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ENGINEERING, CS 2025 HSC 26c

A truss is loaded as shown.
 

Determine the magnitude and nature of the force in member \(\text{M}\).   (6 marks)

Show Answers Only

\(\text{Magnitude} = 5869.7 \ \text{N}\)

\(\text{Nature}\ =\ \text{Tension}\)

Show Worked Solution

Find reactions at supports — take moments about \(\text{A}\):

\(\Sigma M_A\) \(= 0\)  
  \(0 = +(9 \times 1.4)+(15\sin60° \times 2.8)-(15\cos60° \times 0.7)-R_B \times 2.8\)  
  \(0 = 12.6+36.373-5.25-2.8R_B\)  
\(R_B\) \(= \dfrac{43.723}{2.8}\)  
\(R_B\) \(= 15.615 \ \text{kN} \uparrow\)  

 
Consider RHS of section plane — sum vertical forces:

\(\Sigma F_V\) \(= 0\)  
  \(0 = 15.615-15\sin60°-F_M\sin26.565°\)  
\(F_M\) \(= \dfrac{2.625}{\sin26.565°}\)  
\(F_M\) \(= 5.8697 \ \text{kN}\)  
\(F_M\) \(= 5869.7 \ \text{N (Tension)}\)  

♦♦ Mean mark 51%.

Find \(R_B\) as above \(= 15.615 \ \text{kN} \uparrow\)

     

Consider Joint C — sum vertical forces:

\(\Sigma F_V\) \(= 0\)  
  \(0 = 15.615-15\sin60°-F_M\sin26.565°\)  
\(F_M\) \(= \dfrac{2.625}{\sin26.565°}\)  
\(F_M\) \(= 5.8697 \ \text{kN}\)  
\(F_M\) \(= 5869.7 \ \text{N (Tension)}\)  

   

ENGINEERING, CS 2024 HSC 25c

A pin-jointed truss is shown. Member  \(AB\)  has been determined to be 250 N in compression.
 

Complete the table. You must use the method specified to solve each component.   (6 marks)

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Determine}\quad & \quad \textit{Method to} \quad & \quad \quad \quad \quad \quad \quad \quad \quad\textit{Working} \quad \quad \quad \quad \quad \quad \quad \quad\\
\textit{} \rule[-1ex]{0pt}{0pt} & \quad \quad \ \ \  \textit{use} & \textit{}\\
\hline
\rule{0pt}{2.5ex} \text{External} & \text{Mathematical} & \\
\text{reaction at \(E\)} \rule[-1ex]{0pt}{0pt} & \text{} &\\
\\ \\ \\ \\ \\ \\ \\ \\
\text{} & \text{} & \text{........................... N} \\
\text{} & \text{} & \text{Direction: ...........................} \\
\hline
\end{array}

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Determine}\quad & \quad \textit{Method to} \quad & \quad \quad \quad \quad \quad \quad \quad \quad\textit{Working} \quad \quad \quad \quad \quad \quad \quad \quad\\
\textit{} \rule[-1ex]{0pt}{0pt} & \quad \quad \ \ \  \textit{use} & \\
\hline
\rule{0pt}{2.5ex} \text{Internal} & \text{Methods of} & \\
\text{reaction of}  & \text{section} &\\
\text{member of \(CF\)} & &\\
\\ \\ \\ \\ \\ \\ \\ \\
\text{} & \text{} & \text{........................... N} \\
\text{} & \text{} & \text{Nature of force (T or C): ...........................} \\
\hline
\end{array}

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Determine}\quad & \quad \textit{Method to} \quad & \quad \quad \quad \quad \quad \quad \quad \quad\textit{Working} \quad \quad \quad \quad \quad \quad \quad \quad\\
\textit{} \rule[-1ex]{0pt}{0pt} & \quad \quad \ \ \  \textit{use} & \\
\hline
\rule{0pt}{2.5ex} \text{Internal} & \text{Graphical} & \\
\text{reaction of}  &  &\\
\text{member \(BG\)} & &\\
\\ \\ \\ \\ \\ \\ \\ \\
\text{} & \text{} & \text{........................... N} \\
\text{} & \text{} & \text{Nature of force (T or C): ...........................} \\
\hline
\end{array}

Show Answers Only

External reaction at \(E\) (Mathematical method):

\( \Sigma \text{M}_\text{A}\Large{⤸} ^{\small{+}}\) \(=0=(200 \times 2)+(75 \times 10)-(129.9 \times 3.46)-\left(\text{R}_\text{E} \times 12\right) \)  
\(0\) \(=400+750-450-\text{R}_\text{E} \times 12 \)  
\(\text{R}_\text{E}\) \(=\dfrac{700}{12} = 58.33\ \text{N} \uparrow \)  

