Uniformly Distributed Load of Intensity q

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    Uniformly distributed load of intensity qLinearly distributed load of maximum intensity q0

    Concentrated load of magnitude P

    Couple of magnitute M0

    Example

    http://en.wikipedia.org/wiki/File:Fem4.pnghttp://en.wikipedia.org/wiki/File:Fem2.pnghttp://en.wikipedia.org/wiki/File:Fem3.pnghttp://en.wikipedia.org/wiki/File:Fem1.png
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    Example

    The statically indeterminate beam shown in the figure is to be analysed.

    Members AB, BC, CD have the same length.

    Flexural rigidities are EI, 2EI, EI respectively.

    Concentrated load of magnitude acts at a distance from the support A.

    Uniform load of intensity acts on BC.

    Member CD is loaded at its midspan with a concentrated load of magnitude.

    In the following calcuations, counterclockwise moments are positive.

    ]edit] Fixed-end moments

    ]edit] Distribution factors

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9
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    The distribution factors of joints A and D are DAB = 1,DDC = 0.

    ]edit] Carryover factors

    The carryover factors are , except for the carryover factor from D (fixed support) to C which is zero.

    ]edit] Moment distribution

    Joint A Joint B Joint C Joint D

    Distrib.factors

    0 1 0.2727 0.7273 0.6667 0.3333 0 0

    Fixed-end

    moments

    14.700 -6.300 8.333 -8.333 12.500 -12.500

    Step 1 -14.700 -7.350

    Step 2 1.450 3.867 1.934

    Step 3 -2.034 -4.067 -2.034 -1.017

    Step 4 0.555 1.479 0.739

    Step 5 -0.246 -0.493 -0.246 -0.123

    Step 6 0.067 0.179 0.090

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/wiki/File:MomentDistributionMethod2.jpghttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12
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    Step 7 -0.030 -0.060 -0.030 -0.015

    Step 8 0.008 0.022 0.011

    Step 9 -0.004 -0.007 -0.004 -0.002

    Step 10 0.001 0.003

    Sum ofmoment

    s0 -11.569 11.569 -10.186 10.186 -13.657

    Numbers in grey are balaced moments; arrows ( / ) represent the carry-over of moment from one end

    to the other end of a member.

    ]edit] Result

    Moments at joints determined by the moment distribution method

    The conventional engineer's sign convention is used here, i.e. positive moments cause elongation atthe bottom part of a beam member.

    For comparison purposes, the following are the results generated using a matrix method. Note that in theanalysis above, the iterative process was carried to >0.01 precision. The fact that the matrix analysis results

    and the moment distribution analysis results match to 0.001 precision is mere coincidence.

    Moments at joints determined by the matrix method

    The complete shear and bending moment diagrams are as shown. Note that the moment distribution methodonly determines the moments at the joints. Developing complete bending moment diagrams require

    additional calculations using the determined joint moments and internal section equilibrium.

    SFD and BMD

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/wiki/Matrix_methodhttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/wiki/Matrix_method
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    Shear force diagram Bending moment diagram

    ]edit] Notes

    Example

    Example

    The statically indeterminate beam shown in the figure is to be analysed.

    Members AB, BC, CD have the same length.

    Flexural rigidities are EI, 2EI, EI respectively.

    Concentrated load of magnitude acts at a distance from the support A.

    Uniform load of intensity acts on BC.

    Member CD is loaded at its midspan with a concentrated load of magnitude.

    In the following calcuations, counterclockwise moments are positive.

    ]edit] Fixed-end moments

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=14http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpghttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=14http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=14http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9
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    ]edit] Distribution factors

    The distribution factors of joints A and D are DAB = 1,DDC = 0.]edit] Carryover factors

    The carryover factors are , except for the carryover factor from D (fixed support) to C which is zero.

