# Plays : code test 2

## code test 2

`The Analysis:                                                          |---15ft---| ---           +---+       /|\           |   |     |            |---|\   15ft          | | |\\   |            | | | \\  |            | | |  \\ |Force Zombie|14ft   \\                       |----------->| | |    \\                      |   /|\      | | |     \\                    |   5ft      | | |      \\                  \|/  \|/      |\|/|     Ø(\\  Force X        --------------+---+-------|^|<-------                                   ^ ,_                                         /|\|\                                         |   \                                                     Force Y|    \ Force R                                                    |     \                                                          |      \      The Free Body Diagram of the beam is as such: Force Z   /|\---------->/|W\           /|\ \       Sum Fx = 0 = Fz - Fx => Fz = Fx              \ \      Sum Fy = 0 = Fy - Fw => Fy = Fw               \ \     Sum Mw = 0 = 15Fy - 14Fx - 7.5Fw                \|\               = 15Fy - 14Fx - 7.5Fy                 | \              = 7.5Fy - 14Fx                  |\ \          Force W| \ \ => Fy = (14/7.5)Fx where Fx = Fz                 |  \ \                \|/  \G\    Force X                 '   ==^==<----------                       ^ ,_                      /|\|\                       |   \                Force Y|    \Resultant Force                        |     \ (Force R)                       |      \__                       |       \_)ØThe Resultant Force:Fr = FxcosØ + (14/7.5)FxsinØNext we have to look to see the stress the beam is under from the compressive pressure.               |      \       |       \      | Force R\     |Force Y         \    |                               \  \|/                        \| '   Force X -~/|\     Sum Fx = 0 = Fx - (FxcosØ + 14/7.5FxsinØ)cosØ    -------->/|W\            Fx = (FxcosØ + 14/7.5Fxsin)cosØ             /|\ \                   \ \  Sum Fy = 0 = -Fy + (FxcosØ + 14/7.5FysinØ)sinØ                 \ \         Fy = (FxcosØ + 14/7.5Fxsin)sinØ                                                    \ \                                                                                     \ \  Therefore: The resultant force at the top of the member =                     \ \                    ____________________                     \ \         => Fr = \/ Fx^2 + (14/7.5Fx)^2                       \ \                       \G\)Ø                          ,_                           |\                               \ FxcosØ + 14/7.5FxsinØ                              \                                \__                               \_)Ø So the resulting compressive forces on the member results as:                   ___________________                     Force   \/Fx^2 + (14/7.5Fx)^2 + FxcosØ + (14/7.5)FxsinØStress = ----- = -----------------------------------------------                                  Area             Beam Cross Sectional Area Depending on the force of the zombies, it being 1000 lbs or 2000 lbs, we can determine the resultant force which would give us how much stress the beam is under. There are no additional support based on this configuration and the free body diagram shows exactly how this structure configuration will hold against all the forces that can be enacted on the body. The Argument: The wall is stronger with the beams than the beams on the inside. The Analysis:          |---15ft---|                               |O Force Z                   +---+                            /|--------->                   |   |                           /W|O                  /|   |                          / /|                 //|   |                         / /      Sum Fx = 0 = Fz - Fx => Fx = Fz                // |   |                        / /       sum Fy = 0 = -Fy - Fw => Fw = -Fy               //  |   |                       /|/        Sum Mw = 0 = -14Fx - 15Fy - 7.5Fw              //   |   |Force Zombie          / |Force W             = -14Fx - 15Fy + 7.5Fy             //    |   |------------>        / /|                    = -14Fx - 7.5Fy             //     |   | /|\                / /\|/           => Fy = -(14/7.5)Fx where Fx = Fz           //      |   | 5ft               / /  '                           Force X  //)Ø     |   | \|/              /G/)Ø  Force X<--------|^|-------+---+-------        ---^--- <----------                           /|                             /////        / |                              /|Force R/  |                     Force R / |       /   | Force Y                    /  | Force Y     /    |                           /   |   |/_   \|/                        |/_  \|/   '      '                         '     ' The Resultant Force:Fr = FxcosØ + (14/7.5)FxsinØ Next we have to analyze the stress on the beam by defining the resultant forces.                          _,                   ^      /|                  /|\    /            Force Y|    /Force R                    |   /                   |  /                   | /              Sum Fx = 0 = Fx - (FxcosØ + (14/7.5)FxsinØ)cosØ                   |/ Force X               Fx = (FxcosØ + (14/7.5)FxsinØ)cosØ                  /|--------->                 / |                Sum Fy = 0 = Fy - (FxcosØ + (14/7.5)FxsinØ)sinØ                / /                         Fy = (FxcosØ + (14/7.5)FxsinØ)sinØ               / /              / /             / / Therefore: The resultant force at the top of the member =            / /                       ____________________           / /              => Fr = \/ Fx^2 + (14/7.5Fx)^2          / /         / /     ___/_/___        /       /      /     /FxcosØ + (14/7.5)FxsinØ    /  |/_`

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