Structural Support and Materials

77 questions

All types77 Multiple choice39 True / False13 Numeric5 Matching13 Ordering7

A $120$ -pound crate rests on a sturdy $12$ -foot plank spanning two sawhorses placed at opposite ends. The center of the crate is located $3$ feet from the left sawhorse. Assuming the weight of the plank is negligible, what is the downward force, in pounds, exerted by the crate on the left sawhorse?

Show answer & explanation

Answer: 90

Answer

The downward force exerted on the left sawhorse is

90

pounds.

Because the crate is positioned

3

feet from the left sawhorse and

9

feet from the right sawhorse on a

12

-foot span, the left sawhorse must support

\frac{9}{12}

(or

75\%

) of the crate's weight. Multiplying

120

pounds by

0.75

yields exactly

90

pounds.

Step-by-Step Solution

Calculate the distance from the crate to the right sawhorse.

12 - 3 = 9

feet.

The weight supported by one end of a beam is proportional to the distance from the load to the opposite end.

Determine the fraction of the total weight carried by the left sawhorse.

\frac{9}{12}

, which simplifies to

\frac{3}{4}

0.75

This fraction represents the ratio of the opposite distance to the total span, dictating how much load is distributed to the left support.

Calculate the downward force on the left sawhorse.

120 \times 0.75 = 90

pounds.

Multiplying the total crate weight by the calculated proportion gives the specific load carried by that support.

Key Concept

Load Distribution and Balancing

Estimated Time:45s

Question 42Question →

In a mechanical flight control linkage, a solid steel pin is used to connect a single control rod (tang) inside a U-shaped bracket (clevis). A heavy tensile force is applied to the control rod, pulling it directly away from the fixed bracket. Which of the following best describes the primary type of stress experienced by the pin and how its load-bearing capacity compares to a simple two-plate lap joint secured by a single identical pin?

The pin experiences double shear stress, distributing the load across two cross-sectional areas and making it significantly stronger than a lap joint.

The pin experiences single shear stress, concentrating the load at one cross-sectional area and providing the same strength as a lap joint.

The pin experiences pure tensile stress because the control rod is being pulled, making it more susceptible to stretching than a lap joint.

The pin experiences compressive stress as the sides of the bracket squeeze inward, providing greater rigidity than a lap joint.

Show answer & explanation

Answer: The pin experiences double shear stress, distributing the load across two cross-sectional areas and making it significantly stronger than a lap joint.

Answer

The option stating the pin experiences double shear stress, distributing the load across two cross-sectional areas and making it significantly stronger than a lap joint.

The pulling force on the control rod acts perpendicular to the longitudinal axis of the connecting pin, which creates shear stress (a slicing force). Because the U-shaped clevis bracket supports the pin at two distinct points—above and below the control rod—the pin must be sheared in two separate places for the joint to fail. This configuration is known as 'double shear.' Because the load is distributed across two cross-sectional areas of the pin instead of just one, a double shear joint is significantly stronger than a simple lap joint, which only puts the fastener in single shear.

Step-by-Step Solution

Analyze the direction of the applied force relative to the pin.

The control rod is pulled horizontally, which applies a force perpendicular to the longitudinal axis of the vertical connecting pin.

Forces acting perpendicular to the axis of a fastener create shear stress, not tensile stress.

Examine the physical structure of the joint.

The joint consists of three overlapping sections: the upper arm of the clevis bracket, the central control rod (tang), and the lower arm of the clevis bracket.

Identifying the number of overlapping components determines how many places the pin could be severed.

Determine the number of shear planes.

Because the control rod pulls in the middle and the bracket resists on both the top and bottom, the pin would have to be sliced in two separate places to fail.

Two shear planes indicate 'double shear', which distributes the force over twice the cross-sectional area of the pin compared to a standard two-plate lap joint (single shear).

Key Concept

Identifying load distribution and shear stress (single vs. double shear) in mechanical joints.

Question 43Question →

A horizontal steel I-beam is supporting a heavy downward load from a factory roof. If a small hole must be drilled through the beam to route an electrical cable, where is the safest location on the beam's cross-sectional profile to remove material in order to minimize the loss of bending strength?

Through the bottom horizontal flange, because it primarily experiences compressive forces

Directly through the middle of the vertical web

An I-beam cannot be safely drilled; a solid rectangular steel beam of equal weight must be used instead

Through the top horizontal flange, because it primarily experiences tensile forces

Show answer & explanation

Answer: Directly through the middle of the vertical web

Answer

The safest location is directly through the middle of the vertical web.

When a horizontal beam bends under a downward load, the material at the top is compressed, and the material at the bottom is stretched (placed in tension). The exact horizontal center of the profile, called the neutral axis, experiences neither tension nor compression from bending. Because the vertical web at this midpoint is doing no work to resist bending forces, removing a small amount of material there has the smallest possible impact on the beam's overall structural strength.

