 4 EI ) L j

    + ( M F ji )

    M ij (4 EI / L) (2 EI / L) iI F ij = + F M ji (2 EI / L) (4 EI / L) j M ji

    [k] = k ii k ij k ji k jj

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    Stiffness Matrix

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    i

    M ij P

    w j

    i L

    M ji

    j j

    [ M ] = [ K ][ ] + [ FEM ]

    ([ M ] [ FEM ]) = [ K ][ ]

    [ ] = [K ]1[ M ] [FEM ]

    (M F

    M ij M ji

    j

    i

    +) F

    Stiffness matrixFixed-end mommatrix

    ij (M

    +

    ji)

    [ D ] = [ K ]-1([Q] - [FEM]

    ij) P (M F ) Displacement Force matrix (M F Load w ji Load matrix

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    Stiffness Coefficients Derivation: Fixed-End Support

    M i i

    i j L

    M j

    Real beam

    M i + M j L

    M i

    L /3 M j L

    2 EI

    M i + M j L

    M j EI

    Conjugate beam

    EI M L i 2 EI

    + M ' = 0 :

    (

    M i L )(

    L ) + (

    M j L )(

    2 L ) = 0

    From (1) and (2); i 2 EI 3 2 EI 3 M i = (

    4 EI )i

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    M i = 2 M j (1) L

    M L M L M = ( 2 EI ) + F y = 0 : ( i ) + ( i 2 EI j

    2 EI ) = 0 (2) j L i

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    Stiffness Coefficients Derivation: Pinned-End Support

    M i i

    Real beam

    i j j

    L

    M i M i L

    2 L 3

    M i

    EI M L

    L

    Conjugate beam

    i

    i 2 EI j

    + M ' = 0 : ( M i L )( 2 L ) L = 0 + F = 0 : ( M i L ) ( M i L ) + = 0 j 2 EI 3 i y 3 EI 2 EI j i =

    ( M i L )

    3 E

    I

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    = 1 = ( M i L ) 3 EI

    M i = 3 EI L

    =

    (

    M

    i

    L

    )

    j

    6 E I

    14

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    Fixed end moment : Point Load

    P Real beam

    A L B

    M

    Conjugate beam A

    M M EI E

    M EI ML

    M 2 EI

    P

    PL2

    16 EI

    PL

    4 EI

    ML2 EI

    PL2

    16 EI

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    ML ML PL2+ F y = 0 : 2 EI

    2 EI

    + 2 16 EI

    = 0, M =

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    P

    PL PL 8 L 8

    PP /2

    P 2 2

    PL /8 P /2

    -PL /8 -PL /8

    -PL /8

    -

    -PL /16

    -

    -PL /16 P

    L /

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    -PL /8 PL

    + PL + PL = PL

    + 16 16 4 8

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    Uniform load

    w Real beam Conjugate beam

    A

    A L B M M

    M EI E

    M EI ML

    M 2 EI

    wL3

    w 24 EI

    wL2

    8 EI

    ML2 EI

    wL3

    24 EI

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    ML ML

    wL3+ F y = 0 : 2 EI

    2 EI

    + 2 24 EI

    = 0, M =

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    Settlements

    M i = M j Real beamM j MConjugate beam E

    L

    M i + M j

    M L

    A

    M M i + M j

    L EI

    M EI ML

    2 EI M

    ML 2 EI

    + M = 0 : ( ML )( L ) + ( ML )( 2 L

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    B 2 EI

    M =

    3

    6 EI L2

    2 EI 3

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    Pin-Supported End Span: Simple Case P w

    B A

    4 EI L

    + 2 EI A L B A

    A

    L

    +

    B

    P w

    2 EI L

    B

    + 4 EI A L B

    (FEM ) AB (FEM ) BA

    A B M AB

    M BA

    = 0 = (4 EI / L) A + (2 EI / L) B

    = 0 = (2 EI / L) A + (4 EI / L) B

    + ( FEM ) AB

    + ( FEM ) BA

    (1)

    (2)

    2(2) (1) :

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    2 M BA = (6 EI /

    L) B

    + 2(FEM ) BA ( FEM ) BA

    M BA = (3 EI / L) B + ( FEM ) BA (FEM ) BA

    2

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    AB AB

    B

    Pin-Supported End Span: With End Couple and Settlement

    4 EI L

    M A

    A

    + 2 EI A L B A

    A

    P w

    L

    P

    B w

    B

    B

    2 EI L

    F

    + 4 EI A L B

    (M F AB)load (M BA)load

    (M F AB) A B

    (M F A

    BA)

