Maximum Tensile Stress In Relation To Temperature Drop In A Steel Bar
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Maximum Tensile Stress In Relation To Temperature Drop In A Steel Bar 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/in2.
Coefficient of thermal expansion = 0.0000065/oF.

Suppose bar is subjected to a drop in temperature of 50oF. Determine maximum tensile stress in bar.

The strings: S7P3A31 (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 = ∫024 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, σmax = P/[(π/4)(1)2] = 22,900/0.786 = 29.135 lb/ in2.

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