Step-by-Step Solution

Analyze the stresses on a simply supported horizontal beam with a downward load.

The beam bends downward, causing the top to squeeze together (compression) and the bottom to stretch apart (tension).

Understanding load distribution is required to identify areas of high and low stress.

Locate the point of minimum bending stress on the cross-sectional profile.

The transition point between compression (top) and tension (bottom) occurs at the exact center of the vertical web, known as the neutral axis.

The neutral axis is the zone where structural material does the least amount of work resisting bending forces.

Determine the optimal location to drill a hole.

Drilling through the middle of the vertical web removes material from the neutral axis, preserving the highly-stressed flanges.

Removing material from areas of zero bending stress minimizes the overall loss of strength.

Key Concept

Stress Distribution and the Neutral Axis in Beams

Question 44Question →

A $100$ -pound pump is placed on a $10$ -foot scaffold board that spans between two structural beams, Beam X and Beam Y. The board's own weight is negligible. If the pump is located $2$ feet from Beam X, how much of the pump's weight is supported by Beam X?

20

pounds

50

pounds

80

pounds

100

pounds

Show answer & explanation

Answer: $80$ pounds

Answer

The correct answer is

80

pounds.

Because the pump is situated much closer to Beam X, that beam must carry a proportionally larger share of the weight. This can be solved by taking the torque around Beam Y: the total board length (

10

feet) multiplied by the upward force at Beam X must equal the pump's weight (

100

pounds) multiplied by its distance from Beam Y (

8

feet). Solving

(F_X \times 10 = 100 \times 8)

results in a load of

80

pounds on Beam X.

Step-by-Step Solution

Identify the distances of the load from both supports.

The pump is

2

feet from Beam X. Since the board is

10

feet long total, the pump is

8

feet from Beam Y (

10 - 2 = 8

To calculate the load distribution, the distance to the opposite support must be determined.

Apply the principle of moments (torque) using Beam Y as the pivot point.

Torque equation: Force at Beam X

\times

Total distance

=

Pump weight

\times

Distance to Beam Y. Therefore,

F_X \times 10 = 100 \times 8

The sum of torques around any pivot point must be zero for the structural system to remain in static equilibrium.

Solve the equation for the force on Beam X.

10 \times F_X = 800

, which simplifies to

F_X = 80

pounds.

This calculation isolates

F_X

to provide the exact weight supported by Beam X.

Key Concept

Load Distribution and Static Equilibrium

Question 45Question →

A technician is evaluating a mechanical linkage where a solid cylindrical pin passes vertically through a single mounting tab on a bracket and a single tab on a control arm, forming a simple lap joint. During operation, the control arm exerts a strong horizontal pull away from the bracket. What is the primary type of stress acting on the pin in this configuration?

Single shear stress, as the lateral forces attempt to cut the pin along the single interface between the two overlapping tabs.

Double shear stress, since the lateral load is distributed across multiple points along the length of the fastener.

Tensile stress, because the horizontal pulling action forces the pin to stretch along its vertical axis.

Compressive stress, as the horizontal pull causes the mounting tabs to squeeze the pin from top to bottom.

Show answer & explanation

Answer: Single shear stress, as the lateral forces attempt to cut the pin along the single interface between the two overlapping tabs.

Answer

The primary stress acting on the pin is single shear stress.

In a simple lap joint with one tab overlapping another, there is exactly one contact plane where the two parts meet. When horizontal forces pull these parts in opposite directions, they create a scissor-like action across that single plane, subjecting the pin to single shear stress.

Step-by-Step Solution

Analyze the orientation of the applied force relative to the pin.

The control arm pulls horizontally, while the pin is mounted vertically.

Determining the angle of force helps rule out tension, which requires force parallel to the pin's axis.

Determine the number of shear planes in the joint.

The joint consists of a single mounting tab on the bracket and a single tab on the arm, creating exactly one plane of intersection.

The number of mating surfaces determines whether the joint experiences single or double shear.

Identify the primary stress based on the force and planes.

Since a perpendicular force acts across one intersection plane, the pin experiences single shear stress.

Forces attempting to slide overlapping materials in opposite directions cause shear along their contact interface.

Key Concept

Identifying load distribution and shear stress in lap joints.

Estimated Time:1m 0s

Question 46Question →

Structural engineers select specific cross-sectional profiles based on how they distribute internal stresses under different physical loads. Match each structural shape to its defining mechanical behavior.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: I-Beam matches with maximum vertical bending resistance; Hollow Circular Tube matches with uniform torsional resistance; C-Channel matches with flat mounting surface combined with bending resistance; and Angle Iron matches with bidirectional corner stiffness.

