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Bridge Constructor in Concrete

Hover over an equation to highlight matching parentheses.

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.

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 F0x pushing rightward at point 0, F0y pushing upward at point 0, F2x pushing leftward at point 2, and F2y pushing upward at point 2.

Rotational equilibrium of bridge about point 0

F2y ⋅ 64 ft = ∫0 ft64 ft145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft32 ft2 + y121232 ftxdx + 72 000 lb ⋅ xtruck

F2y ⋅ 64 ft = 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft32 ft2 + y121232 ft64 ft2 ÷ 2 + 72 000 lb ⋅ xtruck

F2y = 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft32 ft2 + y1212 + 72 000 lb ⋅ xtruck64 ft

Vertical equilibrium of bridge

F0y = 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft32 ft2 + y121232 ft ⋅ 64 ft + 72 000 lb − F2y

F0y = 2145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft32 ft2 + y1212 + 72 000 lb − F2y

The forces acting on member 01 (in addition to those listed above) are F1x pushing leftward at pin 1 and F1y pushing downward at pin 1.

Vertical equilibrium of member 01

F1y = F0y145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft32 ft2 + y121232 ft ⋅ 32 ft72 000 lb

F1y = F0y145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft32 ft2 + y121272 000 lb

Rotational equilibrium of member 01 about point 0

F1xy1 = F1y ⋅ 32 ft + ∫0 ft32 ft145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft32 ft2 + y121232 ftxdx + 72 000 lb ⋅ xtruck

F1xy1 = F1y ⋅ 32 ft + 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft32 ft2 + y121232 ft32 ft2 ÷ 2 + 72 000 lb ⋅ xtruck

F1x = F1y32 fty1 + 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft32 ft2 + y12122y1 ⋅ 32 ft + 72 000 lb ⋅ xtrucky1

Horizontal equilibrium of member 01

F0x = F1x

Horizontal equilibrium of bridge

F2x = F0x

Segment 0e extends from pin 0 along the bridge to arbitrary solid joint e at (xe, ye). The forces acting on segment 0e (in addition to those listed above) are Fex pushing leftward at point e, Fey pushing downward at point e, and Me pushing counterclockwise at point e.

Vertical equilibrium of segment 0e

Fey = F0y145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft32 ft2 + y121232 ftxe72 000 lb

Horizontal equilibrium of segment 0e

Fex = F0x

Rotational equilibrium of segment 0e about point 0

Me = FeyxeFexye + ∫0 ftxe145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft32 ft2 + y121232 ftxdx + 72 000 lb ⋅ xtruck

Me = FeyxeFexye + 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft32 ft2 + y121232 ftxe2 ÷ 2 + 72 000 lb ⋅ xtruck

Minimum/maximum compression at point e if xe < 32 ft

Pmin/max = FeyFexy132 ft32 ft2 + y121/2 ÷ 12 ft ⋅ t ± 6Met2 ⋅ 12 ft0 lbft2, 576 000 lbft2

Spreadsheet analysis suggests that the optimal values for y1 and t are approximately 0.116 ft and 1.67 ft, respectively, yielding a volume of 1 280 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).

The forces acting on the bridge (in addition to those listed in Premises) are F0x pushing rightward at point 0, F0y pushing upward at point 0, F2x pushing leftward at point 2, and F2y pushing upward at point 2.

Rotational equilibrium of bridge about point 0

F2y ⋅ 96 ft = ∫0 ft96 ft145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft48 ft2 + y121248 ftxdx + 72 000 lb ⋅ xtruck

F2y ⋅ 96 ft = 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft48 ft2 + y121248 ft96 ft2 ÷ 2 + 72 000 lb ⋅ xtruck

F2y = 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft48 ft2 + y1212 + 72 000 lb ⋅ xtruck96 ft

Vertical equilibrium of bridge

F0y = 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft48 ft2 + y121248 ft ⋅ 96 ft + 72 000 lb − F2y

F0y = 2145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft48 ft2 + y1212 + 72 000 lb − F2y

The forces acting on member 01 (in addition to those listed above) are F1x pushing leftward at pin 1 and F1y pushing downward at pin 1.

