Subject: Structural Analysis II

Topic: Deflection of Statically Determinate Structures

Difficulty: Medium / High

To find the horizontal deflection at A, apply fictitious or Dummy force P=1kN horizontal towards left.

(i)Axial force$\overline{(N)}$ and Bending moment $\overline({M})$diagram:

Applying C.O.E:

$\sum m_{D}=0$

$-1\times3-M_{D}=0$

$M_{D}=3kN.m$

$\sum F_{y}=0$

$v_{D}=0$

$\sum F_{X}=0$

$-1+H_{D}=0$

$H_{D}=1kN$

Consider part 'A B':

Consider part BC:-

Consider part CD:-

$\underline{BM_{c}=1kNm}$

$\underline{BM_{D}=-3kNm}$

(ii)Deflection at A:-[considering effect of axial forces]

(a)$\underline{member \ AB}$:-$(10^\circ c \rightarrow 30^{\circ}c)$ . $t_{c}=$

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

$\frac{\Delta t}{h}=\frac{20}{0.2}=100$

$\bar{A}=\frac{1}{2}\times1\times1=0.5{m}^{2}$

$\bar{N}=0$

b)$\underline{member \ BC}:(10^{\circ}c\rightarrow 40^{\circ}c)$

$t_{c}=10+(\frac{40-10}{2})=25^{\circ}c$

$\Delta t=40-10=30^{\circ}c$

$\frac{\Delta t}{h}=\frac{30}{0.2}=150$

$\bar{A}=1\times4=4m^{2}$

$\bar{N}=1kN$

C)$\underline{member \ Cd:}(10^{\circ}c \rightarrow 30^{\circ}c)$

$t_{c}=10+(\frac{30-10}{2})=20^{\circ}c$

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

$\frac{\Delta t}{h}=\frac{20}{0.2}=100$

$\bar{A}=(\frac{-1}{2}\times 3 \times 3+\frac{1}{2}\times 1\times 1)=-4m^{2}$

$\bar{N}=0$

Using Mohr's Equation

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

$1.\int_{A}=12\times10^{-6}\left[(100\times 0.5+20\times 0 \times 1)_{AB}+(150\times 4+25\times 1\times4)_{BC}+(-100\times 4+20\times 0\times 4)_{CD}\right]$

$\int_{A}=12\times 10^{-6}[(50)+(700)+(-400)]$

$\int{A}=12\times 10^{-6}[350]$

$\int_{A}=4.2\times 10^{-3}m$

$\int_{A}=4.2mm(\leftarrow)$

(iii)$\underline{Deflcetion \ at \ A(Neglecting \ axial \ Forces)}:$

$\sum \bar{P}_{j} \int_{j}=\alpha_{t} \sum [\frac{\Delta t}{h}.\bar{A}]$

$1.\int_{A}=12\times10^{-6}\left[(100\times 0.5)+(150\times 4)+(100\times (-4))\right]$

$\int_{A}=12\times 10^{6}(250)$

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

$\int_{A}=3mm(\leftarrow)$