Each cross-section is geometrically optimized for distinct force applications: I-beams isolate material at the extremes to fight bending, circular tubes use symmetry to fight torsion, C-channels offer mounting utility, and angle irons provide simple two-axis corner support.

Step-by-Step Solution

Analyze the I-Beam profile and its typical load.

Matched to vertical bending resistance.

The horizontal flanges of an I-beam are placed far from the neutral axis to absorb the maximum tension and compression forces of a bending moment.

Evaluate the geometric properties of a Hollow Circular Tube.

Matched to uniform torsional resistance.

Because it has a continuous, symmetrical closed shape, it resists twisting forces equally well regardless of the rotational direction.

Identify the functional advantage of a C-Channel.

Matched to the combination of a flat mounting surface and bending resistance.

The flat back (web) allows it to be mounted flush against walls or vehicle chassis, while the protruding flanges provide structural strength.

Assess the role of an Angle Iron.

Matched to bidirectional stiffness for corners.

The L-shape reinforces joints by preventing flex in both the vertical and horizontal planes simultaneously.

Key Concept

Cross-sectional geometry dictates a structural member's ability to handle specific types of stress (bending, torsion, shear).

Estimated Time:1m 30s

Question 47Question →

A construction crew is moving several identical uniform steel I-beams. Each beam is $12$ feet long and weighs $600$ pounds. Two workers will carry each beam by supporting it from beneath at different points along its length.

Rank the following carrying configurations based on the amount of weight supported by the worker positioned on the LEFT side, from the GREATEST load to the LEAST load.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The configuration with the greatest load on the left worker is when the left worker is 3 feet inward (400 lbs), followed by 2 feet inward (360 lbs), then both workers at extreme ends (300 lbs), and finally the least load occurs when the right worker is moved inward (200 lbs).

As a support point moves closer to a uniform object's center of gravity, it takes on a proportionally larger share of the total weight. Moving the opposite support closer to the center transfers weight away from the first support.

Step-by-Step Solution

Determine the location of the center of gravity (CG).

For a uniform

12

-foot beam, the CG is exactly in the middle at the

6

-foot mark.

The weight of a uniform object can be treated as a single downward point force acting at its center.

Establish the formula for calculating load distribution.

Load on Left Worker = Total Weight

\times

(Distance from Right Worker to CG) / (Total Distance Between Workers)

This formula balances the torques around the right worker, which acts as a fulcrum when determining the left worker's lifting force.

Calculate the left worker's load for each configuration.

Item 1:

600 \times (6 / 12) = 300

lbs.
Item 2:

600 \times (3 / 9) = 200

lbs.
Item 3:

600 \times (6 / 10) = 360

lbs.
Item 4:

600 \times (6 / 9) = 400

lbs.

Applying the formula from step 2 translates the distances into exact pounds supported.

Order the results from greatest to least.

400

lbs (Item 4),

360

lbs (Item 3),

300

lbs (Item 1),

200

lbs (Item 2).

The question requests ranking from greatest load supported by the left worker to the least.

Key Concept

Load Distribution and Balancing

Estimated Time:2m 0s

Question 48Question →

Three identical $10$ -foot planks are each placed across two sawhorses, one sawhorse at the left end (Support X) and one at the right end (Support Y). A single $60$ -pound bag of cement is placed on each plank at a different location. Rank the scenarios below based on the amount of downward force carried by Support X, from greatest to least.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The bag closest to Support X places the greatest load on it, followed by the centered bag, while the bag closest to Support Y places the least load on Support X.

Support X must carry the largest portion of the load when the weight is closest to it. As the weight moves further away across the plank, the amount of force supported by X decreases.

Step-by-Step Solution

Understand the principle of load distribution.

The closer a weight is to a given support, the more of that weight the support must carry.

This is a basic rule of mechanical balance and levers.

Evaluate the load on Support X for the bag placed

3

feet from it.

This bag is very close to Support X, so Support X carries more than half the weight.

The distance to X is smaller than the distance to Y.

Evaluate the load on Support X for the centered bag.

This bag is exactly

5

feet from both supports, so Support X carries exactly half the weight.

A centered load is shared evenly.

Evaluate the load on Support X for the bag placed

2

feet from Support Y.

This bag is

8

feet away from Support X, so Support Y carries almost all the weight, leaving very little for Support X.

The distance to X is much larger than the distance to Y.

Key Concept

The proportion of a load carried by a support increases as the center of gravity moves closer to that support.

Question 49Question →

When constructing a building framework in an earthquake-prone region, engineers typically choose structural steel over cast iron. Which of the following best explains why structural steel is the superior building material for this specific application?

Structural steel is highly ductile, allowing it to bend and absorb seismic shock without suddenly fracturing.

