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- The modulus of elasticity of mild steel is equal to .. .? .... .. a. 2 x 10ʻN/m? a. 2 x 10'N/m 2 b. 2 x 10° N/m 4 c. 2 x 10 N/mm 5 d. 2 x 10 N/mm CheckQ/ Find the Maximum stresses in Steel and Aluminum. Est= 210G pa and E al%3D 70 G pa ? W= 15 Kn.M 120mm 5m Al steel 60mm 140 mmPROBLEM 1) An aluminum bar carries the axial loads at the positions shown. If E=70GPA, compute the total deformation of the bar. Assume that the bar is suitably braced to prevent buckling. 0.4m D 10KN 0.8m 0.4m B 5KN 0.6m AL 20KN What is the deformation &pE in mm? A=800 mm² A=1,200 mm²
- SITUATION. A steel plate ABCD of thickness t = 5mm is subjected to uniform stresses: = 100 MPa 0₂ = -60 MPa. Ox E =200GPa, v= 1/3 a = 160mm and b = 200 mm Calculate the change in thickness (mm). -0.0023 mm 0.0023 mm O O 0.0045 mm - 0.0045 mm Ł B AL C D XSituation 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)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 2.5m 2.0m 1.5m
- A composite bar is rigidly attached to the wall at A as shown in the figure. Axial loads are applied at the positions indicated. 3 m 2.5 m C 1200 kN 900 kN 80mmØ 100mmØ E = 83 GPa Bronze E = 200 GPa Steel Which of the following most nearly gives the deformation of C? a. -0.7855 mm О Ъ. +0.8575 mm c. +0.7855 mm o d. - о.8575 mm! Required information A steel bar (Es= 210 GPa) and an aluminum bar (Ea = 70 GPa) are bonded together to form the composite bar shown. 8 mm 8 mm 8 mm Aluminum 24 mm -Steel Determine the maximum stress in steel when the bar is bent about a horizontal axis, with M= 56 N.m. The maximum stress in steel is - MPa.A threaded round steel bar is subjected to a tota axial pull of 8000 lb. If the diameter of the unthreaded portion is 7/8 in., the unit stress in that portion is A. 10,470 psi B. 13,310 psi C. 16,230 psi D. 19,880 psi
- A rectangular steel block is 3.2 inches long in the x-direction, 2.9 inches long in the y-direction, and 4.5 inches long in the z-direction. The block is subjected to a tri-axial loading of three uniformly distributed forces as follows: 48 kips tension in the x-direction, 65 kips compression in the y-direction, and 54 kips tension in the z-direction. If Poisson's ratio is 0.30 and E = 29 x 106 psi, find the uniformly distributed load that must be added in the x-direction to produced no deformation in the z-direction. Your final answer should include two decimal places7. A solid aluminum cylinder is placed within an ASTM A501 steel tube as shown. Compute the stresses in the two materials with an applied load of 45,000 lbs. 45,000 lbs 3.5"Ø 3.0"Ø Steel AluminumA RECTANGULAR ALUMINUM BLOCK IS 40mm LONG IN THE Y DIRECTION,20mm WIDE IN THE Z DIRECTION AND 25mm THICK IN THE XDIRECTION. IT IS SUBJECTED TO A TRIAXIAL LOADING CONSISTING OF AUNIFORMLY DISTRIBUTED FORCE OF Px=3kN (COMPRESSION), Py=4kN(TENSION) AND Pz=7kN (COMPRESSION) IN THE X, Y AND Z DIRECTIONSRESPECTIVELY. IF THE POISSON’S RATIO = 1/4 & E = 70 GPa,DETERMINE THE STRAINS IN THE X, Y AND Z DIRECTIONS. ALSODETRMINE A SINGLE DISTRIBUTED FORCE IN THE Y DIRECTION THEWOULD PRODUCE THE SAME X DEFORMATION AS THE ORIGINAL.