    M AB = M A = 4 EI L

    A

    + 2 EI L

    B

    + ( M F ) load + ( MF ) (1)

    M BA

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    BA BA = 2 EI

    L A + 4 EI

    L B + ( MF

    ) load + ( MF )

    (2)

    E lim inate by 2(2) (1) : M = 3 EI + [( M F ) 1 ( M F ) ] + 1 ( M F ) + M A A 2 BA L B BA load 2 AB load 2 BA 2

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    Fixed-End Moments Fixed-End Moments: Loads

    P

    PL L/ 2 L/ 2 PL 8 8

    P

    L/ 2 L/ 2 PL + ( 1 )[( PL )] = 3PL 8 2 8 16

    wL2

    12

    wL2

    12

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    wL2

    12 + ( 1 )[( 2

    wL2

    12 )] = wL

    2

    8

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    B

    P

    A B

    A B

    B

    Typical Problem C

    P 1 w 2

    A C

    B L1 L2

    PL P

    8

    wL2 PL

    12 w 8

    w

    1

    L L

    0

    M AB = 4 EI L1

    + 2 EI L1

    0

    + 0 + P1 L1 8

    M BA = 2 EI L1

    + 4 EI L1

    + 0 P1 L1 8

    0 2

    M BC = 4 EI L2 + 2 EI

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    C

    B C

    L2 + 0 +

    0

    P2 L2 8 + wL2

    12

    2

    M CB = 2 EI L2

    + 4 EI L2

    + 0 + P2 L2 8

    wL2 12

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    B

    P

    A B

    B C

    C P 1 w 2

    A C

    B L1 L2

    M BA C B M BC

    B

    M BA = 2 EI

    L1

    + 4 EI L1

    + 0 P1 L1 8

    2

    M BC = 4 EI L2

    + 2 EI L2

    + 0 + P2 L2 8

    + wL2 12

    + M B = 0 : C B M BA M BC = 0 Solve for B

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    B

    P

    A B

    A B

    B C

    B C

    C P 1 w 2

    M BA

    A C M CB

    B M BC L1 L2

    Substitute B in M AB, M BA , M BC , M CB

    M AB = 4 EI L1

    0 + 2 EI

    L1 0

    + 0 + P1 L1 8

    M BA = 2 EI L1

    + 4 EI L1

    + 0 P1 L1 8

    0 2

    M BC = 4 EI L2

    + 2 EI L2

    + 0 +

    0

    P2 L2 8

    + wL2 12

    2

    M CB = 2 EI

    L2 + 4 EI

    L2 + 0 + P

    2 L

    2

    8 wL

    2

    12

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    L L

    y L B

    M AB

    A

    P 1 M BA

    C BP

    w 2

    M BC M CB

    C

    1 L2 C y

    B y = B y L + B y R

    M AB

    P 1 M BA A B M BC

    B C P 2

    M CB

    A y B y L 1

    B y R C y 2

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    109/319

  • 8/14/2019 SA - I - Notes

    110/319

    Example of Beams

    Example 1

    Draw the quantitative shear , bending moment diagrams and qualitativedeflected curve for the beam shown. EI is constant.

    10 kN 6 kN/m

    A C

    4 m 4 m B 6 m

  • 8/14/2019 SA - I - Notes

    111/319

  • 8/14/2019 SA - I - Notes

    112/319

    B C

    10 kN 6 kN/m

    A C

    4 mPL P

    4 m B 6 m

    2 8 PL

    8 FEM

    wL w 30 w2

    [ M ] = [ K ][Q] + [FEM]

    M BAM BC

    M AB = 4 EI

    8 A

    0

    + 2 EI 8 B

    + (10)(8) 8

    + M B B

    = 0 : M BA M BC = 02

    M BA = 2 EI

    8

    0 A +

    4 EI 8 B

    (10)(8) 8

    0

    (4 EI

    8 + 4 EI )

    6 B

    10 + (6)(6 30

    = 2.4

    ) =

    M BC = 4 EI + 2 EI (6)(6

    2

    ) + B

    EI 6 6 30 Substitute B in the moment equations:

  • 8/14/2019 SA - I - Notes

    113/319

    M CB = 2 EI 6 B + 4 EI 6 C

    0(6)(6)

    2

    20

    M AB

    = 10.6 kNm,

    M BA = - 8.8 kNm,

    M BC

    = 8.8 kNm

    M CB = -10 kNm

  • 8/14/2019 SA - I - Notes

    114/319

    10 kN 6 kN/m 8.8 kNm

    10.6 kNm A C

    8.8 kNm 10 kNm

    4 m 4 m B 6 m

    M AB = 10.6 kNm,

    M B


'Electrical Relay Logic Diag:Diesel Generator Controls.' · 2018. 8. 31. · .1-3 5ec set at i sfc. notes: i. this drawing is typical'or diesel generators ia-sa 24-3a, 3a-sa, 4a-sa,

'Electrical Relay Logic Diag:Diesel Generator Controls.' · 2018. 8. 31. · .1-3 5ec set at i sfc. notes: i. this drawing is typical'or diesel generators ia-sa 24-3a, 3a-sa, 4a-sa,

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I love blue_-_music_-_gary_more sa

CE2351 SA 2 Lecture Notes

CE2351 SA 2 Lecture Notes

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Formalizing Context (Expanded Notes)jmc.stanford.edu/articles/mccarthy-buvac-98/context.pdf · Formalizing Context (Expanded Notes) John McCarthy and Sa sa Buva c February 28, 2012

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Comp Letter SA&I - 2019

Comp Letter SA&I - 2019

CMD ® Flashtec SA CMD Flashtec SA **** O B D I I ...CMD ® Flashtec SA CMD Flashtec SA **** O B D I I ... ... 3

CMD ® Flashtec SA CMD Flashtec SA **** O B D I I ...CMD ® Flashtec SA CMD Flashtec SA **** O B D I I ... ... 3

Emidio Tribulato Isabella Ialacqua Rosaria Visalli · SE VA SA MA SA NA SU SA Cerca le sa sa pa sa si sa ca sa va si sa so sa se su si pa ... ve sa u sa a sa O sa E VO O sa O vi I

Emidio Tribulato Isabella Ialacqua Rosaria Visalli · SE VA SA MA SA NA SU SA Cerca le sa sa pa sa si sa ca sa va si sa so sa se su si pa ... ve sa u sa a sa O sa E VO O sa O vi I

a o o i s s i i a s s a a s i sa si so si so os is si sa ...ekladata.com/vfrZmSvJBK4O4CPxJ2D5Kb9Ptxg.pdf · sa si so si so os is si sa as si as so is sa 2 . ... le lu il lo li 8 .

a o o i s s i i a s s a a s i sa si so si so os is si sa ...ekladata.com/vfrZmSvJBK4O4CPxJ2D5Kb9Ptxg.pdf · sa si so si so os is si sa as si as so is sa 2 . ... le lu il lo li 8 .

I Il A / 4, / / / ——473 / —SA

I Il A / 4, / / / ——473 / —SA

Important CA Final Notes SA 200 - 299

Important CA Final Notes SA 200 - 299

S mora i sa suva

S mora i sa suva

SA-I: · i) (ii) (iii) ... SA-I: 2- VI- 2 1 (i) (ii) (iii) (iv) (iv)

SA-I: · i) (ii) (iii) ... SA-I: 2- VI- 2 1 (i) (ii) (iii) (iv) (iv)

Saturn I SA-8 Rocket History

Saturn I SA-8 Rocket History

CMD ® Flashtec SA CMD Flashtec SA **** O B D I I ......CMD ® Flashtec SA CMD Flashtec SA **** O B D I I ... ... 11

CMD ® Flashtec SA CMD Flashtec SA **** O B D I I ......CMD ® Flashtec SA CMD Flashtec SA **** O B D I I ... ... 11

10 English a Notes Literature SA 2

10 English a Notes Literature SA 2

83781RWHV - UPTU Notes...(SA/SD) methodology in detail with suitable example. (c) Write short notes on the following : (i) Relation of functional model to object and dynamic models,

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SA 2.3.5 Release Notes

SA 2.3.5 Release Notes

S. VSA SA I SA II LA TOTAL

S. VSA SA I SA II LA TOTAL

SA PCC-Group-I May 2010

SA PCC-Group-I May 2010