Structural steel is extremely brittle, which ensures the framework remains perfectly rigid and will not bend during an earthquake.

Structural steel functions exclusively under compression, whereas cast iron is designed to handle the high tension caused by seismic waves.

Structural steel has significantly more mass than cast iron, anchoring the building more firmly to the ground using its own heavy weight.

Show answer & explanation

Answer: Structural steel is highly ductile, allowing it to bend and absorb seismic shock without suddenly fracturing.

Answer

Structural steel is highly ductile, allowing it to bend and absorb seismic shock without suddenly fracturing.

Structural steel possesses high ductility, which is a material's ability to undergo significant plastic deformation before rupture. During an earthquake, a building framework experiences extreme and shifting lateral loads. A ductile material like structural steel will bend and yield, safely absorbing the seismic energy without experiencing a sudden, catastrophic fracture. Cast iron, by contrast, is brittle and would crack or shatter under the same dynamic, bending stress.

Step-by-Step Solution

Identify the primary forces exerted on a building during an earthquake.

Earthquakes introduce dynamic, unpredictable shocks and vibrations that cause structures to sway, creating rapidly alternating tension, compression, and shear forces.

Understanding the environmental stress is the first step in selecting the appropriate building material.

Compare the mechanical properties of structural steel and cast iron under dynamic stress.

Cast iron is hard but brittle, meaning it snaps or shatters under sudden, bending stress. Structural steel is ductile, meaning it can stretch and deform elastically and plastically before failing.

Matching material behavior to the required load type reveals which material will survive the event.

Determine which property prevents building collapse during seismic activity.

Ductility is the preferred property because the material absorbs the energy of the earthquake by flexing. A building framework that bends is significantly safer than one that snaps abruptly.

This confirms why structural steel is specifically chosen over brittle alternatives for seismic zones.

Key Concept

Ductility vs. Brittleness in Structural Materials

Estimated Time:45s

Question 50Question →

A heavy generator is suspended horizontally by two vertical cables, Cable X on the left and Cable Y on the right. The center of gravity of the generator is located closer to Cable Y. Which of the following describes the load distribution between the two cables?

Cable X carries more of the load than Cable Y.

Cable Y carries more of the load than Cable X.

Cable X and Cable Y carry equal amounts of the load.

The distribution depends entirely on the generator's total mass, not its center of gravity.

Show answer & explanation

Answer: Cable Y carries more of the load than Cable X.

Answer

Cable Y carries more of the load than Cable X.

In mechanical comprehension, when a load is supported by multiple points, the support closer to the object's center of gravity must carry a proportionally larger share of the weight to maintain static equilibrium. Since the generator's center of gravity is closer to Cable Y, Cable Y bears the greater load.

Step-by-Step Solution

Identify the location of the supports and the center of gravity.

Cable X is on the left, Cable Y is on the right, and the center of gravity is closer to Cable Y.

Understanding the geometry of the setup is the first step in determining load distribution.

Apply the principle of load distribution and equilibrium.

The support closer to the center of gravity must provide a larger upward reaction force to balance the weight.

The center of gravity acts as the specific point where the entire downward force of the object's mass is concentrated.

Determine which cable carries the greater load.

Cable Y carries a greater portion of the generator's weight.

Because Cable Y is physically closer to the concentration of the downward force, it must support more of the load than Cable X.

Key Concept

In a system with multiple supports, the support closest to the object's center of gravity carries the greatest proportion of the load.

Question 51Question →

A uniform wooden beam is $10$ feet long and weighs $50$ pounds. It rests horizontally on two sawhorses. Sawhorse A is placed $4$ feet from the left end, and Sawhorse B is placed $9$ feet from the left end. A $50$ -pound bag of cement is placed at the extreme right end of the beam ( $10$ feet from the left end). Assuming the system is in static equilibrium, what is the upward reaction force exerted by Sawhorse B?

30

pounds

50

pounds

60

pounds

70

pounds

Show answer & explanation

Answer: $70$ pounds

Answer

The upward reaction force exerted by Sawhorse B is

70

pounds.

For the system to be in static equilibrium, the moments around any pivot point must sum to zero. Choosing Sawhorse A as the pivot: the beam's center of gravity (

5

ft mark) is

1

ft right of the pivot, creating

50 \times 1 = 50

ft-lbs of clockwise torque. The bag (

10

ft mark) is

6

ft right of the pivot, creating

50 \times 6 = 300

ft-lbs of clockwise torque. To balance the total

350

ft-lbs of clockwise torque, Sawhorse B (located

5

ft right of the pivot) must provide

350 / 5 = 70

pounds of upward force.

Step-by-Step Solution

Identify the pivot point and the locations of all downward forces.

