Question

b) A 6 metre long beam is supported at both ends as shown in Figure 1. QA1. Assuming that E and I are constant over the beam’s length, determine the following:

i- Reaction force at point A

ii- Reaction force at point B

iii- Develop the moment expression for the beam.

iv- Calculate the deflection at the location of the vertical force (15 kN) (point C). [16 marks]

a) Describe the main advantage of Macaulay's method. [4 marks] b) A 6 metre long beam is supported at both ends as shown in Figure 1. QA1. Assuming that E and I are constant over the beam's length, determine the following: 1- Reaction force at point A ii- Reaction force at point B iii- Develop the moment expression for the beam. iv- Calculate the deflection at the location of the vertical force (15 kN) (point C). (16 marks] 15 KN 10 KN B D 2m 2m 2m 6m Fig 1. QA1

answer to both questions if possible please

Answer 1

a.) Main advantage of Macaulay's method is that, we can find sope and deflection atomy point in the beam when multiple loads are acting. Because in this method moment ition is valid for whole (b) (1) Determine reaction force at beam. equal length of ISKN LOKN AA D am Sol em te Emx res RATRB = 25 EMASO is x2 Hoxq = RB X6 RB = 11.67 KN RA = 13.33 KN Ans cio Reactim At B RBS U1.67KN Ang Develop a moment expression for beam. Taking section Xbo at distance * M = 13.33 X - 15(x-2) - 10(9-4) Calculate the deplection at point of Ls en load (C) q Mx = 13.334 - 1S (0-2) 10 (2-4) BId2y = Mx Txe (IV EI dy 13.334-15 (n-1) - 10(x-4) da2

Integrating wirita 8.33x - 15 - 2. EI dy dx 13.33x - 10 (x-4379 Integratty again - Eng - 1992 - 15 (2) Jo Chaptents - 2.5(x - 2)² = 1,67 (2-4) tgh the Applying boundary condition EIXY = 2.223 Now- at a=0 y =D x = L y zo L=6m at no ETXO = 2.22x03 +9 to the at n = 6m EIXO 2,22163² – 2.5 (6-2) ? 1.67(6-433+4x6to G -51.027 yxer = 2.2.233 - 2.5(3-23 -1.67(N-433- Now at x = 2 m 51.62782 Y X EI = 2.22 x 2 s-84.234 x 103 ( KN 103N) m ५