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ENGINEERING, CS 2025 HSC 16 MC

A simply supported beam with a vertical point load is shown.
 

Which row of the table correctly describes the shear force and the bending moment diagrams at point \(\text{A}\)?
 

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \textit{Shear force}\quad \rule[-1ex]{0pt}{0pt}& \ \ \textit{Bending moment}\ \ \\
\hline
\rule{0pt}{2.5ex}\text{Minimum}\rule[-1ex]{0pt}{0pt}&\text{Maximum}\\
\hline
\rule{0pt}{2.5ex}\text{Maximum}\rule[-1ex]{0pt}{0pt}& \text{Maximum}\\
\hline
\rule{0pt}{2.5ex}\text{Maximum}\rule[-1ex]{0pt}{0pt}& \text{Minimum} \\
\hline
\rule{0pt}{2.5ex}\text{Minimum}\rule[-1ex]{0pt}{0pt}& \text{Minimum} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • For a simply supported beam with a central point load, shear force is zero (minimum) at midspan — the shear force diagram crosses zero directly under the point load.
  • Bending moment is maximum at midspan — the bending moment diagram peaks directly under the point load and is zero at both supports.
  • These two facts are fundamental: the point where shear force = 0 always corresponds to maximum bending moment.

\(\Rightarrow A\)


♦ Mean mark 51%.

ENGINEERING, CS 2024 HSC 20 MC

A cross-section of a beam is shown.

Which of the following is the maximum stress in the beam if the maximum bending moment is 200 kN m and \(I_{\text{xx}}\) is  \(168.75 \times 10^{-6}\ \text{m}^4\) ?

  1. 177.8 MPa
  2. 177.8 GPa
  3. 355.6 MPa
  4. 355.6 GPa
Show Answers Only

\(A\)

Show Worked Solution

\(M=200\ \text{kN m}\ = 200\,000\ \text{N m}\)

\(y=\ \text{distance to neutral axis}\ = \dfrac{300}{2} = 150\ \text{mm}\ = 0.15\ \text{m} \)

\(\sigma = \dfrac{My}{I_{\text{xx}}} = \dfrac{200\,000 \times 0.15}{168.75 \times 10^{-6}} = 177.8\ \text{MPa}\) 

\(\Rightarrow A\)

♦ Mean mark 44%.

ENGINEERING, CS 2024 HSC 16 MC

An \(\text{I}\)-beam is loaded as shown.
 

The load is then increased.

Which increase in dimension would provide the most resistance to bending?

  1. \( \textit{A} \)
  2. \( \textit{B} \)
  3. \( \textit{C} \)
  4. \( \textit{D} \)
Show Answers Only

\(D\)

Show Worked Solution
  • The most efficient way to increase bending resistance in an \(\text{I}\)-beam would be to increase the height/depth of the web \((D)\), since resistance to bending increases with the cube of the depth from neutral axis.

\(\Rightarrow D\)

♦ Mean mark 46%.

ENGINEERING, CS 2023 HSC 22b

The diagram shows a child with a mass of 45 kg hanging 2 metres from the left end of a structure, and an adult with a mass of 85 kg hanging 1 metre from the right end.

 

  1. Calculate the reactions at \(\text{R}_\text{L}\) and \(\text{R}_\text{R}\).   (3 marks)
      

\begin{array} {ll}
\text{R}_\text{L} = \text{............................... N} & \text{Direction ...............................} \\
 &  \\
\text{R}_\text{R} = \text{............................... N} & \text{Direction ...............................} \end{array}

 

  1. Complete the shear force and bending moment diagrams of the scenario described.   (3 marks)

 

Show Answers Only

i.    \( \text{R}_{\text{L}}=860\ \text{N} \uparrow \)

\( \text{R}_{\text{L}}= 440\ \text{N} \uparrow \)

  
ii.    

Show Worked Solution

i.    \(  \stackrel {\curvearrowright} {\sum{ \text{M}}{^{+}_\text{L}}}: \)

\(0\) \(=(2 \times 450)+(4 \times 850)-(\text{R}_\text{R} \times 5) \)  
\(5 \times \text{R}_\text{R}\) \(=900 + 3400\)  
\(\text{R}_\text{R}\) \(=860\ \text{N} \uparrow \)  

 

\( \sum \text{F}_\text{V} \uparrow:\)

\(0\) \(= -450-850+860+\text{R}_\text{L} \)  
\( \text{R}_{\text{L}}\) \(= 440\ \text{N} \uparrow \)  

 
ii.    

