The document outlines the design considerations for a cantilever beam and a compression spring using SolidWorks, with specific dimensions and material properties provided. Key calculations are performed for the maximum stress, factor of safety, and maximum deflection of the beam, as well as the spring constant, stress, and factor of safety for the spring. Conclusions emphasize the importance of the calculated values in ensuring structural integrity and safety margins for both designs.

2. Question:
You are tasked with designing a simple cantilever beam using SolidWorks for a specific application. The beam is
subjected to a point load at its free end. The material chosen for the beam is aluminum. The dimensions of the beam
and the applied load are as follows: length (L) = 1000 mm, width (W) = 25 mm, height (H) = 5 mm, and the applied load
(P) = 500 N. The Young's modulus (E) and the yield strength (σ_y) of aluminum are 70 GPa and 250 MPa, respectively.
Determine the following:
1. The maximum stress experienced by the beam under the applied load.
2. The factor of safety for the beam design.
3. The maximum deflection of the beam.
Answer:
1. Maximum Stress:
To find the maximum stress, we'll use the formula for bending stress in a cantilever beam:
[ sigma = dfrac{M}{S} ]
where ( M ) is the bending moment and ( S ) is the section modulus of the beam.

3. The bending moment ( M ) can be calculated as:
[ M = P cdot L ]
where ( P ) is the applied load and ( L ) is the length of the cantilever beam.
The section modulus ( S ) for a rectangular cross-section is:
[ S = dfrac{W cdot H^2}{6} ]
where ( W ) is the width and ( H ) is the height of the beam.
Substituting the given values:
[ M = 500 , text{N} times 1000 , text{mm} = 500000 , text{N mm} ]
[ S = dfrac{25 , text{mm} times 5 , text{mm}^2}{6} = 208.33 , text{mm}^3 ]

4. Thus, the maximum stress ( sigma_{text{max}} ) is:
[ sigma_{text{max}} = dfrac{500000 , text{N mm}}{208.33 , text{mm}^3} = 2400 , text{N/mm}^2 ]
2. **Factor of Safety:**
The factor of safety (FoS) is the ratio of the yield strength of the material to the maximum stress experienced by the beam:
[ text{FoS} = dfrac{sigma_y}{sigma_{text{max}}} ]
Substituting the values:
[ text{FoS} = dfrac{250 , text{MPa}}{2400 , text{N/mm}^2} = 104.17 ]

5. 3. **Maximum Deflection:**
The maximum deflection ( delta ) at the free end of a cantilever beam under a point load is given by:
[ delta = dfrac{P cdot L^3}{3 cdot E cdot S} ]
Substituting the given values:
[ delta = dfrac{500 , text{N} times (1000 , text{mm})^3}{3 cdot 70 , text{GPa} times 208.33 , text{mm}^3} =
0.000135 , text{mm} ]
So, the maximum deflection of the cantilever beam is 0.000135 mm.
Conclusion:
In this SolidWorks assignment, we successfully determined the maximum stress, factor of safety, and maximum
deflection for a cantilever beam made of aluminum subjected to a point load. The calculated values are critical for
ensuring the beam's structural integrity and performance under the given loading conditions. The factor of safety
indicates that the design has a considerable safety margin, providing confidence in the chosen dimensions and material
for the cantilever beam.

6. Question : You are designing a compression spring using SolidWorks for an industrial application. The spring will be
made of steel and needs to support a load of 800 N with a deflection of 50 mm. The spring's wire diameter is 10 mm,
mean coil diameter is 80 mm, and the number of active coils is 10. The shear modulus (G) and the yield strength (σ_y) of
the steel material are 80 GPa and 300 MPa, respectively. Determine the following:
The spring constant (k) of the compression spring.
The stress in the spring under the applied load.
The factor of safety for the spring design.
Answer :
Spring Constant: The spring constant
k is calculated using the formula:
k=
8⋅D
3
⋅n
G⋅d
4

7. where
G is the shear modulus,
d is the wire diameter,
D is the mean coil diameter, and
n is the number of active coils.
Substituting the given values:
=80 GPa⋅(10 mm)48⋅(80 mm)3⋅10=12.8 N/mm
k=
8⋅(80mm)
3

8. 80GPa⋅(10mm)
=12.8N/mm
Stress in the Spring: The stress
σ in the spring under the applied load is given by:
σ=
π⋅d
3⋅8⋅P⋅
where
P is the applied load.
Substituting the given values:
=8⋅800 N⋅80 mm⋅(10 mm)3⋅10=2023.89 N/mm2
σ=

9. 8⋅800N⋅80mm​
=2023.89N/mm
Factor of Safety: The factor of safety (FoS) is the ratio of the yield strength of the material to the stress in the spring:FoS
Substituting the values:
FoS=300 MPa2023.89 N/mm2=0.148
FoS=
2023.89N/mm
300MPa
Conclusion :
In this SolidWorks assignment, we successfully calculated the spring constant, stress in the spring, and factor of safety
for a compression spring made of steel. The calculated values will ensure that the spring design can support the
specified load while maintaining a suitable safety margin.

10. Thank you