200 lbapplied to end D of lever1.5indeter-A single horizontal force P of magnitudeABD. Knowing that portion AB of the lever has a diameter ofmine (a) the normal and shearing stresses on an element located at point Hand having sides parallel to the x and y ares, (b) the principal planes and theprincipal stresses at point H.10 in.4 in.B18 in.1.2 in.

Answered: 200 lb applied to end D of lever 1.5 in…

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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A single horizontal force P with a magnitude of 150 lb is applied to end Dof lever ABD. Knowing that portion AB of the lever has a diameter of 1.2 in., determine (a) the normal and shearing stresses located at point H and having sides parallel to the x and y axes, (b) the principal planes and princi-pal stresses at point H.
1. Determine the equivalent state of stress on an element at the point which represents, the principal stresses and, the maximum in-plane shear stress and the associated average normal stress. Also, determine the corresponding orientations of the element. a. Use Equation Method b. Use Mohr's Circle 28 MPa 14 MPa 7 MPa
Draw the moh'r circle
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Q1: A 6-kips force is applied to the machine element AB as shown. Knowing that the uniform thickness of the element is 0.8 in., determine the combined normal stresses at points (a) and (f). 1.5 in. 1.5 in. B a ob С 8 in. □e 0 f 8 in. 6 kips 35° 8 in.
Ex. 5.3 At a point in the structural member, the stresses are represented as shown below. Employ Mohr's circle to determine (a) the magnitude and orientation of the principal stresses and (b) the magnitude and orientation of the maximum shear stress and associated normal stresses. In each case, show the results on a properly oriented element. Sol. YA 40 MPa 30 MPa 80 MPa
ngure The forces aro applied to the bar shown. Determine the normal and shearing stresses at (a) point a, (b) point b, (c) point c. 80 mm 28 mm 20 mm 170 mm 50 mm 650 N 16 mm 32 mm 40 mm 550 N 10 kN
Q1: A 6-kips force is applied to the machine element AB as shown. Knowing that the uniform thickness of the element is 0.8 in., determine the combined normal stresses at points (a) and (f). 1.5 in. 1.5 in. B a L ob 0 C 8 in. De □ 8 in. A 6 kips 35° 8 in.
Three forces are applied to the machine component ABD as shown. Knowing that the cross section containing point His a 20 x 40-mm rectangle, determine the principal stresses and the maximum shearing stress at point H. Take force Q=2.9 kN. (Round the final answers to two decimal places.) y 50 mm 2.5 kN 20 mm MPa MPa 150 mm 40 mm 0.5 KN 160 mm The principal stresses at point Hare as follows: da ab= The maximum shearing stress at point His( MPa.
For the structure shown, determine the following: a. The normal and shear stresses in section (ksi) b. The maximum normal stress in member BC (ksi) c. The maximum bearing stress on member AB (ksi) Pins A & C have a diameter of 34 in while pin B has a diameter of ½ in. Both members have a thickness of 0.4 in. There are two support plates at A. 1.5 in. B C 60° 8 ft 2 kip/ft A
A pressure vessel of 225-mm inner diameter and 6-mm wall thickness is fabricated from a 1.2 m section of spirally welded pipe AB and is equipped with two rigid end plates. The gage pressure inside the vessel is 2 MPa, and 45-kN centric axial forces P and P' are applied to the end plates. P' A 1.2 m 35⁰ B Determine the shearing stress parallel to the weld. The shearing stress parallel to the weld is P MPa.
Problem 2. Three loads are applied to the short rectangular post shown In Fig. 2(a). The cross- sectional dimensions of the post are shown in Fig. 2(b). Determine: Me (a) The normal and shear stresses and points H and K; (b) The principal stresses and maximum in-plane shear stress at point H, and show the orientation of theses stresses in an appropriate sketch. No. M2 y 50 mm 210 kN %3D 120 mm (Naー 75 mm K 65 kN 30 mm 95 kN 160 mm 150 mm 30 mm Y-axis is the w to the sectin paimt Hond E امقره Fig. 2(b) 文 Section Proputics A= 120 X160=19200 mm Lx=160 - 40960000 mm-4094r L= 160x128 5KNス50 m-) %3D %3D = 23040000mm'= 23; 40Xlʻmm7 Int Faices

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