Vertical Deflection of C, Structural Analysis PDF
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This document contains a detailed analysis of the vertical deflection of point C, considering the effects of temperature and settlement. Calculations use formulas from structural analysis to determine the deflection values. The structural analysis is focused on identifying the effects of non-uniform temperature changes and site settlement on a structure.
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## Determine Vertical Deflection of c **Due to:** - Non-uniform Temp Tin=10, Tout=20 h: 0.6 - Settlement 1.5 cm at support B ### Due to Temp **Formula:** $I\Delta E = \alpha \ T_{in} - T_{out} \ \int_{N}^{L}N, dL + \alpha \ (T_{in}-T_{out}) \int_{M}^{L} M, dL$ **Calculations:** - $I \Delta...
## Determine Vertical Deflection of c **Due to:** - Non-uniform Temp Tin=10, Tout=20 h: 0.6 - Settlement 1.5 cm at support B ### Due to Temp **Formula:** $I\Delta E = \alpha \ T_{in} - T_{out} \ \int_{N}^{L}N, dL + \alpha \ (T_{in}-T_{out}) \int_{M}^{L} M, dL$ **Calculations:** - $I \Delta E = \frac {1}{2} \ 1.416 \ \frac {1}{2} 20 + 20 \ \frac {1}{2} [ \frac{1}{2} \ \frac{1}{2} 416 \times 1.133 \ \frac{1}{2} - (-\frac{1}{2} \ \frac{1}{2} \ 416 \times 1.133 \ \frac{1}{2}) + \frac{1}{2} \ \frac{1}{2} \ 205 \ \frac{1}{2} - \frac{1}{2} \ \frac{1}{2} 1273 \ \frac{1}{2} ]$ - $I\Delta E = -3.664154 \ m = -0.366 \ mm$ ### Due to Settlement **Formula:** $1 \Delta S + R + Settlement = 0 => 1 \Delta S - 1 * 1.5 = 0$ **Calculations:** - $Sc = 1.5 \ cm$ ## Maximum Deflection ### Double Integration **Formulas:** - $M(x) = -1.67x + 6.67 \ (C)(x-6)$ - $y' = EI \ \int [-1.67x + 6.67 \ (C) (x-6) \ dx$ - $y' = EI \ \int [ \frac{1.67}{2}x^2 - 6.67 \ C(x-6)^2 + C_1] \ dx$ - $y = EI \ \int[ \frac{1.67}{6} x^3 - 6.67 \ C(x-6)^3 + Cx + C_2] \ dx$ **Boundary Conditions:** - at x=0 y=0: $C_2=0$ - at x=6 y=0: $C_1=-10$ **Calculations:** - Max deflection occurs at $y'=0$ - $\frac{1.67}{2}x^2 - 10 = 0$ - $X = 3.46 \ m$ - $\int_{x=3.46}^{}y \ dx = EI \ \int_{x=3.46}^{}[ \frac{1.67}{6} (3.46)^3 - 10 \times 3.46] = \frac{-23.07}{EI} \uparrow$ ## Conjugate Beam **Formulas:** - $10 - \frac{1}{2}x + \frac{1}{2}x \ \frac{lo}{6}x = 0$ - $Moment = Deflection: -10 \frac{3.46}{6} + [\frac{1}{2} \frac{1}{2} \frac{416}{6} \ (3.46)^2 \frac{1}{2} \ \frac{1}{2}] \times \frac{3.46}{3} = \frac{-23.09}{EI}$ ## Virtual Work **Formulas:** - $0 = \frac{X}{6} 2 + \frac{X}{6}2 + \frac{lo}{6}X - \frac{(6-x)}{6} \ [ 2 \times \frac{lo}{6} \frac{(6-x)}{6} \ \frac{1}{2} + 10 \times (1-\frac{6-x}{6})] =0$ - $Y = 3.46 \ + 2 \times \frac{416}{6} \times 3.46 + \frac{10}{6} \times 3.46 - \frac{6-3.46}{6} \times [ 2 \times \frac{416}{6} \times 3.46 \ + \frac{10}{6} \times 3.46 + 4166 ] = \frac{-23.09}{EI}$
