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 Forces in Bars

First, we need to find the axial forces in bars \( BC \) and \( AB \). Let's denote the force in \( BC \) as \( F_{BC} \) and in \( AB \) as \( F_{AB} \). From the equilibrium of joint \( B \), resolving forces vertically: \( F_{BC}\sin(30^\circ)=Q \). Given \( Q = 10\,\text{kN}=10\times 10^{3}\,\text{N} \), so \( F_{BC}=\frac{Q}{\sin(30^\circ)}=\frac{10\times 10^{3}}{0.5}=20\times 10^{3}\,\text{N} \) (tensile or compressive? Let's check horizontal: \( F_{AB}=F_{BC}\cos(30^\circ) \), but for stress, we need magnitude for now). Resolving horizontally: \( F_{AB}=F_{BC}\cos(30^\circ)=20\times 10^{3}\times\frac{\sqrt{3}}{2}\approx 17.32\times 10^{3}\,\text{N} \), but wait, actually, let's re - check equilibrium. The load \( Q \) is downward at \( B \). So for joint \( B \), forces: \( F_{BC} \) is from \( B \) to \( C \), \( F_{AB} \) is from \( A \) to \( B \) (assuming \( AB \) is in tension and \( BC \) is in tension? Wait, no, let's do it properly. Let's take upward as positive for vertical. \( F_{BC}\sin30^{\circ}=Q \), so \( F_{BC}=\frac{Q}{\sin30^{\circ}} = 20\,\text{kN} \) (tension, since it pulls \( B \) up). Then horizontal force: \( F_{AB}=F_{BC}\cos30^{\circ}=20\times\frac{\sqrt{3}}{2}=10\sqrt{3}\approx 17.32\,\text{kN} \) (compression? Wait, no, if \( BC \) is pulling \( B \) towards \( C \), then \( AB \) would be pushing \( B \) towards \( A \), so \( F_{AB} \) is compressive? Wait, but stress formula is \( \sigma=\frac{F}{A} \), regardless of tension or compression (magnitude for stress calculation, sign for nature).

Step2: Calculate Stress in \( BC \)

Stress \( \sigma_{BC}=\frac{F_{BC}}{A_1} \). \( A_1 = 100\,\text{mm}^2=100\times 10^{- 6}\,\text{m}^2 \), \( F_{BC}=20\times 10^{3}\,\text{N} \). So \( \sigma_{BC}=\frac{20\times 10^{3}\,\text{N}}{100\times 10^{-6}\,\text{m}^2}=\frac{20\times 10^{3}}{100\times 10^{-6}} = 200\times 10^{6}\,\text{Pa}=200\,\text{MPa} \).

Step3: Calculate Stress in \( AB \)

Stress \( \sigma_{AB}=\frac{|F_{AB}|}{A_2} \). \( F_{AB}=10\sqrt{3}\times 10^{3}\,\text{N}\approx17.32\times 10^{3}\,\text{N} \), \( A_2 = 200\,\text{mm}^2 = 200\times 10^{-6}\,\text{m}^2 \). So \( \sigma_{AB}=\frac{17.32\times 10^{3}}{200\times 10^{-6}}=\frac{17.32\times 10^{9}}{200}=86.6\times 10^{6}\,\text{Pa}=86.6\,\text{MPa} \). Wait, but maybe I made a mistake in force direction. Alternatively, maybe \( AB \) is in compression and \( BC \) is in tension. Let's re - express:

Wait, the formula for normal stress is \( \sigma=\frac{F}{A} \), where \( F \) is the axial force (tensile positive, compressive negative, but stress magnitude is what's often asked, or with sign).

Wait, let's redo the force analysis. Let's consider joint \( B \). The load \( Q = 10\,\text{kN}\) downward. Let \( F_{BC} \) be the force in bar \( BC \) (force exerted by \( BC \) on \( B \)) and \( F_{AB} \) be the force in bar \( AB \) (force exerted by \( AB \) on \( B \)).

