Question

a bracket is shown in the figure, the cross - section area of the bar bc is: a₁ = 100mm², the cross - section area of the bar ab is: a₂ = 200mm², and q = 10kn, determine the normal stress of the bars

Explanation:

Step1: Resolve forces at joint A

Let the force in bar BC be $F_1$ and in bar AB be $F_2$. Taking vertical - direction equilibrium at joint A, $\sum F_y = 0$. We have $F_1\sin30^{\circ}-Q = 0$.
$F_1=\frac{Q}{\sin30^{\circ}}$
Since $Q = 10\times10^{3}N$ and $\sin30^{\circ}=0.5$, then $F_1=\frac{10\times 10^{3}}{0.5}=20\times10^{3}N$. In the horizontal - direction, $\sum F_x = 0$, so $F_2 - F_1\cos30^{\circ}=0$, and $F_2 = F_1\cos30^{\circ}=20\times10^{3}\times\frac{\sqrt{3}}{2}=10\sqrt{3}\times10^{3}N$.

Step2: Calculate normal stress in bar BC

The formula for normal stress is $\sigma=\frac{F}{A}$. For bar BC, $A_1 = 100mm^{2}=100\times10^{-6}m^{2}$ and $F = F_1 = 20\times10^{3}N$.
$\sigma_1=\frac{F_1}{A_1}=\frac{20\times10^{3}}{100\times10^{-6}}=200\times10^{6}Pa = 200MPa$.

Step3: Calculate normal stress in bar AB

For bar AB, $A_2 = 200mm^{2}=200\times10^{-6}m^{2}$ and $F = F_2 = 10\sqrt{3}\times10^{3}N$.
$\sigma_2=\frac{F_2}{A_2}=\frac{10\sqrt{3}\times10^{3}}{200\times10^{-6}}=50\sqrt{3}\times10^{6}Pa\approx86.6MPa$.

Answer:

The normal stress in bar BC is $200MPa$ and in bar AB is approximately $86.6MPa$.