# Bridge Constructor in Concrete

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## Premises

Concrete has density of 145 lbft^{3}, compressive strength of 4 000 lbin^{2} (576 000 lbft^{2}), elastic modulus of 115 ⋅ 10^{6} lbin^{2} (16.6 ⋅ 10^{9} lbft^{2}), and tensile strength of zero. (See AASHTO LRFD Bridge Design Specifications 2012 §§ 3.5.1, 5.4.2.1, 5.4.2.4, and 5.4.2.7.)

The bridge is 12 ft wide and has a rectangular cross-section. Traffic is represented by a uniform load of 640 lbft in conjunction with a point load of 72 000 lb. (See AASHTO LRFD Bridge Design Specifications 2012 §§ 3.6.1.1.1, 3.6.1.2.1, 3.6.1.2.2, and 3.6.1.2.4)

Safety margins, and loads other than concrete and traffic, are ignored. This is just for fun!

Each square in Bridge Constructor's default grid is 8 ft wide. (The in-game value is 2.5 meters.)

The optimal bridge is the bridge that has the minimum volume.

## Region 1

### Bridge 1

The bridge must extend from (0 ft, 0 ft) to (64 ft, 0 ft).

#### Iteration 1

The bridge has thickness `t` and comprises pin 0 at (0 ft, 0 ft), member 01, pin 1 at (32 ft, `y`_{1}), member 12, and pin 2 at (64 ft, 0 ft).

The forces acting on the bridge (in addition to those listed in Premises) are `F`_{0x} pushing rightward at point 0, `F`_{0y} pushing upward at point 0, `F`_{2x} pushing leftward at point 2, and `F`_{2y} pushing upward at point 2.

##### Rotational equilibrium of bridge about point 0

`F`_{2y} ⋅ 64 ft = ∫0 ft64 ft145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft32 ft^{2} + `y`_{1}^{2}^{12}32 ft`x`d`x` + 72 000 lb ⋅ `x`_{truck}

`F`_{2y} ⋅ 64 ft = 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft32 ft^{2} + `y`_{1}^{2}^{12}32 ft64 ft^{2} ÷ 2 + 72 000 lb ⋅ `x`_{truck}

`F`_{2y} = 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft32 ft^{2} + `y`_{1}^{2}^{12} + 72 000 lb ⋅ `x`_{truck}64 ft

##### Vertical equilibrium of bridge

`F`_{0y} = 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft32 ft^{2} + `y`_{1}^{2}^{12}32 ft ⋅ 64 ft + 72 000 lb − `F`_{2y}

`F`_{0y} = 2145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft32 ft^{2} + `y`_{1}^{2}^{12} + 72 000 lb − `F`_{2y}

The forces acting on member 01 (in addition to those listed above) are `F`_{1x} pushing leftward at pin 1 and `F`_{1y} pushing downward at pin 1.

##### Vertical equilibrium of member 01

`F`_{1y} = `F`_{0y} − 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft32 ft^{2} + `y`_{1}^{2}^{12}32 ft ⋅ 32 ft − 72 000 lb

`F`_{1y} = `F`_{0y} − 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft32 ft^{2} + `y`_{1}^{2}^{12} − 72 000 lb

##### Rotational equilibrium of member 01 about point 0

`F`_{1x}`y`_{1} = `F`_{1y} ⋅ 32 ft + ∫0 ft32 ft145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft32 ft^{2} + `y`_{1}^{2}^{12}32 ft`x`d`x` + 72 000 lb ⋅ `x`_{truck}

`F`_{1x}`y`_{1} = `F`_{1y} ⋅ 32 ft + 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft32 ft^{2} + `y`_{1}^{2}^{12}32 ft32 ft^{2} ÷ 2 + 72 000 lb ⋅ `x`_{truck}

`F`_{1x} = `F`_{1y} ⋅ 32 ft`y`_{1} + 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft32 ft^{2} + `y`_{1}^{2}^{12}2`y`_{1} ⋅ 32 ft + 72 000 lb ⋅ `x`_{truck}`y`_{1}