ENGINEERING, CS 2019 HSC 21d

A pictorial sketch of a laminated timber beam used in a bus shelter is shown. This beam has a second moment area of `117 × 10^(-6)\ text{m}^(4)`. The maximum applied bending moment is 17 kNm.
 

  1. Calculate the maximum bending stress present in the beam.   (3 marks)
Show Answers Only

`18.2\ text{MPa}`

Show Worked Solution

Second Moment of Area (`l`)`= 117 times 10^(-6)\ text{m}^(4)`

Maximum Bending Moment (`M`)`= 17 times 10^(3)\ text{N}`

Distance (Neutral axis → outside of beam) = 125 mm

`=> y = 0.125\ text{m}`

`:.\ text{Bending Stress}_text{max}` `=(My)/l`  
  `=(0.125 times 17 times 10^(3))/(117 times 10^(-6)`  
  `=18.2\ text{MPa}`  

ENGINEERING, CS 2018 HSC 23c

The chassis for a motor vehicle could be made from a solid steel bar or hollow steel tube. Both the bar and the tube have the same mass and cost per metre.
 

Justify why a hollow tube would be used in preference to a solid bar to build the vehicle chassis.   (3 marks)

Show Answers Only
  • Hollow tube has a higher second moment of area and is therefore more rigid.
  • Because the chassis is exposed to continual flexural forces and moments when the vehicle is moving, rigidity is important.
  • Stiffness is therefore a major design requirement.
  • The hollow tube also provides damage protection for any electrics cabling.
Show Worked Solution
  • Hollow tube has a higher second moment of area and is therefore more rigid.
  • Because the chassis is exposed to continual flexural forces and moments when the vehicle is moving, rigidity is important.
  • Stiffness is therefore a major design requirement.
  • The hollow tube also provides damage protection for any electrics cabling.
♦♦ Mean mark 35%.

ENGINEERING, CS 2017 HSC 27b

An engineering team has been contracted to design a multi-function lifting device for a coastal container wharf.

The table shows some of the engineering design elements for this lifting device.
 

Explain how the lifting device can be tested and evaluated to determine if the criteria for the listed engineering elements are met.   (6 marks)

Show Answers Only

Show Worked Solution


♦ Mean mark 46%.

ENGINEERING, CS 2019 HSC 23d

A scooter deck, made from aluminium alloy, and its cross-section are shown.

The scooter's manufacturer is concerned that there is too much deflection along the length of the deck when a rider stands on it.

Describe ONE suitable design modification to give the deck greater rigidity without adding extra mass. Use a labelled sketch to support your answer.   (3 marks)

Show Answers Only

Design modifications

  • Add longitudinal ribbing to the deck, ensuring not to create additional mass.
  • This will create an increase in the distance of the neutral axis from the bottom of the deck and a thinner wall structure.
  • The outcome will be a decrease in deflection

Show Worked Solution

Design modifications

  • Add longitudinal ribbing to the deck, ensuring not to create additional mass.
  • This will create an increase in the distance of the neutral axis from the bottom of the deck and a thinner wall structure.
  • The outcome will be a decrease in deflection


 
♦ Mean mark 41%.

ENGINEERING, CS 2021 HSC 14 MC

The diagram shows a beam.

The beam's second moment of area is 76.96 × 106 mm4. It is subjected to a maximum bending moment of 250 kN m.

What is the maximum bending stress of the beam?

  1. 227.4 MPa
  2. 324.8 MPa
  3. 649.7 MPa
  4. 9745.3 MPa
Show Answers Only

`B`

Show Worked Solution

`M=250\ text{kNm}\ = 250 xx10^(-3)\ text{Nm}`

`y=(200)/(2)\ text{mm}\ =100 xx10^(-3)\ text{m}`

`I=76.96 xx10^(6)\ text{mm}^(4)=76.96 xx10^(-6)\ text{m}^(4)`

`S_text{bending}` `=(My)/I`  
  `=(250 xx10^(-3)xx100 xx10^(-3))/(76.96 xx10^(-6))`  
  `=324.8440…\ text{MPa}`  

 
`=>B`