Chapter 3 Trusses - الصفحات الشخصية | الجامعة الإسلامية ...site. For plane truss If b+r = 2j staticallyy determinate If b+r 2j statically indeterminate

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  • Chapter 3Chapter 3TrussesTrusses

    Germany

  • Trusses in BuildingTrusses in Building

  • Trusses Types for BuildingTrusses Types for Building

  • Chapter 2

  • Truss BridgeTruss Bridge

  • Truss Bridge TypesTruss Bridge Types

  • Classification of coplanar TrussClassification of coplanar TrussSimple TrussSimple Truss

  • Compound Trussp

  • Complex Trussp

  • DeterminacyDeterminacyFor plane trussp

    If b+r = 2j statically determinatej yIf b+r > 2j statically indeterminate

    Where b = number of barsfr = number of external support reactionj = number of jointsj f j

  • StabilityStabilityFor plane trusspIf b+r < 2j unstableThe truss can be statically determinate or indeterminate y(b+r >=2j) but unstable in the following cases:External Stability: truss is externally unstable if all of y y f fits reactions are concurrent or parallel

  • Internal stabilit Internal stability:

    Internally stable Internally unstable

    Internally unstable

  • Classify each of the truss as stable , unstable, y , ,statically determinate or indeterminate.

    11319 jb

    22)11(2222

    11 3 19=+

    ===

    jrb

    jrb KKKK

    18)9(2219

    9 4 15=+

    ===

    jrb

    jrb KKKK

    Stable & edeterminat Statically22)11(22 ==j

    Stable & ateindetermin Statically18)9(22 ==j

  • 639 jb 8312 === jrb

    12)6(2212

    6 3 9=+

    ===

    jrb

    jrb KKKK

    16)8(2215

    8 3 12=+

    ===

    jrb

    jrb KKKK

    Stable & edeterminat Statically12)6(22 ==j

    Unstable16)8(22 ==j

  • Stability of Compound TrussStability of Compound Truss

  • The Method of JointsSTEPS FOR ANALYSIS

    1. If the support reactions are not given, draw a FBD of the entire truss and determine all the support reactions using entire truss and determine all the support reactions using the equations of equilibrium.

    2 Draw the free body diagram of a joint with one or two 2. Draw the free-body diagram of a joint with one or two unknowns. Assume that all unknown member forces act in tension (pulling the pin) unless you can determine by (pu g p ) u y u a d yinspection that the forces are compression loads.

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

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

  • The Method of JointsThe Method of Joints

  • Example 1Example 1Solve the following truss

  • ;0F =

    0C 8030sin4

    ;0

    FKNFF

    F

    AGAG

    y

    ==+

    =

    T 93.6030cos8;0

    KNFFF

    ABAB

    x

    ==

    =

  • C620303

    ;0

    KNFF

    Fy

    =

    ;0C 6.2030cos3

    FKNFF

    x

    GBGB

    =

    ==

    C 50.6030sin38 KNFF GFGF ==+

    T 6.2060sin6.260sin

    ;0

    KNFF

    F

    BFBF

    y

    ==

    =

    093.660cos6.260cos6.2;0

    FF

    BC

    x

    =++

    =

    T 33.4 KNFBC =

  • Group WorkGroup WorkDetermine the force in each member

  • Zero Force MemberZero Force Member1- If a joint has only two non-colinear members and there is 1- If a joint has only two non-colinear members and there is no external load or support reaction at that joint, then those two members are zero-force members

  • Zero Force MemberZero Force Member2- If three members form a truss joint for which two of the members are collinear and there is no external load or reaction at that joint, then the third non-collinear member is a zero force

    bmember.

  • Example 2Example 2Find Zero force member of the following truss

  • Method of Section

  • The Method of SectionThe Method of Section

  • The Method of SectionThe Method of SectionSTEPS FOR ANALYSIS

    1. Decide how you need to cut the truss. This is based on: a) where you need to determine forces, and, b) where the total ) y , , )number of unknowns does not exceed three (in general).

    2. Decide which side of the cut truss will be easier to work with (minimize the number of reactions you have to find).

    3. If required, determine the necessary support reactions by d i th FBD f th ti t d l i th E fEdrawing the FBD of the entire truss and applying the EofE.

  • 4. Draw the FBD of the selected part of the cut truss. We need to indicate the unknown forces at the cut members. Initially ywe assume all the members are in tension, as we did when using the method of joints. Upon solving, if the answer is

    i i h b i i i i f positive, the member is in tension as per our assumption. If the answer is negative, the member must be in compression. (Please note that you can also assume forces to be either tension or (Please note that you can also assume forces to be either tension or compression by inspection as was done in the previous example above.))

