Question

What is the ratio of magnitudes of moments in the member BC at the ends B and C in the given figure below

• A. 1 : 3
• B. 1 : 1
• C. 3 : 1
• D. 3 : 4

Answer 1

The correct option is A 1 : 3

$\text{Stiffness of member BC,}{K}_{BC}=\frac{4EI}{L}$

$\text{Stiffness of member BA,}{K}_{BA}=\frac{3EI}{L}$

$\text{Distribution factors},D_{BC}=\frac{K_{BC}}{K_{BC}+K_{BA}}=\frac{4}{7}$

$D_{BA}=\frac{3}{7}$

$\text{Fixed end moment},M_{FBC}=-\frac{w{\ell }^2}{12};F_{MCB}=\frac{w{\ell }^2}{12}$

Releasing moment for at B, the member end moment for AB at end B

$M_{BA}=\frac{3}{7}\frac{w{\ell }^2}{12}$

$M_{BC}=\frac{4}{7}\frac{w{\ell }^2}{12}-\frac{w{\ell }^2}{12}=\frac{-3w{\ell }^2}{84}$

$\text{With carry over factor of}\frac{1}{2}\text{, the moment at end C of member BC is}$

$M_{CB}=\frac{2w{\ell }^2}{7\times 12}+\frac{w{\ell }^2}{12}=\frac{-9w{\ell }^2}{84}$

$=\frac{|M_{BC}|}{|M_{CB}|}=\frac{|-1|}{|3|}=\frac{1}{3}$