If the radius of wire stretched by a lo= doubled then its Young's modulus (a) will be doubled (b) will be halved
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- A cylindrical bar of L meters deforms by I cm. The strain in bar is 1. t 2. 011 0.01 1 3. 100 I 4.Question 3: a. A mass of (Z + 50) kg is hung on a steel wire of diameter (Z + 5) mm and the extension found is (Z + 10) mm. If the Young's Modulus of the wire is 200 GPa, Calculate, i. the cross-sectional area of the fishing rope ii. the tensile strain b. A copper sheet of (Z + 75) cm width, (Z + 100) cm height, and (Z + 0.5) cm thick is acted by a shearing force of 50 × 10³ N, as shown in figure below. If the shearing Modulus of Copper is 4.2 x 101°N/m², find the followings: i. Shear Stress ii. Linear Displacement (Ax) Z=62 F Ax thick Width height1.4-7 The data shown in the table below were obtained from a tensile test of high-strength steel. The test specimen had a diameter of 13 mm and a gage length of 50 mm (see figure for Prob. 1.4-3). At fracture, the elongation between the gage marks was 3.0 mm and the minimum diameter was 10.7 mm. Plot the conventional stress-strain curve for the steefor the steel and determine the proportional limit, modulus of elastics of elastic- ity (i.e., the slope of the initial part of the stress-strain,tress-strain curve), yield stress at 0.1% offset, ultimate stress, percent, elongation in 50 mm, and percent reduction in area. 'ess, percent area. TENSILE-TEST DATA FOR PROB. 1.4-7 Elongation (mm) 0.005 0.015 0.048 Load (kN) 5 10 30 50 0.084 60 0.099 64.5 0.109 67.0 0.119 68.0 0.137 69.0 0.160 70.0 0.229 72.0 0.259 76.0 0.330 84.0 0.584 92.0 0.853 100.0 1.288 112.0 2.814 113.0 Fracture
- This question is from the subject Theory of ElasticityA horizontal bar, fixed at one end (x = 0), has a length of 1 m. and cross-sectional area of 100 mm2 Its elastic modulus varies along its length as given by E(x) = 100e-x GPa, where is the length coordinate (in m) along the axis of the bar. An axial tensile load of 10 KN is applied at the free end (x = 1). The axial displacement of the free end is..... mmIf the radius of a wire stretched by a load is doubled, then its modulus of elasticity will be (A) Doubled (B) Halved (C) Remain unaffected (D) Become 4 times
- Can u help meFigure below shows the bar with three equal elements. Use the finite element method and calculate: 2.1 the global stiffness matrix 2.2 the displacement on node 2, 3, 4 2.3 the Strain in each element 2.4 the stresses in each element using Hook's law and compare with theoretical stresses (o=F/A) (1) (2) (3) -50 N 10 mm 10 mm 10 mm A1=50 mm? A2=20 mm? A:=10 mm? E=200 GPaQ.2) A brass bar 500 mm long and 100 mm × 100 mm in cross-section is subjected to an axial pull in the direction of its length. If the increase in volume of the bar is 50 mm³, then the magnitude of the pull (kN) will be (Take E = 100 GPa and µ = 0.25) (A) 20 (С) 25 (В) 30 (D) 15
- A very thick structure is subjected to certain traction boundary conditions on its surface. The cross-section and the applied load do not vary with the z-coordinate. The following stress function is proposed for this problem: -y p(x,y) = Sin (x) (A x²e + B e") (i) use the biharmonic equation to find restrictions, if any, on values of A and B (ii) calculate all stress components (iii) calculate all strain components in terms of A, B, and C as well as the Young modulus and Poisson's ratio E and y, respectively. (iv) check that the equilibrium equations are satisfied (v) determine the traction boundary conditions at x =± a and y=+bH.W 1. Consider a two degree of freedom bar elements as shown in figure. Using finite element method to formulate the equilibrium equation of it. If the cross sectional area is 12 mm and E=200 GN/m². 20KN +30KN 15KN 300 * 600- + 350 All Dimension in mm 2. Consider a two degree of freedom bar element as shown in figure. Using finite element method to formulate the equilibrium equation of it, and then estimate the stress distributions. If Esteel-200 GN/m, Ecopper 110GN/m and EAL= 120 GN/m?. d=3 Steel AL Соpper 20KN - 30KN →15KN 300 * 200 - 400 * 350 All Dimension in mm5. A strain gauge is used to weigh an object that is mounted on a copper column of 6-in diameter. If the weight of the object is 1 pound (i.e. W=1 lb), find the resistance change AR of the strain gauge. (Note: Young's Modulus Elasticity of Copper E =11.73 × 1010 N/m²)