2. A square plate in the side of a ship with 800 mm sides parallel to the x and yaxes has a uniform thickness h=10 mm and is made of an isotropic steel(E=200,000 MPa and v-0.29).The plate is subjected to a uniform state of stress. If o 0=02y=0 (plane stress),Oxx= 01=500 MPa, and ɛyy-0 for the plate,Determine oy=02, 03, and the three final dimensions Lx, Ly, and L, of the plate,assuming linearly elastic conditions.

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Answered: 2. A square plate in the side of a ship…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Mohr’s circle is shown for a point in a physical object that is subjected to plane stress. Each grid square is 16 MPa in size. (a) Determine the stresses σx, σy, and the magnitude of τxy and show them on a stress element. (b) Determine the stresses σn, σt, and the magnitude of τnt and show them on a stress element that is properly rotated with respect to the x-y element. The sketch must include the magnitude of the angle between the x and n axes and an indication of the rotation direction (i.e., either clockwise or counterclockwise).
At room temperature (20°C), a 0.5-mm gap exists between the ends of the rods shown. The temperature eventually reaches a value of 140°C. Determine the normal stress in the aluminum rod. Determine the change in length of the aluminum rod.
Part 1. A composite assembly of length, L-375 mm, consisting of a steel (G-77 GPa) core (2) connected by rigid plates at the ends of an aluminum [G-35 GPa) tube (1) is shown. The aluminum tube has outside diameter D₁ - 50 mm and inside diameter d₁-30 mm. = 100 MPa, and the The solid steel core has diameter d₂-25 mm. The allowable shear stress of aluminum tube (1) is (rallow) allowable shear stress of steel core (2) is (Tallow)2= 135 MPa. Determine (a) the allowable torque T that can be applied to the composite shaft. (b) the corresponding torques produced in tube (1) and core (2). (c) the angle of twist produced by the allowable torque T. T D₁ L Calculate the polar moment of inertia, J₁, of the aluminum tube. Answer: J₁ (106) mm².
The state of stress at a point in a member is shown on the rectangular stress element in (Figure 6) where the magnitudes of the stresses are |σx|=19 ksi and |σy|=14 ksi. Determine the state of stress on an element rotated such that the +x′ axis is 40∘ below the −x axis of the original stress element. Answer should be as σx′=, σy′=, τx′y′= in ksi
The temperature of an aluminum bar and a steel bar was raised from 40°C to 160°C. If there was a gap 0.4 mm between the two bars at 40°C as show in figure below, determine the normal stress in the aluminum bar and the change in length of the aluminum bar. 0.5 mm 300 mm 250 mm B Aluminum Stainless steel A = 2000 mm² E = 75 GPa A = 800 mm² E = 190 GPa a = 17.3 x 10-6/°C %3D %3D a = 23 X 10-6/°C
-8 in.- -8 in.- (a) (b) Which of the following best explains why cylinder (b) has a higher longitudinal stress in its walls compared to cylinder (a)? Assume the gauge pressure of the fluid inside the cylinder is non-zero. O Cylinder (b) is supported at two locations Cylinder (a) is not free to deform in its longitudinal direction The vertical (longitudinal) pressure in cylinder (a) is transferred directly to the [bottom] support, bypassing the cylinder walls None of the above the longitudinal stress for both cylinders is the same
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