Vertical equilibrium of member 01

F1y = F0y145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft48 ft2 + y121248 ft ⋅ 48 ft72 000 lb

F1y = F0y145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft48 ft2 + y121272 000 lb

Rotational equilibrium of member 01 about point 0

F1xy1 = F1y ⋅ 48 ft + ∫0 ft48 ft145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft48 ft2 + y121248 ftxdx + 72 000 lb ⋅ xtruck

F1xy1 = F1y ⋅ 48 ft + 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft48 ft2 + y121248 ft48 ft2 ÷ 2 + 72 000 lb ⋅ xtruck

F1x = F1y48 fty1 + 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft48 ft2 + y12122y1 ⋅ 48 ft + 72 000 lb ⋅ xtrucky1

Horizontal equilibrium of member 01

F0x = F1x

Horizontal equilibrium of bridge

F2x = F0x

Segment 0e extends from pin 0 along the bridge to arbitrary solid joint e at (xe, ye). The forces acting on segment 0e (in addition to those listed above) are Fex pushing leftward at point e, Fey pushing downward at point e, and Me pushing counterclockwise at point e.

Vertical equilibrium of segment 0e

Fey = F0y145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft48 ft2 + y121248 ftxe72 000 lb

Horizontal equilibrium of segment 0e

Fex = F0x

Rotational equilibrium of segment 0e about point 0

Me = FeyxeFexye + ∫0 ftxe145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft48 ft2 + y121248 ftxdx + 72 000 lb ⋅ xtruck

Me = FeyxeFexye + 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft48 ft2 + y121248 ftxe2 ÷ 2 + 72 000 lb ⋅ xtruck

Minimum/maximum compression at point e if xe < 48 ft

Pmin/max = FeyFexy148 ft48 ft2 + y121/2 ÷ 12 ft ⋅ t ± 6Met2 ⋅ 12 ft0 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 F0x and upward F0y at point 0; leftward F6x and upward F6y at point 6; leftward F7x and upward F7y at point 7; and leftward F8x and upward F8y at point 8.

F0x
F0y
F1x
F1y
M1
F2x12
F2y12
F2x23
F2y23
F3x
F3y
F4x34
F4y34
F4x45
F4y45
F5x
F5y
M5
F6x
F6y
F7x
F7y
F8x
F8y

Rotational equilibrium of bridge about point 0

F6y ⋅ 160 ft − F7x ⋅ 24 ft + F7y ⋅ 48 ft − F8x ⋅ 24 ft + F8y ⋅ 112 ft = 145 lbft3 ⋅ 12 ft ⋅ t01 + 640 lbftx12 + y1212x10 ftx1xdx + ∫160 ft − x1160 ftxdx + 145 lbft3 ⋅ 12 ft ⋅ t12 + 640 lbftx2x12 + y2y1212x2x1x1x2xdx + ∫160 ft − x2160 ft − x1xdx + 145 lbft3 ⋅ 12 ft ⋅ t23 + 640 lbft80 ft − x22 + y3y221280 ft − x2 ⋅ ∫x2160 ft − x2xdx + 72 000 lb ⋅ xtruck

F6y ⋅ 160 ft − F7x ⋅ 24 ft + F7y ⋅ 48 ft − F8x ⋅ 24 ft + F8y ⋅ 112 ft = 145 lbft3 ⋅ 12 ft ⋅ t01 + 640 lbftx12 + y1212x1160 ft2160 ft − x12 + x12 ÷ 2 + 145 lbft3 ⋅ 12 ft ⋅ t12 + 640 lbftx2x12 + y2y1212x2x1x22x12 + 160 ft − x12160 ft − x22 ÷ 2 + 145 lbft3 ⋅ 12 ft ⋅ t23 + 640 lbft80 ft − x22 + y3y221280 ft − x2160 ft − x22x22 ÷ 2 + 72 000 lb ⋅ xtruck

F6y ⋅ 160 ft − F7x ⋅ 24 ft + F7y ⋅ 48 ft − F8x ⋅ 24 ft + F8y ⋅ 112 ft = 145 lbft3 ⋅ 12 ft ⋅ t01 + 640 lbftx12 + y1212x1x1 ⋅ 160 ft + 145 lbft3 ⋅ 12 ft ⋅ t12 + 640 lbftx2x12 + y2y1212x2x1x2x1 ⋅ 160 ft + 145 lbft3 ⋅ 12 ft ⋅ t23 + 640 lbft80 ft − x22 + y3y221280 ft − x280 ft − x2 ⋅ 160 ft + 72 000 lb ⋅ xtruck