Using Sawhorse A (at

4

ft) as the pivot. The beam's weight (

50

lbs) acts at its center (

5

ft). The cement bag (

50

lbs) acts at the right end (

10

ft).

Choosing one support as the pivot eliminates its unknown force from the equilibrium equation.

Calculate the lever arm distances from the pivot for each force.

Beam weight lever arm:

5 - 4 = 1

ft. Cement bag lever arm:

10 - 4 = 6

ft. Sawhorse B lever arm:

9 - 4 = 5

ft.

Torque is determined by multiplying a force by its perpendicular distance from the pivot point.

Calculate the total clockwise torque acting on the beam.

Beam torque:

50 \text{ lbs} \times 1 \text{ ft} = 50 \text{ ft-lbs}

. Cement bag torque:

50 \text{ lbs} \times 6 \text{ ft} = 300 \text{ ft-lbs}

. Total clockwise torque =

350 \text{ ft-lbs}

Both the beam's weight and the cement bag push downward on the right side of the pivot, creating clockwise rotation.

Set the clockwise torque equal to the counterclockwise torque and solve for Sawhorse B's force.

F_B \times 5 \text{ ft} = 350 \text{ ft-lbs}

, which means

F_B = 350 / 5 = 70

pounds.

In static equilibrium, the torques must perfectly balance to prevent the beam from rotating.

Key Concept

Rotational Equilibrium and Center of Gravity

Alternative Method

You can verify this by calculating moments around Sawhorse B instead. The beam's center (

5

ft) is

4

ft to the left (CCW torque:

50 \times 4 = 200

). The cement bag (

10

ft) is

1

ft to the right (CW torque:

50 \times 1 = 50

). Sawhorse A is

5

ft to the left (CW torque:

F_A \times 5

). Balancing torques:

200 = 50 + 5(F_A) \rightarrow F_A = 30

lbs. Since total downward force is

100

lbs,

F_B

must be

100 - 30 = 70

lbs.

Estimated Time:2m 0s

Question 52Question →

When used as a structural building material, timber can support a significantly heavier compressive load if the weight is applied perpendicular to its grain rather than parallel to its grain.

True

False

Show answer & explanation

Answer: False

Answer

False. Timber is much stronger in compression when loaded parallel to the grain.

The statement is false because the fibrous nature of timber means it acts like a series of vertical columns. It possesses its maximum compressive strength when loaded parallel to the grain, not perpendicular.

Step-by-Step Solution

Analyze the physical structure of wood as a building material.

Wood consists of long, hollow cellulose fibers that are aligned in one primary direction, which forms the wood's grain.

Understanding the cellular structure is required to determine how the material handles mechanical stress.

Evaluate the effect of a compressive force relative to the grain orientation.

Compressing wood parallel to the grain pushes against the solid ends of the fibers, providing high load-bearing resistance. Compressing it perpendicular to the grain easily crushes the hollow fiber walls.

This physical behavior proves that the claim of higher strength perpendicular to the grain is incorrect.

Key Concept

Directional strength (anisotropy) and load-bearing properties of timber

Question 53Question →

A footbridge spans a $20$ -foot wide creek, resting on a concrete pillar at each end (Left Pillar and Right Pillar). A person weighing $160$ pounds stands on the bridge, exactly $5$ feet from the Right Pillar. If we ignore the weight of the bridge itself, how many pounds of the person's weight are supported by the Left Pillar?

Show answer & explanation

Answer: 40

Answer

The weight supported by the Left Pillar is inversely proportional to the load's distance from it. Since the person is

15

feet from the Left Pillar and

5

feet from the Right Pillar on a

20

-foot bridge, they are

3

times closer to the Right Pillar. Therefore, the Right Pillar carries

3/4

of the weight (

120

lbs) and the Left Pillar carries

1/4

of the weight (

40

lbs).

Step-by-Step Solution

Determine the relevant distances along the bridge.

Total span =

20

ft. Distance to Right Pillar =

5

ft. Distance to Left Pillar =

15

ft.

To calculate the load on one support, we need the total distance and the distance to the opposite support.

Set up the load distribution calculation for the Left Pillar.

Load on Left Pillar = Total Weight

\times

(Distance to Right Pillar / Total Span)

The proportion of weight supported by one end of a beam is equal to the distance from the load to the opposite end, divided by the total length.

Calculate the supported weight using the values.

160 \times (5 / 20) = 160 \times (1 / 4) = 40

pounds.

Solving the equation gives the exact weight carried by the left support.

Key Concept

Load Distribution and Balancing

Alternative Method

You can also use the concept of torque (moments) in static equilibrium. Taking the Right Pillar as the pivot point: (Force of Left Pillar

\times

20

ft) = (

160

lbs

\times

5

ft). Therefore,

20 \times F_{left} = 800

, which means

F_{left} = 40

lbs.

