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Worksheet Grade 9 Engineering & Design

Culverstone Main Replacement: Sizing a Thrust Block

A grade 9 worksheet on a fictional water main replacement where students find the thrust at a bend under test pressure and size the concrete block that resists it.

Objectives

  1. Students will calculate the thrust force a pressurized bend puts on a buried pipe.
  2. Students will size a concrete bearing block from a thrust force and an allowable soil bearing value.
  3. Students will explain why thrust scales with the square of pipe diameter rather than with diameter itself.

Vocabulary

Key words: thrust block, test pressure, bearing capacity, cross-sectional area, undisturbed soil

Teacher overview

Culverstone is a fictional town replacing a cast iron main that has broken three times in five winters. Where the new pipe turns a corner, the water pushes outward on the bend and would walk the joint apart if nothing held it, so the crew pours a concrete block between the bend and the undisturbed soil behind it. The size of that block comes from two numbers, the thrust the water creates and how much load the soil can carry per square foot. This worksheet has students work that calculation for a bend and for a dead end, then compare two pipe sizes. The error to watch for is scaling the block with the pipe diameter, since thrust follows the pipe's cross-sectional area and therefore the square of the diameter.

Culverstone Main Replacement: Sizing a Thrust Block — Grade 9 Engineering & Design
Name: ____________________________ Date: ______________

Student Worksheet

The town of Culverstone and every figure below are fictional. Use only what is given here.

The new main is 12 inches in diameter. The crew tests it at 150 psi. At a 90-degree bend, the thrust equals 1.414 times the pressure times the pipe's cross-sectional area. At a dead end, the thrust equals the pressure times that same area, with no 1.414 factor. The soil behind the block can safely carry 2,000 pounds per square foot. Use area = 0.785 x diameter squared, and remember that 27 cubic feet make one cubic yard.

  1. The main is 12 inches in diameter. Using area = 0.785 x diameter squared, what is its cross-sectional area in square inches?

  2. The test pressure is 150 psi, which means 150 pounds on every square inch. Using your area from item 1, how many pounds of force does the water push with?

  3. At a 90-degree bend, thrust equals 1.414 times the force you found in item 2. What is the thrust in pounds? Round to the nearest tenth.

  4. The soil can safely carry 2,000 pounds per square foot. Using your thrust from item 3, how many square feet must the block's bearing face cover? Round to the nearest hundredth.

  5. The crew forms the block with a bearing face 4.0 feet wide. Using your area from item 4, how tall must that face be, in feet? Round to the nearest hundredth.

  6. The block is poured 2.5 feet thick, measured from the pipe out to the soil face. Using your rounded bearing area from item 4, how many cubic feet of concrete does it take? Round to the nearest hundredth.

  7. Concrete is ordered by the cubic yard, and 27 cubic feet make one cubic yard. Convert your answer from item 6 to cubic yards. Round to the nearest hundredth.

  8. At a dead end on the same 12-inch main at 150 psi, thrust equals the force from item 2 with no 1.414 factor. How many square feet must a dead-end block cover in the same soil? Round to the nearest hundredth.

  9. Compare the bend block from item 4 with the dead-end block from item 8. By what percent is the bend block larger? Round to the nearest tenth of a percent.

  10. A nearby street has an 8-inch main at the same 150 psi test pressure. What is the thrust at a 90-degree bend on that pipe, in pounds? Round to the nearest tenth.

  11. A crew member says the 12-inch main needs a block one and a half times larger than the 8-inch main, because 12 divided by 8 is 1.5. Compare your answers from items 3 and 10, then explain in two or three sentences why the real ratio is different.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

  1. Answer: 113.04 square inches.
    Area = 0.785 x 12 x 12 = 0.785 x 144 = 113.04 square inches.

  2. Answer: 16,956 pounds.
    150 pounds per square inch x 113.04 square inches = 16,956 pounds of force.

  3. Answer: 23,975.8 pounds.
    1.414 x 16,956 = 23,975.784 pounds, which rounds to 23,975.8 pounds. The 1.414 factor comes from adding two equal forces at right angles.

  4. Answer: 11.99 square feet.
    23,975.784 pounds divided by 2,000 pounds per square foot = 11.9879 square feet, which rounds to 11.99 square feet.

  5. Answer: 3.00 feet.
    11.9879 square feet divided by 4.0 feet of width = 2.997 feet, which rounds to 3.00 feet tall.

  6. Answer: 29.98 cubic feet.
    11.99 square feet x 2.5 feet of thickness = 29.975 cubic feet, which rounds to 29.98 cubic feet. Accept 29.97 from a student who carried the unrounded 11.9879 forward.

  7. Answer: 1.11 cubic yards.
    29.975 cubic feet divided by 27 cubic feet per cubic yard = 1.1102 cubic yards, which rounds to 1.11 cubic yards. A crew would order the next quarter yard up.

  8. Answer: 8.48 square feet.
    At a dead end the thrust is the full 16,956 pounds with no 1.414 factor. 16,956 divided by 2,000 = 8.478 square feet, which rounds to 8.48 square feet.

  9. Answer: 41.4 percent.
    The difference is 11.9879 minus 8.478 = 3.5099 square feet. Then 3.5099 divided by 8.478 = 0.4140, or 41.4 percent. That percent is simply the 1.414 factor minus one, because both blocks sit in the same soil and carry the same pressure on the same pipe.

  10. Answer: 10,655.9 pounds.
    Area = 0.785 x 8 x 8 = 50.24 square inches. Force = 150 x 50.24 = 7,536 pounds. Thrust = 1.414 x 7,536 = 10,655.9 pounds.

  11. Answer: The ratio is 2.25, not 1.5.
    Dividing item 3 by item 10 gives 23,975.8 divided by 10,655.9 = 2.25. Accept any explanation that connects this to area: thrust depends on the pipe's cross-sectional area, and area follows the square of the diameter, so a diameter ratio of 1.5 produces an area ratio of 1.5 squared, which is 2.25. Students who answer 1.5 are scaling by a length when the force scales by an area.

Differentiation

Support

Give these students item 1 already computed and a printed formula strip listing the bend equation, the dead-end equation, and the cubic yard conversion. Provide a labeled sketch of the fictional Culverstone bend showing the pipe, the block, and the arrow of thrust pointing into undisturbed soil, since students often picture the block holding the pipe down rather than sideways. Allow them to stop after item 7.

Extension

Ask students to size the bend block for softer ground where the soil can safely carry only 1,200 pounds per square foot, and to report the new bearing area and block height for the same fictional bend. Then have them explain why a utility sometimes uses restrained joints instead of a poured block, and name one site condition that would make the restrained joint the better choice.

ELL supports

Pre-teach thrust, bearing face, and undisturbed soil with a top-view drawing of a pipe bend, an arrow for the push, and a shaded rectangle for the block. Provide the sentence frame "The thrust is ___ pounds because the pressure is ___ and the area is ___" for item 3. Allow item 11 to be answered with the two numbers, the ratio, and one sentence.

Extensions

Have students find the thrust at a 45-degree bend, where the factor is 0.765 rather than 1.414, on the same fictional 12-inch main at the same test pressure. Ask them to explain why a gentler turn needs a smaller block and to sketch how a crew would lay out two 45-degree bends instead of one 90-degree bend. Students can then write a short paragraph on what happens to a block that was sized for the soil conditions found in dry weather when the trench floods.