 

Internal reaction of \(CF\) (Method of sections):

\( \Sigma \text{F}_\text{V} \uparrow^{+}=0=\left(\sin \, 60^{\circ} \times \text{CF}\right)-75+58.33 \)

\(-\sin \, 60^{\circ} \times \text{CF} \) \(=-75+58.33 \)  
\(\text{CF}\) \(=\dfrac{16.67}{\sin \, 60^{\circ}} =19.25\ \text{N (T)} \)  
♦♦ Mean mark 46%.

Internal reaction of \(BG\) (Graphical):
 

\(BG \approx 19\ \text{N (T)} \)

Show Worked Solution

External reaction at \(E\) (Mathematical method):

\( \Sigma \text{M}_\text{A}\large{⤸} ^{\small{+}}\) \(=0=(200 \times 2)+(75 \times 10)-(129.9 \times 3.46)-\left(\text{R}_\text{E} \times 12\right) \)  
\(0\) \(=400+750-450-\text{R}_\text{E} \times 12 \)  
\(\text{R}_\text{E}\) \(=\dfrac{700}{12} = 58.33\ \text{N} \uparrow \)  

 

Internal reaction of \(CF\) (Method of sections):

\( \Sigma \text{F}_\text{V} \uparrow^{+}=0=\left(\sin \, 60^{\circ} \times \text{CF}\right)-75+58.33 \)

\(-\sin \, 60^{\circ} \times \text{CF} \) \(=-75+58.33 \)  
\(\text{CF}\) \(=\dfrac{16.67}{\sin \, 60^{\circ}} =19.25\ \text{N (T)} \)  
♦♦ Mean mark 46%.

Internal reaction of \(BG\) (Graphical):
 

\(BG \approx 19\ \text{N (T)} \)

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 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)

 

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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 2016 HSC 26b

What are redundant truss members?   (2 marks)

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Redundant truss members:

  • A truss that carries no load under the conditions present.
  • It is important to note that as load conditions change, the truss may not actually remain redundant. 
Show Worked Solution

Redundant truss members:

  • A truss that carries no load under the conditions present.
  • It is important to note that as load conditions change, the truss may not actually remain redundant. 

♦ Mean mark 46%.

ENGINEERING, CS 2016 HSC 26a

A simplified diagram of a pin jointed truss of a hammerhead crane is shown. It is lifting a load of 54 kN.
 

  1. Calculate the reactions at supports `A` and `B`.   (4 marks)
  1. Determine the magnitude and nature of the force in member `M`.   (2 marks)
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i.     Reaction at `A` = 249 kN ↑

Reaction at `B` = 123 kN @ 12.7° acting down and to the right.

ii.   Force in `M` = 121 kN (Compression)

Show Worked Solution

i.

`text{R}_(BH)-27=0`

`text{R}_(BH)=27`

`text{From the diagram,}\ \ text{R}_(BH)=27\ text{kN →}`

+\({\circlearrowleft}\)`sumM_B` `=0`  
`0` `=27xx2+54xx5-75xx1-R_Axx1`  
`R_A` `=54+270-75=249\ text{kN ↑}`  
     
+ ↑`sumF_V` `=0`  
`0` `=249-75-54+R_B`  
`R_B` `=129-249=-120\ text{kN}=120\ text{kN ↓}`  

 

`R_A=249\ text{kN @ 0° ↑}`

`text{Reaction at}\ B:`
 

`\text{Reaction at}\ A = 249\ \text{kN acting vertically, up}`

`\text{Reaction at}\ B = 123\ \text{kN @ 12.7° acting down and to the right.}`
  


Mean mark (i) 52%.

ii.   `theta = tan^(-1)(20/40)=26.6°`

   

 
`F=sqrt(108^2+54^2)=sqrt(14\ 580)=120.747…`

`\text{Force in}\ M = 121\ \text{kN (Compression)}`


♦ Mean mark (ii) 49%.

ENGINEERING, CS 2018 HSC 18 MC

The force `F` is moved from joint `W` to joint `X`, as shown.
 

Which row of the table correctly describes the changes in the internal force in member `XY` and the reaction force at `W` as a result of force `F` moving from `W` to `X` ?
 

Show Answers Only

`B`

Show Worked Solution
  • Initially there is not force in member XY and no reaction at W.
  • When F is moved, a turning moment is created and needs to be countered by the reaction at W.
  • Member XY is also placed in tension, therefore both the force and reaction increase.