Estimated Time:45s

Question 54Question →

Reinforced concrete is a widely used composite building material that combines poured concrete with steel reinforcing bars (rebar). While the steel is primarily added to provide tensile strength, what other critical physical property must the steel and concrete share to prevent the structure from cracking or tearing itself apart during seasonal temperature changes?

They must heat up and cool down at exactly the same rate when exposed to the environment.

They must have nearly identical coefficients of thermal expansion.

They must both be highly ductile materials that stretch easily without fracturing to absorb thermal shock.

They must both act as effective thermal insulators to block heat from penetrating the structure.

Show answer & explanation

Answer: They must have nearly identical coefficients of thermal expansion.

Answer

The steel and concrete must have nearly identical coefficients of thermal expansion so they expand and contract together at the same proportional rate when the temperature changes.

For a composite material like reinforced concrete to remain structurally sound across differing environments, its component materials must expand and contract at the same proportional rate when the temperature changes. Because structural steel and concrete share nearly identical coefficients of thermal expansion, they change size together in harmony. This prevents the creation of internal stresses that would otherwise destroy the bond and crack the brittle concrete.

Step-by-Step Solution

Identify the effect of temperature changes on physical materials.

Almost all building materials expand when heated and contract when cooled (thermal expansion).

Temperature changes are an unavoidable environmental stressor on outdoor structures.

Analyze the mechanics of a composite material like reinforced concrete.

Steel is embedded tightly inside the concrete, creating a solid, bonded structural unit.

The materials must work together to distribute loads.

Evaluate what happens if the two bonded materials react differently to heat.

If steel expanded significantly more than concrete when heated, the internal swelling of the rebar would shatter the brittle concrete from the inside.

Mismatched expansion rates create massive internal shear and tensile stresses.

Conclude the necessary shared property.

They must expand and contract by the exact same amount per degree of temperature change, meaning they have the same coefficient of thermal expansion.

This compatibility ensures the composite material maintains structural integrity year-round.

Key Concept

Thermal expansion compatibility in composite building materials

Estimated Time:45s

Question 55Question →

A single-axle utility trailer has its axle located $8$ feet behind the hitch. The empty trailer weighs $600$ pounds, and its center of gravity is $5$ feet behind the hitch. A machine weighing $1,200$ pounds is loaded onto the trailer so that its center of gravity is $6$ feet behind the hitch. When the trailer is parked horizontally, what is the downward force, in pounds, exerted on the hitch?

Show answer & explanation

Answer: 525

Answer

525

The correct answer is derived by treating the trailer as a lever in static equilibrium. Taking the sum of torques around the axle, the total rotational force from the empty trailer and the loaded machine is

4200

lb-ft. Dividing this by the

8

-foot lever arm to the hitch yields exactly

525

pounds of force.

Step-by-Step Solution

Identify a pivot point and determine all relevant distances.

Using the axle as the pivot point. Distance from axle to trailer CG is 3 ft (

8 - 5

). Distance from axle to machine CG is 2 ft (

8 - 6

). Distance from axle to hitch is 8 ft.

Establishing a pivot point allows us to use the principle of moments (torques) to find the unknown reaction force.

Calculate the total counter-clockwise torque around the axle.

Trailer torque =

600 \text{ lbs} \times 3 \text{ ft} = 1800 \text{ lb-ft}

. Machine torque =

1200 \text{ lbs} \times 2 \text{ ft} = 2400 \text{ lb-ft}

. Total torque =

4200 \text{ lb-ft}

The weight of the trailer and the machine both push down, attempting to rotate the system around the axle.

Set up the static equilibrium equation.

4200 \text{ lb-ft} = F_{hitch} \times 8 \text{ ft}

For the trailer to remain stationary, the clockwise torque provided by the hitch must perfectly balance the total counter-clockwise torque.

Solve for the force at the hitch (

F_{hitch}

F_{hitch} = \frac{4200}{8} = 525 \text{ lbs}

Dividing the total torque by the distance to the hitch gives the required balancing force.

Key Concept

Static Equilibrium and Torque Distribution

Alternative Method

You can also solve this by setting the hitch as the pivot point. The downward torque would be

(600 \times 5) + (1200 \times 6) = 3000 + 7200 = 10200

lb-ft. The upward force from the axle balances this:

F_{axle} \times 8 = 10200

, so

F_{axle} = 1275

lbs. Since the total weight is

1800

lbs (

600 + 1200

), the remaining force supported by the hitch is

1800 - 1275 = 525

lbs.

Estimated Time:2m 0s

Question 56Question →

Two technicians, Technician X and Technician Y, are carrying a heavy transmission using a horizontal lifting bar. Technician X holds the left end of the bar and Technician Y holds the right end. The transmission is secured to the bar much closer to Technician X than to Technician Y. Which of the following statements correctly describes how the weight is distributed between them?

