Applying loads (forces, pressures) in Solidworks - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to apply a force in SolidWorks.

Solidworks

swModel.Extension.SelectByID2("Face1", "FACE", 0, 0, 0, false, 0, null, 0);
swModel.AddForce([1], 0, 0);

Drag options to blanks, or click blank then click option'

A100

BForce

CPressure

D200

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Force' or 'Pressure' instead of a numeric value.

The value 100 represents the magnitude of the force applied in Newtons.

2fill in blank

medium

Complete the code to apply pressure on a surface in SolidWorks.

Solidworks

swModel.Extension.SelectByID2("Surface1", "SURFACE", 0, 0, 0, false, 0, null, 0);
swModel.AddPressure([1]);

Drag options to blanks, or click blank then click option'

AForce

BArea

C5000

DPressure

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Force' or 'Pressure' string instead of a numeric value.

The value 5000 represents the pressure magnitude in Pascals.

3fill in blank

hard

Fix the error in the code to correctly apply a force vector in SolidWorks.

Solidworks

swModel.Extension.SelectByID2("Face2", "FACE", 0, 0, 0, false, 0, null, 0);
swModel.AddForce([1], 0, 100);

Drag options to blanks, or click blank then click option'

AForce

BPressure

C100

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Force' or 'Pressure' as vector components.

The force vector component must be numeric. Using 0 correctly sets the x-component of the force.

4fill in blank

hard

Fill both blanks to apply a force and pressure correctly in SolidWorks.

Solidworks

swModel.Extension.SelectByID2("Face3", "FACE", 0, 0, 0, false, 0, null, 0);
swModel.AddForce([1], 0, 0);
swModel.AddPressure([2]);

Drag options to blanks, or click blank then click option'

A150

BForce

C3000

DPressure

Attempts:

3 left

💡 Hint

Common Mistakes

Using string labels instead of numeric values.

150 is the force magnitude in Newtons, and 3000 is the pressure magnitude in Pascals.

5fill in blank

hard

Fill all three blanks to apply a force vector and pressure in SolidWorks.

Solidworks

swModel.Extension.SelectByID2("Face4", "FACE", 0, 0, 0, false, 0, null, 0);
swModel.AddForce([1], [2], 0);
swModel.AddPressure([3]);

Drag options to blanks, or click blank then click option'

A200

B0

C4000

DForce

Attempts:

3 left

💡 Hint

Common Mistakes

Using string labels instead of numeric values for force and pressure.

The force vector components are 200 (x), 0 (y), and pressure is 4000 Pascals.

Practice

1.

What is the main purpose of applying loads in SolidWorks simulations?

easy

A. To simulate how forces affect the design's strength and behavior

B. To change the color of the model

C. To create new parts automatically

D. To export the design to other software

Solution

1. Step 1: Understand the role of loads in simulation

   Loads represent forces or pressures applied to a model to test its response.

2. Step 2: Identify the purpose of applying loads

   Applying loads helps predict stress, deformation, and failure points in the design.

3. Final Answer:

   To simulate how forces affect the design's strength and behavior -> Option A

4. Quick Check:

   Loads test design behavior = A [OK]

Hint: Loads simulate forces on design to check strength [OK]

Common Mistakes:

• Thinking loads change model appearance
• Confusing loads with part creation
• Assuming loads export data

2.

Which of the following is the correct way to apply a Force load in SolidWorks Simulation?

easy

A. Draw a sketch and convert it to a force

B. Select the entire model and click 'Apply Pressure'

C. Use the 'Mesh' tool to add force values

D. Select the face, then specify magnitude and direction of the force

Solution

1. Step 1: Identify how to apply a force load

   In SolidWorks, you select the face or edge where the force acts.

2. Step 2: Specify force details

   You enter the force magnitude and direction after selecting the location.

3. Final Answer:

   Select the face, then specify magnitude and direction of the force -> Option D

4. Quick Check:

   Force load = select face + magnitude/direction [OK]

Hint: Force load needs face selection and magnitude input [OK]

Common Mistakes:

• Applying pressure instead of force
• Selecting whole model instead of face
• Using mesh tool incorrectly

3.

Given a cantilever beam fixed at one end, if a downward force of 100 N is applied at the free end, what is the expected effect on the beam?

medium

A. The beam will bend downward causing stress near the fixed end

B. The beam will compress uniformly along its length

C. The beam will expand and lengthen

D. The beam will rotate without bending

Solution

1. Step 1: Understand cantilever beam behavior under load

   A force at the free end causes bending with maximum stress at the fixed end.

2. Step 2: Analyze the effect of downward force

   The beam bends downward, creating tension and compression zones, especially near the fixed support.

3. Final Answer:

   The beam will bend downward causing stress near the fixed end -> Option A

4. Quick Check:

   Cantilever + end force = bending stress at fixed end [OK]

Hint: Force at free end bends cantilever near fixed end [OK]

Common Mistakes:

• Assuming uniform compression
• Thinking beam lengthens
• Ignoring bending and only rotation

4.

In a SolidWorks simulation, a user applies a pressure load but the simulation shows zero deformation. What is the most likely error?

medium

A. The pressure magnitude was set too high

B. The pressure was applied to a fixed face with no degrees of freedom

C. The mesh size was too small

D. The material was set to plastic

Solution

1. Step 1: Identify why zero deformation occurs

   If the load is on a fixed face, the model cannot move or deform there.

2. Step 2: Understand fixed constraints effect

   Fixed constraints prevent movement, so pressure on that face causes no deformation.

3. Final Answer:

   The pressure was applied to a fixed face with no degrees of freedom -> Option B

4. Quick Check:

   Pressure on fixed face = zero deformation [OK]

Hint: Check if load is on fixed (immovable) face [OK]

Common Mistakes:

• Assuming high pressure causes zero deformation
• Thinking mesh size causes no deformation
• Confusing material type with deformation

5.

You want to simulate the effect of wind pressure on a car door in SolidWorks. Which combination of load types and locations is best?

hard

A. Apply a force load at the door handle only

B. Apply a torque load on the door hinges

C. Apply a uniform pressure load on the outer surface of the door

D. Apply a pressure load inside the door cavity

Solution

1. Step 1: Identify the nature of wind load

   Wind applies pressure distributed over the outer surface of the door.

2. Step 2: Choose appropriate load type and location

   A uniform pressure load on the door's outer surface best simulates wind effect.

3. Final Answer:

   Apply a uniform pressure load on the outer surface of the door -> Option C

4. Quick Check:

   Wind = uniform pressure on outer surface [OK]

Hint: Wind acts as pressure on outer surface, not force at one point [OK]

Common Mistakes:

• Applying force only at handle
• Using torque on hinges for wind
• Applying pressure inside door cavity