0
1.5kviews
Using Moment Area Method OR Conjugate Beam Method, determine in terms of EI, the slope and deflection at the free end C of an overhang beam loaded as shown.

enter image description here

Subject: Structural Analysis 1

Topic: Deflection Of Statically Determinate Structures Using Geometrical Methods

Difficulty: High

1 Answer
0
25views

\begin{array}{l} \sum V=0 \ R_{A}+R_{B}=30 \mathrm{kN} \ \sum M_{A}=0 \ \left(R_{B} \times 4\right)=30 \times 6 \ R_{B}=45 \mathrm{KN} \ R_{A}=-15 \mathrm{KN} \end{array} BM and M/EI diagram In M/EI diagram as:- \begin{array}{l} R_{A}+R_{B}=\frac{1}{2} \times 30 \times 4+\frac{1}{2} \times 60 \times 2 \ R_{A}+R_{B}=\frac{120}{E I} \end{array} \begin{array}{l} \text M_{BLeft}=0 \ \left(R_{A} \times 4\right)=\frac{1}{2} \times 30 \times 4 \times \frac{4}{3} \ R_{A}=\frac{20}{E I} \mathrm{KN} \ R_{C}=\frac{100}{E I}\mathrm{KN} \end{array} \begin{array}{l} \sum M_{C}=0\ M_{C}+\left(R_{A} \times 6\right)-\left(\frac{1}{2} \times \frac{30}{E^{7}} \times 4 \times\left(2+\frac{4}{3}\right)\right)-\frac{1}{2} \times 60 \times 2 \times \frac{4}{3}=0\ M_{c}=\frac{160}{EI}\ \text { Thus, } \Delta_{c}=M_{c}=\frac{160}{EI} \ \theta_{c}=R_{c}=\frac{100}{EI} \mathrm{CW} \end{array}

Please log in to add an answer.