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Manually draw two Shear Force Diagrams and two Bending moment diagrams for the following structure 3200mm 1. For real st

Posted: Fri May 27, 2022 7:27 am
by answerhappygod
Manually Draw Two Shear Force Diagrams And Two Bending Moment Diagrams For The Following Structure 3200mm 1 For Real St 1
Manually Draw Two Shear Force Diagrams And Two Bending Moment Diagrams For The Following Structure 3200mm 1 For Real St 1 (89.89 KiB) Viewed 18 times
Manually draw two Shear Force Diagrams and two Bending moment diagrams for the following structure 3200mm 1. For real structure 1/3 4.21kN 2. Virtual Structure (where you add the 1kN force to the B с supports for calculation of FAA Legend Real Virtual Structure Structure M6x1) mx) Summary of info is below. M(x2) m(x2) Mx3) m3) Mx4) met) x5 . Please make sure they are manually calculated. and annoted to show workings. x6 x7 MX5) m5) Mx6m) Mx7) mx7) Anything other than what I ask for will be DOWNVOTED Bending Moments on Virtual Structure Bending Moments on Real Structure M(x₁) = 0 M(x₁) = 0 m(x₁) = 0 0≤x≤18 0≤x≤ 1.8 M(x₂) = 0 - M(x₂)-2.105x² Mo m(x₂)=x2-1.8 0 ≤ x₂ ≤ 1.8 0 ≤ x₂ ≤ 1.8 Σ M(x)=0 M(x) = -7.578x6.8202 Mo m(x) = -x3-3.6 0 ≤ x ≤ 1.8 0 ≤ x ≤ 1.8 ΣM(x₁) = 0 M(x4)=0 M(x₁) = -4.149x, -20.461 ΣMa m(x) = -5.4 0 ≤ x ≤ 1.8 0 ≤ x ≤ 1.8 M(x) = 0 M(x) = -2.105(xs)² + 4.149x5 - 24.886 Mo m(x) = -5.4 3.2 3.2 0≤x≤ 3 0 ≤ x ≤ 3 M(x6) = 0 M(x) = 8.639x6-40.921 m(x) = -5.4 ΣMa 3.2 3.2 0≤x≤ 3 0≤x≤ 3 M(x)=0 M(x) = -7.578x m(x7) = -x7 Mo 0 ≤ x ≤ 5.4 0 ≤ x ≤ 5.4 Summary of Reactions ΔΑ 999.78 Ax= FAA 194.40 Ay 4.149kN Dx -2.407kN Dy 8.839kN -5.15KN A 1₂ 4 vs " x1 x2 x3 x4