Question

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A cantilever beam is subjected to a linearly distributed load, with W, = 10 kN/m and to an inclined point load F equal to 20 kN, as shown in the figure. The length of the beam is L=10 m. Make a cut at distance x from the free end of the cantilever, as shown in the figure, and use the method of sections to derive expressions for the internal resultant loadings at the cross-section as functions of x a. Normal force, N(X) b. Shear force, V(x) c. Bending moment, MX) 3 X 4

Answer 1

Draw the free body diagram of the beam as follows: 10 KN 3 4 X 20 KN 10 m The vertical reaction at point A is, Ex10x 10+ 20 3 V32 + 4 4 = 62 kN The horizontal reaction at point A is, 4 =-20% *** 4 =-16 kN The moment reaction at point A is, M.-3x10x10x£x10+ 20% Top 2 x 10 M = 453.33kN-m Calculate the load at section-x, use similar triangles. 10 W 10 (10-x) w = (10-x) Cut the section-x and draw as follows: 10-x 10 kN À 20 KN 4 Calculate the normal force as follows: N = -20%- 132 + 4 N =-16kN Calculate the shear force as follows: 3 10+(10-x) V. = 20 V32 +47 V = 12+10x-0.5x² KN -XX Calculate the bending moment as follows: M. = 20x2 ***«(10-(10-x)xxxşxx+(10-x)x.xx M = 12x+0.33x +5x2 -0.57 M = 12x + 5x2 -0.17%