    5. Apply the equations of equilibrium (EofE) to the selected cut section of the truss to solve for the unknown member forces. Please note that in most cases it is possible to write one equation to solve for one unknown directly.

  • Example 2Example 2Solve the CF & GC members in the truss

  • ;0M E =C 4.3460)93.6(300)12(30sin IbFF CFCF ==+

  • ;0M A =T 4.3460)12(30sin4.346)93.6(300)12( IbFF GCGC

    A

    ==

  • Example 3Example 3Solve the GF & GD members in the truss

  • C8370)3(7)3(6260 kNFFMC 8.10)6(2)3(7)6(3.56sin0

    C 83.70)3(7)3(6.26cos0

    kNFFM

    kNFFM

    GFGDo

    GFGFD

    ==+=

    ==+=

  • Example 4Example 4Solve the ED & EB members in the truss

  • Example 5Example 5Solve the BC & MC members in the K-truss

  • T 21750)20()15(29000 IbFFM BCBCL ==+=

    TIbFF MBy 12000 BJoint At == y

    tttit t lF thTIbFFF MLMLy 29001200120029000

    part cutting totalFor the=+=

  • MJoint At

    FFF

    FFF

    MKMCy

    MKMCx

    0120029000

    00

    132

    132

    133

    133

    =+=

    =+=

    TIbFCIbFMK

    MKMCy

    1532 1532

    1313

    ==

    TIbFMC 1532=

  • Group work 1Group work 1Solve All members

  • Group work 2Group work 2Solve All members

  • Group work 3Group work 3Solve members CH , CI

  • Example 4 Compound TrussExample 4__Compound TrussSolve the truss

  • C4630)60i4()2(4)4(50 kNFFM ++ C 46.30)60sin4()2(4)4(50 kNFFM HGHGC ==++=

  • & FF AIABAJoint

    & FF & FF

    IBIJ

    HJHI

    AIAB

    IJoint HJoint

    F & FF

    JC

    BJBC

    IBIJ

    JJoint BJoint

    JCJJo

  • Example 5 Compound TrussExample 5__Compound Truss

  • Example 5 Compound TrussExample 5__Compound TrussSolve the truss

  • F& FF

    EB

    ABAE

    BJoint AJoint

    AEFSolveAfter

    AGAF

    AE

    & FFF

    AJoint SolveAfter

  • Complex TrussComplex Truss

    iii xsSS +='

    ?0' =+=

    xxsSS ECECEC

    Force in members?=x

  • Complex Trusses P

    DF

    Er + b = 2j,

    3 9 2(6)DF 3 9 (6)

    SAD

    A CDeterminateStable

    =

    B

    F PE E

    =FEC PDF

    sECDF

    + X xA C A C

    SEC + x sEC =

    x = SEC

    A

    B

    C

    = FAD

    A

    B

    C

    sECSi = Si + x si

  • ?0'

    ==+=

    xxsSS ECECEC

    xsSS += '

    ?x

    iii xsSS +=

    Member

    AB

    'iS is ixs iSABACAFFEBEEDFCEC

  • Example 5 Complex TrussExample 5__Complex Truss

  • Space TrussP Determinacy and Stability

    Space TrussDeterminacy and Stability

    b + r < junstable truss b ju stab e t ussb + r = 3jstatically determinate-check stability

    statically indeterminate-check stability b + r > 3jy y j

  • Space Trussx, y, z, Force Components

    Space Trussx, y, z, Force Components

    222 zyxl ++ zyxl ++=

    )(lxFFx = l

    )(lyFFy =

    )(lzFFz =

    222zyx FFFF ++=

  • z z

    Support Reactions

    y yFy

    x xz

    short link

    z

    y

    x F

    y

    ll x

    zFzx roller

    z

    x

    y

    F

    Fxyslotted roller constrained in a cylinder x

    zFzx in a cylinder

    z

    x

    yFx

    Fz

    Fyy

    x ball-and -socket

  • Zero Force MemberZero Force Member1- If all but one of the members connected to a joint lie on the

    l d id d l l d h j i same plane, and provided no external load act on the joint, then the member not lying in the plane of the other members must must subjected to zero force.

    Fz = , FD = z , D

  • Zero Force MemberZero Force Member2- If it has been determined that all but two of several

    b d j i f h h members connected at a joint support zero force, then the two remaining members must also support zero force, provided they dont lie a long the same line and no external load act on they don t lie a long the same line and no external load act on the joint.

    F = F = Fz = , FB =

    Fy = , FD =

  • Example 6 Space TrussExample 6__Space Truss

  • Example 7 Space TrussExample 7__Space Truss