`=>B`

ENGINEERING, CS 2017 HSC 25a

A pin-jointed truss designed to support a roadside sign is shown.
 

  1. Determine the magnitude and direction of the reactions at `A` and `B`.   (4 marks)
  1. Determine the magnitude and nature of the force in member `C`.   (2 marks)
  1. Explain why concrete would be a suitable material to support the roadside sign around point `D`.   (2 marks)
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i.   `R_B=37\ text{N} larr`

`R_A=44.7\ text{kN}, \ theta=26.6^(@)` 

ii.   `F_C=17\ text{kN (compression)}`
iii.   Suitability of concrete

  • Easily poured and formed around the sign.
  • Relatively quick to cure.
  • Weather resistant.
  • Hardens and sets with high strength and hardness.
Show Worked Solution

\[\textbf{i}. \ \ \ce{->[\ce{+}]} \sum M_A = 0 \]

`3xx1.5-20 xx3+R_(B)xx1.5=0`

`4.5-60+1.5R_(B)=0`

`1.5R_(B)` `=55.5`  
`R_(B)` `=(55.5)/(1.5)=37\ text{N} larr`  

  

Mean mark 56%.

`uarr sumF_(V)=0`

`R_(AV)=20\ text{kN} +uarr`

\[\ce{->[\ce{+}]} \sum F_H = 0 \]

`R_(AH)=37+3=40\ text{kN} rarr^(+)`

`R_A=sqrt(20^(2)+40^(2))=sqrt2000=44.7\ text{kN}`

`tan\  theta` `=(R_(AV))/(R_(AH))=20/40=0.5`  
`:.theta` `=26.6^(@)`  

 

\[\textbf{ii}. \]

`text{Taking moments about X:}`

\({+ \circlearrowleft} \Sigma M_X = 0\)

`0` `=-1.5 xxF_(C)+1.5 xx20-1.5 xx3`  
`0` `=-F_(C)+20-3`  
`F_(C)` `=17\ text{kN (compression)}`  

♦♦ Mean mark (ii) 38%.

iii.   Suitability of concrete

  • Easily poured and formed around the sign.
  • Relatively quick to cure.
  • Weather resistant.
  • Hardens and sets with high strength and hardness.

♦ Mean mark (iii) 52%.

ENGINEERING, CS 2019 HSC 22c

The diagram shows some dimensions and forces associated with a telecommunications tower.
 

By considering any necessary reaction, calculate the magnitude of the forces in members `M` and `N`. State the nature of each force. Ignore the weight of the tower.   (6 marks)

Show Answers Only
`M` `=0.324\ text{kN (tension)}`
`N` `=14.33\ text{kN (compression)}`
Show Worked Solution

Forces at Joint `A`

Horizontal forces `=0`

`:.` To calculate vertical force at `A ` → use moments.


♦♦ Mean mark 36%.
\({\circlearrowright}\)`+SigmaM_C` `=0`  
`0` `=-(12xx4)+(R_Axx12)-(10xx7)-(3xx18)`  
`12R_A` `=48+70+54`  
`R_A` `=172/12=14.33\ text{kN}↑`  

    
Forces in Member `N` → method of joints at `A`

  • No horizontal forces
  • Member `AC` redundant and carrying no load
  • `F_(up) = F_(down)`

`:.` Member `AB` in compression (the force acting down on joint `A` from member `AB` is 14.33 kN)

`:.` Force in N = 14.33 kN (compression)

  
Using Method of Sections → take moments about Joint `H`:

Find the perpendicular distance `d`:

`BH^2` `=18^2+6^2`  
`BH` `=sqrt{360}`  
`sin\ 40.6º` `=d/(sqrt{360})`  
`d` `=sqrt{360}xx sin\ 40.6º=12.348\ text{m}`  

 

\({\circlearrowright}\)`+SigmaM_H` `=0`
`0` `=+(12xx2)+(Mxx12.348)-(7xx4)`
`12.348M` `=-24+28`
`M` `=4/12.348=0.324\ text{kN (tension)}`

 

`:.\ ` `M` `=0.324\ text{kN (tension)}`
  `N` `=14.33\ text{kN (compression)}`

ENGINEERING, CS 2022 HSC 26a

  1.  The diagram shows a tower crane being used in the construction of a building.
     

  1. Determine the number of 100 kg concrete blocks required to place the boom arm in equilibrium.   (2 marks)
  1. Under a different set of conditions, a wind force is applied, as shown in the diagram.
     