##### Horizontal equilibrium of member 01

`F`_{0x} = `F`_{1x}

##### Horizontal equilibrium of bridge

`F`_{2x} = `F`_{0x}

Segment 0e extends from pin 0 along the bridge to arbitrary solid joint e at (`x`_{e}, `y`_{e}). The forces acting on segment 0e (in addition to those listed above) are `F`_{ex} pushing leftward at point e, `F`_{ey} pushing downward at point e, and `M`_{e} pushing counterclockwise at point e.

##### Vertical equilibrium of segment 0e

`F`_{ey} = `F`_{0y} − 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft32 ft^{2} + `y`_{1}^{2}^{12}32 ft ⋅ `x`_{e} − 72 000 lb

##### Horizontal equilibrium of segment 0e

`F`_{ex} = `F`_{0x}

##### Rotational equilibrium of segment 0e about point 0

`M`_{e} = `F`_{ey}`x`_{e} − `F`_{ex}`y`_{e} + ∫0 ft`x`_{e}145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft32 ft^{2} + `y`_{1}^{2}^{12}32 ft`x`d`x` + 72 000 lb ⋅ `x`_{truck}

`M`_{e} = `F`_{ey}`x`_{e} − `F`_{ex}`y`_{e} + 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft32 ft^{2} + `y`_{1}^{2}^{12}32 ft`x`_{e}^{2} ÷ 2 + 72 000 lb ⋅ `x`_{truck}

##### Minimum/maximum compression at point e if `x`_{e} < 32 ft

`P`_{min/max} = `F`_{ey}`F`_{ex} ⋅ `y`_{1}32 ft32 ft^{2} + `y`_{1}^{2}^{1/2} ÷ 12 ft ⋅ `t` ± 6`M`_{e}`t`^{2} ⋅ 12 ft ∈ 0 lbft^{2}, 576 000 lbft^{2}

Spreadsheet analysis suggests that the optimal values for `y`_{1} and `t` are approximately 0.116 ft and 1.67 ft, respectively, yielding a volume of 1 280 ft^{3}.

### Bridge 2

The bridge must extend from (0 ft, 0 ft) to (96 ft, 0 ft).

#### Iteration 1

The bridge has thickness `t` and comprises pin 0 at (0 ft, 0 ft), member 01, pin 1 at (48 ft, `y`_{1}), member 12, and pin 2 at (96 ft, 0 ft).

The forces acting on the bridge (in addition to those listed in Premises) are `F`_{0x} pushing rightward at point 0, `F`_{0y} pushing upward at point 0, `F`_{2x} pushing leftward at point 2, and `F`_{2y} pushing upward at point 2.

##### Rotational equilibrium of bridge about point 0

`F`_{2y} ⋅ 96 ft = ∫0 ft96 ft145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft48 ft^{2} + `y`_{1}^{2}^{12}48 ft`x`d`x` + 72 000 lb ⋅ `x`_{truck}

`F`_{2y} ⋅ 96 ft = 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft48 ft^{2} + `y`_{1}^{2}^{12}48 ft96 ft^{2} ÷ 2 + 72 000 lb ⋅ `x`_{truck}

`F`_{2y} = 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft48 ft^{2} + `y`_{1}^{2}^{12} + 72 000 lb ⋅ `x`_{truck}96 ft

##### Vertical equilibrium of bridge

`F`_{0y} = 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft48 ft^{2} + `y`_{1}^{2}^{12}48 ft ⋅ 96 ft + 72 000 lb − `F`_{2y}

`F`_{0y} = 2145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft48 ft^{2} + `y`_{1}^{2}^{12} + 72 000 lb − `F`_{2y}

The forces acting on member 01 (in addition to those listed above) are `F`_{1x} pushing leftward at pin 1 and `F`_{1y} pushing downward at pin 1.

