Question

The magnitude of force is member EF for frame shown below will be

• A. $40kN$
• B. $20\sqrt{2}kN$
• C. $20kN$
• D. $60\sqrt{2}kN$

Answer 1

The correct option is D $60\sqrt{2}kN$



Cut a section through BC, BE and EF

$\Sigma F_y=0;\Rightarrow 20+20+F_{EF}\cos \theta +F_{BC}=0....\left(1\right)$

$\Sigma M_E=0;\Rightarrow F_{BC}\times 8+20\times 8=40\times 8$

$F_{BC}=20kN;(\sin \theta =\cos \theta =\frac{1}{\sqrt{2}})(\because \theta =45{}^{\circ })$

$From\left(1\right);F_{EF}=-60\sqrt{2}kN(-vei.e.compression)$

$\therefore Magnitudeof{F}_{EF}=60\sqrt{2}kN$