Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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- For the simply supported beam subjected to the loading shown, derive equations for the shear force V and the bending moment M for any location in the beam. (Place the origin at point A.) Let a-3.25 m, b=4.75 m, Pg - 35kN, and Pc = 80kN. Construct the shear- force and bending-moment diagrams on paper and use the results to answer the questions in the subsequent parts of this GO exercise. A Ay- 58.66 - Dy- Calculate the reaction forces A, and Dy acting on the beam. Positive values for the reactions are indicated by the directions of the red arrows shown on the free-body diagram below. (Note: Since Ax = 0, it has been omitted from the free-body diagram.) Answers: a 56.33 (a) V= (b) V- (c) V- B i i PB B kN с kN C Determine the shear force acting at each of the following locations: (a) x-2m (b)x - 4 m (c) x-8 m Note that x = 0 at support A. When entering your answers, use the shear-force sign convention detailed in Section 7.2. 3 3 3 KN D b kN D ·x Dyarrow_forwardFor the loaded beam shown in Figure below : 2.1.Determine the reaction forces at the points of support of the beam in Figure Q2 above by applying the principle of equilibrium of moments and start your solution by drawing free body diagram of the beam. 2.2. Develop the equations for shear force and bending moment for the spans AB, BC and CD of the loaded beam shown above and for each equation developed, determine the values of shear forces and bending moment at the points A, B, C and D 2.3. Draw the shear force and bending moment diagrams for the loaded beam shown above.arrow_forwardanswer both partarrow_forward
- Consider the system shown in FIGURE P3-44 of the textbook. The designer has made several changes to the system. These changes are: • the load of 6 kN/m has been changed to 11 kN/m, • the central length of 2 m has been changed to 4 m o this means the load is applied over the beam from 1 m to the right of point 'A' TO point 'C'. What is the magnitude of the resultant of the distributed load? O 5 kN O 55 kN O 0.513 kN O 93.6 kN O 11 kN O 12.1 kN O 91.4 kN O O kNarrow_forwardShort answer pleasearrow_forward
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