##### Vertical equilibrium of member 01

`F`_{1y} = `F`_{0y} − 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft48 ft^{2} + `y`_{1}^{2}^{12}48 ft ⋅ 48 ft − 72 000 lb

`F`_{1y} = `F`_{0y} − 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft48 ft^{2} + `y`_{1}^{2}^{12} − 72 000 lb

##### Rotational equilibrium of member 01 about point 0

`F`_{1x}`y`_{1} = `F`_{1y} ⋅ 48 ft + ∫0 ft48 ft145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft48 ft^{2} + `y`_{1}^{2}^{12}48 ft`x`d`x` + 72 000 lb ⋅ `x`_{truck}

`F`_{1x}`y`_{1} = `F`_{1y} ⋅ 48 ft + 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft48 ft^{2} + `y`_{1}^{2}^{12}48 ft48 ft^{2} ÷ 2 + 72 000 lb ⋅ `x`_{truck}

`F`_{1x} = `F`_{1y} ⋅ 48 ft`y`_{1} + 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft48 ft^{2} + `y`_{1}^{2}^{12}2`y`_{1} ⋅ 48 ft + 72 000 lb ⋅ `x`_{truck}`y`_{1}

##### Horizontal equilibrium of member 01

`F`_{0x} = `F`_{1x}

##### Horizontal equilibrium of bridge

`F`_{2x} = `F`_{0x}

Segment 0e extends from pin 0 along the bridge to arbitrary solid joint e at (`x`_{e}, `y`_{e}). The forces acting on segment 0e (in addition to those listed above) are `F`_{ex} pushing leftward at point e, `F`_{ey} pushing downward at point e, and `M`_{e} pushing counterclockwise at point e.

##### Vertical equilibrium of segment 0e

`F`_{ey} = `F`_{0y} − 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft48 ft^{2} + `y`_{1}^{2}^{12}48 ft ⋅ `x`_{e} − 72 000 lb

##### Horizontal equilibrium of segment 0e

`F`_{ex} = `F`_{0x}

##### Rotational equilibrium of segment 0e about point 0

`M`_{e} = `F`_{ey}`x`_{e} − `F`_{ex}`y`_{e} + ∫0 ft`x`_{e}145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft48 ft^{2} + `y`_{1}^{2}^{12}48 ft`x`d`x` + 72 000 lb ⋅ `x`_{truck}

`M`_{e} = `F`_{ey}`x`_{e} − `F`_{ex}`y`_{e} + 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft48 ft^{2} + `y`_{1}^{2}^{12}48 ft`x`_{e}^{2} ÷ 2 + 72 000 lb ⋅ `x`_{truck}

##### Minimum/maximum compression at point e if `x`_{e} < 48 ft

`P`_{min/max} = `F`_{ey}`F`_{ex} ⋅ `y`_{1}48 ft48 ft^{2} + `y`_{1}^{2}^{1/2} ÷ 12 ft ⋅ `t` ± 6`M`_{e}`t`^{2} ⋅ 12 ft ∈ 0 lbft^{2}, 576 000 lbft^{2}

Spreadsheet analysis suggests that the optimal values for `y`_{1} and `t` are approximately 0.460 ft and 2.30 ft, respectively, yielding a volume of 2 650 ft^{3}.

### Bridge 3

The bridge must extend from (0 ft, 0 ft) to (160 ft, 0 ft). Piers are available at (48 ft, −24 ft) and (112 ft, −24 ft).

#### Iteration 1

The bridge comprises pin 0 at (0 ft, 0 ft), solid joint 1 at (`x`_{1}, `y`_{1}), pin 2 at (`x`_{2}, `y`_{2}), pin 3 at (80 ft, `y`_{3}), pin 4 at (160 ft − `x`_{2}, `y`_{2}), solid joint 5 at (160 ft − `x`_{1}, `y`_{1}), pin 6 at (160 ft, 0 ft), solid joint 7 at (48 ft, −24 ft), solid joint 8 at (112 ft, −24 ft), members 01 and 56 of thickness `t`_{01}, members 12 and 45 of thickness `t`_{12}, members 23 and 34 of thickness `t`_{23}, and members 72 and 84 of thickness `t`_{72}.

