written 6.8 years ago by | • modified 2.9 years ago |
Subject: Structural Analysis 1
Topic: Deflection of Beams using Energy Method
Difficulty: High
written 6.8 years ago by | • modified 2.9 years ago |
Subject: Structural Analysis 1
Topic: Deflection of Beams using Energy Method
Difficulty: High
written 6.8 years ago by | • modified 6.7 years ago |
1. Reaction
2. For horizontal deflection at C
3. For vertical deflection at C
Region | origin | limits | Mu | Mu Hor. | Mu Vert. |
---|---|---|---|---|---|
AB | A | 0-4 | $$8x-\frac{2x^2}{2}-20$$ | $x-4$ | $-2$ |
BC | C | 0-2 | $-2x$ | $-x$ | $-x$ |
$\begin{align} &\Delta H_C= \int_0^4 (8x-x^2-20)(x-4)\frac{dx}{EI}+\int_0^2 (-2x)(0)\frac{dx}{EI} \\&\phantom{\Delta H_C}=\frac{1}{EI} \int_0^4 \left[8x^2-32x-x^3+4x^2-20x+80\right]dx \\&\phantom{\Delta H_C}=\frac{1}{EI} \int_0^4 \left[-x^3+12x^2-\frac{52x^2}{2}+80\right]dx \\& \Delta H_C=\frac{1}{EI} \left[ \frac{-x^4}{4}+\frac{12x^3}{3}-\frac{52x^2}{2}+80x \right]_0^4 \\& \boxed{ \Delta H_C=\frac{96}{EI}}\ (\rightarrow) \\ \\& \Delta V_C= \int_0^4 (8x-x^2-20)(-2) \frac{dx}{EI}+ \int_0^2 (-2x)(-x) \frac{dx}{EI} \\& \phantom{\Delta V_C}=\frac{1}{EI} \left[ \int_0^4 (-16x+2x^2+40)dx+\int_0^2 2x^2 dx \right] \\& \phantom{\Delta V_C}=\frac{1}{EI} \left[-8x^2-\frac{2x^3}{3}+40x \right]_0^4+\left[\frac{2x^3}{3} \right]_0^2 \\& \boxed{ \Delta V_C=\frac{80}{EI}}\ (\downarrow) \end{align}$