smc-3714-70-Compressive stress

ENGINEERING, CS 2023 HSC 26c

A truss is loaded as shown.

Showing working, complete the table. (6 marks)

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \textit{Magnitude} & \textit{Nature of force}\\ & \text{(kN)} & \text{(T or C)} \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ EF \rule[-1ex]{0pt}{0pt} & & \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ CH \rule[-1ex]{0pt}{0pt} & & \\
\hline
\end{array}

Show Answers Only

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \textit{Magnitude} & \textit{Nature of force}\\ & \text{(kN)} & \text{(T or C)} \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ EF \rule[-1ex]{0pt}{0pt} & 73.2\ \text{kN} & \text{Compression} \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ CH \rule[-1ex]{0pt}{0pt} & 71.138\ \text{kN} & \text{Tension} \\
\hline
\end{array}

Show Worked Solution

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \textit{Magnitude} & \textit{Nature of force}\\ & \text{(kN)} & \text{(T or C)} \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ EF \rule[-1ex]{0pt}{0pt} & 73.2\ \text{kN} & \text{Compression} \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ CH \rule[-1ex]{0pt}{0pt} & 71.138\ \text{kN} & \text{Tension} \\
\hline
\end{array}

\(\text{Consider member}\ EF:\)

→ \(\text{Force diagram closes with two collinear forces}\)

→ \(FG\ \text{is a zero force member}\)

→ \(EF = 73.2\ \text{kN (compression)} \)

♦♦♦ Mean mark 36%.

\(\text{Consider member}\ CH:\)

\(+ \uparrow \Sigma F_{V}\)	\(=0\)
\(0\)	\(=-125 + 73.2 \times \sin\,60^{\circ} + CH \times \sin\,60^{\circ}\)
\(CH \times \sin\,60^{\circ}\)	\(=61.607\)
\(CH\)	\(= \dfrac{61.607}{\sin\,60^{\circ}} \)
	\(=71.138\ \text{kN (tension)} \)

ENGINEERING, CS 2018 HSC 26a

A fidget spinner is shown. It is a toy that contains a ball bearing in its centre and is designed to spin on its axis with little effort.

The central components of a fidget spinner are press fitted together using a press punch.

The press punch diameter is 21 mm. The force applied by the press punch to the bearing is 17.3 N.

Calculate the compressive stress in the press punch. (2 marks)

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`50\ 000\ text{Pa}`

Show Worked Solution

`F`	`=17.3\ text{N}`
`a`	`=(pi xx 21^2)/4`
	`=346\ text{mm}^2`
	`=0.000346\ text{m}^2`

`sigma`	`=F/a`
	`=17.3/0.000346`
	`=50\ 000\ text{Pa}`

ENGINEERING, CS 2017 HSC 22d

When a bike goes over a bump, there is a vertical force of 3 kN exerted axially on the main pillar.

The cross-section of the pillar is shown.

Calculate the compressive stress acting in the pillar. (3 marks)

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`15.9\ text{MPa}`

Show Worked Solution

`text{CSA}`	`=(pi(d_1)^2)/4-(pi(d_2)^2)/4`
	`=pi/4(32^2 – 28^2)`
	`=(240pi)/4`
	`=188.5\ text{mm}^2`

`sigma_c`	`=F/A`
	`=3000/188.5`
	`=15.9\ text{MPa}`

ENGINEERING, CS 2019 HSC 23c

The weight of a 650 N rider and the mass of the scooter are evenly distributed between the front and rear pneumatic tyres. The area of contact between each of the two tyres and the ground is 1200 mm². The pressure inside each tyre is 300 kPa.

What is the mass of the scooter? (3 marks)

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`7\ text{kg}`

Show Worked Solution

`A=1200 xx 10^(-6)\ text{m}^(2), \ P=300 xx 10^(3)\ text{Pa}\ \ text{(given)}`

`P`	`=F/A`
`F`	`=PxxA`
	`=300 xx 10^(3) xx 1200 xx 10^(-6)`
	`=300 xx 1200 xx 10^(-3)`
	`=360\ text{N per tyre}`

`text{There are 2 tyres}\ => \ text{Load = 720 N}`

`text{Weight of rider = 650 N}`

`:.\ text{Weight of scooter = 70 N}`

`F`	`=mg`
`m`	`=F/g`
	`=70/10`
	`=7\ text{kg}`

♦ Mean mark 48%.