Subject: Structural Analysis II

Topic: Deflection of Statically Determinate Structures

Difficulty: Medium / High

To find horizontal deflection at D,apply dummy force p=1kN [Horizontal towards right].

(i)$\underline{Axial \ force \ \bar{(N)} and \ Benging \ moment \ \bar{(m)} \ diagram}$

$\underline{Applying \ C.O.E}:$

$\sum m_{A}=0$

$1\times 2-M_{A}=0$

$\underline{M_{A}=2kN.m}$

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

$V_{A}=0$

$\sum F_{X}=0(\rightarrow +ve)$

$-H_{A}=1kN$

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

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

$\underline{Consider \ part \ CD}:-$

(ii)$\underline{Deflection \ at \ D[Considering \ effect \ of \ axial \ forces]}$

(a)$\underline{member \ AB}:-$

$t_{c}=20+(\frac{30-20}{2})=25^{\circ}c$

$\Delta t=30-20=10^{\circ}c$

$\frac{\Delta t}{n}=\frac{10}{0.4}=25$

$A=\frac{-1}{2}\times 2\times 2+\frac{1}{2}\times 3\times 3=2.5m^{2}$

$\bar{N}=0$

(b)$\underline{Member\ at \ Bc}$:

$t_{c}=15+(\frac{25-15}{2})=20^{\circ}c$

$\Delta t=25-15=10^{\circ}c$

$\frac{\Delta t}{h}=\frac{10}{0.4}=25$

$\bar{N}=1kN$

$A=3\times 4=12m^{2}$

(c)$\underline{member \ CD}:$

$t_{c}=15+(\frac{30-15}{2})=22.5^{\circ}c$

$\\Delta t=30-15=15^{\circ}c$

$\frac{\Delta t}{h}\frac{15}{0.4}=37.5$

$\bar{N}=0$

$A=\frac{1}{2}\times 3\times 3=4.5m^{2}$

$\underline{Using \ Mohr's \ equation}$:

$\sum \bar{P}_{j}\int_{j}=\alpha_{t}\sum \left[\frac{\Delta t}{h}.\bar{A}+t_{c} \bar{N}_{L}\right]$

$1\int_{D}=12\times 10^{-6}\left[(25\times 25)+0+(25\times 12+20\times 1\times 4)+(37.5\times4.5)\right]$

$\int_{D}=12\times10^-{6}[62.5+380+198.75]$

$\int_{D}=12\times10^{-6}[611.25]$

$\int_{D}=7.335\times 10^{-3}m$

$\int_{D}=7.335 mm$

(iii)$\underline{Deflection \ at \ D[Neglecting \ effect \ of \ axial \ force]}$

$\int_{D}=\alpha_{t} \sum \left[\frac{\Delta t}{h}. A\right]$

$\int_{D}=12\times 10^{-6}[62.5+300+168.75]$

$\int_{D}=12\times 10^{-6}(531.25)$

$\int_{D}=6.375m\times10^{-3}mm$

$\int_{D}=6.375m(\rightarrow)$