    ]edit] Moment distribution

    Joint A Joint B Joint C Joint D

    Distrib.factors

    0 1 0.2727 0.7273 0.6667 0.3333 0 0

    Fixed-end

    moment

    14.700 -6.300 8.333 -8.333 12.500 -12.500

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/wiki/File:MomentDistributionMethod2.jpghttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12
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    s

    Step 1 -14.700 -7.350

    Step 2 1.450 3.867 1.934

    Step 3 -2.034 -4.067 -2.034 -1.017

    Step 4 0.555 1.479 0.739

    Step 5 -0.246 -0.493 -0.246 -0.123

    Step 6 0.067 0.179 0.090

    Step 7 -0.030 -0.060 -0.030 -0.015

    Step 8 0.008 0.022 0.011

    Step 9 -0.004 -0.007 -0.004 -0.002

    Step 10 0.001 0.003

    Sum ofmoment

    s0 -11.569 11.569 -10.186 10.186 -13.657

    Numbers in grey are balaced moments; arrows ( / ) represent the carry-over of moment from one endto the other end of a member.

    ]edit] Result

    Moments at joints determined by the moment distribution method

    The conventional engineer's sign convention is used here, i.e. positive moments cause elongation atthe bottom part of a beam member.

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13
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    For comparison purposes, the following are the results generated using a matrix method. Note that in theanalysis above, the iterative process was carried to >0.01 precision. The fact that the matrix analysis results

    and the moment distribution analysis results match to 0.001 precision is mere coincidence.

    Moments at joints determined by the matrix method

    The complete shear and bending moment diagrams are as shown. Note that the moment distribution methodonly determines the moments at the joints. Developing complete bending moment diagrams require

    additional calculations using the determined joint moments and internal section equilibrium.

    SFD and BMD

    Shear force diagram Bending moment diagram

    Example

    Example

    The statically indeterminate beam shown in the figure is to be analysed.

    Members AB, BC, CD have the same length.

    Flexural rigidities are EI, 2EI, EI respectively.

    Concentrated load of magnitude acts at a distance from the support A.

    http://en.wikipedia.org/wiki/Matrix_methodhttp://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpghttp://en.wikipedia.org/wiki/Matrix_method
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    Uniform load of intensity acts on BC.

    Member CD is loaded at its midspan with a concentrated load of magnitude.

    In the following calcuations, counterclockwise moments are positive.

    ]edit] Fixed-end moments

    ]edit] Distribution factors

    The distribution factors of joints A and D are DAB = 1,DDC = 0.

    ]edit] Carryover factors

    The carryover factors are , except for the carryover factor from D (fixed support) to C which is zero.

    ]edit] Moment distribution

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12
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    Joint A Joint B Joint C Joint D

    Distrib.factors

    0 1 0.2727 0.7273 0.6667 0.3333 0 0

    Fixed-end

    moments

    14.700 -6.300 8.333 -8.333 12.500 -12.500

    Step 1 -14.700 -7.350

    Step 2 1.450 3.867 1.934

    Step 3 -2.034 -4.067 -2.034 -1.017

    Step 4 0.555 1.479 0.739

    Step 5 -0.246 -0.493 -0.246 -0.123

    Step 6 0.067 0.179 0.090

    Step 7 -0.030 -0.060 -0.030 -0.015

    Step 8 0.008 0.022 0.011

    Step 9 -0.004 -0.007 -0.004 -0.002

    Step 10 0.001 0.003

    Sum ofmoment

    0 -11.569 11.569 -10.186 10.186 -13.657

    http://en.wikipedia.org/wiki/File:MomentDistributionMethod2.jpg
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    s

    Numbers in grey are balaced moments; arrows ( / ) represent the carry-over of moment from one endto the other end of a member.

    ]edit] Result

    Moments at joints determined by the moment distribution method

    The conventional engineer's sign convention is used here, i.e. positive moments cause elongation atthe bottom part of a beam member.

    For comparison purposes, the following are the results generated using a matrix method. Note that in the

    analysis above, the iterative process was carried to >0.01 precision. The fact that the matrix analysis resultsand the moment distribution analysis results match to 0.001 precision is mere coincidence.

    Moments at joints determined by the matrix method

    The complete shear and bending moment diagrams are as shown. Note that the moment distribution methodonly determines the moments at the joints. Developing complete bending moment diagrams require

    additional calculations using the determined joint moments and internal section equilibrium.

    SFD and BMD

    Shear force diagram Bending moment

    Example

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/wiki/Matrix_methodhttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpghttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/wiki/Matrix_method
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    Example

    The statically indeterminate beam shown in the figure is to be analysed.

    Members AB, BC, CD have the same length.

    Flexural rigidities are EI, 2EI, EI respectively.

    Concentrated load of magnitude acts at a distance from the support A.

    Uniform load of intensity acts on BC.

