LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
89 views
in General by (30.0k points)
closed by
The stiffness matrix of a beam is given as \(K = \left[ {\begin{array}{*{20}{c}} {12}&4\\ 4&5 \end{array}} \right]\)  Calculate the flexibility matrix. Flexibility matrix will be ______.
1. \(\frac{K}{{44}}\left[ {\begin{array}{*{20}{c}} {12}&{ - 4}\\ { - 4}&5 \end{array}} \right]\)
2. \(\frac{K}{{44}}\left[ {\begin{array}{*{20}{c}} {12}&4\\ 4&5 \end{array}} \right]\)
3. \(\frac{1}{{44}}\left[ {\begin{array}{*{20}{c}} {12}&{ - 4}\\ { - 4}&5 \end{array}} \right]\)
4. \(\frac{1}{{44}}\left[ {\begin{array}{*{20}{c}} 5&{ - 4}\\ { - 4}&{12} \end{array}} \right]\)

1 Answer

0 votes
by (53.7k points)
selected by
 
Best answer
Correct Answer - Option 4 : \(\frac{1}{{44}}\left[ {\begin{array}{*{20}{c}} 5&{ - 4}\\ { - 4}&{12} \end{array}} \right]\)

Explanation:

If we denote stiffness matrix as M and flexibility matrix as Δ

It is stiffness matrix, and then flexibility matrix is: Δ = K-1

Calculation:

Δ = \(\frac{1}{{\left( {12\times5 - 4\times4} \right)}}\left[ {\begin{array}{*{20}{c}} 5&{ - 4}\\ { - 4}&{12} \end{array}} \right]\)

∴ Δ = \(\frac{1}{{44}}\left[ {\begin{array}{*{20}{c}} 5&{ - 4}\\ { - 4}&{12} \end{array}} \right]\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...