Bridge Constructor in Concrete

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Premises

Concrete has density of 145 lbft³, compressive strength of 4 000 lbin² (576 000 lbft²), elastic modulus of 115 ⋅ 10⁶ lbin² (16.6 ⋅ 10⁹ lbft²), 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.

(Whoops! I forgot about buckling. I'll have to do some recalculation in the future…)

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, y₁), 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 lbft³ ⋅ 12 ft ⋅ t + 640 lbft32 ft² + y₁²¹²32 ftxdx + 72 000 lb ⋅ x_truck

F_2y ⋅ 64 ft = 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft32 ft² + y₁²¹²32 ft64 ft² ÷ 2 + 72 000 lb ⋅ x_truck

F_2y = 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft32 ft² + y₁²¹² + 72 000 lb ⋅ x_truck64 ft

Vertical equilibrium of bridge

F_0y = 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft32 ft² + y₁²¹²32 ft ⋅ 64 ft + 72 000 lb − F_2y

F_0y = 2145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft32 ft² + y₁²¹² + 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 lbft³ ⋅ 12 ft ⋅ t + 640 lbft32 ft² + y₁²¹²32 ft ⋅ 32 ft − 72 000 lb

F_1y = F_0y − 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft32 ft² + y₁²¹² − 72 000 lb

Rotational equilibrium of member 01 about point 0

F_1xy₁ = F_1y ⋅ 32 ft + ∫0 ft32 ft145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft32 ft² + y₁²¹²32 ftxdx + 72 000 lb ⋅ x_truck

F_1xy₁ = F_1y ⋅ 32 ft + 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft32 ft² + y₁²¹²32 ft32 ft² ÷ 2 + 72 000 lb ⋅ x_truck

F_1x = F_1y ⋅ 32 fty₁ + 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft32 ft² + y₁²¹²2y₁ ⋅ 32 ft + 72 000 lb ⋅ x_trucky₁

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 lbft³ ⋅ 12 ft ⋅ t + 640 lbft32 ft² + y₁²¹²32 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_eyx_e − F_exy_e + ∫0 ftx_e145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft32 ft² + y₁²¹²32 ftxdx + 72 000 lb ⋅ x_truck

M_e = F_eyx_e − F_exy_e + 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft32 ft² + y₁²¹²32 ftx_e² ÷ 2 + 72 000 lb ⋅ x_truck

Minimum/maximum compressive pressure at point e if `x`_e < 32 ft

P_min/max = F_eyF_ex ⋅ y₁32 ft32 ft² + y₁²^1/2 ÷ 12 ft ⋅ t ± 6M_et² ⋅ 12 ft ∈ 0 lbft², 576 000 lbft²

Compressive force at point 0

F₀ = F_0x² + F_0y²^1/2 ≤ π² ⋅ 16.6 ⋅ 10⁹ lbft² ⋅ 12 ft ⋅ t³1232 ft² + y₁²

Spreadsheet analysis suggests that the optimal values for y₁ and t are approximately 0.153 ft and 2.24 ft, respectively, yielding a volume of 1 720 ft³.

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, y₁), member 12, and pin 2 at (96 ft, 0 ft).

Rotational equilibrium of bridge about point 0

F_2y ⋅ 96 ft = ∫0 ft96 ft145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft48 ft² + y₁²¹²48 ftxdx + 72 000 lb ⋅ x_truck

F_2y ⋅ 96 ft = 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft48 ft² + y₁²¹²48 ft96 ft² ÷ 2 + 72 000 lb ⋅ x_truck

F_2y = 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft48 ft² + y₁²¹² + 72 000 lb ⋅ x_truck96 ft

Vertical equilibrium of bridge

F_0y = 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft48 ft² + y₁²¹²48 ft ⋅ 96 ft + 72 000 lb − F_2y

F_0y = 2145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft48 ft² + y₁²¹² + 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 lbft³ ⋅ 12 ft ⋅ t + 640 lbft48 ft² + y₁²¹²48 ft ⋅ 48 ft − 72 000 lb

F_1y = F_0y − 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft48 ft² + y₁²¹² − 72 000 lb

Rotational equilibrium of member 01 about point 0

F_1xy₁ = F_1y ⋅ 48 ft + ∫0 ft48 ft145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft48 ft² + y₁²¹²48 ftxdx + 72 000 lb ⋅ x_truck

