27-30 Find the temperature distribution in a laterally = 1.158 cm²/sec), insulated thin copper bar (c² = K/po K/po 50 cm long and of constant cross section with endpoints at x = 0 and 50 kept at 0°C and initial temperature
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- A solid spherical ball of magnesium alloy (E = 6.5 × l0-6 psi, v = 0.35) is lowered into the ocean to a depth of 8000 ft. The diameter of the ball is 9.0 in. (a) Determine the decrease ?d in diameter, the decrease, ?V in volume, and the strain energy U of the ball. (b) At what depth will the volume change be equal to 0.0324% of the original volume?There is a gap c = 1mm between the rods in the picture, the temperature changes evenly quantitatively ΔT = 1.9°C. Calculate the normal force on the cross section of the rods. E1 = 200 GPa, L = 1.0 m, E2 = 80 GPa, A1 = 5000 mm2, A2 = 2000 mm2, α1 = 12*10-61/°C, α2 = 8*10-61/°CA cylindrical component 0.3 m long is solid for 1/3 of its length and hollow with inner diameter ½ the outer diameter for the rest of the length. At 300C a gap of 0.1 mm exists between the component and the rigid supports. The outer diameter is 60 mm and the stress is not to exceed 150 MN/m2. Determine the maximum temperature that will not cause failure of the component. Take α as 1.3x10-5/0c and E as 200 GN/m2
- A steel rod, 5m in length, is 100 sq. mm at the cross-section. The rod is in its original length at 20 degrees Celsius. Determine the thermal strain at 70 degrees Celsius if the rod is unrestricted? Use α= 11.7 x 10^-6 per degree CelsiusFor the given plane strain state, use Mohr's circle to determine the strain state associated with the x' and y' axes rotated to θ indicated in the table: \epsilon_x \epsilon_y \gamma_{xy} θ -500\mu 250\mu 120\mu -15°Determine k, thermal conductivity of a wall if q = 1000 kcal/m2 -hr at thickness, k = 33 mm and ∆t = 30°C.
- The two cylindrical rod segments shown in the figure below (Figure 1) are fixed to the rigid walls such that there is a gap of 0.01 in. between them when T1=75 degrees F . Each rod has a diameter of 1.5 in. Take αal=13(10^-6)/∘F ,Eal=10(10^3)ksi, (σY)al=40ksi, αcu=9.4(10^−6)/∘F, (σY)cu=50ksi, and Ecu=15(10^3)ksi. a) What larger temperature T2 is required in order to just close the gap? b) Determine the magnitude of the average normal stress in each rod if T2=350∘FA pressure difference of 35 pa is available to force -10c air. through a circular sheet metalduct 400 mm in diameter and 20m long. Find thevelocity in m/s. A.11.64 B.38.20 C.12.81 D.20.60 SHOW DETAILED SOLUTIONSSolve the preceding problem for an aluminum plate with h = 10 in.. i = 0.75 in., E = 10,600 ksi, v = 0.33. P = 96 kips. Pt,. = 24 kips. and V =18 kips. For part (b) of Problem 7.5-12, assume that the required strain energy stored is 640 in.-lb. In part (c). the change in volume cannot exceed 0.05%.
- Determine the strain energy per unit volume (units of psi) and the strain energy per unit weight (units of in ) that can be stored in each or the materials listed in the accompanying table, assuming that the material is stressed to the proportional limit. DATA FOR PROBLEM 2.7-5 Material Weight Density (lb/in3) Modulus of Elasticity (ksi) Proportional Limit (psi) Mild sleel 0.284 30,000 36,000 Tool steel 0.284 30,000 75,000 Aluminum 0.0984 10,500 60,000 Rubber (soft) 0.0405 0.300 300Rectangular bars of copper and aluminum are held by pins at their ends, as shown in the figure. Thin spacers provide a separation between the bars. The copper bars have cross-sectional dimensions 0.5 in. × 2.0 in., and the aluminum bar has dimensions 1.0 in. × 2.0 in. Determine the shear stress in the 7/16-in. diameter pins if the temperature is raised by 100°F. (For copper, Et= 18,000 ksi and ac = 9.5 × 10-6/?; for aluminum, Ea= 10,000 ksi and aa= 13 × 10-6/?.) Suggestion: Use the results of Example 2-10The statically indeterminate structure shown in the figure consists of a horizontal rigid bar AB supported by five equally spaced springs. Springs l, 2, and 3 have stiff nesses 3k, 5k. and k, respectively. When unstressed, the lower ends of all Five springs lie along: a horizontal line. Bar AB. which has weight W, causes the springs to elongate by an amount S. (a) Obtain a formula For the total strain energy of the springs in terms of the downward displacement d of the bar. (b) Obtain a formula for the displacement S by equating the strain energy of the springs to the work done by the weight W.