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
(there is a figure showing a bracket structure with points a, b, c, a 30° angle, and force analysis diagrams with $sum x = 0$ leading to $-f_1cos30^{circ}+f_2 = 0$, $sum y = 0$ leading to $f_1cos60^{circ}-q = 0$, and then $f_1 = 2q = 20kn$, $f_2=\frac{1}{2}sqrt{3}f_1 = 17.32kn$)

Explanation:

Step1: Recall the formula for normal stress

The formula for normal stress \(\sigma\) is \(\sigma=\frac{F}{A}\), where \(F\) is the axial force and \(A\) is the cross - sectional area.

Step2: Calculate the normal stress of bar BC (\(\sigma_1\))

For bar BC, the axial force \(F_1 = 20\mathrm{KN}=20\times10^{3}\mathrm{N}\) and the cross - sectional area \(A_1 = 100\mathrm{mm}^2 = 100\times10^{-6}\mathrm{m}^2\).
Using the stress formula \(\sigma_1=\frac{F_1}{A_1}\), we substitute the values:
\(\sigma_1=\frac{20\times 10^{3}\mathrm{N}}{100\times 10^{-6}\mathrm{m}^2}=\frac{20\times 10^{3}}{100\times 10^{-6}}\mathrm{Pa}=200\times 10^{6}\mathrm{Pa} = 200\mathrm{MPa}\)

Step3: Calculate the normal stress of bar AB (\(\sigma_2\))

For bar AB, the axial force \(F_2=17.32\mathrm{KN} = 17.32\times 10^{3}\mathrm{N}\) and the cross - sectional area \(A_2 = 200\mathrm{mm}^2=200\times 10^{-6}\mathrm{m}^2\).
Using the stress formula \(\sigma_2=\frac{F_2}{A_2}\), we substitute the values:
\(\sigma_2=\frac{17.32\times 10^{3}\mathrm{N}}{200\times 10^{-6}\mathrm{m}^2}=\frac{17.32\times 10^{3}}{200\times 10^{-6}}\mathrm{Pa}=86.6\times 10^{6}\mathrm{Pa}=86.6\mathrm{MPa}\)

Answer:

The normal stress of bar BC is \(\boldsymbol{200\mathrm{MPa}}\) and the normal stress of bar AB is \(\boldsymbol{86.6\mathrm{MPa}}\)