Technician X supports a greater portion of the weight than Technician Y.

Technician Y supports a greater portion of the weight than Technician X.

Both technicians support exactly equal portions of the weight.

Technician Y supports a greater portion of the weight due to the mechanical advantage of the longer bar section.

Show answer & explanation

Answer: Technician X supports a greater portion of the weight than Technician Y.

Answer

Technician X supports a greater portion of the weight than Technician Y.

In load balancing, the support that is physically closer to the center of gravity of the load carries a proportionally larger share of the weight. Because the transmission is secured closer to Technician X, Technician X must exert more upward force to maintain equilibrium.

Step-by-Step Solution

Identify the locations of the two supports and the applied load.

The supports are at each end of the bar (Technician X and Technician Y), and the load (the transmission) is located closer to Technician X.

Understanding the physical arrangement is necessary to analyze the forces.

Apply the general rule of load distribution for offset weights.

When a weight is placed between two supports but off-center, the support closest to the weight carries a proportionally larger share of the total load.

This is derived from the principles of static equilibrium and rotational moments; a shorter lever arm from the load requires a greater supporting force.

Conclude which technician bears more weight based on their proximity to the load.

Since the transmission is closer to Technician X, Technician X bears a greater portion of the load.

The weight distribution is inversely proportional to the distance from the load to the support.

Key Concept

In a system with two supports, an offset load distributes its weight unevenly, with the support closest to the load's center of gravity bearing the larger portion of the weight.

Question 57Question →

Review the following mechanical fastening scenarios. Match each scenario to the primary type of stress or characteristic failure mode it is most likely to experience.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Overlapping pulled plates experience single shear stress; over-tightened bolts suffer excessive tension; keyed pulleys transfer torsional shear stress; and alternating bending cycles lead to metal fatigue.

Each scenario describes a distinct physical loading condition. Overlapping plates pulled in opposite directions slice across the bolt (single shear). Torquing a nut stretches the bolt shaft along its length (tension). A rotating shaft transferring power through a key creates a twisting sideways force (torsional shear). Continuous bending back and forth causes microscopic damage over time (fatigue).

Step-by-Step Solution

Analyze the overlapping plates scenario.

Identified that parallel pulling forces on overlapping joined plates create a slicing or cutting action (shear) across the fastener.

This is the classic definition of a lap joint in single shear.

Analyze the over-tightening scenario.

Identified that tightening a nut draws it up the bolt threads, pulling the bolt shaft lengthwise.

Applying torque translates into clamping force, which places the bolt shaft under tensile stress (tension).

Analyze the keyed pulley scenario.

Identified that the key prevents the pulley from slipping on the spinning shaft, transferring rotational force.

The transfer of rotational motion creates a twisting shear force known as torsional stress on the key.

Analyze the alternating bending scenario.

Identified that continuous, repeating stress cycles cause microscopic wear within the metal structure.

Cyclic loading is the primary cause of metal fatigue, especially where stress risers like threads exist.

Key Concept

Identifying the specific types of mechanical stress (shear, tension, torsion, fatigue) applied to fasteners in different operating environments.

Question 58Question →

A uniform wooden plank is $16$ feet long and weighs $120$ pounds. It rests horizontally on two supports. The left support is positioned at the extreme left end of the plank, and the right support is positioned $12$ feet away from the left end. If a $60$ -pound weight is placed on the extreme right end of the plank, how much upward force must the right support provide to maintain equilibrium?

90

pounds

120

pounds

160

pounds

180

pounds

Show answer & explanation

Answer: $160$ pounds

Answer

160

pounds

The system is in equilibrium, meaning the torques must balance. Taking the left support as the pivot, the downward forces create a clockwise torque: the plank's weight (

120

lbs) acts at its center (

8

ft away) creating

960 \text{ lb-ft}

of torque, and the box (

60

lbs) acts at the end (

16

ft away) creating another

960 \text{ lb-ft}

. The total downward torque is

1920 \text{ lb-ft}

. The right support is located

12

feet away and must push upward to create an equal counter-clockwise torque. Therefore,

12 \times \text{Force} = 1920

, which means the required force is

160

pounds.

Step-by-Step Solution

Set the left support as the pivot point (fulcrum).

All distances for torque calculations will be measured from the

0

-foot mark.

Choosing one of the unknown supports as the pivot eliminates its force from the torque equation, allowing us to solve directly for the other support.

Calculate the total clockwise torque acting on the plank.