Vertical equilibrium: \( F_{BC}\sin(30^{\circ})=Q \)
\( F_{BC}=\frac{Q}{\sin(30^{\circ})}=\frac{10\times 10^{3}\,\text{N}}{0.5}=20\times 10^{3}\,\text{N} \) (tensile, since it has to balance \( Q \) upward)

Horizontal equilibrium: \( F_{AB}=F_{BC}\cos(30^{\circ}) \) (but direction: if \( F_{BC} \) is pulling \( B \) towards \( C \), then \( F_{AB} \) must be pushing \( B \) towards \( A \), so \( F_{AB} \) is compressive, so \( F_{AB}=-F_{BC}\cos(30^{\circ}) \) (negative for compression). But for stress, \( \sigma=\frac{F}{A} \), where \( F \) is the axial force (compressive force is negative, tensile is positive, but stress magnitude is \( |\frac{F}{A}| \), or with sign depending on convention).

Now, stress in \( BC \): \( \sigma_{BC}=\frac{F_{BC}}{A_1}=\frac{20\times 10^{3}\,\text{N}}{100\,\text{mm}^2}=\frac{20\times 10^{3}\,\text{N}}{100\times 10^{-6}\,\text{m}^2}=200\times 10^{6}\,\text{Pa}=200\,\text{MPa} \) (tensile stress, since \( F_{BC} \) is tensile)

Stress in \( AB \): \( \sigma_{AB}=\frac{F_{AB}}{A_2}=\frac{- 20\times 10^{3}\times\cos(30^{\circ})\,\text{N}}{200\,\text{mm}^2}=\frac{-20\times 10^{3}\times\frac{\sqrt{3}}{2}\,\text{N}}{200\times 10^{-6}\,\text{m}^2}=\frac{-10\sqrt{3}\times 10^{3}\,\text{N}}{200\times 10^{-6}\,\text{m}^2}\approx\frac{-17.32\times 10^{3}\,\text{N}}{200\times 10^{-6}\,\text{m}^2}=- 86.6\times 10^{6}\,\text{Pa}=- 86.6\,\text{MPa} \) (compressive stress, the negative sign indicates compression)

Step4: Recall Normal Stress Formula

Normal stress \( \sigma=\frac{F}{A} \), where \( F \) is the axial force in the bar and \( A \) is the cross - sectional area.

For bar \( BC \):
\( F_{BC} = 20\,\text{kN}=20\times 10^{3}\,\text{N} \), \( A_1 = 100\,\text{mm}^2 = 100\times 10^{-6}\,\text{m}^2 \)
\( \sigma_{BC}=\frac{F_{BC}}{A_1}=\frac{20\times 10^{3}}{100\times 10^{-6}} = 200\times 10^{6}\,\text{Pa}=200\,\text{MPa} \)

For bar \( AB \):
\( F_{AB}=F_{BC}\cos(30^{\circ})=20\times 10^{3}\times\frac{\sqrt{3}}{2}=10\sqrt{3}\times 10^{3}\,\text{N}\approx17.32\times 10^{3}\,\text{N} \) (compressive, so \( F_{AB}=- 17.32\times 10^{3}\,\text{N} \))
\( A_2 = 200\,\text{mm}^2=200\times 10^{-6}\,\text{m}^2 \)
\( \sigma_{AB}=\frac{F_{AB}}{A_2}=\frac{- 17.32\times 10^{3}}{200\times 10^{-6}}\approx - 86.6\times 10^{6}\,\text{Pa}=- 86.6\,\text{MPa} \) (the negative sign indicates compressive stress, magnitude is \( 86.6\,\text{MPa} \))

Answer:

For bar \( BC \): Normal stress \( \boldsymbol{\sigma_{BC}=200\,\text{MPa}} \) (tensile)
For bar \( AB \): Normal stress \( \boldsymbol{\sigma_{AB}\approx - 86.6\,\text{MPa}} \) (compressive, or magnitude \( 86.6\,\text{MPa} \))

(If we consider magnitude only, for \( BC \): \( 200\,\text{MPa} \), for \( AB \): \( 86.6\,\text{MPa} \))