This structure has 24 unknown forces.

The forces acting on the bridge (in addition to those listed in Premises) are:
rightward `F`_{0x} and upward `F`_{0y} at point 0;
leftward `F`_{6x} and upward `F`_{6y} at point 6;
leftward `F`_{7x} and upward `F`_{7y} at point 7; and
leftward `F`_{8x} and upward `F`_{8y} at point 8.

⋅

F_{0x} |

F_{0y} |

F_{1x} |

F_{1y} |

M_{1} |

F_{2x12} |

F_{2y12} |

F_{2x23} |

F_{2y23} |

F_{3x} |

F_{3y} |

F_{4x34} |

F_{4y34} |

F_{4x45} |

F_{4y45} |

F_{5x} |

F_{5y} |

M_{5} |

F_{6x} |

F_{6y} |

F_{7x} |

F_{7y} |

F_{8x} |

F_{8y} |

##### Rotational equilibrium of bridge about point 0

`F`_{6y} ⋅ 160 ft − `F`_{7x} ⋅ 24 ft + `F`_{7y} ⋅ 48 ft − `F`_{8x} ⋅ 24 ft + `F`_{8y} ⋅ 112 ft
= 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{01} + 640 lbft`x`_{1}^{2} + `y`_{1}^{2}^{12}`x`_{1}∫0 ft`x`_{1}`x`d`x` + ∫160 ft − `x`_{1}160 ft`x`d`x`
+ 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{12} + 640 lbft`x`_{2} − `x`_{1}^{2} + `y`_{2} − `y`_{1}^{2}^{12}`x`_{2} − `x`_{1}∫`x`_{1}`x`_{2}`x`d`x` + ∫160 ft − `x`_{2}160 ft − `x`_{1}`x`d`x`
+ 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{23} + 640 lbft80 ft − `x`_{2}^{2} + `y`_{3} − `y`_{2}^{2}^{12}80 ft − `x`_{2} ⋅ ∫`x`_{2}160 ft − `x`_{2}`x`d`x`
+ 72 000 lb ⋅ `x`_{truck}

`F`_{6y} ⋅ 160 ft − `F`_{7x} ⋅ 24 ft + `F`_{7y} ⋅ 48 ft − `F`_{8x} ⋅ 24 ft + `F`_{8y} ⋅ 112 ft
= 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{01} + 640 lbft`x`_{1}^{2} + `y`_{1}^{2}^{12}`x`_{1}160 ft^{2} − 160 ft − `x`_{1}^{2} + `x`_{1}^{2} ÷ 2
+ 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{12} + 640 lbft`x`_{2} − `x`_{1}^{2} + `y`_{2} − `y`_{1}^{2}^{12}`x`_{2} − `x`_{1}`x`_{2}^{2} − `x`_{1}^{2} + 160 ft − `x`_{1}^{2} − 160 ft − `x`_{2}^{2} ÷ 2
+ 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{23} + 640 lbft80 ft − `x`_{2}^{2} + `y`_{3} − `y`_{2}^{2}^{12}80 ft − `x`_{2}160 ft − `x`_{2}^{2} − `x`_{2}^{2} ÷ 2
+ 72 000 lb ⋅ `x`_{truck}

`F`_{6y} ⋅ 160 ft − `F`_{7x} ⋅ 24 ft + `F`_{7y} ⋅ 48 ft − `F`_{8x} ⋅ 24 ft + `F`_{8y} ⋅ 112 ft
= 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{01} + 640 lbft`x`_{1}^{2} + `y`_{1}^{2}^{12}`x`_{1} ⋅ `x`_{1} ⋅ 160 ft
+ 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{12} + 640 lbft`x`_{2} − `x`_{1}^{2} + `y`_{2} − `y`_{1}^{2}^{12}`x`_{2} − `x`_{1}`x`_{2} − `x`_{1} ⋅ 160 ft
+ 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{23} + 640 lbft80 ft − `x`_{2}^{2} + `y`_{3} − `y`_{2}^{2}^{12}80 ft − `x`_{2}80 ft − `x`_{2} ⋅ 160 ft
+ 72 000 lb ⋅ `x`_{truck}

