Thomas Adcock
MSD I
Preliminary Structural Analysis
2/8/13
Problem Formulation:
The problem modeled is a housing for a device undergoing a specified acceleration. The forces and
stress on the casing will be modeled and compared to specifications provided.
𝜎 = 𝐹/𝐴
Equation 1. Simplified 1 dimensional stress
∑𝐹 = 0
Equation 2. D'Alembert’s principle
Mathematical Analysis:
Free Body Diagram:
m*a
m
F
F=m*a
Equation 3. Force Equation
Combining Equation 3 with Equation 1:
𝐹 𝑚∗𝑎
=
𝐴
𝐴
Equation 4. Stress
𝜎=
For a hollow box of internal dimensions X by Y and thickness t, the cross sectional area is:
𝐴 = (𝑋 + 2𝑡) ∗ (𝑌 + 2𝑡) − 𝑋 ∗ 𝑌
Equation 5. Area
Combining Equation 4 and Equation 5:
𝜎=
𝐹 𝑚∗𝑎
𝑚∗𝑎
=
=
(𝑋 + 2𝑡) ∗ (𝑌 + 2𝑡) − 𝑋 ∗ 𝑌
𝐴
𝐴
Equation 6. Stress
The specifies acceleration is 20G. Substituting this in to Equation 6 gives:
𝜎=
𝐹 𝑚∗𝑎
𝑚∗𝑎
𝑚 ∗ 20 ∗ 𝑔
=
=
=
(𝑋 + 2𝑡) ∗ (𝑌 + 2𝑡) − 𝑋 ∗ 𝑌 (𝑋 + 2𝑡) ∗ (𝑌 + 2𝑡) − 𝑋 ∗ 𝑌
𝐴
𝐴
Equation 7. Stress
Because the mass is unknown, a maximum mass will be solved for using the yield stress of the plastic as
the stress 𝜎. Solving for mass:
𝜎 ∗ [(𝑋 + 2𝑡) ∗ (𝑌 + 2𝑡) − 𝑋 ∗ 𝑌]/[20 ∗ 𝑔] = 𝑚
Equation 8. Mass
Assumed Key Parameters:
𝜎 = 6500 𝑃𝑆𝐼 = 44.82𝑀𝑃𝑎 = 44.82 ∗ 106 𝑃𝑎
X=70mm=.07m
Y=65mm=.065m
t=3mm=.003m
g=9.81m/s2
Quantitative Outcome of Model:
𝜎∗
[(𝑋 + 2𝑡) ∗ (𝑌 + 2𝑡) − 𝑋 ∗ 𝑌]
[20 ∗ 𝑔]
= 44.82 ∗ 106 𝑃𝑎 ∗
= 193𝑘𝑔
Stress Strain Curves:
Conclusion:
[(. 07𝑚 + 2 ∗ .003𝑚) ∗ (. 065 + 2 ∗ .003𝑚) − .07𝑚 ∗ .065𝑚]
=𝑚
𝑚
[20 ∗ 9.81 2 ]
𝑠
The maximum mass that can be accelerated safely by the casing far exceeds a reasonable mass for the
product indicating that structural failure is extremely unlikely. The linear elastic region is around 15% of
the UTS. This corresponds to 10% of the calculated mass or 19.3kg.
http://www.azom.com/work/kWfiTWyN68bIo815M9HJ_files/image003.gif
http://www.machinist-materials.com/comparison_table_for_plastics.htm