ENGR 111 Technical Memorandum

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  • 8/18/2019 ENGR 111 Technical Memorandum

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    APPENDIX

    Material that is to be included in the appendix are: derivation of equations that are presented in methods,

    computations used to establish initial conditions or other system parameters, computational code used,

    and graphs or other types of outputs. The material in the appendix can be referred to in the main body of 

    the report. Derivations and computations are to be presented in a neat and orderly format with

    appropriate comment statements. These do not need to be typed out, but it is recommended. Comment

    statements need to be provided so that the reader, or even yourself if you need to revisit the problem, canfollow the train of thought for an analysis. igures need to be presented and variables defined. !e sure to

    show proper handling and presentation of units. The presence of this material will also allow for 

    identification of where an error may have occurred if your solution is not correct.

    "ach #ppendix should be labeled with a separate letter $#ppendix #%, $#ppendix !%, etc.

    inally, the main body of the report should be within a maximum length of & to '( pages depending on

    how many graphs are included. The appendix may ma)e the report a bit longer.

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    #**"+D- # /oad /imit Calculations

     

     D :∑  F Y = ADsin (45° )+  P4=0

     D :∑  F  X = ADcos(45 °)+ DE=0

     A :∑  F  X = ADcos ( 45° )+ AB=0

     A :∑  F Y = ADsin (45° )+ AE+ P

    4=0

     E :∑  F Y = AE+BEsin (45 ° )=0

     E :∑  F  X = DE+BEcos ( 45 ° )+ EF 

     F :∑  F Y =BF =0

     AD=CH =− P4

    ×  2

    √ 2=(−.354) P

     DE=GH =−− P2√ 2

    ×√ 2

    2=( .250) P

     AB=BC =−− P2√ 2

    × √ 22=( .250 )  P

    *01*01

    CA   B

    *01 *01

    D   HFE   G

    2.2.3415*

    2.6

    2.62.64(5*

    2.64(5* 2.62.64(5*

    2(52(5*2(5*

    E

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     AE=CG=−− P2√ 2

    ×√ 2

    2− P

    4=(0 ) P

    BE=BG=−0×  2

    √ 2=(0)  P

     EF = FG=− P4−0× √ 

    2

    2=(−.250 )  P

    BF =(0 ) P

    #**"+D- # 7 /oad /imit Calculations

    2.2.3415*

    2.6

    2.62.64(5*

    2.64(5* 2.62.64(5*

    2(52(5*2(5*

    E

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     Load Limit = Pif ( .250 ) P=1.17 lbs

     P=4.68

    #**"+D- ! 7 8isuals