User:Egm3520.s12.Team 5.Shankwitz/sandbox

Problem Statement
The maximum allowable stress for the beam pictured is 24MPa in tension and 30MPa in compression.



For d=40mm, find the largest couple M which can be applied to the beam.

Given
Maximum stress in compression: $$ \sigma_{max,C}=30\, MPa $$

Maximum stress in tension: $$ \sigma_{max,T}=24\, MPa $$

$$ d=40\, mm $$

Analysis
Free body diagram of the beam:



Centroid
Splitting the bar's cross section into two rectangles, 1 and 2, with 1 being the upper section and 2 being the lower section

$$ \displaystyle A_1 = 15\, mm \times 40\, mm = 600\, mm^2 $$

$$ A_2 = 20\, mm \times 25\, mm = 500\, mm^2 $$

$$ \bar y_1 = 25 + \frac{15}{2} = 32.5\, mm $$

$$ \bar y_2 = 25/2 = 12.5\, mm $$

We can use these to find the neutral axis (centroid) for the entire cross section with this relationship
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$$ \displaystyle \bar Y \sum A = \sum \bar y A $$ (5.4-1) Yielding an answer of
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$$ \displaystyle \bar Y = 23.409\, mm $$ (5.4-2)
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Centroidal Moment of Inertia
Using the parallel axis theorem, the centroidal moment of inertia can be represented as
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$$ \displaystyle I=\sum (I + Ad^2) $$ (5.4-3) where $$ I $$ for a rectangle is
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$$ \displaystyle I=\frac{bh^3}{12} $$ (5.4-4) Now, inserting the values for the entire bar
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$$ \displaystyle I=[\frac{(40)(15^3)}{12}+(600)(32.5-23.409)^2]+[\frac{(20)(25^3)}{12}+(500)(12.5-23.409)^2] $$     (5.4-5)
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$$ \displaystyle I=146.4 \times 10^{-9}\, m^4 $$ (5.4-6)
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Stress
Because of the direction the bar is being bent due to the applied moment, the top of the beam will undergo compression and the bottom of the bar will undergo tension. So, when calculating the maximum stress due to compression, the distance from the neutral axis to the top of the bar will be used. Likewise, the distance from the neutral axis to the bottom of the bar will be used in the calculation for maximum stress due to tension.
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$$ \displaystyle \sigma_{max,C}=\frac{Mc}{I} \Rightarrow M=\frac{\sigma_{max,C}I}{c} $$ (5.4-7) This gives us $$ M=187.6\, N\cdot m $$
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For tension:
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$$ \displaystyle \sigma_{max,T}=\frac{Mc}{I} \Rightarrow M=\frac{\sigma_{max,T}I}{c} $$ (5.4-8) Which gives us $$ M=211.8\, N\cdot m $$
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The smaller of these two values is our answer.