Plank torque =

120 \text{ lbs} \times 8 \text{ ft} = 960 \text{ lb-ft}

. Weight torque =

60 \text{ lbs} \times 16 \text{ ft} = 960 \text{ lb-ft}

. Total clockwise torque =

1920 \text{ lb-ft}

The uniform plank's weight acts at its center of gravity (

8

feet from the left), and the external weight acts at the extreme right end (

16

feet from the left). Both create a downward, clockwise rotation around the left support.

Equate the clockwise torque to the counter-clockwise torque to solve for the right support's force.

F_{right} \times 12 \text{ ft} = 1920 \text{ lb-ft}

. Dividing by

12

gives

F_{right} = 160 \text{ lbs}

For the system to remain in static equilibrium, the upward (counter-clockwise) torque provided by the right support must exactly balance the downward (clockwise) torques.

Key Concept

Static Equilibrium and Principle of Moments

Question 59Question →

A mechanic is constructing a heavy-duty tow linkage. Two thick steel plates are overlapped to form a lap joint and are secured together using three high-strength bolts. When a tractor pulls a heavy load using this linkage, the plates are pulled in opposite directions, parallel to their touching surfaces. Which of the following best describes the primary mechanical stress exerted on the bolts?

Shear stress, because the opposing lateral forces attempt to slice through the cross-section of the bolts.

Tensile stress, because the pulling force from the tractor attempts to stretch the bolts lengthwise.

Compressive stress, because the plates squeeze the body of the bolts as the load is applied.

Torsional stress, because the pulling action causes the bolts to twist violently inside their mounting holes.

Show answer & explanation

Answer: Shear stress, because the opposing lateral forces attempt to slice through the cross-section of the bolts.

Answer

Shear stress, because the opposing lateral forces attempt to slice through the cross-section of the bolts.

When overlapping plates in a lap joint are pulled in opposite directions (parallel to their mating surfaces), the fasteners holding them together must resist a lateral force. This force tries to cut the fasteners in half straight across their diameter. In mechanical engineering, a force that attempts to slice through a cross-section of material is known as shear stress.

Step-by-Step Solution

Analyze the direction of the applied forces relative to the fasteners.

The force pulls the steel plates in opposite directions, parallel to their touching surfaces. This means the force acts perpendicular to the length (axis) of the bolts.

Identifying the orientation of force vectors relative to the fastener is critical for determining the type of stress.

Evaluate the physical effect of this perpendicular force on the bolts.

Because the top plate pushes one way and the bottom plate pushes the other way at the exact point where they meet, it creates a cutting or slicing action across the middle of the bolts.

Visualizing how the material deforms or fails under load defines the mechanical nature of the stress.

Match the physical action to the correct mechanical terminology.

A force that attempts to cut or slice a material parallel to its cross-section is formally known as shear stress.

Standard engineering terminology must be applied to correctly classify joint loads.

Key Concept

Fastener Joint Loading: Shear vs. Tension

Question 60Question →

A mechanic is securing a critical structural joint using a drilled bolt, a castellated nut, and a cotter pin. Arrange the following steps in the correct chronological order to properly install and lock this fastener assembly according to standard mechanical practices.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence is to first clean and inspect the components, apply the initial baseline torque, check the hole alignment, tighten the nut further if necessary to align the slots, and finally install and secure the cotter pin.

Proper installation of a castellated nut and cotter pin requires a strict sequence. First, components must be cleaned for accurate torque application. Second, the nut is tightened to the specified baseline torque to achieve the necessary joint tension. Third, alignment is checked. Fourth, because the nut's slots and the bolt's drilled hole rarely align perfectly at the exact baseline torque, standard mechanical practice dictates tightening the nut slightly further until alignment is achieved; backing off the nut is strictly prohibited as it would compromise the minimum required clamping force. Finally, the cotter pin is inserted and its prongs are bent to provide a positive mechanical lock against vibration.

Step-by-Step Solution

Identify the prerequisite step before any fastener assembly.

Cleaning and inspecting the threads is placed first.

Damaged or dirty threads can lead to false torque readings and subsequent joint failure.

Determine the action that provides the primary clamping force.

Threading the nut and applying the initial specified torque is placed second.

The joint must be secured to its baseline tension before any locking mechanism is engaged.

Evaluate what must be checked before installing the positive locking mechanism.

Checking the alignment of the nut's slots and the bolt's hole is placed third.

The cotter pin requires a clear, unobstructed path through both components.

Determine the correct procedure if the locking holes are not perfectly aligned.

Tightening the nut further until the next slot aligns is placed fourth.

Loosening the nut would compromise the minimum required clamping force (tension) on the joint, so it must only be tightened.

Identify the final locking step.

Inserting and bending the cotter pin is placed fifth.

This provides a positive mechanical lock, completing the assembly process.

Key Concept

Proper installation sequence and torque application for positive-locking fasteners (castellated nuts and cotter pins).

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