thermal stress

Solution to Problem 266 Thermal Stress

Problem 266
Calculate the increase in stress for each segment of the compound bar shown in Fig. P-266 if the temperature increases by 100°F. Assume that the supports are unyielding and that the bar is suitably braced against buckling.
 

Figure P-266

 

Solution to Problem 265 Thermal Stress

Problem 265
A bronze bar 3 m long with a cross sectional area of 320 mm2 is placed between two rigid walls as shown in Fig. P-265. At a temperature of -20°C, the gap Δ = 2.5 mm. Find the temperature at which the compressive stress in the bar will be 35 MPa. Use α = 18.0 × 10-6 m/(m·°C) and E = 80 GPa.
 

Figure P-265

 

Solution to Problem 264 Thermal Stress

Problem 264
A steel rod 3 feet long with a cross-sectional area of 0.25 in.2 is stretched between two fixed points. The tensile force is 1200 lb at 40°F. Using E = 29 × 106 psi and α = 6.5 × 10-6 in./(in.·°F), calculate (a) the temperature at which the stress in the bar will be 10 ksi; and (b) the temperature at which the stress will be zero.
 

Solution to Problem 263 Thermal Stress

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.
 

Solution to Problem 262 Thermal Stress

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.
 

Solution to Problem 261 Thermal Stress

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.
 

Thermal Stress

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

$\delta_T = \alpha L \, (T_f \, - \, T_i) = \alpha L \, \Delta T$

where α is the coefficient of thermal expansion in m/m°C, L is the length in meter, Ti and Tf 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.
 

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