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User: stanzaa37

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stanzaa370
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Posts by stanzaa37

<prev • 19 results • page 1 of 2 • next >
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Answer: A: Analysis the Continuous beam loaded and supported as shown in fig 1)Using clapey
... ![enter image description here][1] 1)Free Bending moment For AB, BM at $E=\frac{wl^{2}}{8}=\frac{24\times3^{2}}{8}=27kN.m$ For BC ![enter image description here][2] $\ \ \ M_{L}=-7.5\times2 =-15kN$ $M_{R}=7.5\times2$ $=15kN.M$ For CD BM at G=$\frac{wab}{l}=\frac{15\times2\times3}{5}=30kN ...
written 2 days ago by stanzaa370
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Analysis the Continuous beam loaded and supported as shown in fig 1)Using clapeyron's theorem of three moments method Note that support C settles by 8mm during loading Take $EI=1600kN/M^{2}$
... ![enter image description here][1] **Subject**: Structural Analysis II **Topic**: Flexibility Method **Difficulty**: Medium / High [1]: https://i.imgur.com/jTgUvg9.png ...
sa2(75) written 2 days ago by stanzaa370
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Answer: A: A Propped Cantilever of Span 4m carries a UDL of 5kN/m over entire Span calculat
... As a support is fixed support. Therefore we have to use an imaginary span near the fixed end as shown in fig 1) Free bending moment For Span AB BM at C=$\frac{wL^{2}}{8}=\frac{5\times(4)^{2}}{8}=10kN.m$ 2) Free Bending Moment diagram ![enter image description here][1] 3) Applying theorem of th ...
written 2 days ago by stanzaa370
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A Propped Cantilever of Span 4m carries a UDL of 5kN/m over entire Span calculate fixed end moment by using clapeyron's theorem of three moment
... ![enter image description here][1] **Subject**: Structural Analysis II **Topic**: Flexibility Method **Difficulty**: Medium / High [1]: https://i.imgur.com/xOpgYRq.png ...
sa2(75) written 2 days ago by stanzaa370
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Answer: A: Analysed the beam by using Three-moment theorem suppport B sinks by 5mm $I=9300c
... Free Bending Moment: For Span AB BM at $E=\frac{wab}{l}=\frac{80\times2\times4}{6}=106.67KN.m$ For Span BC BM at $F=\frac{wl^{2}}{8}=\frac{20\times5^{2}}{8}=62.5kN.m$ For Span CD BM at $G=\frac{wab}{l}=\frac{60\times3\times2}{5}=72kN.m$ Support B sinks by 5mm =0.005m=-0.005(Downward) $ ...
written 2 days ago by stanzaa370
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Analysed the beam by using Three-moment theorem suppport B sinks by 5mm $I=9300cm^{4} E=2.1\times10^{5}N/mm^{2}$
... **Subject**: Structural Analysis II **Topic**: Flexibility Method **Difficulty**: Medium / High ![enter image description here][1] [1]: https://i.imgur.com/PzcbdYt.png ...
sa2(75) written 4 days ago by stanzaa370
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Answer: A: Analyse the contineous beam by using three moment theorem
... 1)Free Bending moment - For Span BC at $F=\frac{wl^{2}}{8}=\frac{2\times(6)^{2}}{8}=9kN.m$ - For Span CD at $G=\frac{wl^{2}}{8}=\frac{2\times(8)^{2}}{8}=16kN.m$ 2) Free Bending moment diagram ![enter image description here][1] 3) Find Reaction at Support:- ![enter image description he ...
written 4 days ago by stanzaa370
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Analyse the contineous beam by using three moment theorem
... ![enter image description here][1] **Subject**: Structural Analysis II **Topic**: Flexibility Method **Difficulty**: Medium / High [1]: https://i.imgur.com/nA87an9.png ...
sa2(75) written 4 days ago by stanzaa370
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Answer: A: Analyse the contineous beam by using three moment theorem
... 1) Free Bending moment: [considering each span simply supported] - For Span AB B mat $E=\frac{wab}{l}$ =$\frac{80(2)(4)}{6}=\underline{106.6 kN.m}$ - For Span BC B mat $f=\frac{wl^{2}}{8}$=$\frac{20(5)^{2}}{8}=\underline{62.5 kN.m}$ - For Span CD B mat $G=\frac{wab}{l}$=$\frac{60(2)(3)}{5}= ...
written 4 days ago by stanzaa370
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Analyse the contineous beam by using three moment theorem
... ![enter image description here][1] **Subject**: Structural Analysis II **Topic**: Flexibility Method **Difficulty**: Medium / High [1]: https://i.imgur.com/rQfWc7C.png ...
sa2(75) written 4 days ago by stanzaa370

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