Determine the vertical displacement of joint C.The members of the truss are 2014-T6 aluminum, 40 mm diameter rods.

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Normal Forces. The normal forces developed in each member of the truss can be determined using method of joints.
Joint A (Fig. a )

[latex]\stackrel{+}{ ightarrow} \Sigma F_{x}=0 ; \quad F_{A C}\left(\frac{3}{5} ight)-15=0 \quad F_{A C}=25 kip ( C ) \+\uparrow \Sigma F_{y}=0 ; \quad 25\left(\frac{4}{5} ight)-F_{A B}=0 \quad F_{A B}=20 kip ( T )[/latex]

Joint B (Fig. b )

[latex]\stackrel{+}{ ightarrow} \Sigma F_{x}=0 ; \quad F_{B C}=0 \+\uparrow \Sigma F_{y}=0 ; \quad 20-F_{B D}=0 \quad \quad F_{B D}=20 kip ( T )[/latex]

Joint C (Fig. c )

[latex]+ earrow \Sigma F_{x^{\prime}}=0 ; \quad F_{C D}=0 \+ warrow \Sigma F_{y^{\prime}}=0 ; \quad F_{C E}-25=0 \quad F_{C E}=25 \operatorname{kip}( C )[/latex]

Axial Strain Energy. [latex]A=\frac{\pi}{4}\left(2^{2} ight)=\pi in ^{2}[/latex] and [latex]L_{A C}=L_{C E}=L_{C D}=\sqrt{3^{2}+4^{2}}=5 ft[/latex].

[latex]\left(U_{i} ight)_{a} =\Sigma \frac{N^{2} L}{2 A E}=\frac{1}{2(\pi)\left[29\left(10^{3} ight) ight]}\left[20^{2}(4)(12)+25^{2}(5)(12)+20^{2}(4)(12)+25^{2}(5)(12) ight] \=0.6224 in \cdot kip[/latex]

External Work. The external work done by the 15 kip force is

[latex]U_{e}=\frac{1}{2} P \Delta=\frac{1}{2}(15)\left(\Delta_{h} ight)_{A}=7.5\left(\Delta_{h} ight)_{A}[/latex]

Conservation of Energy.

[latex]U_{e}=\left(U_{i} ight)_{a} \7.5\left(\Delta_{h} ight)_{A}=0.6224 \\left(\Delta_{h} ight)_{A}=0.08298 in =0.0830 in[/latex]

Question 14.28: Determine the vertical displacement of joint C.The members o...

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