Using the slope deflection method, determine the final joint moments for the frame shown below. The frame is

manufactured using a constant E$-$value of $206{10}^3$Mpa and an I$-$value of $23\mathrm{.}5{10}^{6}m{m}^4.$ Member A$-$B has an I$-$

value thats $1\mathrm{.}23$ x the I$-$value of member BC$.$

$-6EI\frac{}{L^2}$

Slope Deflection Equations:

MODIFICATION FACTORS:

$\frac{3}{4}$ FOR PIN ENDS & M'ij $=$ Mij$-$Mj$/2$

$M_{LJ}=\frac{3EI}{L}({}_{\text{I}}-)+M_{IJ}^{F^{\text{'}}}$ For pin enus

$M_{AB}=\frac{2EI}{(L)}(2{}_A+{}_B-3)+M_{AB}^F,M_{BA}=\frac{2EI}{(L)}({}_A+2{}_B-3)+M_{BA}^F$

$V_{AB}=\frac{-6EI}{L^2}({}_A+{}_B-2)+V_{AB}^F,V_{BA}=\frac{-6EI}{L^2}({}_A+{}_B-2)+V_{BA}^F$

$3/2$ FOR ANTI $-$SYMMETRY

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