`F`_{6y} ⋅ 160 ft − `F`_{7x} ⋅ 24 ft + `F`_{7y} ⋅ 48 ft − `F`_{8x} ⋅ 24 ft + `F`_{8y} ⋅ 112 ft
= 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{01} + 640 lbft`x`_{1}^{2} + `y`_{1}^{2}^{12}
+ 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{12} + 640 lbft`x`_{2} − `x`_{1}^{2} + `y`_{2} − `y`_{1}^{2}^{12}
+ 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{23} + 640 lbft80 ft − `x`_{2}^{2} + `y`_{3} − `y`_{2}^{2}^{12} ⋅ 160 ft
+ 72 000 lb ⋅ `x`_{truck}

##### Vertical equilibrium of bridge

`F`_{0y} + `F`_{6y} + `F`_{7y} + `F`_{8y}
= 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{01} + 640 lbft`x`_{1}^{2} + `y`_{1}^{2}^{12}`x`_{1} ⋅ 2`x`_{1}
+ 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{12} + 640 lbft`x`_{2} − `x`_{1}^{2} + `y`_{2} − `y`_{1}^{2}^{12}`x`_{2} − `x`_{1} ⋅ 2`x`_{2} − `x`_{1}
+ 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{23} + 640 lbft80 ft − `x`_{2}^{2} + `y`_{3} − `y`_{2}^{2}^{12}80 ft − `x`_{2} ⋅ 280 ft − `x`_{2}
+ 72 000 lb

`F`_{0y} + `F`_{6y} + `F`_{7y} + `F`_{8y}
= 2145 lbft^{3} ⋅ 12 ft ⋅ `t`_{01} + 640 lbft`x`_{1}^{2} + `y`_{1}^{2}^{12}
+ 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{12} + 640 lbft`x`_{2} − `x`_{1}^{2} + `y`_{2} − `y`_{1}^{2}^{12}
+ 145 lbft^{3} ⋅ 12 ft ⋅ `t`_{23} + 640 lbft80 ft − `x`_{2}^{2} + `y`_{3} − `y`_{2}^{2}^{12}
+ 72 000 lb

##### Horizontal equilibrium of bridge

`F`_{0x} − `F`_{6x} − `F`_{7x} − `F`_{8x} = 0 lb

### Bridge 4

The bridge must extend from (0 ft, 0 ft) to (128 ft, 0 ft). A pier is available at (88 ft, −24 ft).

## Region 5

### Bridge 1

The bridge must extend from (0 ft, 0 ft) to (320 ft, 0 ft).

#### Iteration 1

The bridge has thickness `t` and comprises pin 0 at (0 ft, 0 ft), member 01, pin 1 at (160 ft, `y`_{1}), member 12, and pin 2 at (320 ft, 0 ft).

The forces acting on the bridge (in addition to those listed in Premises) are `F`_{0x} pushing rightward at point 0, `F`_{0y} pushing upward at point 0, `F`_{2x} pushing leftward at point 2, and `F`_{2y} pushing upward at point 2.