    Member CD is loaded at its midspan with a concentrated load of magnitude.

    In the following calcuations, counterclockwise moments are positive.

    ]edit] Fixed-end moments

    ]edit] Distribution factors

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10
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    The distribution factors of joints A and D are DAB = 1,DDC = 0.

    ]edit] Carryover factors

    The carryover factors are , except for the carryover factor from D (fixed support) to C which is zero.

    ]edit] Moment distribution

    Joint A Joint B Joint C Joint D

    Distrib.factors

    0 1 0.2727 0.7273 0.6667 0.3333 0 0

    Fixed-end

    moments

    14.700 -6.300 8.333 -8.333 12.500 -12.500

    Step 1 -14.700 -7.350

    Step 2 1.450 3.867 1.934

    Step 3 -2.034 -4.067 -2.034 -1.017

    Step 4 0.555 1.479 0.739

    Step 5 -0.246 -0.493 -0.246 -0.123

    Step 6 0.067 0.179 0.090

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/wiki/File:MomentDistributionMethod2.jpghttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12
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    Step 7 -0.030 -0.060 -0.030 -0.015

    Step 8 0.008 0.022 0.011

    Step 9 -0.004 -0.007 -0.004 -0.002

    Step 10 0.001 0.003

    Sum ofmoment

    s0 -11.569 11.569 -10.186 10.186 -13.657

    Numbers in grey are balaced moments; arrows ( / ) represent the carry-over of moment from one end

    to the other end of a member.

    ]edit] Result

    Moments at joints determined by the moment distribution method

    The conventional engineer's sign convention is used here, i.e. positive moments cause elongation atthe bottom part of a beam member.

    For comparison purposes, the following are the results generated using a matrix method. Note that in theanalysis above, the iterative process was carried to >0.01 precision. The fact that the matrix analysis results

    and the moment distribution analysis results match to 0.001 precision is mere coincidence.

    Moments at joints determined by the matrix method

    The complete shear and bending moment diagrams are as shown. Note that the moment distribution methodonly determines the moments at the joints. Developing complete bending moment diagrams require

    additional calculations using the determined joint moments and internal section equilibrium.

    SFD and BMD

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/wiki/Matrix_methodhttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/wiki/Matrix_method
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    Shear force diagram Bending moment diagram

    Example

    Example

    The statically indeterminate beam shown in the figure is to be analysed.

    Members AB, BC, CD have the same length.

    Flexural rigidities are EI, 2EI, EI respectively.

    Concentrated load of magnitude acts at a distance from the support A.

    Uniform load of intensity acts on BC.

    Member CD is loaded at its midspan with a concentrated load of magnitude.

    In the following calcuations, counterclockwise moments are positive.

    ]edit] Fixed-end moments

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpghttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9
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    ]edit] Distribution factors

    The distribution factors of joints A and D are DAB = 1,DDC = 0.

    ]edit] Carryover factors

    The carryover factors are , except for the carryover factor from D (fixed support) to C which is zero.

    ]edit] Moment distribution

    Joint A Joint B Joint C Joint D

    Distrib.factors

    0 1 0.2727 0.7273 0.6667 0.3333 0 0

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/wiki/File:MomentDistributionMethod2.jpghttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12
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    Fixed-end

    moments

    14.700 -6.300 8.333 -8.333 12.500 -12.500

    Step 1 -14.700 -7.350

    Step 2 1.450 3.867 1.934

    Step 3 -2.034 -4.067 -2.034 -1.017

    Step 4 0.555 1.479 0.739

    Step 5 -0.246 -0.493 -0.246 -0.123

    Step 6 0.067 0.179 0.090

    Step 7 -0.030 -0.060 -0.030 -0.015

    Step 8 0.008 0.022 0.011

    Step 9 -0.004 -0.007 -0.004 -0.002

    Step 10 0.001 0.003

    Sum ofmoment

    s

    0 -11.569 11.569 -10.186 10.186 -13.657

    Numbers in grey are balaced moments; arrows ( / ) represent the carry-over of moment from one endto the other end of a member.

    ]edit] Result

    Moments at joints determined by the moment distribution method

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13
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    The conventional engineer's sign convention is used here, i.e. positive moments cause elongation atthe bottom part of a beam member.

