2.64A 2.75-kN tensile load is applied to a test coupon made from 1.6-mm flat steel plate (E = 200 GPa, v = 0.30).Determine the resulting change in (a) the 50-mm gage length, (b) the width of portion AB of the test coupon,(c) the thickness of portion AB, (d) the cross-sectional area of portion AB.2.75 kNA12 mm50 mmB2.75 kN 2.43Each of the rods BD and CE is made of brass (E=105 GPa) and has a cross-sectional area of 200 mm².Determine the deflection of end A of the rigid member ABC caused by the 2-kN load.Answer0.536 mm↓.A2 kNFig. P2.43DOBFEO225 mmC550 mm75 mm100 mm

Answered: 2.64 A 2.75-kN tensile load is applied…

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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1.66 Three forces, each of magnitude P = 4 kN, are applied to the mecha- nism shown. Determine the cross-sectional area of the uniform portion of rod BE for which the normal stress in that portion is +100 MPa. A E Fig. P1.66 B 0.100 m HAI P 0.150 m 0.300 m PIP C 0.250 m D
Do not round intermediate answers. Give your final answer(s) to three decimal places. Check your units. A 24-mm-diameter rod made of an experimental plastic is subjected to a tensile force of magnitude P=9 kN. Knowing that an elongation of 17 mm and a decrease in diameter of 0.57 mm are observed in a 300-mm length, determine: (a) the modulus of elasticity, (b) the modulus of rigidity, and (c) Poisson's ratio for the material. E = G = nu = MPa MPa
A 2.75-kN tensile load is applied to a test coupon made from 1.6-mm flat steel plate (E=200 GPa, ν= 0.30). Determine the resulting change (a) in the 50-mm gage length, (b) in the width of portion ABof the test coupon, (c) in the thickness of portion AB, (d)in the cross-sectional area of portion AB.
An 6 mm diameter aluminum rod has lines scribed on it that are exactly 10 cm apart. After a 2.08 kN force has been applied, the separation between the lines has increased to l' = 10.009 cm. 30 cm 10 cm -F l' a. Calculate the stress and strain in the rod. (Express your answer for stress using two significant figures.) |MPa (Express your answer for strain using four decimal spaces.) E = b. To what total length has the rod stretched? (Express your answer using three decimal places.) L final cm
In a standard tensile test, an aluminum rod of 20-mm diameter is subjected to a tension force of P= 30 kN. It is given that v = 0.35 and E= 70 GPa. 20-mm diameter 150 mm P' Determine the change in diameter of the rod. (Round the final answer to six decimal places.) The change in diameter of the rod is - 0.009207 mm.
A steel bolt has a diameter of 8 mm. The bolt is encased by a bronze tube that has an outer diameter of 15 mm and inner diameter of 8 mm. A force is applied to the bolt. F d3 d1 F Values for the figure are given in the following table. Note the figure may not be to scale. Variable Value F₁ 25 kN Esteel 200 GPa Ebronze 105 GPa a. Determine the magnitude of the normal force in the steel bolt, Nsteel. b. Determine the magnitude of the normal force in the bronze tube, Nbronze. c. Determine the magnitude of the normal stress in the steel bolt, σsteel. d. Determine the magnitude of the normal stress in the bronze tube, σbronze Round your final answers to 3 significant digits. Nsteel = KN Nbronze = KN σ steel = MPa MPa Obronze
1.4 (A). Two forkend pieces are to be joined together by a single steel pin of 25mm diameter and they are required to transmit 50 kN. Determine the minimum cross-sectional area of material required in one branch of either fork if the stress in the fork material is not to exceed 180 MN/m2. What will be the maximum shear stress in the pin? C1.39 x 10e4mZ; 50.9MN/mZ.] 10:41
Two portions of member AB are glued together along a plane forming an angle with the horizontal. The ultimate stress for the glued joint is 3.3 ksi in tension and 2.2 ksi in shear. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 2.0 in. B 2.4 kips 1.25 in. Determine the value of 0 for which the factor of safety of the member is maximum. (Hint: Equate the expressions obtained for the factors of safety with respect to the normal and shearing stresses.)
Aluminum Steel A weight is suspended from two wires. One wire is steel and the other aluminium. The elastic modulus for steel is 192GPA and Aluminium 68GP.. Both wires have the same cross-sectional area 4.2mm² and initial length 2.75m. If the suspended weight is 580N, determine the magnitude of the stress in the steel wire. Enter your answer in MPa to one decimal place. Your Answer: Answer
solve i and ii please.
1200 N 2. For the bone shown to the left, assume the cross-section at C can be approximated as annular, with an outer diameter of 25mm. When the 1200 N forces are applied, the average normal stress in the bone at point C is determined to be 3.80 MPa. Determine the inner diameter of the bone cross section at C. Ans: d; = 14.93 mm 4 in. P 3 in. B 1200 N "UI 8
1.35 A steel pipe of 400-mm outer diameter is fabricated from 10-mm- thick plate by welding along a helix that forms an angle of 20° with a plane perpendicular to the axis of the pipe. Knowing that a 300-kN axial force P is applied to the pipe, determine the nor- mal and shearing stresses in directions respectively normal and tangential to the weld. 10 mm Weld -20° Fig. P1.35 and P1.36

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