I need assistance with the 'Repeat this step with h=0,3,6,9,12' part, I have attached what I think is the needed data. P

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answerhappygod
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I need assistance with the 'Repeat this step with h=0,3,6,9,12' part, I have attached what I think is the needed data. P

Post by answerhappygod »

I need assistance with the 'Repeat this step with
h=0,3,6,9,12' part, I have attached
what I think is the needed data. Please and Thank you for the
assistance.
I Need Assistance With The Repeat This Step With H 0 3 6 9 12 Part I Have Attached What I Think Is The Needed Data P 1
I Need Assistance With The Repeat This Step With H 0 3 6 9 12 Part I Have Attached What I Think Is The Needed Data P 1 (203.3 KiB) Viewed 39 times
I Need Assistance With The Repeat This Step With H 0 3 6 9 12 Part I Have Attached What I Think Is The Needed Data P 2
I Need Assistance With The Repeat This Step With H 0 3 6 9 12 Part I Have Attached What I Think Is The Needed Data P 2 (1.57 MiB) Viewed 39 times
The frame shown below has a span of 12 m with 4 m high columns. It has a pitched roof of a height h = 2.5 m. Supports A and Care pin supports and B is a pin. Draw the free body diagram of this frame. Then calculate the reactions and draw the shear force and bending moment diagrams of the frame. Repeat the above steps assuming h = 0, 3, 6, 9, and 12 m. Determine the optimum h values to give you (a) the smallest shear force at the top of the columns and (b) the smallest bending moment at the top of the columns. Comment on the results. Repeat the analyses using any analysis computer program. 6 kN/m 6 kN/m B. h = 2.5 m 2 kN/m 4 m G 6 m 6 m Marking scheme Free Body Diagram and Calculation of Loads (2 marks) Analysis (2 marks) Analyses for different h values (2 marks) Optimum h values (1 mark) Comment (1 mark) Computer analyses (2 marks)
0 Given data from the queetion: 6 kN/m 1 L L LT T 2.5m B 6 kN/m IV Le Jet € um 2 kN/m thec Gm BC Free body diagram (F.B.D) of portion 646.8=390W AX + D IN ay 28 Alc ta x 4 cy BD = 1 6242.52 = 6.6 tqr & Mc=0 --C8xYus)-(Byxb) + (34.01.22.62"x3) (Bx X6 x6 Cosa ° + 39 bin 92.62 (u +9)=0 - E -6.5 B4 – bfy +10€ + 78.9820 Bx loe 6.5 Bx + 6By = 186.75 186.75 - f
EG-o I Cx-39 Sin 22.62 + Bx=0 Cx215-Bx - 1 ( ) ۔ Ery=0 0 Cy-39 608 22.62°+ et By = 0 (lit) Cy = 36-By B4 F.B.D B.D of of the portion AB By BX 32 KN V Iz.com 2x4-3kry um Ax 1 "Ay
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