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truss rigid frames

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    Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 6.1-6.3 2

    Week 5

    TRUSSESMETHOD OF JOINTS

    Week 5

    Homeworkproblems

    Chapter 6

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    Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 6.1-6.3 3

    TRUSSES METHOD OF JOINTS

    Todays Objectives:

    Students will be able to:

    a) Define a simple truss.

    b) Determine the forces inmembers of a simpletruss.

    c) Identify zero-forcemembers.

    APPLICATIONS

    Trusses are commonly used to supporta roof.

    For a given truss geometry and load,how can we determine the forces inthe truss members and select theirsizes?

    A more challenging question is thatfor a given load, how can we designthe trusses geometry to minimizecost?

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    Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 6.1-6.3 4

    Trusses are also used in avariety of structures like cranesand the frames of aircraft orspace stations.

    How can we design a lightweight structure that will meetload, safety, and cost

    specifications?

    APPLICATIONS

    DEFINING A SIMPLE TRUSS

    Atruss is a structure composed of slender

    members joined together at their end points.

    If a truss, along with the imposed load, lies in a singleplane (as shown at the top right), then it is called aplanar truss.

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    Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 6.1-6.3 5

    DEFINING A SIMPLE TRUSS

    A simple truss is a planar trusswhich begins with a triangularelement and can be expanded byadding two members and a joint.

    For these trusses, the number ofmembers (M), the number ofreactions (R) and the number ofjoints (J) are related by theequation

    M = 2 J R Note: R = 3 for statically determinate trusses

    ANALYSIS and DESIGN ASSUMPTIONS

    When designing both the member and the joints of a truss,first it is necessary to determine the forces in each trussmember. This is called the force analysis of a truss.When doing this, two assumptions are made:

    1. All loads are applied at the joints. The weight of the trussmembers is often neglected as the weight is usually small ascompared to the forces supported by the members.

    2. The members are joined together by smooth pins. Thisassumption is satisfied in most practical cases where the jointsare formed by bolting or welding.

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    Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 6.1-6.3 6

    ANALYSIS and DESIGN ASSUMPTIONS

    With these two assumptions,the members act as two-forcemembers.

    They are loaded in eithertension or compression.

    Often compressive membersare made thicker to

    prevent buckling.

    THE METHOD OF JOINTS

    In this method of solving for the forces in truss members, theequilibrium of ajoint (pin) is considered. All forces acting atthe joint are shown in a FBD.

    This includes all external forces (including support reactions) aswell as the forces acting in the members. Equations ofequilibrium (

    FX= 0 and FY= 0) are used to solve for theunknown forces acting at the joints.

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    Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 6.1-6.3 7

    QUIZ

    1. One of the assumptions used when analysing a simple trussis that the members are joined together by __________.

    A) weldingB) boltingC) rivetingD) smooth pinsE) super glue

    2. In the method of joints, typically _________ equations ofequilibrium are applied at every joint.

    A) two

    B) threeC) fourD) six

    STEPS FOR ANALYSIS

    1. If the support reactions are not given, draw a FBD of the entire trussand determine all the support reactions using the equations ofequilibrium.

    2. Draw the free-body diagram of a joint with one or two unknowns.Assume that all unknown member forces act in tension (pulling thepin) unless you can determine by inspection that the forces arecompression loads.

    3. Apply the scalar equations of equilibrium, FX = 0 and FY= 0, to determine the unknown(s). If the answer is positive, thenthe assumed direction (tension) is correct, otherwise it is in theopposite direction (compression).

    4. Repeat steps 2 and 3 at each joint in succession until all the requiredforces are determined.

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    Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 6.1-6.3 8

    ZERO-FORCE MEMBERS

    If a joint has only two non-colinearmembers and there is no external loador support reaction at that joint, thenthose two members are zero-forcemembers. In this example membersDE, CD, AF, and AB are zero forcemembers.

    You can easily prove these results byapplying the equations of equilibriumto joints D and A.

    Zero-force members can be removed(as shown in the figure) whenanalysing the truss.

    If three members form a truss joint forwhich two of the members are collinear andthere is no external load or reaction at that

    joint, then the third non-collinear member isa zero force member.

    Again, this can easily be proven. One canalso remove the zero-force member, as

    shown, on the left, for analysing the trussfurther.

    Please note that zero-force members areused to increase stability and rigidity of thetruss, and to provide support for variousdifferent loading conditions.

    ZERO-FORCE MEMBERS

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    Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 6.1-6.3 9

    QUIZ

    1. Truss ABC is changed by decreasing its height from H to0.9 H. Width W and load P are kept the same. Whichone of the following statements is true for the revisedtruss as compared to the original truss?

    A) Force in all its members have decreased.B) Force in all its members have increased.C) Force in all its members have remained the same.D) None of the above.

    H

    P

    A

    B C

    W

    2. For this truss, determine the number of zero-

    force members.A) 0B) 1C) 2D) 3E) 4

    F F

    F

    QUIZ

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    Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 6.1-6.3 10

    SOLUTION

    Analysing pin A:

    FY + = 0

    = 100 (5 / 13) FAB

    FAB = 260 kN (C)

    FX+ = 240 FAC (12 / 13) ( 260) = 0

    FAC = 480 kN (T)

    FBD of pin A

    y

    FAB

    AFAC 240 kN

    x13 5

    12

    100 kN

    QUIZ1. Using this FBD, you find that FBC = 500 N. Member

    BC must be in __________.

    A) tensionB) compressionC) can not be determined

    2. For the same magnitude of force to be carried, truss members incompression are generally made _______ as compared to members intension.

    A) thickerB) thinnerC) the same size

    FBD

    FBC

    B

    BY

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    Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 6.1-6.3 1

    TRUSSESMETHOD OF SECTIONS

    TRUSSESTHE METHOD OF SECTIONS

    Todays Objectives:

    Students will be able todetermine forces intruss members usingthe method of sections.

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    THE METHOD OF SECTIONS


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