F6y ⋅ 160 ft − F7x ⋅ 24 ft + F7y ⋅ 48 ft − F8x ⋅ 24 ft + F8y ⋅ 112 ft = 145 lbft3 ⋅ 12 ft ⋅ t01 + 640 lbftx12 + y1212 + 145 lbft3 ⋅ 12 ft ⋅ t12 + 640 lbftx2x12 + y2y1212 + 145 lbft3 ⋅ 12 ft ⋅ t23 + 640 lbft80 ft − x22 + y3y2212 ⋅ 160 ft + 72 000 lb ⋅ xtruck

Vertical equilibrium of bridge

F0y + F6y + F7y + F8y = 145 lbft3 ⋅ 12 ft ⋅ t01 + 640 lbftx12 + y1212x1 ⋅ 2x1 + 145 lbft3 ⋅ 12 ft ⋅ t12 + 640 lbftx2x12 + y2y1212x2x1 ⋅ 2x2x1 + 145 lbft3 ⋅ 12 ft ⋅ t23 + 640 lbft80 ft − x22 + y3y221280 ft − x2 ⋅ 280 ft − x2 + 72 000 lb

F0y + F6y + F7y + F8y = 2145 lbft3 ⋅ 12 ft ⋅ t01 + 640 lbftx12 + y1212 + 145 lbft3 ⋅ 12 ft ⋅ t12 + 640 lbftx2x12 + y2y1212 + 145 lbft3 ⋅ 12 ft ⋅ t23 + 640 lbft80 ft − x22 + y3y2212 + 72 000 lb

Horizontal equilibrium of bridge

F0xF6xF7xF8x = 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).

The forces acting on the bridge (in addition to those listed in Premises) are F0x pushing rightward at point 0, F0y pushing upward at point 0, F2x pushing leftward at point 2, and F2y pushing upward at point 2.

Rotational equilibrium of bridge about point 0

F2y ⋅ 320 ft = ∫0 ft320 ft145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft160 ft2 + y1212160 ftxdx + 72 000 lb ⋅ xtruck

F2y ⋅ 320 ft = 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft160 ft2 + y1212160 ft320 ft2 ÷ 2 + 72 000 lb ⋅ xtruck

F2y = 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft160 ft2 + y1212 + 72 000 lb ⋅ xtruck320 ft

Vertical equilibrium of bridge

F0y = 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft160 ft2 + y1212160 ft ⋅ 320 ft + 72 000 lb − F2y

F0y = 2145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft160 ft2 + y1212 + 72 000 lb − F2y

The forces acting on member 01 (in addition to those listed above) are F1x pushing leftward at pin 1 and F1y pushing downward at pin 1.

Vertical equilibrium of member 01

F1y = F0y145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft160 ft2 + y1212160 ft ⋅ 160 ft72 000 lb

F1y = F0y145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft160 ft2 + y121272 000 lb

Rotational equilibrium of member 01 about point 0

F1xy1 = F1y ⋅ 160 ft + ∫0 ft160 ft145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft160 ft2 + y1212160 ftxdx + 72 000 lb ⋅ xtruck

F1xy1 = F1y ⋅ 160 ft + 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft160 ft2 + y1212160 ft160 ft2 ÷ 2 + 72 000 lb ⋅ xtruck

F1x = F1y160 fty1 + 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft160 ft2 + y12122y1 ⋅ 160 ft + 72 000 lb ⋅ xtrucky1

Horizontal equilibrium of member 01

F0x = F1x

Horizontal equilibrium of bridge

F2x = F0x

Segment 0e extends from pin 0 along the bridge to arbitrary solid joint e at (xe, ye). The forces acting on segment 0e (in addition to those listed above) are Fex pushing leftward at point e, Fey pushing downward at point e, and Me pushing counterclockwise at point e.

Vertical equilibrium of segment 0e

Fey = F0y145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft160 ft2 + y1212160 ftxe72 000 lb

Horizontal equilibrium of segment 0e

Fex = F0x

Rotational equilibrium of segment 0e about point 0

Me = FeyxeFexye + ∫0 ftxe145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft160 ft2 + y1212160 ftxdx + 72 000 lb ⋅ xtruck

Me = FeyxeFexye + 145 lbft3 ⋅ 12 ft ⋅ t + 640 lbft160 ft2 + y1212160 ftxe2 ÷ 2 + 72 000 lb ⋅ xtruck

Minimum/maximum compression at point e if xe < 160 ft

Pmin/max = FeyFexy1160 ft160 ft2 + y121/2 ÷ 12 ft ⋅ t ± 6Met2 ⋅ 12 ft0 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.