  1. Determine the magnitude and nature of the internal reaction in member A.   (6 marks)
Show Answers Only

i.   `12`

ii.  `A=25\ text{kN in compression}`

Show Worked Solution

i.   Let the total weight of the concrete blocks `= x`

`50x` `=9.23xx65`  
`50x` `=600`  
`x` `=600/50=12\ text{kN ↓}=12\ 000\ text{N ↓}`  
`m` `=1200\ text{kg}`  

 
∴ 12 × 100 kg concrete blocks are needed for the counterweight.


♦ Mean mark (i) 46%.

ii.   Magnitude and nature of internal reaction

\( \circlearrowright+\Sigma M_R \) `=0`  
`0` `=-(6xx5.5)+(R_Lxx1)+(10xx6)-(10.4xx7)`  
`R_L` `=45.8\ text{kN}↑`  

  
Taking the horizontal section shown:


  

\( \circlearrowright + \Sigma M_P \) \(= 0\)  
`0` `=(Axx1)+(45.8xx1)-(10.4xx2)`  
`A` `=-25\ text{kN}`  

  
∴ `A=25\ text{kN in compression}`


♦ Mean mark (ii) 43%.

ENGINEERING, CS 2020 HSC 25c

 

A cycle bridge has been constructed using a Warren girder truss loaded as shown. The diagram is drawn to scale.
 

By considering necessary loads and reactions, calculate the magnitude and nature of the force in member `C`.   (6 marks)

Show Answers Only

`C` = 6.24 kN in tension

Show Worked Solution

Let the length of a member be 2 units

`sin60°` `=d/2`  
`d` `=2 xx sin60°=1.732\ text{units}`  

 

\({+ \circlearrowleft \Sigma M_B}\) `=0`  
`0` `=(3000xx1.732)+(20\ 000xx4)+(-R_Axx6)+(1200xx8)`  
`0` `=5196+80\ 000-6R_A+9600`  
`6R_A` `=94\ 796`  
`R_A` `=15\ 799.33=15.80\ text{kN} ↑`  

`sin60°` `= y/C`  
`y` `= C\ sin60°`  
`+↑ SigmaF_Y` `=  -1200 + 15\ 799.33-20\ 000-C\ sin60°`  
`0` `=  -5400.667-C\ sin60°`  
`C\ sin60°` `=  -5401`  
`C` `=  -5401/(sin60°)=  -6236.54`  

 
`=>` Assumed direction was incorrect and reaction in `C` is away from the joint.

∴  `C= 6.24\ \text{kN in tension}`


♦ Mean mark 48%.

ENGINEERING, CS 2021 HSC 25b

A truss is fixed to a wall at `A` and `B` as shown. Ignore the mass of the truss.
 


 

  1. Determine the horizontal reaction at `A`.   (2 marks)
  1. Determine, using method of joints, the internal reaction in member `AC`. Indicate the nature of the force in the member.   (3 marks)
  1. Determine, using method of sections, the internal reaction in member `CE`. Indicate the nature of the force.   (3 marks)
Show Answers Only

i.    `3.75\ text{kN ←}`

ii.    `AC=4802\ text{N}`

iii.   `CE = 3\ text{kN (tension)}`

Show Worked Solution

i.   Horizontal reaction at `A`

\(\circlearrowright \Sigma \text{M}_\text{B}\) \(= 0\)  
`0` `= -(A_H xx 4\ text{m}) + (1500\ text{N} xx  10\ text{m})`  
`A_H` `= 3750\ text{N}= 3.75\ text{kN ←}`  

 

ii.   Method of joints at A
 
     

→Σ`F_H` `=0`  
`0` `= -3750+AC xx sin(51.34°)`  
`AC` `=3750/sin(51.34°)=4802\ text{N}`  

♦♦ Mean mark (ii) 41%.

iii.  Method of Sections

\(\circlearrowright \Sigma \text{M}_\text{B}\) \(=0\)  
`0` `= (1500 xx 10)-(CE xx 5)`  
`CE` `=(15\ 000)/5=3000\ text{N}=3\ text{kN (tension)}`  

♦♦♦ Mean mark (iii) 30%.

ENGINEERING, CS 2021 HSC 7 MC

The diagram shows a section of a pin jointed truss in equilibrium. The force in Member 1 is 200 kN in compression.
 

Which row of the table identifies the magnitude and nature of the forces in Member 2 and Member 3?
 

Show Answers Only

`A`

Show Worked Solution
  • The sum of horizontal forces and the sum of vertical forces must both be equal to 0.

`=>A`