For the rigid jointed frame subjected to temp variation as in fig. Determine total deflection at 'c'.Assume $\alpha=12\times 10^{-6}/^{\circ}c$ depth of members as 600mm.Neglect effect of axial force

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teamques10 ★ 68k

To find vertical difflection at 'c' apply dummy force of 1kN=p. vertically towards down

$\underline{Axial \ force \ \bar(N)\ and \ bending \ moment \ \bar(M) \ diagram \ Applying \ C.O.E}$

$\sum m_{A}=0$ enter image description here

$1\times 4M_{A}=0$

$M_{A}=4kN.m$

$\sum F_{y}=0(\uparrow +ve)$

$V_{A}-1=0$

$V_{A}=1kN$

$\sum F_{x}=0(\rightarrow +Ve)$

$\sum F_{x}=0 \ \ \ \ H_{A}=0$ enter image description here

$\underline{consider \ part \ AB}$:

enter image description here

$\underline{consider \ part \ BC}$:

enter image description here

$\underline{Deflection \ at \ 'C' [neglecting \ axial \ force]}$

$\underline{For \ member \ at \ AB}$

$\Delta t=30-10=20^{\circ}c \ \ \ \ \bar{A}=3\times 4=-12m^{2} \ \ \ \ h=0.6m$

$\underline{For \ member \ at \ BC}$

$\Delta t=40-20=20^{\circ}c \ \ \ \ h=0.6m \ \ \ \ A=\frac{1}{2}\times 4\times 4=-8m^{2}$

$\sum P_{j}d_{j}=\alpha_{t}\left[\frac{\Delta t}{h}.A\right]$

$1\int_{c}=12\times 10^{-6}\left[\frac{20}{0.6}\times(-12)+\frac{20}{0.6}(-8)\right]$

$\int_{c}=12\times 10^{-6}(-4-25.33)$

$\int_{c}=-0.3500$

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