F_1xy₁ = F_1y ⋅ 48 ft + 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft48 ft² + y₁²¹²48 ft48 ft² ÷ 2 + 72 000 lb ⋅ x_truck

F_1x = F_1y ⋅ 48 fty₁ + 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft48 ft² + y₁²¹²2y₁ ⋅ 48 ft + 72 000 lb ⋅ x_trucky₁

Horizontal equilibrium of member 01

F_0x = F_1x

Horizontal equilibrium of bridge

F_2x = F_0x

Vertical equilibrium of segment 0e

F_ey = F_0y − 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft48 ft² + y₁²¹²48 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_eyx_e − F_exy_e + ∫0 ftx_e145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft48 ft² + y₁²¹²48 ftxdx + 72 000 lb ⋅ x_truck

M_e = F_eyx_e − F_exy_e + 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft48 ft² + y₁²¹²48 ftx_e² ÷ 2 + 72 000 lb ⋅ x_truck

Minimum/maximum compression at point e if `x`_e < 48 ft

P_min/max = F_eyF_ex ⋅ y₁48 ft48 ft² + y₁²^1/2 ÷ 12 ft ⋅ t ± 6M_et² ⋅ 12 ft ∈ 0 lbft², 576 000 lbft²

Spreadsheet analysis suggests that the optimal values for y₁ and t are approximately 0.460 ft and 2.30 ft, respectively, yielding a volume of 2 650 ft³.

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 (x₁, y₁), pin 2 at (x₂, y₂), pin 3 at (80 ft, y₃), pin 4 at (160 ft − x₂, y₂), solid joint 5 at (160 ft − x₁, y₁), 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 t₀₁, members 12 and 45 of thickness t₁₂, members 23 and 34 of thickness t₂₃, and members 72 and 84 of thickness t₇₂.

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

M₁

F_2x12

F_2y12

F_2x23

F_2y23

F_3x

F_3y

F_4x34

F_4y34

F_4x45

F_4y45

F_5x

F_5y

M₅

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 lbft³ ⋅ 12 ft ⋅ t₀₁ + 640 lbftx₁² + y₁²¹²x₁∫0 ftx₁xdx + ∫160 ft − x₁160 ftxdx + 145 lbft³ ⋅ 12 ft ⋅ t₁₂ + 640 lbftx₂ − x₁² + y₂ − y₁²¹²x₂ − x₁∫x₁x₂xdx + ∫160 ft − x₂160 ft − x₁xdx + 145 lbft³ ⋅ 12 ft ⋅ t₂₃ + 640 lbft80 ft − x₂² + y₃ − y₂²¹²80 ft − x₂ ⋅ ∫x₂160 ft − x₂xdx + 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 lbft³ ⋅ 12 ft ⋅ t₀₁ + 640 lbftx₁² + y₁²¹²x₁160 ft² − 160 ft − x₁² + x₁² ÷ 2 + 145 lbft³ ⋅ 12 ft ⋅ t₁₂ + 640 lbftx₂ − x₁² + y₂ − y₁²¹²x₂ − x₁x₂² − x₁² + 160 ft − x₁² − 160 ft − x₂² ÷ 2 + 145 lbft³ ⋅ 12 ft ⋅ t₂₃ + 640 lbft80 ft − x₂² + y₃ − y₂²¹²80 ft − x₂160 ft − x₂² − x₂² ÷ 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 lbft³ ⋅ 12 ft ⋅ t₀₁ + 640 lbftx₁² + y₁²¹²x₁ ⋅ x₁ ⋅ 160 ft + 145 lbft³ ⋅ 12 ft ⋅ t₁₂ + 640 lbftx₂ − x₁² + y₂ − y₁²¹²x₂ − x₁x₂ − x₁ ⋅ 160 ft + 145 lbft³ ⋅ 12 ft ⋅ t₂₃ + 640 lbft80 ft − x₂² + y₃ − y₂²¹²80 ft − x₂80 ft − x₂ ⋅ 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 lbft³ ⋅ 12 ft ⋅ t₀₁ + 640 lbftx₁² + y₁²¹² + 145 lbft³ ⋅ 12 ft ⋅ t₁₂ + 640 lbftx₂ − x₁² + y₂ − y₁²¹² + 145 lbft³ ⋅ 12 ft ⋅ t₂₃ + 640 lbft80 ft − x₂² + y₃ − y₂²¹² ⋅ 160 ft + 72 000 lb ⋅ x_truck

