
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Using the LRFD procedure, select the lightest W10, 16 ft long beam-column in an unbraced frame. Based on a first-order analysis, the member supports the following factored loads: Pnt load of 148 k and Plt load of 106 k. The member also supports factored moments:Mntx moment of 92 ft-k and Mltx moment of 64 ft-k.The moments are equal at each end and the member is bent in single curvature. There are no transverse loads between the ends.Kx = 1.5,Ky =1.0,Pstory = 2800 k,Pe storyx = 72,800 k,Cb = 1.0 and Fy = 50 ksi.
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- A thin beam with a solid rectangular cross-section is loaded as shown in Figure Q7 below. W L/3 L/3 L/3 Figure Q7 QUESTION 10 You have been asked to design a beam to be made from a material with a Yield Strength of 360 MPa, using a Factor of Safety of 4 according to the Tresca Criterion. What should the maximum allowable shear stress be when the beam is loaded according to the design specification? Oa. 90 MPa O b.360 MPa Oc 45 MPa Od.180 MPa Oe. None of the provided answers are correct O1. 22.5 MPa Og 720 MPaarrow_forward1) A long span open-web steel joist with a span of 70 feet is required to support a floor. The joists are spaced at 3.5 ft apart, the dead load is 15 lb/ft² (not including the self weight), the live load is 80 lb/ft? and the live load deflection is limited to L/360 (which is that used to determine the live load limit - in red/light - based on deflection in the SJI catalogue tables and those limits must be multiplied by 1.5 if L/240 is used). Using the LRFD table provided, select the most economical joist that can be used considering the self weight. The lighter values are the limiting unfactored live load. (Note: longer spans that can support the load can also be used.)arrow_forwardA) Calculate the stress in each link when a force of 600 lbs is applied to the rigid element AF, employing manual calculations and finite element methods. B) Determine the corresponding deflection at point A, manually applying the finite element penalty method to model and solve the constraints effectively. Available Data: The links BC and DE are made of steel. Each link has dimensions of 1/2 inch in width and 1/4 inch in thickness. Modulus of elasticity (E): 29 x 10^6 psi. Instructions: For this problem, manual calculation is crucial. Utilize the finite element method with the penalty approach to handle boundary conditions and constraints effectively. This method involves incorporating penalty factors into the system equations to enforce the constraints strictly.arrow_forward
- c. Determine critical load and critical stress for the column/axial member. Justify selection of the formulas for the calculationsInformation provided: i) Boundary/end conditions: In this instance, the column is assumed to be fixed at both ends, simulating conditions where the column is integrally connected to the floor slabs above and below, a common condition in many mechanical design contexts. ii) External axial load: Assuming the column supports a floor with a uniformly distributed load, we've calculated the total axial load on the column to be 453.589 kN. iii) Properties of the column: The cylindrical column is composed of steel, a commonly used material with a known Young's modulus (E) of approximately 200 GPa. The steel has a yield stress of around 250 MPa and ultimate stress near 400 MPa. The diameter (d) of the column is 0.3048 m, providing a cross-sectional area (A) of π*(d/2)² = 0.073 m².arrow_forwardNeed help with parts a-earrow_forwardDesign a wide flange W column 8 m in height to support an axial dead load of 1000 kN and a live load of 1200kN in the interior of a building. The column base is rigidly fixed to the footing and the top of the column isrigidly framed to very stiff girders. Assume that bracing is provided to prevent side sway in the weak axis/planeof the column (you don’t have to consider slenderness in this direction), but side sway is not prevented aroundits strong axis. Select an economical W shape.arrow_forward
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