Quiz & test questions Grade 10 Engineering & Design

Kilnbury Water Main Thrust Blocks: A Grade 10 Design Test

A grade 10 engineering unit test on thrust restraint: students find the force at a capped end and at a bend, then size the concrete block that holds the fitting.

Grade: Grade 10
Subject: Engineering & Design
Material: Quiz & test questions
Time: 50 min
Prep: No prep
Materials: Calculator

Standards

HSG-MG.A.1 Common Core Math

Use geometric shapes, their measures, and their properties to describe objects (e.g., modeling a tree trunk or a human torso as a cylinder).

How this material aligns: Items 3 and 10 model the bore of a water main as a circle and apply the circle-area relationship to find the area each test pressure acts on.

All materials for HSG-MG.A.1 →

HSG-MG.A.3 Common Core Math

Apply geometric methods to solve design problems (e.g., designing an object or structure to satisfy physical constraints or minimize cost).

How this material aligns: Items 7 and 8 design a concrete block that delivers the required bearing area within a fixed trench-face height, the design-under-constraint task this standard names.

All materials for HSG-MG.A.3 →

HSN-Q.A.1 Common Core Math

Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

How this material aligns: Items 4, 5, 6, and 9 carry pressure and area units through a chain from pounds per square inch to a thrust in pounds to a bearing area in square feet.

All materials for HSN-Q.A.1 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will calculate the thrust force at a capped end and at a bend in a pressurized water main.
Students will size a concrete thrust block so its bearing area matches the soil's allowable bearing capacity.
Students will explain how pipe diameter and test pressure change the thrust a fitting must resist.

Vocabulary

Key words: thrust, thrust block, restrained joint, bearing capacity, fitting factor, pressure test

Teacher overview

This 50-minute unit test (42 points) checks whether students can calculate the thrust that water pressure creates at a fitting and then size a concrete block to resist it. It mixes two multiple-choice items, two written-response items, and eight computation items, all built around a new main in the fictional town of Kilnbury. The step students most often miss is the unit change from a thrust in pounds and a soil capacity in pounds per square foot to a bearing area in square feet, so ask them to write the units beside every number before they divide. A second common error is squaring the radius instead of the diameter in the area formula, which produces a thrust one-quarter of the correct value and a block far too small to hold the bend.

Kilnbury Water Main Thrust Blocks: A Grade 10 Design Test — Grade 10 Engineering & Design

Name: ____________________________ Date: ______________

Assessment

Kilnbury is a fictional town. It is installing a new water main along Foundry Road. The main has an inside diameter of 12.0 inches. Before the main goes into service, the crew will pressure-test it at 150 psi. Water under pressure pushes on every surface inside the pipe. Along a straight run those pushes balance each other. At a bend or a capped end they do not, and the leftover force is called thrust. Thrust can pull a joint apart if nothing holds the fitting in place. Crews have two common ways to hold a fitting. They can pour a concrete thrust block that presses against the undisturbed soil at the side of the trench. They can also install restrained joints, which grip the pipe on both sides of the fitting. Kilnbury has chosen thrust blocks. The soil at the side of the trench can safely carry 2,000 pounds per square foot.

Use these relationships and values:

Area of a circle: Area = 0.785 × Diameter squared
Thrust = Pressure × Area × Fitting factor
Fitting factors: capped dead end 1.000, 90-degree bend 1.414, 45-degree bend 0.765
Required bearing area = Thrust ÷ Soil bearing capacity
There are 27 cubic feet in 1 cubic yard

(2 points) Thrust appears at a bend in a pressurized main because of which force?
- A) The weight of the water pressing down on the floor of the trench
- B) Friction between the water and the pipe wall along a straight run
- C) The pressure of the water pushing outward where the pipe turns
- D) The suction that forms in the main when a pump shuts down
(3 points) Name the two ways described above that a crew can hold a fitting against thrust. For each one, give a site condition that would make it the better choice.
(4 points) Find the cross-sectional area of the 12.0-inch main in square inches. Use Area = 0.785 × Diameter squared.
(4 points) The main ends in a cap at the edge of the property. Using the area from item 3, the 150 psi test pressure, and the dead-end fitting factor of 1.000, find the thrust on that cap in pounds.
(4 points) Foundry Road turns at a 90-degree bend. Using the thrust from item 4 and the 90-degree fitting factor of 1.414, find the thrust at that bend in pounds.
(4 points) The block at the 90-degree bend must press on soil that can carry 2,000 pounds per square foot. Using the thrust from item 5, find the bearing area the block face needs, in square feet.
(4 points) The trench face allows a block no taller than 4.0 feet. Using the bearing area from item 6, find the width of the block face in feet.
(4 points) The block will be poured 1.5 feet thick, measured along the pipe. Using the face dimensions from items 6 and 7, find the volume of concrete in cubic yards.
(4 points) The same main also turns at a 45-degree bend. Using the dead-end thrust from item 4, the 45-degree fitting factor of 0.765, and the soil capacity of 2,000 pounds per square foot, find the bearing area that bend's block needs, in square feet.
(4 points) A branch line off the same main has an inside diameter of 8.0 inches and ends in a cap. Find the thrust on that cap at the same 150 psi test pressure.
(3 points) The 8.0-inch cap in item 10 carries much less thrust than the 12.0-inch cap in item 4, even though the test pressure is the same. Explain why, using the way area depends on diameter.
(2 points) Which change would most reduce the bearing area required at the 90-degree bend?
- A) Testing the main at 100 psi instead of 150 psi
- B) Using a 16.0-inch main instead of the 12.0-inch main at the same test pressure
- C) Pouring the block against loose backfill instead of undisturbed soil
- D) Pouring the block 2.0 feet thick instead of 1.5 feet thick

