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Plot the shear and moment diagrams for the beam subjected to the two work. forces and combination of distributed loads. Afterwhen x = 20.6 ft, V = N·m when x 25.9 ft, V lb The maximum (absolute value) shear force in the beam is: The maximum (absolute

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Answer #1

Giun beam uling beami equillibnium 3 KN KN quakion」 Ο.GI KNIN) RA Re 3m 1am 01() (0.5x X2) ( x 12 x 2 x- 3(16) + 3 (27) + dnS5.12 0.61 (x) 빕6 A 3mc 0.61 KN/m SF (X 6.5)5.12 0.61(3,5) 6.5 (0.41) 3.52 SF (X-6.5) m 2.18 KN sr()-14) 5.12 + 0.Gl (11)-14SF(0.6)0.49 KN Bri BF 트 -+ 5.12 (A) + 10.39 (x-1. Brn (x-20·りー 5.12 (20,c) + 10.3จ ( 9.1) _ 3 (4.6)-12 (16.6) 5) -3(n-K)-12(n

Check :

0.00KB/snO 12:31 VOLTE 36% FA Frame1 4. 3.00290.00 .00290.00 5.12 2.51 -4 0.00 1412:31 VOLTE 36% FA Frame1 5.12kN 3.00kN .00290.00 1.05kN 0.49kN 0.49kN -0 2 ENT.39kN -1.95kN 5.12 2.51 4 00 - 2.00 l0.00 2.0012:31 ї.ul 0.00KB/s R 9 田Frame1 3 1664Nm00kNm 11.45k8ms 9.86 00290.00° 3.00290.00 2 2.00 2.51 5.12 7-1 kdm 4 2.00 4.0 00 6.00

Answers :

X in m Shear force in N Bending moment in N - m
1.9 1609 6298
6.5 - 2180 2560
14 120 - 11520
19 120 - 13640
20.6 490 - 13170
25.9 3000 5120
Max 5120 - 13660

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