##### Rotational equilibrium of bridge about point 0

`F`_{2y} ⋅ 320 ft = ∫0 ft320 ft145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft160 ft^{2} + `y`_{1}^{2}^{12}160 ft`x`d`x` + 72 000 lb ⋅ `x`_{truck}

`F`_{2y} ⋅ 320 ft = 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft160 ft^{2} + `y`_{1}^{2}^{12}160 ft320 ft^{2} ÷ 2 + 72 000 lb ⋅ `x`_{truck}

`F`_{2y} = 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft160 ft^{2} + `y`_{1}^{2}^{12} + 72 000 lb ⋅ `x`_{truck}320 ft

##### Vertical equilibrium of bridge

`F`_{0y} = 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft160 ft^{2} + `y`_{1}^{2}^{12}160 ft ⋅ 320 ft + 72 000 lb − `F`_{2y}

`F`_{0y} = 2145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft160 ft^{2} + `y`_{1}^{2}^{12} + 72 000 lb − `F`_{2y}

The forces acting on member 01 (in addition to those listed above) are `F`_{1x} pushing leftward at pin 1 and `F`_{1y} pushing downward at pin 1.

##### Vertical equilibrium of member 01

`F`_{1y} = `F`_{0y} − 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft160 ft^{2} + `y`_{1}^{2}^{12}160 ft ⋅ 160 ft − 72 000 lb

`F`_{1y} = `F`_{0y} − 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft160 ft^{2} + `y`_{1}^{2}^{12} − 72 000 lb

##### Rotational equilibrium of member 01 about point 0

`F`_{1x}`y`_{1} = `F`_{1y} ⋅ 160 ft + ∫0 ft160 ft145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft160 ft^{2} + `y`_{1}^{2}^{12}160 ft`x`d`x` + 72 000 lb ⋅ `x`_{truck}

`F`_{1x}`y`_{1} = `F`_{1y} ⋅ 160 ft + 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft160 ft^{2} + `y`_{1}^{2}^{12}160 ft160 ft^{2} ÷ 2 + 72 000 lb ⋅ `x`_{truck}

`F`_{1x} = `F`_{1y} ⋅ 160 ft`y`_{1} + 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft160 ft^{2} + `y`_{1}^{2}^{12}2`y`_{1} ⋅ 160 ft + 72 000 lb ⋅ `x`_{truck}`y`_{1}

##### Horizontal equilibrium of member 01

`F`_{0x} = `F`_{1x}

##### Horizontal equilibrium of bridge

`F`_{2x} = `F`_{0x}

Segment 0e extends from pin 0 along the bridge to arbitrary solid joint e at (`x`_{e}, `y`_{e}). The forces acting on segment 0e (in addition to those listed above) are `F`_{ex} pushing leftward at point e, `F`_{ey} pushing downward at point e, and `M`_{e} pushing counterclockwise at point e.

##### Vertical equilibrium of segment 0e

`F`_{ey} = `F`_{0y} − 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft160 ft^{2} + `y`_{1}^{2}^{12}160 ft ⋅ `x`_{e} − 72 000 lb

##### Horizontal equilibrium of segment 0e

`F`_{ex} = `F`_{0x}

##### Rotational equilibrium of segment 0e about point 0

`M`_{e} = `F`_{ey}`x`_{e} − `F`_{ex}`y`_{e} + ∫0 ft`x`_{e}145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft160 ft^{2} + `y`_{1}^{2}^{12}160 ft`x`d`x` + 72 000 lb ⋅ `x`_{truck}

`M`_{e} = `F`_{ey}`x`_{e} − `F`_{ex}`y`_{e} + 145 lbft^{3} ⋅ 12 ft ⋅ `t` + 640 lbft160 ft^{2} + `y`_{1}^{2}^{12}160 ft`x`_{e}^{2} ÷ 2 + 72 000 lb ⋅ `x`_{truck}

##### Minimum/maximum compression at point e if `x`_{e} < 160 ft

`P`_{min/max} = `F`_{ey}`F`_{ex} ⋅ `y`_{1}160 ft160 ft^{2} + `y`_{1}^{2}^{1/2} ÷ 12 ft ⋅ `t` ± 6`M`_{e}`t`^{2} ⋅ 12 ft ∈ 0 lbft^{2}, 576 000 lbft^{2}

Spreadsheet analysis suggests that the optimal values for `y`_{1} and `t` are approximately 6.53 ft and 10.9 ft, respectively, yielding a volume of 41 800 ft^{3}.