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

Answer Key

Answer: C) The pressure of the water pushing outward where the pipe turns.
Pressure acts on every inside surface of the pipe. On a straight run the push on one wall is matched by the push on the wall opposite it, so the forces cancel. At a bend the wall on the outside of the turn has no matching wall across from it, so a net outward force remains. Choices A, B, and D describe real effects in a pipeline, yet none of them produces that unbalanced force at a fitting.
Answer: Concrete thrust blocks and restrained joints.
Accept any response that names both. A thrust block is the better choice where firm, undisturbed soil stands at the side of the trench and there is room to pour against it. Restrained joints are the better choice where the soil is soft or the trench is crowded with other utilities, because they lock the joints over a length of pipe on each side of the fitting so the soil along that whole length resists the force, instead of relying on one block against the trench wall, and because the crew does not have to wait for concrete to cure.
Answer: 113.04 square inches.
Area = 0.785 × 12.0 squared = 0.785 × 144 = 113.04 square inches.
Answer: 16,956 pounds.
Thrust = 150 × 113.04 × 1.000 = 16,956 pounds.
Answer: about 23,976 pounds.
Thrust = 16,956 × 1.414 = 23,975.8, which rounds to about 23,976 pounds.
Answer: about 12.0 square feet.
Bearing area = 23,975.8 ÷ 2,000 = 11.99, which rounds to about 12.0 square feet. Students who divide a force in pounds by a capacity in pounds per square foot should be able to name square feet as the unit left over.
Answer: 3.0 feet.
Width = 12.0 ÷ 4.0 = 3.0 feet, so the block face measures 3.0 feet wide by 4.0 feet tall.
Answer: about 0.67 cubic yards.
Volume = 3.0 × 4.0 × 1.5 = 18.0 cubic feet, and 18.0 ÷ 27 = 0.667, which rounds to about 0.67 cubic yards.
Answer: about 6.5 square feet.
Thrust = 16,956 × 0.765 = 12,971.3 pounds. Bearing area = 12,971.3 ÷ 2,000 = 6.486, which rounds to about 6.5 square feet. The shallower bend turns the water less, so it needs a block roughly half the size of the one at the 90-degree bend.
Answer: 7,536 pounds.
Area = 0.785 × 8.0 squared = 0.785 × 64 = 50.24 square inches. Thrust = 150 × 50.24 × 1.000 = 7,536 pounds.
Answer: Area grows with the square of the diameter.
Accept any response explaining that the 8.0-inch pipe has two-thirds the diameter of the 12.0-inch pipe, so its cross-sectional area is two-thirds squared, or four-ninths, of the larger area. Thrust is pressure multiplied by area, and the pressure is the same in both cases, so the cap thrust also falls to about four-ninths of the larger value: 7,536 ÷ 16,956 = 0.444.
Answer: A) Testing the main at 100 psi instead of 150 psi.
Bearing area is proportional to thrust, and thrust is proportional to pressure, so a test at 100 psi rather than 150 psi cuts the required area to two-thirds of 12.0 square feet, or about 8.0 square feet. Choice B raises the requirement, because a 16.0-inch main has about 1.78 times the cross-sectional area of a 12.0-inch main. Choice C lowers the soil's allowable bearing, which makes the block larger rather than smaller. Choice D adds concrete without changing the face area that presses on the soil.

Differentiation

Support (below grade level)

Give students a formula card with Area = 0.785 × Diameter squared, Thrust = Pressure × Area × Fitting factor, and Required bearing area = Thrust ÷ Soil bearing capacity written out, alongside a labeled plan view of the bend with the fictional Kilnbury values of 150 psi and 2,000 pounds per square foot already marked. Work item 3 with the class so that every student begins the chain with a correct area, and award full credit on items 5 through 9 for correct method with an arithmetic error carried forward.

Extension (above grade level)

Ask students to size the block for a 22.5-degree bend on the same fictional Kilnbury main, using a fitting factor of 0.390, and then to write two sentences explaining why a utility might use restrained joints rather than a poured block at such a shallow bend.

ELL supports

Provide a labeled plan-view sketch of the bend showing the thrust arrow, the block, the trench wall, and the pipe, with the terms thrust, bearing area, fitting factor, and restrained joint glossed beside the drawing. Allow items 2 and 11 to be answered with a labeled sketch plus one sentence, and offer the sentence frame "The thrust on the smaller pipe is less because its area is ___."

Extensions

Suppose the contractor for the fictional Kilnbury project asks to use restrained joints at every bend instead of poured blocks. Have students list the questions an engineer would need answered before approving that change, including the soil type along Foundry Road, the length of pipe that must be restrained on each side of each fitting, and the effect on the construction schedule. Students can then write a short memo recommending one approach for a bend that sits close to an existing gas line, where there is no room to pour concrete against undisturbed soil.