2.70-7120-2x.6251.25.753251.95-R.752xR.375.75S.00-.756.00BRACKETMATERIAL -MILD STEEL

Orthographic Projection It is a form of presenting a three-dimensional object into Two dimensional.…

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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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The wooden frame is constructed from wooden 2” X 4” Lumber (Nominal size; Actual size: 1 ½” X 3 ½”) and bears a uniform distributed load. All connections are pin connected. If wo = 100 lb/ft, and L = 9 ft. Note that the distributed load is the force per unit length in the horizontal direction. A) Find the vertical reaction force at A. 747 450 248 900 563 B) Find the normal stress perpendicular to the grain at point D. -26.4 -33.9 -85.7 -21.4 -66.9 C) Find the shear force, V, on the cross-section at E. 106 900 159 150 35
load (KN) Elongationc PAL 0.5 3-8 7-9 12-9 14-3 15-9 16.9 17-8 19.1 21-2 24-2 26-9 30 32.0 NO S a. 1 0-3 0-5 0-7 0-9 1 1-1 11-3 1-6 1-9 n(mm) 2-2 2-5 28 calculation -compression Stress (6c) = A compression strain (c) = Lo-Lf Lo -young's Modulus (E)=0C Ec hight = 6-8 mm. d = 5.9 mm.
d) A leather round belt transmit power from small pulley rotates 1200 rpm. The diameter of small pulley is one third of the diameter big pulley. The center distance between the pulleys is twice the diameter of big pulley. The stresses in the belt should not exceed 2 MPa and belt density is 950 kg/m?. Due to the friction of rotation, the ratio tension on the tight side to the slack side is 2:1 and coefficient of friction is 0.35. For design requirement, the ratio of diameter leather round belt to diameter of small pulley is 1:18. Determine, i) Contact angle. i) Diameter belt. iii) Belt tension. iv) Torque.
S2 - 9 and 10

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