Question

Using slope deflection method, the end rotations at A and B for a horizontal member 4m long with flexural regidity EI are found as EI/100 and - EI/2000 respectively. The beam carries a 80 kN vertical downward load of 1m from A. The support moments at A and B are (assuming clockwise moments and rotations as positive)
$takingEI=5\times {10}^{4}Nm{m}^{2}arerespectively.$

• A. -45 kN.m, +15 kN.m
• B. -7.5 kN.m, +15 kN.m
• C. +30 kN.m, -15 kN.m
• D. +15 kN.m, -15 kN.m

Answer 1

The correct option is A -45 kN.m, +15 kN.m

$M_{AB}=M_{FAB}+\frac{2EI}{L}(2{\theta }_A+{\theta }_B-\frac{3\delta }{L})$

$M_{FAB}=\frac{-80\times 1\times 3^2}{4^2}-45kNm$

$M_{FAB}=\frac{-80\times 30\times 1^2}{4^2}=15kNm$

$M_{AB}=-45+\frac{2EI}{4}[2{\theta }_A-\frac{{\theta }_A}{2}]({\theta }_B=\frac{-{\theta }_A}{2})=-45+\frac{EI}{2}\left(\frac{3}{2}{\theta }_A\right)$

$=-45+\frac{EI}{2}(\frac{3}{2}\times \frac{EI}{1000})$

$=-45+\frac{3}{4}\left(\frac{EI}{1000}\right)EI$

$=-45+\frac{3}{4}\frac{(5\times {10}^4{)}^2}{1000}=-43.125\simeq 45$

$M_{BA}=15+\frac{2EI}{4}(2{\theta }_B+{\theta }_A)$

$=15+\frac{2EI}{4}(-\frac{2{\theta }_A}{2}+{\theta }_A)=15kNm$