Vertical equilibrium of bridge

F_0y + F_6y + F_7y + F_8y = 145 lbft³ ⋅ 12 ft ⋅ t₀₁ + 640 lbftx₁² + y₁²¹²x₁ ⋅ 2x₁ + 145 lbft³ ⋅ 12 ft ⋅ t₁₂ + 640 lbftx₂ − x₁² + y₂ − y₁²¹²x₂ − x₁ ⋅ 2x₂ − x₁ + 145 lbft³ ⋅ 12 ft ⋅ t₂₃ + 640 lbft80 ft − x₂² + y₃ − y₂²¹²80 ft − x₂ ⋅ 280 ft − x₂ + 72 000 lb

F_0y + F_6y + F_7y + F_8y = 2145 lbft³ ⋅ 12 ft ⋅ t₀₁ + 640 lbftx₁² + y₁²¹² + 145 lbft³ ⋅ 12 ft ⋅ t₁₂ + 640 lbftx₂ − x₁² + y₂ − y₁²¹² + 145 lbft³ ⋅ 12 ft ⋅ t₂₃ + 640 lbft80 ft − x₂² + y₃ − y₂²¹² + 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, y₁), member 12, and pin 2 at (320 ft, 0 ft).

Rotational equilibrium of bridge about point 0

F_2y ⋅ 320 ft = ∫0 ft320 ft145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft160 ft² + y₁²¹²160 ftxdx + 72 000 lb ⋅ x_truck

F_2y ⋅ 320 ft = 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft160 ft² + y₁²¹²160 ft320 ft² ÷ 2 + 72 000 lb ⋅ x_truck

F_2y = 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft160 ft² + y₁²¹² + 72 000 lb ⋅ x_truck320 ft

Vertical equilibrium of bridge

F_0y = 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft160 ft² + y₁²¹²160 ft ⋅ 320 ft + 72 000 lb − F_2y

F_0y = 2145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft160 ft² + y₁²¹² + 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 lbft³ ⋅ 12 ft ⋅ t + 640 lbft160 ft² + y₁²¹²160 ft ⋅ 160 ft − 72 000 lb

F_1y = F_0y − 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft160 ft² + y₁²¹² − 72 000 lb

Rotational equilibrium of member 01 about point 0

F_1xy₁ = F_1y ⋅ 160 ft + ∫0 ft160 ft145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft160 ft² + y₁²¹²160 ftxdx + 72 000 lb ⋅ x_truck

F_1xy₁ = F_1y ⋅ 160 ft + 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft160 ft² + y₁²¹²160 ft160 ft² ÷ 2 + 72 000 lb ⋅ x_truck

F_1x = F_1y ⋅ 160 fty₁ + 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft160 ft² + y₁²¹²2y₁ ⋅ 160 ft + 72 000 lb ⋅ x_trucky₁

Horizontal equilibrium of member 01

F_0x = F_1x

Horizontal equilibrium of bridge

F_2x = F_0x

Vertical equilibrium of segment 0e

F_ey = F_0y − 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft160 ft² + y₁²¹²160 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_eyx_e − F_exy_e + ∫0 ftx_e145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft160 ft² + y₁²¹²160 ftxdx + 72 000 lb ⋅ x_truck

M_e = F_eyx_e − F_exy_e + 145 lbft³ ⋅ 12 ft ⋅ t + 640 lbft160 ft² + y₁²¹²160 ftx_e² ÷ 2 + 72 000 lb ⋅ x_truck

Minimum/maximum compression at point e if `x`_e < 160 ft

P_min/max = F_eyF_ex ⋅ y₁160 ft160 ft² + y₁²^1/2 ÷ 12 ft ⋅ t ± 6M_et² ⋅ 12 ft ∈ 0 lbft², 576 000 lbft²

Spreadsheet analysis suggests that the optimal values for y₁ and t are approximately 6.53 ft and 10.9 ft, respectively, yielding a volume of 41 800 ft³.