    For comparison purposes, the following are the results generated using a matrix method. Note that in theanalysis above, the iterative process was carried to >0.01 precision. The fact that the matrix analysis results

    and the moment distribution analysis results match to 0.001 precision is mere coincidence.

    Moments at joints determined by the matrix method

    The complete shear and bending moment diagrams are as shown. Note that the moment distribution methodonly determines the moments at the joints. Developing complete bending moment diagrams require

    additional calculations using the determined joint moments and internal section equilibrium.

    SFD and BMD

    Shear force diagram Bending moment diagram

    Example

    http://en.wikipedia.org/wiki/Matrix_methodhttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpghttp://en.wikipedia.org/wiki/Matrix_method
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    Example

    The statically indeterminate beam shown in the figure is to be analysed.

    Members AB, BC, CD have the same length.

    Flexural rigidities are EI, 2EI, EI respectively.

    Concentrated load of magnitude acts at a distance from the support A.

    Uniform load of intensity acts on BC.

    Member CD is loaded at its midspan with a concentrated load of magnitude.

    In the following calcuations, counterclockwise moments are positive.

    ]edit] Fixed-end moments

    ]edit] Distribution factors

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10
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    The distribution factors of joints A and D are DAB = 1,DDC = 0.

    ]edit] Carryover factors

    The carryover factors are , except for the carryover factor from D (fixed support) to C which is zero.

    ]edit] Moment distribution

    Joint A Joint B Joint C Joint D

    Distrib.factors

    0 1 0.2727 0.7273 0.6667 0.3333 0 0

    Fixed-end

    moments

    14.700 -6.300 8.333 -8.333 12.500 -12.500

    Step 1 -14.700 -7.350

    Step 2 1.450 3.867 1.934

    Step 3 -2.034 -4.067 -2.034 -1.017

    Step 4 0.555 1.479 0.739

    Step 5 -0.246 -0.493 -0.246 -0.123

    Step 6 0.067 0.179 0.090

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/wiki/File:MomentDistributionMethod2.jpghttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12
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    Step 7 -0.030 -0.060 -0.030 -0.015

    Step 8 0.008 0.022 0.011

    Step 9 -0.004 -0.007 -0.004 -0.002

    Step 10 0.001 0.003

    Sum ofmoment

    s0 -11.569 11.569 -10.186 10.186 -13.657

    Numbers in grey are balaced moments; arrows ( / ) represent the carry-over of moment from one end

    to the other end of a member.

    ]edit] Result

    Moments at joints determined by the moment distribution method

    The conventional engineer's sign convention is used here, i.e. positive moments cause elongation atthe bottom part of a beam member.

    For comparison purposes, the following are the results generated using a matrix method. Note that in theanalysis above, the iterative process was carried to >0.01 precision. The fact that the matrix analysis results

    and the moment distribution analysis results match to 0.001 precision is mere coincidence.

    Moments at joints determined by the matrix method

    The complete shear and bending moment diagrams are as shown. Note that the moment distribution methodonly determines the moments at the joints. Developing complete bending moment diagrams require

    additional calculations using the determined joint moments and internal section equilibrium.

    SFD and BMD

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/wiki/Matrix_methodhttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/wiki/Matrix_method
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    Shear force diagram Bending moment diagram

    Example

    Example

    The statically indeterminate beam shown in the figure is to be analysed.

    Members AB, BC, CD have the same length.

    Flexural rigidities are EI, 2EI, EI respectively.

    Concentrated load of magnitude acts at a distance from the support A.

    http://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpg
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    Uniform load of intensity acts on BC.

    Member CD is loaded at its midspan with a concentrated load of magnitude.

    In the following calcuations, counterclockwise moments are positive.

    ]edit] Fixed-end moments

    ]edit] Distribution factors

    The distribution factors of joints A and D are DAB = 1,DDC = 0.

    ]edit] Carryover factors

    The carryover factors are , except for the carryover factor from D (fixed support) to C which is zero.

