Question

Consider the deformable pin-jointed truss with loading, geometry and section properties as shown in the figure.


Given that
$E=2\times {10}^{11}N/m^2,A=10m{m}^2,L=1m\text{and}P=1kN$

The horizontal displacement of joint C (in mm, up to one decimal place) is______

• A. 2.7

Answer 1

The correct option is A 2.7

Analysis of truss under external loading


Analysis of truss under unit load in desired directions

$\begin{array}{cccccc}& \text{Force due}& \text{Force due}& & & \\ \text{Member}& \text{to external}& \text{to unit load}& \text{Length}\left(l\right)& \text{Axial}& \frac{F{F}^{\prime }l}{AE}\\ & \text{loading}\left(F\right)& \text{in the desired}& & \text{rigidity}& \\ & & \text{direction}\left(F^{\prime }\right)& & & \\ BC& -3P& -1& L& AE& \frac{3PL}{AE}\\ AC& P\sqrt{2}& \sqrt{2}& L\sqrt{2}& 2AE& \frac{\sqrt{2}PL}{AE}\\ AB& -P& -1& L& AE& \frac{PL}{AE}\end{array}$

$\sum \frac{F{F}^{\prime }l}{AE}=\frac{5.414PL}{AE}$

$\therefore \text{Horizontal displacement joint C}$

$=\frac{5.414PL}{AE}$

$(P=1kN,L=1m,A=10m{m}^2,E=2\times {10}^{11}N/m^2)$

$\therefore \Delta =\frac{5.414\times {10}^3\times {10}^3}{10\times 2\times {10}^5}=2.7mm$