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Determine reactions AY, BY, AX, BX. Calculate normal force, shear force and bending moments for diagrams and draw it. All needed informations: Given: L= 7meter H1= 1meter H2= 3meter a= 1meter b= 3meter F1= -10kN F2= 6kN P1= -3kN/meter P2= 2kN/meter

Determine reactions AY, BY, AX, BX. Calculate normal force, shear force and bending moments for diagrams and draw it. All needed informations: Given: L= 7meter H1= 1meter H2= 3meter a= 1meter b= 3meter F1= -10kN F2= 6kN P1= -3kN/meter P2= 2kN/meter

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Determine reactions AY, BY, AX, BX. Calculate normal force, shear force and bending moments for diagrams and draw it.

All needed informations:

Given:

L= 7meter

H1= 1meter

H2= 3meter

a= 1meter

b= 3meter

F1= -10kN

F2= 6kN

P1= -3kN/meter

P2= 2kN/meter

P₂ F₂₁ L b LL~ F₂ ✓² P₁ H1 *
Transcribed Image Text:P₂ F₂₁ L b LL~ F₂ ✓² P₁ H1 *
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Is this right because the answers is different?

2K Ay -Ax 10 उन 16 1.5 1.5 1.5 1.5 74. Br 3E Bending moment at Hinge = 0 >-(Ayx 4) + (A₂x4) = (2x4) 1 - (10x1.5) and, EMB: 0 > - (Ayx 7) + (A₂x¹) = -(2x4) (1-1)-(10×4.5) + 6× 1·5) ..Ay = 5.125 KN, A₂ = 5.375 km Fy=0 ⇒ 5.125+ 6-10 = 1.125 KN = EF=0> B₂ = (3x3) + 5.375-(2x4) - 6.375 KN.
Transcribed Image Text:2K Ay -Ax 10 उन 16 1.5 1.5 1.5 1.5 74. Br 3E Bending moment at Hinge = 0 >-(Ayx 4) + (A₂x4) = (2x4) 1 - (10x1.5) and, EMB: 0 > - (Ayx 7) + (A₂x¹) = -(2x4) (1-1)-(10×4.5) + 6× 1·5) ..Ay = 5.125 KN, A₂ = 5.375 km Fy=0 ⇒ 5.125+ 6-10 = 1.125 KN = EF=0> B₂ = (3x3) + 5.375-(2x4) - 6.375 KN.
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Follow-up Question

Can you please simlify that how you solve these equations to got ay and ax, step by step, thanks. (ay 5.125 kN and AX 5.375 kN)

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Follow-up Question

This is different task, but can u show how to solve like this way?

Ax, Ay: ➡ → Bx, By: ➡ Mnv - Ay 4 m + Ax EMB=- Ay 6 m + Ax Ay 14 m 90 kNm Ay = +6,43 kN (1) 3 m + 30 kN 1 m - 30 kN Mn,v - Ay 4 m + Ax 3 m + 75 kNm = 0 EMB= -Ay · 18 m + Ax 3 m + 165 kNm = 0 Mn,o= + By 2 m + Bx ZMA + By 6 m - Bx 1,5 m +20 kN 1,5 m = 0 0,5 m +20 kN - 3,5 m = 0 sij. Mnv 6,434 m + Ax 3 m + 75 kNm = 0 Ax= -16,43 kN (+) 10 kN/m Mn,o= + By 2 m + Bx 2 m = 0 EMA= - By 12 m + Bx2m + 190 kNm = 0 By 14 m 190 kNm By = + 13,57 kN (1) sij. Mn,o 13,57 kN 2 m + Bx 2 m = 0 Bx=13,57 kN (+) 2 m = 0 1 m - 20 kN 2,5 m - 30 kN - 1,5 m = 0 TAY 1m * 30 kN ·3 Ax TAY • (-2) 1m 1,5 m 10 kN/m C α 20 KN 1,5 m 20 KN 1,5 m 1,5 m * 20 KN E Ax= 16,43 kN Bx 2 m 2 m Bx = 13,57 kN * By = 13,57 kN 1m 2m
Transcribed Image Text:Ax, Ay: ➡ → Bx, By: ➡ Mnv - Ay 4 m + Ax EMB=- Ay 6 m + Ax Ay 14 m 90 kNm Ay = +6,43 kN (1) 3 m + 30 kN 1 m - 30 kN Mn,v - Ay 4 m + Ax 3 m + 75 kNm = 0 EMB= -Ay · 18 m + Ax 3 m + 165 kNm = 0 Mn,o= + By 2 m + Bx ZMA + By 6 m - Bx 1,5 m +20 kN 1,5 m = 0 0,5 m +20 kN - 3,5 m = 0 sij. Mnv 6,434 m + Ax 3 m + 75 kNm = 0 Ax= -16,43 kN (+) 10 kN/m Mn,o= + By 2 m + Bx 2 m = 0 EMA= - By 12 m + Bx2m + 190 kNm = 0 By 14 m 190 kNm By = + 13,57 kN (1) sij. Mn,o 13,57 kN 2 m + Bx 2 m = 0 Bx=13,57 kN (+) 2 m = 0 1 m - 20 kN 2,5 m - 30 kN - 1,5 m = 0 TAY 1m * 30 kN ·3 Ax TAY • (-2) 1m 1,5 m 10 kN/m C α 20 KN 1,5 m 20 KN 1,5 m 1,5 m * 20 KN E Ax= 16,43 kN Bx 2 m 2 m Bx = 13,57 kN * By = 13,57 kN 1m 2m
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Follow-up Question

Is there any simpler way to got AY,AX,BY,BX?

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Follow-up Question

Is this way simplier to got AY, BY, AX,BX find out? Can u simplier that equations, i don't got 1.88kN and 0.38 kN. I only got 1.77 and 2 kN

Thanks

ΣΜΑ = From beam ACDE, 0 0 = Ex4 + Ex4 + (−10) ×2.5 +1.88×- X 9.02 = 4 Ex + 4 Ey ΣΜΒ = 0 From beam EFB, E E = 0 = -Ex3 + Ex3 - 6x1.5 - · (-3) × ²32²2 (−3) ×- -4.5 = - 3 Ex 3 E + 3 E y = 4.123² 2 ...(1) Solving eqn 1 and 2, 1.88 KN (→) 0.38 KN (↓) ..(2)
Transcribed Image Text:ΣΜΑ = From beam ACDE, 0 0 = Ex4 + Ex4 + (−10) ×2.5 +1.88×- X 9.02 = 4 Ex + 4 Ey ΣΜΒ = 0 From beam EFB, E E = 0 = -Ex3 + Ex3 - 6x1.5 - · (-3) × ²32²2 (−3) ×- -4.5 = - 3 Ex 3 E + 3 E y = 4.123² 2 ...(1) Solving eqn 1 and 2, 1.88 KN (→) 0.38 KN (↓) ..(2)
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