In the figure shown, `A` and `B` are two short steel rods each of cross-sectional area `5 cm^(2)`. The lower ends of `A` and `B` are welded to a fixed plate `CD`. The upper end of `A` is welded to the `L`-shaped piece `EFG`, which can slide without friction on upper end of `B`. A horizontal pull of `1200 N` is exerted at `G` as shown. Neglect the weight of `EFG`.

Longitudinal stress in `B` is

Answer

C

Explanation

Here first of all draw the free body diagram of the steel rods and the `L`-shaped structure as shown in the figure.
The direction of `N_(1)` and `N_(2)` can be found by using the equilibrium codition.
For `L`-shaped structure
For vertical equlibrium `N_(1)=N_(2)`

For horizontal eqilibrium `F=1200N`
For rotational equilibriumj, torque about `E` (or any other point) `=0`
`impliesN_(2)xx4-1200xx3=0`
`N_(2)=900N`
Shearing stress is `A=F/A=1200/5N//m^(2)=240N/cm^2`
Shearing stress in `B=0`
Longitudinal stress in `A=(N_(1))/A=180N//cm^(2)`
Longitudinal stress in `B=(N_(2))/A=180N//cm^(2)`
(compressive in nature)