Question

Draw the shear and moment diagrams for the loaded beam. After you have the diagrams, answer the questions as a check on your work.

Question 7 Draw the shear and moment diagrams for the loaded beam. After you have the diagrams, answer the questions as a check on your work. 9 KN 7 KN 6 kN/m 4 kN/m e B- ham tomt9m— 5m Questions: When x = 2.3 m, V = KNm When x = 8.6 m, V = kNm When x = 12.5 m, V = KN-m When x = 22.2 m, V = | KN M= KN.m The maximum (absolute value) shear force in the beam is:

Answer 1

SKN 6 KN/m 7 KN 4 kN/m 4m 6 m 9m sm Vin) ven) 30:37 | 30-37 kn 6-37 -2.63 -56-63 M(N) 231-68 KN-m 121.47 Support reactions: EFy=o: EMA. 20 Ay + By - (4xG) + 9 + (6*9) + 7 = porn (4X6)*7 + (x10) + (946) (14.5) +(7x 24) - Bulien By = 63-63 kW Ays 30.37 KN

Shear 4 bending equations! For span osh.com Van 30.37 KN V(2-3) = 30-37 KN MCHU = 30-37 01 M (2-3) = 69.851 K N-m Far span 4 ms H2<lom V (72) E 30.37 - 4(02-4) (8.6) = 11-37 KN M(N2) = 30-3702 - 2 (012-4) M(8-6) = 218.862 KN-m *) For span som s oz<iom V (03) = 30-31 - 4(6) - 9 -6 (013-10) M(83) = 30-37013 - 4 (6) [063-73 9 (013-10) – 6 (023-1012 V(12-5) = -17-63 KN M (12-5) = 206.375 kN-m For sban 19ms r4 < 24 m V (114) = 30-37 - 4(6--6 (9) +63.63 = 7 KN M(N) = 30-37014 - 4(6) [04-7] - 9 (074-10) - 6(9) (na-uuri +63-63 (04-137

V (22-2) = 7 KN M (22-2) = -12.57 KN-m Final answers! : When en r=2-3 m > lv = 30. 37KNI AM= 69.851 KN-m When no 8.6 m > V = 11.97 KN 4= 218-862 kW-m when r = 12-5 m = v= -17.63 KN Me 206-375 KNM When r= 22. 2m - Tv= 7KN ME-12-57 K N-m The max.(absolute value) shear force: - I Vmax l = 56-63 KN . The max. (absolute value) bending moment. 1 Mmax) = 231-68 kN-ml