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- An aluminum rod is rigidly attached between a steel rod and a bronze as shown. Axial loads are applied at the positions indicated. Find the stress of each rod in MPa. Steel Aluminum Bronze A- 500mm A = 400mm? A- 200mm 40KN 10KN 20KN 1.5m 2.5m 2.0m Determine the change in length of steel rod having a length of 800mm and a diameter of 6.5mm. The rod is subjected to a force P equal to 110KN. Young's Modulus is 200GPA Compute normal strain of the rodSituation 1: A rigid bar ABC weighs 10 kN/m. The compressive stress of the copper bar is 20 MPa. Answer the following questions: 3m B 2 m Copper L=3m A = 1500 mm² Brass C L = 2 m A = 1200 mm² A. What is the stress of the brass bar? Indicate if it will experience tension (T) or compression (C) (MPa) B. What is the minimum required diameter of the pin at A assuming that the pin connection is in double shear? Consider the allowable shearing stress of the pin as 15 MPa. Round your answer in a multiple of 5. (mm)2. A rod is composed of an aluminum section rigidly attached between steel and bronze sections as shown in the figure. Axial loads are applied at the positions indicated. The cross-sectional area of the rod is 300sq.mm. Express your answer in MPa. Determine the stress in the steel. b. a. Determine the stress in the aluminum. Determine the stress in the bronze. C. STEEL ALUMINUM BRONZE 56kN 14kN -0.6m- -0.9m- -0.76m-
- 2. The bronze bar 3 m long with a cross-sectional area of 350mm2 is placed between two rigid walls. At a temperature of -20 oC, there is a gap of 2.2 mm. Find the temperature atwhich the compressive stress in the bar will be 30 MPa ifthe bar will experience:a. No tensile.b. With an tensile of 5,000 N.Use α = 18.0 x 10-6 /oC and E = 80 GPa.2. Axial loads are applied to the compound rod that is composed of an aluminum segment rigidly connected between steel and bronze segments. What is the stress in each material given that P = 10 kN? Specify whether tension or compression. Bronze Aluminum Steel A =400 mm2 A= 600 mm2 A = 300 mm2 2P ЗР 4P P -3 m 5 m 4 m Stress in Bronze (MPa) %D Stress in Aluminum (MPa) %3D Stress in Steel (MPa)1/2 Problem 1: Two solid cylindrical rods AB and BC are welded together at B and loaded as shown in. If the stress must not exceed 14OMPA in either rod, determine the smallest allowable value of the diameter of rod AB and ВС. 300mm B 40 kN 250mm - d2 30 kN
- GIVEN: Bar has a cross-sectional area of 4 in? and is placed between two rigid immoveable supports when it is 30.00 inches long at a temperature of -20°F. REQ’D: A) The stress in the bar at 90°F if it were made of steel. Esteel = 10 x 10° psi asteel = 13 x 106/°F B) The final length of the aluminum barAn aluminum rod is rigidly attached between a steel rod and a bronze as shown. Axial loads are applied at the positions indicated. Find the stress of each rod in MPa. Steel Aluminum Bronze A- 500mm? A = 400mm? A - 200mm 40KN 10KN 20KN 2.5m 2.0m 1.5mThe light rigid bar ABCD shown below is pinned at B and connected to two vertical rods. Assuming that the bar was initially horizontal and the rods stress-free, determine the stress in each rod after the load P = 20 kN is applied. steel L=1.5 m A=300 mm2 2 m- E= 200 GPa B D Aluminun L=2 m P = 20 kN - 1m A= 500 mm2 E= 70 GPa
- CLO 1: A rigid beam rests on two bronze bars 700 mm long. Another aluminum bar is placed between them which is 699.93 mm long. If the weight of the beam is 375 KN, determine the stress in the aluminum bar (in MPa). Bronze Bronze Alum Alum Bronze A in mm2 850 1500 E in Gpa 70 83 103.56 O 79.45 451.30 33.473) An bronze rod is rigidly attached between a aluminum rod and a steel rod as shown. Axial loads are applied at the positions indicated. Find the maximum allowable value of P that will not exceed a stress in steel of 140MPA, in aluminum of 90MPA or in bronze of 100MPа. Aluminum A = 500mm? Bronze A = 200mm? Steel A = 150mm2 2P ЗР L st = 1.2m L br = 2m Lal = 3.5mSituation 2 For the assembly shown, determine the stress in each of the two vertical rods if the temperature rises 40°C after the load P=50 KN is applied. Neglect the deformation and the mass of the horizontal bar AB. Compute the vertical movement under the load. Steel Aluminum E=200 GPa E=70 GPa L-4m A=300 mm? a-11.7x10*/°C L-3m A=900 mm a=23x10“ /°C 3 m 3 m 3 m Situation 3