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Draw the shear and moment diagrams for the beam shown below. 10 kN 4 kN/m 50 kN · m 5 т 3 т

Draw the shear and moment diagrams for the beam shown below. 10 kN 4 kN/m 50 kN · m 5 т 3 т

Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter5: Stresses In Beams (basic Topics)
Section: Chapter Questions
Problem 5.6.13P: A two-axle carriage that is part of an over head traveling crane in a testing laboratory moves...
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EXAMPLE 6.4 Draw the shear and moment diagrams for the beam shown in Fig. 6-7a. 15 kN 5 kN/m 80 kN-m SOLUTION Support Reactions. The reactions at the supports are shown on the free-body diagram of the beam, Fig. 6-7d. 5m -5 m (a) Shear and Moment Functions. Since there is a discontinuity of distributed load and also a concentrated load at the beam's center, two 00 LNam regions of x must be considered in order to describe the shear and moment functions for the entire beam. 5.75 kN (b) 0 s x, < 5 m, Fig. 6-7b: +1£F, = 0; 5.75 kN - V = 0 15 kN 5 kN/m(x, - 5) 80 kN-m V = 5.75 kN (1) (+EM = 0; -80 kN • m – 5.75 kN x1 + M = 0 5m -5x M = (5.75x1 + 80) kN• m (2) 5.75 kN (C) 15 kN 5 kN/m 5 m < x2 s 10 m, Fig. 6-7c: 80 kN-m +1EF, = 0; 5.75 kN – 15 kN – 5 kN/m(x, - 5 m) - V = 0 R| V = (15.75 - 5x2) kN (3) 5.75 kN 34.25 kN (+EM = 0; -80 kN • m – 5.75 kN x2 + 15 kN(x2 – 5 m) V (kN) 5.75 X2 - 5 m x (m) + 5 kN/m(x2 - 5 m) + M = 0 -9.25 M = (-2.5x + 15.75x, + 92.5) kN •m (4) M (kN-m) -34.25 l108.75 Shear and Moment Diagrams. Equations 1 through 4 are plotted in Fig. 6-7d. -x (m) (d) Draw the shear and moment diagrams for the beam shown below. 10 kN 4 kN/m 50 kN · m 5 т 3 т
Transcribed Image Text:EXAMPLE 6.4 Draw the shear and moment diagrams for the beam shown in Fig. 6-7a. 15 kN 5 kN/m 80 kN-m SOLUTION Support Reactions. The reactions at the supports are shown on the free-body diagram of the beam, Fig. 6-7d. 5m -5 m (a) Shear and Moment Functions. Since there is a discontinuity of distributed load and also a concentrated load at the beam's center, two 00 LNam regions of x must be considered in order to describe the shear and moment functions for the entire beam. 5.75 kN (b) 0 s x, < 5 m, Fig. 6-7b: +1£F, = 0; 5.75 kN - V = 0 15 kN 5 kN/m(x, - 5) 80 kN-m V = 5.75 kN (1) (+EM = 0; -80 kN • m – 5.75 kN x1 + M = 0 5m -5x M = (5.75x1 + 80) kN• m (2) 5.75 kN (C) 15 kN 5 kN/m 5 m < x2 s 10 m, Fig. 6-7c: 80 kN-m +1EF, = 0; 5.75 kN – 15 kN – 5 kN/m(x, - 5 m) - V = 0 R| V = (15.75 - 5x2) kN (3) 5.75 kN 34.25 kN (+EM = 0; -80 kN • m – 5.75 kN x2 + 15 kN(x2 – 5 m) V (kN) 5.75 X2 - 5 m x (m) + 5 kN/m(x2 - 5 m) + M = 0 -9.25 M = (-2.5x + 15.75x, + 92.5) kN •m (4) M (kN-m) -34.25 l108.75 Shear and Moment Diagrams. Equations 1 through 4 are plotted in Fig. 6-7d. -x (m) (d) Draw the shear and moment diagrams for the beam shown below. 10 kN 4 kN/m 50 kN · m 5 т 3 т
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