    ]edit] Moment distribution

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12
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    Joint A Joint B Joint C Joint D

    Distrib.factors

    0 1 0.2727 0.7273 0.6667 0.3333 0 0

    Fixed-end

    moments

    14.700 -6.300 8.333 -8.333 12.500 -12.500

    Step 1 -14.700 -7.350

    Step 2 1.450 3.867 1.934

    Step 3 -2.034 -4.067 -2.034 -1.017

    Step 4 0.555 1.479 0.739

    Step 5 -0.246 -0.493 -0.246 -0.123

    Step 6 0.067 0.179 0.090

    Step 7 -0.030 -0.060 -0.030 -0.015

    Step 8 0.008 0.022 0.011

    Step 9 -0.004 -0.007 -0.004 -0.002

    Step 10 0.001 0.003

    Sum ofmoment

    0 -11.569 11.569 -10.186 10.186 -13.657

    http://en.wikipedia.org/wiki/File:MomentDistributionMethod2.jpg
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    s

    Numbers in grey are balaced moments; arrows ( / ) represent the carry-over of moment from one endto the other end of a member.

    ]edit] Result

    Moments at joints determined by the moment distribution method

    The conventional engineer's sign convention is used here, i.e. positive moments cause elongation atthe bottom part of a beam member.

    For comparison purposes, the following are the results generated using a matrix method. Note that in the

    analysis above, the iterative process was carried to >0.01 precision. The fact that the matrix analysis resultsand the moment distribution analysis results match to 0.001 precision is mere coincidence.

    Moments at joints determined by the matrix method

    The complete shear and bending moment diagrams are as shown. Note that the moment distribution methodonly determines the moments at the joints. Developing complete bending moment diagrams require

    additional calculations using the determined joint moments and internal section equilibrium.

    SFD and BMD

    Shear force diagram Bending moment diagram

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/wiki/Matrix_methodhttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpghttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/wiki/Matrix_method
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    Example

    Example

    The statically indeterminate beam shown in the figure is to be analysed.

    Members AB, BC, CD have the same length.Flexural rigidities are EI, 2EI, EI respectively.

    Concentrated load of magnitude acts at a distance from the support A.

    Uniform load of intensity acts on BC.

    Member CD is loaded at its midspan with a concentrated load of magnitude.

    In the following calcuations, counterclockwise moments are positive.

    ]edit] Fixed-end moments

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethod.jpghttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=9
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    ]edit] Distribution factors

    The distribution factors of joints A and D are DAB = 1,DDC = 0.

    ]edit] Carryover factors

    The carryover factors are , except for the carryover factor from D (fixed support) to C which is zero.

    ]edit] Moment distribution

    Joint A Joint B Joint C Joint D

    Distrib.factors

    0 1 0.2727 0.7273 0.6667 0.3333 0 0

    Fixed-end

    moments

    14.700 -6.300 8.333 -8.333 12.500 -12.500

    Step 1 -14.700 -7.350

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/wiki/File:MomentDistributionMethod2.jpghttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=10http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=11http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=12
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    Step 2 1.450 3.867 1.934

    Step 3 -2.034 -4.067 -2.034 -1.017

    Step 4 0.555 1.479 0.739

    Step 5 -0.246 -0.493 -0.246 -0.123

    Step 6 0.067 0.179 0.090

    Step 7 -0.030 -0.060 -0.030 -0.015

    Step 8 0.008 0.022 0.011

    Step 9 -0.004 -0.007 -0.004 -0.002

    Step 10 0.001 0.003

    Sum ofmoments

    0 -11.569 11.569 -10.186 10.186 -13.657

    Numbers in grey are balaced moments; arrows ( / ) represent the carry-over of moment from one endto the other end of a member.

    ]edit] Result

    Moments at joints determined by the moment distribution method

    The conventional engineer's sign convention is used here, i.e. positive moments cause elongation atthe bottom part of a beam member.

    For comparison purposes, the following are the results generated using a matrix method. Note that in theanalysis above, the iterative process was carried to >0.01 precision. The fact that the matrix analysis results

    and the moment distribution analysis results match to 0.001 precision is mere coincidence.

    Moments at joints determined by the matrix method

    http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/wiki/Matrix_methodhttp://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/w/index.php?title=Moment_distribution_method&action=edit&section=13http://en.wikipedia.org/wiki/Matrix_method
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    The complete shear and bending moment diagrams are as shown. Note that the moment distribution methodonly determines the moments at the joints. Developing complete bending moment diagrams require

    additional calculations using the determined joint moments and internal section equilibrium.

    SFD and BMD

    Shear force diagram Bending moment diagram

    http://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodBMD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpghttp://en.wikipedia.org/wiki/File:MomentDistributionMethodSFD.jpg