Question

a bracket is shown in the figure, the cross - section area of the bar bc is: $a_1 = 100mm^2$, the cross - section area of the bar ab is: $a_2 = 200mm^2$, and $q = 10kn$, determine the normal stress of the bars

Explanation:

Step1: Analyze the force on bar BC

First, we need to find the axial force on bar BC. From the equilibrium of joint B, the vertical component of the force in BC ($F_{BC}$) should balance the load Q. So, $F_{BC}\sin30^{\circ}=Q$. Given $Q = 10\mathrm{KN}$, we can solve for $F_{BC}$:
$F_{BC}=\frac{Q}{\sin30^{\circ}}=\frac{10\mathrm{KN}}{0.5}=20\mathrm{KN}$ (tension, since it's pulling joint B)

Step2: Analyze the force on bar AB

The horizontal component of the force in BC balances the force in AB ($F_{AB}$). So, $F_{AB}=F_{BC}\cos30^{\circ}$. Substituting $F_{BC}=20\mathrm{KN}$:
$F_{AB}=20\mathrm{KN}\times\frac{\sqrt{3}}{2}=10\sqrt{3}\mathrm{KN}\approx17.32\mathrm{KN}$ (compression, since it's pushing joint B)

Step3: Calculate normal stress for bar BC

Normal stress $\sigma$ is given by $\sigma=\frac{F}{A}$, where $F$ is the axial force and $A$ is the cross - sectional area. For bar BC, $F = F_{BC}=20\mathrm{KN}=20\times10^{3}\mathrm{N}$, $A = A_{1}=100\mathrm{mm}^{2}=100\times10^{-6}\mathrm{m}^{2}$.
$\sigma_{1}=\frac{F_{BC}}{A_{1}}=\frac{20\times10^{3}\mathrm{N}}{100\times10^{-6}\mathrm{m}^{2}} = 2\times10^{8}\mathrm{Pa}=200\mathrm{MPa}$

Step4: Calculate normal stress for bar AB

For bar AB, $F = F_{AB}=10\sqrt{3}\times10^{3}\mathrm{N}\approx17.32\times10^{3}\mathrm{N}$, $A = A_{2}=200\mathrm{mm}^{2}=200\times10^{-6}\mathrm{m}^{2}$.
$\sigma_{2}=\frac{F_{AB}}{A_{2}}=\frac{17.32\times10^{3}\mathrm{N}}{200\times10^{-6}\mathrm{m}^{2}}=8.66\times10^{7}\mathrm{Pa}=86.6\mathrm{MPa}$ (compressive stress, but stress magnitude is what we calculate here)

Answer:

Normal stress of bar BC: $\boldsymbol{200\mathrm{MPa}}$ (tensile), Normal stress of bar AB: $\boldsymbol{86.6\mathrm{MPa}}$ (compressive)