Designing the Optimum Cast Iron T-Beam

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The discussion focuses on designing an optimal cast iron T-beam, emphasizing the relationship between compressive and tensile stresses, which should have a ratio of 4:1. The participants outline the equations for tensile and compressive stress based on moment and centroid calculations. An attempt to solve for the dimensions of the T-beam leads to a deadlock, resulting in the equation simplifying to 0=0. The proposed solution involves reformulating the equation into a quadratic form to find the height of the bottom rectangle (H2). Ultimately, using the quadratic formula with specific values for thickness and top length will yield the necessary dimensions for the beam design.
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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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