Maximum Tensile Stress In Relation To Temperature Drop In A Steel Bar

**Strings (S _{i}P_{j}A_{jk}) = S_{7}P_{3}A_{31} Base Sequence = 12735 String Sequence = 12735 - 3 - 31**

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Maximum Tensile Stress In Relation To Temperature Drop In A Steel Bar

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A steel bar in the form of a frustum (truncated cone) is rigidly fixed at both of its ends (figure 125.2). Its dimensions are:

length = 24 inches; diameter of circular section at one end = 1 inch.

diameter of circular section at the other end = 3 inches.

Modulus of elasticity, E = 30,000,000 lb/in^{2}.

Coefficient of thermal expansion = 0.0000065/^{o}F.

Suppose bar is subjected to a drop in temperature of 50^{o}F. Determine maximum tensile stress in bar.

**The strings**:
S_{7}P_{3}A_{31} (Force - Pull).
**The math**:

Pj Problem of Interest is of type *force* (pull).

Assumptions: bar is rigidly supported; bar is free to contract; proportional limit of steel bar ≥ maximum stress; *superposition* applicable.

Temperature drop caused a shortening of bar. Tensile Stress of interest is the maximum required to produce an elongation equal to the shortening (i.e stretch bar back to its original length).

shortening, *e* due to temperature drop = 50(0.0000065)24 = 0.00780 in.

Consider a circular cross-sectional area, A with diameter d, and distance x inches from the end with the smaller diameter (figure 125.2). Then by symmetry:

(d -1)/x = (3-1)/24. So, d = 1 + x/12.

So, Area A = (π/4)(1 + x/12)^{2}

So, *e* = elongation = ∫_{0}^{24} P/[(πE/4)(1 + x/12)^{2}] = 0.00000034P = 0.00780.

So, P = 22,900 lb

Maximum stress occurs where cross-sectional area is minimum, that is, cross-sectional area with diameter 1.

So, maximum stress, σ_{max} = P/[(π/4)(1)^{2}] = 22,900/0.786 = 29.135 lb/ in^{2}.

The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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