1 A rod is composed of aluminum section with diameter of 25mm is rigidlyattached between hallow steel with outside diameter (do) = 25mm and insidediameter (di) = 10mm and solid bronze section with diameter of 25mm asshown in the figure and carries the axial loads. IF P- 3000 N find the axialstresses in each section.600 mm900 mm750 mmSP2PSteelAluminumdo 25mmBranzed= 25mmd. 10mmd-25mm2. From problem no. 1 compute the STRAIN (E) of each section. Where E steet-200,000 MPa, E Alumanum = 70,000 MPa and E tronse = 83,000 MPa

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Answered: LA rod is composed of aluminum section…

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 thick-walled cylinder with closed ends has internal and external radii 100 mm and 180 mm respectively, is subject to internal and external pressures of 148 MPa and 58 MPa respectively. Apply Lame's equations to find the following: Radial and hoop stresses at radii 120 mm, 140 mm and 160 mm respectively. (a) (b) Axial stress of the cylinder.
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2. Refer to the figure below. In addition to the load at point D, if another load Q (Q=10F) along x direction acts at the center of the right end section of bar OABC, solve for the maximum load F with a safety factor of 2.0 using the maximum shear stress theory. Find the stresses at the origin point O. Assume the yield strength of the steel is 42 kpsi with v=0.3. Hints: The centroid of a semicircle lies at a distance of 4r/(3) measured along the vertical radius (r) or y-direction. 2 in. 1-in D. -in R. 1-in D. 12 in 15 in 1-in D.
Q8(): For the truss shown in the Figure, if the diameter of pins at A and G are 3 in and 2 in, respectively. Both stresses in A and 70 kips 50 kips 4 ft B 4 ft are double shear. Determine the shear stresses in pins at A and G and the stresses in 9 ft members AG; AF member GF if their diameter is 4 and in.
Vertical load is applied to point B of a 10 mm radius cylindrical bar as shown in the figure. Calculate the shear stress occurring in the point.
Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in the figure. One load of P=33P=33 kipskips is applied to the assembly at A, two loads of Q=30Q=30 kipskips are applied at B, and two loads of R=35R=35 kipskips are applied at C. The normal stress magnitude in aluminum rod (1) must be limited to 1717 ksiksi. The normal stress magnitude in steel rod (2) must be limited to 3030 ksiksi. The normal stress magnitude in brass rod (3) must be limited to 2121 ksiksi. Determine the minimum diameter required for each of the three rods.
The figure below shows a section of a column subjected to load P, which is applied 50 mm from the midline of the section's web. The column is made from material, which has an allowable normal stress of 180 MPa (both in tension and compression). The area of the section is 10050 mm?; the principal moments of inertia Imar = 146.554x106 mm and Imin 20.076x10° mm*. !! Determine the maximum allowable load P. 15 mm 150 mm 15 mm 150 mm 50 mm 15 mm 100 mm 100 mm 1194.6 kN 363.6 kN 1809.0 kN O 516.4 kN O 189.5 kN
In the figure, there is a steel shaft where moment is applied from the C point. The modulus of elasticity of this shaft is 210 GPa and its shear modulus is 80 GPa. The material of the shaft is Ç1010 and the yield stress is 240 MPa.Find the shaft diameters Dab, Dbc, Dcd, Dde and the torsion angle between A and E according to the Vonmises hypothesis by taking the safety coefficient as 2
The flat bar shown in the figure is subjected to tensile forces P = 3000 lb. The bar has thickness t = 0.25 in. a) If the designer wants to make a stepped bar with shoulder fillets as shown in the figure, determine R the maximum stresses for fillet radii R = 0.25 in. P and R = 0.5 in. if the bar widths are b = 4.0 in. and c = 2.5 in. b) Comment on the relation between the stress and the fillet radius each time it is increased. c) What is the factor of safety for each fillet radius if the part is made of cast iron Sut = 30 ksi? Solution: 3.0 2.8 D/d = 2 1.15 D. 2.6 1.5 2.4 2.2 - 1.05 O nom %3D %3D td к, 2.0 3.0 1.8 E 1.6 E 1.01 1.4 E 1.2 1.0 0.05 0.10 0.15 0.20 0.25 0.30 r/d
So 1 2 and 3
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