Problem 263
Steel railroad reels 10 m long are laid with a clearance of 3 mm at a temperature of 15°C. At what temperature will the rails just touch? What stress would be induced in the rails at that temperature if there were no initial clearance? Assume α = 11.7 µm/(m·°C) and E = 200 GPa.
 

Problem 262
A steel rod is stretched between two rigid walls and carries a tensile load of 5000 N at 20°C. If the allowable stress is not to exceed 130 MPa at -20°C, what is the minimum diameter of the rod? Assume α = 11.7 µm/(m·°C) and E = 200 GPa.
 

Problem 261
A steel rod with a cross-sectional area of 0.25 in² is stretched between two fixed points. The tensile load at 70°F is 1200 lb. What will be the stress at 0°F? At what temperature will the stress be zero? Assume α = 6.5 × 10^-6 in/(in·°F) and E = 29 × 10⁶ psi.
 

For two numbers x and y, let x, a, y be a sequence of three numbers. If x, a, y is an arithmetic progression then 'a' is called arithmetic mean. If x, a, y is a geometric progression then 'a' is called geometric mean. If x, a, y form a harmonic progression then 'a' is called harmonic mean.
 

Let AM = arithmetic mean, GM = geometric mean, and HM = harmonic mean. The relationship between the three is given by the formula
 

Below is the derivation of this relationship.
 

Temperature changes cause the body to expand or contract. The amount δ_T, is given by
 

where α is the coefficient of thermal expansion in m/m°C, L is the length in meter, T_i and T_f are the initial and final temperatures, respectively in °C. For steel, α = 11.25 × 10^-6 m/m°C.
 

If temperature deformation is permitted to occur freely, no load or stress will be induced in the structure. In some cases where temperature deformation is not permitted, an internal stress is created. The internal stress created is termed as thermal stress.
 

Problem 257
Three bars AB, AC, and AD are pinned together as shown in Fig. P-257. Initially, the assembly is stress free. Horizontal movement of the joint at A is prevented by a short horizontal strut AE. Calculate the stress in each bar and the force in the strut AE when the assembly is used to support the load W = 10 kips. For each steel bar, A = 0.3 in.² and E = 29 × 10⁶ psi. For the aluminum bar, A = 0.6 in.² and E = 10 × 10⁶ psi.
 

Problem 256
Three rods, each of area 250 mm², jointly support a 7.5 kN load, as shown in Fig. P-256. Assuming that there was no slack or stress in the rods before the load was applied, find the stress in each rod. Use E_st = 200 GPa and E_br = 83 GPa.
 

Geometric Progression, GP
Geometric progression (also known as geometric sequence) is a sequence of numbers where the ratio of any two adjacent terms is constant. The constant ratio is called the common ratio, r of geometric progression. Each term therefore in geometric progression is found by multiplying the previous one by r.
 

Eaxamples of GP:

• 3, 6, 12, 24, … is a geometric progression with r = 2
• 10, -5, 2.5, -1.25, … is a geometric progression with r = -1/2

Arithmetic Progression, AP
Definition

Pythagorean Theorem
In any right triangle, the sum of the square of the two perpendicular sides is equal to the square of the longest side. For a right triangle with legs measures $a$ and $b$ and length of hypotenuse $c$, the theorem can be expressed in the form