Designing the Optimum Cast Iron T-Beam

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SUMMARY

The discussion focuses on designing an optimum T-section for a cast iron beam, emphasizing the relationship between compressive and tensile stress, which is defined by a factor of 4. Key equations include TStress = (M*c1)/I and CStress = -(M*c2)/I, where M represents the moment, c1 and c2 are distances to centroids, and I is the moment of inertia. The solution involves deriving a quadratic equation to solve for the height of the bottom rectangle (H2) based on the established relationships and dimensions of the T-beam.

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Homework Statement


Cast iron has a strength in compression of about three to four times the strength in tension, depending upon the grade.
using the stress as a prime consideration, design the optimum T section for a cast iron beam using a uniform section thickness such that the compressive stress will be related to the tensile stress by a factor of 4


2. Homework Equations
TStress = (M*c1)/I
CStress = -(M*c2)/I


The Attempt at a Solution

 
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TSectionStress
Please see attached Diagram of T Beam

Let 1 = Top Rectangle
Let 2 = Bottom Rectangle
Let T = Thickness = 10
Let H1 = T = Height of top
Let H2 = height of bottom
Let HT = Total height of T section = T + H2
Let B1 = Length of Top
Let B2 = length of bottom = T
Let A1 = Top Area = B1 * T
Let A2 = Bottom area = H2 * T
Let AT = Total Area = A1 + A2
Let C1 = centroid from the top of 1 = ((T * B1 * (T / 2)) + ((T * H2) * (T + (H2 / 2)))) / AT
Let C2 = Centroid from the bottom of 2 = HT - C1
Let I1 = Second moment of area of top rectangle = (((B1 / 10) * ((H1 ^ 3) / 10)) / 12) / 100
Let I2 = Second moment of area of bottom rectangle = (((B2 / 10) * ((H2 ^ 3) / 10)) / 12) / 100
Let D1 = Distance from C1 to 1 axis = C1 - (T / 2)
Let D2 = Distance from C2 to 2 axis = C2 - (H2 / 2)
Let IX1 = Second moment of area about any parallel axis to the c1 axis a distance d1 removed = (I1 + ((A1 / 100) * ((D1 / 10) ^ 2)))
Let IX2 = Second moment of area about any parallel axis to the c2 axis a distance d2 removed = (I2 + ((A2 / 100) * ((D2 / 10) ^ 2)))
let I = IX1 + IX2
let StressT = Tensile Stress = ((1600 * (C1 / 10)) / I)
Let StressC = Compressive Stress = ((1600 * (C2 / 10)) / I)




Attempt at a solution

StressC/StressT = 4

((1600 * (C2 / 10)) / I)/((1600 * (C1 / 10)) / I) = 4
Therefore
C1 = C2/4 and C2 =4*C1

But from above C2 = HT-C1
So
((T * B1 * (T / 2)) + ((T * H2) * (T + (H2 / 2)))) / AT= HT-C1/4

The above equation should give me H2
but I can only solve it to get 0=0
What am I doing wrong.
 

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  • T Diagram.png
    T Diagram.png
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Solution
((T * B1 * (T / 2)) + ((T * H2) * (T + (H2 / 2)))) / (B1 * T) + (H2 * T)
Put the above equation for C1 into the form of a quadratic equation in terms of H2
Then put in values for T and B1.
Use the Quadratic formula the find H2.
 

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